ENGINEERING MECHANICS DYNAMICS 6TH EDITION BY ANTHONY BEDFORD, WALLACE FOWLER SOLUTIONS MANUAL by QuentinAxel

Engineering Mechanics Dynamics 6e Anthony Bedford, Wallace Fowler (Solu�ons Manual All Chapters, 100% Original Veriﬁed, A+ Grade) (Chapter 12-21)

Chapter 12 Problem 12.1 In 1967, the International Committee of Weights and Measures defined one second to be the time required for 9,192,631,770 cycles of the transition between two quantum states of the cesium-133 atom. Express the number of cycles in two seconds to four significant digits.

Solution:

Problem 12.2 The base of natural logarithms is e = 2.71828183... . (a) Express e to three significant digits. (b) Determine the value of e 2 to three significant digits. (c) Use the value of e you obtained in part (a) to determine the value of e 2 to three significant digits.

Solution:

[Comparing the answers of parts (b) and (c) demonstrates the hazard of using rounded-off values in calculations.]

Problem 12.3 The base of natural logarithms (see Problem 12.2) is given by the infinite series 1 1 1 e = 2+ + + + . 2! 3! 4! Its value can be approximated by summing the first few terms of the series. How many terms are needed for the approximate value rounded off to five digits to be equal to the exact value rounded off to five digits?

The number of cycles 2(9,192,631, 770) = 18,385, 263, 540.

two

seconds

Expressed to four significant digits, this is 18,390, 000, 000 or 1.839E10 cycles. 18,390,000,000 or 1.839E10 cycles.

(a) The rounded-off value is e = 2.72. (b) e 2 = 7.38905610..., so to three significant digits it is e 2 = 7.39. (c) Squaring the three-digit number we obtained in part (a) and expressing it to three significant digits, we obtain e 2 = 7.40. (a) e = 2.72. (b) e 2 = 7.39. (c) e 2 = 7.40.

Solution: The exact value rounded off to five significant digits is e = 2.7183. Let N be the number of terms summed. We obtain the results N

Sum

2.5

2.666...

2.708333...

2.71666...

2.7180555...

2.7182539...

We see that summing seven terms gives the rounded-off value 2.7183. Seven.

Problem 12.4 The opening in the soccer goal is 24 ft wide and 8 ft high, so its area is 24 ft × 8 ft = 192 ft 2 . What is its area in m 2 to three significant digits? Solution: A = 192 ft 2

1m ( 3.281 ) = 17.8 m ft

A = 17.8 m 2 .

Problem 12.5 In 2020, teams from China and Nepal, based on their independent measurements using GPS satellites, determined that the height of Mount Everest is 8848.86 meters. Determine the height of the mountain to three significant digits (a) in kilometers; (b) in miles.

Solution: (a) The height of the mountain in kilometers to three significant digits is 8848.86 m = 8848.86 m

1 km ( 1000 ) = 8.85 km. m

(b) Its height in miles to three significant digits is 8848.86 m = 8848.86 m

ft 1 mi ( 3.281 )( 5280 ) = 5.50 mi. 1m ft

(a) 8.85 km. (b) 5.50 mi.

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Problem 12.6 The distance D = 6 in (inches). The magnitude of the moment of the force F = 12 lb (pounds) about point P is defined to be the product M P = FD. What is the value of M P in N-m (newton-meters)? F

Solution:

The value of M P is

M P = FD = (12 lb)(6 in) = 72 lb-in. Converting units, the value of M P in N-m is 72 lb-in = 72 lb-in

1N ( 0.2248 )( 1 m ) = 8.14 N-m. lb 39.37 in

8.14 N-m. P D

Problem 12.7 The length of this Boeing 737 is 110 ft 4 in and its wingspan is 117 ft 5 in. Its maximum takeoff weight is 154,500 lb. Its maximum range is 3365 nautical miles. Express each of these quantities in SI units to three significant digits.

Solution:

Because 1 ft = 12 in, the length in meters is

m (110 + 124 ft )( 0.3048 ) = 33.6 m. 1 ft In the same way, the wingspan in meters is m (117 + 125 ft )( 0.3048 ) = 35.8 m. 1 ft The weight in newtons is N ( 4.448 ) = 687,000 N. 1 lb

( 154, 500 lb )

One nautical mile is 1852 meters. Therefore, the range in meters is (3365 nautical miles)

1852 m ( 1 nautical ) = 6.23E6 m. mile

Length = 33.6 m, wingspan = 35.8 m, weight = 687 kN, range = 6230 km.

Problem 12.8 The maglev (magnetic levitation) train from Shanghai to the airport at Pudong reaches a speed of 430 km/h. Determine its speed (a) in mi/h; (b) in ft/s. Solution: (a)

(

)

km 0.6214 mi = 267mi/h . h 1 km v = 267 mi/h.

v = 430

(b)

(

km 1000 m h 1 km = 392 ft/s.

v = 430

1 ft )( 0.3048 )( 1 h ) m 3600 s

v = 392 ft/s. Source: Courtesy of Qilai Shen/EPA/Shutterstock.

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Problem 12.9 In the 2006 Winter Olympics, the men’s 15-km cross-country skiing race was won by Andrus Veerpalu of Estonia in a time of 38 minutes, 1.3 seconds. Determine his average speed (the distance traveled divided by the time required) to three significant digits (a) in km/h; (b) in mi/h.

Solution:  60 min  15 km   1.3 min  1 h  38 + 60 = 23.7 km/h.

(a) v =

(

)

v = 23.7 km/h.  1 mi   = 14.7 mi/h. (b) v = (23.7 km/h)  1.609 km  v = 14.7 mi/h.

Problem 12.10 The Porsche’s engine exerts 229 ft-lb (foot-pounds) of torque at 4600 rpm. Determine the value of the torque inN-m (newton-meters). Solution: T = 229 ft-lb

1N ( 0.2248 )( 1 m ) = 310 N-m. lb 3.281 ft

T = 310 N-m.

Problem 12.10

Problem 12.11 The kinetic energy of the man in Practice Example 12.1 is defined by 12 mv 2, where m is his mass and υ is his velocity. The man’s mass is 68 kg and he is moving at 6 m/s, so his kinetic energy is 12 (68 kg)(6 m/s) 2 = 1224 kg-m 2 /s 2. What is his kinetic energy in US customary units?

Solution:

Problem 12.12 The acceleration due to gravity at sea level in SI units is g = 9.81 m/s 2 . By converting units, use this value to determine the acceleration due to gravity at sea level in US customary units.

Solution:

Problem 12.13 The value of the universal gravitational constant in SI units is G = 6.67E−11 m 3 /kg-s 2 . Use this value and convert units to determine the value of G in US customary units.

Solution:

2  1 slug  1 ft T = 1224 kg-m 2 /s 2    14.59 kg  0.3048 m

(

)

= 903 slug-ft 2 /s. T = 903 slug-ft 2 /s.

Use Table 12.2. The result is:

1ft ( sm )( 0.3048m ) = 32.185...( sft ) = 32.2( sft ).

g = 9.81

Converting units,

6.67E − 11 m 3 /kg-s 2 = 6.67E − 11

m 3 3.281 ft 3 kg-s 2 1m

(

)

  1 kg  0.0685 slug  = 3.44E − 8 ft 3 /slug-s 2 . G = 3.44E − 8 ft 3 /slug-s 2 .

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Problem 12.14 The density (mass per unit volume) of aluminum is 2700 kg/m 3 . Determine its density in slug/ft 3 .

Solution:

Converting units, the density is

 0.0685 slug  1 m 3 2700 kg/m 3 = 2700 kg/m 3     3.281 ft 1 kg

(

)

5.24 slug/ft 3 .

Problem 12.15 The cross-sectional area of the C12×30 American Standard Channel steel beam is A = 8.81 in 2 . What is its cross-sectional area in mm 2 ?

Solution: A = 8.81 in 2

( 25.41 inmm ) = 5680 mm . 2

Problem 12.16 A pressure transducer measures a value of 300 lb/in 2. Determine the value of the pressure in pascals. A pascal (Pa) is one newton per square meter.

Solution: Convert the units using Table 12.2 and the definition of the Pascal unit. The result: 300

N 12 in 1ft ( inlb )( 4.448 )( 1 ft ) ( 0.3048 ) 1 lb m

= 2.0683...(10 6 )

( mN ) = 2.07(10 ) Pa. 2

Problem 12.17 A horsepower is 550 ft-lb/s. A watt is 1 N-m/s. Determine how many watts are generated by the engines of the passenger jet if they are producing 7000 horsepower. Solution:  550 ft-lb/s  1 m P = 7000 hp    1 hp  3.28 1ft

(

1N )( 0.2248 ) lb

= 5.22 × 10 6 W. P = 5.22 × 10 6 W.

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Problem 12.18 A typical speed of the head of a driver on the PGA Tour is 100 miles per hour. Determine the speed in ft/s, m/s, and km/h. Solution:

The speed in ft/s is

100 mi/h = 100 mi/h

ft 1h ( 5280 )( 3600 ) = 147 ft/s. 1 mi s

The speed in m/s is 147 ft/s = 147 ft/s

m ( 0.3048 ) = 44.7 m/s. 1 ft

The speed in km/h is 100 mi/h = 100 mi/h

km ( 1.609 ) = 161 km/h. 1 mi

147 ft/s, 44.7 m/s, 161 km/h.

Source: Courtesy of Takanakai/123RF.

Problem 12.19 In 2000, a carbon nanotube was shown to support a tensile stress of 63 GPa (gigapascals). A pascal is 1 N/m 2 . Determine the value of this tensile stress in lb/in 2 . (This is the amount of force in pounds that could theoretically be supported in tension by a bar of this material with a cross-sectional area of one square inch.)

Solution:

Problem 12.20 In dynamics, the moments of inertia of an object are expressed in units of ( mass ) × ( length ) 2. If one of the moments of inertia of a particular object is 600 slug-ft 2, what is its value in kg-m 2?

Solution:

Converting units, the tensile stress is

(

N 1 lb m 2 4.448 N = 9.14E6 lb/in 2 .

63E9 N/m 2 = 63E9

1m )( 39.37 ) in

9.14E6 lb/in 2 .

Converting units,

2  14.59 kg  1 m 600 slug-ft 2 = 600 slug-ft 2    1 slug  3.281 ft

(

)

= 813 kg-m 2 . 813 kg-m 2 .

Problem 12.21 defined by

The airplane’s drag coefficient C D is

D , S 12 ρ v 2 where D is the drag force exerted on the airplane by the air, S is the wing area, ρ is the density (mass per unit volume) of the air, and v is the magnitude of the airplane’s velocity. At the instant shown, the drag force D = 5300 N, the wing area is S = 100 m 2 , the air density is ρ = 1.226 kg/m 3 , and the velocity is v = 60 m/s. (a) What is the value of the drag coefficient? (b) Determine the values of D, S, ρ, and v in terms of US customary units and use them to calculate the drag coefficient. CD =

Solution: (a) The drag coefficient is CD =

D 1 S ρv 2 2

5300 N 1 ( 100 m 2 ) ( 1.226 kg/m 3 )( 60 m/s ) 2 2 = 0.0240.

Source: Courtesy of Fasttailwind/Shutterstock. (b) Converting units, the drag force is D = 5300 N

the reference area is

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lb ( 0.2248 ) = 1190 lb, 1N

S = 100 m 2

ft ( 3.281 ) = 1080 ft , 1m 2

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12.21 (Continued) the density is  0.0685 slug  1 m 3 ρ = 1.226 kg/m 3     3.281 ft 1 kg

(

)

0.00238 slug/ft 3 ,

and the velocity is v = 60 m/s

ft ( 3.281 ) = 197 ft/s. 1m

Therefore, the drag coefficient is CD = =

D 1 S ρv 2 2 1 2

1190 lb

( 1080 ft 2 ) ( 0.00238 slug/ft 3 )( 197 ft/s ) 2

= 0.0240. (a) C D = 0.0240. (b) D = 1190 lb, S = 1080 ft 2 , ρ = 0.00238 slug/ft 3 , v = 197 ft/s, C D = 0.0240.

Problem 12.22 A person has a mass of 72 kg. (a) What is their weight at sea level in newtons? (b) What is their mass in slugs? (c) What is their weight at sea level in pounds?

Solution:

W = mg = ( 4.93 slugs )( 32.2 ft/s 2 ) = 159 lb. Or we could use a conversion factor to convert their weight from newtons to pounds, obtaining

(a) From Eq. (12.6), their weight is

lb ( 0.2248 ) 1N

W = ( 706 N )

W = mg = ( 72 kg )( 9.81 m/s 2 ) = 706 N.

= 159 lb. (a) 706 N. (b) 4.93 slugs. (c) 159 lb.

(b) Using a unit conversion factor, their mass in slugs is  0.0685 slug  m = ( 72 kg )    1 kg = 4.93 slugs.

Problem 12.23 The 1 ft × 1 ft × 1 ft cube of iron weighs 490 lb at sea level. Determine the weight in newtons of a 1 m × 1 m × 1 m cube of the same material at sea level.

1 ft

The weight density is γ = 490 lb 1 ft 3

Solution:

The weight of the 1 m 3 cube is: 1 ft 3 490 lb 1 ft 1N W = γV = (1 m) 3 1 ft 3 0.3048 m 0.2248 lb = 77.0 kN.

(

)

(

)(

1 ft

)

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Problem 12.24 The area of the Pacific Ocean is 64,186,000 square miles and its average depth is 12,925 ft. Assume that the weight per unit volume of ocean water is 64 lb/ft 3. Determine the mass of the Pacific Ocean (a) in slugs; (b) in kilograms.

Solution:

The volume of the ocean is

V = (64,186, 000 mi 2 )(12, 925 ft)

( 5,1280mi ft )

= 2.312 × 10 19 ft 3 . (a) m = ρV =

64 lb/ft ( 32.2 )(2.312 × 10 ft ) ft/s 3

= 4.60 × 10 19 slugs. (b)  14.59 kg  m = (4.60 × 10 19 slugs)   1 slug  = 6.71 × 10 20 kg.

Problem 12.25 Two oranges weighing 8 oz (ounces) each at sea level are 3 ft apart. (a) What is the magnitude of the gravitational force F they exert on each other? (b) If a stationary object of mass m is subjected to a constant force F and no other forces, the time required for it to reach a speed v is mv t = . F If you subjected one of the oranges to a constant force equal to the force you determined in part (a) and no other forces, how long would it take to reach a speed of 1 ft/s?

3 ft

Solution: (a) One pound equals 10 ounces, so the weight of each orange is W = 0.5 lb, and its mass is m =

W 0.5 lb = = 0.0155 slug. g 32.2 ft/s 2

In US customary units, the universal gravitational constant is G = 3.44E − 8 ft 3 /slug-s 2 . From Eq. (12.1), the gravitational force the oranges exert on each other is Gm 2 r2 ( 3.44E − 8 ft 3 /slug-s 2 )( 0.0155 slug ) 2 = ( 3 ft ) 2 = 9.22E − 13 lb.

F =

(b) From the given equation, the time required is mv F ( 0.0155 slug )( 1 ft/s ) = 9.22E − 13 lb = 1.68E10 s (534 years).

t =

(a) 9.22E − 13 lb. (b) 1.68E10 s (534 years).

Problem 12.26 A person weighs 180 lb at sea level. The radius of the Earth is 3960 mi. What force is exerted on the person by the gravitational attraction of the Earth if he is in a space station in orbit 200 mi above the surface of the Earth?

Solution:

Use Eq. (12.5). 2

( Rr ) =  Wg  g R R+ H  3960 = W ( 3960 + 200 )

W = mg

= (180)(0.90616) = 163 lb.

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Problem 12.27 The acceleration due to gravity on the surface of the Moon is 1.62 m/s 2 . The Moon’s radius is R M = 1738 km. (a) What is the weight in newtons on the surface of the Moon of an object that has a mass of 10 kg? (b) Using the approach described in Practice Example 12.6, determine the force exerted on the object by the gravity of the Moon if the object is located 1738 km above the Moon’s surface.

Solution: (a) W = mg M = (10 kg)(1.26 m/s 2 ) = 12.6 N W = 12.6 N 2

( ) . The force is then

(b) Adapting Eq. 12.4, we have a M = g M R M r F = ma M = (10 kg)(1.62 m/s 2 )

1738 km ( 1738 km + 1738 km )

= 4.05 N F = 4.05 N.

Problem 12.28 If an object is near the surface of the Earth, the variation of its weight with distance from the center of the Earth can often be neglected. The acceleration due to gravity at sea level is g = 9.81 m/s 2. The radius of the Earth is 6370 km. The weight of an object at sea level is mg, where m is its mass. At what height above the surface of the Earth does the weight of the object decrease to 0.99 mg ?

Solution:

Problem 12.29 A person has a mass of 72 kg. The radius of the Earth is 6370 km. (a) What is his weight at sea level in newtons and in pounds? (b) If he is in a space station 420 km above the surface of the Earth, what force is exerted on him by the Earth’s gravitational attraction in newtons and in pounds?

(b) To apply Eq. (12.5), we need his distance from the center of the Earth:

 R E  2 W = mg   = 0.99mg  R E + h  Solve for the radial height, h = RE

r = 6370 km + 420 km = 6790 km. Then from Eq. (12.5), the gravitational force on him is W = mg

(a) From Eq. (12.6), his weight at sea level is

RE 2 r2

= (72 kg)(9.81 m/s 2 )

(6,370, 000 m) 2 (6, 790, 000 m) 2

= 622 N.

W = mg = (72 kg)(9.81 m/s 2 ) = 706 N. Using a conversion factor, his weight at sea level in pounds is lb ( 0.2248 ) 1N

= 159 lb.

1 − 1 ) = (6370)(1.0050378 − 1.0) ( 0.99

= 32.09 ... km = 32,100 m = 32.1 km.

Solution:

W = 706 N

Use a variation of Eq. (12.5).

The force on him in pounds is W = 622 N

lb ( 0.2248 ) 1N

= 140 lb. (a) 706 N, 159 lb. (b) 622 N, 140 lb.

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Problem 12.30 If you model the Earth as a homogeneous sphere, at what height above sea level does your weight decrease to one-half of its value at sea level? Determine the answer (a) in kilometers; (b) in miles. (The radius of the Earth is 6370 km. ) Solution:

Gm E , r2 where m E is the Earth’s mass. The acceleration due to gravity at sea level is a =

(1)

where R E is the Earth’s radius. Let H be the height above sea level at which the acceleration due to gravity equals 0.5g. That is,

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Gm E . ( RE + H ) 2

Dividing this equation by Eq. (1) yields 0.5 =

R E2 . ( RE + H ) 2

Solving for H, we obtain

(a) Your weight will decrease to 0.5 times its value at sea level at the height at which the Earth’s acceleration due to gravity decreases to 0.5 times its value at sea level. From Eq. (12.3), the acceleration due to gravity at a distance r from the center of the Earth is

Gm E g = , R E2

0.5g =

1 − 1) R ( 0.5 1 = ( − 1 ) (6370 km) 0.5

H =

= 2640 km. (b) In terms of miles, the value of H is H = 2640 km

1 mi ( 1.609 ) km

= 1640 mi. (a) 2640 km. (b) 1640 mi.

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Chapter 13 Problem 13.1 The position of an object in straight-line motion is given as a function of time by s = 6t + bt 2 m, where b is a constant. At time t = 2 s, the object’s velocity is v = 14 m/s. What are the object’s velocity and acceleration at t = 4 s?

Solution: v =

The object’s velocity is ds = 6 + 2bt m/s. dt

Setting v = 14 m/s at t = 2 s, v = 14 m/s = 6 + 2b(2 s), we see that the constant b = 2. Then the velocity at t = 4 s is v = 6 + 4t m/s = 6 + 4(4 s) m/s = 22 m/s, and the acceleration is dv a = = 4 m/s 2 . dt

v = 22 m/s, a = 4 m/s 2 .

Problem 13.2 The milling machine is programmed so that during the interval of time from t = 0 to t = 2 s, the position of its head (in inches) is given as a function of time by s = 4t − 2t 2 . What are the velocity (in in/s) and acceleration (in in/s 2 ) of the head at t = 1 s?

Solution:

The motion is governed by the equations

s = (4 in/s)t − (2 in/s 2 )t 2 , v = (4 in/s) − 2(2 in/s 2 )t, a = −2(2 in/s 2 ). At t = 1 s, we have v = 0, a = −4 in/s 2 .

Problem 13.3 Suppose you throw a ball straight up and you want it to reach a height of 20 ft above the point where you released it. With what initial velocity must you throw it? Determine the velocity in ft/s and in mi/h. The acceleration due to gravity is 32.2 ft/s 2 .

We measure the ball’s position s (positive upward) from the point where it is released. When the ball reaches its highest position s = 20 ft, its velocity v = 0. Let v 0 be the ball’s unknown initial upward velocity at time t = 0. The ball’s downward acceleration a = −32.2 ft/s 2 is constant. We can integrate to determine the velocity as a function of time: a =

Solution:

dv = −32.2 ft/s 2 , dt

∫v dv = ∫0 −32.2 dt,

v = v 0 − 32.2 t ft/s. s 5 20 ft v 5 0

(1)

Integrating again, we determine the position as a function of time: v = s

ds = v 0 − 32.2 t ft/s, dt t

∫0 ds = ∫0 (υ 0 − 32.2t ) dt, s 5 0 v 5 vo

s = v 0t −

32.2 2 t . 2

(2)

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13.3 (Continued) From Eq. (1), the time at which the ball slows to zero velocity (its highest point) is t = v 0 /32.2 s. Substituting this time and s = 20 ft into Eq. (2), 32.2 2 s = v 0t − t : 2 υ0 32.2 υ 0 2 20 ft = υ 0 , − 32.2 2 32.2

(

)

(

)

we can solve for the initial velocity, obtaining v 0 = 35.9 ft/s (24.5 miles/hr). 35.9 ft/s, 24.5 miles/hr.

Problem 13.4 Suppose you hold an object 5 ft above the floor and release it from rest. The acceleration due to gravity is 32.2 ft/s 2 . (a) How long does the object take to reach the floor? (b) How fast is it moving just before it hits the floor? Solution: s50 v50

5 ft

Floor s

where c is an integration constant. If we measure time from the instant the object is dropped, we know that v = 0 when t = 0, so the constant c = 0. Integrating again, we determine the position as a function of time: ds = 32.2 t ft/s, dt 32.2 2 s = t + d ft. 2

v =

We measure s from the position where the object is dropped (positive downward). Then s = 0 when t = 0 and the constant d = 0. We know the distance to the floor, so we can solve for the time at which it arrives there. Setting 32.2 2 s = 5 ft = t , 2 and solving, we obtain t = 0.557 s. (b) Knowing the time of arrival, we can solve for the velocity just before it hits: v = 32.2 t ft/s

(a) The object’s acceleration a = 32.2 ft/s 2 is constant. We can integrate to determine its velocity as a function of time: dv a = = 32.2 ft/s 2 , dt v = 32.2 t + c ft/s,

M13_BEDF4016_06_SE_C13_SEC_13.2.indd 11

= 32.2(0.557 s) ft/s = 17.9 ft/s. (a) 0.557 s. (b) 17.9 ft/s.

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Problem 13.5 The rocket starts from rest at t = 0 and travels straight up. Its height above the ground as a function of time can be approximated by s = bt 2 + ct 3 , where b and c are constants. At t = 10 s, the rocket’s velocity and acceleration are v = 229 m/s and a = 28.2 m/s 2 . Determine the time at which the rocket reaches supersonic speed (325 m/s). What is its altitude when that occurs?

Solution:

The governing equations are

s = bt 2 + ct 3 , v = 2bt + 3ct 2 , a = 2b + 6ct. Using the information that we have allows us to solve for the constants b and c. (229 m/s) = 2b(10 s) + 3c(10 s) 2 , (28.2 m/s 2 ) = 2b + 6c(10 s). Solving these two equations, we find b = 8.80 m/s 2 , c = 0.177 m/s 3 . When the rocket hits supersonic speed we have (325 m/s) = 2(8.80 m/s 2 )t + 3(0.177 m/s 3 )t 2 ⇒

t = 13.2 s.

The altitude at this time is s = (8.80 m/s 2 )(13.2 s) 2 + (0.177 m/s 3 )(13.2 s) 3

s = 1940 m.

Problem 13.6 The position of a point during the interval of time from t = 0 to t = 6 s is given by s = − 12 t 3 + 6t 2 + 4t m. (a) What is the maximum velocity during this interval of time, and at what time does it occur? (b) What is the acceleration when the velocity is a maximum?

Solution: 1 s = − t 3 + 6t 2 + 4t m 2 3 2 v = − 2 t + 12t + 4 m/s a = −3t + 12 m/s 2 dv = 0 (it could be a minimum) dt da This occurs at t = 4 s. At this point = −3 so we have a maximum. dt (a) Max velocity is at t = 4 s. where v = 28 m/s and (b) a = 0 m/s 2 . Maximum velocity occurs where a =

Problem 13.7 Let t = 0 denote the time at which the floatplane touches down. Its velocity in ft/s is given by the function v = 110 − 30t + 5t 2 ft/s. (a) What is the magnitude of the plane’s acceleration at the instant it touches down? (b) What distance does the plane travel from t = 0 to t = 3 s? Solution: (a) The plane’s acceleration is dv a = = −30 + 10t ft/s 2 . dt At t = 0, it is a = −30 ft/s 2 . (b) Integrating to determine the plane’s position, ds = v = 110 − 30t + 5t 2 ft/s, dt 5 s = 110t − 15t 2 + t 3 + b ft, 3

Source: Courtesy of Flyver/Alamy Stock Photo.

where b is an integration constant. If we let the position where it touches down at t = 0 be s = 0, the constant b = 0. At t = 3 s, we obtain s = 240 ft. (a) 30 ft/s 2 . (b) 240 ft.

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Problem 13.8 The rocket lifts off at time t = 0, and at t = 20 s the distance s = 929 m. If you assume that its acceleration is constant, what is its velocity at t = 20 s? Solution: Let a 0 denote the rocket’s constant acceleration. We can integrate to determine the velocity and position: dv = a0, dt v

∫0 dv = ∫0 a 0 dt, v = a 0t. s

∫0 ds = ∫0 a 0t dt, s =

1 a t 2. 2 0

Setting s = 929 m at t = 20 s, we obtain a 0 = 4.65 m/s 2 . Then the velocity at t = 20 s is

v = a 0t = (4.65 m/s 2 )(20 s) = 92.9 m/s. v = 92.9 m/s.

Source: Courtesy of Oleg_Yakovlev/Shutterstock.

Problem 13.9 The rocket lifts off at time t = 0. Its acceleration is given as a function of time in seconds by a =

7.6E6 − 9.81 m/s 2 . 5.4E5 − 2100t

What is the rocket’s velocity at t = 20 s?

Solution:

The acceleration is

T a = − g, m 0 − ct where T = 7.6E6, m 0 = 5.4E5, c = 2100, and g = 9.81 m/s 2 . We can integrate to determine the velocity as a function of time: a =

dv T = − g, dt m 0 − ct t



∫0 dv = ∫0  m 0 − ct − g  dt, t

v = ∫ 0

T dt − gt. m 0 − ct

To evaluate this integral, we make the change of variable z = m 0 − ct. The differential dz = −c dt, and we obtain 1 m 0 − ct T dz − gt c ∫m0 z m 0 1 T dz = ∫ − gt c m 0 − ct z T  m 0  = ln   − gt. c  m 0 − ct 

v = −

Substituting the numerical values, at t = 20 s we obtain v = 96.8 m/s. In the solution to Problem 13.5, in which the acceleration was assumed to be constant, we obtained v = 92.9 m/s. v = 96.8 m/s.

Source: Courtesy of Oleg_Yakovlev/Shutterstock.

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Problem 13.10 During a strong earthquake, the horizontal position of the ground due to the dominant P wave is given as a function of time in seconds by s = 40 sin(1.25t ) mm. (a) What are the resulting maximum magnitudes of the velocity and acceleration of the ground? (b) When the magnitude of the velocity is a maximum, what is the magnitude of the acceleration?

If we look at the behavior of the cosine and sine,

Solution:

–0.5

We can differentiate the position to determine the velocity and acceleration:

1.5 1 0.5

–1.5

ds = 40(1.25) cos1.25t mm/s, dt dv = −40(1.25) 2 sin1.25t mm/s 2 . a = dt

The magnitude of the velocity is a maximum when cos1.25t = ±1. Its value is

cos

–1

s = 40 sin1.25t mm, v =

sin

6 8 Angle, radians

when the cosine equals plus or minus one, the sine equals zero. Therefore the acceleration equals zero when the magnitude of the velocity is a maximum.

v max = 40(1.25) mm/s = 50 mm/s.

(a) v max = 50 mm/s, a max = 62.5 mm/s 2 . (b) Zero.

The magnitude of the acceleration is a maximum when sin1.25t = ±1. Its value is a max = 40(1.25) 2 mm/s 2 = 62.5 mm/s 2 .

Whenever it’s practical, it’s good idea to look at the graph of the velocity in a situation like this: 14 12 10 Velocity, mm/s

Problem 13.11 In an assembly operation, the robot’s arm moves along a straight horizontal line. During the interval of time from t = 0 to t = 2 s, the position of the arm is given by s = 6t 3 − 2t 4 mm. (a) Determine the maximum magnitude of the velocity during this interval of time. (b) What is the position when the velocity is a maximum? s

8 6 4 2 0

0.5

1.5

Time, seconds (b) Substituting t = 1.5 s into the expression for the position, we obtain s = 10.1 mm.

Solution:

(a) 13.5 mm/s. (b) s = 10.1 mm.

(a) We can differentiate to determine the velocity and acceleration as functions of time: s = 6t 3 − 2t 4 mm, ds = 18t 2 − 8t 3 mm/s, dt dv = 36t − 24t 2 mm/s 2 . a = dt

v =

Setting the acceleration equal to zero, we find that the derivative of the velocity is zero at t = 0 and at t = 1.5 s. Substituting these times into the expression for the velocity, we obtain v = 0 and v = 13.5 mm/s. We also check the magnitude of the velocity at the end of the interval t = 2 s, and find it is v = 8 mm/s. The maximum velocity is 13.5 mm/s.

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Problem 13.12 In an assembly operation, the robot’s arm moves along a straight horizontal line. During the interval of time from t = 0 to t = 2 s, the acceleration of the arm is given by a = 36t − 24t 2 mm/s 2 . At t = 0, its position is s = 0 and its velocity is 20 mm/s. What are the position and velocity at t = 2 s? s

Solution:

We integrate the acceleration to determine the velocity,

obtaining dv = 36t − 24t 2 mm/s 2 , dt v = 18t 2 − 8t 3 + c mm/s, a =

where c is an integration constant. From the given value of the velocity at t = 0, we see that c = 20 mm/s. Integrating again, the position is ds = 18t 2 − 8t 3 + 20 mm/s, dt s = 6t 3 − 2t 4 + 20t + d mm,

v =

where d is another integration constant. From the given initial condition, d = 0. From the obtained expressions, v = 28 mm/s, s = 56 mm. s = 56 mm, v = 28 mm/s.

Problem 13.13 A Porsche starts from rest at time t = 0. During the first 10 seconds of its motion, its velocity in km/h is given by v = 22.8t − 0.88t 2 , where t is in seconds. (a) What is the car’s maximum acceleration in m/s 2 , and when does it occur? (b) What distance in km does the car travel during the 10 seconds?

Solution:

First convert the numbers into meters and seconds

(

1 hr )( 3600 ) = 6.33 m/s s

(

1 hr )( 3600 ) = 0.244 m/s s

22.8

km 1000 m hr 1 km

0.88

km 1000 m hr 1 km

The governing equations are then 1 1 (6.33 m/s)(t 2 /s) − (0.88 m/s)(t 3 /s 2 ), 2 3

s =

v = (6.33 m/s)(t /s) − (0.88 m/s)(t 2 /s 2 ), a = (6.33 m/s 2 ) − 2(0.88 m/s)(t /s 2 ), d a = −2(0.88 m/s)(1/s 2 ) = −1.76 m/s 3 . dt The maximum acceleration occurs at t = 0 (and decreases linearly from its initial value). a max = 6.33 m/s 2@ t = 0. In the first 10 seconds the car travels a distance

(

)

(

)

(

1 1 km (10 s) 2 km (10 s) 3  1 hr s =  22.8 − 0.88  s 2 hr 3 hr s 2  3600 s

)

s = 0.235 km.

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Problem 13.14 The acceleration of a point is a = 20t m/s 2 . When t = 0, s = 40 m and v = −10 m/s. What are the position and velocity at t = 3 s?

where C 2 is the constant of integration.

Solution:

At t = 0, s = 40 m, thus C 2 = 40. The position is

The velocity is

s = ∫ (10t 2 − 10)dt + C 2 =

( 103 )t − 10t + C . 3

( 103 )t − 10t + 40 m.

v = ∫ ad t + C1 ,

s =

where C1 is the constant of integration. Thus

At t = 3 seconds,

v = ∫ 20t dt + C1 = 10t 2 + C1 .

10 s =  t 3 − 10t + 40  = 100 m.  3  t =3

At t = 0, v = −10 m/s, hence C1 = −10 and the velocity is v = 10t 2 − 10 m/s. The position is

The velocity at t = 3 seconds is

s = ∫ v dt + C 2 ,

v = [10t 2 − 10] t = 3 = 80 m/s.

Problem 13.15 The accelerations of gravity at the surfaces of the Earth, the Moon, and Mars are 9.81 m/s 2 , 1.62 m/s 2 , and 3.72 m/s 2 , respectively. Suppose that an object is projected upward from the surfaces of each of the three celestial bodies at a speed of 25 m/s. How high does the object rise in each case? (The gravitational accelerations can be treated as constants.)

From Eq. (1), the time at which the object’s velocity decreases to zero (its highest point) is t = v 0 /g. Substituting this time into Eq. (2), the height reached is  v  1  v 2 s = v 0  0  − g  0   g 2  g

Solution:

Let g denote the acceleration due to gravity and let v 0 be the initial velocity. We can integrate the constant acceleration to determine the velocity and position as functions of time: a =

dv = −g, dt

From this expression with v 0 = 25 m/s and the three given values of the gravitational acceleration, we obtain s Earth = 31.9 m (105 ft), s Moon = 193 m (633 ft), s Mars = 84.0 m (276 ft). Earth: 31.9 m, Moon: 193 m, Mars: 84.0 m.

∫v dv = −g ∫0 dt, 0

v = s

ds = v 0 − gt, dt

v 02 . 2g

(1)

∫0 ds = ∫0 (v 0 − gt ) dt, s = v 0t −

1 2 gt . (2) 2

Problem 13.16 As a first approximation, a bioengineer studying the mechanics of bird flight assumes that the gull takes off with constant acceleration. Video measurements indicate that it requires a distance of 4.3 m to take off and is moving horizontally at 6.1 m/s when it does. What is its acceleration?

Solution:

The governing equations are

a = constant, v = at, s =

1 2 at . 2

Using the information given, we have 6.1 m/s = at, 4.3 m =

1 2 at . 2

Solving these two equations, we find t = 1.41 s and a = 4.33 m/s 2 .

Source: Courtesy of Janet Griffin/Shutterstock.

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Problem 13.17 Progressively developing a more realistic model, the bioengineer next models the acceleration of the gull (in m/s 2 ) by the equation a = C (0.5 + sin ωt ), where C and ω are constants. From video measurements of it taking off, he estimates that ω = 14 rad/s and determines that the bird requires 1.42 s to take off and is moving horizontally at 6.1 m/s when it does. What is the constant C ?

Solution:

Writing

dv a = = C (0.5 + sin wt ) dt and integrating, we have 6.1 m/s

∫0

d v = 6.1 m/s = C ∫

1.42 s 0

(0.5 + sin wt ) dt.

Evaluating the integral, we obtain C = 8.19 m/s 2 . C = 8.19 m/s 2 .

Source: Courtesy of Janet Griffin/Shutterstock.

Problem 13.18 Missiles designed for defense against ballistic missiles have attained accelerations in excess of 100 g’s, or 100 times the acceleration due to gravity. Suppose that the missile shown lifts off from the ground and has a constant acceleration of 100 g’s. How long does it take to reach an altitude of 3000 m? How fast is it going when it reaches that altitude? Solution:

The governing equations are

a = constant, v = at, s =

1 2 at 2

Using the given information we have 3000 m =

1 100(9.81 m/s 2 )t 2 ⇒ 2

t = 2.47 s.

The velocity at that time is v = 100(9.81 m/s 2 )(2.47 s) = 243 m/s.

v = 2430 m/s.

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Problem 13.19 The missile shown lifts off from the ground and, because it becomes lighter as its fuel is expended, its acceleration (in g’s) is given as a function of time in seconds by a =

100 . 1 − 0.15t

What is its velocity in kilometers per hour 2 s after liftoff? Solution:

We can integrate the constant acceleration to determine the velocity and position as functions of time: 100 g dv = , dt 1 − 0.15t v 2 s 100(9.81 m/s 2 ) ∫0 dv = ∫0 1 − 0.15t dt, = 2330 m/s. a =

This is 2330 m/s

1 km s ( 1000 )( 3600 ) = 8400 km/h (5218 mi/h). m 1 hr

8400 km/h.

Problem 13.20 The airplane releases its drag parachute at time t = 0. Its velocity is given as a function of time by 80 m/s. v = 1 + 0.32t What is the airplane’s acceleration at t = 3 s? Solution: v =

−25.6 80 dv ;a = ⇒ a(3 s) = −6.66 m/s 2 . = 1 + 0.32t dt (1 + 0.32t ) 2

Problem 13.21 How far does the airplane in Problem 13.20 travel during the interval of time from t = 0 to t = 10 s? Solution: v =

(

)

10 s 80 80 1 + 3.2 ;s = ∫ dt = 250 ln = 359 m. 0 1 + 0.32t 1 + 0.32t 1

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Problem 13.22 The velocity of a bobsled is v = 10t ft/s. When t = 2 s, the position of the sled is s = 25 ft. What is its position when t = 10 s?

Solution: The equation for straight line displacement under constant acceleration is a(t − t 0 ) 2 s = + v (t 0 )(t − t 0 ) + s(t 0 ). 2 Choose t 0 = 0. At t = 2, the acceleration is dv (t )  a =  = 10 ft/s 2 ,  dt  t = 2 the velocity is v (t 0 ) = 10(2) = 20 ft/s, and the initial displacement is s(t 0 ) = 25 ft. At t = 10 seconds, the displacement is s =

Problem 13.23 In September 2003, Tony Schumacher started from rest and drove a quarter mile (1320 ft) in 4.498 s in a National Hot Rod Association race. His speed as he crossed the finish line was 328.54 mi/h. Assume that the car’s acceleration can be expressed by a linear function of time a = b + ct. (a) Determine the constants b and c. (b) What was the car’s speed 2 s after the start of the race?

10 (10 − 2) 2 + 20(10 − 2) + 25 = 505 ft. 2

Solution: a = b + ct, v = bt +

Both constants of integration are zero.  88 ft/s  c (a) 328.54 mph   = b(4.498 s) + (4.498 s) 2  60 mph  2 b c (4.498 s) 2 + (4.498 s) 3 2 6

1320 ft = ⇒

b = 177 ft/s 2 c = −31.16 ft/s 3 .

(b) v = b(2 s) +

Problem 13.24 The velocity of an object is v = 200 − 2t 2 m/s. When t = 3 s, the position of the object is s = 600 m. What are the position and acceleration of the object at t = 6 s?

ct 2 bt 2 ct 3 , s = + 2 2 6

Solution:

c (2 s) 2 = 292 ft/s. 2

The acceleration is

dv (t ) = −4t m/s 2 . dt At t = 6 seconds, the acceleration is a = −24 m/s 2 . Choose the initial conditions at t 0 = 3 seconds. The position is obtained from the velocity: 6

6 2 s(t − t 0 ) = ∫ v (t ) dt + s(t 0 ) =  200t − t 3  + 600 = 1070 m.  3 3 3

Problem 13.25 An inertial navigation system measures the acceleration of a vehicle from t = 0 to t = 6 s and determines it to be a = 2 + 0.1t m/s 2 . At t = 0, the vehicle’s position and velocity are s = 240 m and v = 42 m/s, respectively. What are the vehicle’s position and velocity at t = 6 s?

Solution: a = 2 + 0.1t m/s 2 v 0 = 42 m/s s 0 = 240 m Integrating v = v 0 + 2t + 0.1t 2 /2 s = v 0 t + t 2 + 0.1t 3 /6 + s 0 Substituting the known values at t = 6 s, we get v = 55.8 m/s s = 531.6 m.

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Problem 13.26 You step out of your car at a game park and are seen by a resting lion 50 m away. The lion runs toward you, has constant acceleration until it reaches its maximum speed of 22 m/s after 3 s, and then runs with constant speed. How long do you have from the instant the lion started to get back into the car?

Solution: Let a 0 denote the lion’s initial constant acceleration. We integrate to determine the velocity as a function of time during the first three seconds: dv = a0, dt v

∫0 dv = ∫0 a 0 dt, v = a 0t. We know that v = 22 m/s at t = 3 s, so the acceleration is 22 m/s 3s = 7.33 m/s 2 .

a0 =

Now we can integrate to determine the position as a function of time: v =

ds = a 0t, dt

∫v ds = ∫0 a 0t dt, s =

Source: Courtesy of Bucky_za/E+/Getty Images.

1 a t 2. 2 0

At t = 3 s, the distance the lion has traveled is 1 a t2 2 0 1 = (7.33 m/s 2 )(3 s) 2 2 = 33 m.

s =

At this point the lion is 17 meters from you and traveling at his maximum speed. The time it takes him to travel that 17 meters is 17 m 22 m/s = 0.773 s.

t =

We see that you have 3.77 seconds to get into the car. 3.77 s.

Problem 13.27 The graph shows the airplane’s acceleration during takeoff. What is the airplane’s velocity when it rotates (lifts off) at t = 30 s? Solution: v =

Velocity = Area under the curve

1 (3 ft/s 2 + 9 ft/s 2 )(5 s) + (9 ft/s 2 )(25 s) = 255 ft/s. 2

a 9 ft/s2

3 ft/s2 0

30 s

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Problem 13.28 The graph shows the airplane’s acceleration during takeoff. What distance has the airplane traveled when it lifts off at t = 30 s? Solution:

for 0 ≤ t ≤ 5 s

a =

( 6 5ft/ss )t + (3 ft/s ), v = ( 6 5ft/ss ) t2 + (3 ft/s )t

s =

( 6 5ft/ss ) t6 + (3 ft/s ) t2

v(5 s) = 30 ft/s, s(5 s) = 62.5 ft

for 5 s ≤ t ≤ 30 s

9 ft/s2

a = 9 ft/s 2 , v = (9 ft/s 2 )(t − 5 s) + 30 ft/s, s = (9 ft/s 2 )

(t − 5 s) 2 + (30 ft/s)(t − 5 s) + 62.5 ft 2

3 ft/s2 0

⇒ s(30 s) = 3625 ft.

Problem 13.29 A car is traveling at 15 mi/h when a traffic light 60 ft ahead turns yellow. The driver requires 1 s to react before applying the brakes. What constant rate of deceleration will cause the car to come to a stop just as it reaches the light?

30 s

From Eq. (1), the time at which the velocity decreases to zero is t = −

22 . a

(Notice from this expression that the acceleration a is negative.) Substituting this expression and s = 38 ft into Eq. (2) gives

( 22a ) + 12 a( − 22a ) .

38 = 22 − 15 mi/h

Solving for a, we obtain a = −12.7 ft/s 2 . 12.7 ft/s 2 . 60 ft

Solution:

The car’s initial velocity in ft/s is

15 mi/h = 15 mi/h

ft 1h ( 5280 )( 3600 ) = 22 ft/s. 1 mi s

The car travels 22 ft before the brakes are applied, so it is 60 ft − 22 ft = 38 ft from the light when it begins decelerating. We can integrate the (unknown) constant acceleration to determine the velocity and position as functions of time. Let t = 0 and s = 0 correspond to the point at which it starts decelerating: dv = a, dt v

∫22 ft/s dv = ∫0 a dt, ds = 22 ft/s + at, dt

v = s

(1)

∫0 ds = ∫0 (22 ft/s + at ) dt, s = 22t +

1 2 at ft. 2

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Problem 13.30 A car is traveling at 15 mi/h when a traffic light 60 ft ahead turns yellow. The driver requires 1 s to react before stepping on the gas. If the car accelerates at a constant rate of 8 ft/s 2 and the light stays yellow for 2 s, will the driver reach the light before it turns red? How fast is the car moving when it reaches the light?

Solution:

The car’s initial velocity in ft/s is

15 mi/h = 15 mi/h

ft 1h ( 5280 )( 3600 ) = 22 ft/s. 1 mi s

The car travels 22 ft before it begins accelerating, so it is 60 ft − 22 ft = 38 ft from the light. We can integrate the constant acceleration to determine the velocity and position as functions of time. Let t = 0 and s = 0 correspond to the point at which it starts accelerating: dv = a, dt

15 mi/h

∫22 ft/s dv = ∫0 a dt, ds = 22 ft/s + at, dt

v = 60 ft s

(1)

∫0 ds = ∫0 (22 ft/s + at ) dt, s = 22t +

1 2 at ft. 2

(2)

Setting s = 38 ft and a = 8 ft/s 2 in Eq. (2), we obtain a quadratic equation for the time it takes for the car to reach the light after he starts accelerating. The positive root of that equation is t = 1.38 s. We see that he reaches the light 2.38 seconds after it turns yellow, and does not beat the light. From Eq. (1), the car’s velocity when it reaches the light is v = 22 ft/s + at = 22 ft/s + (8 ft/s 2 )(1.38 s) = 33.0 ft/s (22.5 mi/h). No, 33.0 ft/s (22.5 mi/h).

Problem 13.31 A high-speed rail transportation system has a top speed of 100 m/s. For the comfort of the passengers, the magnitude of the acceleration and deceleration is limited to 2 m/s 2 . Determine the minimum time required for a trip of 100 km. Strategy: A graphical approach can help you solve this problem. Recall that the change in the position from an initial time t 0 to a time t is equal to the area defined by the graph of the velocity as a function of time from t 0 to t.

Solution:

Divide the time of travel into three intervals: The time required to reach a top speed of 100 m/s, the time traveling at top speed, and the time required to decelerate from top speed to zero. From symmetry, the first and last time intervals are equal, and the distances traveled during these intervals are equal. The initial time is obtained from v (t 1 ) = at 1 , from which t 1 = 100/2 = 50 s. The distance traveled during this time is s(t 1 ) = at 12 /2 from which s(t 1 ) = 2(50) 2 /2 = 2500 m. The third time interval is given by v (t 3 ) = −at 3 + 100 = 0, from which t 3 = 100/2 = 50 s. a Check. The distance traveled is s(t 3 ) = − t 32 + 100t 3 , from 2 which s(t 3 ) = 2500 m. Check. The distance traveled at top speed is s(t 2 ) = 100000 − 2500 − 2500 = 95000 m = 95 km. The time of travel is obtained from the distance traveled at zero acceleration: s(t 2 ) = 95000 = 100t 2 , from which t 2 = 950. The total time of travel is t total = t 1 + t 2 + t 3 = 50 + 950 + 50 = 1050 s = 17.5 minutes . A plot of velocity versus time can be made and the area under the curve will be the distance traveled. The length of the constant speed section of the trip can be adjusted to force the length of the trip to be the required 100 km.

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Problem 13.32 The graph closely models the sprinter’s acceleration. After 4.2 s, his acceleration remains approximately zero. (a) What maximum speed does he attain? (b) How long does it take him to run 100 m?

From t = 3 s to t = 4.2 s, his acceleration is also a linear function of time. We can write it in the general form a = pt + q, where p and q are constants. Using the conditions that a = 3b at t = 3 s and a = 0 at t = 4.2 s to determine p and q, we obtain the function a = b(10.5 − 2.5t ). We integrate to determine the velocity. dv = b(10.5 − 2.5t ), dt 2.5 2 t + c. v = b 10.5t − 2 a =

(

)

Using the fact that v = 4.5b m/s at t = 3 s to evaluate the constant c, we obtain 2.5 2 v = b −15.75 + 10.5t − t . 2

(

)

Source: Courtesy of Image Source Trading Ltd/Shutterstock.

Now we integrate to determine the position.

6.6 m/s2

v =

(

s = b −15.75t +

4.2 s

)

ds 2.5 2 = b −15.75 + 10.5t − t , dt 2

Using the fact that s = 4.5b m at t = 3 s to evaluate the constant d gives

(

s = b 15.75 − 15.75t +

Solution: From t = 0 to t = 3 s, his acceleration is described by the linear function a = bt m/s 2 , where the constant b = 2.2. Integrating to obtain his velocity, we obtain dv = bt, dt b v = t 2 + c. 2 a =

)

10.5 2 2.5 3 t + d. t − 2 6

)

10.5 2 2.5 3 t − t . 2 6

Using these expressions, his velocity and position at t = 4.2 s are v t = 4.2 s = 6.3b m/s = 13.9 m/s, s t = 4.2 s = 11.34 b m = 24.9 m. After 4.2 s, his velocity is constant, v = 13.9 m/s (31.0 miles/hr). That is his maximum speed. His position after 4.2 s is

He starts from rest, so the constant c = 0. Integrating again to obtain his position, we obtain a = −3.86 ft/s 2 .

s = s t = 4.2 s + v t = 4.2 s (t − 4.2 s). = 24.9 m + (13.9 m/s)(t − 4.2 s).

b ds = t 2, dt 2 b 3 s = t + d, 6

Setting s = 100 m in this expression and solving for t, we obtain t = 9.62 s.

v =

(a) v = 13.9 m/s. (b) 9.62 s.

and if we measure his position from where he starts, the constant d = 0. At t = 3 s, his velocity and position are b (3 s) 2 = 4.5b m/s = 9.9 m/s, 2 b s t = 3 s = (3 s) 3 = 4.5b m = 9.9 m. 6

v t =3 s =

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Problem 13.33 A race car starts from rest and accelerates at a = 5 + 2t ft/s 2 for 10 s. The brakes are then applied, and the car has a constant acceleration a = −30 ft/s 2 until it comes to rest. Determine (a) the maximum velocity, (b) the total distance traveled, and (c) the total time of travel.

(b) The distance traveled in the first interval is s(10) = ∫

10 0

5 1 10 (5t + t 2 ) dt =  t 2 + t 3  = 583.33 ft. 2 3 0

The time of travel in the second interval is v (t 2 − 10) = 0 = a(t 2 − 10) + v (10), t 2 ≥ 10 s, from which

Solution:

(t 2 − 10) = −

(a) For the first interval, the velocity is v (t ) = ∫ (5 + 2t )dt + v (0) = 5t + t 2

since v (0) = 0. The velocity is an increasing monotone function; hence the maximum occurs at the end of the interval, t = 10 s, from which

from which (b)

v max = 150 ft/s.

s(5) =

Problem 13.34 When t = 0, the position of a point is s = 6 m, and its velocity is v = 2 m/s. From t = 0 to t = 6 s, the acceleration of the point is a = 2 + 2t 2 m/s 2 . From t = 6 s until it comes to rest, its acceleration is a = −4 m/s 2 . (a) What is the total time of travel? (b) What total distance does the point move? Solution:

For the first interval the velocity is

2 v (t ) = ∫ (2 + 2t 2 )dt + v (0) =  2t + t 3  + 2 m/s.  3 

(

−30 2 5 + 150(5) + 583.33 = 958.33 ft. 2

The total time of travel is (a)

t total = 39.5 + 6 = 45.5 seconds.

(b) The distance traveled is −4 (t − 6) 2 + v (6)(t − 6) + s(6) 2 = −2(39.5) 2 + 158(39.5) + 270,

s(t − 6) =

from which the total distance is s total = 3390 m.

)

2 3 1 t + 2 dt + 6 =  t 2 + t 4 + 2t  + 6.   3 6

Problem 13.35 Zoologists studying the ecology of the Serengeti Plain estimate that the average adult cheetah can run 100 km/h and the average springbok can run 65 km/h. If the animals run along the same straight line, start at the same time, are each assumed to have constant acceleration, and reach top speed in 4 s, how close must a cheetah be when the chase begins to catch a springbok in 15 s?

a (t − 10) 2 + v (10)(t 2 − 10) + s(10), 2 2

The displacement at the end of the interval is s(6) = 270 m. For the second interval, the velocity is v (t − 6) = a(t − 6) + v (6) = 0, t ≥ 6, from which v (6) 158 (t − 6) = − = − = 39.5. a −4

The velocity at the end of the interval is v (6) = 158 m/s. The displacement in the first interval is s(t ) = ∫ 2t +

150 = 5, and −30

Solution: The top speeds are Vc = 100 km/h = 27.78 m/s for the cheetah, and Vs = 65 km/h = 18.06 m/s. The acceleration is V V a c = c = 6.94 m/s 2 for the cheetah, and a s = s = 4.513 m/s 2 4 4 for the springbuck. Divide the intervals into the acceleration phase and the chase phase. For the cheetah, the distance traveled in the first 6.94 2 is s c (t ) = (4) = 55.56 m. The total distance traveled at the 2 end of the second phase is s total = Vc (11) + 55.56 = 361.1 m. For the springbuck, the distance traveled during the acceleration phase is 4.513 2 s s (t ) = (4) = 36.11 m. The distance traveled at the end of the 2 second phase is s s (t ) = 18.06(11) + 36.1 = 234.7 m. The permissible separation between the two at the beginning for a successful chase is d = s c (15) − s s (15) = 361.1 − 234.7 = 126.4 m.

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Problem 13.36 Suppose that a person unwisely drives 75 mi/h in a 55 mi/h zone and passes a police car going 55 mi/h in the same direction. If the police officers begin constant acceleration at the instant they are passed and increase their velocity to 80 mi/h in 4 s, how long does it take them to be even with the pursued car?

The distance traveled during acceleration is

Solution:

The separation between the two cars at 4 seconds is

1.467

The conversion from mi/h to ft/s is

9.169 2 (4) + 55(1.467)(4) = 396 ft. 2

The distance traveled by the pursued car during this acceleration is s c (t 1 ) = 75(1.467)t 1 = 110(4) = 440 ft.

d = 440 − 396 = 44 ft.

h ( mi ft− −second ).

This distance is traversed in the time

The acceleration of the police car is a =

s(t 1 ) =

t2 =

(80 − 55)(1.467) ft/s = 9.169 ft/s 2 . 4s

44 = 6. (80 − 75)(1.467)

The total time is t total = 6 + 4 = 10 seconds.

Problem 13.37 If θ = 1 rad and d θ /dt = 1 rad/s, what is the velocity of P relative to O? Strategy: You can write the position of P relative to O as s = (2 ft) cos θ + (2 ft) cos θ and then take the derivative of this expression with respect to time to determine the velocity.

Solution:

The distance s from point O is

s = (2 ft) cos θ + (2 ft) cos θ. The derivative is ds dθ = −4 sin θ . dt dt For θ = 1 radian and

dθ = 1 radian/second, dt

ds = v (t ) = −4(sin(1 rad)) = −4(0.841) = −3.37 ft/s. dt 2 ft

2 ft

u O P

Problem 13.38 If θ = 1 rad, d θ /dt = −2 rad/s, and d 2 θ /dt 2 = 0, what are the velocity and acceleration of P relative to O?

Solution:

The velocity is

ds dθ = −4 sin θ = −4(sin (1 rad))(−2) = 6.73 ft/s . dt dt The acceleration is

( )

2 ft

( )

d 2s dθ 2 d 2θ = −4 cos θ − 4 sin θ , dt 2 dt dt 2

2 ft

from which d 2s = a = −4 cos(1 rad)(4) = −8.64 ft/s 2 . dt 2

u O s

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Problem 13.39* If θ = 1 rad and d θ /dt = 1 rad/s, what is the velocity of P relative to O?

For θ = 1 radian, α = 0.4343 radians. The position relative to O is. s = 200 cos θ + 400 cos α. The velocity is

( )

ds dθ dα = v (t ) = −200 sin θ − 400 sin α . dt dt dt From the expression for the angle

400 mm

200 mm

α, cos α u

( ddtα ) = 0.5cos θ ( ddtθ ),

from which the velocity is P

v (t ) = (−200 sin θ − 200 tan α cos θ)

( ddtθ ).

substitute: v (t ) = −218.4 mm/s.

Solution:

The acute angle formed by the 400 mm arm with the horizontal is given by the sine law: 200 400 = , sin α sin θ

for which sin α =

( 200 )sin θ. 400

Problem 13.40 In Practice Example 13.5, determine the time required for the plane’s velocity to decrease from 50 m/s to 10 m/s. Solution:

From Active Example 13.4 we know that the acceleration is given by a = −(0.004/m)v 2 .

We can find an expression for the velocity as a function of time by integrating dv dv = −(0.004/m)v 2 ⇒ 2 = −(0.004/m)t dt v

a =

10 m/s dv

1 10 m/s

∫50 m/s v 2 = ( − v )50 m/s = ( −10 m + 50 m ) = ( − 50 m ) 1s

= −(0.004/m)t 1m ( 504 sm )( −0.004 ) = 20 s.

t = −

t = 20 s.

Problem 13.41 An engineer designing a system to control a router for a machining process models the system so that the router’s acceleration (in in/s 2 ) during an interval of time is given by a = −0.4v , where v is the velocity of the router in in/s. When t = 0, the position is s = 0 and the velocity is v = 2 in/s. What is the position at t = 3 s?

Solution:

We will first find the velocity at t = 3.

0.4 dv = − v, s dt v2 3 s − 0.4 dv dt, ∫ 2 in/s v = ∫0 s v2 −0.4 ln (3 s) = −1.2 = 2 in/s s υ 2 = (2 in/s)e −1.2 = 0.602 in/s.

( ) ( ) ) ( )

a =

(

Now we can find the position a =

v dv −0.4 = v, ds s

(

0.602 in/s

∫ 2 in/s

)

s2 v dv −0.4 ds = ∫ 0 s v

(

)

(0.602 in/s) − (2 in/s) = s

s2 =

( −0.4 )s s

−1.398 in/s = 3.49 in. −0.4/s

s 2 = 3.49 in.

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Problem 13.42 The boat is moving at 10 m/s when its engine is shut down. Due to hydrodynamic drag, its subsequent acceleration is a = −0.05v 2 m/s 2 , where v is the velocity of the boat in m/s. What is the boat’s velocity 4 s after the engine is shut down?

Solution: a =

dv = −(0.05 m −1 )v 2 dt dv

∫10 m/s v 2 = −(0.05 m −1 ) ∫0 dt ⇒ − 1v 10 m/s = −(0.05 m −1 )t v =

10 m/s 1 + (0.5 s −1 )t

v(4 s) = 3.33 m/s.

Problem 13.43 In Problem 13.42, what distance does the boat move in the 4 s following the shutdown of its engine?

Solution: v =

From Problem 13.42 we know

4s ds 10 m/s 10 m/s = ⇒ s(4 s) = ∫ dt 0 dt 1 + (0.5 s −1 )t 1 + (0.5 s −1 )t

2 + (1 s −1 )(4 s)  s(4 s) = (20 m) ln   = 21.97 m.  2

Problem 13.44 A steel ball is released from rest in a container of oil. Its downward acceleration is a = 2.4 − 0.6v in/s 2 , where v is the ball’s velocity in in/s. What is the ball’s downward velocity 2 s after it is released?

Solution: dv = (2.4 in./s) − (0.6 s −1 )v dt v t dυ ∫0 (2.4 in./s) − (0.6 s −1 )υ v = ∫0 dt

a =

5 v + 4 in./s = t ⇒ v = (4 in./s)(1 − e −(0.6 s−1 )t ) − ln 3 4 in./s

(

)

v (2 s) = 2.795 in./s.

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Problem 13.45 In Problem 13.44, what distance does the ball fall in the first 2 s after its release?

Solution: v =

From 13.44 we know

ds = (4 in./s )(1 − e −(0.6 s−1 )t ) dt t

s(2 s) = ∫ (4 in./s)(1 − e −(0.6 s−1 )t ) dt 0

20 in. (−0.6 s−1 )t (e − 1) + (4 in./s)t 3

s(2 s) = 3.34 in.

Problem 13.46 The greatest ocean depth yet discovered is the Marianas Trench in the western Pacific Ocean. A steel ball released at the surface requires 64 min to reach the bottom. The ball’s downward acceleration is a = 0.9 g − cv , where g = 9.81 m/s 2 and the constant c = 3.02 s −1 . What is the depth of the Marianas Trench in kilometers?

v =

0.9 g ds = (1 − e −ct ). dt c

Integrating, t 0.9 g

∫0 ds = ∫0

(1 − e −ct ) dt.

We obtain 0.9 g 1 e −ct t+ s = − . c c c At t = (64)(60) = 3840 s, we obtain

(

Solution: a =

Integrating and solving for v ,

dv = 0.9 g − cv . dt

)

s = 11,225 m.

Separating variables and integrating, v

∫0 0.9g − cv = ∫0 dt = t. Problem 13.47 The acceleration of a regional airliner during its takeoff run is a = 14 − 0.0003v 2 ft/s 2 , where v is its velocity in ft/s. How long does it take the airliner to reach its takeoff speed of 200 ft/s?

Solution: a =

dv = (14 ft/s 2 ) − (0.0003 ft −1 )v 2 dt

200 ft/s

∫0

t dv = ∫ dt 0 (14 ft/s 2 ) − (0.0003 ft −1 )v 2

t = 25.1 s.

Problem 13.48 In Problem 13.47, what distance does the airliner require to take off?

Solution: a = v

dv = (14 ft/s 2 ) − (0.0003 ft −1 )v 2 ds

200 ft/s

∫0

v dv

(14 ft/s 2 ) − (0.0003 ft −1 )v 2

= ∫ ds 0

s = 3243 ft.

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Problem 13.49 A sky diver jumps from a helicopter and is falling straight down at 30 m/s when her parachute opens. From then on, her downward acceleration is approximately a = g – cv 2 , where g = 9.81 m/s 2 and c is a constant. After an initial “transient” period, she descends at a nearly constant velocity of 5 m/s. (a) What is the value of c, and what are its SI units? (b) What maximum deceleration is the sky diver subjected to? (c) What is her downward velocity when she has fallen 2 m from the point at which her parachute opens?

Solution: (a)

Assume c > 0.

After the initial transient, she falls at a constant velocity, so that the acceleration is zero and cv 2 = g, from which c =

g 9.81 m/s 2 = = 0.3924 m −1. v2 (5) 2 m 2 /s 2

(b) The maximum acceleration (in absolute value) occurs when the parachute first opens, when the velocity is highest: a max = g − cv 2 = g − c(30) 2 = 343.4 m/s 2 . (b)

Choose is coordinates such that distance is measured positive downward. The velocity is related to position by the chain rule: dv dv ds dv = = v = a, dt ds dt ds from which v dv = ds. g − cv 2 Integrate:

( − 21c )ln g − cv

= s + C.

When the parachute opens s = 0 and v = 30 m/s, from which

( 21c )ln g − 900c = −7.4398.

C = −

The velocity as a function of distance is ln g − cv 2 = −2c(s + C ). For s = 2 m, v = 14.4 m/s.

Problem 13.50 On December 10, 1954, Colonel John Stapp rode a rocket sled to a land record speed of 282.5 m/s. The sled accelerated for 5 s and then was brought to a stop in 1.4 s. Assuming that the acceleration and deceleration were constant, determine the number of g’s (the magnitudes of the acceleration and deceleration divided by the acceleration due to gravity) he endured during the ride.

Solution:

If an object’s acceleration during an interval of time is constant, the magnitude of the acceleration is equal to the magnitude of the change in its velocity divided by the time. In this case the magnitudes of the acceleration and deceleration were Acceleration: Deceleration:

282.5 m/s = 56.5 m/s 2 , 5s 282.5 m/s = 202 m/s 2 . 1.4 s

The corresponding g levels were Acceleration: Deceleration:

56.5 m/s 2 = 5.76, 9.81 m/s 2 2 202 m/s = 20.6. 9.81 m/s 2

Acceleration: 5.76. Deceleration: 20.6.

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Problem 13.51 An accelerometer on the rocket sled in Problem 13.50 measured a maximum deceleration level of 46.2 g’s during the 1.4 s required for the velocity to decrease from 282.5 m/s to zero. Assume that the sled’s acceleration during that period is given by a = −kv 2 m/s 2 , where k is a constant and v is the sled’s velocity. Determine the value of k that results in the measured maximum deceleration. Then determine the distance traveled during the deceleration from the top speed down to a speed of 10 m/s. Solution:

The given equation indicates that the maximum deceleration occurs at the maximum velocity, which is 282.5 m/s. Therefore we set

To determine the distance traveled, we write the acceleration as a =

dv dv ds dv = = v = −kv 2 , dt ds dt ds

then we can separate variables and integrate:

( 1k ) dvv = ds,

−

s 1 10 m/s dv = ∫ ds, 0 k ∫ 282.5 m/s v 1[ 282.5 m/s = s, ln v ]10 m/s k 1 282.5 m/s ln = s. 10 m/s k

−

(

)

This result gives s = 588 m. k = 0.005679 m −1 , distance is 588 m.

46.2 g = kv 2 : 46.2(9.81 m/s 2 ) = k (282.5 m/s) 2 , obtaining k = 0.005679 m −1 .

Problem 13.52 A car’s acceleration is related to its position by a = 0.01 s m/s 2 . When s = 100 m, the car is moving at 12 m/s. How fast is the car moving when s = 420 m?

Solution: a= v

dv = 0.01 m/s 2 ds

420

∫12 v dv = 0.01∫100 s ds v s 2 420 m v2  f = 0.01    2   2  100 m 12 m/s

v 2f (420 2 − 100 2 ) 12 2 = + 0.01 2 2 2 v f = 42.5 m/s.

Problem 13.53 Engineers analyzing the motion of a linkage determine that the velocity of an attachment point is given by v = A + 4 s 2 ft/s, where A is a constant. When s = 2 ft, the acceleration of the point is measured and determined to be a = 320 ft/s 2 . What is the velocity of the point when s = 2 ft?

Solution:

The velocity as a function of the distance is

dv v = a ds Solve for a and carry out the differentiation. a = v

dv = ( A + 4 s 2 )(8s) ds

When s = 2 ft, a = 320 ft/s 2 , from which A = 4. The velocity at s = 2 ft is v = 4 + 4(2 2 ) = 20 ft/s.

Problem 13.54 The acceleration of an object is given as a function of its position in feet by a = 2s 2 ft/s 2 . When s = 0, its velocity is v = 1 ft/s. What is the velocity of the object when s = 2 ft?

Solution: a =

We are given

v dv 2 = s 2, ds ft-s 2

(

)

∫1 ft/s v dv = ( ft-s 2 ) ∫0 s 2 ds v

2 ft

(1 ft/s) 2 v2 2 (2 ft) 3 − = 2 2 ft-s 2 3

(

)

v = 3.42 ft/s.

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Problem 13.55 An object in straight-line motion has acceleration given as a function of its position s by 4 a = − 2 m/s 2 . s Suppose that at time t = 0, the object’s position is s 0 = 2 m and it is given an initial velocity v 0 = 1 m/s. Its negative acceleration causes it to slow down. What is its position s when the velocity decreases to zero? Solution:

Separating variables and integrating, 4 v dv = − 2 ds, s v s 4 ∫v 0 v dv = ∫s0 − s 2 ds, v v2  =  4 s    s  2 v0 s0

1 1 1 2 (v − v 0 2 ) = 4  − . s 2 s0  Setting v = 0 and solving for the position s, we obtain

We can write the acceleration as

dv dv ds dv 4 a = = = v = − 2 m/s 2 . dt ds dt ds s

s =

8 . 8 − v02 s0

With s 0 = 2 m and v 0 = 1 m/s, we obtain s = 2.67 m. s = 2.67 m.

Problem 13.56 The object in Problem 13.55 was given an initial velocity v 0 = 1 m/s and reached position s = 2.67 m before its velocity decreased to zero. If you were to give it a slightly larger initial velocity, it would travel farther before stopping. If you continue to increase the initial velocity, it reaches the escape velocity for this problem, at which the object would travel infinitely far. What is that initial velocity? Solution:

We can write the acceleration as

dv dv ds dv 4 a = = = v = − 2 m/s 2 . dt ds dt ds s Separating variables and integrating, 4 v dv = − 2 ds, s v s 4 ∫v 0 v dv = ∫s0 − s 2 ds,

Setting v = 0 and solving for the position s, we obtain s =

8 . 8 − v02 s0

The denominator in this expression decreases, and s increases, as v 0 is increased. The denominator approaches zero, and the position s approaches infinity, when the initial velocity approaches the value v0 =

8 s0

8 2m = 2 m/s. =

v 0 = 2 m/s.

v v2  =  4 s    s  2 v0 s0

1 1 1 2 (v − v 0 2 ) = 4  − . s 2 s0 

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Problem 13.57 A spring–mass oscillator consists of a mass and a spring connected as shown. The coordinate s measures the displacement of the mass relative to its position when the spring is unstretched. If the spring is linear, the mass is subjected to a deceleration proportional to s. Suppose that a = −4 s m/s 2 and that you give the mass a velocity v = 1 m/s in the position s = 0. (a) How far will the mass move to the right before the spring brings it to a stop? (b) What will be the velocity of the mass when it has returned to the position s = 0?

Solution: (a)

When the acceleration is a function of position, it is often advantageous to use the chain rule to express it as dv dv ds dv a = = = v. dt ds dt ds In this form, you can separate variables and integrate to determine the velocity as a function of position: dv v = − 4 s, ds v dv = −4 s ds, v

∫1 m/s v dv = ∫0 −4s ds, υ

 1 υ2  = −[2s 2 ] 0s ,  2  1 m/s v 2 = 1 − 4s 2 . From this expression we see that v = 0 when s = 1/2 m. When s = 0, we obtain v = ±1 m/s. The negative sign applies to the case when the mass is returning from moving to the right.

(b) s

Problem 13.58 In Problem 13.57, suppose that at t = 0 you release the mass from rest in the position s = 1 m. What is the magnitude of the velocity of the mass when s = 0.5 m?

(a) s = 0.5 m. (b) v = −1 m/s.

Solution: a =

We express the acceleration as

dv , dt

then separate variables and integrate to determine the velocity as a function of time: dv v = − 4 s, ds v dv = −4 s ds, v

∫0 v dv = ∫1 m −4s ds,  1v 2   2 

υ 0

= −[2s 2 ]1s m ,

v = ±2 1 − s 2 m/s. When s = 0.5 m, we obtain v = ±1.73 m/s. 1.73 m/s.

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Problem 13.59* In Problem 13.57, the mass is given a velocity v = 1 m/s in the position s = 0. How long does it take the spring to bring the mass to a stop? Strategy: First determine the velocity of the mass as a function of s. Then express the velocity as v = ds /dt, separate variables, and integrate to determine the position as a function of time.

then separate variables and integrate to determine the velocity as a function of position: dv v = − 4 s, ds v dv = −4 s ds, v

∫1 m/s v dv = ∫0 −4s ds, υ

 1v 2  = −[2s 2 ] 0s ,  2  1 m/s ds v = = 1 − 4s 2 . dt We separate variables in this expression and integrate, obtaining t ds = ∫ dt, 0 1 − 4s 2 s  1 arcsin(2s)  = t,  2  0 arcsin(2s) = 2t. s

∫0

We see that 1 s = sin(2t ), 2 so the velocity as a function of time is v =

Solution:

We express the acceleration as

dv dv ds dv a = = = v, dt ds dt ds

ds = cos(2t ) m/s. dt

The velocity decreases to zero when 2t =

π , giving t = 0.785 s. 2

0.785 s.

Problem 13.60 The mass is released from rest with the springs unstretched. Its downward acceleration is a = 32.2 − 50 s ft/s 2 , where s is the position of the mass measured from the position in which it is released. (a) How far does the mass fall? (b) What is the maximum velocity of the mass as it falls?

Solution: a =

The acceleration is given by

dv dv ds dv = = v = 32.2 − 50 s ft/s 2 . dt ds dt ds

Integrating, we get v

∫0 v dv = ∫0 (32.2 − 50s) ds or

v2 = 32.2s − 25s 2 . 2

The mass falls until v = 0. Setting v = 0, we get 0 = (32.2 − 25s)s. We find v = 0 at s = 0 and at s = 1.288 ft. Thus, the mass falls 1.288 ft before coming to rest. (b) From the integration of the equation of motion, we have v 2 = 2(32.2s − 25s 2 ). The maximum velocity occurs where dv = 0. From the original equation for acceleration, we ds dv have a = v = (32.2 − 50 s) ft/s 2 . Since we want maxids mum velocity, we can assume that v ≠ 0 at this point. Thus, (a)

0 = (32.2 − 50 s), or s = (32.2/50) ft when v = v MAX . Substituting this value for s into the equation for v , we get 2 v MAX = 2

(25)(32.2) − ( (32.2) ), 50 50 2

or v MAX = 4.55 ft/s. (a) 1.29 ft; (b) 4.55 ft/s.

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Problem 13.61 Suppose that the mass in Problem 13.60 is in the position s = 0 and is given a downward velocity of 10 ft/s. (a) How far does the mass fall? (b) What is the maximum velocity of the mass as it falls?

Solution: dv a = v = (32.2 ft/s 2 ) − (50 s −2 )s ds v

∫10 ft/s v dv = ∫0 [(32.2 ft/s 2 ) − (50 s −2 )s] ds (10 ft/s) 2 v2 s2 − = (32.2 ft/s 2 )s − 2 2 2 v 2 = (10 ft/s) 2 + (64.4 ft/s 2 )s − (50 s −2 )s 2 (a)

The mass falls until v = 0 0 = (10 ft/s) 2 + (64.4 ft/s 2 )s − (50 s −2 )s 2 ⇒ s = 2.20 ft.

(b) s

The maximum velocity occurs when a = 0 0 = (32.2 ft/s 2 ) − (50 s −2 )s ⇒ s = 0.644 ft v 2 = (10 ft/s) 2 + (64.4 ft/s 2 )(0.644 ft) − (50 s −2 )(0.644 ft) 2 v = 10.99 ft/s.

Problem 13.62 If a spacecraft is 300 km above the surface of the Earth, what initial velocity v 0 straight away from the Earth would be required for the vehicle to reach the Moon’s orbit 383,000 km from the center of the Earth? The radius of the Earth is 6378 km. Neglect the effect of the Moon’s gravity. (See Example 13.6.)

Solution: Example 13.5 gives an expression for the initial velocity v 0 necessary to reach a distance h from the center of the Earth, 1 1 2 gR E2  − ,  s0 h

v0 =

where g is the acceleration due to gravity at the surface of the Earth, R E is the radius of the Earth, and s 0 is the initial distance from the center of the Earth. Substituting the given values, we obtain

300 km 1 1 2 gR E2  −   s0 h

v0 =

2(9.81 m/s 2 )(6.378E6 m) 2

1 ( 6.378E6 m1+ 0.3E6 m − 383E6 m)

= 10,800 m/s.

383,000 km

10,800 m/s.

Problem 13.63 The Moon’s radius is 1738 km. The magnitude of the acceleration due to gravity of the Moon at a distance s from the center of the Moon is 4.89E12 m/s 2 . s2 Suppose that a spacecraft is launched straight up from the Moon’s surface with a velocity of 2000 m/s. (a) What will the magnitude of its velocity be when it is 1000 km above the surface of the Moon? (b) What maximum height above the Moon’s surface will it reach?

Solution: Set v 0 = 2000 m/s a = v

G = 4.89 × 10 12 m 3 /s 2 , r0 = 1.738 × 10 6 m,

v r G 1 v 2 v2 dv G 1 = − 2 ⇒ ∫ v d v = −∫ 2 ds ⇒ − 0 = G  −  v0 r0 s r ds s 2 2 r0 

 r − r  v 2 = v 02 + 2G  0   rr0  (a)

v (r0 + 1.0 × 10 6 m) = 1395 m/s.

(b)

The maximum velocity occurs when v = 0 r =

2G = 6010 km ⇒ h = r − r0 = 4272 km. 2G − r0v 02

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Problem 13.64* The velocity of an object subjected only to the Earth’s gravitational field is 1/ 2 1  1  v =  v 0 2 + 2 gR E 2  −   , s s 0   

where s is the object’s position relative to the center of the Earth, v 0 is the object’s velocity at position s 0 , and R E is the Earth’s radius. Using this equation, show that the object’s acceleration is given as a function of s by a = −gR E 2 /s 2 .

Solution:

1/ 2 1 1 v = v 02 + 2 gR E2  −  s s0 

a =

dv dv = v dt ds

Rewrite the equation given as v 2 = v 0 2 + Take the derivative with respect to s. 2 gR 2 dv = − 2E ds s Thus

a = v

Problem 13.65 Suppose that a tunnel could be drilled straight through the Earth from the North Pole to the South Pole and the air was evacuated. An object dropped from the surface would fall with acceleration a = −gs /R E , where g is the acceleration of gravity at sea level, R E is the radius of the Earth, and s is the distance of the object from the center of the Earth. (The acceleration due to gravity is equal to zero at the center of the Earth and increases linearly with distance from the center.) What is the magnitude of the velocity of the dropped object when it reaches the center of the Earth?

2 gR E2 2 gR E2 − s s0

gR 2 dv − 2E . ds s

Solution:

The velocity as a function of position is given by

gs dv v = − . ds RE Separate variables and integrate:  g  2 v 2 = −  s + C.  R E  At s = R E , v = 0, from which C = gR E . Combine and reduce:  s2  v 2 = gR E  1 − 2   RE  At the center of the Earth s = 0, and the velocity is v =

gR E .

N Tunnel s RE

Problem 13.66 Two homogeneous spheres, each with mass m = 1 kg and 0.1-m radius, float in space and are affected only by their own gravity. They are initially stationary relative to each other. The coordinate r measures the distance of each sphere from the point midway between them. Gravity causes the acceleration of each sphere toward the midpoint to be Gm , a = 4r 2 where the value of the universal gravitational constant is G = 6.67E − 11 N-m 2 /kg 2 . If the initial distance r = 1 m, what is the velocity of each sphere toward the midpoint just before they come into contact with each other?

Solution: Let v denote the velocity of one of the spheres toward the midpoint. (Notice that, because r is decreasing, dr /dt = −v . ) From the given acceleration, dv Gm = . dt 4r 2 Applying the chain rule, dv dr dv Gm dv = = − v = − 2, dt dr dt dr 4r we can separate variables and integrate, obtaining v

1 2 1 Gm  1  r Gm  1 = − , v = 2 4  r  r0 4  r r0  v =

r Gm

∫0 v dv = −∫r 4r 2 dr, Gm  1 1 − . r0  2  r

Setting m = 1 kg, v = 1.73E − 5 m/s.

r0 = 1 m,

and

r = 0.1 m,

get

1.73E − 5 m/s.

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Problem 13.67 In a second test, the coordinates of the position (in m) of the helicopter in Practice Example 13.7 are given as functions of time by

Solution:

x = 4 + 2t,

a x = 0,

We have

x = (4 m) + (2 m/s)t, v x = (2 m/s),

y = (4 m) + (4 m/s)t + (1 m/s 2 )t 2 ,

v y = (4 m/s) + 2(1 m/s 2 )t,

a y = 2(1 m/s 2 ).

At t = 3 s, we have

y = 4 + 4t + t 2 . (a) What is the magnitude of the helicopter’s velocity at t = 3 s? (b) What is the magnitude of the helicopter’s acceleration at t = 3 s?

x = 10 m,

v x = 2 m/s,

a x = 0,

y = 25 m,

v y = 10 m/s,

a y = 2 m/s 2 .

Thus v =

v x2 + v y2 = 10.2 m/s

v = 10.2 m/s.

a =

0 2 + (2 m/s 2 ) 2 = 2 m/s 2

a = 2 m/s 2 .

Problem 13.68 In terms of a particular reference frame, the position of the center of mass of the F-14 at the time shown (t = 0) is r = 10 i + 6 j + 22k (m). The velocity from t = 0 to t = 4 s is v = (52 + 6t ) i + (12 + t 2 ) j − (4 + 2t 2 )k (m/s). What is the position of the center of mass of the plane at t = 4 s? Solution: r0 = 10 i + 6 j + 22k m v = (52 + 6t ) i + (12 + t 2 ) j − (4 + 2t 2 )k m/s 4

x 4 = ∫ v x dt = 52t + 3t 2 + x 0 0

x 4 = (52)(4) + 3(4) 2 + 10 m = 266.0 m

Source: Courtesy of Frank Rossoto Stocktrek/Getty Images.

y 4 = ∫ v y dt = 12t + t 3 /3 + y 0 0

y 4 = 12(4) + (4) 3 /3 + 6 m = 75.3 m 4

z 4 = ∫ v z dt = −(4t + 2t 3 /3) + z 0 0

z 4 = −4(4) − 2(4) 3 /3 + 22 = −36.7 m r | t = 4 s = 266 i + 75.3 j − 36.7k (m)

Problem 13.69 A projectile is launched from ground level with initial velocity v 0 = 40 ft/s. Determine its range R and the maximum height it reaches if (a) θ 0 = 25 ° and (b) θ 0 = 55 °.

Solution: Let t = 0 be the time at which the projectile is launched. The initial components of the velocity are v x = v 0 cos θ 0 , v y = v 0 sin θ 0 and the components of the acceleration are a x = 0, a y = −g. We can integrate the components of acceleration with respect to time to determine the velocity and position as functions of time. In the x direction, ax =

dv x = 0, dt

∫v cos θ dv x = 0, 0

x R

and in the y direction,

ay =

(1)

vx = x

vy 0

(2)

(3)

(4)

dy = v 0 sin θ 0 − gt, dt

∫0 dy = ∫0 (v 0 sin θ0 − gt ) dt, y = (v 0 sin θ 0 )t −

vy =

x = (v 0 cos θ 0 )t,

∫v sin θ dv y = ∫0 −g dt,

dx = v 0 cos θ 0 , dt

∫0 dx = ∫0 v 0 cos θ0 dt,

dv y = −g, dt

1 2 gt . 2

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13.69 (Continued)

Substituting this value into Eq. (4), the maximum height is

To determine the range, we can use Eq. (4) to calculate the time at which y = 0, obtaining 2v 0 sin θ 0 t = . g

 v sin θ 0  1  v 0 sin θ 0  2 y max = (v 0 sin θ 0 ) 0  − g     2   g g =

Then substituting this value into Eq. (2) gives the range: R =

2v 02 sin θ 0 cos θ 0 . (5) g

The maximum height is reached at the time the y component of the velocity is zero. From Eq. (3), t =

v 0 sin θ 0 . g

v 02 sin 2 θ 0 . 2g

(6)

(a) Substituting g = 32.2 ft/s 2 , v 0 = 40 ft/s and θ 0 = 25 ° into Eqs. (5) and (6), we obtain R = 38.1 ft, y max = 4.44 ft. (b) Substituting g = 32.2 ft/s 2 , v 0 = 40 ft/s and θ 0 = 55 ° into Eqs. (5) and (6), we obtain R = 46.7 ft, y max = 16.7 ft. (a) R = 38.1 ft, height = 4.44 ft. (b) R = 46.7 ft, height = 16.7 ft.

Problem 13.70 A projectile is launched from ground level with initial velocity v 0 = 20 m/s. Determine its range R if (a) θ 0 = 30 °, (b) θ 0 = 45 °, and (c) θ 0 = 60 °. Solution:

9.81 m/s 2 , v

Set g =

= 20 m/s 1 a y = −g, v y = −gt + v 0 sin θ 0 , s y = − gt 2 + v 0 sin θ 0t 2 a x = 0, v x = v 0 cos θ 0 , s x = v 0 cos θ 0 t 0

u0 x

When it hits the ground, we have

2v 0 sin θ 0 1 0 = − gt 2 + v 0 sin θ 0t ⇒ t = 2 g v 02 sin 2θ 0 R = v 0 cos θ 0t ⇒ R = g (a) θ 0 = 30 ° ⇒ R = 35.3 m. (b) θ 0 = 45 ° ⇒ R = 40.8 m.

⇒

Problem 13.71 Immediately after its first bounce, the velocity of the golf ball is 12 ft/s and the angle between its path and the horizontal is 80 °. What maximum height does it reach?

Solution: Let t = 0 be the time at which the ball first bounces. In terms of the ball’s initial velocity v 0 and the initial angle θ 0 between its path and the horizontal, the initial components of its velocity are v x = v 0 cos θ 0 , v y = v 0 sin θ 0 . The components of the acceleration are a x = 0, a y = −g. We integrate to determine the y components of the velocity and position as functions of time: ay = vy

dv y = −g, dt

(1)

vy =

dy = v 0 sin θ 0 − gt, dt

y = (v 0 sin θ 0 )t −

1 2 gt . 2

(2)

The maximum height is reached at the time the y component of the velocity is zero. From Eq. (1), that time is t =

 v sin θ 0  1  v 0 sin θ 0  2 y max = (v 0 sin θ 0 ) 0  − g    g 2 g v 02 sin 2 θ 0 2g (12 ft/s) 2 sin 2 80 ° = 2(32.2 ft/s 2 ) = 2.17 ft. =

∫0 dy = ∫0 (v 0 sin θ0 − gt ) dt,

Source: Courtesy of Canbedone/Shutterstock.

Substituting this value into Eq. (2), the maximum height is

∫v sin θ dv y = ∫0 −g dt, 0

2.17 ft.

v 0 sin θ 0 . g

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Problem 13.72 Suppose that you are designing a mortar to launch a rescue line from coast guard vessels to ships in distress. The light line is attached to a weight fired by the mortar. Neglect aerodynamic drag and the weight of the line for your preliminary analysis. If you want the line to be able to reach a ship 300 ft away when the mortar is fired at 45° above the horizontal, what muzzle velocity is required? Solution: R =

458

From 13.70 we know that

v 02 sin 2θ 0 ⇒ υ0 = g

v0 =

Rg sin 2θ 0

(300 ft)(32.2 ft/s 2 ) = 98.3 ft/s. sin(90 °)

Problem 13.73 In Problem 13.72, what maximum height above the point from which it was fired is reached by the weight?

Putting in the numbers we have h =

(98.3 ft/s) 2 sin 2 (45 °) = 75 ft. 2(32.2 ft/s 2 )

Solution:

From Problem 13.70 we have 1 v y = −gt + v 0 sin θ 0 , s y = − gt 2 + v 0 sin θ 0 t 2 When we reach the maximum height, 0 = −gt + v 0 sin θ 0 ⇒ t =

y 458

v 0 sin θ 0 g

2 1 1  v sin θ 0  h = − gt 2 + v 0 sin θ 0t ⇒ h = − g  0    g 2 2   v 0 sin θ 0  + v 0 sin θ 0    g v 2 sin 2 θ 0 h = 0 2g

Problem 13.74 When the athlete releases the shot, it is 1.82 m above the ground and its initial velocity is v 0 = 13.6 m/s. Determine the horizontal distance the shot travels from the point of release to the point where it hits the ground.

308

Solution: Let t = 0 be the time at which the shot is released. The initial components of the velocity are v x = v 0 cos30 °, v y = v 0 sin 30 ° , where v 0 is the initial velocity, and the components of the acceleration are a x = 0, a y = −g. We can integrate the components of acceleration with respect to time to determine the velocity and position as functions of time. In the x direction, dv x ax = = 0, dt vx

∫v cos 30° dv x = 0,

(1)

vx = x

dx = v 0 cos30 °, dt

∫0 dx = ∫0 v 0 cos30° dt, x = (v 0 cos30 °)t,

(2)

and in the y direction, dv y ay = = −g, dt vy

∫v sin 30° dv y = ∫0 −g dt,

(3)

vy =

dy = v 0 sin 30 ° − gt, dt

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13.74 (Continued) y

and solve for t, obtaining t = 1.62 s. Then from Eq. (2), the horizontal distance to the point where it hits the ground is

∫0 dy = ∫0 (v 0 sin 30° − gt ) dt,

(4) 1 2 gt . 2 Equation (4) is the y position relative to the point where the shot was released. To determine the time at which it hits the ground, we set 1 y = −1.82 m = (v 0 sin 30 °) t − gt 2 2 1 = (13.6 m/s)sin 30 °t − (9.81 m/s 2 ) t 2 2 y = (v 0 sin 30 °)t −

x = (v 0 cos30 °)t = (13.6 m/s) cos30°(1.62 s) = 19.0 m. 19.0 m.

Problem 13.75 A pilot wants to drop survey markers at remote locations in the Australian outback. If he flies at a constant velocity v 0 = 40 m/s at altitude h = 30 m and the marker is released with zero velocity relative to the plane, at what horizontal distance d from the desired impact point should the marker be released? Solution:

We want to find the horizontal distance traveled by the marker before it strikes the ground (y goes to zero for t > 0.)

a y = −g

ax = 0 v x = v x0

v y = v y 0 − gt

x = x 0 + v x0 t

y = y 0 + v y 0 t − gt 2 /2

From the problem statement, x 0 = 0, v y 0 = 0, v x 0 = 40 m/s, and y 0 = 30 m. The equation for y becomes y = 30 − (9.81)t 2 /2 Solving with y = 0, we get t f = 2.47 s. Substituting this into the equation for x, we get x f = 40 t f = 98.9 m.

Problem 13.76 If the pitching wedge the golfer is using gives the ball an initial angle θ 0 = 50 °, what range of velocities v 0 will cause the ball to land within 3 ft of the hole? (Assume that the hole lies in the plane of the ball’s trajectory.) Solution:

Set the coordinate origin at the point where the golfer strikes the ball. The motion in the horizontal ( x ) direction is given by a x = 0, V x = V0 cos θ 0 , x = (V0 cos θ 0 )t. The motion in the vertical (y) direction is given by a y = −g, V y = V0 sin θ 0 − gt, y = (V0 sin θ 0 ) t −

3 ft 30 ft

gt 2 . 2

From the x equation, we can find the time at which the ball reaches the required value of x (27 or 33 feet). This time is t f = x f /(V0 cos θ 0 ). We can substitute this information the equation for Y with Y f = 3 ft and solve for V0 . The results are: For hitting (27,3) feet, V0 = 31.2 ft/s. For hitting (33,3) feet, V0 = 34.2 ft/s.

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Problem 13.77 A batter strikes a baseball 3 ft above home plate and pops it up. The second baseman catches it 6 ft above second base 3.68 s after it was hit. What was the ball’s initial velocity, and what was the angle between the ball’s initial velocity vector and the horizontal?

90 ft

Solution:

Second base

The equations of motion g = 32.2 ft/s 2

ax = 0

a y = −g

v x = v 0 cos θ 0

v y = −gt + v 0 sin θ 0

s x = v 0 cos θ 0 t

1 s y = − gt 2 + v 0 sin θ 0 t + 3 ft 2

When the second baseman catches the ball we have

127.3 ft = v 0 cos θ 0 (3.68 s) 1 6 ft = − (32.2 ft/s 2 )(3.68 s) 2 + v 0 sin θ 0 (3.68 s) 2 Solving simultaneously we find

Home plate

v 0 = 70.02 ft/s, θ 0 = 60.4 °.

Problem 13.78 The crossbar of the goalposts in American football is yc = 10 ft above the ground. To make a field goal, the kicker must send the ball between the two vertical supports of the crossbar and above the crossbar. Suppose the kicker is attempting a 45-yd field goal ( x c = 135 ft) and he kicks the ball with initial velocity v 0 = 72 ft/s and angle θ 0 = 43 °. Does he score? If so, by what vertical distance does the ball clear the crossbar? Solution: Let t = 0 be the time at which the ball is kicked. In terms of its initial velocity v 0 and elevation angle θ 0 , the initial components of its velocity are v x = v 0 cos θ 0 , v y = v 0 sin θ 0 . The components of its acceleration are assumed to be a x = 0, a y = −g. We can integrate the components of acceleration with respect to time to determine the velocity and position as functions of time. In the x direction, ax = vx

Substituting this value into Eq. (4), the ball’s height when it reaches the crossbar is 1 2 gt 2  xc  1  x c  2 = (v 0 sin θ 0 )  − g    v 0 cos θ 0  2  v 0 cos θ 0 

y = (v 0 sin θ 0 )t −

(1)

dx = v 0 cos θ 0 , dt

 2 (135 ft)sin 43 ° 1 135 ft = − (32.2 ft/s 2 )    (72 ft/s) cos 43 °  cos 43 ° 2 = 20.1 ft.

vx =

dv x = 0, dt

∫v cos θ dv x = 0, 0

Yes, it clears the crossbar by 10.1 ft.

∫0 dx = ∫0 v 0 cos θ0 dt, (2) x = (v 0 cos θ 0 )t,

and in the y direction, dv y = −g, ay = dt vy

∫v sin θ dv y = ∫0 −g dt, 0

vy = y

dy = v 0 sin θ 0 − gt, dt

(3)

∫0 dy = ∫0 (v 0 sin θ0 − gt ) dt, y = (v 0 sin θ 0 )t −

1 2 gt . 2

(4)

From Eq. (2), the time at which x = x c is t =

xc . v 0 cos θ 0

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Problem 13.79 A projectile is launched from a tower 30 m above ground level with an initial speed v 0 = 20 m/s. The launch angle θ 0 ranges from 10° to 60 °. Draw a graph of the range (the horizontal distance from the launch point to where the projectile hits the ground) versus the launch angle. Use your graph to estimate the launch angle that results in the maximum range.

ay = vy

dv y = −g, dt t

∫v sin θ dv y = ∫0 −g dt, 0

vy =

(3)

(4)

dy = v 0 sin θ 0 − gt, dt t

∫0 dy = ∫0 (v 0 sin θ0 − gt ) dt, y = (v 0 sin θ 0 ) t −

1 2 gt . 2

Equation (4) is the y position relative to the point where the projectile was launched. To determine the time at which it hits the ground, we set

v0 u0

y = −30 m = (v 0 sin θ 0 )t −

1 2 gt . 2

For a given value of θ 0 , we can solve this quadratic equation for the time and substitute it into Eq. (2) to obtain the range. We show the graph of the range as a function of the launch angle:

30 m

70 65

Let t = 0 be the time at which the projectile is launched. The initial components of the velocity are v x = v 0 cos θ 0 , v y = v 0 sin θ 0 and the components of the acceleration are a x = 0, a y = −g. We can integrate the components of acceleration with respect to time to determine the velocity and position as functions of time. In the x direction,

Range, meters

Solution:

55 50

dv x = 0, ax = dt vx

∫v cos θ dv x = 0, 0

(1)

vx = x

40 10

dx = v 0 cos θ 0 , dt

∫0 dx = ∫0 v 0 cos θ0 dt, x = ( v 0 cos θ 0 ) t,

25 30 35 40 45 Launch angle, degrees

(2)

and in the y direction,

From the graph we estimate that the maximum range occurs at θ 0 = 32.5 °. θ 0 = 32.5 °.

Problem 13.80 A zoology graduate student is provided with a bow and an arrow tipped with a syringe of tranquilizer and is assigned to measure the temperature of a black rhinoceros (Diceros bicornis). His bow shoots an arrow at 35 m/s. An irritated rhino suddenly charges straight toward him at 12 m/s. If he aims his bow 20° above the horizontal, how far away should the animal be when he releases the arrow?

Solution:

Let t = 0 be the time at which the arrow is released. In terms of the arrow’s initial velocity v 0 and elevation angle θ 0 , the initial components of its velocity are v x = v 0 cos θ 0 , v y = v 0 sin θ 0 . The components of its acceleration are assumed to be a x = 0, a y = −g. We can integrate the components of acceleration with respect to time to determine the velocity and position as functions of time. In the x direction, ax = vx

∫v cos θ dv x = 0, 0

(1)

∫0 dx = ∫0 v 0 cos θ0 dt,

(2)

vx = 208

dv x = 0, dt

dx = v 0 cos θ 0 , dt t

x = (v 0 cos θ 0 )t,

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13.80 (Continued)

2v 0 sin θ 0 g 2(35 m/s)sin 20 ° = (9.81 m/s 2 ) = 2.44 s.

t =

and in the y direction, ay = vy

dv y = −g, dt t

∫v sin θ dv y = ∫0 −g dt, 0

(3)

Then substituting this value into Eq. (2) gives the arrow’s range:

vy = y

dy = v 0 sin θ 0 − gt, dt

2v 02 sin θ 0 cos θ 0 g 2(35 m/s) 2 sin 20 ° cos 20 ° = (9.81 m/s 2 ) = 80.3 m.

R =

∫0 dy = ∫0 (v 0 sin θ0 − gt ) dt, y = (v 0 sin θ 0 )t −

1 2 gt . 2

(4)

To determine the arrow’s time of flight, we can use Eq. (4) to calculate the time at which y = 0, obtaining

During the time of flight of the arrow, the rhino travels (12 m/s)(2.44 s) = 29.3 m. Therefore it should be 80.3 m + 29.3 m = 110 m away when he releases the arrow. 110 m.

Problem 13.81 An American football quarterback stands at A. At the instant he throws the ball, the receiver is at B running at 25 ft/s toward C, the point where he catches the ball. The ball is thrown at a 45° angle relative to the horizontal, and it is thrown and caught at the same height above the ground. Determine the ball’s initial velocity and the length of time it is in the air.

x C

z B

C ax =

dv x = 0, dt

∫v cos θ dv x = 0, 0

vx = Receiver’s path

(1)

dx = v 0 cos θ 0 , dt t

∫0 dx = ∫0 v 0 cos θ0 dt, (2) x = (v 0 cos θ 0 ) t,

Path of the ball

and in the y direction, ay = vy

∫v sin θ dv y = ∫0 −g dt,

908

(3)

vy = B

dv y = −g, dt

dy = v 0 sin θ 0 − gt, dt t

∫0 dy = ∫0 (v 0 sin θ0 − gt ) dt,

30 ft

y = (v 0 sin θ 0 )t −

1 2 gt . 2

(4)

From Eq. (4), the time of flight of the ball ( y = 0) is

Solution:

Let t = 0 be the time at which the ball is thrown. Orient a coordinate system with the x axis directed from A toward C and the y axis vertical. In terms of its initial velocity v 0 and elevation angle θ 0 = 45 °, the initial components of the ball’s velocity are v x = v 0 cos θ 0 , v y = v 0 sin θ 0 . The components of its acceleration are assumed to be a x = 0, a y = −g. We can integrate the components of acceleration with respect to time to determine the velocity and position as functions of time. In the x direction,

tf =

2v 0 sin θ 0 . (5) g

The distance the receiver travels in time t f is (25 ft/s)t f . From the triangle ABC, we see that this distance and the distance x traveled by the ball are related by x 2 = [(25 ft/s)t f ] 2 + (30 ft) 2 .

(6)

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13.81 (Continued) Solving Eq. (2) for v 0 and substituting it into Eq. (5) yields the equation x =

gt f 2 cos θ 0 . 2sin θ 0

Solving this, we obtain the time of flight of the ball, t f = 1.85 s. From Eq. (5), the ball’s initial velocity is v 0 = 42.1 ft/s. 42.1 ft/s, 1.85 s.

Then substituting this result into Eq. (6) gives a quadratic equation for the time t f : g 2t f 4 cos 2 θ 0 − (25 ft/s) 2 t f 2 − (30 ft) 2 = 0. 4 sin 2 θ 0

Problem 13.82 Suppose that the quarterback in Problem 13.81 throws the ball with a flatter trajectory, at an angle of 30° above the horizontal. Determine the ball’s initial velocity and the length of time it is in the air.

x C

z B

and in the y direction, Receiver’s path

ay =

dv y = −g, dt

∫v sin θ dv y = ∫0 −g dt,

Path of the ball

(3)

vy =

dy = v 0 sin θ 0 − gt, dt

∫0 dy = ∫0 (v 0 sin θ0 − gt ) dt, y = (v 0 sin θ 0 )t −

908

1 2 gt . 2

(4)

From Eq. (4), the time of flight of the ball ( y = 0) is A

tf =

30 ft

2v 0 sin θ 0 . (5) g

The distance the receiver travels in time t f is (25 ft/s)t f . From the triangle ABC, we see that this distance and the distance x traveled by the ball are related by

Solution:

Let t = 0 be the time at which the ball is thrown. Orient a coordinate system with the x axis directed from A toward C and the y axis vertical. In terms of its initial velocity v 0 and elevation angle θ 0 = 30 °, the initial components of the ball’s velocity are v x = v 0 cos θ 0 , v y = v 0 sin θ 0 . The components of its acceleration are assumed to be a x = 0, a y = −g. We can integrate the components of acceleration with respect to time to determine the velocity and position as functions of time. In the x direction, ax =

dv x = 0, dt

∫v cos θ dv x = 0, 0

(1)

vx = x

dx = v 0 cos θ 0 , dt

x 2 = [(25 ft/s)t f ] 2 + (30 ft) 2 .

(6)

Solving Eq. (2) for v 0 and substituting it into Eq. (5) yields the equation x =

gt f 2 cos θ 0 . 2sin θ 0

Then substituting this result into Eq. (6) gives a quadratic equation for the time t f : g 2t f 4 cos 2 θ 0 − (25 ft/s) 2 t f 2 − (30 ft) 2 = 0. 4 sin 2 θ 0 Solving this, we obtain the time of flight of the ball, t f = 1.25 s. From Eq. (5), the ball’s initial velocity is v 0 = 40.1 ft/s. 40.1 ft/s, 1.25 s.

∫0 dx = ∫0 v 0 cos θ0 dt, (2) x = (v 0 cos θ 0 )t,

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Problem 13.83 The cliff divers of Acapulco, Mexico, must time their dives so that they enter the water at the crest (high point) of a wave. The crests of the waves are 1 m above the mean water depth h = 4 m. The horizontal velocity of the waves is equal to gh. The diver’s aiming point is 2 m out from the base of the cliff. Assume that his velocity is horizontal when he begins the dive. (a) What is the magnitude of the diver’s velocity when he enters the water? (b) How far from his aiming point must a wave crest be when he dives in order for him to enter the water at the crest?

Solution: t = 0, v y∆ = 0, y = 27 m, x 0 = 0 a y = −g = −9.81 m/s V y = V y00 − gt y = y 0 − gt 2/ 2 y = 1 m at t IMPACT for an ideal dive to hit the crest of the wave t 1 = t IMPACT = 2.30 s V y (t 1 ) = 22.59 m/s ax = 0 Vx = Vx0 X I = Vx 0 t1 + X 0 At impact X I = 8.4 m. For impact to occur as planned, then V x = 8.4/t 1 = 3.65 m/s = constant

26 m

The velocity at impact is (a) V =

(V x ) 2 + [V y (t 1 )] 2 = 22.9 m/s

The wave moves at

gh = 6.26 m/s.

The wave crest travels 2.30 seconds while the diver is in their s = ght 1 = 14.4 m.

1m h 6.4 m

2m 27 m

8.4 m

Problem 13.84 A projectile is launched at 10 m/s from a sloping surface. The angle α = 80 °. Determine the range R. 10 m /s 308

Solution:

Set g = 9.81 m/s 2 , v 0 = 10 m/s.

The equations of motion are a x = 0, v x = v 0 cos(80 ° − 30°), s x = v 0 cos 50 °t 1 a y = −g, v y = −gt + v 0 sin (80 ° − 30 °)t, s y = − gt 2 + v 0 sin 50 °t 2 When the projectile hits we have

R cos30 ° = v 0 cos 50 °t R

⇒ t = 2.32 s, R = 17.21 m.

1 −R sin 30 ° = − gt 2 + v 0 sin 50 °t 2

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Problem 13.85 A projectile is launched at 100 ft/s at 60° above the horizontal. The surface on which it lands is described by the equation shown. Determine the x coordinate of the point of impact. y

100 ft /s

Solution:

The motion in the x direction is a x = 0, v x = V0 cos θ 0 , x = (V0 cos θ 0 )t, and the motion in the y direction is given by a y = −g, v y = (V0 sin θ 0 ) − gt, y = (V0 sin θ 0 )t − gt 2 /2. We know that V0 = 100 ft/s and θ 0 = 60 °. The equation of the surface upon which the projectile impacts is y = −0.001x 2 . Thus, the time of impact, t I , can be determined by substituting the values of x and y from the motion equations into the equation for the surface. t2 Hence, we get (V0 sin θ 0 )t I − g I = −0.001(V0 cos θ 0 ) 2 t I2 . Evaluating 2 with the known values, we get t I = 6.37 s Substituting this value into the motion equations reveals that impact occurs at ( x, y ) = (318.4, −101.4) ft.

608 x y 5 20.001x2

Problem 13.86 At t = 0, a steel ball in a tank of oil is given a horizontal velocity v = 2 i (m/s). The components of the ball’s acceleration, in m/s 2 , are a x = −1.2v x , a y = −8 − 1.2v y , and a z = −1.2v z . What is the velocity of the ball at t = 1 s?

Solution:

Assume that the effect of gravity is included in the given accelerations. The equations for the path are obtained from: (1)

dv x = a x = −1.2v x . Separate variables and integrate: dt dv x = −1.2 dt, vx

from which ln(v x ) = −1.2t + V x . At t = 0, v x (0) = 2, from which ln

( v2 ) = −1.2t x

Inverting: v x (t ) = 2e −1.2t .

x (2)

dv y = a y = −8 − 1.2v y . Separate variables and integrate: dt dv y = −1.2 dt, 8 + vy 1.2 from which ln

( 1.28 + v ) = −1.2t + V . y

At t = 0, v y (0) = 0, from

(

ln 1 +

1.2 v 8 y

) = −1.2t.

Inverting: v y (t ) = (3)

8 −1.2t − 1). (e 1.2

dv z = a z = −1.2v z , from which ln(v z ) = −1.2t + Vz . Invert to dt obtain v z (t ) = V z e −1.2t . At t = 0, v z (0) = 0, hence V z = 0 and v z (t ) = 0. At t = 1 second, v x (1) = 2e −1.2 = 0.6024 m/s , and

( 128 )(1 − e

v y (1) = −

−1.2 ) = − 4.66 m/s

, or

v = 0.602 i − 4.66 j (m/s) .

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Problem 13.87 In Problem 13.86, what is the position of the ball at t = 1 s relative to its position at t = 0?

Solution:

Use the solution for the velocity components from Problem 13.86. The equations for the coordinates: (1)

dx = v x = 2e −1.2t , from which dt

( 1.22 )e

x (t ) = −

−1.2 t

+ Cx.

At t = 0, x (0) = 0, from which x (t ) = (2)

( 1.22 )(1 − e

−1.2 t ).

( )

dy 8 = (e −1.2t − 1) , from which dt 1.2

(3) Since v z (0) = 0 and z(0) = 0, then z (t ) = 0. At t = 1,

( 1.28 )( 1.2 + t ) + C . e −1.2t

y (t ) = −

At t = 0, y (0) = 0, from which

( 1.28 )( e1.2 + t − 1.21 ).

x (1) =

( 1.22 )(1 − e

y (1) =

( 1.28 )( e1.2 + 1 − 1.21 ) = −2.784 m , or

−1.2 t

y (t ) = −

−1.2 ) = 1.165 m .

−1.2

r = 1.165i − 2.784 j (m) .

Problem 13.88 The point P moves along a circular path with radius R. Show that the magnitude of its velocity is v = R d θ dt . Strategy: Use Eqs. (13.23).

Solution: x = R cos θ y = R sin θ

( ddtθ ) dθ v = R cos θ ( ) dt v x = −R sin θ

P u

Problem 13.89 If y = 150 mm, dy dt = 300 mm/s, and d 2 y dt 2 = 0, what are the magnitudes of the velocity and acceleration of point P ?

V =

V x2 + V y2

V =

R 2 sin 2 θ

V =

V = R

dθ dt

( ddtθ ) + R cos θ ( ddtθ )

( ddtθ ) (sin θ + cos θ)

Solution:

The equation for the location of the point P is 1

R 2 = x 2 + y 2 , from which x = ( R 2 − y 2 ) 2 = 0.2598 m, and

( )( ) ( )

y dy dx = − = −0.1732 m/s, dt x dt y dx dy y 1 dy 2 d 2x = − + 2 − dt 2 x dt x dt dt x = −0.4619 m/s 2 .

( )( ) ( )( ddt y ) 2

The magnitudes are: 2

vP =

( dxdt ) + ( dydt ) = 0.3464 m/s.

ap =

( ddt x ) + ( ddt y ) = 0.4619 m/s .

y 300 mm

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Problem 13.90* A car travels at a constant speed of 100 km/h on a straight road of increasing grade whose vertical profile can be approximated by the equation shown. When the car’s horizontal coordinate is x = 400 m, what is the car’s acceleration?

Solution:

Denote C = 0.0003 and V = 100 km/h = 27.78 m/s. The magnitude of the constant velocity is 2

( dydt ) + ( dxdt )

V =

The equation for the road is y = Cx 2 from which

( )

dy dx = 2Cx . dt dt

Substitute and solve: y = 0.0003x2

dx V = = 27.01 m/s. dt (2Cx ) 2 + 1 dx is positive (car is moving to right in sketch). The acceleration is dt  d 2x d  V −4C 2Vx dx = = 3 dt 2 dt  (2Cx ) 2 + 1  dt . 2 ((2Cx ) + 1) 2

( )

= −0.0993 m/s 2 d 2y dx 2 d 2x = 2C + 2Cx = 0.4139 m/s 2 , dt 2 dt dt 2

(

( )

)

a = −0.099 i + 0414 j(m/s 2 ).

Problem 13.91* Suppose that a projectile has the initial conditions shown in Fig. 13.12. Show that in terms of the x ′ y ′ coordinate system with its origin at the highest point of the trajectory, the equation describing the trajectory is g y′ = − ( x ′) 2 . 2v 0 2 cos 2 θ 0

At the peak, dy = 0, dx peak from which v 2 cos θ 0 sin θ 0 , xp = 0 g and y p =

v 02 sin 2 θ 0 . 2g

The primed coordinates: y ′ = y − y p reduce: x9

y′ = −

g( x ′ + x p ) 2 + ( x ′ + x p ) tan θ 0 − y p . 2v 02 cos 2 θ 0

y′ = −

g (( x ′) 2 + x 2p + 2 x ′x p ) + ( x ′ + x p ) tan θ 0 2v 02 cos 2 θ 0

−

v 02 sin 2 θ 0 . 2g

Substitute x p = y′ = −

Solution:

The initial conditions are t = 0, x (0) = 0, y (0) = 0, v x (0) = v 0 cos θ 0 , and v y (0) = v 0 sin θ 0 . The accelerations are a x (t ) = 0, a y (t ) = −g. The path of the projectile in the x, y system is obtained by solving the differential equations subject to the initial conditions: g x (t ) = (v 0 cos θ 0 )t, y (t ) = − t 2 + (v 0 sin θ 0 )t 2

x ′ = x − x p . Substitute and

v 02 cos θ 0 sin θ 0 , g

v 2 sin 2 θ 0 v 2 sin 2 θ 0 g( x ′) 2 − 0 − x ′ tan θ 0 + x ′ tan θ 0 + 0 2v 02 cos 2 θ 0 2g g

v 2 sin 2 θ 0 − 0 2g y′ = −

g ( x ′) 2 . 2v 02 cos 2 θ 0

Eliminate t from the equations by substituting t =

x v 0 cos θ 0

to obtain y( x) = −

gx 2 2v 02 cos 2 θ 0

+ x tan θ 0 .

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Problem 13.92* The acceleration components of a point are a x = −4 cos 2t, a y = −4 sin 2t, and a z = 0. At t = 0, the position and velocity of the point are r = i and v = 2 j. Show that (a) the magnitude of the velocity is constant, (b) the velocity and acceleration vectors are perpendicular, (c) the magnitude of the acceleration is constant and points toward the origin, and (d) the trajectory of the point is a circle with its center at the origin. Solution:

(3) F or a z = 0 and zero initial conditions, it follows that v z (t ) = 0 and z (t ) = 0. (a) The magnitude of the velocity is v =

(−2sin (2t )) 2 + (2 cos(2t )) 2 = 2 = const.

(b) The velocity is v(t ) = −i 2sin (2t ) + j2 cos (2t ). The acceleration is a (t ) = −i4 cos(2t ) − j4 sin (2t ). If the two are perpendicular, the dot product should vanish: a (t ) ⋅ v(t ) = (−2sin (2 t ))(−4 cos(2 t )) + (2 cos(2 t )) (−4 sin(2 t )) = 0, and it does (c) The magnitude of the acceleration:

The equations for the path are

a =

dv (1) x = a x = −4 cos(2t ), from which v x (t ) = −2 sin(2t ) + V x . dt dx At t = 0, v x (0) = 0, from which V x = 0. = v x = −2sin (2t ), dt from which x (t ) = cos(2t ) + C x . At t = 0, x (0) = 1, from which

(−4 cos(2t )) 2 + (−4 sin (2t )) 2 = 4 = const .

The unit vector parallel to the acceleration is a e = = −i cos(2t ) − j sin (2t ), a

C x = 0.

which always points to the origin.

dv y (2) = a y = −4 sin (2t ), from which v y (t ) = 2 cos(2t ) + V y . dt dy At t = 0, v y (0) = 2, from which V y = 0. = v y = 2 cos(2t ), dt from which y (t ) = sin(2t ) + C y . At t = 0, y (0) = 0, from which C y = 0.

Problem 13.93 When an airplane touches down at t = 0, a stationary wheel is subjected to a constant angular acceleration α = 110 rad/s 2 until t = 1 s. (a) What is the wheel’s angular velocity at t = 1 s? (b) At t = 0, the angle θ = 0. Determine θ in radians and in revolutions at t = 1 s.

(d) T he trajectory path is x (t ) = cos(2t ) and y (t ) = sin(2t ). These satisfy the condition for a circle of radius 1: 1 = x 2 + y 2.

Solution: α = 110 rad/s 2 ω = αt + ω 0 θ =

( 12 αt ) + ω t + θ 2

From the problem statement, ω 0 = θ 0 = 0 (a) At t = 1 s, ω = (110)(1) + 0 = 110 rad/s (b) At t = 1 s, θ = 110(1) 2 /2 = 55 radians (8.75 revolutions). u

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Problem 13.94 The wheel shown is part of the main landing gear of a Boeing 737. Its diameter is 44.5 in. When the airplane touches down at 150 knots, assume that the stationary wheel is subjected to a constant angular acceleration for 1 s until it is rolling. (When the wheel is rolling, its angular velocity is equal to the airplane’s velocity divided by the wheel’s radius. Express the velocity in ft/s and the radius in ft.) The airplane’s velocity is nearly constant during the second the wheel is skidding. What is the magnitude of the wheel’s angular acceleration? Solution:

The airplane’s velocity in feet per second is

150 knots = 150 knots

ft/s ( 1.689 ) 1 knot

= 253 ft/s. The wheel’s radius in feet is 1 ft R = 22.25 in = 1.85 ft. 12 in The wheel’s angular velocity after one second when it has begun rolling is v ω = R 253 ft/s = 1.85 ft = 137 rad/s.

(

)

Problem 13.95 The angular acceleration of the line L relative to the line L 0 is given as a function of time by α = 2.5 − 1.2t rad/s 2 . At t = 0, θ = 0 and the angular velocity of L relative to L 0 is ω = 5 rad/s. Determine θ and ω at t = 3 s.

To achieve this angular velocity in one second, its constant angular acceleration is α = 137 rad/s 2 . 137 rad/s 2 .

Solution: α = 2.5 − 1.2t ω = 2.5t − 0.6t 2 + 5 θ = 1.25t 2 − 0.2t 3 + 5t ⇒

θ(3) = 1.25(3) 2 − 0.2(3) 3 + 5(3) = 20.85 rad ω (3) = 2.5(3) − 0.6(3) 2 + 5 = 7.1 rad/s.

u L0

Problem 13.96 The figure views the Earth from above the north pole. Let L be a line from the center of the Earth to a fixed point on the equator, and let L 0 be a fixed reference direction. (a) Is d θ /dt positive or negative? (Remember that the sun rises in the east.) (b) Assume that the Earth rotates relative to L 0 once in 24 hours. Determine the angular velocity of the Earth in radians per second and in degrees per hour. Solution: (a) R emember that the north pole points out of the page in the figure. Imagine that you’re standing at the equator where it is intersected by L, and you are facing east. That means you are facing upward in the figure. The sun rises in front of you, which means that you (fixed relative to the Earth) are moving upward. Therefore d θ /dt is positive.

L u

(b) The angular velocity in radians per second is 2π rad 2π rad 1 h ω = = = 7.27E-5 rad/s. 24 h 24 h 3600 s

(

)

The angular velocity in degrees per hour is ω =

360 ° = 15 deg/h. 24 h

(a) Positive. (b) 7.27E − 5 rad/s, 15 deg/h.

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Problem 13.97 The astronaut is not rotating. He has an orientation control system that can subject him to a constant angular acceleration of 0.1 rad/s 2 about the vertical axis in either direction. If he wants to rotate 180° about the vertical axis (that is, rotate so that he is facing toward the left) and not be rotating in his new orientation, what is the minimum time in which he could achieve the new orientation? Solution: He could achieve the rotation in minimum time by accelerating until he has turned 90 ° and then decelerating for the same time while he rotates the final 90 °. Thus the time needed to turn 90 ° (π /2 rad) is α = 0.1 rad/s 2 , ω = αt θ =

1 2 αt ⇒ t = 2

2(π /2) = 5.605 s. 0.1 rad/s 2

2θ = α

2t = 11.2 s.

The total maneuver time is 2t.

Problem 13.98 The astronaut is not rotating. He has an orientation control system that can subject him to a constant angular acceleration of 0.1 rad/s 2 about the vertical axis in either direction. Refer to Problem 13.97. For safety, the control system will not allow his angular velocity to exceed 15° per second. If he wants to rotate 180° about the vertical axis (that is, rotate so that he is facing toward the left) and not be rotating in his new orientation, what is the minimum time in which he could achieve the new orientation? Solution:

He could achieve the rotation in minimum time by accelerating until he has reached the angular velocity limit, then coasting until he has turned 90 °. He would then continue to coast until he needed to decelerate to a stop at the 180 ° position. The maneuver is symmetric in the spin up and the spin down phases. We will first find the time needed to reach the angular velocity limit and also find the angle through which he has rotated in this time. α = 0.1 rad/s 2 ω = αt 1 ⇒ t 1 = θ1 =

(15/180)π rad/s ω = = 2.62 s. α 0.1 rad/s 2

1 2 1 αt = (0.1 rad/s 2 )(2.62 s) 2 = 0.343 rad. 2 2

Now we need to find the coast time (constant angular velocity)

(

)

π 15 ° rad − 0.343 rad = π rad/s t 2 ⇒ t 2 = 4.69 s 2 180 ° Thus the total time to complete a 90 ° turn is t = t 1 + t 2 = 2.62 s + 4.69 s = 7.31 s. The time for the full 180 ° turn is 2t = 14.6 s.

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Problem 13.99 The rotor of an electric generator is rotating at 200 rpm when the motor is turned off. Due to frictional effects, the angular acceleration of the rotor after the motor is turned off is α = −0.01ω rad/s 2 , where ω is the angular velocity in rad/s. (a) What is the rotor’s angular velocity one minute after the motor is turned off? (b) After the motor is turned off, how many revolutions does the rotor turn before it comes to rest? Strategy: To do part (b), use the chain rule to write the angular acceleration as

Solution:

Let α = k ω, where k = −0.01 s −1 . Note that 200 rpm =

20.9 rad/s. (a) One minute after the motor is turned off α =

t ω dω  ω  dω = kω ⇒ ∫ = ∫ k dt ⇒ ln   = kt 0 ω0 ω  ω0  dt

ω = ω 0 e kt = (20.9 rad/s)e (−0.01/ s)(60 s) = 11.5 rad/s. ω = 11.5 rad/s (110 rpm). (b) When the rotor comes to rest θ 0 dω = k ω ⇒ ∫ d ω = ∫ k d θ ⇒ −ω 0 = kθ ω0 0 dθ 1 1 (−20.9 rad/s) θ = (−ω 0 ) = k −0.01 s −1 1 rev θ = 2094 rad = 333 rev. 2π rad θ = 333 rev.

α = ω

dω d ω dθ dω α = = = ω. dt d θ dt dθ

(

)

Problem 13.100 The needle of a measuring instrument is connected to a torsional spring that gives it an angular acceleration α = −4θ rad/s 2 , where θ is the needle’s angular position in radians relative to a reference direction. The needle is given an angular velocity ω = 2 rad/s in the position θ = 0. (a) What is the magnitude of the needle’s angular velocity when θ = 30 ° ? (b) What maximum angle θ does the needle reach before it rebounds?

Solution: ω θ dω ω2 22 = − 4θ ⇒ ∫ ω d ω = − 4 ∫ θ d θ ⇒ − = −2θ 2 2 0 dθ 2 2 ω = 2 1 − θ2

α = ω

(a)

ω = 2 1 − (π /6) 2 = 1.704 rad/s.

(b) Maximum angle means ω = 0.

θ = 1 rad = 57.3 °.

Problem 13.101 The angle θ measures the direction of the unit vector e relative to the x-axis. The angular velocity of e is ω = d θ /dt = 2 rad/s, a constant. Determine the derivative d e/dt when θ = 90 ° in two ways: (a) Use Eq. (13.33). (b) Express the vector e in terms of its x and y components and take the time derivative of e.

e u

Solution: (a) d e = d θ n = ω n dt dt when θ = 90 °, n = −i de = −2 i rad/s dt

when θ = 90 °

(b) e = (1)cos θi + (1)sin θ j

( )

de dθ dθ = −sin θ i + cos θ j dt dt dt Evaluating at θ = 90 ° de dθ = − i = −2 i rad/s. dt dt

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e u

v 5 du 5 2 rad/s dt x

e u 5 908 x

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Problems 13.102 The angle θ measures the direction of the unit vector e relative to the x-axis. The angle θ is given as a function of time by θ = 2t 2 rad. What is the vector d e/dt at t = 4 s?

e u

Solution:

By definition:

( )

de dθ = n, dt dt where

(

)

(

)

π π + j sin θ + 2 2 is a unit vector in the direction of positive θ. The angular rate of change is n = i cos θ +

 dθ  = [4t ] t = 4 = 16 rad/s.  dt  t=4

The angle is θ = [mod(2t 2 , 2π )] t = 4 = mod(32, 2π ) = 0.5841 rad, where mod(x, y ) (“modulus”) is a standard function that returns the remainder of division of the first argument by the second. From which,

(

)

(

π π  de  = 16 i cos 0.5841 + + j sin 0.5841 +  dt  2 2 t=4 = −8.823i + 13.35 j.

Problem 13.103 The rotor of an electric generator is rotating at 100 rpm. Due to frictional effects, the angular acceleration of the rotor when the motor is turned off is α = −0.02ω rad/s 2 , where ω is the angular velocity in rad/s. What is the rotor’s angular velocity 30 s after it is turned off?

Solution: 100

))

The initial angular velocity in rad/s is

(

revolutions 2 π rad min 1 revolution

)( 160mins ) = 10.5 rad/s.

Let t = 0 be the time at which the generator is turned off. We can write the angular acceleration as dω = −0.02ω rad/s 2 , dt separate variables and integrate: ω

dω

30 s

∫10.5 rad/s ω = −0.02∫0

dt,

ω [ln ω ]10.5 rad/s

= −0.02(30 s), ω = −0.02(30 s), ln 10.5 rad/s ω = (10.5 rad/s)e −0.02(30 s)

(

)

= 5.75 rad/s. 5.75 rad/s (54.9 rpm).

Problem 13.104 The center O of the motor armature is stationary. The armature is rotating at constant angular velocity ω = 1000 rpm (revolutions per minute). What are the magnitudes of the velocity and acceleration of point P relative to point O?

Solution:

The angular velocity in rad/s is

(

revolutions 2 π rad ω = 1000 min 1 revolution

)( 160mins ) = 105 rad/s.

From Eq. (13.43), the velocity is v = Rω = (0.08 m)(105 rad/s) = 8.38 m/s. From Eqs. (13.44) and (13.45), the tangential component of the acceleration is zero and the normal component of the acceleration is

P O

a n = Rω 2 = (0.08 m)(105 rad/s) 2 = 877 m/s 2 .

80 mm

8.38 m/s, 877 m/s 2 .

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Problem 13.105 The armature starts from rest at t = 0 and has constant angular acceleration α = 2 rad/s 2 . At t = 4 s, what are the velocity and acceleration of point P relative to point O in terms of normal and tangential components?

P O

Solution:

We can find the angular velocity at t = 4 s.

α = 2 rad/s 2

80 mm

ω = (2 rad/s 2 )(4 s) = 8 rad/s. Then v = r ω = (0.08 m)(8 rad/s) = 0.64 m/s, a t = rα = (0.08 m)(2 rad/s 2 ) = 0.16 m/s 2 , (0.064 m/s 2 ) v2 = = 5.12 m/s 2 r 0.08 m v = (0.64 m/s)e t

an =

a = (0.16 m/s 2 )e t + (5.12 m/s 2 )e n .

Problem 13.106 Suppose that you want to design a medical centrifuge to subject samples to normal accelerations of 1000 g’s. (a) If the distance from the center of the centrifuge to the sample is 300 mm, what speed of rotation in rpm is necessary? (b) If you want the centrifuge to reach its design rpm in 1 min, what constant angular acceleration is necessary? Solution: (a) T he normal acceleration at a constant rotation rate is a n = Rω 2 , giving an = R

ω =

300 mm

(1000)9.81 = 180.83 rad/s. 0.3

The speed in rpm is N = ω

( rads )( 21πrev )( 60 s ) = 1730 rpm. rad 1 min

(b) The angular acceleration is α =

180.83 ω = = 3.01 rad/s 2 . t 60

Problem 13.107 The medical centrifuge starts from rest at t = 0 and is subjected to a constant angular acceleration α = 3 rad/s 2 . (a) What is the magnitude of the acceleration to which the samples are subjected at t = 1 s? (b) At what time t do the samples reach an acceleration magnitude of 500 g’s? Solution: (a) D ue to the constant angular acceleration, the angular velocity as a function of time is ω = αt. From Eqs. (13.44) and (13.45), the components of acceleration of a sample as functions of time are

300 mm

a t = Rα, a n = Rω 2 = Rα 2t 2 ,

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13.107 (Continued) where R = 0.3 m. Therefore the magnitude of the acceleration as a function of time is a = =

at 2 + an 2 ( Rα) 2 + ( Rα 2t 2 ) 2 .

(1)

Setting t = 1 s, we obtain a = 0.285 m/s 2 . (b) 500 g ’s is (500)(9.81 m/s 2 ). Setting a in Eq. (1) equal to this value and solving for the time, we obtain t = 135 s. (a) 0.285 m/s 2 . (b) t = 135 s.

Problem 13.108 A centrifuge used to subject engineering components to high acceleration has a radius of 8 m. It starts from rest at t = 0, and during its 2-min acceleration phase it is programmed so that its angular acceleration is given as a function of time in seconds by α = 0.192 − 0.0016t rad/s 2 . At t = 120 s, what is the magnitude of the acceleration a component is subjected to?

Solution: ω= ∫

120 s 0

We will first calculate the angular velocity

([0.192 rad/s 2 ] − [0.0016 rad/s 3 ]t ) dt

= [0.192 rad/s][120 s] −

1 [0.0016 rad/s 3 ][120 s] 2 2

= 11.52 rad/s The normal and tangential components of acceleration are a t = rα = (8 m)([0.192 rad/s 2 ] − [0.0016 rad/s 3 ][120 s]) = 0 a n = r ω 2 = (8 m)(11.52 rad/s) 2 = 1060 m/s 2 Since the tangential component is zero, then the total acceleration is the same as the normal acceleration a = 1060 m/s 2 (108g′s).

Source: Courtesy of Sandia National Laboratories.

Problem 13.109 A powerboat being tested for maneuverability is started from rest at t = 0 and driven in a circular path 12 m in radius. The tangential component of the boat’s acceleration as a function of time is a t = 0.4t m/s 2 . (a) What are the boat’s velocity and acceleration in terms of normal and tangential components at t = 4s? (b) What distance does the boat move along its circular path from t = 0 to t = 4 s? Solution: (a) a t = 0.4t m/s 2

a n = +v 2 /r

v = 0.2t 2 m/s At t = 4 s, a = 0.4te t + v 2 /re n a = 1.6e t + 0.853e n v = 3.2e t m/s. (b) s = 0.2t 3 /3 s| 4 s = 4.27 m.

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Problem 13.110 The angle θ = 2t 2 rad. (a) What are the velocity and acceleration of point P in terms of normal and tangential components at t = 1 s? (b) What distance along the circular path does point P move from t = 0 to t = 1 s?

u O

Solution: θ = 2t 2 dθ = 4t = ω dt 2 d θ rad = 4 2 = α dt 2 s s = rθ = 4θ = 8t 2

v t = 16t m/s v = r ω = 4(4t ) = 16t at = (a)

dv = 16 m/s 2 dt

v = 16(1)e t m/s = 16 e t (m/s). 4m

a = Rαe t + Rω 2 e N

a = (4)(4)e t + (4)(4 2 )e N (m/s 2 ) a = 16e t + 64 e N (m/s 2 ). (b) s = Rθ = 8t 2 = 8(1) 2 = 8 m.

Problem 13.111 The angle θ = 4t 2 rad. What are the magnitudes of the velocity and acceleration of point P when P has gone one revolution around the circular path starting at t = 0?

u 4m

Solution:

The angular velocity and angular acceleration of the line from the center to P are dθ = 8t rad/s, dt dω α = = 8 rad/s 2 . dt

To determine the time at which P has made one revolution, we set

The velocity of P is

θ = 4t 2 = 2π rad,

v = Rωe t = R(8t )e t ,

obtaining t = 1.25 s. Using this value, we obtain v = 40.1 m/s,

and the components of the acceleration are

a = 403 m/s 2 .

a t = Rα = 8 R,

v = 40.1 m/s, a = 403 m/s 2 .

ω =

a n = −Rω 2 = −(8t ) 2 R. Their magnitudes are v = 8 Rt, a =

at 2 + an 2 =

(8 R) 2 + (64 Rt 2 ) 2 .

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Problem 13.112 At the instant shown, the crank AB is rotating with a counterclockwise angular velocity of 5000 rpm. Determine the velocity of point B (a) in terms of normal and tangential components and (b) in terms of cartesian components.

Solution: ω = (5000 rpm)

( 2πrevrad )( 60minsec ) = 524 rad/s ft ( 121 in. )(524 rad/s)e = (87.3e ) ft/s.

458 B

(a)

VB = (2 in.)

(b)

VB = (87.3 ft/s)(−cos 45 °i − sin 45 ° j) = (−61.71i − 61.71 j) ft/s.

A 2

Problem 13.113 The crank AB is rotating with a constant counterclockwise angular velocity of 5000 rpm. Determine the acceleration of point B (a) in terms of normal and tangential components and (b) in terms of cartesian components.

Solution: ω = (5000 rpm)

( 2πrevrad )( 60minsec ) = 524 rad/s

a t = 0, a n = (2 in)

458 B

a P = (45,700 e n ) ft/s 2 . a p = (45,700 ft/s 2 )(cos 45 °i − sin 45 ° j) =

Problem 13.114 Suppose that a circular tunnel of radius R could be dug beneath the equator. In principle, a satellite could be placed in orbit about the center of the Earth within the tunnel. The acceleration due to gravity in the tunnel would be gR /R E , where g is the acceleration due to gravity at sea level and R E is the Earth’s radius. Determine the velocity of the satellite and show that the time required to complete one orbit is independent of the radius R. (See Example 13.11.) Equator

(32,300 i − 32,300 j) ft/s 2 .

(b)

(a)

( 121 ftin )(524 rad/s) = 45,693 ft/s ≈ 45,700

Solution: an =

To be in orbit we must have

gR g v2 = ⇒ v = R R RE RE

The velocity of the satellite is given by the distance it travels in one orbit divided by the time needed to complete that orbit. 2π R v v = ⇒ t = = t 2π R

g RE = R

g . RE

Notice that the time does not depend on the radius R.

R Tunnel

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Problem 13.115 At the instant shown, the magnitude of the airplane’s velocity is 130 m/s, its tangential component of acceleration is a t = −4 m/s 2 , and the rate of change of its path angle is d θ /dt = 5 °/s. (a) What are the airplane’s velocity and acceleration in terms of normal and tangential components? (b) What is the instantaneous radius of curvature of the airplane’s path?

Solution: ω = (5 °/s)

( π180rad° ) = ( 36π )rad/s

a Pt = −4 m/s 2 , a n = (130 m/s)ω = 11.34 m/s 2 (a)

v p = (130 e t ) m/s a p = (−4 e t + 11.34 e n ) m/s 2 .

(b) ρ =

(130 m/s) 2 v2 = = 1490 m. an 11.34 m/s 2

Problem 13.116 In the preliminary design of a sunpowered car, a group of engineering students estimates that the car’s acceleration will be 0.6 m/s 2 . Suppose that the car starts from rest at A, and the tangential component of its acceleration is a t = 0.6 m/s 2 . What are the car’s velocity and acceleration in terms of normal and tangential components when it reaches B?

50 m

A 200 m

Solution: at = v

s v dv = 0.6 m/s 2 ⇒ ∫ v d v = ∫ (0.6 m/s 2 ) ds 0 0 ds

v 2 = 2(0.6 m/s 2 )s At point B

(

S B = 200 +

v B2 50 π m ⇒ v B = 18.28 m/s, a Bn = = 6.68 m/s 2 2 50 m

)

Thus v B = (18.28e t ) m/s, a B = (0.6e t + 6.68e n ) m/s 2 .

Problem 13.117* After subjecting the car’s design to wind-tunnel testing, the students estimate that the tangential component of the car’s acceleration will be a t = 0.6 − 0.002v 2 m/s 2 , where v is the car’s velocity in m/s. If the car starts from rest at A, what are its velocity and acceleration in terms of normal and tangential components when it reaches B?

Solution:

(

At point B S B = 200 + at = v

)

50 π m 2

sB vB dv v dv = 0.6 − 0.002v 2 ⇒ ∫ = ∫ ds 0 0.6 − 0.002v 2 0 ds

v B = 14.20 m/s, a Bn =

v 2B = 4.03 m/s 2 50 m

a t = 0.6 − 0.002(14.20 m/s) 2 = 0.197 m/s 2 50 m

Thus B v B = (14.20 e t ) m/s,

a B = (0.197e t + 4.03e n ) m/s 2 . 200 m

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Problem 13.118* Suppose that the tangential component of acceleration of a car is given in terms of the car’s position by a t = 0.4 − 0.001s m/s 2 , where s is the distance the car travels along the track from point A. What are the car’s velocity and acceleration in terms of normal and tangential components at point B?

Solution: at =

dv dv ds = = 0.4 − 0.001s m/s 2 dt ds dt

dv = 0.4 − 0.001s ds

∫0 v dv = ∫0 (0.4 − 0.001s) ds v2 0.001s 2  S B =  0.4 s −   2 2 0

50 m

From Fig. P13.116, S B = 200 + 2πρ /4 where ρ = 50 m, so S B = 278.5 m Solving for v ,

A 200 m

v = 12.05 m/s v = 12.05e t (m/s). a = (0.4 − 0.001s B )e t + v 2 /ρe N (m/s 2 ) Solving, a = 0.121e t + 2.905e N (m/s 2 ).

Problem 13.119 The car increases its speed at a constant rate from 40 mi/h at A to 60 mi/h at B. What is the magnitude of its acceleration 2 s after it passes point A? Solution:

y 120 ft

Use the chain rule to obtain

308

dv v = a, ds

308

where a is constant. Separate variables and integrate: v 2 = 2as + C. At

(

80 ft

100 ft

)

5280 s = 0, v (0) = 40 = 58.67 ft/s, 3600 from which C = 3441.78. The acceleration is a =

v2 − C 2s

The distance traveled from A to B is s = 2(80) + (30)

π ( 180 )(120 + 100) = 275 ft,

and the speed in [v (s)] s = 275 = 60

5280 ( 3600 ) = 88 ft/s,

from the constant acceleration is a =

(88) 2 − 3441.78 = 7.817 ft/s 2 . 2(275)

The velocity is as a function of time is v (t ) = v (0) + at = 58.67 + 7.817t ft/s. The distance from A is s(t ) = 58.67t +

7.817 2 t . 2

At a point 2 seconds past A, the distance is s(2) = 132.97 ft, and the velocity is v (2) = 74.3 ft/s. The first part of the hill ends at 142.83, so that at this point the car is still in the first part of the hill. The tangential acceleration is a t = 7.817 ft/s 2 . The normal acceleration is an =

(74.3) 2 v2 = = 46.0 ft/s 2 . R 120

The magnitude of the acceleration is a =

7.817 2 + 46.0 2 = 46.66 ft/s 2 .

Note: This is a large acceleration–the driver (and passengers) would no doubt be uncomfortable.

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Problem 13.120 The car increases its speed at a constant rate from 40 mi/h at A to 60 mi/h at B. Determine the magnitude of its acceleration when it has traveled along the road a distance (a) 120 ft from A and (b) 160 ft from A.

y 120 ft 308

Solution:

Use the solution in Problem 13.119.

308

(a) The velocity at a distance 120 ft from A is v (120) =

2as + C =

80 ft

(v (120)) 2 (72.92) 2 = = 44.3 ft/s 2 . R 120

(v (160)) 2 5943.1 = = 59.43 ft/s 2 . 100 R

The magnitude of the acceleration is

7.817 2 + 44.3 2 = 45 ft/s 2 .

a =

(b) The velocity at distance 160 ft from A is v (160) =

100 ft

At 160 ft the car is on the second part of the hill. The tangential acceleration is unchanged: a t = 7.817 ft/s 2 . The normal acceleration is an =

The magnitude of the acceleration is a =

(2)(7.817)(120) + 3441.78

= 72.92 ft/s. At 120 ft the car is in the first part of the hill. The tangential acceleration is a t = 7.817 ft/s 2 from Problem 13.119. The normal acceleration is an =

80 ft

7.817 2 + 59.43 2 = 59.94 ft/s 2 .

[Note: The car will “lift off” from the road.]

2(7.817)(160) + C = 77.1 ft/s.

Problem 13.121 Astronaut candidates are to be tested in a centrifuge with 10-m radius that rotates in the horizontal plane. Test engineers want to subject the candidates to an acceleration of 5 g’s, or five times the acceleration due to gravity. Earth’s gravity effectively exerts an acceleration of 1 g in the vertical direction. Determine the angular velocity of the centrifuge in revolutions per second so that the magnitude of the total acceleration is 5 g’s.

10 m

Solution: a n 2 + g 2 = (5g) 2 ⇒ a n = an = ω =

rω 2

⇒ ω =

24 g

a n /r

24(9.81 m/s 2 ) = 2.19 rad/s. 10m

Problem 13.122 If the centrifuge accelerates from rest so that its angle of rotation is θ = 0.03t 2 rad, at what time t is the candidate subjected to a total acceleration of 5 g’s? Solution: Let a denote the magnitude of the horizontal acceleration due to the rotation of the centrifuge. Accounting for the effective vertical acceleration due to gravity, we want the candidate’s total acceleration to equal 5 g: a 2 + (1 g) 2 = 5g. We want the horizontal acceleration to be 4.90(9.81 m/s 2 ).

a = 4.90 g =

The angular velocity and angular acceleration of the centrifuge as functions of time are dθ = 0.06t rad/s, dt dω α = = 0.06 rad/s 2 . dt

ω =

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10 m

The components of the horizontal acceleration are a t = Rα = 0.06 R, a n = −Rω 2 = −(0.06t ) 2 R. The magnitude of the horizontal acceleration is a t2 + a n2 =

(0.06 R) 2 + [(0.06t ) 2 R] 2 .

Setting this equal to the horizontal acceleration a and solving for t, we obtain t = 36.5 s. t = 36.5 s.

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Problem 13.123 The athlete releases the shot with velocity v = 16 m/s. (a) What are the velocity and acceleration of the shot in terms of normal and tangential components when it is at the highest point of its trajectory? (b) What is the instantaneous radius of curvature of the shot’s path when it is at the highest point of its trajectory?

v 208

Solution: et

16 m/s 208

(a) W e know that the shot’s acceleration is g downward. The horizontal component of the acceleration is zero, so the horizontal component of velocity, (16 m/s) cos 20 ° = 15.0 m/s, is constant. When it is at the highest point of its trajectory, the direction of its motion is horizontal, so its velocity in terms of normal and tangential components at that instant is v = 15.0 e t (m/s). Its acceleration at that instant is g in the direction normal to its path and in the direction of the concave side, so in terms of normal and tangential components, a = g e n = 9.81 e n m/s 2 . (b) From Eq. (13.40), we have v2 : ρ (15.0 m/s) 2 , 9.81 m/s 2 = ρ an =

which yields ρ = 23.0 m. (a) v = 15.0 e t (m/s), a = g e n = 9.81e n m/s 2 . (b) 23.0 m.

Problem 13.124 An athlete releases a shot at a speed v 0 = 42 ft/s . What are the velocity and acceleration of the shot in terms of normal and tangential components at t = 1 s? Strategy: First analyze the motion in terms of cartesian coordinates.

Let t = 0 be the time at which the shot is released. The initial components of the velocity are v x = v 0 cos 20 °, v y = v 0 sin 20 °, where v 0 is the initial velocity, and the components of the acceleration are a x = 0, a y = −g. We can integrate the components of acceleration with respect to time to determine the velocity as a function of time. In the x direction, ax =

208

dv x = 0, dt

∫v cos 20° dv x = 0, 0

vx =

dx = v 0 cos 20 °, dt

and in the y direction, ay = vy

dv y = −g, dt t

∫v sin 20° dv y = ∫0 −g dt, 0

vy =

dy = v 0 sin 20 ° − gt. dt

At t = 1 s, v x = 39.5 ft/s, v y = −17.8 ft/s. The magnitude of the velocity at that time is v x 2 + v y 2 = 43.3 ft/s.

Solution:

Using the components of the velocity, the angle between the path and the horizontal is arctan(17.8/39.5) = 24.3 °:

42 ft/s

42 ft/s 208

24.38 et

208 x

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13.124 (Continued) The velocity is v = 43.3 e t (ft/s) and the acceleration is a = g sin 24.3 ° e t + g cos 24.3 ° e n = 13.3 e t + 29.3 e n (ft/s 2 ). v = 43.3 e t (ft/s), a = 13.3 e t + 29.3 e n (ft/s 2 ).

Problem 13.125 An athlete releases a shot at a speed v 0 = 42 ft/s . What is the instantaneous radius of curvature of its path at t = 1 s?

and in the y direction, ay =

dv y = −g, dt

∫v sin 20° dv y = ∫0 −g dt,

208

vy =

dy = v 0 sin 20 ° − gt, (2) dt

∫0 dy = ∫0 (v 0 sin 20° − gt ) dt, y = (v 0 sin 20 °)t −

1 2 gt . 2

At t = 1 s, v x = 39.5 ft/s, v y = −17.8 ft/s. The magnitude of the velocity at that time is v x 2 + v y 2 = 43.3 ft/s. The angle between the path and the horizontal is arctan(17.8/ 39.5) = 24.3 °:

42 ft/s

24.38 et

208 en

Solution: y

The normal component of the acceleration is a n = g cos 24.3 ° = 29.3 ft/s 2 . From Eq. (13.40),

42 ft/s 208

v2 : ρ (43.3 ft/s) 2 29.3 ft/s 2 = , ρ an =

we obtain ρ = 63.9 ft. Let t = 0 be the time at which the shot is released. The initial components of the velocity are v x = v 0 cos 20 °, v y = v 0 sin 20 °, where v 0 is the initial velocity, and the components of the acceleration are a x = 0, a y = −g. We can integrate the components of acceleration with respect to time to determine the velocity as a function of time. In the x direction, ax =

dv x = 0, dt

∫v cos 20° dv x = 0, 0

vx = x

dx = v 0 cos 20 °, (1) dt t

∫0 dx = ∫0 v 0 cos 20° dt, x = (v 0 cos 20 °)t,

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An alternative approach is to use Eq. (13.42). From Eq. (1), at t = 1 s the distance x = 39.467 ft. Solving Eq. (1) for t and substituting it into Eq. (2), we obtain the equation for the path: y = tan 20 ° x −

g x 2. 2v 0 2 cos 2 20 °

Its first two derivatives are dy g = tan 20 ° − 2 x = −0.45190, (3) dx v 0 cos 2 20 ° d 2y g = − 2 = −0.020672 ft −1 . (4) dx 2 v 0 cos 2 20 ° Substituting these values into Eq. (13.42), we obtain ρ = 63.9 ft. ρ = 63.9 ft.

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Problem 13.126 The cartesian coordinates of a point moving in the x–y plane are x =

20 + 4t 2 m

and

= 10 − t 3 m.

What is the instantaneous radius of curvature of the path of the point at t = 3 s?

Solution:

The components of the velocity: v = 8ti − (3t 2 ) j.

At t = 3 seconds, the magnitude of the velocity is | v | t = 3 = (8t ) 2 + (−3t 2 ) 2 = 36.12 m/s. The components of the acceleration are vy −3t 2 a = 8i − (6t ) j. The instantaneous path angle is tan β = = . 8t vx At t = 3 seconds, β = −0.8442 rad. The unit vector parallel to the path is e t = i cos β + j sin β . The unit vector normal to the path pointing toward the instantaneous radial center is π π e n = i cos β − + j sin β − = i sin β − j cos β . 2 2 The normal acceleration is the component of acceleration in the direction of e n . Thus, a n = e n ⋅ a or a n = 8sin β + (6t ) cos β . At t = 3 seconds, a n = 5.98 m/s 2 . The radius of curvature at t = 3 seconds is

(

ρ =

Problem 13.127 The helicopter starts from rest at t = 0. The cartesian components of its acceleration are a x = 0.6t m/s 2 and a y = 1.8 − 0.36t m/s 2 . Determine the tangential and normal components of the acceleration at t = 6 s. y

)

(

)

|v| 2 = 218 m. an

Solution:

The solution will follow that of Example 13.11, with the time changed to t = 6 s. The helicopter starts from rest (v x , v y ) = (0, 0) at t = 0. Assume that motion starts at the origin (0, 0). The equations for the motion in the x direction are a x = 0.6t m/s 2 , v x = 0.3t 2 m/s, x = 0.1t 3 m, and the equations for motion in the y direction are a y = 1.8 − 0.36t m/s 2 , v y = 1.8t − 0.18t 2 m/s, and 2 y = 0.9t − 0.06t 3 m. At t = 6 s, the variables have the values a x = 3.6 m/s 2 , a y = −0.36 m/s 2 , v x = 10.8 m/s, v y = 4.32 m/s, x = 21.6 m, and y = 19.44 m. The magnitude of the velocity is given by v =

v x 2 + v y 2 = 11.63 m/s.

The unit vector in the tangential direction is given by eT =

v xi + v y j v = = 0.928i + 0.371 j. v v

The tangential acceleration component is given by a T = a ⋅ e T = 0.928a x + 0.371a y = 3.21 m/s 2 . The magnitude of the acceleration is given by a = x

The normal acceleration component is given by aN =

Problem 13.128 Suppose that when the centrifuge in Example 13.13 is turned on, its motor and control system give it an angular acceleration (in rad/s 2 ) α = 12 − 0.02ω, where ω is the centrifuge’s angular velocity. Determine the tangential and normal components of the acceleration of the samples at t = 0.2 s. Solution:

a x 2 + a y 2 = 3.62 m/s 2 .

| a | 2 − a T2 = 1.67 m/s 2 .

[12 rad/s 2 ] − [0.02 s −1 ]ω = [12 rad/s 2 ]e −0.004 ω =

1s (12 rad/s 2 )(1 − e −0.004 ) = 2.40 rad/s 0.02

At this time, the angular acceleration is α = (12 rad/s 2 ) − (0.02 s −1 )(2.40 rad/s) = 11.95 rad/s 2 . The components of acceleration are

We will first integrate to find the angular velocity at

t = 0.2 s. dω α = = ([12 rad/s 2 ] − [0.02 s −1 ]ω ) dt ω 0.2 s dω ∫0 = [12 rad/s 2 ] − [0.02 s −1 ]ω ∫0 dt

(

a t = rα = (0.3 m)(11.95 rad/s 2 ) = 3.59 m/s 2 α n = r ω 2 = (0.3 m)(2.40 rad/s) 2 = 1.72 m/s 2 . a = (3.59e t + 1.72e n ) m/s 2 .

)

−1 s [12 rad/s 2 ] − [0.02 s −1 ]ω = 0.2 s ln 0.02 [12 rad/s 2 ]

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Problem 13.129* For astronaut training, the airplane shown is used to achieve “weightlessness” for a short period of time by flying along a path such that its acceleration is a x = 0 and a y = −g. If the velocity of the plane at O at time t = 0 is v = v 0 i, show that the autopilot must fly the airplane so that its tangential component of acceleration as a function of time is gt /v 0 . at = g 1 + ( gt /v 0 ) 2

Solution: angle is

vy −gt , = vx v0 −gt sin β = . v 0 2 + ( gt ) 2

β : tan β =

The unit vector parallel to the velocity vector is e = i cos β + j sin β . The acceleration vector is a = − jg. The component of the acceleration tangent to the flight path is a t = −g sin β ., from which gt . v 0 2 + ( gt ) 2

at = g

The velocity of the path is v(t ) = v 0 i − gtj . The path

Divide by v 0 , −

  gt  2  2  gt  a t = g  1 +     .   v 0    v 0  x O

Problem 13.130* In Problem 13.129, what is the airplane’s normal component of acceleration as a function of time? y

Solution:

From Problem 13.129, the velocity is v(t ) = v 0 i − gtj. The flight path angle is β, from which v0 . v 0 2 + ( gt ) 2

cos β =

The unit vector parallel to the flight path is e = i cos β + j sin β . The unit vector normal to e is

(

)

(

π π + j sin β − 2 2 = i sin β − j cos β ,

e n = i cos β −

x O

)

pointing toward the instantaneous radial center of the path. The acceleration is a = − jg. The component parallel to the normal component is a n = g cos β , from which −

Problem 13.131 If y = 100 mm, dy /dt = 200 mm/s, and d 2 y /dt 2 = 0, what are the velocity and acceleration of P in terms of normal and tangential components?

  gt  2  2 v0 = g  1 +    . 2 2   v 0   v 0 + ( gt ) 

an = g

Solution:

The equation for the circular guide is R 2 = x 2 + y 2 , from which x = R 2 − y 2 = 0.283 m, and

( )

y dy dx = − = v x = −0.0707 m/s. dt x dt The velocity of point P is vp = iv x + jv y , from which the velocity is v = v x 2 + v y 2 = 0.212 m/s. The angular velocity ω =

v = 0.7071 rad/s. R

The angle is P

( yx ) = 19.5° dv y dy d = − a = dt dt ( x dt ) y dx dy 1 dy = − ( ) + ( ) − ( yx )( ddt y ) x dt x dt ( dt ) β = tan −1 x

y 300 mm

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= −0.1591 m/s 2

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13.131 (Continued) The unit vector tangent to the path (normal to the radius vector for a circle) is e p = −i sin β + j cos β , from which a t = −a x sin β = 53.0 mm/s 2 . Since a y = 0 a n = −Rω 2 = −0.150 m/s 2 . Check: a n = a x cos β = −0.15 m/s 2 check.

Problem 13.132* Suppose that the point P moves upward in the slot with velocity v = 300 e t (mm/s). When y = 150 mm, what are dy /dt and d 2 y /dt 2 ?

Solution: θ =

sin −1

The position in the guide slot is y = R sin θ, from which y = sin −1 (0.5) = 30 °. R

( )

x = R cos θ = 259.8 mm. From the solution to Problem 13.131,

( yx ) dydt = −( yx )v .

vx = −

The velocity is v = 300 = P

−

v x2 + v y2 = v y

( yx ) + 1, from which

 2  y 2 v y = 300  + 1  = 259.8 mm/s   x

( )

and v x = −150 mm/s (Since the point is moving upward in the slot, v y is positive.). The velocity along the path in the guide slot is assumed constant, hence a t = 0. The normal acceleration is

y 300 mm

an =

| v |2 = 300 mm/s 2 R

directed toward the radius center, from which d 2y = −a n sin θ = −150 mm/s 2 . dt 2

Problem 13.133* A car travels at 100 km/h on a straight road of increasing grade whose vertical profile can be approximated by the equation shown. When the car’s horizontal coordinate is x = 400 m, what are the tangential and normal components of the car’s acceleration? y

Solution:

The strategy is to use the acceleration in cartesian coordinates found in the solution to Problem 13.90, find the angle with respect to the x-axis, dy θ = tan −1 , dx and use this angle to transform the accelerations to tangential and normal components. From the solution to Problem 13.90 the accelerations are a = −0.0993i + 0.4139 j (m/s 2 ). The angle at d θ = tan −1 Cx 2 = tan −1 (6 x × 10 −4 ) x = 400 = 13.5 °. dx x = 400 From trigonometry (see figure) the transformation is a t = a x cos θ + a y sin θ, a n = −a x sin θ + a y cos θ, from which

( ) (

y 5 0.0003x2

)

a t = 0.000035 ... = 0 ,

a n = 0.4256 m/s 2 . Check: The velocity is constant along the path, so the tangential component dv of the acceleration is zero, a t = = 0, check. dt

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Problem 13.134 A boy rides a skateboard on the concrete surface of an empty drainage canal described by the equation shown. He starts at y = 20 ft, and the magnitude of his velocity is approximated by v = 2(32.2)(20 − y ) ft/s. (a) Use Eq. (13.42) to determine the instantaneous radius of curvature of the boy’s path when he reaches the bottom. (b) What is the normal component of his acceleration when he reaches the bottom?

y y 5 0.03x 2

Solution: (a) y =

The acceleration is

0.03 x 2 , so

dy = 0.06 x and 2 = 0.06. dx dx

[1 + (0.06 x ) 2 ] 3/ 2 ft. 0.06

( dydt )

1 1 dy 2 dx 2 + = v = K (20 − y ) 2 = K (20 − Cx 2 ) 2 , dt dt

( ) ( )

where K = 8.025, C = 0.03. From y

= 0,

= Cx 2 ,

( ddt x ) 2

= 2C

x=0

θ = tan −1

( ) + 2Cx ( ddt x ). 2

( dxdt )

= 0,

= 77.28 ft/s 2 . x=0

( dydx )

= 0.

x=0

From the solution to Problem 2.133, the tangential and normal accelerations are a t = a x cos θ + a y sin θ, a n = −a x sin θ + a y cos θ, from which

Substitute: 1

a t = 0 , and a n = 77.28 ft/s 2 .

K (20 − Cx 2 ) 2 dx = 1 . dt (4C 2 x 2 + 1) 2

Check: The velocity is constant along the path, so the tangential dv component of the acceleration is zero, a t = = 0. check. By dt inspection, the normal acceleration at the bottom of the canal is identical to the y component of the acceleration. check.

Since the boy is moving the right, dx dx = . dt dt

Problem 13.135 In Problem 13.134, what is the normal component of the boy’s acceleration when he has passed the bottom and reached y = 10 ft?

Solution:

Use the results of the solutions to Problems 13.133 and 13.134. From the solution to Problem 13.134, at y = 10 ft,  y  x =   = 18.257 ft,  c 

( dxdt )

y y 5 0.03x 2

( ddt x )

y =10

(

−

= −K

( dxdt )

)

y =10

= 17.11 ft /s.

 Cx  1 1 y =10   (20 − Cx 2 ) 2 (4C 2 x 2 + 1) 2

1  (4C 2 x )(20 − Cx 2 ) 2   3  (4C 2 x 2 + 1) 2 

y =10

= −24.78 ft/s.

( ddt x ) 2

y =10

( ( dxdt ) + 2Cx( ddt x ))

= 2C

The angle is θ = tan −1

from which

= K (20 − C 2 ) 2 (4C 2 x 2 + 1) 2

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x=0

The angle with respect to the x axis at the bottom of the canal is

( )

and

x=0

( ddt y )

dy dx = 2Cx , dt dt

dx > 0, dt

( )

K (4C 2 x )(20 − Cx 2 ) 2 dx . 3 dt (4C 2 x 2 + 1) 2

( )

(b) The magnitude of the velocity is

d 2y dx = 2C dt 2 dt

( dxdt )

At the bottom of the canal the values are dx = K 20 = 35.89 ft/s. dt x = 0

At x = 0, ρ = 16.7 ft.

1 + 1) 2

1 (20 − Cx 2 ) 2 (4C 2 x 2

−

From Eq (13.42), ρ =

−KCx

d 2x = dt 2

d 2y

( dxdt )

y =10

= −9.58 ft /s 2 . y =10

= 47.61 °.

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13.135 (Continued) From the solution to Problem 13.133, a t = a x cos θ + a y sin θ, a n = −a x sin θ + a y cos θ, from which a t = −23.78 ft/s 2 , a n = 11.84 ft/s 2 .

Problem 13.136* By using Eqs. (13.41): (a) show that the relations between the cartesian unit vectors and the unit vectors e t and e n are i = cos θ e t − sin θ e n j = sin θ e t + cos θ e n . (b) Show that

de t dθ e = dt dt n

de n dθ = − et. dt dt

and

Problem 13.137 For an interval of time, the polar coordinates of the collar A are given as functions of time in seconds by r = 2t − 0.8t 2 m and θ = 0.6t 2 rad. What is the magnitude of the velocity of the collar at t = 1 s?

Solution:

Equations e n = − sin θ i + cos θ j.

(13.41)

are

e t = cos θ i + sin θ j

and

(a) M ultiplying the equation for e t by cos θ and the equation for e n by (− sin θ) and adding the two equations, we get i = cos θe t − sin θe n . Similarly, by multiplying the equation for e t by sin θ and the equation for e n by cos θ and adding, we get j = sin θ e t + cos θ e n . de (b) Taking the derivative of e t = cos θ i + sin θ j, we get t = dt dθ dθ (− sin θ i + cos θ j) = en . dt dt Similarly, taking the derivative of e n = − sin θ i + cos θ j, we get d e n /dt = −(d θ /dt )e t .

Solution:

From Eq. (13.57), the radial component of the velocity at

t = 1 s is dr = 2 − 2(0.8)t m/s. dt = 0.4 m/s.

vr =

The transverse component is dθ = (2t − 0.8t 2 )[2(0.6)t ] m/s dt = 1.44 m/s.

vθ = r A

The magnitude of the velocity is u

v =

v r 2 + v θ 2 = 1.49 m/s.

1.49 m/s.

Problem 13.138 In Practice Example 13.14, suppose that the robot arm is reprogrammed so that the point P traverses the path described by r = 1 − 0.5sin 2πt m, θ = 0.5 − 0.2 cos 2πt rad. What is the velocity of P in terms of polar coordinates at t = 0.8 s?

Solution:

We have

( ) ( )

(

)

t t r =  1 − 0.5sin 2π  m, θ =  0.5 − 0.2 cos 2π  rad,   s  s  dr t dθ t rad/s. = −π cos 2π m/s, = 0.4 π sin 2π dt s dt s

(

)

At time t = 0.8 s, r = 1.48 m,

dr dθ = −0.971 m/s, θ = 0.438 rad, = −1.20 rad/s. dt dt

The velocity is vr =

dr dθ = −0.971 m/s, v θ = r = (1.48 m)(−1.20 rad/s) dt dt

= −1.76 m/s. v = (−0.971e r − 1.76e θ ) m/s.

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Problem 13.139 The length and elevation of the crane boom are given as functions of time by r = 3 + 0.2t m, θ = 0.5t rad. (a) What is the velocity of A at t = 1 s in terms of polar coordinates? (b) What is the velocity of A at t = 1 s in terms of cartesian coordinates? Strategy: Part (a) tells you the velocity in two directions. Use that information to obtain the answer to part (b).

Solution:

y A

(a) The radial component of the velocity is

dr = 0.2 m/s. dt The transverse component of the velocity at t = 1 s is dθ vθ = r = (3 + 0.2t )(0.5) dt = 1.6 m/s. vr =

(b) A t t = 1 s the angle θ = 0.5 rad = 28.6 °. From the orientation of the unit vector directions,

u u

we can see that v x = v r cos θ − v θ sin θ = −0.592 m/s, v y = v r sin θ + v θ cos θ = 1.50 m/s. Alternatively, we can start with the equations for the cartesian coordinates of A, x = r cos θ, y = r sin θ, and take derivatives: dx dr dθ cos θ − (r sin θ) = vx = dt dt dt = v r cos θ − v θ sin θ = −0.592 m/s, dy dr dθ sin θ + (r cos θ) = vy = dt dt dt = v r sin θ + v θ cos θ = 1.50 m/s. (a) v = 0.2 e r + 1.6 e θ (m/s). (b) v = −0.592 i + 1.50 j (m/s).

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Problem 13.140 The length and elevation of the crane boom are given as functions of time by r = 3 + 0.2t m, θ = 0.5t rad. (a) What is the acceleration of A at t = 1 s in terms of polar coordinates? (b) What is the acceleration of A at t = 1 s in terms of cartesian coordinates?

we can see that a x = a r cos θ − a θ sin θ = −0.798 m/s 2 , a y = a r sin θ + a θ cos θ = −0.208 m/s 2 . Alternatively, we can start with the equations for the cartesian coordinates of A,

x = r cos θ,

y = r sin θ, and take derivatives:

dx dr dθ cos θ − (r sin θ) , = dt dt dt d 2x d 2r dr dθ dr dθ cos θ − sin θ sin θ = ax = − dt 2 dt 2 dt dt dt dt dθ 2 d 2θ − (r cos θ) − (r sin θ) 2 dt dt = a r cos θ − a θ sin θ

(

)

(

)

( )

u x

= −0.798 m/s 2 , dy dr dθ sin θ + (r cos θ) , = dt dt dt d 2y d 2r dr dθ dr dθ sin θ + cos θ cos θ = ay = + dt 2 dt 2 dt dt dt dt dθ 2 d 2θ − (r sin θ) + (r cos θ) 2 dt dt = a r sin θ + a θ cos θ

(

)

(

)

( )

Solution:

= −0.208 m/s 2 .

(a) Evaluating the required terms,

(a) a = −0.8 e r + 0.2 e θ (m/s 2 ). (b) a = −0.798 i − 0.208 j (m/s 2 ).

r = 3 + 0.2t = 3.2 m, dr = 0.2 m/s, dt 2 d r = 0, dt 2 dθ = 0.5 rad/s, dt 2 d θ = 0. dt 2 From Eqs. (13.59), the radial component of the acceleration is d 2r dθ 2 −r = 0 − (3 + 0.2t )(0.5) 2 2 dt dt = −0.8 m/s 2 .

ar =

( )

The transverse component is d 2θ dr d θ + 2 = 0 + 2(0.2)(0.5) dt 2 dt dt = 0.2 m/s 2 .

aθ = r

(b) A t t = 1 s the angle θ = 0.5 rad = 28.6 °. From the orientation of the unit vector directions,

u u

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Problem 13.141 The radial line rotates with a constant angular velocity of 2 rad/s. Point P moves along the line at a constant speed of 4 m/s. Determine the magnitudes of the velocity and acceleration of P when r = 2 m. (See Example 13.15.) y 2 rad/s

4 m/s P

Solution:

The angular velocity of the line is

dθ = ω = 2 rad/s, dt 2 d θ from which 2 = 0. dt The radial velocity of the point is dr = 4 m/s, dt d 2r from which 2 = 0. dt The vector velocity is dθ dr e = 4 e r + 4 e θ (m/s). v = e +r dt r dt θ The magnitude is

( )

v = x

( )

4 2 + 4 2 = 5.66 m/s.

The acceleration is a = [ −2(4) ] e r + [ 2(4)(2) ] e θ = −8e r + 16e θ (m/s 2 ). The magnitude is a =

Problem 13.142 At the instant shown, the coordinates of the collar A are x = 2.3 ft, y = 1.9 ft. The collar is sliding on the bar from B toward C at a constant speed of 4 ft/s. (a) What is the velocity of the collar in terms of polar coordinates? (b) Use the answer to part (a) to determine the angular velocity of the radial line from the origin to the collar A at the instant shown.

8 2 + 16 2 = 17.89 m/s 2 .

y C

608 x

Solution:

We will write the velocity in terms of cartesian coordinates.

v x = (4 ft/s) cos60 ° = 2 ft/s, v y = (4 ft/s)sin 60 ° = 3.46 ft/s. The angle θ is y 1.9 ft θ = tan −1 = tan −1 = 39.6 °. x 2.3 ft Now we can convert to polar coordinates v r = v x cos θ + v y sin θ = (2 ft/s) cos39.6 ° + (3.46 ft/s)sin 39.6 °,

( )

(

)

v θ = −v x sin θ + v y cos θ = −(2 ft/s)sin 39.6 ° + (3.46 ft/s) cos39.6 °. (a)

v = (3.75e r + 1.40 e θ ) ft/s.

(b)

v dθ = θ = dt r

1.40 ft/s dθ = 0.468 rad/s. = 0.468 rad/s. dt (2.3 ft) 2 + (1.9 ft) 2

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Problem 13.143 A boat searching for underwater archaeological sites in the Aegean Sea moves at 4 knots and follows the path r = 10θ m, where θ is in radians. (A knot is one nautical mile, or 1852 meters, per hour.) When θ = 3π rad, what is the boat’s velocity in terms of polar coordinates?

Solution: 4(1852 m/h)

The boat’s velocity in m/s is 1h ( 3600 ) = 2.06 m/s. s

From the equation for the path, r = 10θ, we have dr dθ = 10 . dt dt The boat’s velocity is v = v r e r + v θe θ = =

dr dθ e + r eθ dt r dt

dr r dr e + e . dt r 10 dt θ

(1)

We know the magnitude of the velocity is 2.06 m/s, so 2

( drdt ) + ( 10r drdt ) = (2.06 m/s) . (2) 2

When θ = 3π rad, r = 10θ = 10(3π rad) = 94.2 m, and we can solve Eq. (2), obtaining dr /dt = 0.217 m/s. Then from Eq. (1), the velocity is 94.2 m v = (0.217 m/s)e r + (0.217 m/s)e θ 10 m = 0.217 e r + 2.05e θ (m/s). v = 0.217 e r + 2.05e θ (m/s).

Problem 13.144* A boat moves at 4 knots and follows the path r = 10θ m, where θ is in radians. (A knot is one nautical mile, or 1852 meters, per hour.) When θ = 3π rad, what is the boat’s acceleration in terms of polar coordinates?

When θ = 3π rad, r = 10θ = 10 ( 3π rad ) = 94.2 m, and we can solve Eq. (2), obtaining dr /dt = 0.217 m/s. The boat’s velocity is dr r dr 94.2 m e + e = (0.217 m/s)e r + (0.217 m/s)e θ dt r (3) 10 dt θ 10 m = 0.217 e r + 2.05e θ (m/s).

v =

From Eq. (1), d 2r d 2θ = 10 2 . 2 dt dt The boat’s acceleration is

( ( )

)

 d 2r dθ 2  d 2θ dr d θ a = ar e r + aθe θ =  2 − r e  er + r 2 + 2  dt dt  dt dt dt θ (4)  d 2r  r d 2r r dr 2  2 dr 2  e . =  2 − +  er +   θ  dt  10 dt 2 100 dt  10 dt 

( )

Here’s a very subtle part. Although we know that the magnitude of the boat’s velocity is constant, 4 knots, its acceleration is not zero because the boat is traveling a curved path. What we do know is that the component of the acceleration tangent to its path is zero. By dividing the velocity by its magnitude, we can obtain a unit vector tangent to the path:

Solution: 4(1852 m/h)

The boat’s velocity in m/s is 1h ( 3600 ) = 2.06 m/s. s

From the equation for the path, r = 10θ, we have dr dθ = 10 . (1) dt dt The boat’s velocity is v = v r e r + v θe θ =

dr dθ e + r eθ dt r dt

dr r dr e + e . dt r 10 dt θ We know the magnitude of the velocity is 2.06 m/s, so =

e =

0.217 e r + 2.05e θ (m/s) v = = 0.106 e r + 0.994 e θ . v 2.06 (m/s)

What we know is that e ⋅ a = 0.106 a r + 0.994 a θ = 0. From Eq. (4), we can see that only unknown in this equation is d 2r /dt 2 . Solving, we obtain d 2r /dt 2 = −0.000495 m/s 2 . Then from Eq. (4), the boat’s acceleration is a = −0.0449e r + 0.00477e θ (m/s 2 ). a = −0.0449e r + 0.00477e θ (m/s 2 ).

( drdt ) + ( 10r drdt ) = (2.06 m/s) . (2)

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Problem 13.145 The collar A slides on the circular bar. The radial position of A (in meters) is given as a function of θ by r = 2 cos θ. At the instant shown, θ = 25 ° and d θ /dt = 4 rad/s. Determine the velocity of A in terms of polar coordinates.

Solution: r = 2 cos θ, r = −2sin θθ , r = −2sin θθ − 2 cos θθ 2 Using the given data we have θ = 25 °, θ = 4, θ = 0 r = 1.813, r = −3.381, r = −29.00

v = r e r + rθ e θ = (−3.381e r + 7.25e θ ) m/s. A r u x

Problem 13.146 In Problem 13.145, d 2 θ /dt 2 = 0 at the instant shown. Determine the acceleration of A in terms of polar coordinates.

Solution:

See Problem 13.145

a = (r − rθ 2 )e r + (rθ + 2r θ )e θ = (−58.0 e r − 27.0 e θ ) m/s 2 .

A r u x

Problem 13.147 The radial coordinate of the Earth satellite is related to its angular position θ by the equation 1.91E7 m. r = 1 + 0.5cos θ The product of its radial position and the transverse component of its velocity is equal to the constant rv θ = 8.72E10 m 2 /s. What is the satellite’s velocity in terms of polar coordinates when θ = 120 ° ?

Solution:

Let C = 1.91E7 and D = 8.72E10. Then

r = C(1 + 0.5cos θ) −1 , (1) dθ rv θ = r 2 = D m 2 /s. (2) dt The radial component of the velocity is vr =

dr dr d θ = . (3) dt d θ dt

From Eq. (1), dr = 0.5C (1 + 0.5cos θ) −2 sin θ, (4) dθ and using Eqs. (1), (2) and (4), Eq. (3) becomes

Satellite

0.5CD sin θ r 2 (1 + 0.5cos θ) 2 (5) 0.5 D sin θ = . C

vr =

From Eqs. (1), (2), and (5), the components of velocity at at θ = 120 ° from Eq. (13.57) are v r = 1977 m/s, v θ = 3424 m/s. v r = 1977 m/s, v θ = 3424 m/s.

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Problem 13.148* For the satellite described in Problem 13.147, what is the acceleration in terms of polar coordinates when θ = 120 ° ? Satellite

dr = 0.5C (1 + 0.5cos θ) −2 sin θ, (4) dθ and using Eqs. (1), (2) and (4), Eq. (3) becomes dr 0.5CD sin θ = 2 dt r (1 + 0.5cos θ) 2 (5) 0.5 D sin θ = . C Applying the chair rule again and using Eq. (5), we obtain

From Eq. (1),

( )

d 2r d dr d dr d θ = = dt 2 dt dt d θ dt dt (6) 0.5 D cos θ d θ = . C dt Finally, from Eq. (2) we have

Solution:

Let C = 1.91E7 and D = 8.72E10. Then

r = C(1 + 0.5cos θ) −1 , (1) dθ = D m 2 /s. (2) dt To evaluate the components of the acceleration for a given angle, we need to determine the values of r, dr /dt, d 2r /dt 2 , d θ /dt, and d 2θ /dt 2 . The terms r and d θ /dt are given by Eqs. (1) and (2). We begin by determining dr /dt. Applying the chain rule, rv θ = r 2

d 2θ d d dr ( Dr −2 ) = = ( Dr −2 ) dt 2 dt dr dt (7) dr − 3 = −2 Dr . dt Applying Eqs. (1), (2), (5), (6) and (7), the components of the acceleration at θ = 120 ° from Eqs. (13.59) are a r = −0.307 m/s 2 , a θ = 0. a r = −0.307 m/s 2 , a θ = 0.

dr dr d θ = . (3) dt d θ dt

Problem 13.149 A bead slides along a wire that rotates in the x–y plane with constant angular velocity ω 0 . The radial component of the bead’s acceleration is zero. The radial component of its velocity is v 0 when r = r0 . Determine the polar components of the bead’s velocity as a function of r. Strategy: The radial component of the bead’s velocity is dr , vr = dt and the radial component of its acceleration is ar =

dv r d 2r dθ 2 −r = − rω 0 2 . 2 dt dt dt

( )

By using the chain rule, dv r dv r dr dv r = = v , dt dr dt dr r you can express the radial component of the acceleration in the form a r =

Solution:

From the strategy:

ar = 0 = v r

dv r − ω 02r . dr

Separate variables and integrate: v r dv r = ω 02rdr, from which v r2 r2 = ω 02 + C. 2 2 At r = r0 , v r = v 0 , from which C =

v 02 − ω 02r02 , and 2

vr =

v 02 + ω 02 (r 2 − r02 ).

The transverse component is vθ = r

( ddtθ ) = rωθ, from which

v =

v 02 + ω 02 (r 2 − r02 )e r + r ω 0 e θ .

dv r v − rω 0 2 . dr r

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Problem 13.150 If the motion of a point in the x–y plane is such that its transverse component of acceleration a θ is zero, show that the product of its radial position and its transverse velocity is constant: rv θ = constant.

Solution:

We are given that a θ = rα + 2v r ω = 0. Multiply the entire relationship by r. We get

( ( ddtω ) + 2r ( drdt ) ω ) = dtd (r ω).

0 = (r 2α + 2rv r ω ) = r 2

d 2 (r ω ) = 0, then r 2 ω = constant. Now note that v θ = r ω. dt We have r 2 ω = r (r ω ) = rv θ = constant. This was what we needed to prove. Note that if

Problem 13.151* From astronomical data, Johannes Kepler deduced that the line from the sun to a planet traces out equal areas in equal times (Fig. a). Show that this result follows from the fact that the transverse component a θ of the planet’s acceleration is zero. [When r changes by an amount dr and θ changes by an amount dθ (Fig. b), the resulting differential element of area is dA = 12 r (r d θ).]

Solution:

From the solution to Problem 13.150, a θ = 0 implies that

dθ = constant. dt The element of area is 1 dA = r (rd θ), 2 r 2ω = r 2

(

)

dA 1 dθ 1 = r r = r 2 ω = constant. dt 2 dt 2 dA Thus, if = constant, then equal areas are swept out in equal times. dt

t2 t2 1 Dt

t1 1 Dt

t2 + ∆t

t1 + ∆t

A t1

(a)

y y

r 1 dr du

r u

r 1 dr

dA du x

(b)

r u

x (b)

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Problem 13.152 The bar rotates in the x–y plane with constant angular velocity ω 0 = 12 rad/s. The radial component of acceleration of the collar C (in m/s 2 ) is given as a function of the radial position in meters by a r = −8r . When r = 1 m, the radial component of velocity of C is v r = 2 m/s. Determine the velocity of C in terms of polar coordinates when r = 1.5 m. Strategy: Use the chain rule to write the first term in the radial component of the acceleration as dv r dv r dr dv r d 2r = = = v . 2 dt dt dr dt dr r

r x

Solution:

We have

d 2r dθ 2 −r = −(8 rad/s 2 ) r , ar = dt 2 dt   dθ 2 d 2r =    − (8 rad/s 2 )  r = ([12 rad/s] 2 − [8 rad/s 2 ])r = ([136 rad/s 2 )r    dt  dt 2

( )

Using the supplied strategy we can solve for the radial velocity dv d 2r = v r r = (136 rad/s 2 )r dt 2 dr vr

1.5 m

∫2 m/s v r dv r = (136 rad/s 2 ) ∫1 m r dr v r2 (2 m/s) 2 [1.5 m] 2 [1 m] 2 − = (136 rad/s 2 ) − 2 2 2 2

(

)

Solving we find v r = 13.2 m/s. dθ = (1.5 m)(12 rad/s) = 18 m/s. dt Thus v = (13.2e r + 18e θ ) m/s. We also have v θ = r

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Problem 13.153 The hydraulic actuator moves the pin P upward with velocity v = 2 j (m/s). Determine the velocity of the pin in terms of polar coordinates and the angular velocity of the slotted bar when θ = 35 °.

Solution: v P = 2 j (m/s) r = re r v = r e r + rθ e θ Also, r = 2 i + yj (m)

v = y j (m/s) = 2 j (m/s) y r = y sin θ tan θ = x rθ = y cos θ

r =

x2 + y2

θ = 35 °, Solving, we get y = 1.40 m, u

r = 1.15 m/s, r = 2.44 m,

rad θ = 0.671 s Hence V = r e r + rθ e θ V = 1.15e r + 1.64 e θ (m/s)

u . y

Problem 13.154 The hydraulic actuator moves the pin P upward with constant velocity v = 2 j (m/s). Determine the acceleration of the pin in terms of polar coordinates and the angular acceleration of the slotted bar when θ = 35 °.

Solution:

From Problem 13.153

V = 2 j m/s, constant dv ≡ 0 dt θ = ω y = 2 m/s

a =

θ = ω = 0.671 rad/s θ = 35 °

y = 0 θ = α y y tan θ = = x 2 y = 2 tan θ y = 2 sec 2 θθ

y = 2(2sec θ)(sec θ tan θ)θ 2 + 2 sec 2 θθ ..

θ =

θ = 0.631 rad/s 2

[−2sec θ tan θ](θ ) 2 sec θ

x y

u x 5 2m

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Problem 13.155 In Example 13.16, determine the velocity of the cam follower when θ = 135 ° (a) in terms of polar coordinates and (b) in terms of cartesian coordinates.

Solution: (a) θ = 135 °, ω = d θ /dt = 4 rad /s, and α = 0. r = 0.15(1 + 0.5cos θ) −1 = 0.232 m. dr dθ = 0.075 sin θ(1 + 0.5cos θ) −2 dt dt = 0.508 m/s. dr dθ e + r eθ dt r dt = 0.508e r + 0.928e θ (m/s).

v =

(b) v x = v r cos θ − v θ sin θ = −1.015 m/s. v y = v r sin θ + v θ cos θ = −0.297 m/s.

Problem 13.156* In Example 13.16, determine the acceleration of the cam follower when θ = 135° (a) in terms of polar coordinates and (b) in terms of cartesian coordinates.

Solution:

See the solution of Problem 13.155.

d 2r dθ 2 (a) cos θ(1 + 0.5cos θ) −2 = 0.075 2 dt dt dθ 2 sin 2 θ(1 + 0.5cos θ) −3 + 0.075 dt = 0.1905 m/s 2 .

( ) ( )

 d 2r dθ 2  dr d θ   d 2θ a =  2 −r e e + r 2 + 2  dt  dt dt  r dt dt  θ = −3.52e r + 4.06e θ (m/s 2 ).

( )

(b) a x = a r cos θ − a θ sin θ = −0.381 m/s 2 a y = a r sin θ + a θ cos θ = −5.362 m/s 2 .

Problem 13.157 In the cam–follower mechanism, the slotted bar rotates with constant angular velocity ω = 10 rad/s and the radial position of the follower A is determined by the profile of the stationary cam. The path of the follower is described by the polar equation

Solution: (a) θ = 30 °, ω = d θ /dt = 10 rad/s, and α = 0. r = 1 + 0.5cos 2θ = 1.25 ft dr dθ = − sin 2θ. dt dt = −8.66 ft/s.

r = 1 + 0.5cos 2θ ft. Determine the velocity of the cam follower when θ = 30 ° (a) in terms of polar coordinates and (b) in terms of cartesian coordinates. y

dr dθ e + r eθ dt r dt = −8.66e r + 12.5e θ (ft/s).

v =

(b) v x = v r cos θ − v θ sin θ = −13.75 ft/s, v y = v r sin θ + v θ cos θ = 6.50 ft/s. A

r u

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Problem 13.158* In Problem 13.157, determine the acceleration of the cam follower when θ = 30 ° (a) in terms of polar coordinates and (b) in terms of cartesian coordinates. y

Solution: (a)

d 2r

See the solution of Problem 13.157.

= −2θ 2 cos 2θ

dt 2

= −100 ft/s 2  d 2r dθ 2  dr d θ   d 2θ a =  2 −r e . e + r 2 + 2  dt  dt dt  r dt dt  θ = −225e r − 173e θ (ft/s 2 ).

( )

(b) a x = a r cos θ − a θ sin θ = −108 ft/s 2 , a y = a r sin θ + a θ cos θ

r u

= −263 ft/s 2 . x

Problem 13.159* The cartesian coordinates of a point P in the x–y plane are related to the polar coordinates of the point by the equations x = r cosθ and y = r sin θ. (a) Show that the unit vectors i and j are related to the unit vectors e r and e θ by i = cos θ e r − sin θ e θ and

Solution: (a) F rom geometry (see Figure), the radial unit vector is e r = i cos θ + j sin θ, and since the transverse unit vector is at right angles: π π e θ = i cos θ + + j sin θ + = −i sin θ + j cos θ. 2 2 Solve for i by multiplying e r by cos θ, e θ by sin θ, and subtracting the resulting equations:

(

)

(

)

i = e r cos θ − e θ sin θ.

j = sin θ e r + cos θ e θ . (b) Beginning with the expression for the position vector of P in terms of cartesian coordinates, r = xi + yj, derive Eq. (13.52) for the position vector in terms of polar coordinates. (c) By taking the time derivative of the position vector of point P expressed in terms of cartesian coordinates, derive Eq. (13.55) for the velocity in terms of polar coordinates.

Solve for j by multiplying e r by sin θ, and e θ by cos θ, and the results: j = e r sin θ + e θ cos θ. (b) T he position vector is r = xi + yj = (r cos θ) i + (r sin θ) j = r ( i cos θ + j sin θ). Use the results of Part (a) expressing i, j in terms of e r , e θ : r = r (e r cos 2 θ − e θ cos θ sin θ + e r sin 2 θ + e θ sin θ cos θ) = re r . (c) The time derivatives are:

(

dr dr dθ cos θ − r sin θ = v = i dt dt dt dr dθ sin θ + r cos θ , +j dt dt

(

)

from which eu

v = er

r u

Substitute the results of Part (a)

v =

dr dθ dr e + r eθ = e + r ωe θ . dt r dt dt r

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dr dθ ( i cos θ + j sin θ) + r (−i sin θ + j cos θ). dt dt

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Problem 13.160 The airplane flies in a straight line at 400 mi/h. The radius of its propeller is 5 ft, and the propeller turns at 2000 rpm in the counterclockwise direction when seen from the front of the airplane. Determine the velocity and acceleration of a point on the tip of the propeller in terms of cylindrical coordinates. (Let the z-axis be oriented as shown in the figure.)

Solution: v = 400

The speed is

1h ( mile )( 3600 )( 5280 ft ) = 586.7 ft/s h s 1 mile

The angular velocity is ω = 2000

rad 1 min ( 21πrev )( 60 s ) = 209.4 rad/s.

The radial velocity at the propeller tip is zero. The transverse velocity is v θ = ωr = 1047.2 ft/s. The velocity vector in cylindrical coordinates is v = 1047.2e θ + 586.7e z (ft/s).

5 ft

The radial acceleration is a r = −r ω 2 = −5(209.4) 2 = −219324.5 ft/s 2 . The transverse acceleration is

( )( )

d 2θ dr d θ + 2 = 0, dt 2 dt dt since the propeller rotates at a constant angular velocity. The acceleration a z = 0, since the airplane travels at constant speed. Thus aθ = r

a = −219324.5e r (ft/s 2 ).

Problem 13.161 A charged particle P in a magnetic field moves along the spiral path described by r = 1 m, θ = 2 z rad, where z is in meters. The particle moves along the path in the direction shown with constant speed v = 1 km/s. What is the velocity of the particle in terms of cylindrical coordinates?

Solution:

The radial velocity is zero, since the path has a constant radius. The magnitude of the velocity is v =

( ddtθ ) + ( dzdt ) = 1000 m/s.

The angular velocity is

y Substitute:

v =

dθ dz = 2 . dt dt 2

( ddtθ ) + 14 ( ddtθ )

( ddtθ ) r + 14 = 1.25, 2

dθ 1000 = = 894.4 rad/s, dt 1.25 from which the transverse velocity is

from which P z 1 km/s

vθ = r

( ddtθ ) = 894.4 m/s.

The velocity along the cylindrical axis is

( )

dz 1 dθ = = 447.2 m/s. dt 2 dt The velocity vector: v = 894.4 e θ + 447.2e z .

y x

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Problem 13.162 At t = 0, two projectiles A and B are simultaneously launched from O with the initial velocities and elevation angles shown. Determine the velocity of projectile A relative to projectile B (a) at t = 0.5 s and (b) at t = 1 s.

Solution: v A = −(9.81 m/s 2 j)t + (10 m/s)(cos60 °i + sin 60 ° j) v B = −(9.81 m/s 2 j)t + (10 m/s)(cos30 °i + sin 30 ° j ) v A /B = v A − v B = (10 m/s)(−0.366 i + 0.366 j) v A /B = (−3.66 i + 3.66 j) m/s

Since v A /B doesn’t depend on time, the answer is the same for both times v A /B = (−3.66 i + 3.66 j) m/s. 10 m/s A B

10 m/s

608 308

Problem 13.163 Relative to the Earth-fixed coordinate system, the disk rotates about the fixed point O at 10 rad/s. What is the velocity of point A relative to point B at the instant shown?

Solution: v A = −(10 rad/s)(2 ft) i = −(20 ft/s) i v B = (10 rad/s)(2 ft) j = (20 ft/s) j v A /B = v A − v B = (−20 i − 20 j) ft/s.

y A

10 rad/s

B O

2 ft

Problem 13.164 Relative to the Earth-fixed coordinate system, the disk rotates about the fixed point O with a constant angular velocity of 10 rad/s. What is the acceleration of point A relative to point B at the instant shown?

Solution: a A = −(10 rad/s) 2 (2 ft) j = −(200 ft/s 2 ) j a B = −(10 rad/s) 2 (2 ft) i = −(200 ft/s 2 ) i a A /B = a A − a B = (200 i − 200 j) ft/s 2 .

y A

10 rad/s

B 2 ft

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Problem 13.165 The train on the circular track is traveling at 50 ft/s. The train on the straight track is traveling at 20 ft/s. In terms of the Earth-fixed coordinate system shown, what is the velocity of passenger A relative to passenger B?

Solution: v A = (−20 j) ft/s, v B = (50 j) ft/s

v A /B = v A − v B = (−70 j) ft/s.

50 ft/s

Problem 13.166 The train on the circular track is traveling at a constant speed of 50 ft/s. The train on the straight track is traveling at 20 ft/s and is increasing its speed at 2 ft/s 2 . In terms of the Earth-fixed coordinate system shown, what is the acceleration of passenger A relative to passenger B?

20 ft/s

Solution: a A = (−2 j) ft/s 2 , a B = −

(50 ft/s) 2 i = (−5i) ft/s 2 500 ft

a A /B = a A − a B = (5i − 2 j) ft/s 2 .

50 ft/s

20 ft/s

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Problem 13.167 In Practice Example 13.17, suppose that the velocity of the uniform current is 4 knots flowing east. (a) If the helmsman wants to travel northwest relative to the Earth, how many degrees west of north must he point the ship? What is the resulting magnitude of the ship’s velocity relative to the Earth? (b) Draw a graph of the number of degrees west of north the helmsman should point the ship as a function of the ship’s speed relative to the water for a range of ship speeds from 5 to 10 knots.

Solution: (a) Let A denote the ship and B the current: y

vA/B

vA 458

Denote the magnitudes of the velocity vectors of the ship, the current, and the ship relative to the current by v A = a, v B = b = 4 knots, and v A /B = c = 5 knots. From the triangle we see that b + a cos 45 ° = c sin β , (1) a sin 45 ° = c cos β , (2) where β is the angle west of north to the direction of the ship relative to the water. Squaring Eqs. (1) and (2) and summing them results in the equation a 2 + b 2 + 2ab cos 45 ° = c 2 . (3) Notice that this equation can also be obtained through the law of cosines (Appendix A). We can solve this quadratic equation for a, obtaining a = v A = 1.29 knots. Then from Eq. (2), the angle β = 79.4 °. (b) S olving Eq. (3) for the range of values of the ship speed c and then obtaining the angle β from Eq. (2), we obtain the desired graph: 80 78 76

b, Degrees

74 72 70 68 66 64 62 60

5.5

6.5

7.5

8.5

9.5

Ship Speed, knots (a) 79.4 °, 1.29 knots.

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Problem 13.168 A private pilot wishes to fly from a city P to a city Q that is 200 km directly north of city P. The airplane will fly with an airspeed of 290 km/h. At the altitude at which the airplane will be flying, there is an east wind (that is, the wind’s direction is west) with a speed of 50 km/h. What direction should the pilot point the airplane to fly directly from city P to city Q? How long will the trip take?

Solution:

Let A denote the plane and B the wind: y

vA/B

N E

200 km

50 km/h

The wind velocity is v B = 50 km/h and the plane’s airspeed is v A /B = 290 km/h. From the triangle we see that  v  β = arcsin  B  = 9.93 °,  v A /B  v A = v A /B cos β = 286 km/h, where β is the angle east of north to the direction of the plane relative to the air. The time required for the trip is 200 km = 0.700 hr (42.0 minutes). vA

9.93 ° east of north, 42 minutes.

Problem 13.169 The river flows north at 3 m/s (Assume that the current is uniform.) If you want to travel in a straight line from point C to point D in a boat that moves at a constant speed of 10 m/s relative to the water, in what direction should you point the boat? How long does the crossing take?

Solution:

Let A denote the boat and B the current:

y D vA/B b

3 m/s

vA 458

Denote the magnitudes of the velocity vectors of the boat, the current, and the boat relative to the current by N

500 m

v A = a, v B = b = 3 m/s, and v A /B = c = 10 m/s. E

W S

From the triangle we see that a sin 45 ° − b = c sin β , (1) a cos 45 ° = c cos β . (2) Squaring Eqs. (1) and (2) and summing them results in the equation

C 500 m

a 2 − 2ab sin 45 ° + b 2 = c 2 . We can solve this quadratic a = v A = 11.9 m/s. Then from

equation

for

obtaining

Eq. (2), the angle β = 32.8 °. The distance from C to D is (500 m) 2 + (500 m) 2 = 707 m, so the time for the crossing is 707 m t = = 59.5 s. vA 32.8 ° north of east, 59.5 s.

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Problem 13.170 The river flows north at 3 m/s (Assume that the current is uniform.) (a) What minimum speed must a boat have relative to the water to travel in a straight line from point C to point D? How long does it take to make the crossing? (b) Traveling at the same speed, can the boat make a return trip from point D to point C?

Solution: (a) L et A denote the boat and B the current. If we consider the vector diagram showing the relationships of the velocities,

y D vA/B

3 m/s D

vB 458

500 m

C E

W S C

It is clear that the magnitude of the boat’s speed relative to the water, v A /B , is a minimum when v A /B is perpendicular to the line from C to D:

500 m

y D

vA/B vB

458

vA C

From this diagram we can see that v A 2 + v A /B 2 = v B 2 = (3 m/s) 2 and also that v A = v A /B , so we obtain v A = v A /B = = 2.12 m/s. The distance From C to D is

1 vB 2

(500 m) 2 + (500 m) 2 = 707 m, so the time for the crossing is 707 m t = = 333 s. vA (b) C learly the boat can’t make the return trip moving at the same speed, it isn’t traveling as fast as the current. (a) 2.12 m/s, 333 s. (b) No.

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Problem 13.171* Relative to the Earth, the sailboat sails north with speed v 0 = 6 knots (nautical miles per hour) and then sails east at the same speed. The telltale indicates the direction of the wind relative to the boat. Determine the direction and magnitude of the wind’s velocity (in knots) relative to the Earth.

v0 608

v0 Telltale

N E

W S

Solution:

Using either position one or position two we have

v wind /ground = v wind /boat + v boat /ground

v wind /ground = (2.536 i + 6 j) knots,

In position one we have

v wind /ground =

v wind /ground = v wind /boat1i + (6 knots) j

direction = tan −1

In position two we have

(2.536) 2 + (6) 2 knots = 6.51 knots,

( 2.536 ) = 22.91° east of north. 6

v wind /ground = v wind /boat 2 (− cos60 °i + sin 60 ° j) + (6 knots) i Since the wind has not changed these two expressions must be the same. Therefore v wind /boat1 = −v wind /boat 2 cos60 ° + 6 knots    6 knots = v wind /boat 2 sin 60°   v wind /boat 1 = 2.536 knots ⇒   v wind /boat 2 = 6.928 knots 

Problem 13.172 Suppose that you throw a ball straight up at 10 m/s and release it at 2 m above the ground. (a) What maximum height above the ground does the ball reach? (b) How long after you release it does the ball hit the ground? (c) What is the magnitude of its velocity just before it hits the ground?

Solution:

The equations of motion for the ball are

a y = −g = −9.81 m/s 2 , v y = v y 0 − gt = 10 − 9.81t (m/s), and y = y 0 + v y 0 t − gt 2 /2 = 2 + 10t − 9.81t 2 /2 (m). (a) T he maximum height occurs when the velocity is zero. Call this time t = t 1 . It can be obtained by setting velocity to zero, i.e., v y = 0 = 10 − 9.81t 1 (m/s). Solving, we get t 1 = 1.02 s. Substituting this time into the y equation, we get a maximum height of y MAX = 7.10 m. (b) The ball hits the ground when y = 0 m. To find out when this occurs, we set y = 0 m into the y equation and solve for the time(s) when this occurs. There will be two times, one positive and one negative. Only the positive time has meaning for us. Let this time be t = t 2 . The equation for t 2 is y = 0 = 2 + 10t 2 − 9.81t 2 2 /2 (m). Solving, we get t 2 = 2.22 s. (c) The velocity at impact is determined by substituting t 2 = 2.22 s into the equation for v y . Doing this, we find that at impact, v y = −11.8 m/s

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Problem 13.173 The Porsche starts from rest at time t = 0. At time t = 10 s, its velocity is 70 mi/h. If you assume that its acceleration is constant, what distance does it travel in 10 s (in ft)?

Solution: Let a 0 denote the car’s constant acceleration. We can integrate to determine its velocity and position as functions of time: dv = a0, dt v

∫0 dv = ∫0 a 0 dt, (1) ds = a 0t. dt

v = s

∫0 ds = ∫0 a 0t dt, s =

1 a t 2. 2 0

(2)

At time t = 10 s, the car’s velocity in ft/s was ft 1 hr ( 70 miles )( 5280 )( 3600 ) = 103 ft/s. hr mile s From Eq. (1), its acceleration was 103 ft/s 10 s = 10.3 ft/s 2 .

a0 =

Then from Eq. (2) the distance traveled was 1 (10.3 ft/s 2 )(10 s) 2 2 = 513 ft.

s =

s = 513 ft.

Problem 13.174 The Porsche starts from rest at time t = 0. Its acceleration can be approximated by a function of the form a = b − ct 2 , where b and c are constants. Measurements indicate that at t = 5 s, the car has traveled 47.5 m from the point where it started, and at t = 10 s, the car has traveled 175 m from the point where it started. At t = 10 s, what is the car’s velocity in m/s and in mi/h?

Solution:

We can integrate the given expression for the acceleration to determine the velocity and position as functions of time and the constants b and c: a = v

dv = b − ct 2 , dt t

∫0 dv = ∫0 (b − ct 2 ) dt, (1) v =

ds 1 = bt − ct 3 , dt 3

∫0 ds = ∫0 ( bt − 3 ct 3 ) dt, s

(2) 1 2 1 4 ct . bt − 2 12 Using the measurements made at t = 5 s and at t = 10 s, Eq. (2) yields two equations: s =

1 1 b(5s) 2 − c(5 s) 4 , 2 12 1 1 175 m = b(10 s) 2 − c(10 s) 4 . 2 12

47.5 m =

Solving, we obtain b = 3.9, c = 0.024. Using these values, at t = 10 s Eq. (1) gives the velocity v = 3.9(10 s) −

1 (0.024)(10 s) 3 3

= 31.0 m/s. In miles/hr this is 0.6214 miles v = 31.0 m/s 1000 m = 69.3 miles/hr.

(

s )( 3600 ) 1 hr

31.0 m/s, 69.3 miles/hr.

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Problem 13.175 The angle θ = 4t + t 2 rad, where t is time in seconds. Determine the angular velocity ω of the line L relative to the line L 0 at time t = 2 s, in rad/s and in revolutions per minute (rpm).

Solution:

The angular velocity as a function of time is

dθ dt = 4 + 2t rad/s.

ω =

At t = 2 s, it is

ω = 4 + 2(2 s) = 8 rad/s = 8 rad/s

60 s ( 1 revolution )( 1 minute ) 2π rad

= 76.4 rpm.

u L0

Problem 13.176 The angular acceleration of the line L relative to the line L 0 is α = 2 rad/s 2 . At t = 1 s, the angle θ = 5 rad, and at t = 3 s, the angle θ = 21 rad. Determine the angular velocity ω at time t = 2 s, in rad/s and in revolutions per minute (rpm). L

8 rad/s, 76.4 rpm.

Solution: We can integrate the angular acceleration to determine the angular velocity and angle as functions of time: dω = 2 rad/s 2 , dt dθ ω = = 2t + b rad/s, dt θ = t 2 + bt + c rad, α =

where b and c are integration constants. From the given values of θ at t = 1 s and t = 3 s, we obtain the two equations 5 rad = (1 s) 2 + b(1 s) + c rad, 21 rad = (3 s) 2 + b(3 s) + c rad.

u L0

Solving, we obtain b = 4, c = 0. The angular velocity as a function of time is ω = 2t + 4 rad/s. At t = 2 s, it is ω = 2(2 s) + 4 = 8 rad/s = 8 rad/s

60 s ( 1 revolution )( 1 minute ) 2π rad

= 76.4 rpm.

8 rad/s, 76.4 rpm.

Problem 13.177 The angular acceleration of the line L relative to the line L 0 is given as a function of the angle θ in radians by α = 2θ rad/s 2 . When θ = 2 rad, the angular velocity ω = 6 rad/s. Determine the angular velocity when θ = 4 rad, in rad/s and in degrees per second. Strategy: Express the angular acceleration as d ω dθ dω dω α = = = ω. dt d θ dt dθ

Solution: As in the case of straight-line motion, when the acceleration depends on the displacement the chain rule can be used to express the acceleration in terms of the displacement instead of the time: α =

dω d ω dθ dω = = ω. dt d θ dt dθ

Using this expression, we can separate variables and integrate, dω ω = 2θ, dθ ω d ω = 2θ d θ, ω

4 rad

∫6 rad/s ω d ω = ∫2 rad 2θ d θ, ω

 1ω2  = [θ 2 ] 42 rad rad ,  2  6 rad/s 1 2 1 ω − (6 rad/s) 2 = (4 rad) 2 − (2 rad) 2 , 2 2

obtaining ω = 7.75 rad/s (444 deg/s). u L0

ω = 7.75 rad/s (444 deg/s).

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Problem 13.178 Water leaves the nozzle at 20 ° above the horizontal and strikes the wall at the point indicated. What is the velocity of the water as it leaves the nozzle? Strategy: Determine the motion of the water by treating each particle of water as a projectile.

Solution: Denote θ = 20°. The path is obtained by integrating the equations: dv y dv = −g and x = 0, from which dt dt dy dx = −gt + Vn sin θ, = Vn cos θ. dt dt g y = − t 2 + (Vn sin θ)t + y 0 . 2 x = (Vn cos θ)t + x 0 . Choose the origin at the nozzle so that y 0 = 0, and x 0 = 0. When the stream is y (t impact ) = (20 − 12) = 8 ft, the time is g 0 = − (t impact ) 2 + (Vn sin θ)t impact − 8. 2 At this same time the horizontal distance is

208

20 ft

x (t impact ) = 35 = (Vn cos θ)t impact , from which t impact = Substitute:

12 ft

2 g  35  0 = −   + 35 tan θ − 8, 2  Vn cos θ 

35 ft

from which Vn =

Problem 13.179 In practice, the quarterback throws the football with velocity v 0 at 45 ° above the horizontal. At the same instant, the receiver standing 20 ft in front of him starts running straight downfield at 10 ft/s and catches the ball. Assume that the ball is thrown and caught at the same height above the ground. What is the velocity v 0? v0 10 ft/s

458

35 . Vn cos θ

g 35 = 68.62 ft/s. ( cos ) θ 2(35tan θ − 8)

Substitute: 2v 0 sin θ 20 = , from which (v 0 cos θ − 10) g 400 g = 2v 0 sin θ(v 0 cos θ − 10) 2 . The function f (v 0 ) = 2v 0 sin θ(v 0 cos θ − 10) 2 − 400 g was graphed to find the zero crossing, and the result refined by iteration: v 0 = 36.48 ft/s . Check: The time of flight is t = 1.27 s and the distance down field that the quarterback throws the ball is d = 12.7 + 20 = 32.7 ft = 10.6 y ds, which seem reasonable for a short, “lob” pass. check.

v0 20 ft

458

10 ft/s

Solution:

Denote θ = 45 °. The path is determined by integrating

the equations; d 2y d 2x = −g, = 0, from which dt 2 dt 2 dy dx = −gt + v 0 sin θ, = v 0 cos θ. dt dt g y = − t 2 + (v 0 sin θ)t, 2 x = (v 0 cos θ)t, where the origin is taken at the point where the ball leaves the quarterback’s hand. When the ball reaches the receiver’s hands, 2v 0 sin θ y = 0, from which t fight = . g At this time the distance down field is the distance to the receiver: x = 10t flight + 20. But also x = (v 0 cos θ)t flight , from which t flight =

20 ft Zero crossing 100 80 60 40 20 f(v) 0 –20 –40 –60 –80 –100 35 35.5 36 36.5 37 37.5 38 Velocity, ft /s

20 . (v 0 cos θ − 10)

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Problem 13.180 The constant velocity v = 2 m/s. What are the magnitudes of the velocity and acceleration of point P when x = 0.25 m?

Solution:

Let x = 2t m/s. Then x = 0.25 m at t = 0.125 s. We know that v x = 2 m/s and a x = 0.

From y = 0.2sin(2πt ), we obtain

dy = 0.4 π cos(2πt ) and dt d 2y = −0.8π 2 sin(2πt ). dt 2 At t = 0.125 s,

y 5 0.2 sin px P x

y = 0.141 m and

dy = v y = 0.889 m/s and dt 2 d y = a y = −5.58 m/s. dt 2

Therefore

Problem 13.181 The constant velocity v = 2 m/s. What is the acceleration of point P in terms of normal and tangential components when x = 0.25 m? y

v =

v x2 + v y2 = 2.19 m/s,

a =

a x2 + a y2 = 5.58 m/s 2 .

Solution:

See the solution of Problem 13.180. The angle θ between the x axis and the path is vy  0.889 = 24.0 °. Then θ = arctan   = arctan vx  2

(

)

a t = a x cos θ + a y sin θ = 0 + (−5.58)sin 24.0 ° = −2.27 m/s 2 , y 5 0.2 sin px

a N = a x sin θ − a y cos θ = 0 − (−5.58) cos 24.0 ° = 5.10 m/s 2 . The instantaneous radius is

P x

ρ =

v x2 + v y2 (2) 2 + (0.889) 2 = = 0.939 m. aN 5.10

v ay

at u

Problem 13.182 The constant velocity v = 2 m/s. What is the acceleration of point P in terms of polar coordinates when x = 0.25 m?

Solution:

See the solution of Problem 13.192. The polar angle θ is

( )

(

)

y 0.141 θ = arctan = arctan = 29.5 °. Then 0.25 x a r = a x cos θ + a y sin θ = 0 + (−5.58)sin 29.5 ° = −2.75 m/s 2 , a θ = −a x sin θ + a y cos θ = 0 + (−5.58) cos 29.5 ° = −4.86 m/s 2 .

y y 5 0.2 sin px

ay au

x u

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Problem 13.183 A point P moves along the spiral path r = (0.1)θ ft, where θ is in radians. The angular position θ = 2t rad, where t is in seconds, and r = 0 at t = 0. Determine the magnitudes of the velocity and acceleration of P at t = 1 s. P

Solution:

The path r = 0.2t ft, θ = 2t rad. The velocity components are

dr dθ vr = = 0.2 ft/s, v θ = r = (0.2t )2 = 0.4t. dt dt At t = 1 seconds the magnitude of the velocity is v r2 + v θ2 =

v =

0.2 2 + 0.4 2 = 0.447 ft/s.

The acceleration components are: d 2r dθ 2 −r = −(0.2t )(2 2 ) ft/s 2 , dt 2 dt d 2θ dr d θ aθ = r + 2 = 2(0.2)(2) = 0.8 ft/s 2 . dt 2 dt dt ar =

r u

( ) ( ) ( )( )

The magnitude of the acceleration is a =

Problem 13.184 In the cam–follower mechanism, the slotted bar rotates with constant angular velocity ω = 12 rad/s, and the radial position of the follower A is determined by the profile of the stationary cam. The slotted bar is pinned a distance h = 0.2 m to the left of the center of the circular cam. The follower moves in a circular path 0.42 m in radius. Determine the velocity of the follower when θ = 40 ° (a) in terms of polar coordinates and (b) in terms of cartesian coordinates.

Solution: (a) T he first step is to get an equation for the path of the follower in terms of the angle θ. This can be most easily done by referring to the diagram at the right. Using the law of cosines, we can write R 2 = h 2 + r 2 − 2hr cos θ. This can be rewritten as r 2 − 2hr cos θ + (h 2 − R 2 ) = 0. We need to find the components of the velocity. These are v r = r and v θ = rθ . We can differentiate Carrying out this the relation derived from the law of cosines to get r. differentiation, we get 2rr − 2hr cos θ + 2hrθ sin θ = 0. Solving we get for r, r =

r u

a r2 + a θ2 = 1.13 ft/s 2 .

hrθ sin θ . (h cos θ − r )

Recalling that ω = θ and substituting in the numerical values, i.e., R = 0.42 m, h = 0.2 m, ω = 12 rad/s, and θ = 40 °, we get r = 0.553 m, v r = −2.13 m/s, and v θ = 6.64 m/s (b) The transformation to cartesian coordinates can be derived from e r = cos θ i + sin θ j, and e θ = − sin θ i + cos θ j. Substituting these into v = v r e r + v θ e θ , we get v = (v r cos θ − v θ sin θ) i + (v r sin θ + v θ cos θ) j. Substituting in the numbers, v = −5.90 i + 3.71 j (m/s).

P r

u O

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Problem 13.185* In Problem 13.184, determine the acceleration of the follower when θ = 40 ° (a) in terms of polar coordinates and (b) in terms of cartesian coordinates. y

r u

Solution: (a) I nformation from the solution to Problem 13.184 will be used in this solution. In order to determine the components of the acceleration in polar coordinates, we need to be able to determine all of the variables in the right hand sides of a r = r − rθ 2 and that a θ = rθ + 2 r θ . We already know everything except r and θ. Since ω is constant, θ = ω = 0. We need only to find the value for r and the value for r at θ = 40 °. Substituting into the original equation for r, we find that r = 0.553 m at this position on the cam. we start with r = v r . Taking a derivative, we start with To find r, rr − hr cos θ + hrθ sin θ = 0 from Problem 13.184 (we divided through by 2). Taking a derivative with respect to time, we get r 2 + 2hrθ sin θ + hrθ 2 cos θ + hrθ sin θ r = . (h cos θ − r )

Evaluating, we get r = −46.17 m/s 2 . Substituting this into the equation for a r and evaluating a n , we get a r = −125.81 m/s 2 and a θ = −51.2 m/s 2 . (b) The transformation of cartesian coordinates can be derived from e r = cos θ i + sin θ j, and e θ = sin θ i + cos θ j. Substituting these into a = a r e r + a e e θ , we get a = a r e r + a θ e θ , we get a = (a r cos θ − a θ sin θ) i + (a r sin θ + a θ cos θ) j. Substituting in the numbers, we get a = −63.46 i − 120.1 j (m/s 2 ).

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Chapter 14 Problem 14.1 A person weighs 150 lb. Case (a) shows him standing on a trapdoor. Case (b) shows him at the instant the trapdoor springs open. What does Newton’s second law tell you in each case?

Solution:

The free-body diagrams are

y W

W x

N (a)

(b)

In Case (a), the trapdoor exerts a force N on him. We know that his acceleration is zero, so Newton’s second law is ΣFx = W − N = ma x = 0. Newton’s second law simply states that the sum of the forces acting on him is zero. The normal force N = W = 150 lb. In Case (b), the only force acting on him is his weight, and Newton’s second law is ΣFx = W = ma x . His weight W = mg, so his acceleration is a x = g = 32.2 ft/s 2 . (a)

(b)

Source: Courtesy of Aila Images/Shutterstock.

(a) His acceleration is zero. The trapdoor exerts an upward 150-lb force on him. (b) He accelerates downward at 32.2 ft/s 2 .

Problem 14.2 The mass of the Sikorsky UH-60A helicopter is 9300 kg. It takes off vertically with its rotor exerting a constant upward thrust of 112 kN. (a) How fast is the helicopter rising 3 s after it takes off? (b) How high has it risen 3 s after it takes off? Strategy: Be sure to draw the free-body diagram of the helicopter.

Solution:

The equation of motion is ΣF : 112 kN − 9.3(9.81) kN = (9,300 kg)a Solving, we find that a = 2.23 m/s 2 . Using kinematics we can answer the questions a = 2.23 m/s 2 , v = at = (2.23 m/s 2 )(3 s) = 6.70 m/s, 1 1 h = at 2 = (2.23 m/s 2 )(3 s) 2 = 10.0 m. 2 2 (a) 6.70 m/s, (b) 10.0 m.

Source: Courtesy of Zigmunds Dizgalvis/Shutterstock.

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Problem 14.3 The mass of the Sikorsky UH-60A helicopter is 9300 kg. It takes off vertically at t = 0. The pilot advances the throttle so that the upward thrust of its engine (in kN) is given as a function of time in seconds by T = 100 + 2t 2 . (a) How fast is the helicopter rising 3 s after it takes off? (b) How high has it risen 3 s after it takes off?

Solution:

The equation of motion is

ΣF : 100 kN + 2 kN

( st )

− 9.3(9.81) kN = (9,300 kg)a Solving, we find that a = (0.943 m/s 2 ) + (0.215 m/s 4 )t 2 . Using kinematics we can answer the questions a = (0.943 m/s 2 ) + (0.215 m/s 4 )t 2 . 1 υ = (0.943 m/s 2 )t + (0.215 m/s 4 )t 3 3 1 1 2 2 h = (0.943 m/s )t + (0.215 m/s 4 )t 4 2 12 Evaluating these expressions at t = 3 s, (a) v = 4.76 m/s, (b) d = 5.69 m.

Source: Courtesy of Zigmunds Dizgalvis/Shutterstock.

Problem 14.4 The 50-lb box is at rest at time t = 0. The horizontal surface is smooth. The magnitude of the force is given as a function of time in seconds by F = 30 + 6t lb. What is the box’s velocity at t = 2 s?

Applying Newton’s second law, ΣFx = F cos 20 ° = ma x , ΣFy = N − F sin 20 ° − W = 0. From the first equation, the acceleration is 1 F cos 20 ° m 1 = (30 + 6t lb) cos 20 °. m

ax =

F 208

We integrate to determine the velocity as a function of time,

Solution:

dv x = ax, dt

The mass of the box is

∫0 dv x = ∫0 a x dt,

W 50 lb m = = g 32.2 ft/s 2 = 1.55 slugs.

vx = ∫

2s 1 0

(30 + 6t lb) cos 20 ° dt,

obtaining v x = 43.6 ft/s.

The free-body diagram of the box is

43.6 ft/s.

F 208

W y

N x

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Problem 14.5 The 50-lb box is at rest at time t = 0. The coefficients of friction between the box and the horizontal surface are µ s = 0.2 and µ k = 0.18. The magnitude of the force is given as a function of time in seconds by F = 30 + 6t lb. What is the box’s velocity at t = 2 s? F 208

Applying Newton’s second law, ΣFx = F cos 20 ° − µ k N = ma x , ΣFy = N − F sin 20 ° − W = 0. Solving the second equation for N and substituting it into the first equation, we can solve the resulting equation for the acceleration as a function of time: 1 a x = [(cos 20 ° − µ k sin 20 °) F − µ kW ] m 1 = [(cos 20 ° − µ k sin 20 °)(30 + 6t lb) − µ kW ]. m We integrate to determine the velocity as a function of time, dv x = ax, dt

Solution: m =

∫0 dv x = ∫0 a x dt,

The mass of the box is

W 50 lb = g 32.2 ft/s 2

vx = ∫

2s 1 0

[(cos 20 ° − µ k sin 20 °)(30 + 6t lb) − µ kW ] dt,

obtaining υ x = 29.1 ft/s.

= 1.55 slugs. The free-body diagram of the box is

29.1 ft/s.

F 208

W y

mkN

N x

Problem 14.6 The 50-lb box is at rest at time t = 0. The inclined surface is smooth. The magnitude of the force is given as a function of time in seconds by F = 30 + 6t lb. What is the box’s velocity at t = 2 s?

Applying Newton’s second law, ΣFx = F cos 20 ° − W sin 20 ° = ma x , ΣFy = N − F sin 20 ° − W cos 20 ° = 0. The acceleration is ax =

1 ( F cos 20 ° − W sin 20 °). m

Substituting the value of m and the given expression for F, we obtain the acceleration as a function of time: a x = d + et ft/s 2 ,

208

where d = 7.14 ft/s 2 and e = 3.63 ft/s 3 . We integrate to determine the velocity as a function of time.

Solution:

dv x = a x = d + et ft/s 2 , dt

The mass of the box is

W 50 lb m = = g 32.2 ft/s 2 = 1.55 slugs.

∫0 dv x = ∫0 (d + et ) dt v x = dt +

The free-body diagram of the box is

1 2 et ft/s 2 . 2

Substituting the values of d and e, at t = 2 s we obtain v x = 21.5 ft/s.

208

y W

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Problem 14.7 The coefficient of kinetic friction between the 14-kg box and the inclined surface is µ k = 0.1. The velocity of the box is zero when it is subjected to a constant horizontal force F = 20 N. What is the velocity of the box 2 s later?

Solution:

From the governing equations and the slip equation, we can find the acceleration of the box (assumed to be down the incline).

ΣF : 14(9.81) N sin 20 ° − f − (20 N) cos 20 ° = (14 kg)a, ΣF : N − 14(9.81) N cos 20 ° − (20 N)sin 20 ° = 0. Slip: f = (0.1) N .

Solving, we have a = 1.04 m/s 2 , N = 136 N, f = 13.6 N. 208

Since a > 0, our assumption is correct. Using kinematics, v = at = (1.04 m/s 2 )(2 s) = 2.08 m/s. v = 2.08 m/s down the incline.

Problem 14.8 The 170-lb skier is schussing on a 25 ° slope. At the instant shown, he is moving at 40 ft/s. The kinetic coefficient of friction between his skis and the snow is µ k = 0.08. If he makes no attempt to check his speed, how long does it take for it to increase to 60 ft/s?

Solution:

The governing equations and the slip equation are used to find the acceleration

ΣF : N − (170 lb) cos 25 ° = 0, ΣF : (170 lb)sin 25 ° − f =

170 lb ( 32.2 )a. ft/s 2

Slip: f = (0.08) N . Solving yields a = 11.3 ft/s 2 , N = 154 lb, f = 12.3 lb. Using kinematics, we find 60 ft/s = 40 ft/s + (11.3 ft/s 2 )t Solving yields

t = 1.77 s.

Source: Courtesy of Blazar SLU/Shutterstock.

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Problem 14.9 A 250,000-kg Airbus 330 starts from rest at time t = 0 and accelerates down the runway. The net force on the aircraft in newtons is given as a function of time in seconds by F = 1000 + 66000t − 1200t 2 N. The airplane requires 36 s to take off. Find (a) its velocity at takeoff and (b) the distance required.

Solution:

Applying Newton’s second law, the airplane’s acceleration in terms of the airplane’s mass m is a =

1 1 F = (1000 + 66000t − 1200t 2 ) m/s 2 . m m

We can integrate to determine the plane’s velocity and position as functions of time: 1 dv = (1000 + 66000t − 1200t 2 ) m/s 2 , dt m v t 1 ∫0 dv = ∫0 m (1000 + 66000t − 1200t 2 ) dt, 1 ds = (1000t + 33000t 2 − 400t 3 ), v = dt m s t 1 ∫0 ds = ∫0 m (1000t + 33000t 2 − 400t 3 ) dt, 1 s = (500t 2 + 11000t 3 − 100t 4 ). m a =

At t = 36 s we obtain v = 96.6 m/s and s = 1380 m.

Source: Courtesy of Senohrabek/Shutterstock.

Problem 14.10 The total external force on the 10-kg object is constant and equal to ΣF = 80 i + 20 j − 60 k (N). At time t = 0, its velocity is v = 14 i + 16 j + 20 k (m/s). What is the straight-line distance from its position at t = 0 to its position at t = 4 s?

(a) 96.6 m/s. (b) 1380 m.

Solution:

The object’s acceleration is

80 i + 20 j − 60 k (N) ΣF = a = m 10 kg = 8i + 2 j − 6k (m/s 2 ). We can integrate to determine the components of velocity as functions of time: dv x = 8 m/s 2 , dt

∫14 m/s dv x = ∫0 8 dt, v x = 14 + 8t m/s, dv y = 2 m/s 2 , dt vy

∫16 m/s dv y = ∫0 2 dt,

v y = 16 + 2t m/s, dv z = −6 m/s 2 , dt

x z

∫20 m/s dv z = −∫0 6 dt, v z = 20 − 6t m/s. Defining its position at t = 0 to be x = 0, y = 0, z = 0, we integrate to determine the components of position as functions of time: dx = 14 + 8t m/s, dt x

∫0 dx = ∫0 (14 + 8t ) dt, x = 14 t + 4t 2 m/s, dy = 16 + 2t m/s, dt y

∫0 dy = ∫0 (16 + 2t ) dt, y = 16 t + t 2 m/s, dz = 20 − 6t m/s, dt z

∫0 dz = ∫0 (20 − 6t ) dt, z = 20 t − 3t 2 m/s. At t = 4 s we obtain x = 120 ft, y = 80 ft, z = 32 ft. The straightline distance between the two positions is x 2 + y 2 + z 2 = 148 ft. 148 ft.

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Problem 14.11 The total external force on the 10-kg object shown in Problem 14.10 is given as a function of time by ΣF = (−20t + 90) i − 60 j + (10t + 40)k (N). At time t = 0, its position is r = 40 i + 30 j − 360 k (m) and its velocity is v = −14 i + 26 j + 32k (m/s). What is its position at t = 4 s?

Solution: 1 [(−20t + 90) i − 60 j + (10t + 40)k]N (10 kg) a = [(−2t + 9) i − 6 j + (t + 4)k] m/s 2

a =

Integrate to get the velocity v = ∫ a dt + v 0

(

)

1 2 t + 4t + 32 k  m/s v =  (−t 2 + 9t − 14) i + (−6t + 26) j +   2 Integrate again to get the position r = ∫ v dt + r0

(

)

1 9 r =  − t 3 + t 2 − 14t + 40 i + (−3t 2 + 26t + 30) j  3 2 1 3 + t + 2t 2 + 32t − 360 k m   6

(

)

At the time indicated (t = 4 s) we have r = [34.7 i + 86 j − 189.3k] m.

Problem 14.12 Measurements of the 10-kg object’s position determine that it is given as a function of time in seconds by r = (15t 2 − 80) i + 30t 2 j − (2t 3 + 50) k (m). What is the magnitude of the total external force on the object at t = 4 s? y

Solution:

We can take derivatives of the position to determine the acceleration as a function of time: r = (15t 2 − 80)i + 30t 2 j − (2t 3 + 50)k (m), dr = 30ti + 60tj − 6t 2 k (m/s), dt dv a = = 30 i + 60 j − 12tk (m/s 2 ). dt The total external force on the object is ΣF = ma = 300 i + 600 j − 120tk (N). Its magnitude at t = 4 s is ΣF = 825 N. v =

825 N.

x z

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Problem 14.13 Before the 18,000-lb plane touches down, the sum of the external forces acting on it, which are its weight, the aerodynamic force, and its thrust T = 2000 lb, is zero. At the instant it touches down, the runway exerts a 420-lb frictional force on the skidding wheels and the thrust is decreased to near zero. (Assume that the aerodynamic force is the same before and just after touching down.) What is the horizontal component of the plane’s acceleration immediately after it touches down?

Solution: y

T 158

x Before the plane touches down, let T, A, W denote the thrust, aerodynamic, and weight force vectors respectively. Their sum equals zero:

y T

T + A + W = 0. In terms of components,

158

T (cos15 °i + sin15 ° j) + ( A x i + A y j) − Wj = 0 . x

We see that the horizontal component of the aerodynamic force is A x = −T cos15 °. Immediately after the plane touches down, the only horizontal forces acting on it are the horizontal component of the aerodynamic force and the friction force f = 420 lb. From Newton’s second law, W  −T cos15 ° − f =   a x : g −(2000 lb) cos15 ° − 420 lb =

lb ( 18,000 )a . 32.2 ft/s 2

This yields a x = −4.21 ft/s 2 . a x = −4.21 ft/s 2 .

Problem 14.14 The 18,000-lb plane is moving at 200 ft/s when it touches down. If the pilot does not apply the brakes, the only significant force acting on the plane is aerodynamic drag. Its magnitude is D = 0.04v x 2 lb, where v x is the plane’s velocity in ft/s. How long would it take for the plane’s velocity to decrease to 100 ft/s?

Solution:

Newton’s second law is

W  −D =   a x : g  W  dv −0.04 v x2 =   x .  g  dt We can separate variables and integrate to determine the time:  W  dv dt = −25  2x ,  g  vx

W 

100 ft/s dv

∫0 dt = −25 g  ∫200 ft/s v x2x ,

158

 W  1  t = 25     g   v x  200 ft/s

100 ft/s

= 25

lb ( 18,000 )( 1001ft/s − 2001ft/s ) 32.2 ft/s 2

= 70.0 s. 70.0 s.

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Problem 14.15 At the instant shown, the rocket is traveling straight up at 100 m/s. Its mass is 90,000 kg and the thrust of its engine is 2400 kN. Aerodynamic drag exerts a resisting force (in newtons) of magnitude 0.8v 2 , where v is the magnitude of the velocity. How long does it take for the rocket’s velocity to increase to 200 m/s? Solution:

The equation of motion is

ΣF : (2400 kN) − (90,000 kg)(9.81 m/s 2 ) − (0.8 kg/m)v 2 = (90,000 kg)a Solving for the acceleration we have dv = (16.9 m/s 2 ) − (8.89 × 10 −6 m −1 )v 2 dt 200 m/s t dv ∫100 m/s (16.9 m/s 2 ) − (8.89 × 10 −6 m −1 )v 2 = ∫0 dt = t

a =

Source: Courtesy of Oleg_Yakovlev/Shutterstock.

Carrying out the integration, we find t = (81.7 s)(tanh −1[(0.000726)(200)] − tanh −1[(0.000726)(100)]) t = 6.01 s.

Problem 14.16 The rocket lifts off at time t = 0 and moves straight up. The constant thrust of its engines is T = 1.3E6 lb. Due to the fuel burned by its engines, its weight is a function of time given by W = W0 − ct, where W0 = 1.2E6 lb and c = 4700 lb/s. What is the rocket’s velocity at t = 60 s in ft/s and in mi/h? (Neglect aerodynamic drag.)

Solution:

Consider the free-body diagram of the rocket:

x y W

T Newton’s second law is W ΣFx = T − W = a , g x so the acceleration as a function of time is ax =

  dv x T T = g − 1 = g  − 1 .  W0 − ct  dt W

(

)

We can integrate to determine the velocity: v





∫0 dv x = ∫0 g  W0 − ct − 1: T dt − gt. − ct To evaluate the integral, make the change of variable z = W0 − ct. The differential dz = −cdt, and we obtain v x = g∫

0 W0

gT W0 − ct dz − gt c ∫W 0 z gT W0 dz = − gt c ∫W0 − ct z  T  W0   = g  ln   − t  .  c  W0 − ct  

vx = −

Source: Courtesy of Oleg_Yakovlev/Shutterstock.

Substituting the values, at t = 60 s we get v x = 454 ft/s, which is 1 mile 3600 s 454 ft/s = 309 miles/hr. 5280 ft 1 hr

(

)(

)

454 ft/s, 309 mi/hr.

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Problem 14.17 The combined weight of the motorcycle and rider is 360 lb. The coefficient of kinetic friction between the tires and the road is µ k = 0.8. The rider starts from rest, spinning the rear wheel. Neglect the horizontal force exerted on the front wheel by the road. In 2 s, the motorcycle moves 35 ft. What was the normal force between the rear wheel and the road?

Solution:

Kinematics

a = constant, v = at, s = Dynamics: Fr = Friction:

1 2 1 at , 35 ft = a(2 s) 2 ⇒ a = 17.5 ft/s 2 2 2

360 lb ( 32.2 )(17.5 ft/s ) = 195.6 lb ft/s 2

Fr = (0.8) N ⇒ N =

195.6 lb = 245 lb. 0.8

Source: Courtesy of Arthur Bargan/Shutterstock.

Problem 14.18 The excavator’s 1800-kg bucket B undergoes motion in the x – y plane. From t = 0 to t = 2 s, the coordinates of the center of mass of the bucket are x = −0.8t 2 + 0.5t + 7 m, y = 0.2t 3 + 0.4t 2 + 5 m. Determine the x and y components of the force exerted on the bucket by its supports at t = 1 s. y B

Solution:

We can differentiate to determine the components of the

acceleration: x = −0.8t 2 + 0.5t + 7 m, dx = −1.6t + 0.5 m/s, dt dv x ax = = −1.6 m/s 2 , dt y = 0.2t 3 + 0.4t 2 + 5 m,

vx =

dy = 0.6t 2 + 0.8t m/s, dt dv y ay = = 1.2t + 0.8 m/s 2 . dt At t = 1 s, the components of the acceleration are a x = −1.6 m/s 2 , a y = 2 m/s 2 . vy =

Let S = S x i + S y j denote the force exerted on the bucket by its supports. From Newton’s second law, ΣFx = S x = ma x , ΣFy = S y − mg = ma y , we obtain S x = −2880 N, S y = 21,300 N. −2880 i + 21,300 j (N). x

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Problem 14.19 During a test flight in which a 9000-kg helicopter starts from rest at t = 0, the acceleration of its center of mass from t = 0 to t = 10 s is

Solution:

a = 0.6t i + (1.8 − 0.36t ) j (m/s 2 ).

ΣF =

From Newton’s second law: ΣF = ma. The sum of the external forces is ΣF = F − W = 9000[(0.6t ) i + (1.8 − 0.36t ) j] t = 6 = 32400 i − 3240 j, from which the magnitude is 32400 2 + 3240 2 = 32562 (N).

What is the magnitude of the total external force on the helicopter (including its weight) at t = 6 s? y

L T

Pat

W x

Problem 14.20 The engineers conducting the test described in Problem 14.19 want to express the total force on the helicopter at t = 6 s in terms of three forces: the weight W, a component T tangent to the path, and a component L normal to the path. What are the values of W, T, and L?

Solution: Integrate the acceleration: v = (0.3t 2 )i + (1.8t − 0.18t 2 ) j, since the helicopter starts from rest. The instantaneous flight path angle dy dy dx −1 (1.8t − 0.18t 2 ) is tan β = = = . At t = 6 s, dx dt dt (0.3t 2 )  (1.8(6) − 0.18(6) 2 )  β t = 6 = tan −1   = 21.8 °. A unit vector tangent  0.3(6) 2

W = − j(9000)(9.81) = −88.29 j (kN).

( )( )

to this path is e t = i cos β + j sin β . A unit vector normal to this path e n = −i sin β + j cos β . The weight acts downward:

From Newton’s second law, F − W = ma, from which F = W + ma = 32400 i + 85050 j (N). The component tangent to the path is L

T = F ⋅ e t = 32400 cos β + 85050 sin β = 61669.4 (N). T

The component normal to the path is L = F ⋅ e n = −32400 sin β + 85050 cos β = 66934 (N).

Pat

W x

100

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Problem 14.21 At the instant shown, the forces acting on the 11,000-kg airplane are its weight, the thrust T = 90 kN, the lift L = 170 kN, and the drag D = 34 kN. Determine the magnitude of the airplane’s acceleration.

Solution:

Newton’s second law for the plane is

ΣFx = T cos15 ° − D − mg sin15 ° = ma x , ΣFy = T sin15 ° + L − mg cos15 ° = ma y . Solving, we obtain a x = 2.27 m/s 2 , a y = 8.10 m/s 2 . The magnitude of the acceleration is a =

a x 2 + a y 2 = 8.41 m/s 2 .

8.41 m/s 2 .

T 158

158 Horizontal

Problem 14.22 The 11,000-kg airplane’s path is described by the equation y = cx 2 , where the constant c = 8.0E−5. At the instant shown, its velocity is v = 300 i (m/s) and the component of its acceleration in the direction parallel to its path is a x = 5 m/s 2 . The number of g’s to which the pilot is subjected is given in terms of the airplane’s acceleration a and the acceleration due to gravity g by N =

a + gj . g

How many g’s does the pilot experience at the instant shown? Strategy: Use the equation for the airplane’s path to determine its y component of acceleration.

Solution:

We need to determine the airplane’s y component of acceleration. From the equation for the airplane’s path, we see that dy d d dx dx = = 2cx (cx 2 ) = (cx 2 ) dt dt dx dt dt

and d 2y d 2x d dx dx 2 = = 2c + 2cx 2 . 2cx 2 dt dt dt dt dt

(

)

( )

At the instant shown, x = 0. The first equation just confirms that the airplane’s y component of velocity is zero at that instant. In the second equation, dx /dt = v x = 300 m/s is the airplane’s velocity, so we can solve for the y component of its acceleration: d 2y dx 2 = 2c dt 2 dt = 2cv x 2

( )

ay =

= 2(8.0E-5)(300 m/s) 2 = 14.4 m/s 2 . The number of g’s the pilot is subjected to is

N = L

T 158

158 Horizontal

a + gj g

a x i + a y j + gj g (5 m/s 2 ) i + (14.4 m/s 2 ) j + (9.81 m/s 2 ) j = 9.81 m/s 2 = 2.52. =

2.52 g’s.

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Problem 14.23 The coordinates in meters of the 360-kg sport plane’s center of mass relative to an Earthfixed reference frame during an interval of time are x = 20t − 1.63t 2 ,

y = 35t − 0.15t 3 , and z = −20t − 1.38t 2 , where t is the time in seconds. The y-axis points upward. The forces exerted on the plane are its weight, the thrust vector T exerted by its engine, the lift force vector L, and the drag force vector D. At t = 4 s, determine

T + L + D. z

Solution:

Denoting the airplane’s mass by m, Newton’s second law for the plane is T + L + D − mg j = ma.

dz = −20 − 2(1.38)t, dt

d 2z = a z = −2(1.38), . dt 2

We can use the given equations for the airplane’s position to determine its acceleration at t = 4 s and solve for (T + L + D) Deriving the components of the acceleration as functions of time,

at t = 4 s we obtain a = −3.26 i − 3.6 j − 2.76k (m/s 2 ). Then

dx = 20 − 2(1.63)t, dt

d 2x = a x = −2(1.63), dt 2

T + L + D = −1170 i + 2240 j − 994 k (N).

dy = 35 − 3(0.15)t 2 , dt

d 2y = a y = −6(0.15) t, dt 2

Problem 14.24 The force (in newtons) exerted on the 360-kg sport plane by its engine, the lift force, and the drag force as functions of time (in seconds) is T + L + D = −(2000 + 470t ) i + (3000 − 1000t ) j + (900 + 600t )k (N). If the coordinates of the plane’s center of mass are (0, 0, 0) and its velocity is 34 i + 40 j + 10 k (m/s) at t = 0, what are the coordinates of its center of mass at t = 4 s?

T + L + D = mg j + ma = −1170 i + 2240 j − 994 k (N).

Then we can integrate to determine the position: x





∫0 dx = ∫0  34 − m (2000t + 235t 2 )  dt, x = 34t −

(

)

1 235 3 1000t 2 + t m. 3 m

It the y direction, the velocity is dv y 1 = (3000 − 1000t ), dt m t 1 vy ∫40 m/s dv y = ∫0 m (3000 − 1000t ), dy 1 = 40 + (3000t − 500t 2 ) m/s, vy = dt m ay =

and the acceleration is





∫0 dy = ∫0  40 + m (3000t − 500t 2 )  dt, y = 40t +

(

)

1 500 3 1500t 2 − t m. 3 m

In the z direction, we have dv z 1 = (900 + 600t ), dt m t 1 vz ∫10 m/s dv z = ∫0 m (900 + 600t ), 1 dz = 10 + (900t + 300t 2 ) m/s. vz = dt m and az =

Solution:

We can integrate the x component of the plane’s acceleration to determine the x component of its velocity as a function of time: dv x 1 ax = = − (2000 + 470t ), dt m vx t 1 ∫34 m/s dv x = −∫0 m (2000 + 470t ), 1 dx = 34 − (2000t + 235t 2 ) m/s. vx = dt m

102





∫0 dz = ∫0  10 + m (900t + 300t 2 )  dt, z = 10t +

1 (450t 2 + 100t 3 ) m. m

At t = 4 s we obtain x = 77.6 m, y = 197 m, z = 77.8 m. x = 77.6 m, y = 197 m, z = 77.8 m.

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Problem 14.25 The robot manipulator is programmed so that x = 40 + 24t 2 mm, y = 4t 3 mm, and z = 0 during the interval of time from t = 0 to t = 4 s. The y-axis points upward. What are the x and y components of the total force exerted by the jaws of the manipulator on the 2-kg widget A at t = 3 s?

Solution: x = 40 + 24t 2 mm

y = 4t 3 mm

v x = 48t mm/s

v y = 12t 2 mm/s

a x = 48 mm/s 2

a y = 24t mm/s 2

At t = 3 s a x = 48 × 10 −3 m/s 2 , a y = 72 × 10 −3 m/s 2

Fx = ma x

m = 2 kg

Fy − mg = ma y Solving, Fx = 0.096 N

Fy = 19.764 N mg

y Fx x

Problem 14.26 The robot manipulator is programmed so that it is stationary at t = 0 and the components of the acceleration of A are a x = 400 − 0.6v x mm/s 2 and a y = 200 − 0.2v y mm/s 2 from t = 0 to t = 4 s, where v x and v y are the components of the velocity in mm/s. The y-axis points upward. What are the x and y components of the force exerted on the 2-kg widget A at t = 2 s?

Solving for the velocity yields 400 vx = (1 − e −0.6t ). 0.6 From this result we can see that 200 vy = (1 − e −0.2t ). 0.2 At t = 2 s we obtain v x = 466 mm/s, v y = 330 mm/s. Converting to meters, the components of the acceleration are a x = 0.120 m/s 2 , a y = 134 m/s 2 . Let Fx and Fy be the components of the force exerted on the widget by the manipulator. From Newton’s second law,

Fx = ma x , Fy − mg = ma y , we obtain Fx = 0.241 N, Fy = 19.9 N.

Fx = 0.241 N, Fy = 19.9 N. y

mg x

x Fx

Solution: To determine the acceleration at t = 2 s, and thereby the force on the widget, we need to determine the velocity. In the x direction we write dv x ax = = 400 − 0.6v x mm/s 2 , dt

separate variables and integrate, obtaining dv x = dt, 400 − 0.6v x t vx dv ∫0 400 − x0.6v x = ∫0 dt, v  − 1 ln(400 − 0.6v )  x = t, x   0.6 0   400 ln   = 0.6t.  400 − 0.6v x 

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Problem 14.27 In the sport of curling, the object is to slide a “stone” weighing 44 lb onto the center of a target located 31 yards from the point of release. In terms of the coordinate system shown, the point of release is at x = 0, y = 0. Suppose that the shot comes to rest at x = 31 yards, y = 1.5 yards. Assume that the coefficient of kinetic friction is µ k = 0.008. What were the x and y components of the stone’s velocity at release?

Solution: y

4.5 ft

93 ft

The dashed line is the α = arctan(4.5 ft/93 ft) = 2.77 °, L =

mk N

stone’s path. The angle and the length of the path is

(93 ft) 2 + (4.5 ft) 2 = 93.1088 ft.

Let s, v , and a denote the stone’s position, velocity and acceleration along its path. Applying Newton’s second law, the acceleration is dv 1 a = = − µk N, dt m

31 yd

where m = (44 lb)/(32.2 ft/s 2 ) = 1.37 slugs and the normal force N = 44 lb. Let v 0 be the stone’s initial velocity. Integrating, we obtain Curling Stone

t 1

∫v dv = −∫0 m µ k N dt, 0

v =

ds 1 = v 0 − µ k N t, (1) dt m

∫0 ds = ∫0 ( v 0 − m µ k N t ) dt, L

L = v 0t −

1 µ N t 2. 2m k

(2)

Setting v = 0, we can solve Eq. (1) for t and substitute it into Eq. (2), obtaining the result x

2µ k NL . m From this equation we obtain v 0 = 6.9260 ft/s. The components of the velocity at release are

v02 =

v x = v 0 cos α = 6.918 ft/s, v y = v 0 sin α = 0.3347 ft/s. v x = 6.918 ft/s, v y = 0.3347 ft/s.

Problem 14.28 The two masses are released from rest. How fast are they moving at t = 0.5 s? (See Example 14.4.)

Solving, we find T = 28.0 N, a = 4.20 m/s 2 . To find the velocity, we integrate the acceleration v = at = (4.20 m/s)(0.5 s) = 2.10 m/s. v = 2.10 m/s.

2 kg

Solution:

5 kg

The free-body diagrams are shown. The governing

equations are ΣFy left : T − 2(9.81) N = (2 kg)a ΣFy right : T − 5(9.81) N = −(5 kg)a

104

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Problem 14.29 The two weights are released from rest. The horizontal surface is smooth. How far does the 10-lb weight fall in one-half second?

Let s denote the displacement of the weights. The mass of the 5-lb weight is 5 lb/32.2 m/s 2 = 0.155 slugs. In the horizontal direction we have T = (0.155 slugs)

dv , dt v

∫0 T dt = (0.155 slugs)∫0 dv ,

5 lb

T t = (0.155 slugs)v = (0.155 slugs) t

ds , (1) dt

∫0 T t dt = (0.155 slugs)∫0 ds, 1 2 T t = (0.155 slugs)s. 2 The mass of the 10-lb weight is 10 lb/32.2 m/s 2 = 0.311 slugs. For it we have 10 lb − T = (0.311 slugs)

10 lb

dv , dt v

Solution:

We can draw individual free-body diagrams for the two

∫0 [10 lb − T ] dt = (0.311 slugs)∫0 dv , [10 lb − T ] t = (0.311 slugs)v = (0.311 slugs)

weights: t

5 lb

ds , (2) dt

∫0 [10 lb − T ] t dt = (0.311 slugs)∫0 ds, T

1 [10 lb − T ] t 2 = (0.311 slugs)s. 2

If we set t = 0.5 s, Eqs. (1) and (2) contain two unknowns, s and the tension T. Solving, we obtain T = 3.33 lb, s = 2.68 ft.

2.68 ft.

10 lb

Problem 14.30 The two weights are released from rest. The coefficient of kinetic friction between the horizontal surface and the 5-lb weight is µ k = 0.25. How far does the 10-lb weight fall in one-half second?

Let s denote the displacement of the weights. The mass of the 5-lb weight is 5 lb/32.2 m/s 2 = 0.155 slugs. Because the acceleration of the 5-lb weight is zero in the vertical direction, N = 5 lb. In the horizontal direction we have T − µ k (5 lb) = (0.155 slugs)

5 lb

dv , dt v

∫0 [T − µ k (5 lb)] dt = (0.155 slugs)∫0 dv , [T − µ k (5 lb)] t = (0.155 slugs)v = (0.155 slugs) t

ds , (1) dt

∫0 [T − µ k (5 lb)] t dt = (0.155 slugs)∫0 ds, 1 [T − µ k (5 lb)] t 2 = (0.155 slugs)s. 2

10 lb

The mass of the 10-lb weight is 10 lb/32.2 m/s 2 = 0.311 slugs. For it we have dv 10 lb − T = (0.311 slugs) , dt v

Solution:

We can draw individual free-body diagrams for the two

∫0 [10 lb − T ] dt = (0.311 slugs)∫0 dv , [10 lb − T ] t = (0.311 slugs)v = (0.311 slugs)

weights: 5 lb

mk N

ds , (2) dt

∫0 [10 lb − T ] t dt = (0.311 slugs)∫0 ds,

1 [10 lb − T ] t 2 = (0.311 slugs)s. 2

N T

If we set t = 0.5 s, Eqs. (1) and (2) contain two unknowns, s and the tension T. Solving, we obtain T = 4.17 lb, s = 2.35 ft. 2.35 ft.

10 lb

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Problem 14.31 The mass of each box is 20 kg. The coefficient of kinetic friction between the boxes and the surface is µ k = 0.1. If the system is released from rest, how fast are the boxes moving 1 s later?

For the left box we can see that the normal force N A = mg. Applying Newton’s second law, we have T − µ k mg = ma, where a is the box’s acceleration toward the right. For the right box the normal force N B = mg cos30 °. Newton’s second law is mg sin 30 ° − µ k mg cos30 ° − T = ma, where a is the box’s acceleration down the inclined surface. We have two equations in terms of T and a. Solving them, we obtain T = 50.4 N, a = 1.54 m/s 2 . After 1 second, the boxes will be moving with velocity

308

Solution:

We draw individual free-body diagrams for the two

boxes:

v = at = (1.54 m/s 2 )(1 s) = 1.54 m/s. T = 50.4 N, a = 1.54 m/s 2 .

1.54 m/s 2 .

T T

mk NA

NA mk NB

Problem 14.32* The masses m A = 15 kg and m B = 35 kg. The coefficients of friction between all the surfaces are µ s = 0.45 and µ k = 0.35. The blocks are stationary when the constant force F is applied. Determine the resulting acceleration of block B if (a) F = 300 N; (b) F = 500 N.

Block B does slide in both cases (a) and (b). Also, we now know that the friction force f B = µ k N B = µ k (m A g + m B g). The next question is whether block A slides or not. Let us first assume that it doesn’t. Then we can apply Newton’s second law to the two blocks treated as a single object and determine their acceleration. Knowing the acceleration, we can apply Newton’s second law to block A to determine the friction force f A . Then we will know that block A slips if f A > µ s N A = µ s m A g = 66.2 N. (1)

A B

Newton’s second law for the two blocks together if block A doesn’t slide is

F − µ k (m A g + m B g) = (m A + m B )a. (2) Substituting the values for case (a), we obtain a = 2.57 m/s 2 . Substituting this acceleration into Newton’s second law for block A,

Solution:

f A = m Aa

(a) Consider the free-body diagrams of the individual blocks:

= (15 kg)(2.57 m/s 2 ) = 38.5 N.

mAg NA fA

f A = m Aa

m Bg fB

We see from Eq. (1) that in case (a), block A does not slip, and the acceleration we have obtained is the answer, a = 2.57 m/s 2 . (b) For case (b), we solve Eq. (2) and obtain a = 6.57 m/s 2 . Substituting this acceleration into Newton’s second law for block A yields

F NB

The most boring possibility is that block B doesn’t slide when F is applied, in which case its acceleration is zero. The normal force between block B and the floor is N B = m A g + m B g. The question is whether F > µ s N B . Substituting the given values, we see that µ s N B = µ s (m A g + m B g) = 221 N.

106

= (15 kg)(6.57 m/s 2 ) = 98.5 N. According to Eq. (1), we see that in this case block A does slip. That means the friction force f A = µ k N A = µ k m A g. Newton’s second law for block B is therefore F − f A − f B = F − µ k m A g − µ k (m A g + m B g) = m B a, which yields a = 7.91 m/s 2 . (a) 2.57 m/s 2 . (b) 7.91 m/s 2 .

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Problem 14.33 The crane’s trolley at A moves to the right with constant acceleration, and the 800-kg load moves without swinging. (a) What is the acceleration of the trolley and load? (b) What is the sum of the tensions in the parallel cables supporting the load?

Solution: (a) From Newton’s second law, T sin θ = ma x , and T cos θ − mg = 0. Solve a x =

mg T sin θ ,T = , from which m cos θ

a = g tan θ = 9.81 (tan 5 °) = 0.858 m/s 2 . (b)

T =

800(9.81) = 7878 N. cos 5 °

58 T

58 mg

Problem 14.34 The mass of A is 30 kg and the mass of B is 6 kg. The horizontal surface is smooth. The constant force F causes the system to accelerate. The force exerted on B by the slender bar is parallel to the bar. The angle θ = 25 ° is constant. Determine F. A

F − T sin θ = m A a, (1) T sin θ = m B a. (2) For B we also write Newton’s second law in the vertical direction, T cos θ − m B g = 0. (3)

We have three equations in the three unknowns F , T , and a. We can solve Eq. (3) for T, then solve Eq. (2) for a, and then solve Eq. (1) for F. We obtain T = 64.9 N, a = 4.57 m/s 2 and F = 165 N.

F = 165 N.

Solution:

Let a denote their acceleration. We write Newton’s second law for each object:

Consider the free-body diagrams of the individual

objects:

mAg F T

mBg

Problem 14.35 The mass of A is 30 kg and the mass of B is 5 kg. The coefficient of kinetic friction between A and the horizontal surface is µ k = 0.2. The constant force F causes the system to accelerate. The angle θ = 20 ° is constant. Determine F.

A F

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14.35 (Continued) Solution: We have four unknowns (F , T , N , a) and four equations of motion: ΣFxA : F − T sin θ − µ k N = (30 kg)a, ΣFyA : N − T cos θ − (30 kg)(9.81 m/s 2 ) = 0, ΣFxB : T sin θ = (5 kg)a, ΣFyB : T cos θ − (5 kg)(9.81 m/s 2 ) = 0. Solving, we find T = 52.2 N, a = 3.57 m/s 2 , N = 343 N, F = 194 N.

Problem 14.36 The 100-lb crate is initially stationary. The coefficients of friction between the crate and the inclined surface are µ s = 0.2 and µ k = 0.16. Determine how far the crate moves from its initial position in 2 s if the horizontal force F = 90 lb.

308

Solution:

Denote W = 100 lb, g = 32.17 ft/s 2 , F = 90 lb, and θ = 30 °. Choose a coordinate system with the positive x axis parallel to the inclined surface. (See free body diagram next page.) The normal force exerted by the surface on the box is N = F sin θ + W cos θ = 131.6 lb. The sum of the non-friction forces acting to move the box is Fc = F cos θ − W sin θ = 27.9 lb. Slip only occurs if Fc ≥ N µ s which 27.9 > 26.32 (lb), so slip occurs.

g ( F − sgn(Fc )µ k N ) = 2.125 ft/s 2 . The velocity is W c a v (t ) = at ft/s, since v (0) = 0. The displacement is s = t 2 ft, since 2 s(0) = 0. The position after 2 s is s(2) = 4.43 ft up the inclined which a =

surface.

The direction of slip is determined from the sign of the sum of the non friction forces: Fc > 0, implies that the box slips up the surface, and Fc < 0, implies that the box slips down the surface (if the condition for slip is met). Since Fc > 0 the box slips up the surface. After the box slips, the sum of the forces on the box parallel to the surface is ΣFx = Fc − sgn(Fc )µ k N , where sgn(Fc ) =

F f

Fc W  . From Newton’s second law, ΣFx =   a, from g Fc

Problem 14.37 The 100-lb crate is initially stationary. The coefficients of friction between the crate and the inclined surface are µ s = 0.2 and µ k = 0.12. Determine how far the crate moves from its initial position in 2 s if the horizontal force F = 30 lb.

Solution:

Consider the free-body diagram of the crate: y

W f

308 N

308

Writing Newton’s second law in the y direction, we have N − F sin 30 ° − W cos30 ° = 0. 308

108

We can solve this, obtaining N = 102 lb.

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14.37 (Continued) The component of the force F and the weight in the x direction is F cos30 ° − W sin 30 ° = −24.0 lb, so those forces tend to push the crate down the plane. The friction force will resist that (as shown in the diagram), so we must consider whether the crate will slip down the surface or not. It will if the force down the surface W sin 30 ° − F cos30 ° > µ s N . Checking, we find that µ s N = 20.3 lb, so it does slip. Writing Newton’s second law for the acceleration down the surface with f = µ k N, W W sin 30 ° − F cos30 ° − µ k N = a, g we can solve for the crate’s acceleration up the surface, obtaining a = 3.81 ft/s 2 . We can now integrate the constant acceleration to determine the velocity and position as functions of time:

Problem 14.38 The crate has a mass of 120 kg, and the coefficients of friction between it and the sloping dock are µ s = 0.6 and µ k = 0.5. (a) What tension must the winch exert on the cable to start the stationary crate sliding up the dock? (b) If the tension is maintained at the value determined in part (a), what is the magnitude of the crate’s velocity when it has moved 2 m up the dock?

dv = a, dt v

∫0 dv = ∫0 a dt, v =

ds = at, dt

∫0 ds = ∫0 at dt, s =

1 2 at . 2

At t = 2 s we obtain s = 7.62 ft. 7.62 ft down the inclined surface.

Solution: Choose a coordinate system with the x axis parallel to the surface. Denote θ = 30 °. (a) The normal force exerted by the surface on the crate is N = W cos θ = 120(9.81)(0.866) = 1019.5 N. The force tending to move the crate is Fc = T − W sin θ, from which the tension T = W (sin θ) + µ s N = 1200.3 N.

required to start slip is

(b) After slip begins, the force acting to move the crate is F = T − W sin θ − µ k N = 101.95 N. From Newton’s second F 101.95 law, F = ma, from which a = = = 0.8496 m/s 2 . m 120 The velocity is v (t ) = at = 0.8496t m/s, since v (0) = 0. The

( )

a 2 t , since s(0) = 0. When the crate has moved 2 2(2) 2 m up the slope, t 10 = = 2.17 s and the velocity is a v = a(2.17) = 1.84 m/s.

position is s(t ) =

308

f W

Problem 14.39 The coefficients of friction between the load A and the bed of the utility vehicle are µ s = 0.4 and µ k = 0.36. If the floor is level (θ = 0), what is the largest acceleration (in m/s 2 ) of the vehicle for which the load will not slide on the bed?

Solution:

The load is on the verge of slipping. There are two unknowns (a and N). The equations are ΣFx : µ s N = ma,

ΣFy : N − mg = 0

Solving we find a = µ s g = (0.4) (9.81 m/s 2 ) = 3.92 m/s 2 . a = 3.92 m/s 2 .

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Problem 14.40 The coefficients of friction between the load A and the bed of the utility vehicle are µ s = 0.4 and µ k = 0.36. The angle θ = 20 °. Determine the largest forward and rearward accelerations of the vehicle for which the load will not slide on the bed.

Solution:

The load is on the verge of slipping. There are two unknowns (a and N).

Forward: The equations are ΣFx : µ s N − mg sin θ = ma, ΣFy : N − mg cos θ = 0 Solving we find a = g(µ s cos θ − sin θ) = (9.81 m/s 2 )([0.4]cos 20 ° − sin 20 °) a = 0.332 m/s 2 . Rearward: The equations are ΣFx : −µ s N − mg sin θ = −ma, ΣFy : N − mg cos θ = 0 Solving we find a = g(µ s cos θ + sin θ) = (9.81 m/s 2 )([0.4]cos 20 ° + sin 20 °) a = 7.04 m/s 2 .

Problem 14.41 The package starts from rest and slides down the smooth ramp. The hydraulic device B exerts a constant 2000-N force and brings the package to rest in a distance of 100 mm from the point at which it makes contact. What is the mass of the package?

Solution: mg

308

A N

B 308

308 2000 N N

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14.41 (Continued) Set g = 9.81 m/s 2

Now analyze the motion after it hits point B.

First analyze the motion before it gets to point B.

ΣF : mg sin 30 ° − 2000 N = ma

ΣF : mg sin 30 ° = ma

a = v

a = g sin 30 °,

v = ( g sin 30 °)t,

t2 s = ( g sin 30 °) 2

When it gets to B we have

2000 N dv = g sin 30 ° − ds m

( g sin 30° − 2000m N ) ds (4.43 m/s) 2000 N 0− = ( g sin 30 ° − (0.1 m) m ) 2 0

0.1 m

∫4.43 m/s v dv = ∫0 2

t2 s = 2 m = ( g sin 30 °) ⇒ t = 0.903 s 2 v = ( g sin 30 °)(0.903 s) = 4.43 m/s

Solving the last equation we find

Problem 14.42 The force exerted on the 10-kg mass by the linear spring is F = −ks, where k is the spring constant and s is the displacement of the mass relative to its position when the spring is unstretched. The value of k is 40 N/m. The mass is in the position s = 0 and is given an initial velocity of 4 m/s toward the right. Determine the velocity of the mass as a function of s. Strategy: Use the chain rule to write the acceleration as dv dv ds dv = = v. dt ds dt ds

m = 19.4 kg.

Solution:

The equation of motion is −ks = ma s k v v2 dv k v2 k s2 a = v = − s ⇒ ∫ v d v = −∫ − 0 = − ds ⇒ 0 m v0 ds m 2 2 m 2

Thus v = ± v 02 −

40 N/m 2 k 2 s = ± (4 m/s) 2 − s 10 kg m

v = ±2 4 − s 2 m/s.

s k

Problem 14.43 The 450-kg boat is moving at 10 m/s when its engine is shut down. The magnitude of the hydrodynamic drag force (in newtons) is 40v 2 , where v is the magnitude of the velocity in m/s. When the boat’s velocity has decreased to 1 m/s, what distance has it moved from its position when the engine was shut down?

Solution:

The equation of motion is

40 2 v 450 To integrate, we write the acceleration as F = −40v 2 = (450 kg)a ⇒ a = −

1 m/s dv  1 m/s  40 2 40 s dv = − v ⇒ ∫ = − ds ⇒ ln   10 m/s  10 m/s v 450 450 ∫ 0 ds 40 = − s 450 450 s = − ln(0.1) = 25.9 m. 40

a = v

s = 25.9 m.

Source: Courtesy of Zdorov Kirill Vladimirovich / Shutterstock.

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Problem 14.44 A sky diver and his parachute weigh 200 lb. He is falling vertically at 100 ft/s when his parachute opens. With the parachute open, the magnitude of the drag force (in pounds) is 0.5v 2. (a) What is the magnitude of the sky diver’s acceleration at the instant the parachute opens? (b) What is the magnitude of his velocity when he has descended 20 ft from the point where his parachute opens?

Solution: Choose a coordinate system with s positive downward. For brevity denote C d = 0.5, W = 200 lb, g = 32.17 ft/s 2 . From  W  dv Newton’s second law W − D =   , where D = 0.5v 2 . Use the  g  dt

( )

chain rule to write v

dv = − ds

(

0.5v 2 g

+ g = g 1−

)

C dv 2 . W

(a) At the instant of full opening, the initial velocity has not decreased, and the magnitude of the acceleration is a init

(

= g 1−

Cd 2 v W

) = 24 g = 772.08 ft/s.

(b) Separate variables and integrate:

v dv = gds, from which C dv 2 W

1−

(

ln 1 −

C dv 2 W

) = −( 2CW g )s + C. Invert and solve: d

2C g  W  − d s v 2 =   1 − Ce W . At s = 0, v (0) = 100 ft/s, from  C d  

which C = 1 −

C d 10 4 = −24, and W

2C g  W  − d s v 2 =   1 + 24 e W . At s = 20 ft the velocity is  C d  

v (s = 20) =

g  − (20)   = 28.0 ft/s. 2W  1 + 24 e W  

Problem 14.45 The Panavia Tornado, with a mass of 18,000 kg, lands at a speed of 213 km/h. The decelerating force (in newtons) exerted on it by its thrust reversers and aerodynamic drag is 60,000 + 2.5v 2 , where v is the magnitude of the airplane’s velocity in m/s. What is the length of the airplane’s landing roll? Strategy: Use the chain rule to express the airplane’s acceleration in terms of its displacement (see Example 14.5).

Solution:

The airplane’s initial velocity in m/s is

v 0 = 213 km/h

m 1h ( 1000 )( 3600 ) 1 km s

= 59.2 m/s. Applying Newton’s second law, the airplane’s acceleration is a = −

1 (60,000 + 2.5v 2 ) m/s 2 , m

where m = 18,000 kg is its mass. We use the chain rule to express the plane’s acceleration in terms of its displacement s: a =

dv dv ds dv = = v. dt ds dt ds

Then we can separate variables and integrate. a = v m∫

v0 0

1 dv = − (60, 000 + 2.5v 2 ), ds m

s v dv = ∫ ds = s. 0 60, 000 + 2.5v 2

(Notice that we have reversed the limits on the integral on the left, which eliminated the minus sign.) We can evaluate the integral on the left by introducing the change of variable

Source: Courtesy of InsectWorld/Shutterstock.

u = 60,000 + 2.5v 2 , du = 5v dv .

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14.45 (Continued) The integral becomes v dv

60,000 + 2.5v 0 2 du

∫0 60, 000 + 2.5v 2 = 5 ∫60,000 =

(

)

60, 000 + 2.5v 0 2 1 ln . 5 60, 000

The distance the airplane rolls is s =

(

60, 000 + 2.5v 0 2 m ln 5 60, 000

)

18,000 kg  60, 000 + 2.5(59.2 m/s) 2  ln   5 60, 000   = 490 m. =

490 m.

Problem 14.46 The 18,000-kg airplane lands at a speed of 213 km/h. The decelerating force (in newtons) exerted on it by its thrust reversers and aerodynamic drag is 60,000 + 2.5v 2 , where v is the magnitude of the airplane’s velocity in m/s. How long does it take for the airplane to roll to a stop?

Solution:

The airplane’s initial velocity in m/s is

v 0 = 213 km/h

m 1h ( 1000 )( 3600 ) 1 km s

= 59.2 m/s. Applying Newton’s second law, the airplane’s acceleration is a =

dv 1 = − (60,000 + 2.5v 2 ) m/s 2 , dt m

where m = 18,000 kg is its mass. We can separate variables and integrate to determine the time: dv = dt, 60, 000 + 2.5v 2 v0 t dv m∫ = ∫ dt = t. 0 60, 000 + 2.5v 2 0

−m

(Notice that we have reversed the limits of the integral on the left, which eliminated the minus sign.) Writing this equation as

Source: Courtesy of InsectWorld/Shutterstock.

t = m∫

v0 0

dv m v 0 dv = , (1) 2 60, 000 + 2.5v 2.5 ∫ 0 c + v 2

where c = 60,000/2.5, we have from Appendix A v0

∫0 c + v 2 = c 1/ 2 arctan c 1/02 . Using this expression, from Eq. (1) we obtain t = 17.0 s. 17.0 s.

Problem 14.47 The Aerospatiale Puma helicopter weighs 31,000 lb. It takes off vertically from sea level, and after a brief initial acceleration its upward velocity in ft/s is given as a function of its altitude h in feet above the ground by v = 66 − 0.01h. (a) How long does it take the helicopter to climb to an altitude of 3000 ft? (b) The vertical forces on the helicopter include its weight, the upward force exerted by its rotor, and aerodynamic forces. What is the sum of the rotor and aerodynamic forces when its altitude is 3000 ft?

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Source: Courtesy of Paul Horia Malaianu/Shutterstock.

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14.47 (Continued) Solution: (a) Let x denote the altitude. Writing the upward velocity as

(b) From the given equation for the velocity, the helicopter’s vertical acceleration is

dx v = = 66 − 0.01x ft/s, dt

dv dx = −0.01 dt dt = −0.01 v ft/s 2 .

a =

we can separate variables and integrate to determine the time required to reach 3000 ft:

At 3000 ft the velocity is v = 66 − 0.01(3000) = 36 ft/s, so its vertical acceleration is a = −0.01(36) = −0.36 ft/s 2 . Let F denote the sum of the rotor and aerodynamic forces and let W denote the weight. From Newton’s second law,

dx = dt, 66 − 0.01x 3000

∫0

t dx = ∫ dt = t. 0 66 − 0.01x

F −W =

To evaluate the integral on the left, we make the change of variable

W W a = (−0.36 ft/s 2 ), g g

we obtain F = 30,650 lb.

u = 66 − 0.01x,

(a) 60.6 s. (b) 30,650 lb.

du = −0.01dx, obtaining 3000

∫0

66 du dx 1 = 66 − 0.01x 0.01 ∫36 u

( )

1 66 ln 0.01 36

= 60.6 s.

Problem 14.48 In a cathode-ray tube, an electron (mass = 9.11E−31 kg) is projected at O with velocity v = (2.2E7) i (m/s). While the electron is between the charged plates, the electric field generated by the plates subjects it to a force F = −eEj, where the charge of the electron e = 1.6E−19 C (coulombs) and the electric field strength E = 15 kN/C. External forces on the electron are negligible when it is not between the plates. Where does the electron strike the screen?

Solution:

For brevity denote L = 0.03 m, D = 0.1 m. The time

spent between the charged plates is t p =

L 3 × 10 −2 m = = V 2.2 × 10 7 m/s

1.3636 × 10 −9 s. From Newton’s second law, F = m e a p . The accele ration due to the charged plates is ap =

(1.6 × 10 −19 )(15 × 10 3 ) −eE j = − j = j2.6345 × 10 15 m/s 2 . me 9.11 × 10 −31

The velocity is v y = −a pt and the displacement is y =

a pt 2p = − j2.4494 × 2 − 3 10 (m). The velocity is v yp = −a pt = − j3.59246 × 10 6 m/s. The

Screen

exit from the plates the displacement is y p = −

time spent in traversing the distance between the plates and the screen is

2222 x

t ps =

D 10 −1 m = = 4.5455 × 10 −9 s. The vertical displaceV 2.2 × 10 7 m/s

ment at the screen is

1111 30 mm

ap 2 t . At the 2

100 mm

y s = v ypt ps + y p = − j(3.592456 × 10 6 )(4.5455 × 10 −9 ) − j2.4494 × 10 −3 = −18.8 j (mm).

Screen

– – – –

O++++ 30 mm

114

x 100 mm

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Problem 14.49 In Problem 14.48, determine where the electron strikes the screen if the electric field strength is E = 15 sin (ωt ) kN/C, where the frequency ω = 2.0E9 s −1 .

Screen y 2222

Solution:

eE a = − j = − me

(1.6 × 10 −19 C)(15000 sin ωt N/C)

×10 15 ) sin ωt (m/s 2 ).

9.11 × 10 −31 kg The

velocity

vy = j

(2.6345 × 10 15 ) ω

2.6345 × 10 15 j = − j1.3172 2 × 10 9 15 (2.6345 × 10 ) sin ωt + Ct, ×10 6. The displacement is y = j ω2 since y(0) = 0. The time spent between the charged plates is (see 14.48)

t p = 1.3636 × 10 −9 s,

1111 30 mm

100 mm

j = −j(2.6345

cos ωt + C. Since v y (0) = 0 C = −

Problem

Use the solution to Problem 14.48. Assume that the electron enters the space between the charged plates at t = 0, so that at that instant the electric field strength is zero. The acceleration due to the charged plates is

from

2.6345 × 10 15 sin(ωt p ) + Ct p = − j1.531 4 × 10 18 ×10 −3 (m). The time spent between the plates and the screen The displacement is y p = j

is t ps = 4.5455 × 10 −9 s. The vertical deflection at the screen is y s = y p + v ypt ps = −13 j (mm).

which

ωt p = 2.7273 rad. At exit from the plates, the vertical velocity is 2.6345 × 10 15 v yp = j cos(ωt p ) + C = − j2.523 × 10 6 (m/s). 2 × 10 9

Problem 14.50 An astronaut wants to travel from a space station to a satellite S that needs repair. She departs the space station at O. A spring-loaded launching device gives her maneuvering unit an initial velocity of 1 m/s (relative to the space station) in the y direction. At that instant, the position of the satellite is x = 70 m, y = 50 m, z = 0, and it is drifting at 2 m/s (relative to the station) in the x direction. The astronaut intercepts the satellite by applying a constant thrust parallel to the x-axis. The total mass of the astronaut and her maneuvering unit is 300 kg. (a) How long does it take the astronaut to reach the satellite? (b) What is the magnitude of the thrust she must apply to make the intercept? (c) What is the astronaut’s velocity relative to the satellite when she reaches it?

Solution: The path of the satellite relative to the space station is x s (t ) = 2t + 70 m, y s (t ) = 50 m. From Newton’s second law, T = ma x , 0 = ma y . Integrate to obtain the path of the astronaut, using the initial conditions v x = 0, v y = 1 m/s, x = 0, y = 0. y a (t ) = t, x a (t ) =

T 2 t . (a) When the astronaut intercepts the x path 2m

of the satellite, y a (t int ) = y s (t int ), from which t int = 50 s . (b) The intercept of the y-axis path occurs when x a (t int ) = x s (t int ), from which T 2 t = 2t int + 70, from which 2m int  2t + 70  170 T = (2m) int 2 = 40.8 N.  = 2(300)   t int 2500

(

)

(c) The velocity of the astronaut relative to the space station is T v = i t + j = 6.8i + j. The velocity of the satellite relative to m int

( )

the space station is v s = 2 i. The velocity of the astronaut relative to the

satellite is v a /s = i(6.8 − 2) + j = 4.8i + j (m/s).

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Problem 14.51 What is the acceleration of the 8-kg collar A relative to the smooth bar?

Solution:

For brevity, denote θ = 20 °, α = 45 °, F = 200 N, m = 8 kg. The force exerted by the rope on the collar is Frope = 200( i sin θ + j cos θ) = 68.4 i + 187.9 j (N). The force due to

gravity is Fg = −gmj = −78.5 j N. The unit vector parallel to the bar, positive upward, is e B = −i cos α + j sin α. The sum of the forces acting to move the collar is ΣF = Fc = e B ⋅ Frope + e B ⋅ Fg = Frope

208

sin(α − θ) − gm sin α = 29.03 N. The collar tends to slide up the bar since Fc > 0. From Newton’s second law, the acceleration is 200 N

a =

Fc = 3.63 m/s 2 . m

458 208

Frope

N Fg

Problem 14.52 Determine the acceleration of the 8-kg collar A relative to the bar if the coefficient of kinetic friction between the collar and the bar is µ k = 0.1.

Solution:

Use the solution to Problem 14.51. Fc = Frope sin(α − θ) − gm sin α = 29.03 N. The normal force is perpendicular to the bar, with the unit vector e N = i sin α + j cos α. The normal force is N = e N ⋅ Frope + e N ⋅ Fg = Frope cos(α − θ) −gm cos α = 125.77 N. The collar tends to slide up the bar since Fc > 0. The friction force opposes the motion, so that the sum of the forces on the collar is ΣF = Fc − µ k N = 16.45 N. From Newton’s

208

second law, the acceleration of the collar is a =

16.45 = 2.06 m/s 2 8

up the bar. 200 N

A 458

Problem 14.53 The force F = 50 lb. What is the magnitude of the acceleration of the 20-lb collar A along the smooth bar at the instant shown? y

Solution:

Frope = (50 lb)

(5 − 2) i + (3 − 2) j + (0 − 2)k (5 − 2) 2 + (3 − 2) 2 + (0 − 2) 2

The force of gravity is Fgrav = −(20 lb) j (5, 3, 0) ft

The unit vector along the bar is e bar =

A (2, 2, 2) ft

(2 − 2) i + (2 − 0) j + (2 − 4)k (2 − 2) 2 + (2 − 0) 2 + (2 − 4) 2

The component of the total force along the bar is

Fbar = F · e bar = 14.2 lb x

Thus

(2, 0, 4) ft

116

The force in the rope is

14.2 lb =

lb ( 32.220 ft/s )a ⇒ a = 22.9 ft/s . 2

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Problem 14.54* In Problem 14.53, determine the magnitude of the acceleration of the 20-lb collar A along the bar at the instant shown if the coefficient of static friction between the collar and the bar is µ k = 0.2.

Solution:

Use the results from Problem 14.53.

The magnitude of the total force exerted on the bar is F = Frope + Fgrav

= 48.6 lb

The normal force is N =

F 2 − Fbar 2 = 46.5 lb

The total force along the bar is now Fbar − 0.2 N = 4.90 lb

y (5, 3, 0) ft

Thus 4.90 lb =

lb ( 32.220 ft/s )a ⇒ a = 7.89 ft/s . 2

A (2, 2, 2) ft

F x

z (2, 0, 4) ft

Problem 14.55 The 6-kg collar starts from rest at position A, where the coordinates of its center of mass are (400, 200, 200) mm, and slides up the smooth bar to position B, where the coordinates of its center of mass are (500, 400, 0) mm, under the action of a constant force F = −40 i + 70 j − 40 k (N). How long does the collar take to go from A to B?

Solution:

The unit vector from A toward B is e AB = e t = 0.333i +0.667 j − 0.667k and the distance from A to B is 0.3 m. The free body diagram of the collar is shown at the right. There are three forces acting on the collar. These are the applied force F, the weight force W = −mg j = −58.86 j (N), and the force N which acts normal to the smooth bar. Note that N, by its definition, will have no component tangent to the bar. Thus, we need only consider F and W when finding force components tangent to the bar. Also note that N is the force that the bar exerts on the collar to keep it in line with the bar. This will be important in the next problem. The equation of motion for the collar is ΣFcollar = F + W + N = ma. In the direction tangent to the bar, the equation is (F + W) ⋅ e AB = ma t .

The projection of (F + W) onto line AB yields a force along AB which is F AB = 20.76 N. The acceleration of the 6-kg collar caused by this force is a t = 3.46 m/s 2 . We now only need to know how long it takes the collar to move a distance of 0.3 m, starting from rest, with this acceleration. The kinematic equations are v t = a t t, and s t = a t t 2 /2. We set s t = 0.3 m and solve for the time. The time required is t = 0.416 s.

A x

W = –mgJ

F z

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Problem 14.56* In Problem 14.55, how long does the collar take to go from A to B if the coefficient of kinetic friction between the collar and the bar is µ k = 0.2?

Solution:

We use the unit vector e AB from Problem 14.55. The free body diagram for the collar is shown at the right. There are four forces acting on the collar. These are the applied force F, the weight force W = −mg j = −58.86 j (N), the force N which acts normal to the smooth bar, and the friction force f = −µ k N e AB . The normal force must be equal and opposite to the components of the forces F and W which are perpendicular (not parallel) to AB. The friction force is parallel to AB. The magnitude of F + W is calculate by adding these two known forces and then finding the magnitude of the sum. The result is that F + W = 57.66 N. From Problem 14.55, we know that the component of F + W tangent to the bar is F AB = 20.76 N. Hence, knowing the total force and its component tangent to the bar, we can find the magnitude of its component normal to the bar. Thus, the magnitude of the component of F + W normal to the bar is 53.79 N. This is also the magnitude of the normal force N. The equation of motion for the collar is ΣFcollar = F + W + N − µ k N e AB = ma. In the direction tangent to the bar, the equation is (F + W) ⋅ e AB − µ k N = ma t .

W= – mgJ

A x

F z

The force tangent to the bar is FAB = (F + W) ⋅ e AB − µ k N = 10.00 N. The acceleration of the 6-kg collar caused by this force is a t = 1.667 m/s 2 . We now only need to know how long it takes the collar to move a distance of 0.3 m, starting from rest, with this acceleration. The kinematic equations are v t = a t t, and s t = a t t 2 /2. We set s t = 0.3 m and solve for the time. The time required is t = 0.600 s.

mkN

Problem 14.57 The 120-kg crate is drawn across the smooth floor by a winch that retracts the cable at a constant rate of 0.4 m/s. (a) At the instant shown, what is the tension in the cable? (b) What is the tension in the cable when the 4-m horizontal distance has been reduced to 2 m ?

Solution: L

From the figure we see that L2 = x 2 + (2 m) 2 , so for a given value of x we know the value of L. Differentiating this equation with respect to time, we have 2L

dL dx = 2 x , (1) dt dt

and differentiating again gives 2

( dL ) + L ddt L = ( dxdt ) + x ddt x . (2) dt 2

We are told that dL /dt = −0.4 m/s and that d 2 L /dt 2 = 0, so for a given value of x we can use Eqs. (1) and (2) to determine dx /dt and d 2 x /dt 2 . Notice that the acceleration of the crate toward the right is a = −d 2 x /dt 2 .

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14.57 (Continued) Now let’s consider the free-body diagram of the crate:

Newton’s second law is T

( Lx ) = ma.

With this equation we can determine the tension T for a given value of x. (a) When x = 4 m, we obtain a = 0.01 m/s 2 and T = 1.34 N. (b) When x = 2 m, we obtain a = 0.08 m/s 2 and T = 13.6 N.

x mg

(a) 1.34 N. (b) 13.6 N.

Problem 14.58 If y = 100 mm, dy /dt = 600 mm/s, and d 2 y /dt 2 = −200 mm/s 2 , what horizontal force is exerted on the 0.4-kg slider A by the smooth circular slot?

Solution:

displacement is x 2 = R 2 − y 2 . dy dx 2 dx d 2x Differentiate twice with respect to time: x = −y , + x 2 dt dt dt dt 2 2 2 2x 2   y dy d 1 dy d y  from which. = − + 1  = − − y , 2  dt dt x  x dt dt 2 2 y d 2 y Substitute: d x = −1.3612 m/s 2 . From Newton’s second − . dt 2 x dt 2

( ) ( )

The

horizontal

( ) ( )( ) ( )

( )

law, Fh = ma x = −1.361(0.4) = −0.544 N.

y 300 mm

Problem 14.59 The 1-kg collar P slides on the vertical bar and has a pin that slides in the curved slot. The vertical bar moves with constant velocity v = 2 m/s. The y -axis points upward. What are the x and y components of the total force exerted on the collar by the vertical bar and the slotted bar when x = 0.25 m?

Solution: vx =

dx = 2 m/s, constant dt

ax = 0 =

d 2x dt 2

y = 0.2sin(π x ) v y = 0.2π cos(π x )

dx dt

a y = −0.2π 2 sin(π x )

y 5 0.2 sin px

( dxdt ) + 0.2π cos(πx) ddt x 2

when x = 0.25 m,

P x v 1m

ax = 0

( )

π a y = −0.2π 2 sin (2) 2 m/s 4 a y = −5.58 m/s 2

ΣFx : Fx = m(0) ΣFy : Fy − mg = ma y Solving, Fx = 0, Fy = 4.23 N.

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Problem 14.60* The 1200-kg car travels along a straight road of increasing grade whose vertical profile in meters is given by the equation shown. Cruise control keeps the car at a constant speed of 120 km/h. When x = 200 m, what are the x and y components of the total force acting on the car (including its weight)?

y 5 0.0003x2

Solution:

The car’s speed in m/s is

v = 120 km/h

m 1h ( 1000 )( 3600 ) = 33.3 m/s. 1 km s

We write the given equation for the car’s path as y = cx 2 . Taking derivatives with respect to time, we get dy dx = 2cx , (1) dt dt d 2y dx 2 d 2x = 2c + 2cx 2 . (2) dt dt 2 dt

( )

so we know that d 2y d 2x sin α = 0. (5) cos α + 2 dt dt 2 For a given value of x, the angle α can be obtained from Eq. (4). Then Eqs. (1), (2), (3) and (5) can be solved for dx /dt, dy /dt, d 2 x /dt 2 , and d 2 y /dt 2 . Finally, Newton’s second law gives us the components of force: Fx = ma x = m

d 2x , dt 2

Fy = ma y = m

d 2y . dt 2

We know the magnitude of the car’s velocity, so dy 2 dx 2 + = v 2 . (3) dt dt

( )

We also know that the magnitude of its velocity is constant, so the component of its acceleration tangent to the path is zero. The angle between the path and the horizontal is α = arctan

( dydx ) = arctan(2cx), (4)

Problem 14.61 The two 40-kg blocks are released from rest. The contacting surfaces are smooth. What is the horizontal velocity of block B after 0.4 s? Strategy: Use the fact that the components of the accelerations of the two blocks perpendicular to their mutual interface must be equal.

At x = 200 m, we obtain α = 6.84 °, dx /dt = 33.1 m/s, dy /dt = 3.97 m/s, d 2 x /dt 2 = −0.0777 m/s 2 , d 2 y /dt 2 = 0.648 m/s 2 , Fx = −93.3 N, and Fy = 777 N. Fx = −93.3 N, Fy = 777 N.

Solution: aA

NB B Let the accelerations of the blocks be denoted by a A and a B . Applying Newton’s second law to each block, 708

mg − N cos 70 ° = ma A , N sin 70 ° = ma B . We know that the components of acceleration normal to their contacting surfaces must be equal: a A cos 70 ° = a B sin 70 °. We have three equations in terms of a A , a B , and N . Solving, we obtain a A = 8.66 m/s 2 , a B = 3.15 m/s 2 , N = 134 N. After 0.4 s, the velocity of block B is v B = a Bt = (3.15 m/s 2 )(0.4 s) = 1.26 m/s. 1.26 m/s.

120

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Problem 14.62* The two 40-kg blocks are released from rest. The coefficient of kinetic friction between all contacting surfaces is µ k = 0.12. What is the horizontal velocity of block B after 0.4 s?

A B

Solution: 708

NAmk

Nmk N

aB We know that the components of acceleration normal to their contacting surfaces must be equal:

Nmk NBmk

a A cos 70 ° = a B sin 70 °.

Let the accelerations of the blocks be denoted by a A and a B . Applying Newton’s second law to each block, N A − N sin 70 ° + N µ k cos 70 ° = 0,

We have five equations in terms of a A , a B , N , N A , and N B . Solving, we obtain a A = 7.26 m/s 2 , a B = 2.64 m/s 2 , N A = 163 N, N B = 475 N, N = 181 N. After 0.4 s, the velocity of block B is v B = a Bt = (2.64 m/s 2 )(0.4 s) = 1.06 m/s.

mg − N Aµ k − N cos 70 ° − N µ k sin 70 ° = ma A ,

1.06 m/s.

N sin 70 ° − N µ k cos 70 ° − N Bµ k = ma B , −mg − N cos 70 ° − N µ k sin 70 ° + N B = 0.

Problem 14.63 The 3000-lb vehicle has left the ground after driving over a rise. At the instant shown, it is moving horizontally at 30 mi/h and the bottoms of its tires are 24 in above the (approximately) level ground. The Earth-fixed coordinate system is placed with its origin 30 in above the ground, at the height of the vehicle’s center of mass when the tires first contact the ground. (Assume that the vehicle remains horizontal.) When that occurs, the vehicle’s center of mass initially continues moving downward and then rebounds upward due to the flexure of the suspension system. While the tires are in contact with the ground, the force exerted on them by the ground is −2400 i − 18,000 y j (lb), where y is the vertical position of the center of mass in feet. When the vehicle rebounds, what is the vertical component of the velocity of the center of mass at the instant the wheels leave the ground? (The wheels leave the ground when the center of mass is at y = 0.)

24 in 30 in

Solution: This analysis follows that ofExample 14.3. The equation for velocity used to determine how far down the vehicle compresses its springs also applies as the vehicle rebounds. From Example 14.3, we know that the vehicle comes to rest with v Y = 0 and y = 1 ft. Following the Example, the velocity on the rebound is given by vY

∫0 v Y dv Y =∫1 32.2(1 − 6 y) dy. Evaluation, we get v Y = 11.3 ft/s. (+ y is down). Note that the vertical velocity component on rebound is the negative of the vertical velocity of impact.

30 in 24 in

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Problem 14.64* A steel sphere in a tank of oil is given an initial velocity v = 2 i (m/s) at the origin of the coordinate system shown. The radius of the sphere is 15 mm. The density of the steel is 8000 kg/m 3 and the density of the oil is 980 kg/m 3 . If V is the sphere’s volume, the (upward) buoyancy force on the sphere is equal to the weight of a volume V of oil. The magnitude of the hydrodynamic drag force D on the sphere as it falls is D = 1.6 v N, where v is the magnitude of the sphere’s velocity in m/s. What are the x and y components of the sphere’s velocity at t = 0.1 s? Solution:

The volume of the sphere is

V = (4/3)π (0.015 m) 3 = 1.41E − 5 m 3 , so its mass is m = (8000 kg/m 3 )V = 0.113 kg. The net downward force on the sphere due to the effects of gravity and buoyancy is F = = 0.974 N.

With these equations we can separate variables and integrate. In the x direction, m dv ( 1.6 ) v = −dt, x

(8000 − 980 kg/m 3 )V (9.81 m/s 2 )

Consider the free-body diagram of the sphere showing the forces exerted by gravity, buoyancy and hydrodynamic drag. The angle α is the angle between the horizontal and the path and v is the velocity. y

t m v x dv x = −∫ dt, ∫ 0 v 1.6 0 v x

m  v x  ln   = −t, 1.6  v 0  where v 0 = 2 m/s is the initial velocity in the x direction. This equation can be solved for the x component of velocity as a function of time:

v x = v 0 e −(1.6/m )t . (1) In the y direction, m dv y = −dt, 1.6v y + F

a F

m∫

Applying Newton’s second law, we have

vy 0

t dv y = −∫ dt, 0 1.6v y + F

m  1.6v y + F  ln  = −t,  F 1.6 

− D cos α = m

dv x , dt

which can be solved for the y component of velocity as a function of time:

D sin α − F = m

dv y . dt

vy =

The key to the solution is to notice that v cos α = v x and v sin α = −v y . (The y component of the velocity will be negative.) These results mean that, using the given equation for the drag, D cos α = 1.6 v cos α = 1.6v x N,

F −(1.6/m )t [e − 1 ]. (2) 1.6

At t = 0.1 s, from Eqs. (1) and (2) we obtain v x = 0.729 m/s, v y = −0.461 m/s. v x = 0.729 m/s, v y = −0.461 m/s.

D sin α = 1.6 v sin α = −1.6v y N, and Newton’s second law become

122

−1.6v x = m

dv x , dt

−1.6v y − F = m

dv y . dt

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Problem 14.65* In Problem 14.64, what are the x and y coordinates of the sphere at t = 0.1 s?

Solution:

The volume of the sphere is V = (4/3)π (0.015 m) 3 = 1.41E − 5 m 3 , so its mass is

m = (8000 kg/m 3 )V = 0.113 kg. The net downward force on the sphere due to the effects of gravity and buoyancy is F = (8000 − 980 kg/m 3 )V (9.81 m/s 2 ) = 0.974 N. Consider the free-body diagram of the sphere showing the forces exerted by gravity, buoyancy and hydrodynamic drag. The angle α is the angle between the horizontal and the path and v is the velocity.

where v 0 = 2 m/s is the initial velocity in the x direction. This equation can be solved for the x component of velocity as a function of time, v x = v 0 e −(1.6/m )t .

Integrating again yields the x coordinate as a function of time: D

vx =

dx = v 0 e −(1.6/m )t , dt

∫0 dx = ∫0 v 0 e −(1.6/m)t dt, a

x = − v

Applying Newton’s second law, we have − D cos α = m

dv x , dt

D sin α − F = m

dv y . dt

x =

(1)

mv 0 −(1.6/m )t t [e ]0 , 1.6

mv 0 [ 1 − e −(1.6/m )t ]. 1.6

In the y direction, m dv y = −dt, 1.6v y + F m∫

vy 0

t dv y = −∫ dt, 0 1.6v y + F

m  1.6v y + F  ln  = −t,  F 1.6  which can be solved for the y component of velocity as a function of time: F −(1.6/m )t [e − 1 ]. 1.6

D sin α = 1.6 v sin α = −1.6v y N,

vy =

and Newton’s second law become

This expression can be integrated to determine y as a function of time:

−1.6v x = m

dv x , dt

−1.6v y − F = m

dv y . dt

dy F −(1.6/m )t [e = − 1 ], dt 1.6 y

With these equations we can separate variables and integrate. In the x direction, m dv ( 1.6 ) v = −dt, x

t m v x dv x = −∫ dt, ∫ 0 v 1.6 0 v x

m  v x  ln   = −t, 1.6  v 0 

BandF_6e_ISM_C14_Pt1.indd 123

∫ dy = ∫ 1.6 [ e −(1.6/m)t − 1 ]dt, 0

t F  e −(1.6/m )t − t , y =   0 1.6  −(1.6/m) y =

(2)

mF  −(1.6/m)t 1.6t − + 1  . −e  m (1.6) 2 

At t = 0.1 s, from Eqs. (1) and (2) we obtain x = 161 mm, y = −28.3 mm. x = 161 mm, y = −28.3 mm.

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Problem 14.66 The powerboat is moving at a constant speed of 8 m/s in a circular path with radius R = 30 m. The mass of the boat with its passengers is 420 kg. Determine the tangential and normal components of force acting on the boat.

Solution:

We are told that the tangential component of acceleration dv /dt = a t = 0. The normal component of acceleration is (8 m/s) 2 v2 = R 30 m = 2.13 m/s 2 .

an =

The components of force acting on the boat are Ft = ma t = 0, Fn = ma n = (420 kg)(2.13 m/s 2 ) = 896 N (201 lb). Ft = 0, Fn = 896 N.

Source: Courtesy of Sven Hansche/123RF.

Problem 14.67 In preliminary design studies for a sun-powered car, it is estimated that the mass of the car and driver will be 100 kg and the torque produced by the engine will result in a 60-N tangential force on the car. Suppose that the car starts from rest on the track at A and is subjected to a constant 60-N tangential force. Determine the magnitude of the car’s velocity and the normal component of force on the car when it reaches B.

Solution:

We first find the tangential acceleration and use that to find the velocity at B. Ft = ma t ⇒ 60 N = (100 kg) a t ⇒ a t = 0.6 m/s 2 , s v v2 dv ⇒ ∫ v d v = ∫ a t ds ⇒ = a t s, 0 0 ds 2 π v B = 2a t s B = 2(0.6 m/s 2 ) 200 m + [50 m] = 18.3 m/s. 2

at = v

(

)

The normal component of the force is Fn = ma n = m 50 m

(18.3 m/s) 2 v2 = (100 kg) = 668 N. R 50 m

v B = 18.3 m/s, Fn = 668 N.

200 m

Problem 14.68 In a test of a sun-powered car, the mass of the car and driver is 100 kg. The car starts from rest on the track at A, moving toward the right. The tangential force exerted on the car (in newtons) is given as a function of time by Σ Ft = 20 + 1.2t. Determine the magnitude of the car’s velocity and the normal component of force on the car at t = 40 s.

Solution:

We first find the tangential acceleration and use that to find the velocity v and distance s as functions of time. ΣFt = (20 + 1.2t ) N = (100 kg) a t dv = 0.2 + 0.012t dt v = 0.2t + 0.006t 2

at =

s = 0.1t 2 + 0.002t 3 At t = 40 s, we have v = 17.6 m/s, s = 288 m.

50 m

For this distance the car will be on the curved part of the track. The normal component of the force is B

Fn = ma n = m

A 200 m

124

(17.6 m/s) 2 v2 = (100 kg) = 620 N. R 50 m

v B = 17.6 m/s, Fn = 620 N.

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Problem 14.69 An astronaut candidate with a mass of 72 kg is tested in a centrifuge with a radius of 10 m. The centrifuge rotates in the horizontal plane. It starts from rest at time t = 0 and has a constant angular acceleration of 0.2 rad/s 2 . Determine the magnitude of the horizontal force exerted on the candidate by the centrifuge (a) at t = 0; (b) at t = 10 s.

Solution: (a) At t = 0 he’s stationary, there’s no horizontal force on him. (b) Because the angular acceleration α = 0.2 rad/s 2 is constant, the angular velocity at t = 10 s is ω = αt = (0.2 rad/s 2 )(10 s) = 2 rad/s. The tangential component of his acceleration is a t = rα = (10 m)(0.2 rad/s 2 ) = 2 m/s 2 , and the normal component is a n = r ω 2 = (10 m)(2 rad/s) 2 = 40 m/s 2 . His total acceleration in the horizontal plane is a t2 + a n2 =

Source: Courtesy of NASA Ames/Dominic Hart.

(2 m/s 2 ) 2 + (40 m/s 2 ) 2

= 40.0 m/s 2 . The horizontal force exerted on him is m a = (72 kg)(40.0 m/s 2 ) = 2880 N (648 lb). (a) Zero. (b) 2880 N.

Problem 14.70 The circular disk lies in the horizontal plane. At the instant shown, the disk rotates with a counterclockwise angular velocity of 4 rad/s and a counterclockwise angular acceleration of 2 rad/s 2 . The 0.5-kg slider A is supported horizontally by the smooth slot and the string attached at B. Determine the tension in the string and the magnitude of the horizontal force exerted on the slider by the slot.

2 rad/s2 4 rad/s

A B 0.6 m

Solution: ω = 4 rad/s α = 2 rad/s 2 R = 0.6 m m = 0.5 kg ΣFn : T = m Rω 2 ΣFt : F = m Rα Solving, T = 4.8 N, F = 0.6 N.

4 rad/s 0.6 m

0.2 rad/s 2 F

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Problem 14.71 The circular disk lies in the horizontal plane and rotates with a constant counterclockwise angular velocity of 4 rad/s. The 0.5-kg slider A is supported horizontally by the smooth slot and the string attached at B. Determine the tension in the string and the magnitude of the horizontal force exerted on the slider by the slot.

Solution: R = 0.6 m ω = 4 rad/s m = 0.5 kg α = 0 ΣFn : N cos 45 ° + T sin 45 ° = mRω 2 ΣFt : −N sin 45 ° + T cos 45 ° = mRα = 0 Solving, N = T = 3.39 N.

0.6 m T

908

4 rad/s

v = 4 rad/s

0.6 m

Problem 14.72 The 62,000-lb Gulfstream is flying in the vertical plane at 420 ft/s. At the instant shown the angle θ = 35 ° and the cartesian components of the plane’s acceleration are a x = 4 ft/s 2 , a y = 40 ft/s 2 . (a) What are the tangential and normal components of the total force acting on the airplane (including its weight)? (b) What is d θ /dt in degrees per second?

Solution: (a) Knowing the cartesian components of the acceleration and the angle θ, we can determine the normal and tangential components: a t = a x cos θ + a y sin θ = 26.2 ft/s 2 , a n = −a x sin θ + a y cos θ = 30.5 ft/s 2 . The tangential and normal components of the total force are W a = 50,500 lb, g t W ΣFn = a = 58,700 lb. g n ΣFt =

(b) In terms of the magnitude v of the plane’s velocity, the x component of the velocity is v x = v cos θ. Taking the derivative of this expression with respect to time, we obtain

dv x dv dθ cos θ − v sin θ: = dt dt dt dθ a x = a t cos θ − v sin θ. dt We can solve this equation for d θ /dt, obtaining d θ /dt = 0.0726 rad/s = 4.16 deg/s.

(a) ΣFt = 50,500 lb, ΣFn = 58,700 lb. (b) 4.16 deg/s.

Problem 14.73 A person with a mass of 52 kg is in the space station described in Example 14.8. When she is in the outer ring with radius R = 100 m, her simulated weight is 200 N. (a) What is the station’s angular velocity in rad/s? (b) Suppose that she climbs a ladder through one of the radial tunnels leading to the center of the station. Let r be her distance from the center of the station. What is her simulated weight as a function of r in meters?

a n = v 2 /r = r ω 2 . The floor would press “up” on her, her simulated weight, with force N = ma n = mω 2r. If R denotes the radius of the station’s outer ring, her weight there is N = mω 2 R : 200 N = (52 kg)ω 2 (100 m). Solving, the angular velocity of the station is ω = 0.196 rad/s. (This is one revolution every 32.0 seconds.) (b) Her simulated weight at radial distance r is N = mω 2r

Solution: (a) When she is at a radial distance r, the magnitude of her velocity due to her circular motion in the rotating station is v = ωr, where ω is the station’s angular velocity. The normal component of her acceleration, toward the center of the station, is

126

= (52 kg)(0.196 rad/s) 2 r = 2r N. Notice that her simulated weight at the center of the station is (of course) zero. (a) 0.196 rad/s. (b) 2 r newtons.

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Problem 14.74 Small parts on a conveyer belt moving with constant velocity v are allowed to drop into a bin. Show that the angle θ at which the parts start sliding on the belt satisfies the equation cos θ −

v2 1 sin θ = , gR µs

where µ s is the coefficient of static friction between the parts and the belt.

Solution:

The condition for sliding is ΣFt = −mg sin θ + f = 0, where −mg sin θ is the component of weight acting tangen tially to the belt, and f = µ s N is the friction force tangential to the belt. From Newton’s second law the force perpendicular v2 to the belt is N − mg cos θ = −m , from which the condiR v2 tion for slip is −mg sin θ + µ s mg cos θ − µ s m = 0. Solve: R 1 v2 cos θ − sin θ = µs gR

Problem 14.75 The 1-slug mass m rotates around the vertical pole in a horizontal circular path. The angle θ = 30 ° and the length of the string is L = 4 ft. What is the magnitude of the velocity of the mass? Strategy: Notice that the vertical acceleration of the mass is zero. Draw the free-body diagram of the mass and write Newton’s second law in terms of tangential and normal components.

u W

Solution: ΣF↑ : T cos30 ° − mg = 0 ΣFn : T sin 30 ° = m

v2 v2 = m L sin 30 ° ρ

Solving we have T = v =

mg , v 2 = g( L sin 30 °) tan 30 ° cos 30 ° (32.2 ft/s 2 )(4 ft) sin 2 30 ° = 6.10 ft/s. cos30 ° T

L 308

Problem 14.76 The 12-kg mass m swings around the vertical pole in a horizontal circular path with a constant speed of 3 m/s. The length of the string is L = 1.5 m. Determine the angle θ and the tension in the string.

Solution:

There is no acceleration in the vertical direction, so

T cos θ = mg.

In terms of the length L = 1.5 m of the string, the radius of the circular path of the mass is R = L sin θ.

(2)

There is no force in the tangential e t direction, so the magnitude of the velocity of the mass is constant: dv /dt = 0. Newton’s second law in the normal e n direction is v2 T sin θ = ma n = m . R Substituting Eq. (2), this becomes

BandF_6e_ISM_C14_Pt2.indd 127

(1)

TL sin 2 θ = mv 2 .

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14.76 (Continued) Solving Eq. (1) for T and substituting the result into this equation gives gL sin 2 θ

v 2 cos θ,

gL (1 − cos 2 θ) = v 2 cos θ.

The velocity is given, so we can solve this quadratic equation for cos θ, obtaining cos θ = 0.740. The angle θ = 42.3 °. Equation (1) gives the tension T = 159 N. θ = 42.3 °, 159 N.

Problem 14.77 The 10-kg mass m rotates around the vertical pole in a horizontal circular path of radius R = 1 m. If the magnitude of the velocity of the mass is v = 3 m/s, what are the tensions in the strings A and B?

Solution:

Choose a Cartesian coordinate system in the vertical plane that rotates with the mass. The weight of the mass is W = − jmg = − j98.1 N. The radial acceleration is by definition v2 directed inward: a n = −i = −9 i m/s 2 . The angles from the R horizontal are θ A = 90 ° + 35 ° = 125 °, θ B = 90 ° + 55 ° = 145 °. The unit vectors parallel to the strings, from the pole to the mass, are: e A = +i cos θ A + j sin θ A . e B = +i cos θ B + j sin θ B . From Newton’s

( )

second law for the mass, T − W = ma n , from which T A e A + TB e B v2 − jmg = −i m . Separate components to obtain the two simultaneR ous equations: T A cos125 ° + TB cos145 ° = −90 N T A sin 55 ° +

(

358 A B

)

TB sin 35 ° = 98.1 N. Solve: T A = 84 N.

558

TB = 51 N. y

R TA

TB uB uA x –j mg

Problem 14.78 The 10-kg mass m rotates around the vertical pole in a horizontal circular path of radius R = 1 m. For what range of values of the velocity v of the mass will the mass remain in the circular path described?

Solution: The minimum value of v will occur when the string B is impending zero, and the maximum will occur when string A is impending zero. From the solution to Problem 14.77, T A cos125 ° + TB cos145 ° = −m

( vR ), 2

T A sin125 ° + TB sin145 ° = mg. These equations are to be solved for the velocity when one of the string tensions is set equal to zero. For T A = 0, v = 3.743 m/s. For TB = 0, v = 2.621 m/s. The range: 2.62 ≤ v ≤ 3.74 m/s.

358 A B 558

R m

128

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Problem 14.79 Suppose you are designing a monorail transportation system that will travel at 50 m/s, and you decide that the angle θ that the cars swing out from the vertical when they go through a turn must not be larger than 20 °. If the turns in the track consist of circular arcs of constant radius R, what is the minimum allowable value of R? (See Practice Example 14.7.)

Solution:

The equations of motion are

ΣFy : T cos θ − mg = 0 ΣFn : T sin θ = ma n = m

v2 R

Solving we have R =

(50 m/s) 2 v2 = = 700 m g tan θ (9.81 m/s 2 ) tan 20 °

R = 700 m.

Problem 14.80 An airplane of weight W = 200, 000 lb makes a turn at constant altitude and at constant velocity v = 600 ft/s. The bank angle is 15 °. (a) Determine the lift force L. (b) What is the radius of curvature of the plane’s path?

Solution: (a) The plane’s acceleration in the vertical direction is zero, so L cos15 ° − W = 0, giving L = 207,000 lb. (b) Newton’s second law in the direction normal to the plane’s path is

158

L sin15 ° =

W a , g n

which yields a n = 8.63 ft/s 2 . From the relation

an =

v2 , R

we obtain R = 41,700 ft. W

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(a) 207,000 lb. (b) 41,700 ft.

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Problem 14.81 The suspended 2-kg mass m is stationary. (a) What are the tensions in the strings A and B? (b) If string A is cut, what is the tension in string B immediately afterward?

Solution: (a)

458

∑ Fx = TB cos 45° − T A = 0, ∑ Fy = TB sin 45° − mg = 0. Solving yields T A = 19.6 N, TB = 27.7 N.

(b) Use Normal and tangential components.

∑ Fn = ma n :

v2 . ρ But v = 0 at the instant of release, so TB − mg cos 45 ° = m

458

TB = mg cos 45 ° = 13.9 N.

mg TB

et mg

Problem 14.82 The airplane flies with constant velocity v along a circular path in the vertical plane. The radius of the airplane’s circular path is 2000 m. The mass of the pilot is 72 kg. (a) The pilot will experience “weightlessness” at the top of the circular path if the airplane exerts no net force on him at that point. Draw a free-body diagram of the pilot and use Newton’s second law to determine the velocity v necessary to achieve this condition. (b) Suppose that you don’t want the force exerted on the pilot by the airplane to exceed four times his weight. If he performs this maneuver at v = 200 m/s, what is the minimum acceptable radius of the circular path? Solution:

The FBD assumes that the seat pushes up on the pilot. If the seat (or shoulder straps) pushes down, we will us a negative sign for N. v2 Dynamics: ∑ F↑ : N − mg = −m ρ (a)

N = 0 ⇒ v =

gρ =

(9.81 m/s 2 ) (2000 m) = 140 m/s.

(b) The force will push down on the pilot v2 v2 N = −4 mg ⇒ −5mg = −m ⇒ ρ = ρ = 815 m. 5g ρ

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Problem 14.83 The smooth circular bar rotates with constant angular velocity ω 0 about the vertical axis AB. The radius R = 0.5 m. The mass m remains stationary relative to the circular bar at β = 40 °. Determine ω 0 .

Solution:

N mg The forces on the free-body diagram of the mass are its weight mg and the normal force N exerted on it by the bar. The mass moves in a circular path with constant speed, so its acceleration in the direction tangent to its path is zero. That means the bar exerts no force on the mass in the tangential direction (normal to the circular bar); the normal force N points toward the center of the bar.

The mass has no acceleration in the vertical direction, so

N cos β = mg.

(1)

The mass moves in a circular path of radius R sin β , so its velocity is

v = ω 0 R sin β . Newton’s second law in the direction normal to the path is v2 N sin β = ma n = m . R sin β Using Eqs. (1) and (2) to eliminate N and v from Eq. (3), we get g = ω0 2. R cos β

(2)

(3)

From this equation we obtain ω0 = =

g R cos β 9.81 m/s 2 ( 0.5 m ) cos 40 °

= 5.06 rad/s. ω 0 = 5.06 rad/s.

Problem 14.84 The force exerted on a charged particle by a magnetic field is F = q v × B, where q and v are the charge and velocity vector of the particle and B is the magnetic field vector. A particle of mass m and positive charge q is projected at O with velocity v = v 0 i into a uniform magnetic field B = B 0 k. Using normal and tangential components, show that (a) the magnitude of the particle’s velocity is constant and (b) the particle’s path is a circle with radius mv 0 /qB 0 . y

v0 O

BandF_6e_ISM_C14_Pt2.indd 131

Solution:

(a) The force F = q(v × B) is everywhere normal to the velocity and the magnetic field vector on the particle path. Therefore the tangential component of the force is zero, hence from Newton’s secdv ond law the tangential component of the acceleration = 0, from dt which v (t ) = C = v 0 , and the velocity is a constant. Since there is no component of force in the z-direction, and no initial z-component of the velocity, the motion remains in the x -y plane. The unit vector k is positive out of the paper in the figure; by application of the right hand rule the cross product v × B is directed along a unit vector toward the instantaneous center of the path at every instant, from which F = − F e n , where e n is a unit vector normal to the path. The normal component of the acceleration is a n = −(v 02 /ρ )e n , where ρ is the radius of curvature of the path. From Newton’s second law, F = ma n , from which − F = −m(v 02 /ρ ). The magnitude of the cross product can be written as v × B = v 0 B 0 sin θ = v 0 B 0 , since v2 θ = 90 ° is the angle between v and B. From which: qv 0 B 0 = m 0 , ρ mv 0 from which the radius of curvature is ρ = . Since the term on the qB 0 right is a constant, the radius of curvature is a constant, and the path is a mv 0 circle with radius . qB 0

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Problem 14.85 The mass m is attached to a string that is wrapped around the fixed post of radius R. At t = 0, the mass is given a velocity v 0 as shown. Neglect external forces on m other than the force exerted by the string. Determine the tension in the string as a function of the angle θ. Strategy: The velocity vector of the mass is perpendicular to the string. Express Newton’s second law in terms of normal and tangential components.

Solution:

Make a hypothetical cut in the string and denote the tension in the part connected to the mass by T. The acceleration normal to the v2 path is . The instantaneous radius of the path is ρ = L 0 − Rθ, from ρ v 02 which by Newton’s second law, ∑ Fn = T = m , from which L 0 − Rθ 2 v0 T = m . L 0 − Rθ L0

m u

L0 R

(b)

Problem 14.86 The mass m is attached to a string that is wrapped around the fixed post of radius R. At t = 0, the mass is given a velocity v 0 as shown. Neglect external forces on m other than the force exerted by the string. Determine the angle θ as a function of time.

R v0

Solution:

Use the solution to Problem 14.85. The angular velocity is v x (t ) = 26.7 ft/s, x (t ) = 26.7t. From Problem 14.85, v0 dθ ρ = L 0 − Rθ, from which = . Separate variables: dt ( L 0 − Rθ) R ( L 0 − Rθ)d θ = v 0 dt. Integrate: L 0 θ − θ 2 = v 0 t, since θ(0) = 0. 2 2v 0 t L In canonical form θ 2 + 2bθ + c = 0, where b = − , and c = . R R 2 L L0 2v t The solution: θ = −b ± b 2 − c = 0 ± − 0 . R R R The angle increases with time, so the negative sign applies. Reduce: L  2 Rv 0 t  θ = 0  1 − 1 −  Check: When Rθ = L 0 , the string has been R L20  2 Rv 0 t fully wrapped around the post. Substitute to obtain: 1 − = 0, L20 2 Rv 0 t from which = 1, which is the value for impending failure as t L20 increases because of the imaginary square root. Thus the solution behaves as expected. check.

( )

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Problem 14.87 The Ferrari 488 with its driver weighs 3200 lb. The static coefficient of friction between its tires and the road is µ s = 1.28. If you assume that the normal force between its tires and the road equals the car’s weight, what is the minimum theoretical distance the car requires to stop from 60 mi/h?

Solution: If N = W = 3200 lb denotes the normal force, the road exerts a braking force µ s N. Newton’s second law is W −µ s N = a, g so the car’s deceleration is a = −µ s g. We can integrate the constant acceleration to obtain the velocity and position as functions of time: a =

dv = −µ s g, dt

∫v d v = ∫0 −µ s g dt,

(1)

v =

ds = v 0 − µ s gt, dt

∫0 ds = ∫0 (v 0 − µ s gt ) dt, s = v 0t −

1 µ gt 2 . 2 s

(2)

Setting v = 0 in Eq. (1), the time required to stop is t = v 0 / ( µ s g ). Substituting this into Eq. (2), the distance required is

Source: Courtesy of Art Konovalov/Shutterstock.

 v  1  v 2 s = v 0  0  − µ s g  0   µsg  2  µsg  =

v0 2 . 2µ s g

Setting v 0 = 88 ft/s, g = 32.2 ft/s 2 , and µ s = 1.28, we obtain s = 93.9 ft. 93.9 ft.

Problem 14.88 Using the data in Problem 14.87 and making the same assumptions, what is the highest constant speed at which the car could drive on a flat, circular track of 600-ft radius without skidding?

Solution: Let R = 600 ft denote the track’s radius and v the car’s velocity. Newton’s second law in the normal direction is W Wv2 Fn = a = . g n gR Setting Fn = µ s N = µ sW, g = 32.2 ft/s 2 , and µ s = 1.28, we obtain v = 157 ft/s (107 mi/h). 157 ft/s.

Source: Courtesy of Art Konovalov/Shutterstock.

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Problem 14.89 The freeway off-ramp is circular with 60-m radius (Fig. a). The off-ramp has a slope β = 10 ° (Fig. b). If the coefficient of static friction between the tires of a car and the road is µ s = 0.45, what is the maximum constant speed at which it can enter the ramp without losing traction? (See Example 14.9.)

Solution:

As a car enters the ramp at higher and higher speeds, a larger and larger friction force toward the concave side of the road is needed to prevent the car from sliding outward. The maximum speed is the one for which the friction force has its maximum possible value:

msN

N mg

The car has no acceleration in the vertical direction, so we have N cos β − mg − µ s N sin β = 0. (1) In terms of the radius R of the road, the car’s normal component of acceleration is v2 an = . R Newton’s second law in the normal direction is v2 N sin β + µ s N cos β = ma n = m . (2) R Dividing Eqs. (1) and (2) by the mass m gives

60 m

(a)

( mN )cos β − g − µ ( mN )sin β = 0, (3) ( Nm )sin β + µ ( Nm )cos β = vR . (4) s

We can solve Eq. (3) for N /m, then solve Eq. (4) for the velocity, obtaining v = 20.0 m/s. v = 20.0 m/s.

(b)

Problem 14.90 The freeway off-ramp is circular with 60-m radius (Fig. a). The off-ramp has a slope β = 10 ° (Fig. b). A car enters the ramp when it is icy, and as a result the coefficient of static friction between the tires of the car and the road is reduced to µ s = 0.12. What is the minimum constant speed at which the car can negotiate the ramp without sliding?

Solution: The danger is that if the car is going too slowly it will lose traction and slide in the down-slope direction. A friction force in the opposite direction resists that. The minimum velocity is when the friction force has its maximum value: msN

N mg The car has no acceleration in the vertical direction, so we have N cos β − mg + µ s N sin β = 0. (1) In terms of the radius R of the road, the car’s normal component of acceleration is v2 an = . R Newton’s second law in the normal direction is

60 m

N sin β − µ s N cos β = ma n = m

(a)

v2 . (2) R

Dividing Eqs. (1) and (2) by the mass m gives

( Nm )cos β − g + µ ( Nm )sin β = 0, (3) ( mN )sin β − µ ( Nm )cos β = vR . (4) s

We can solve Eq. (3) for N /m, then solve Eq. (4) for the velocity, obtaining v = 5.70 m/s. v = 5.70 m/s.

(b)

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Problem 14.91 A car traveling at 30 m/s is at the top of a hill. The coefficient of kinetic friction between the tires and the road is µ k = 0.8. The instantaneous radius of curvature of the car’s path is 200 m. If the driver applies the brakes and the car’s wheels lock, what is the resulting deceleration of the car in the direction tangent to its path?

Solution:

Consider the free-body diagram of the car after the brakes

are applied:

mkN

Newton’s second law in the direction tangent to the car’s path is −µ k N = ma t . (1) The concave side of the car’s path is downward, so Newton’s second law in the normal direction is v2 mg − N = ma n = m . (2) ρ We can solve Eq. (2) for N and substitute the result into Eq. (2) to obtain a t = −4.25 m/s 2 . 4.25 m/s 2 .

Problem 14.92 A car traveling at 30 m/s is at the bottom of a depression. The coefficient of kinetic friction between the tires and the road is µ k = 0.8. The instantaneous radius of curvature of the car’s path is 200 m. If the driver applies the brakes and the car’s wheel’s lock, what is the resulting deceleration of the car in the direction tangential to its path? Compare your answer to that of Problem 14.91.

Solution:

Consider the free-body diagram of the car after the brakes

are applied:

mkN

Newton’s second law in the direction tangent to the car’s path is −µ k N = ma t . (1) The concave side of the car’s path is upward, so Newton’s second law in the normal direction is v2 N − mg = ma n = m . (2) ρ We can solve Eq. (2) for N and substitute the result into Eq. (1) to obtain a t = −11.4 m/s 2 . The difference in the curvatures of the road causes the normal force N to be substantially larger in this case compared to Problem 14.91, resulting in a much larger friction force. 11.4 m/s 2 .

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Problem 14.93 The combined mass of the motorcycle and rider is 160 kg. The motorcycle starts from rest at t = 0 and moves along a circular track with a 400-m radius. The tangential component of acceleration of the motorcycle as a function of time is a t = 2 + 0.2t m/s 2 . The coefficient of static friction between the tires and the track is µ s = 0.8. How long after it starts does the motorcycle reach the limit of adhesion, which means that its tires are on the verge of slipping? How fast is the motorcycle moving when that occurs? Strategy: Draw a free-body diagram showing the tangential and normal components of force acting on the motorcycle.

Solution: m = 160 kg R = 400 m Along track motion: a t = 2 + 0.2t m/s 2 Vt = V = 2t + 0.1t 2 m/s s = t 2 + 0.1t 3 /3 m Forces at impending slip F + f = µ k N at impending slip F+ f =

F 2 + f 2 since f ⊥ F

Force eqns. ∑ Ft : F = ma t

R = 400 m

∑ Fn : f =

m = 160 kg

mv 2 /R

∑ FZ : N − mg = 0

g = 9.81 m/s 2 µ s = 0.8

+ = µs N a t = 2 + 0.2t F2

v = 2t + 0.1t 2 Six eqns, six unknowns ( F , f , v , a t , N , t ) Solving, we have t = 14.4 s

F = 781 N

v = 49.6 m/s

f = 983 N N = 1570 N a t = 4.88 m/s 2

Source: Courtesy of ATTILA Barsan/Shutterstock.

(at t = 14.4 s) mg

en P

400 m V 2/R

f N

f et

136

V 2/ R en

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Problem 14.94 The center of mass of the 12-kg object moves in the x – y plane. Its polar coordinates are given as functions of time by r = 12 − 0.4t 2 m, θ = 0.02t 3 rad. Determine the polar components of the total force acting on the object at t = 2 s. y

Solution:

Using the given expressions, we evaluate the necessary

derivatives: d 2r = −0.8 m/s 2 . dt 2 dθ d 2θ θ = 0.02t 3 rad, = 0.06t 2 rad/s, = 0.12t rad/s 2 . dt dt 2 Newton’s second law in the radial direction is r = 12 − 0.4t 2 m,

dr = −0.8t m/s, dt

ΣFr = ma r  d 2r dθ 2  = m 2 − r   dt dt  = m[−0.8 − (12 − 0.4t 2 )(0.06t 2 ) 2 ] N.

( )

At t = 2 s we obtain ΣFr = −16.8 N. r

Newton’s second law in the transverse direction is ΣFθ = ma θ

d 2θ dr d θ  = m  r 2 + 2  dt dt dt  = m[(12 − 0.4t 2 )(0.12t ) + 2(−0.8t )(0.06t 2 )] N.

At t = 2 s we obtain ΣFθ = 20.7 N. ΣFr = −16.8 N, ΣFθ = 20.7 N.

Problem 14.95 A 150-lb person walks on a large disk that rotates with constant angular velocity ω 0 = 0.5 rad/s. He walks at a constant speed (relative to the disk) v 0 = 4 ft/s along a straight radial line painted on the disk. If the coefficient of static friction between his shoes and the polished disk is µ s = 0.14, what maximum radial distance does he reach before he slips?

Solution:

We know that

dr d 2r = 4 ft/s, = 0. dt dt 2 dθ d 2θ = ω 0 = 0.5 rad/s, = 0. dt dt 2 Newton’s second law in the radial direction is ΣFr = ma r  d 2r dθ 2  = m 2 − r   dt dt  2 = −mr ω 0 lb.

( )

Newton’s second law in the transverse direction is ΣFθ = ma θ d 2θ dr d θ  = m  r 2 + 2  dt dt dt  = 2m (4 ft/s) ω 0 lb. The magnitude of the total horizontal force is (ΣFr ) 2 + (ΣFθ ) 2 = m r 2 ω 0 4 + 64 ω 0 2 . The only way this horizontal force can be exerted on him is as a friction force exerted on his feet by the disk. He will slip if this force exceeds µ s N, where N = mg is the normal force exerted on him by the disk. Therefore the maximum radius to which he can walk is can be obtained the equation (ΣFr ) 2 + (ΣFθ ) 2 = m r 2 ω 0 4 + 64 ω 0 2 = µ s mg, which we can solve for r: r =

µ s 2 g 2 − 64 ω 0 2 . ω0 4

r = 8.32 ft.

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Problem 14.96 The robot is programmed so that the 0.4-kg part A describes the path r = 1 − 0.5cos 2πt m,

ar =

d 2r dt 2

−r

The radial component of the acceleration is 2

( ddtθ ) .

The derivatives:

θ = 0.5 − 0.2sin 2πt rad. Determine the polar components of force exerted on A by the robot’s jaws at t = 2 s.

Solution:

A u

dr d = (1 − 0.5cos 2πt ) = π sin 2πt, dt dt 2 d r d = (π sin 2πt ) = 2π 2 cos 2πt; dt 2 dt dθ d = (0.5 − 0.2sin 2πt ) = −0.4 π cos 2πt. dt dt d 2θ d = (−0.4 π cos 2πt ) = 0.8π 2 sin 2πt. dt 2 dt From which [a r ] t = 2 = 2π 2 cos 4 π − (1 − 0.5cos 4 π )(−0.4 π cos 4 π ) 2 , = 2π 2 − 0.08π 2 = 18.95 m/s 2 ; θ(t = 2) = 0.5 rad. From Newton’s second law, Fr − mg sin θ = ma r , and Fθ − mg cosθ = ma θ , from which Fr = 0.4 a r + 0.4 g sin θ = 9.46 N. The transverse component of the acceleration is aθ = r

( ddt θ ) + 2( drdt )( ddtθ ), 2

from which [a θ ] t = 2 = (1 − 0.5cos 4 π )(0.8π 2 sin 4 π ) + 2(π sin 4 π ) (−0.4 π sin 4 π ) = 0, and Fθ = 3.44 N.

Fr u

Problem 14.97 A 50-lb object P moves along the spiral path r = (0.1)θ ft, where θ is in radians. Its angular position is given as a function of time by θ = 2t rad, and r = 0 at t = 0. Determine the polar components of the total force acting on the object at t = 4 s.

Solution: θ = 2t, θ = 2, θ = 0, r = 0.1θ = 0.2t, r = 0.2, r = 0 At t = 4 s, θ = 8, θ = 2, θ = 0, r = 0.8, r = 0.2, r = 0 Thus

 50 lb  Fr = m(r − rθ 2 ) =  (0 − [0.8 ft][2 rad/s] 2 )  32.2 ft/s 2  = −4.97 lb.

r u

 50 lb  Fθ = m(rθ + 2r θ ) =  (0 + 2[0.2 ft/s][2 rad/s])  32.2 ft/s 2  = 1.24 lb.

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Problem 14.98 The smooth bar rotates in the horizontal plane with constant angular velocity ω 0 = 60 rpm. If the radial velocity of the 1-kg collar A is v r = 10 m/s when its radial position is r = 1 m, what is its radial velocity when r = 2 m? (See Practice Example 14.10.) v0

Solution: Notice that no radial force acts on the collar, so the radial acceleration is zero. Write the term dv r dv r dr dv d 2r = = = vr r dt 2 dt dr dt dr The angular velocity is ω = 60 rpm

( 2πrevrad )( 160mins ) = 6.28 rad/s.

Then d 2r d 2r − rω 2 = 0 ⇒ = rω 2 2 dt dt 2 vr 2m dv d 2r = v r r = rω 2 ⇒ ∫ v r d v r = ∫ ω 2r dr 10 m / s 1m dt 2 dr (10 m/s) 2 [1 m] 2  v r2  [2 m] 2 − = (6.28 rad/s) 2  −   2 2 2 2  ar =

Problem 14.99 The smooth bar rotates in the horizontal plane with constant angular velocity ω 0 = 60 rpm. The spring constant is k = 20 N/m and the unstretched length of the spring is 3 m. If the radial velocity of the 1-kg collar A is v r = 10 m/s when its radial position is r = 1 m, what is its radial velocity when r = 2 m? (See Practice Example 14.10.)

v r = 14.8 m/s.

Solution:

Notice that the only radial force comes from the spring.

Write the term dv r dv r dr dv d 2r = = = vr r dt 2 dt dr dt dr The angular velocity is ω = 60 rpm

( 2πrevrad )( 160mins ) = 6.28 rad/s.

The equation of motion in the radial direction is v0 ΣFrm : − kr = ma r ⇒ a r = − k

Then

(

)

d 2r k d 2r k − rω 2 = − r ⇒ = r ω2 − dt 2 m dt 2 m vr 2m dv d 2r k k = vr r = r ω2 − ⇒ ∫ rdr v r dv r = ∫ ω2 − 2 10 m/s 1 m dt dr m m  (10 m/s) 2 20 N/m  [2 m] 2 [1 m] 2 v r2  − − =  (6.28 rad/s) 2 −  2 2 1 kg  2 2 ar =

k r m

)

(

)

v r = 12.6 m/s.

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Problem 14.100 The 2-kg mass m is released from rest with the string horizontal. The length of the string is L = 0.6 m. By using Newton’s second law in terms of polar coordinates, determine the magnitude of the velocity of the mass and the tension in the string when θ = 45 °. Strategy: Use the chain rule to express the second derivative of the angular position in terms of the angular velocity and the angle:

L u

d 2θ dω d ω dθ dω = = = ω. dt 2 dt d θ dt dθ Solution:

Consider the free-body diagram of the mass:

( )

d 2θ d dθ dω = = , dt 2 dt dt dt

and then we applied the chain rule to change the independent variable from time to position:

d 2θ dω d ω dθ dω = = = ω. dt 2 dt d θ dt dθ

The radial distance r = L is constant, so the polar components of the acceleration are d 2r dθ 2 dθ 2 −r = −L , 2 dt dt dt d 2θ dr d θ d 2θ aθ = r 2 + 2 = L 2. dt dt dt dt

( )

ar =

Notice that Eq. (2) has the form of an acceleration that depends on position. In the study of straight-line motion, we analyzed such problems by writing the acceleration as the time derivative of the velocity, which in this case looks like this:

( )

We use the free-body diagram to write Newton’s second law in the radial and transverse directions: dθ 2 ΣFr = mg sin θ − T = ma r = −mL , (1) dt d 2θ ΣFθ = mg cos θ = ma θ = mL 2 . (2) dt

( )

Substituting this expression into Eq. (2), we can separate variables and integrate: d 2θ dω g cos θ = L 2 = L ω, dt dθ g cos θ d θ = ω d ω, L ω g θ cos θ d θ = ∫ ω d ω, 0 L ∫0 g 1 sin θ = ω 2 . 2 L For a given angle, this determines the angular velocity. Knowing the angular velocity, Eq. (1) determines the tension. For θ = 45 °, we obtain ω = 4.81 rad/s, T = 41.6 N. The velocity of the mass is v = ω L = 2.89 m/s. Velocity is 2.89 m/s, tension is 41.6 N.

Problem 14.101 The 1-lb block A is given an initial velocity v 0 = 14 ft/s to the right when it is in the position θ = 0, causing it to slide up the smooth circular surface. By using Newton’s second law in terms of polar coordinates, determine the magnitude of the velocity of the block when θ = 60 °.

Solution:

Consider the free-body diagram of the block:

4 ft

A The radial distance R = 4 ft is constant, so the polar components of the acceleration are d 2r dθ 2 dθ 2 −r = −R , 2 dt dt dt d 2θ dr d θ d 2θ aθ = r 2 + 2 = R 2. dt dt dt dt ar =

140

( )

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14.101 (Continued)

−g sin θ = R

We use the free-body diagram to write Newton’s second law in the radial and transverse directions: W dθ 2 ΣFr = W cos θ − N = ma r = − R , (1) g dt W d 2θ ΣFθ = −W sin θ = ma θ = R . (2) g dt 2 We apply the chain rule to change the independent variable from time to position:

( )

d 2θ dω d ω dθ dω = = = ω. dt 2 dt d θ dt dθ Substituting this expression into Eq. (2), we can separate variables and integrate:

Problem 14.102 The 1-lb block A is given an initial velocity to the right when it is in the position θ = 0, causing it to slide up the smooth circular surface. Determine the minimum initial velocity needed for the block to reach θ = 90 °.

d 2θ dω = R ω, dt 2 dθ

g − sin θ d θ = ω d ω, R ω g θ − ∫ sin θ d θ = ∫ ω d ω, ω0 R 0 g 1 − ( 1 − cos θ ) = ( ω 2 − ω 0 2 ), 2 R where ω 0 is the initial angular velocity. Writing this equation in terms of the velocity of the block v = Rω, we write it as 1 gR ( 1 − cos θ ) = ( v 0 2 − v 2 ). 2 For a given angle, this equation determines the velocity. For θ = 60 °, we obtain v = 8.20 ft/s. Velocity is 8.20 ft/s.

Solution:

Consider the free-body diagram of the block:

R 4 ft u

eu N

The radial distance R = 4 ft is constant, so the polar components of the acceleration are d 2r dθ 2 dθ 2 −r = −R , 2 dt dt dt 2 2 d θ dr d θ d θ aθ = r 2 + 2 = R 2. dt dt dt dt ar =

( )

d 2θ dω d ω dθ dω = = = ω. dt 2 dt d θ dt dθ Substituting this expression into Eq. (2), we can separate variables and integrate: d 2θ dω −g sin θ = R 2 = R ω, dt dθ g − sin θ d θ = ω d ω, R ω g θ − ∫ sin θ d θ = ∫ ω d ω, ω0 R 0 g 1 − ( 1 − cos θ ) = ( ω 2 − ω 0 2 ), 2 R where ω 0 is the initial angular velocity. Writing this equation in terms of the velocity of the block v = Rω, we write it as 1 gR(1 − cos θ) = (v 0 2 − v 2 ). 2 Setting v = 0 and θ = 90 °, we obtain v 0 = 16.0 ft/s. 16.0 ft/s.

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Problem 14.103 The skier passes point A going 17 m/s. From A to B, the radius of his circular path is 6 m. By using Newton’s second law in terms of polar coordinates, determine the magnitude of the skier’s velocity as he leaves the jump at B. Neglect tangential forces other than the tangential component of his weight.

So (1) becomes g cos θ = r

dw w. dθ

Separating variables, w dw =

g cos θd θ. (2) r

At A, θ = 45 ° and w = v A /r = 17/6 = 2.83 rad/s. Integrating (2), wB

90 °

∫ 2.83wdw = r ∫ 45° cos θdθ, we obtain w B = 3.00 rad/s. His velocity at B is rw B = 18.0 m/s. v B = 18.0 m/s. 458 A B

Solution:

In terms of the angle θ shown, the transverse component of his weight is mg cos θ. Therefore ∑ Fθ = ma θ :

0  d 2θ dr d θ  . mg cos θ = m  r 2 + 2  dt dt dt  Note that

(1)

d 2θ dw dw d θ dw = = = w, dt 2 dt d θ dt dθ

er mg

Problem 14.104* A 2-kg mass rests on a flat horizontal bar. The bar begins rotating in the vertical plane about O with a constant angular acceleration of 1 rad/s 2 . The mass is observed to slip relative to the bar when the bar is 30 ° above the horizontal. What is the static coefficient of friction between the mass and the bar? Does the mass slip toward or away from O? 2 kg

1 rad/s2 O 1m

Solution:

From Newton’s second law for the radial component −mg sin θ ± µ s N = −mRω 2 , and for the normal component: N − mg cos θ = m Rα. Solve, and note that dω dω α = = ω = 1 = const, ω 2 = 2θ, since ω (0) = 0, to dt dθ obtain −g sin θ ± µ s ( g cos θ + Rα) = −2 Rθ. For α = 1, R = 1, this reduces to ±µ s (1 + g cos θ) = −2θ + g sin θ. Define the quantity FR = 2θ − g sin θ. If FR > 0, the block will tend to slide away from O, the friction force will oppose the motion, and the negative sign is to be chosen. If FR < 0 , the block will tend to slide toward O, the friction force will oppose the motion, and the positive sign is to be chosen. The equilibrium condition is derived from the equations of motion: sgn(FR )µ s (1 + g cos θ) = (2θ − g sin θ), from which µ s = sgn(FR )

2θ − g sin θ = 0.406 . Since Fr = −3.86 < 0, the 1 + g cos θ

block will slide toward O.

u f

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Problem 14.105* The 1/4-lb slider A is pushed along the circular bar by the slotted bar. The circular bar lies in the horizontal plane. The angular position of the slotted bar is θ = 10t 2 rad. Determine the polar components of the total external force exerted on the slider at t = 0.2 s. Solution: The interior angle β is between the radius from O to the slider A and the horizontal, as shown in the figure. The interior angle formed by the radius from C to the slider A and the line from O to the slider is β − θ. The angle β is found by applying the law of sines: 2 2 = from which sin θ = sin (β − θ) which is satissin θ sin (β − θ) fied by β = 2θ. The radial distance of the slider from the hinge point r 2 2sin 2θ is also found from the sine law: = ,r = , sin (180 − β ) sin θ sin θ from which r = 4 cos θ. The radial component of the acceleration dθ 2 d 2r dθ d is a r = −r (10t 2 ) = 20t. = . The derivatives: 2 dt dt dt dt 2 2 d θ d r dr dθ = 20. = −4 sin θ = −(80 sin θ)t. 2 = −80 sin θ dt 2 dt dt dt

( )

A u

2 ft 2 ft

The transverse acceleration is a θ = r

[a θ ] t = 0.2 = [a θ = (4 cos(10t 2 ))(20) + 2(−80sin(10t 2 ))(t )(20)(t )] t = 0.2 = 23.84 ft/s 2 .

( )

− (1600 cos θ)t 2 . Substitute:

( ddt θ ) + 2( drdt )( ddtθ ). Substitute:

The transverse component of the external force is W  Fθ =   a θ = 0.185 lb. g

[a r ] t = 0.2 = [a r = −80sin(10t 2 ) − (1600 cos(10t 2 ))t 2 − (4 cos(10t 2 ))(20t ) 2 ] t = 0.2 = −149.0 ft/s 2 .

From Newton’s second law, the radial component of the external force is W  Fr =   a r = −1.158 lb. g

b O 2 ft

Problem 14.106* The 1/4-lb slider A is pushed along the circular bar by the slotted bar. The circular bar lies in the vertical plane. The angular position of the slotted bar is θ = 10t 2 rad. Determine the polar components of the total force exerted on the slider by the circular and slotted bars at t = 0.25 s.

2 ft

Solution: Assume that the orientation in the vertical plane is such that the θ = 0 line is horizontal. Use the solution to Problem 14.105. For positive values of θ the radial component of acceleration due to gravity acts toward the origin, which by definition is the same direction as the radial acceleration. The transverse component of the acceleration due to gravity acts in the same direction as the transverse acceleration. From which the components of the acceleration due to gravity in the radial and transverse directions are g r = g sin θ and g θ = g cos θ. These are to be added to the radial and transverse components of acceleration due to the motion. From Problem 14.105, θ = 10t 2 rad [a r ] t = 0.25 = [−80sin θ − (1600 cos θ)t 2

− (4 cos θ) (20t ) 2 ] t = 0.25 = −209 ft/s 2 . u

2 ft

From

Newton’s

second law for the radial component  W   Fr − mg sin θ =   a r , from which Fr = −1.478 lb The transverse g component of the acceleration is (from Problem 14.105)

[a θ ] t = 0.25 = [(4 cos θ)(20)

2 ft

+ 2(−80sin θ)(t )(20)(t )] t = 0.25 = −52.14 ft/s 2 . From

second law for transverse component  W  Fθ − mg cos θ =   a θ , from which Fθ = −0.2025 lb. g

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Newton’s

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Problem 14.107* The slotted bar rotates in the horizontal plane with constant angular velocity ω 0 . The mass m has a pin that fits in the slot of the bar. A spring holds the pin against the surface of the fixed cam. The surface of the cam is described by r = r0 (2 − cos θ). Determine the polar components of the total external force exerted on the pin as functions of θ. v0

m Cam

Solution:

The angular velocity is constant, from which θ = ∫ ω 0 dt + C = ω 0 t + C . Assume that θ(t = 0) = 0, from dθ 2 d 2r which C = 0. The radial acceleration is a r = −r . The deriv2 dt dt d 2θ dθ d dr d atives: (ω t ) = ω 0 , = (r (2 − cos θ)) = = = 0. dt dt 0 dt 2 dt dt 0 dθ d 2r d r0 sin θ = ω 0 r0 sin θ, 2 = (ω r sin θ) = ω 02r0 cos θ. dt dt dt 0 0 Substitute: a r = ω 02r0 cos θ − r0 (2 − cos θ)(ω 02 ) = 2r0 ω 02 (cos θ − 1). From Newton’s second law the radial component of the external force is

( )

Fr = ma r = 2mr0 ω 02 (cos θ − 1). d 2θ The transverse component of the acceleration is a θ = r 2 + dt dr d θ 2 . Substitute: a θ = 2r0 ω 02 sin θ. From Newton’s second law, dt dt the transverse component of the external force is

( )( )

k u

Fθ = 2mr0 ω 02 sin θ. r0

Problem 14.108* In Problem 14.107, suppose that the unstretched length of the spring is r0 . Determine the smallest value of the spring constant k for which the pin will remain on the surface of the cam. v0

m Cam

Solution:

The spring force holding the pin on the surface of the cam is Fr = k (r − r0 ) = k (r0 (2 − cos θ) − r0 ) = kr0 (1 − cos θ). This force acts toward the origin, which by definition is the same direction as the radial acceleration, from which Newton’s second law for the pin is ΣF = kr0 (1 − cos θ) = −ma r . From the solution to Problem 14.107, kr0 (1 − cos θ) = −2mr ω 02 (cos θ − 1). Reduce and solve: k = 2mω 02 . Since cos θ ≤ 1, kr0 (1 − cos θ) ≥ 0, and 2mr0 ω 02 (cos θ − 1) ≤ 0. If k > 2mω 02 , Define Feq = kr0 (1 − cos θ) + 2mr ω 02 (cos θ − 1). If Feq > 0 the spring force dominates over the range of θ, so that the pin remains on the cam surface. If k < 2mω 02 , Feq < 0 and the radial acceleration dominates over the range of θ, so that the pin will leave the cam surface at some value of θ. Thus k = 2mω 02 is the minimum value of the spring constant required to keep the pin in contact with the cam surface.

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Problem 14.109 A electron in a magnetic field moves along a spiral path described by the equations r = 1 m, θ = 3z rad, where z is in meters. It moves along the path in the direction shown, with a constant speed v = 1 km/s. The mass of the electron is 9.11E−31 kg. Determine the force acting on the electron in terms of cylindrical coordinates. (Neglect its weight.)

dr dθ dz dz , = 0, vθ = r = 3r , vz = dt dt dt dt d 2r dθ 2 dz 2 , −r = −9r ar = dt 2 dt dt 2 2 d θ dr d θ d z = 3r 2 , aθ = r 2 + 2 dt dt dt dt d 2z . az = dt 2 Let v = 1000 m/s denote the magnitude of the velocity. We know that vr =

( )

v 2 = vr 2 + vθ 2 + v z 2

( dzdt ) + ( dzdt ) dz = ( 9r + 1 )( ) , dt = 3r

dz = dt

1 v. 9r 2 + 1

Because r is constant, we see from this result that

d 2z = 0, dt 2

1 km/s

and the components of the acceleration become (setting r = 1 m )

Solution:

a r = −0.9v 2 m/s 2 ,

Because r is constant, the derivatives

a z = 0.

The only nonzero component of force is

dr d 2r = = 0. dt dt 2

ΣFr = ma r = m (−0.9v 2 )

From the given expression for θ, we see that dθ dz = 3 rad/s, dt dt

a θ = 0,

= −0.9(9.11E − 31 kg)(1000 m/s) 2

d 2θ d 2z = 3 2 rad/s 2 . dt 2 dt

= −8.20E − 25 N.

Using these expressions, the components of the electron’s velocity and acceleration are

ΣF = −( 8.20E − 25 ) e r ( N ).

Problem 14.110 At the instant shown, the cylindrical coordinates of the 4-kg part A held by the robotic manipulator are r = 0.6 m, θ = 25 °, and z = 0.8 m. (The coordinate system is fixed with respect to the Earth, and the y-axis points upward.) A’s radial position is increasing at dr /dt = 0.2 m/s, and d 2 r /dt 2 = −0.4 m/s 2 . The angle θ is increasing when d θ /dt = 1.2 rad/s, and d 2 θ /dt 2 = 2.8 rad/s 2 . The base of the manipulator arm is accelerating in the z direction at d 2 z /dt 2 = 2.5 m/s 2 . Determine the force vector exerted on A by the manipulator in terms of cylindrical coordinates.

The total force acting on part A in cylindrical coord 2r − r ω 2 , ∑ Fθ = ma θ dt 2 dr d 2z = m rα + 2 ω , and ∑ Fz = ma z = m 2 . We are given the dt dt values of every term in the right hand side of these equations. (Recall

Solution:

dinates is given by ∑ Fr = ma r = m

(

( ) )

To find the forces exerted on the part by the manipulator, we need to draw a free body diagram of the part and resolve the weight into components along the various axes. We get ∑ F = ∑ Fmanip + W = ma where the components of ∑ F have already been determined above. In cylindrical coordinates, the weight is given as W = −mg sin θe r − mg cos θe θ . From the previous equation, ∑ F = ∑ Fmanip − mg sin θe r − mg cos θe θ . Substituting in terms of the components, we get ( ∑ Fmanip ) z = 11.5 ( newtons), ( ∑ Fmanip )θ = 44.2 ( newtons) and ( ∑ Fmanip ) z = 10.0 ( newtons). y

258

BandF_6e_ISM_C14_Pt2.indd 145

)

the definitions of ω and α. Substituting in the known values, we get ∑ Fr = −5.06 N, ∑ Fθ = 8.64 N, and ∑ Fz = 10.0 N. These are the total forces acting on Part A, including the weight.

Fmanip. z

(

W5–mgj x

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Problem 14.111 Suppose that the robotic manipulator is used in a space station to investigate zero-g manufacturing techniques. During an interval of time, the manipulator is programmed so that the cylindrical coordinates of the 4-kg part A are θ = 0.15t 2 rad, r = 0.5(1 + sin θ) m, and z = 0.8(1 + θ) m. Determine the force vector exerted on A by the manipulator at t = 2 s in terms of cylindrical coordinates. y

Solution: θ = 0.15t 2 rad, dθ = 0.3t rad/s, dt d 2θ = 0.3 rad/s 2 . dt 2 r = 0.5 (1 + sin θ) m, dr dθ = 0.5 cos θ m/s, dt dt d 2r d 2θ dθ 2 sin θ m/s 2 . = 0.5 2 cos θ − 0.5 dt 2 dt dt z = 0.8(1 + θ) m,

( )

Evaluating these expressions at t = 2 s, the acceleration is

dz dθ = 0.8 m/s, dt dt d 2z d 2θ 0.8 m/s 2 . = dt 2 dt 2

(

)

 d 2r dθ 2  d 2θ dr d θ d 2z a =  2 −r e + 2 ez e + r 2 + 2  dt dt  r dt dt dt θ dt = −0.259e r + 0.532e θ + 0.24 e z (m/s 2 ).

( )

Therefore ΣF = ma

= −1.04 e r + 2.13e θ + 0.96e z (N).

Problem 14.112* In Problem 14.111, draw a graph of the magnitude of the force exerted on part A by the manipulator as a function of time from t = 0 to t = 5 s, and use the graph to estimate the maximum force during that interval of time.

Solution:

Use a numerical solver to work problem 14.111 for a series of values of time during the required interval and plot the magnitude of the resulting force as a function of time. From the graph, the maximum force magnitude is approximately 8.4 N and it occurs at a time of about 4.4 seconds. Fmag (newtons) vs t (s)

F 8 m a 7 g 6 – 5 n e 4 w t 3 o n 2 s 1 0

1.5

2.5 3 Time (s)

3.5

4.5

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Problem 14.113 The International Space Station is in a circular orbit 225 miles above the Earth’s surface. (a) What is the magnitude of the velocity of the space station? (b) How long does it take to complete one revolution? Solution:

The radius of the orbit is

r0 = R E + 225 mi = 3960 + 225 mi = 2.21 × 10 7 ft. (a) From Eq (14.24), the velocity is v0 = =

g R E2 r0 (32.2)[(3960)(5280)] 2 2.21 × 10 7

Source: Courtesy of 3Dsculptor / Shutterstock.

= 25200ft/s(17200 mi/h). (b) Let T be the time required. Then v 0T = 2π r0 , 2π r0 so T = = 5500 s (1.53 h). v0

Problem 14.114 The Moon is approximately 383, 000 km from the Earth. Assume that the Moon’s orbit around the Earth is circular with velocity given by Eq. (14.24). (a) What is the magnitude of the Moon’s velocity? (b) How long does it take to complete one revolution around the Earth?

Solution: R E = 6370 km = 6.37 × 10 6 m r0 = 383000 km = 3.83 × 10 8 m v0 =

gR E2 = 1020 m/s r0 EARTH

MOON

383000 km

Period = 2πr0 /v 0 Period = 2.36 × 10 6 s = 27.3 days.

Problem 14.115 Suppose that you place a satellite into an elliptic Earth orbit with an initial radius r0 = 6700 km and an initial velocity v 0 such that the maximum radius of the orbit is 13,400 km. (a) Determine v 0 . (b) What is the magnitude of the satellite’s velocity when it is at its maximum radius? (See Practice Example 14.11.) v0

Solution: (a) Substituting the initial and maximum radii into Eq. (14.25), 1+ε 13,400,000 m = ( 6,700,000 m ) 1−ε and solving, the eccentricity is ε = 0.333. Then from Eq. (14.23),

(

ε = 0.333 =

)

r0v 0 2 − 1: gR E 2 ( 6, 700, 000 m ) v 0 2

( 9.81 m/s 2 )( 6,370, 000 m ) 2

− 1,

we obtain v 0 = 8900 m/s. (b) From Eq. (14.15), we can see that the product of the radius and the velocity when the satellite is at its maximum radius is equal to the product of the initial radius and the initial velocity. Let v denote the velocity at the maximum radius: vrmax = v 0 r0 : v (13,400,000 m) = (8900 m/s)(6,700,000 m). This gives v = 4450 m/s. (a) 8900 m/s. (b) 4450 m/s.

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Problem 14.116 A satellite is given an initial velocity v 0 = 6700 m/s at a distance r0 = 2 R E from the center of the Earth as shown in Fig. 14.8a. Draw a graph of the resulting orbit.

Solution:

The graph is shown.

2.88RE 2.88RE 5.13RE

Problem 14.117 The time required for a satellite in a circular Earth orbit to complete one revolution increases as the radius of the orbit increases. If you choose the radius properly, the satellite will complete one revolution while the Earth revolves 360 °. If a satellite is placed in such an orbit directly above the equator and it moves from west to east, it will remain permanently above the same point on the Earth as the Earth rotates beneath it. This type of orbit, conceived by Arthur C. Clarke, is called geosynchronous, and is used for some communication and television broadcast satellites. Determine the radius of a geosynchronous orbit in km. [Note: The Earth rotates, not 360 °, but nearly 361 ° in 24 hours, due to the Earth’s motion in its orbit around the sun. The Earth actually rotates 360 ° in 86,164.09 s. You should use this value to determine the radius of a geosynchronous orbit.] Problem 14.118* You can send a spacecraft from the Earth to the Moon in the following way: First, launch the spacecraft into a circular “parking” orbit of radius r0 around the Earth (Fig. P14.118a). Then increase its velocity in the direction tangent to the circular orbit to a value v 0 such that it will follow an elliptic orbit whose maximum radius is equal to the radius rM of the Moon’s orbit around the Earth (Fig. P14.118b). The radius rM = 238, 000 mi. Let r0 = 4160 mi. What velocity v 0 is necessary to send a spacecraft to the Moon? (This description is simplified in that it disregards the effect of the Moon’s gravity.) Moon’s orbit

2RE

Solution:

Let R denote the orbit radius and let T = 86,164.09 s. The circumference of the circular orbit is 2πR, so the velocity of the satellite must be v =

2π R . T

From Eq. (14.24), we know that v =

gR E 2 . R

We have two equations in v and R. Solving them, we obtain v = 3070 m/s and R = 42,100 km (26,200 miles). 42,100 km.

Solution:

Note that

R E = 3960 mi, r0 = 4160 mi, rM = 238,000 mi First find the eccentricity: r − r0 1+ε rmax = rM = r0 ⇒ ε = M 1−ε rM + r0

(

)

Then use eq. 14.23 r v2 2 grM (1 + ε) g ε = 0 02 − 1 ⇒ v 0 = R E = RE gR E r0 r0 (r0 + rM ) Putting in the numbers we have v 0 = 35,500 ft/s.

Elliptic orbit

Parking orbit

r0 rM

(a)

148

(b)

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Problem 14.119* On September 26, 2022, the DART (Double Asteroid Redirection Test) mission demonstrated the feasibility of altering the orbit of an asteroid by impacting it with a spacecraft. Dimorphos is a small asteroid that orbits a larger asteroid, Didymos. Assume that the orbit is circular, with a radius of 1.2 km, and that one orbit is completed in 11.92 hours. The Dart spacecraft strikes the asteroid moving in the direction tangential to its orbit and decreases its velocity by 0.132 mm/s. What is the minimum radius of the resulting elliptic orbit of Dimorphos about Didymos? (Assume that the mass of the spacecraft is negligible compared to that of Dimorphos, and the mass of Dimorphos is negligible compared to Didymos.) Strategy: Knowing the radius of the initial circular orbit and the time required for one orbit, you can determine the equivalent value of the term gR E 2 for Didymos.

Solution: Let r0 = 1200 m. The velocity of Dimorphos in its initial circular orbit is 2πr0 1 hour v0 = = 0.175704287 m/s. 11.92 hours 3600 s

(

)

Substituting r0 and v 0 into Eq. (14.24), we obtain for Didymos g D R D2 = 37.0 m 3 /s 2 . The velocity of Dimorphos after its collision with the DART spacecraft is v 0 = 0.175704287 m/s − 0.000132 m/s = 0.175572287 m/s. Using this value and r0 , we can determine the eccentricity of the new elliptic orbit from Eq. (14.23): ε =

r0 v 02 − 1 = −0.00150196. g D R D2

Then from Eq. (14.22) with θ = 180 °, the minimum radius is r = r0

( 11 +− εε ) = 1196.4 m,

which is 3.6 m smaller than r0 . 1.1964 km. DART spacecraft

Didymos

Dimorphos

Problem 14.120 The weight of the helicopter is W = 31,000 lb. It lifts off at time t = 0 and moves straight up. At t = 2 s it has risen 12 ft. Assume that its acceleration is constant. (a) How fast is the helicopter moving at t = 2 s? (b) What upward thrust T does its rotor exert on it? (Neglect aerodynamic forces.)

Setting x = 12 ft at t = 2 s in Eq. (2), 1 12 ft = a x ( 2 s ) 2 , 2 we can solve for the acceleration, obtaining a x = 6 ft/s 2 . Then from Eq. (1), the velocity at t = 2 s is v x = a xt = (6 ft/s 2 )(2 s) = 12 ft/s. (b) The free-body diagram of the helicopter is

T x y W

Source: Courtesy of Paul Horia Malaianu/Shutterstock. Knowing the acceleration, we can write Newton’s second law, ΣFx = ma x :

Solution: (a) We can integrate the (unknown) constant acceleration to determine the velocity and position as functions of time: dv x = ax, dt vx

a   T = W  1 + x   g 

∫0 d v x = ∫0 a x dt, (1) vx = x

dx = a x t. dt

x =

1 a t 2. 2 x

6 ft/s 2 32.2 ft/s 2

)

= 36,800 lb.

(2)

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(

= (31, 000 lb) 1 +

(a) 12 ft/s. (b) 36,800 lb.

∫0 dx = ∫0 a xt dt,

W  T − W = ma x =   a x , g and use it to determine the thrust:

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Problem 14.121 The weight of the helicopter is W = 31,000 lb. It lifts off at time t = 0 and moves straight up. The pilot gradually advances the throttle so that the upward thrust exerted by the rotor is given as a function of time in seconds by T = W (1 + 0.12t ) lb. (a) How fast is the helicopter moving at t = 2 s? (Neglect aerodynamic forces.) (b) What distance does it rise in 2 s?

Source: Courtesy of Paul Horia Malaianu/Shutterstock.

Solution:

Integrating to determine the velocity and position, we have dv x = ax, dt

(a) The free-body diagram of the helicopter is vx

∫0 dv x = ∫0 0.12gt dt,

vx =

x x

(1)

dx = 0.06 gt 2 . dt t

∫0 dx = ∫0 0.06gt 2 dt, (2)

x = 0.02 gt 3 .

From Eq. (1), the velocity at t = 2 s is v x = 0.06 gt 2

Newton’s second law is

= 0.06(32.2 ft/s 2 )(2 s) 2 = 7.73 ft/s. (b) From Eq. (2), the position at t = 2 s is

W  ΣFx = T − W = ma x =   a x . g

x = 0.02 gt 3

Using the given expression for the thrust, the acceleration is

( Wg ) T − W T = ( − 1) g W

ax =

(

= 0.02(32.2 ft/s 2 )(2 s) 3

)

= 5.15 ft. (a) 7.73 ft/s. (b) 5.15 ft.

= 0.12 gt.

Problem 14.122 A “cog” engine hauls three cars of sightseers to a mountaintop in Bavaria. The mass of each car, including its passengers, is 10, 000 kg, and the friction forces exerted by the wheels of the cars are negligible. Determine the forces in the couplings 1, 2, and 3 if (a) the engine is moving at constant velocity and (b) the engine is accelerating up the mountain at 1.2 m/s 2 .

Solution: (a) The force in coupling 1 is F1 = 10,000 g sin 40 ° = 631 kN. The force on coupling 2 is F2 = 20,000 g sin(40) = 126.1 kN. The force on coupling 3 is F3 = 30,000 g sin 40 ° = 189.2 kN. (b) From Newton’s second law, F1a − mg sin 40 ° = ma. Under constant acceleration up the mountain, the force on coupling 1 is F1a = 10,000 a + 10,000 g sin 40 ° = 75.1 kN. The force on coupling 2 is F2 a = 2 F1a = 150.1 kN. The force on coupling 3 is F2 a = 3 F1a = 225.2 kN.

3 F3 2

F2 F1

1 458

150

mg N

N mg

mg N

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Problem 14.123 In a future mission, a spacecraft approaches the surface of an asteroid passing near the Earth. Just before it touches down, the spacecraft is moving downward at a constant velocity relative to the surface of the asteroid and its downward thrust is 0.01 N. The computer decreases the downward thrust to 0.005 N, and an onboard laser interferometer determines that the acceleration of the spacecraft relative to the surface becomes 5 × 10 −6 m/s 2 downward. What is the gravitational acceleration of the asteroid near its surface? Solution:

Assume that the mass of the spacecraft is negligible compared to mass of the asteroid. With constant downward velocity, the thrust balances the gravitational force: 0.01 − mg s = 0, where m is the mass of the space craft. With the change in thrust, this becomes 0.005 − mg s = m(−5 × 10 −6 ) N/kg 2 . Multiply the first equation by 0.005, the second by 0.01, and subtract: The result: gs =

Source: Courtesy of NASA.

× 10 ) = 1 × 10 N/kg . ( 0.01(5 0.01 − 0.005 ) −6

(

−5

)

mgs

Problem 14.124 A car with a mass of 1470 kg, including its driver, is driven at 130 km/h over a slight rise in the road. At the top of the rise, the driver applies the brakes. The coefficient of static friction between the tires and the road is µ s = 0.9, and the radius of curvature of the rise is 160 m. Determine the car’s deceleration at the instant the brakes are applied, and compare it with the deceleration on a level road.

Solution: mg et en

mkN

The car’s velocity in m/s is 130 km/h

m 1h ( 1000 )( 3600 ) = 36.1 m/s. 1 km s

Newton’s second law in the tangential and normal directions is ΣFt = −µ k N = ma t , v 2 (1) . ρ We have two equations in terms of N and a t . Solving, we obtain a t = −1.49 m/s 2 , N = 2440 N. ΣFn = mg − N = ma n = m

On a level road, N = mg and Eq. (1) gives a t = −8.83 m/s 2 . 1.49 m/s 2 . On a level road, 8.83 m/s 2 .

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Problem 14.125 The car drives at constant velocity up the straight segment of road on the left. If the car’s tires continue to exert the same tangential force on the road after the car has gone over the crest of the hill and is on the straight segment of road on the right, what will be the car’s acceleration?

Solution:

The tangential force on the left is, from Newton’s second law, Ft − mg sin(5 °) = ma = 0. On the right, from Newton’s second law: Ft + mg sin(8 °) = ma from which the acceleration is a = g(sin 5 ° + sin8 °) = 0.2264 g.

mg N

Problem 14.126 The aircraft carrier Nimitz weighs 91, 000 tons. (A ton is 2000 lb.) Suppose that it is traveling at its top speed of approximately 30 knots (a knot is 6076 ft/h ) when its engines are shut down. If the water exerts a drag force of magnitude 20, 000v lb, where v is the carrier’s velocity in feet per second, what distance does the carrier move before coming to rest?

Solution:

Ft N

mg 88

The force on the carrier is F = −Kv, where K = 20,000.

gK F dv = − v . Use the chain rule to write v = m W dx gK gK − v Separate variables and integrate: dv = − dx, v ( x ) = W W gK 6076 ft 1h − x + C. The initial velocity: v(0) = 30 W 1h 3600 s gK = 50.63 ft/s, from which C = v (0) = 50.63, and v ( x ) = v ( 0 ) − x, W from which, at rest, The acceleration is a =

(

x =

Problem 14.127 If m A = 10 kg, m B = 40 kg, and the coefficient of kinetic friction between all surfaces is µ k = 0.11, what is the acceleration of B down the inclined surface?

)(

)

W v (0) = 14,321 ft = 2.71 mi. gK

Solution:

We draw individual free-body diagrams of the two boxes:

mkNA

mAg

ax mBg

A B 208

mkNA

208

mkNB

208

Here a x is the acceleration of box B down the surface. We apply Newton’s second law to each box: ΣFx = m B g sin 20 ° − µ k N A − µ k N B − T = m B a x , (1) ΣFy = −m B g cos 20 ° − N A + N B = 0, (2) ΣF− x = T − m A g sin 20 ° − µ k N A = m A a x , (3) ΣFy = −m A g cos 20 ° + N A = 0. (4) We can determine the normal forces from Eqs. (2) and (4). Then Eqs. (1) and (3) become two equations in T and a x . Solving, we obtain N A = 92.2 N, N B = 461 N, T = 49.6 N, and a x = 0.593 m/s 2 . 0.593 m/s 2 .

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Problem 14.128 The system is released from rest. If A weighs 20 lb, B weighs 80 lb, and the coefficient of kinetic friction between all surfaces is µ k = 0.12, how long does it take box B to slide 1 ft down the inclined surface? Solution:

A We draw individual free-body diagrams of the two

boxes:

y 208

mkNA

mAg

ax mBg

mkNA

208

Integrating the constant acceleration to determine the velocity and position as functions of time, dv x = ax, dt

mkNB

208

vx =

∫0 dv x = ∫0 a x dt, x

∫0 dx = ∫0 a xt dt, 1 a t 2. 2 x The time required for B to slide 1 ft down the surface is x =

2x = ax = 1.15 s.

t =

ΣFy = −m A g cos 20 ° + N A = 0. (4) We can determine the normal forces from Eqs. (2) and (4). Then Eqs. (1) and (3) become two equations in T and a x . Solving, we obtain N A = 18.8 lb, N B = 94.0 lb, T = 10.0 lb, and a x = 1.52 ft/s 2 .

1.15 s.

Problem 14.129 A gas gun is used to accelerate projectiles to high velocities for research on material properties. The projectile is held in place while gas is pumped into the tube to a high pressure p 0 on the left and the tube is evacuated on the right. The projectile is then released and is accelerated by the expanding gas. Assume that the pressure p of the gas is related to the volume V it occupies by pV γ = constant, where γ is a constant. If friction can be neglected, show that the velocity of the projectile at the position x is

Solution:

v =

2 p 0 Ax 0 γ  1 1  − γ −1 ,  γ − 1  m(γ − 1)  x 0 x

where m is the mass of the projectile and A is the cross-sectional area of the tube. Projectile p0 x0

dx = a x t, dt

2(1 ft) 1.52 ft/s 2

The force acting on the projectile is F = pA where p is the instantaneous pressure and A is the area. From pV γ = K, where K = p 0V0γ is a constant, and the volume V = Ax, it follows that K p = , and the force is F = KA 1− γ x −γ . The acceleration is ( Ax ) γ F K 1− γ −γ a = = A x . m m The equation to be integrated: dv K 1− γ −γ dv dv dx dv = A x , where the chain rule = = v has been dx m dt dx dt dx used. Separate variables and integrate:

v2 = 2

( Km ) A ∫ x dx + C = 2( Km ) A 1− γ

−γ

1− γ  

x 1− γ   1 − γ  + C .

When x = x 0 , v 0 = 0, therefore v2 = 2

( Km ) 1A− γ (x 1− γ

1− γ

− x 10− γ ).

Substitute K = p 0V0γ = p 0 A γ x 0γ and reduce: v2 =

2 p 0 Ax 0γ 1− γ (x − x 10− γ ). Rearranging: m(1 − γ )

v =

2 p 0 Ax 0γ  1 1  − γ −1 .  m(γ − 1)  x 0γ −1 x

p x

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Problem 14.130 The weights of the blocks are W A = 120 lb and W B = 20 lb, and the surfaces are smooth. Determine the acceleration of block A and the tension in the cord.

Solution:

The two free-body diagrams described in the strategy are

a T T WA WB

WB a

NA B Here a denotes the acceleration of block A to the right and the equal downward acceleration of block B. We apply Newton’s second law to the left free-body diagram,  W + W B  T =  A  a,   g and to the left one, WB − T =

WB a. g

We have two equations in T T = 17.5 lb, a = 4.02 ft/s 2 .

and

Solving,

obtain

Acceleration is 4.02 ft/s 2 , tension is 17.5 lb.

Problem 14.131 The 100-Mg space shuttle is in orbit when its engines are turned on, exerting a thrust force T = 10 i − 20 j + 10 k (kN) for 2 s. Neglect the resulting change in mass of the shuttle. At the end of the 2-s burn, fuel is still sloshing back and forth in the shuttle’s tanks. What is the change in the velocity of the center of mass of the shuttle (including the fuel it contains) due to the 2-s burn?

Solution:

At the completion of the burn, there are no external forces on the shuttle (it is in free fall) and the fuel sloshing is caused by internal forces that cancel, and the center of mass is unaffected. The change in velocity is 2(10 4 ) 2(2 × 10 4 ) 2(10 4 ) i− j+ k dt = 10 5 10 5 10 5 m = 0.2 i − 0.4 j + 0.2k (m/s).

∆v = ∫

Problem 14.132 The water skier contacts the ramp with a velocity of 25 mi/h parallel to the surface of the ramp. Neglecting friction and assuming that the tow rope exerts no force on him once he touches the ramp, estimate the horizontal length of the skier’s jump from the end of the ramp. Solution: Break the path into two parts: (1) The path from the base to the top of the ramp, and (2) from the top of the ramp until impact with the water. Let u be the velocity parallel to the surface of the ramp, and let z be the distance along the surface of the ramp. du 8 From the chain rule, u = − g sin θ, where θ = tan −1 = 21.8 °. 20 dz Separate variables and integrate: u 2 = −(2 g sin θ) z + C . At the base of the ramp

( )

(

mi 5280 ft u(0) = 25 h 1 mi

)(

)

1h = 36.67 ft/s, 3600 s

from which C = (36.67 2 ) = 1344.4 and u =

8 ft

20 ft g − t 2 + 10.7t + 8 ft. v x (t ) = 26.7 ft/s, x (t ) = 26.7t. When 2 y (t impact ) = 0, the skier has hit the water. The impact time is 10.7 16 2 t impact + 2bt impact + c = 0 where b = − , c = − . The solug g tion t impact = −b ±

b 2 − c The = 1.11 s, = −0.45 s. negative

values has no meaning here. The horizontal distance is x (t impact ) = 26.7t impact = 29.7 ft.

C − (2 g sin θ) z .

At z = 8 2 + 20 2 = 21.54 ft u = 28.8 ft/s. (2) In the second part of the path the skier is in free fall. The equations to be integrated dv y dy are = −g, = v y , with v (0) = u sin θ = 28.8(0.3714) = dt dt dv x dx = 0, = v x , with v x (0) = u cos θ = 10.7 ft/s, y (0) = 8 ft. dt dt 26.7 ft/s, x (0) = 0 Integrating: v y (t ) = −gt + 10.7 ft/s. y (t ) =

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Problem 14.133 Suppose you are designing a rollercoaster track that will take the cars through a vertical loop of 40-ft radius. If you decide that, for safety, the downward force exerted on a passenger by their seat at the top of the loop should be at least one-half the passenger’s weight, what is the minimum safe velocity of the cars at the top of the loop?

Solution:

Consider the free-body diagram of a passenger at the top

of the track:

N et W

Newton’s second law in the normal direction is W + N =

W W v2 a = . g n g ρ

If the force exerted by the seat is N = 0.5W and ρ = 40 ft, we obtain v = 43.9 ft/s. 43.9 ft/s.

40 ft

Problem 14.134 As the smooth bar rotates in the horizontal plane, the string winds up on the fixed cylinder and draws the 1-kg collar A inward. The bar starts from rest at t = 0 in the position shown and rotates with constant angular acceleration. What is the tension in the string at t = 1 s? 2

6 rad/s

Solution: The angular velocity of the spool relative to the bar is α = 6 rad/s 2 . The acceleration of the collar relative to the bar is α = 6 rad/s 2 . The acceleration of the collar relative to the bar is d 2r = −Rα = −0.05(6) = −0.3 m/s 2 . The take up velocity of the dt 2 spool is v s = ∫ Rα dt = −0.05(6)t = −0.3t m/s. The velocity of the collar relative to the bar is dr = −0.3t m/s. dt The velocity of the collar relative to the bar is dr /dt = −0.3t m/s. The position of the collar relative to the bar is r = −0.15t 2 + 0.4 m. d 2θ The angular acceleration of the collar is = 6 rad/s 2 . The angudt 2 dθ lar velocity of the collar is = 6t rad/s. The radial acceleration is dt 2 2 dθ d r ar = −r = −0.3 − (−0.15t 2 + 0.4)(6t ) 2 . At t = 1 s the dt 2 dt radial acceleration is a r = −9.3 m/s 2 , and the tension in the string is

( )

10 0

T = ma r = 9.3 N.

A T

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Problem 14.135* In Problem 14.134, suppose that the coefficient of kinetic friction between the collar and the bar is µ k = 0.2. What is the tension in the string at t = 1 s? 6 rad/s2

Solution:

Use the results of the solution to Problem 14.134 At t = 1 s, the horizontal normal force is

( ddt θ + 2( drdt )( ddtθ )) = 2.1 N,

N H = ma θ = m r

from which the total normal force is N = Newton’s

second

law:

( −T + µ k

N H2 + (mg) 2 From

N H2 + (mg 2 ) ) e r + N H e θ =

ma r e r + ma θ e θ , from which −T + µ k N H2 + (mg) 2 = ma r . From

T = 11.306 N.

the solution to Problem 14.152, a r = −9.3 m/s 2 . Solve: The tension is

Problem 14.136 If you want to design the cars of a train to tilt as the train goes around curves in order to achieve maximum passenger comfort, what is the relationship between the desired tilt angle θ, the velocity v of the train, and the instantaneous radius of curvature, ρ, of the track?

Solution:

For comfort, the passenger should feel the total effects of acceleration down toward his feet, that is, apparent side (radial) accelerations should not be felt. This condition is achieved when the tilt θ is such that mg sin θ − m(v 2 /ρ ) cos θ = 0, from which tan θ =

v2 . ρg

Problem 14.137 To determine the coefficient of static friction between two materials, an engineer at the U.S. National Institute of Standards and Technology places a small sample of one material on a horizontal disk whose surface is made of the other material and then rotates the disk from rest with a constant angular acceleration of 0.4 rad/s 2 . If she determines that the small sample slips on the disk after 9.903 s, what is the coefficient of friction?

156

200 mm

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14.137 (Continued)

Because the angular acceleration is constant, the angular velocity ω = αt, so we can write the acceleration as

Solution:

a = Rα e t + Rα 2t 2 e n . Its magnitude is ( Rα ) 2 + ( Rα 2t 2 )

a =

= Rα 1 + α 2t 4 .

en R

The only horizontal force acting on the sample (neglecting air resistance) is the friction force exerted by the disk. Denoting the magnitude of the friction force by f, Newton’s second law is f = m a = mRα 1 + α 2t 4 . The sample slips when the friction force exceeds µ k N, and the normal force N = mg, so we obtain an equation for the coefficient of friction in terms of the time of slip:

In terms of the angular acceleration α = and angular velocity ω of the disk, the acceleration of the sample in terms of normal and tangential components is 0.4 rad/s 2

µk =

Rα 1 + α 2t 4 g

(0.2 m)(0.4 rad/s 2 ) 1 + (0.4 rad/s 2 ) 2 (9.903 s) 4 9.81 m/s 2 = 0.320. =

a = Rα e t + Rω 2 e n .

µ k = 0.320.

Problem 14.138* The 1-kg slider A is pushed along the curved bar by the slotted bar. The curved bar lies in the horizontal plane, and its profile is described by r = 2(θ /2π + 1) m, where θ is in radians. The angular position of the slotted bar is θ = 2t rad. Determine the polar components of the total external force exerted on the slider when θ = 120 °.

Solution: dr 2 = . dt π

The radial position is r = 2

( πt + 1). The radial velocity:

dθ = 2. The angudt lar acceleration is zero. At θ = 120 ° = 2.09 rad. From Newton’s second law, the radial force is Fr = ma r , from which The radial acceleration is zero. The angular velocity:

 dθ 2  Fr = −  r  e = −10.67e r N.  dt  r

( )

The transverse force is Fθ = ma θ , from which

( )( ddtθ )  e = 2.55e N.

dr Fθ = 2   dt

Problem 14.139* In Problem 14.138, suppose that the curved bar lies in the vertical plane. Determine the polar components of the total force exerted on A by the curved and slotted bars at t = 0.5 s.

Solution:

Assume that the curved bar is vertical such that the line θ = 0 is horizontal. The weight has the components: W = (W sin θ)e r + (W cos θ)e θ . From Newton’s second law: Fr − W sin θ = ma r , and Fθ − W cos θ = ma θ ., from which Fr − g sin 2te r = −r (d θ /dt ) 2 e r , from which t Fr = −2 + 1 (2 2 ) + g sin 2t , π

( (

)

t = 0.5 s, Fr = −1.02 N .

( )

(

The

transverse

component

)

2 8 Fθ = 2 (2) + g cos 2t = + g cos 2t . At t = 0.51 s, π π Fθ = 7.85 N. u

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Chapter 15 Problem 15.1 In Practice Example 15.2, suppose that the mass of the container A is 120 kg. What is the velocity of the container when it has reached the position s = 3 m?

Solution:

The 120 kg container starts from rest at t = 0 The h orizontal force (in newtons) is F = 700 − 150s. The coefficient of kinetic friction is µ k = 0.26. U 12 = T2 − T1 = (m / 2)[ v 22 − v 12 ] 3

∫0 (700 − 150s − 0.26(120 ∗ 9.81)) ds = (120 / 2)v 22 v 2 = 2.91 m/s.

Problem 15.2 The mass of the Sikorsky UH-60A helicopter is 9300 kg. It takes off vertically with its rotor exerting a constant upward thrust of 112 kN. Use the principle of work and energy to determine how far it has risen when its velocity is 6 m/s. Strategy: Be sure to draw the free-body diagram of the helicopter.

Solution: T

The work done as the helicopter moves from its initial position to height s is U 12 = ∫

∑ Ft ds

= ∫ (T − mg) ds 0

= (T − mg)s. Equating the work to the change in kinetic energy, U 12 = (T − mg)s =

1 2 mv − 0, 2

we obtain mv 2 2(T − mg) (9300 kg)(6 m/s) 2 = 2[112,000 N − (9300 kg)(9.81 m/s 2 )] = 8.06 m.

s =

Source: Courtesy of Zigmunds Dizgalvis/Shutterstock.

8.06 m.

Problem 15.3 The 10-kg box is at rest on the horizontal surface when the constant force F = 35 N is applied. The coefficient of kinetic friction between the box and the surface is µ k = 0.25. Determine how fast the box is moving when it has moved 2 m from its initial position (a) by applying Newton’s second law; (b) by applying the principle of work and energy. F

Solution: mg

mkN

(a) The normal force N = mg. Applying Newton’s second law, F − µ k N = F − µ k mg = ma, The acceleration of the box is 1 ( F − µ k mg) m 1 = [35 N − 0.25(10 kg)(9.81 m/s 2 )] 10 kg = 1.05 m/s 2 .

a =

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15.3 (Continued) We integrate the constant acceleration to determine the velocity and position as functions of time:

(b) The work done as the box moves a distance s is s

U 12 = ∫ ( F − µ k mg) ds

dv = a, dt v

= ( F − µ k mg)s. t

∫0 dv = ∫0 a dt, (1)

Equating the work to the change in kinetic energy,

ds = at, v = dt s

U 12 = ( F − µ k mg)s = we obtain

∫0 ds = ∫0 at dt,

1 2 mv − 0, 2

Solving Eq. (2) for t and substituting it into Eq. (1), we obtain an equation for the velocity in terms of the position:

2( F − µ k mg)s m 2[35 N − 0.25(10 kg)(9.81 m/s 2 )](2 m) = 10 kg = 2.05 m/s.

v =

(a), (b) 2.05 m/s.

1 2 at . 2

s =

(2)

v =

2as 2(1.05 m/s 2 )(2 m)

= 2.05 m/s.

Problem 15.4 At the instant shown, the 40-lb box is moving up the inclined surface at 3 ft/s. The constant force F = 25 lb. Use the principle of work and energy to determine how fast the box will be moving when it has moved 2 ft up the surface from its present position (a) if the inclined surface is smooth; (b) if the coefficient of kinetic friction between the box and the inclined surface is µ k = 0.12.

Equating the work to the change in kinetic energy, 1W 2 1W 2 v − v : 2 g 2 2 g 1 1 (40 lb) [(25 lb) cos 20 ° − (40 lb)sin 20 °](2 ft) = v 2 2 (32.2 ft/s 2 ) 2 1 (40 lb) − (3 ft/s) 2 2 (32.2 ft/s 2 )

U 12 = ( F cos 20 ° − W sin 20°)s =

and solving, we obtain v 2 = 6.37 ft/s. (b)

208

mkN

There is no acceleration normal to the surface, so

Solution:

N − F sin 20 ° − W cos 20 ° = 0.

(a) F

208

This equation determines N. The work done as the box moves a distance s is s

U 12 = ∫ ( F cos 20 ° − µ k N − W sin 20 °) ds 0

The work done as the box moves a distance s is s

U 12 = ∫ ( F cos 20 ° − W sin 20 °) ds 0

= ( F cos 20 ° − W sin 20 °)s.

= ( F cos 20 ° − µ k N − W sin 20 °)s. Equating the work to the change in kinetic energy, 1W 2 1W 2 v − v : 2 g 2 2 g 1 [(25 lb) cos 20 ° − (0.12)(46.1 lb) − (40 lb)sin 20 °](2 ft)

U 12 = ( F cos 20 ° − µ k N − W sin 20 °)s =

1 (40 lb) 1 (40 lb) v 2 − (3 ft/s) 2 2 (32.2 ft/s 2 ) 2 2 (32.2 ft/s 2 )

and solving, we obtain v 2 = 4.77 ft/s. (a) v 2 = 6.37 ft/s. (b) v 2 = 4.77 ft/s.

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Problem 15.5 The 0.43-kg soccer ball is 1 m above the ground when it is kicked straight up at 20 m/s. By using the principle of work and energy, determine (a) how high above the ground the ball goes and (b) the magnitude of the ball’s velocity just before it hits the ground.

Solution: Let s denote upward displacement, at let s = 0 be the position of the ball when it is kicked. The ball’s weight mg is constant, so we can use Eq. (15.8) to evaluate the work done on it. The principle of work and energy is 1 1 −mg(s 2 − s1 ) = mv 2 2 − mv 1 2 : 2 2 1 1 −mg(s 2 − 0) = mv 2 2 − m(20 m/s) 2 . 2 2 (a) Setting v 2 = 0, we determine the ball’s greatest height above where it was kicked to be s 2 = 20.4 m, so the greatest height above the ground is 21.4 m. (b) Setting s 2 = −1 m, the ball’s velocity just before it hits the ground is v 2 = 20.5 m/s. (a) 21.4 m. (b) 20.5 m/s.

Problem 15.6 Assume that the 0.43-kg soccer ball is stationary just before it is kicked upward at 20 m/s. The duration of the kick is 0.05 s. What average power is transferred to the ball during the kick?

Solution: 1 Pav = 2 1 = 2

From Eq. (15.10), the average power during the kick is 1 mv 2 2 1 t 2 − t1

mv 2 2 −

(0.43 kg)(20 m/s) 2 −

0.05 s = 1720 N-m/s (watts).

1 (0.43 kg)(0) 2 2

1720 N-m/s. 1m

Problem 15.7 The 2000-lb drag racer accelerates from rest. When it has traveled 800 ft, it is moving at 350 ft/s. Assume that the horizontal force acting on the car is constant and use the principle of work and energy to determine its value. Solution:

Let F denote the constant force and s = 800 ft. The work done is Fs, so 1 2 mv : 2 1 2000 lb F (800 ft) = (350 ft/s) 2 , 2 32.2 ft/s 2 Fs =

(

)

giving F = 4760 lb. 4760 lb.

160

Source: Courtesy of Dan_prat/E+/Getty Images.

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Problem 15.8 The 2000-lb drag racer accelerates from rest at position s = 0. The horizontal force acting on the car is given as a function of s in feet by 11,000 − 20 s + 0.01s 2 lb. Use the principle of work and energy to determine the magnitude of the car’s velocity when it has traveled 800 ft. Solution: U 12 = ∫

800 ft 0

Using the given expression, the work done on the car is (11, 000 − 20 s + 0.01s 2 lb) ds

= 4.107E6 ft-lb. Then 1 2 mv : 2 1 2000 lb v 2, 4.107E6 ft-lb = 2 32.2 ft/s 2 U 12 =

(

Source: Courtesy of Dan_prat/E+/Getty Images.

)

which yields v = 364 ft/s (248 mi/h). 364 ft/s.

Problem 15.9 As the 28,000-lb airplane takes off, assume that the total horizontal force acting on it is constant and equal to F = 25,000 lb. Use the principle of work and energy to determine how much runway is required for its velocity to reach its takeoff speed of 240 ft/s. Solution:

Let s denote the length of runway required. Applying the principle of work and energy, 1 1 mv 2 − mv 1 2 : 2 2 2 1 28,000 lb (25,000 lb)s = v 2 − 0, 2 32.2 ft/s 2 2 U 12 =

(

)

we obtain s = 1000 ft.

Source: Courtesy of VanderWolf Images/Shutterstock.

1000 ft.

Problem 15.10 As the 28,000-lb airplane takes off, assume that the total horizontal force acting on it is constant and equal to F = 25,000 lb. It requires 1000 ft of runway to take off. One horsepower (hp) is 550 ft-lb/s. Use the principle of work and energy to determine the maximum horsepower transferred to the airplane during its takeoff. Solution:

Applying the principle of work and energy,

1 1 mv 2 − mv 1 2 : 2 2 2 1 28,000 lb (25,000 lb)(1000 ft) = v 2 − 0, 2 32.2 ft/s 2 2 U 12 =

(

)

we find that the velocity at takeoff is v 2 = 240 ft/s. The maximum power transferred during takeoff is

Source: Courtesy of VanderWolf Images/Shutterstock.

P = (25,000 lb)(240 ft/s) = 6.00E6 ft-lb/s. This is 1 hp P = 6.00E6 ft-lb/s = 10,900 hp. 550 ft-lb/s

(

)

10,900 hp.

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Problem 15.11 As the 28,000-lb airplane takes off from rest at position s = 0, the total horizontal force acting on it given as a function of s in feet is F = Te −cs , where T = 26,000 lb and c = 0.0001 ft −1 . It requires 1000 ft of runway to take off. Use the principle of work and energy to determine the magnitude of the airplane’s velocity at takeoff. Solution: s2

U 12 = ∫

= ∫

The work done during takeoff is F ds

1000 ft

(26,000 lb)e −(0.0001 ft −1 )s ds

= 2.47E7 ft-lb.

Source: Courtesy of VanderWolf Images/Shutterstock.

Applying the principle of work and energy, 1 1 mv 2 − mv 1 2 : 2 2 2 1 28,000 lb 2.47E7 ft-lb = v 2 − 0, 2 32.2 ft/s 2 2 U 12 =

(

)

we find that the velocity at takeoff is v 2 = 239 ft/s. 239 ft/s.

Problem 15.12 The spring (k = 20 N/m) is unstretched when s = 0. The 5-kg cart is moved to the position s = −1 m and released from rest. What is the magnitude of its velocity when it is in the position s = 0?

Solution:

First we calculate the work done by the spring and by

gravity U 12 = ∫

−1 m

(−ks + mg sin 20 °) ds

1 = k (−1 m) 2 + mg sin 20 °(1 m) 2 1 = (20 N/m)(−1 m) 2 + (5 kg)(9.81 m/s 2 )sin 20 ° (1 m) 2 = 26.8 N-m.

Now using work and energy U 12 = 208

1 2 mv ⇒ v = 2

2U 12 = m

2(26.8 N-m) = 3.27 m/s. 5 kg

v = 3.27 m/s.

Problem 15.13 The spring (k = 20 N/m) is unstretched when s = 0. The 5-kg cart is moved to the position s = −1 m and released from rest. What maximum distance down the sloped surface does the cart move relative to its initial position?

Solution:

The cart starts from a position of rest, and when it reaches the maximum position, it is again at rest. Therefore, the total work must be zero.

U 12 = ∫

−1 m

208

(−ks + mg sin 20 °) ds

1 = − k (s 2 − [−1 m] 2 ) + mg sin 20°(s − [−1 m]) 2 1 = − (20 N/m)(s 2 − [−1 m] 2 ) + (5 kg)(9.81 m/s 2 )sin 20 ° (s − [−1 m]) = 0 2 This is a quadratic equation that has the two solutions s1 = −1 m, s 2 = 2.68 m. The distance relative to the initial is s = s 2 + 1 m. s = 3.68 m.

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Problem 15.14 The force exerted on a car by a prototype crash barrier as the barrier crushes is F = −(120 s + 40 s 3 ) lb, where s is the distance in feet from the initial contact. The effective length of the barrier is 18 ft. How fast can a 5000-lb car be moving and be brought to rest within the effective length of the barrier?

Solution: 18

U 12 = ∫ −(120 s + 40 s 3 ) ds 0

1 1 = − (120)(18) 2 − (40)(18) 4 = −1.07 × 10 6 ft-lb. 2 4 Using work and energy, we have U 12 = 0 −

The barrier can provide a maximum amount of work

given by

1 2 mv 2

−1.07 × 10 6 ft-1b = −

(

)

1 5000 lb v2 2 32.2 ft/s 2

Solving for the velocity, we find v = 117 ft/s (80.0 mi/h).

Problem 15.15 A 5000-lb car hits the crash barrier at 80 mi/h and is brought to rest in 0.11 s. What average power is transferred from the car during the impact?

Solution:

The average power is equal to the change in kinetic energy divided by the time P =

1 2

88 ft/s 2 1 5000 lb 80 mi/h 2 mv 2 60 mi/h 2 32.2 ft/s = = 9.72 × 10 6 ft-1b/s. ∆t 0.11 s

(

)(

)

P = 9.72 × 10 6 ft-lb/s (17,700 hp).

Problem 15.16 A group of engineering students constructs a sun-powered car and tests it on a circular track with a 1000-ft radius. The car, with a weight of 460 lb including its occupant, starts from rest. The total tangential component of force on the car is ΣFt = 30 − 0.2s lb, where s is the distance (in ft) the car travels along the track from the position where it starts. (a) Determine the work done on the car when it has gone a distance s = 120 ft. (b) Determine the magnitude of the total horizontal force exerted on the car’s tires by the road when it is at the position s = 120 ft.

BandF_6e_ISM_C15_Pt1.indd 163

Solution: (a) U = ∫

120 ft 0

[30 − 0.2 s] lb ds = 2160 ft-lb.

(b) 2160 ft-1b =

(

)

1 460 lb v 2 ⇒ v = 17.4 ft/s 2 32.2 ft/s 2

Ft = [30 − 0.2(120)] = 6 lb Fn = m F =

v2 460 lb (17.4 ft/s) 2 = = 4.32 lb 32.2 ft/s 2 1000 ft ρ

(

Ft2 + Fn2 =

)

(6 lb) 2 + (4.32 lb) 2 = 7.39 lb.

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Problem 15.17 At the instant shown, the 72-kg vaulter’s center of mass is 2.6 m above the ground and the vertical component of his velocity is 1.2 m/s. As his pole straightens, the combined effects of the pole and his arms exert an upward force on him of magnitude 800 + 160 y 3 N, where y is the vertical position of his center of mass relative to its position at the instant shown. This force is exerted on him from y = 0 to y = 1.8 m, after which only his weight acts on him. What maximum height above the ground is reached by the vaulter’s center of mass?

Solution:

The vaulter’s weight is mg = (72 kg)(9.81 m/s 2 ) = 706 N. As he rises from y = 0 to y = 1.8 m, the upward force acting on him is ∑ Fy = 800 + 160 y 3 − 706 = 94 + 160 y 3 N.

Applying the principle of work and energy to his motion as he rises from y = 0 to y = 1.8 m, U 12 = ∫

1.8 m 0

(94 + 160 y 3 N) dy =

1 1 (72 kg)v 2 2 − (72 kg)(1.2 m/s) 2 , 2 2

we obtain v 2 = 4.22 m/s. Above y = 1.8 m, the only force acting on him is his weight. Let y max denote the maximum height he reaches above y = 0. Applying work and energy to his motion from y = 1.8 m to the maximum height (where his velocity is zero), 1 U 12 = −mg( y max − 1.8 m) = 0 − (72 kg)(4.22 m/s) 2 2 gives y max = 2.71 m. His maximum height above the ground is y max + 2.6 m = 5.31 m. 5.31 m.

Problem 15.18 The springs (k = 25 lb/ft) are unstretched when s = 0. The 50-lb weight is released from rest in the position s = 0. (a) When the weight has fallen 1 ft, how much work has been done on it by each spring? (b) What is the magnitude of the velocity of the weight when it has fallen 1 ft?

Solution: k

(a) The work done by each spring U 12 = ∫

1 ft 0

1 −ks ds = − (25 lb/ft)(1 ft) 2 = −12.5 ft-lb. 2

(b) The velocity is found from the work-energy equation. The total work includes the work done by both springs and by gravity U 12 = (50 lb)(1 ft) − 2(12.5 ft-lb) = 25 ft-lb. U 12 =

(

)

1 2 1 50 lb mv = v2 2 2 32.2 ft/s 2

Solving for the velocity we find v = 5.67 ft/s. (a) −12.5 ft-lb, (b) 5.67 ft/s.

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Problem 15.19 The coefficients of friction between the 90-kg crate and the ramp are µ s = 0.35 and µ k = 0.25. (a) What cable tension T0 must the winch exert to start the crate moving up the ramp? (b) If the tension remains at the value T0 after the crate starts sliding, what is the magnitude of the crate’s velocity after it has moved 2 m ?

188

Solution: we obtain N = 840 N, T0 = 567 N. Then using these values, setting f = µ k N and applying the principle of work and energy,

188

U 12 = (T0 − mg sin18 ° − µ k N )(2 m) =

1 mv 2 − 0, 2 2

we obtain v 2 = 1.93 m/s.

v 2 = 1.93 m/s.

x To determine the value of T0 necessary to make the crate slip, assume it is in equilibrium and that the friction force f = µ s N. From the equilibrium equations ∑ Fx = T0 − mg sin18 ° − µ s N = 0, ∑ Fy = N − mg cos18 ° = 0,

Problem 15.20 The kinetic coefficient of friction between the 90-kg crate and the ramp is µ k = 0.25. The crate starts from rest at position s = 0. If the cable tension exerted by the winch is T = T0 (1 + 0.2s), where T0 = 600 N, what is the magnitude of the crate’s velocity when it has moved 2 m ?

188

Solution: Applying the principle of work and energy, T mg

1 mv 2 − 0, 2 2 we obtain v 2 = 3.25 m/s.

U 12 = 188

v 2 = 3.25 m/s.

The friction force f = µ k N. There is no acceleration normal to the inclined surface, so N = mg cos18 ° = 840 N. The work done as the crate moves 2 m is U 12 = ∫

2m 0

[T0 (1 + 0.2s) − mg sin18 ° − µ k N ] ds

= 474 N-m.

Problem 15.21 The 200-mm -diameter gas gun is evacuated on the right of the 8-kg projectile. On the left of the projectile, the tube contains gas with pressure p 0 = 1 × 10 5 Pa (N/m 2 ). The force F is slowly increased, moving the projectile 0.5 m to the left from the position shown. The force is then removed, and the projectile accelerates to the right. If you neglect friction and assume that the pressure of the gas is related to its volume by pV = constant, what is the velocity of the projectile when it has returned to its original position?

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Gas

Projectile F

Solution:

The constant is K = pV = 1 × 10 5 (1)(0.1) 2 π = 3141.6 N-m. The force is F = pA. The volume is V = As, from K which the pressure varies as the inverse distance: p = , from which As K F = . s

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15.21 (Continued) The work done by the gas is U = ∫

1 0.5

F ds = ∫

0.5 x

ds = [ K ln(s)]1.0 0.5 = K ln(2).

From the principle of work and energy, the work done by the gas is equal to the gain in kinetic energy: K ln(2) = v =

1 2 2K mv , and v 2 = ln(2), 2 m

2K ln(2) = 23.33 m/s. m

Note: The argument of ln(2) is dimensionless, since it is ratio of two distances.

Problem 15.22 In Problem 15.21, if you assume that the pressure of the gas is related to its volume by pV = constant while the gas is compressed (an isothermal process) and by pV 1.4 = constant while it is expanding (an isentropic process), what is the velocity of the projectile when it has returned to its original position? Solution: The isothermal constant is K = 3141.6 N-m from the solution to Problem 15.21. The pressure at the leftmost position is K p = = 2 × 10 5 N/m 2 . A(0.5) The isentropic expansion constant is K e = pV 1.4 = (2 × 10 5 )( A 1.4 )(0.51.4 ) = 596.5 N-m The pressure during expansion is p =

Ke K e −1.4 = 1.4 s . ( As) 1.4 A

Problem 15.23 In Example 15.3, suppose that the angle between the inclined surface and the horizontal is increased from 20 ° to 30 °. What is the magnitude of the velocity of the crates when they have moved 400 mm?

Gas

Projectile F

1m The force is F = pA = K e A −0.4 s −1.4 . The work done by the gas during expansion is 1.0 1.0 s −0.4  1.0 U = ∫ F ds = ∫ K e A −0.4 s −1.4 ds = K e A −0.4  0.5 0.5  −0.4  0.5 = 1901.8 N-m From the principle of work and energy, the work done is equal to the gain in kinetic energy, 1 1 ∫0.5 F ds = 2 mv 2 , from which the velocity is v =

2(1901.8) = 21.8 m/s. m

Solution:

Doing work–energy for the system

0.4

∫0 (m A g sin 30° − µ k m A g cos30° + m B g) ds = 2 (m A + m B )v 22 [40 sin 30 ° − (0.15)(40) cos30 ° + 30](9.81)(0.4) =

1 (70)v 22 2

Solving for the velocity we find v 2 = 2.24 m/s.

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Problem 15.24 The system is released from rest. The 4-kg mass slides on the smooth horizontal surface. By using the principle of work and energy, determine the magnitude of the velocity of the masses when the 20-kg mass has fallen 1 m.

4 kg

Solution: (4 kg)(9.81 m/s2) 20 kg N magnitude of its velocity. Note that v is also the magnitude of the vertical velocity of the 4-kg mass. The only work done by external forces on the system is that due to the weight of the 20-kg mass, which is U 12 = (20 kg)(9.81 m/s 2 )s. (20 kg)(9.81 m/s2)

We equate this to the change in the total kinetic energy of the system, (20 kg)(9.81 m/s 2 )s =

We could solve this problem by analyzing the motion of each mass separately, but it saves a bit of time to apply the principle of work and energy to the two masses and the cable considered as a single system. Let s be the downward displacement of the 20-kg mass, and let v be the

1 1 (4 kg)v 2 2 + (20 kg)v 2 2 − 0. 2 2

Setting s = 1 m, we obtain v 2 = 4.04 m/s. 4.04 m/s.

Problem 15.25 Solve Problem 15.24 if the coefficient of kinetic friction between the 4-kg mass and the horizontal surface is µ k = 0.4.

4 kg

Solution: (4 kg)(9.81 m/s2) mkN

N 20 kg forces on the system is that due to the weight of the 20-kg mass and the friction force on the 4-kg mass, which is s

(20 kg)(9.81 m/s2)

We apply the principle of work and energy to the two masses and the cable considered as a single system. Let s be the downward displacement of the 20-kg mass, and let v be the magnitude of its velocity. Note that s is also the horizontal displacement of the 4-kg mass and v is the magnitude of its vertical velocity. The only work done by external

BandF_6e_ISM_C15_Pt1.indd 167

U 12 = [(20 kg)(9.81 m/s 2 ) − µ k N ] s. The normal force N = (4 kg)(9.81 m/s 2 ) = 39.2 N. We equate the work to the change in the total kinetic energy of the system, [(20 kg)(9.81 m/s 2 ) − 0.5(39.2 N)] s =

1 1 (4 kg)v 2 2 + (20 kg)v 2 2 − 0. 2 2

Setting s = 1 m, we obtain v 2 = 3.88 m/s. 3.88 m/s.

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Problem 15.26 Each box weighs 50 lb and the inclined surfaces are smooth. The system is released from rest. Determine the magnitude of the velocities of the boxes when they have moved 1 ft. 308

Solution:

458

308

U 12 = [W sin 45 ° − W sin 30°]s.

work done by external forces on the system is that due to the weights of the boxes, which is

We can apply the principle of work and energy to the two boxes and the cable considered as a single system. We denote the two boxes as A and B. It’s clear that A, the left box, will move downward. Let s be their displacement, and let v be the magnitude of their velocity. The only

We equate this to the change in the total kinetic energy of the system, 1 1 U 12 = (W /g)v 2 2 + (W /g)v 2 2 − 0, 2 2 obtaining v 2 = 2.58 ft/s. 2.58 ft/s.

Problem 15.27 The masses of the left and right boxes are 80 kg and 60 kg, respectively. The coefficient of kinetic friction between the boxes and the inclined surfaces is µ k = 0.2. The system is released from rest. Determine the magnitude of their velocity when they have moved 1 m.

308 458

Solution: on them. Note that the normal forces are N A = m A g cos 45 ° = 555 N and N B = m B g cos30 ° = 510 N. The work done is 458

mAg

mkNA NA

mBg

308

The normal force N = (4 kg)(9.81 m/s 2 ) = 39.2 N. We equate the work to the change in the total kinetic energy of the system,

NB mkNB

We apply the principle of work and energy to the two boxes and the cable considered as a single system. Denote the two boxes as A and B. It’s clear that A, the left box, will move downward. Let s be their displacement, and let v be the magnitude of their velocity. Work is done on the system by the weights of the boxes and by the friction forces acting

168

U 12 = [m A g sin 45 ° − µ k N A − m B g sin 30 ° − µ k N B ] s.

U 12 =

1 1 m v 2 + m Bv 2 2 − 0, 2 A 2 2

obtaining v 2 = 0.825 m/s. 0.825 m/s.

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Problem 15.28 The weights of the three blocks are W A = 100 lb, W B = 30 lb, and WC = 20 lb. All the sur faces are smooth. By applying the principle of work and energy to blocks A and B individually, determine the magnitude of their velocity when they have moved 1 ft. C

Solution:

T A

NC WB

458 458

The principle of work and energy for box B is

U 12 = (T − W B cos 45 °)s =

We have two equations in terms of T and v 2 . Setting s = 1 ft and solving, we obtain T = 32.6 lb, v 2 = 4.95 ft/s.

Work is done on the individual boxes by their weights and by the cable. Let s be the displacement of box A down the inclined surface. The principle of work and energy for it is U 12 = (W A cos 45 ° − T )s =

1 WB 2 v . 2 g 2

4.95 ft/s.

1 WA 2 v . 2 g 2

Problem 15.29 Solve Problem 15.28 by applying the principle of work and energy to blocks A and B, the cable connecting them, and the pulley considered as a single system.

Solution:

WB WA NA

C B A

458

The only work done by external forces on the system is that due to the weights of the boxes. Let s be the displacement of box A down the inclined surface. The principle of work and energy for the system is 1 WA 2 1 WB 2 U 12 = (W A cos 45 ° − W B cos 45 °)s = v + v . 2 g 2 2 g 2 Setting s = 1 ft and solving, we obtain v 2 = 4.95 ft/s. 4.95 ft/s.

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Problem 15.30 The weights of the three blocks are W A = 100 lb, W B = 30 lb, and WC = 20 lb. The coefficient of kinetic friction between all surfaces is µ k = 0.2. Determine the magnitude of the velocity of blocks A and B when they have moved 1 ft. (See Example 15.4.)

Solution:

When friction forces between parts of a system are involved, work and energy needs to be applied to the parts of the system separately, because although internal friction forces cancel out, the work done by them may not (see Example 15.3). The free-body diagrams of the individual boxes are

mkNC

T NC

C B

mkNB

A NB

mkNB

mkNC 458

mkNA

458

In the direction normal to the surface, we have the equilibrium equations N C = WC cos 45 ° = 14.1 lb, N B = W B cos 45 ° + N C = 35.4 lb, N A = W A cos 45 ° + N B = 106 lb. Let s denote the displacement of box A down the inclined surface. We apply the principle of work and energy to boxes A and B: 1 WA 2 v , 2 g 2 1 WB 2 (T − W B cos 45 ° − µ k N B − µ k N C )s = v . 2 g 2 (W A cos 45 ° − T − µ k N A − µ k N B )s =

We have two equations in terms of T and v 2 . Setting s = 1 ft and solving, we obtain T = 33.7 lb, v 2 = 2.37 ft/s. 2.37 ft/s.

Problem 15.31 In Example 15.6, suppose that the skier is moving at 20 m/s when he is in position 1. Determine the horizontal component of his velocity when he reaches position 2, 20 m below position 1.

Solution: 1 2 1 mv − mv 12 2 2 2 1 1 − m (9.81)(0 − 20) = mv 22 − m (20) 2 2 2

U 12 = −mg ( y 2 − y1 ) =

Solving for v 2 we find v 2 = 28.1 m/s.

Problem 15.32 Suppose that you stand at the edge of a 200-ft cliff and throw rocks at 30 ft/s in the three directions shown. Neglecting aerodynamic drag, use the principle of work and energy to determine the magnitude of the velocity of the rock just before it hits the ground in each case. Solution:

In all three cases, the work done by the weight of the rock is positive (its height decreases) and equal to the change in height times the weight: U 12 = W (200 ft). We equate this to the change in the rock’s kinetic energy, 1 W  1 W  U 12 = W (200 ft) =  v 2 2 −   (30 ft/s) 2 , 2 g  2 g  obtaining v 2 = 117 ft/s.

(a) 308 308

(b) (c)

200 ft

(a), (b), (c) 117 ft/s.

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Problem 15.33 The 30-kg box is sliding down the smooth surface at 1 m/s when it is in position 1. Determine the magnitude of the box’s velocity at position 2 in each case.

The normal force exerted on the box by the inclined surface does no work, so work is done only by its weight. The work is positive and equal to the change in height times the weight. In both cases,

U 12 = mg(2 m). We equate this to the change in the box’s kinetic energy, 1 1 U 12 = mg(2 m) = mv 2 2 − m(1 m/s) 2 , 2 2

Solution:

obtaining v 2 = 6.34 m/s. (a),(b) 6.34 m/s.

2m 2

2 608

408

(a)

(b)

Problem 15.34 Solve Problem 15.33 if the coefficient of kinetic friction between the box and the inclined surface is µ k = 0.2.

Solution:

In each case, the work the work done by the weight of the box is positive and equal to the change in height times the weight:

U weight = (2 m) mg. The question is, what work is done by the friction force?

2m 2

s u

608 (a)

mkN

408 (b)

From the free-body diagram (left figure), we see that N = mg cos θ. The friction force points in the direction opposite to the box’s motion, so the work done will be negative, and its magnitude is the product of the friction force and the distance s the box travels down the surface. From the right figure, we see that s = (2 m)/ sin θ, so the work done by friction is U friction = −µ k N s = −µ k mg cos θ

2m ( sin ) θ

(2 m) µ k mg . tan θ We equate the total work to the change in the box’s kinetic energy: 1 1 U weight + U friction = mv 2 2 − m (1 m/s) 2 . 2 2 In case (a) this yields v 2 = 5.98 m/s, and in case (b) it yields v 2 = 5.56 m/s. = −

(a) 5.98 m/s. (b) 5.56 m/s.

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Problem 15.35 In case (a), a 5-lb ball is released from rest at position 1 and falls to position 2. In case (b), the ball is released from rest at position 1 and swings to position 2. For each case, use the principle of work and energy to determine the magnitude of the ball’s velocity at position 2. [In case (b), notice that the force exerted on the ball by the string is perpendicular to the ball’s path.] Solution:

2 ft

The work is independent of the path, so both cases are

the same. U = −m (32.2 ft/s 2 )(0 − 2 ft) =

1 mv 2 − 0 ⇒ v 2 = 11.3 ft/s. 2 2

(a)

Problem 15.36 The length of the string is L = 1.5 m. When the 2-kg ball is in position 1 with the string horizontal, it is moving downward at 1 m/s. What is the magnitude of its velocity when it is in position 2?

(b)

1 408 L

Solution:

The force exerted on the ball by the string is normal to its path, and so does no work. The work done by the weight of the ball is positive and equal to the weight times the change in height: 2

U 12 = mgL sin 40 °. We equate this to the change in the ball’s kinetic energy, U 12 = mgL sin 40 ° =

1 1 mv 2 − m (1 m/s) 2 , 2 2 2

obtaining v 2 = 4.46 m/s. 4.46 m/s.

Problem 15.37 The length of the string is L = 1.5 m. The 2-kg ball is released from rest in position 1 with the string horizontal. What is the tension in the string when the ball is in position 2? Strategy: Use work and energy to determine the ball’s velocity, then apply Newton’s second law in terms of normal and tangential components.

Solution:

The force exerted on the ball by the string is normal to its path, and so does no work. The work done by the weight of the ball is positive and equal to the weight times the change in height:

U 12 = mgL sin 40 °. We equate this to the change in the ball’s kinetic energy, 1 mv 2 − 0, 2 2 obtaining v 2 = 4.35 m/s.

U 12 = mgL sin 40 ° =

Now consider the free-body diagram of the ball when it is in position 2:

408 L

408

We write Newton’s second law in the normal direction: v 2 ∑ Fn = T − mg sin 40 ° = ma n = m 2 . L From this equation we obtain T = 37.8 N. 37.8 N.

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Problem 15.38 The 400-lb wrecker’s ball swings at the end of a 25-ft cable. If the magnitude of the ball’s velocity at position 1 is 4 ft/s, what is the magnitude of its velocity just before it hits the wall at position 2?

658 958

Solution: U = −(400 lb)(−25 ft sin 95 ° − [−25 ft sin 65 °]) U =

(

)

1 400 lb (v 2 2 − [4 ft/s] 2 ) 2 32.2 ft/s 2

⇒ v 2 = 12.7 ft/s.

Problem 15.39 The 400-lb wrecker’s ball swings at the end of a 25-ft cable. If the magnitude of the ball’s velocity at position 1 is 4 ft/s, what is the maximum tension in the cable as the ball swings from position 1 to position 2?

658 958

Solution:

From the solution to Problem 15.37, the tension in the cable is T = mg sin α + mv 2 / L. Using conservation of energy, we have W (L sin α − L sin(65 °)) = (m / 2)(v 2 2 − v 1 2 ) where v1 = 4 ft / s. The maximum value for v 2 occurs when α = 90 °. Evaluating, we obtain v 2 = 12.9 ft / s and

T = 482.9 lb.

Problem 15.40 A stunt driver wants to drive a car through the circular loop of radius R = 5 m. He enters the loop at 25 m/s and coasts through. (a) What is the magnitude of the car’s velocity at the top of the loop? (b) What downward force is exerted on the 70-kg driver by the car seat when he is at the top of the loop?

Solution: (a) The force exerted on the car by the track is normal to its path, and so does no work. The work done by the car’s weight as it moves from the bottom to the top of the loop is negative and its magnitude is equal to the weight times the change in height: U 12 = −2 Rmg. We equate this to the change in the car’s kinetic energy, 1 1 mv 2 − mv 1 2 : 2 2 2 1 1 −2(5 m)(9.81 m/s 2 ) = v 2 2 − (25 m/s) 2 , 2 2

We write Newton’s second law in the normal direction: v 2 ∑ Fn = N + m d g = m d a n = m d 2 . R From this equation we obtain N = 5320 N (1200 lb). (a) 20.7 m/s. (b) 5320 N.

U 12 =

obtaining v 2 = 20.7 m/s. (b) Let m d = 70 kg, and let N denote the downward force exerted on the driver by the seat. Consider the free-body diagram of the driver when he is at the top of the loop: N

md g

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Problem 15.41 The 2-kg collar starts from rest at position 1 and slides down the smooth rigid wire. The y-axis points upward. What is the magnitude of the velocity of the collar when it reaches position 2? y

Solution:

The work done by the weight is U weight = mgh, where h = y1 − y 2 = 5 − (−1) = 6 m. From the principle of work and 1 energy, mgh = mv 2 , from which 2 v =

2 gh = 10.85 m/s.

(5, 5, 2) m

2 kg

x 2 z

(3, 21, 3) m

Problem 15.42 The 4-lb collar slides down the smooth rigid wire from position 1 to position 2. When it reaches position 2, the magnitude of its velocity is 24 ft/s. What was the magnitude of its velocity at position 1?

Solution: U = (4 lb)(6 − [−1]) ft =

(

)

(22, 6, 4) ft

1 4 lb ([24 ft/s] 2 − v 12 ) 2 32.2 ft/s 2

4 lb

⇒ v 1 = 11.2 ft/s. x

The lift is normal to the path and does no work. Because we know the initial and final velocities, we can use the principle of work and energy to determine the total work done. By subtracting from that the work done by the plane’s weight, we will know the work done by the thrust and drag.

(4, 21, 4) ft

Problem 15.43 The forces acting on the 30,000-lb airplane are the thrust T and drag D, which are parallel to the airplane’s path, the lift L, which is perpendicular to the path, and its weight W. The airplane climbs from an altitude of 3000 ft to an altitude of 15,000 ft. During the climb, the magnitude of its velocity increases from 800 ft/s to 1400 ft/s. (a) What work is done on the airplane by its lift during the climb? (b) What work is done by the thrust and drag combined? Solution:

D W We see that the work done by the thrust and drag is U 12 − U weight = 6.15E8 ft-lb − (−3.60E8 ft-lb) = 9.75E8 ft-lb. (a) Zero. (b) 9.75E8 ft-lb.

The airplane’s mass is m = W /g = = 932 slugs. The principle of work and energy is 1 1 U 12 = mv 2 2 − mv 1 2 2 2 1 1 = (932 slugs)(1400 ft/s) 2 − (932 slugs)(800 ft/s) 2 2 2 = 6.15E8 ft-lb. The work done by the plane’s weight is 30,000 lb/32.2 ft/s 2

U weight = −(30,000 lb)(15,000 ft − 3000 ft) = −3.60E8 ft-lb.

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Problem 15.44 The 2400-lb car is traveling 40 mi/h at position 1. If the combined effect of the aerodynamic drag on the car and the tangential force exerted on its wheels by the road is that they exert no net tangential force on the car, what is the magnitude of the car’s velocity at position 2?

Solution:

The initial velocity is

(

)

5280 v 1 = 40 = 58.67 ft/s 3600 The change in elevation of the car is h = 120(1 − cos30 °) + 100(1 − cos30 °) = 220(1 − cos30 °) = 29.47 ft The initial kinetic energy is 1  W  2 v = 128,265 ft-1b 2  g  1

120 ft

The work done by gravity is

308

U gravity = ∫ (−W ) ds = −Wh = −2400(h) = −70, 738.6 ft-lb. 0

From the principle of work and energy the work done is equal to the gain in kinetic energy:

1  W  2 1 W  v −  v 12 , 2  g  2 2 g 

U gravity =

100 ft

from which

308

v2 =

Problem 15.45 The 2400-lb car is traveling 40 mi/h at position 1. If the combined effect of the aerodynamic drag on the car and the tangential force exerted on its wheels by the road is that they exert a constant 400-lb tangential force on the car in the direction of its motion, what is the magnitude of the car’s velocity at position 2?

2 g(−70,738.6 + 128,265) = 39.3 ft/s = 26.8 mph. W

Solution:

From the solution to Problem 15.44, the work done by gravity is U gravity = −70,738.6 N-m due to the change in elevation of 1 W  the car of h = 29.47 ft, and  v 12 = 128,265 ft-1b. 2 g  The length of road between positions 1 and 2 is s = 120(30 °)

( 180π ° ) + 100(30°)( 180π ° ) = 115.2 ft.

The work done by the tangential force is s

U tgt = ∫ 400 ds = 400(115.2) = 46076.7 ft-lb. 0

From the principle of work and energy

120 ft 308

U gravity + U tgt =

1  W  2 1 W  v −  v 12 , 2  g  2 2 g 

from which 2(32.17)(−70,738.6 + 46,076.7 + 128,265) 2400 = 52.73 ft/s = 36 mph.

v =

100 ft

308

Problem 15.46 The mass of the rocket is 250 kg. Its engine has a constant thrust of 45 kN. The length of the launching ramp is 10 m. If the magnitude of the rocket’s velocity when it reaches the end of the ramp is 52 m/s, how much work is done on the rocket by friction and aerodynamic drag? Solution: Because we know the initial and final velocities, we can use the principle of work and energy to determine the total work done on the rocket. By subtracting from that the work done by its thrust and by its weight, we will know the work done by friction and drag.

BandF_6e_ISM_C15_Pt1.indd 175

The principle of work and energy is 1 1 U 12 = mv 2 2 − mv 1 2 2 2 1 = (250 kg)(52 m/s) 2 − 0 2 = 338,000 N-m.

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15.46 (Continued) The work done by the thrust is U thrust = (45,000 N)(10 m) = 450,000 N-m. The work done by the rocket’s weight is U weight = −(250 kg)(9.81 m/s 2 )(2 m) = −4910 N-m. Therefore the work done by friction and drag is U 12 − U thrust − U weight = 338,000 N-m − 450,000 N-m − (−4910 N-m) = −107,000 N-m. −107 kN-m.

Problem 15.47 A bioengineer interested in the energy requirements of sports determines from videotape that when the athlete begins his motion to throw the 7.25-kg shot (Fig. P15.47a), the shot is stationary and 1.50 m above the ground. At the instant the athlete releases it (Fig. P15.47b), the shot is 2.10 m above the ground. The shot reaches a maximum height of 4.60 m above the ground and travels a horizontal distance of 18.66 m from the point where it was released. How much work does the athlete do on the shot from the beginning of his motion to the instant he releases it?

Solution: Let v x0 and v y0 be the velocity components at the instant of release. Using the chain rule, dv y dv y dy dv y ay = = = v = −g, dt dy dt dy y and integrating, 0

4.6

∫v v y dv y = −g ∫2.1 dy. y0

1 − v y0 2 = −g(4.6 − 2.1), we find that v y0 = 7.00 m/s. The shot’s 2 1 x and y coordinates are given by x = v x 0 t, y = 2.1 + v y 0 t − gt 2 . 2 Solving the first equation for t and substituting it into the second,  x  1  x  2 y = 2.1 + v y 0   − g  v x 0  2  v x 0  Setting x = 18.66 m, y = 0 in this equation and solving for v x0 gives v x0 = 11.1 m/s. The magnitude of the shot’s velocity at release is U 2 = v x 0 2 + v y 0 2 = 13.1 m/s. Let U A be the work he does 1 1 U A − mg( y 2 − y1 ) = mv 22 − mv 12 2 2 U A − (7.25 kg)(9.81 m/s 2 )(2.1 − 1.5) = 12 (7.25)(13.1) 2 − 0, or U A = 666 N-m. y

(a)

vy0

(b) 2.1 m

vx0

4.6 m x 18.66 m

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Problem 15.48 A small pellet of mass m = 0.2 kg starts from rest at position 1 and slides down the smooth surface of the cylinder to position 2, where θ = 30 °. (a) What work is done on the pellet as it slides from position 1 to position 2? (b) What is the magnitude of the pellet’s velocity at position 2?

1 2 R

308 R cos 308

1 2 m

0.8 m

(b) Applying the principle of work and energy, 1 U 12 = (0.2 kg)v 2 2 − 0, 2 we obtain v 2 = 1.45 m/s. (a) 0.210 N-m. (b) 1.45 m/s.

Solution: (a) The force exerted by the surface of the cylinder is normal to the path and does no work. Let R = 0.8 m. The work done on the pellet by its weight as it goes to position 2 is positive and equals the product of the weight and the change in height: U 12 = mg( R − R cos θ) = (0.2 kg)(9.81 m/s 2 )(0.8 m)(1 − cos30°) = 0.210 N-m.

Problem 15.49* A small pellet of mass m = 0.2 kg starts from rest at position 1 and slides down the smooth surface of the cylinder. Determine the value of the angle θ at which the pellet loses contact with the cylindrical surface.

Consider the free-body diagram of the pellet as it falls:

en N

1 2 m

0.8 m

et mg

u Newton’s second law in the normal direction is v 2 ∑ Fn = mg cos θ − N = ma n = m 2 . (2) R Solving Eq. (1) for v 2 2 , substituting the result into Eq. (2) and solving for N, we obtain N = mg(3cos θ − 2). (3)

Solution:

The force exerted by the surface of the cylinder is normal to the path and does no work. Let R = 0.8 m. The work done on the pellet by its weight as it goes down is positive and equals the product of the weight and the change in height: U 12 = mg( R − R cos θ). Applying the principle of work and energy, U 12 = mgR (1 − cos θ) =

θ = 48.2 °.

1 mv 2 − 0. (1) 2 2

BandF_6e_ISM_C15_Pt1.indd 177

When θ = 0, the normal force N = mg. As θ increases, cos θ decreases, and the normal force becomes equal to zero when 3cos θ = 2, which is θ = 48.2 °. At greater angles, the normal force obtained from Eq. (3) becomes negative, which is not possible. At θ = 48.2 ° the particle separates from the cylindrical surface.

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Problem 15.50 Suppose that you want to design a bumper that will bring a 50-lb package moving at 10 ft/s to rest 6 in from the point of contact with the bumper. If friction is negligible, what is the necessary spring constant k ?

10 ft/s k

The package’s weight and the normal force exerted on it by the floor are normal to its path and so do no work. From Eq. (15.14), the work done by the spring as it is compressed 0.5 ft is 1 U 12 = − k (S 2 2 − S1 2 ) 2 1 = − k[(0.5 ft) 2 − 0]. 2 Equating this to the change in kinetic energy of the box, 1 1 U 12 = − k (0.5 ft) 2 = 0 − (1.55 slugs)(10 ft/s) 2 2 2 and solving, we obtain k = 621 lb/ft. k = 621 lb/ft.

Solution: 0.5 ft

N The mass of the package is m =

W 50 lb = = 1.55 slugs. g 32.2 ft/s 2

Problem 15.51 At the instant shown, the space between the 50-lb package moving at 10 ft/s and the spring is 4 in. You want to design the spring to bring the package to rest 3 in from the point of contact with the spring. The coefficient of kinetic friction between the package and the floor is µ k = 0.2. What is the necessary spring constant k ?

Solution: 0.333 ft

0.25 ft

W k mkN

The mass of the package is

10 ft/s k

m =

W 50 lb = = 1.55 slugs. g 32.2 ft/s 2

The package’s weight and the normal force exerted on it by the floor are normal to its path and so do no work. The work done by the friction force as the package moves from its initial position to where it comes to rest is U friction = −µ k N (0.583 ft) = −µ kW (0.583 ft). From Eq. (15.14), the work done by the spring as it is compressed 0.25 ft is 1 U spring = − k (S 2 2 − S1 2 ) 2 1 = − k[(0.25 ft) 2 − 0]. 2 Equating the total work to the change in kinetic energy of the box, 1 U friction + U spring = −µ kW (0.583 ft) − k (0.25 ft) 2 2 1 = 0 − m(10 ft/s) 2 2 and solving, we obtain k = 2300 lb/ft. k = 2300 lb/ft.

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Problem 15.52 The 50-lb package starts from rest, slides down the smooth ramp, and is stopped by the spring. (a) If you want the package to be brought to rest 6 in from the point of contact, what is the necessary spring constant k ? (b) What maximum deceleration is the package subjected to? Solution:

4 ft

308

(a) Find the spring constant 1 U 12 = mgh − kx 2 = 0 2 2mgh 2(50 lb)(4.5 ft) sin 30 ° = k = x2 (0.5 ft) 2 k = 900 lb/ft. (b) The maximum deceleration occurs when the spring reaches the maximum compression (the force is then the largest). kx − mg sin θ = ma k x − g sin θ m (900 lb/ft) a = (0.5 ft) − (32.2 ft/s 2 ) sin 30 ° 50 lb 32.2 ft/s 2 a =

(

)

a = 274 ft/s 2 .

Problem 15.53 The 50-lb package starts from rest, slides down the ramp, and is stopped by the spring. The coefficient of static friction between the package and the ramp is µ k = 0.12. If you want the package to be brought to rest 6 in from the point of contact, what is the necessary spring constant k ?

4 ft

Solution:

Find the spring constant 1 2 kx = 0 2

U 12 = mgd sin θ − µ k mg cos θ d −

308

2mgd (sin θ − µ k cos θ) x2 2(50 lb)(4.5 ft) k = (sin 30 ° − 0.12 cos30 °) (0.5 ft) 2

k =

k = 713 lb/ft.

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Problem 15.54 The system is released from rest with the spring unstretched. The spring constant is k = 150 N/m. Determine the magnitude of the velocity of the masses when the right mass has fallen 2 m.

Solution:

We consider the masses, cable and spring as a system. Let s denote the downward displacement of the right mass, which also equals the upward displacement of the left mass and the stretch of the spring. The work done by the weights of the masses is U weight = (20 kg) g s − (4 kg) g s. The work done by the spring is 1 1 U spring = − k (S 2 2 − S1 2 ) = − k s 2 . 2 2 We equate the total work to the change in kinetic energy of the masses, U weight + U spring = (20 kg) g s − (4 kg) g s − =

4 kg

20 kg

1 2 ks 2

1 1 (20 kg) v 2 2 + (4 kg) v 2 2 − 0. 2 2

Setting s = 2 m and solving, we obtain v 2 = 1.08 m/s. 1.08 m/s.

Problem 15.55 The system is released from rest with the spring unstretched. The spring constant is k = 200 N/m. What maximum downward velocity does the right mass attain as it falls?

We equate the total work to the change in kinetic energy of the masses: U weight + U spring = (20 kg) g s − (4 kg) g s − =

1 2 ks 2

1 1 (20 kg) v 2 2 + (4 kg) v 2 2 − 0. 2 2

To determine the value of s at which the maximum velocity occurs, we set the derivative of the left side of this equation with respect to s equal to zero, (20 kg) g − (4 kg) g − k s = 0, obtaining s = 0.785 m. Substituting this value into Eq. (1) and solving, we obtain v 2 = 2.27 m/s. 2.27 m/s.

4 kg

20 kg k

Solution:

180

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Problem 15.56 The system is released from rest with the spring unstretched. The 4-kg mass slides on the smooth horizontal surface. The spring constant is k = 90 N/m. Determine the magnitude of the velocity of the masses when the 20-kg mass has fallen 0.5 m.

Solution: 2 k (4 kg)(9.81 m/s )

4 kg s

(20 kg)(9.81 m/s2)

We consider the masses, cable and spring as a system. Let s denote the downward displacement of the 20-kg mass, which also equals the horizontal displacement of the 4-kg mass and the stretch of the spring. The work done by the weight of the 20-kg mass is U weight = (20 kg) g s.

20 kg

From Eq. (15.14), the work done by the spring as it is stretched is 1 U spring = − k (S 2 2 − S1 2 ) 2 1 = − k s 2. 2 The principle of work and energy is 1 k s2 2 1 1 = (20 kg) v 2 2 + (4 kg) v 2 2 − 0. 2 2

U weight + U spring = (20 kg) g s −

Setting s = 0.5 m and solving, we obtain v 2 = 2.69 m/s. 2.69 m/s.

Problem 15.57 The system is released from rest with the spring unstretched. The coefficient of kinetic friction between the 4-kg mass and the horizontal surface is µ k = 0.6. The spring constant is k = 90 N/m. Determine the magnitude of the velocity of the masses when the 20-kg mass has fallen 0.5 m.

Solution: 2 k (4 kg)(9.81 m/s )

mkN

4 kg

(20 kg)(9.81 m/s2)

20 kg

The normal force exerted on the 4-kg mass by the surface equals its weight, so the work done by the friction force is U friction = −µ k (4 kg) g s. From Eq. (15.14), the work done by the spring as it is stretched is 1 U spring = − k (S 2 2 − S1 2 ) 2 1 = − k s 2. 2 The principle of work and energy is U weight + U friction + U spring = (20 kg) g s − µ k (4 kg) g s − =

1 2 ks 2

1 1 (20 kg) v 2 2 + (4 kg) v 2 2 − 0. 2 2

Setting s = 0.5 m and solving, we obtain v 2 = 2.50 m/s. 2.50 m/s.

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Problem 15.58 The 40-lb crate is released from rest on the smooth inclined surface with the spring unstretched. The spring constant is k = 8 lb/ft. (a) How far down the inclined surface does the crate slide before it stops? (b) What maximum velocity does the crate attain on its way down?

(a) We can determine the maximum distance by setting v 2 = 0 , 2W sin 30 ° = 5 ft. obtaining x 2 = k (b) By taking the derivative of the kinetic energy with respect to x 2 , we can determine the position at which the velocity is a maximum.

(

)

(

)

d 1 d 1 = mv 2 2 = W x 2 sin30 ° − kx 2 2 = W sin30 ° − kx 2 2 dx 2 2 dx 2 Setting this derivative equal to zero, we obtain

x2 =

W sin 30 ° = 2.5 ft. k

Then substituting this position into Eq. (1), we obtain the maximum velocity v 2 = 6.34 ft/s. (a) 5 ft/s; (b) 6.34 ft/s. 308

Solution:

The mass of the crate is m = 40 lb/32.2 ft/s 2 = 1.24 slugs. Let x be the displacement of the crate down the inclined surface relative to its initial position. Work is done on the crate by its weight and the force exerted by the spring. When the crate has moved from x1 = 0 to a position x 2 it has moved downward a distance x 2 sin30 ° and stretched the spring an amount x 2 , so the work done is 1 U 12 = W ( x 2 sin30°) − kx 2 2 . 2 It starts from rest, so the principle of work and energy is 1 1 U 12 = W ( x 2 sin30°) − kx 2 2 = mv 2 2 − 0 (1) 2 2

Problem 15.59 Solve Problem 15.58 if the coefficient of kinetic friction between the 40-lb mass and the inclined surface is µ k = 0.2.

308

(a) We can determine the maximum distance by setting v 2 = 0, 2( W sin 30 ° − f ) = 3.27 ft. obtaining x 2 = k (b) By taking the derivative of the kinetic energy with respect to x 2 , we can determine the position at which the velocity is a maximum.

(

)

(

1 d 1 d W x 2 sin30 ° − fx 2 − kx 2 2 = mv 2 2 = dx 2 2 dx 2 2 k

)

= W sin30 ° − f − kx 2 . Setting this derivative equal to zero, we obtain x2 =

W sin 30 ° − f = 1.63 ft. k

Then substituting this position into Eq. (1), we obtain the maximum velocity v 2 = 4.15 ft/s.

308

(a) 3.27 ft/s; (b) 4.15 ft/s.

Solution:

The mass of the crate is m = 40 lb/32.2 ft /s 2 = 1.24 slugs. There is no acceleration normal to the inclined surface, so ∑ Fy = N − W cos30 ° = 0.

308

The friction force is f = µ k N = µ k W cos30 ° = 6.93 lb. Let x be the displacement of the crate down the inclined surface relative to its initial position. Work is done on the crate by its weight and the force exerted by the spring. When the crate has moved from x1 = 0 to a position x 2 it has moved downward a distance x 2 sin30 ° and stretched the spring an amount x 2 , The total work done by the weight, friction force, and spring is 1 U 12 = W ( x 2 sin30 ° ) − fx 2 − kx 2 2 . 2

W f N

The crate starts from rest, so the principle of work and energy is U 12 = W ( x 2 sin30 ° ) − fx 2 −

182

1 1 kx 2 = mv 2 2 − 0. 2 2 2

(1)

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Problem 15.60 The 4-kg collar starts from rest in position 1 on the smooth bar with the spring unstretched. The spring constant is k = 100 N/m. How far does the collar fall relative to position 1?

Solution: V0 = V f = 0 Let position 2 be the location where the collar comes to rest Ks 2 U 12 = − + mgs 2 1 1 Also U 12 = mV f2 − mV02 = 0 2 2 Ks 2 Thus 0 = mgs − 2 s(2mg − Ks) = 0 Solving, s = 0.785 m.

1 k

s 1 mg

Problem 15.61 In position 1 on the smooth bar, the 4-kg collar has a downward velocity of 1 m/s and the spring is unstretched. The spring constant is k = 100 N/m. What maximum downward velocity does the collar attain as it falls?

where m = 4 kg and V1 = 1 m/s Thus, 1 1 Ks 2 mV22 − mV12 = − + mgs 2 2 2 Finding

Solution:

The work is

Ks 2 U 12 = − + mgs 2 Also, U 12 =

1 1 mV22 − mV12 2 2

BandF_6e_ISM_C15_Pt1.indd 183

mV2

(1)

dV2 , and setting it to zero, ds

dV2 = −Ks + mg = 0 ds s = mg /k = 0.392 m

Solving (1) for V2 we get V2 = 2.20 m/s.

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Problem 15.62 The 4-kg collar starts from rest in position 1 on the smooth bar. The tension in the spring in position 1 is 20 N. The spring constant is k = 100 N/m. How far does the collar fall relative to position 1? k

Solution:

For this problem, we need a new reference for the spring. If the tension in the spring is 20 N T = kδ 0

K = 100 N/m

and

δ 0 = 0.2 m.

(the initial stretch)

In this case, we have s + 0.2

U 12 = −

Ks 2 s + 0.2 + mgs 2 0.2 0.2

Also V0 = V f = 0 ∴ U 12 = 0 K (s + 0.2) 2 K (0.2) 2 + + mg(s + 0.2) − mg(0.2) 2 2

0 = −

Solving, we get s = 0.385 m.

Problem 15.63 The 4-kg collar is released from rest in position 1 on the smooth bar. If the spring constant is k = 6 kN/m and the spring is unstretched in position 2, what is the velocity of the collar when it has fallen to position 2? Solution: Denote d = 200 mm, h = 250 mm. The stretch of the spring in position 1 is S1 = h 2 + d 2 − d = 0.120 m and at 2S 2 = 0. The work done by the spring on the collar is U spring = ∫

1 (−ks) ds =  − ks 2  = 43.31 N-m.  2  0.120

0 0.12

250 mm

The work done by gravity is U gravity = ∫

−h 0

(−mg) ds = mgh = 9.81 N-m.

1 From the principle of work and energy U spring + U gravity = mv 2 , 2 from which v =

( m2 )(U

spring

200 mm

+ U gravity ) = 5.15 m/s.

Problem 15.64 The 4-kg collar is released from rest in position 1 on the smooth bar. The spring constant is k = 4 kN/m. The tension in the spring in position 2 is 500 N. What is the velocity of the collar when it has fallen to position 2?

Solution:

Denote d = 200 mm, h = 250 mm. The stretch of the spring at position 2 is S2 =

T 500 = = 0.125 m. k 4000

250 mm

The unstretched length of the spring is L = d − S 2 = 0.2 − 0.125 = 0.075 m. The stretch of the spring at position 1 is S1 = h 2 + d 2 − L = 0.245 m. The work done by the spring is U spring = ∫

S2 S1

(−ks) ds =

The work done by gravity is U gravity = mgh = 9.81 N-m. From the 1 principle of work and energy is U spring + U gravity = mv 2 , from 2 2(U spring + U gravity ) which v = = 7.03 m/s. m

184

1 k (S12 − S 22 ) = 88.95 N-m. 2

200 mm

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Problem 15.65 The 4-kg collar starts from rest in position 1 on the smooth bar. Its velocity when it has fallen to position 2 is 4 m/s. The spring is unstretched when the collar is in position 2. What is the spring constant k ?

1 2 mv = 32 N-m. 2 From the solution to Problem 15.63, the stretch of the spring in position 1 is S1 = h 2 + d 2 − d = 0.120 m. The potential of the spring is

Solution:

The kinetic energy at position 2 is

U spring = ∫ (−ks) ds = S1

1 2 kS . 2 1

The work done by gravity is U gravity = mgh = 9.81 N-m. From the 1 principle of work of work and energy, U spring + U gravity = mv 2 . 2 Substitute and solve: k =

250 mm

2( 12 mv 2 − U gravity) = 3082 N/m. S12

2 200 mm

Problem 15.66 The 10-kg collar starts from rest at position 1 and slides along the smooth bar. The y-axis points upward. The spring constant is k = 100 N/m, and the unstretched length of the spring is 2 m. What is the velocity of the collar when it reaches position 2? y

2 (4, 4, 2) m

Solution:

The work done by the weight of the collar as it moves from position 1 to position 2 is

U weight = −mg (4 m − 1 m). The lengths of the spring at positions 1 and 2 are L1 =

(6 m − 1 m) 2 + (2 m − 1 m) 2 + (1 m − 0) 2 = 5.20 m,

L2 =

(6 m − 4 m) 2 + (2 m − 4 m) 2 + (1 m − 2 m) 2 = 3 m.

The work done by the spring as the collar moves from position 1 to position 2 is 1 U spring = − k (S 2 2 − S1 2 ) 2 1 = − k[( L 2 − 2 m) 2 − ( L1 − 2 m) 2 ]. 2

(6, 2, 1) m 1 (1, 1, 0) m

Applying the principle of work and energy, U weight + U spring = −mg (4 m − 1 m) −

1 k[( L 2 − 2 m) 2 2

−( L1 − 2 m) 2 ] 1 = 0 − mv 2 2 , 2

and solving, we obtain v 2 = 5.77 m/s. 5.77 m/s.

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Problem 15.67 A spring-powered mortar is used to launch 5-lb packages of fireworks into the air. The package starts from rest with the spring compressed to a length of 6 in. The unstretched length of the spring is 30 in. If the spring constant is k = 250 lb/ft, what is the magnitude of the velocity of the package as it leaves the mortar?

30 in

Solution:

Let 1 denote the initial position of the package and 2 its position when the spring is unstretched. The work done by its weight while the package moves from 1 to 2 is

U weight = W (2 ft)sin 60 °. The work done by the spring is 1 U spring = − k (S 2 2 − S1 2 ) 2 1 = − k[0 − (2 ft) 2 ]. 2 Equating this to the change in kinetic energy of the package, 1 U weight + U spring = W (2 ft)sin 60 ° + k (2 ft) 2 2 1W 2 = v 2 − 0, 2 g and solving, we obtain v 2 = 80.9 ft/s. 80.9 ft/s.

6 in

608

Problem 15.68 A spring-powered mortar is being designed to launch 5-lb packages of fireworks into the air. The package starts from rest with the spring compressed to a length of 6 in. The unstretched length of the spring is 30 in. If you want to design the mortar to send the package to a height of 100 ft above its position when the spring is unstretched, what is the necessary spring constant k ? (Neglect aerodynamic drag.)

30 in

Solution:

Let 1 denote the initial position of the package and 2 its position when the spring is unstretched.

6 in

The work done by its weight while the package moves from 1 to 2 is U weight = W (2 ft)sin 60 °.

608

The work done by the spring is 1 U spring = − k (S 2 2 − S1 2 ) 2 1 = − k[0 − (2 ft) 2 ]. 2

We equate the total work to the change in kinetic energy of the package: U weight + U spring = W (2 ft)sin 60 ° + 1W 2 = v − 0. 2 g 2

1 k (2 ft) 2 2

(1)

We need to determine the value of v 2 that will send the package 100 ft above position 2. Then we can solve Eq. (1) for the necessary spring constant. The vertical component of the velocity in position 2 is v 2 sin 60 °. Let y denote the upward displacement of the package relative to its position when the spring is unstretched. We can integrate the constant acceleration to determine the velocity and position as functions of time: dv y ay = = −g, dt vy

∫v sin 60° dv y = ∫0 −g dt,

∫0 dy = ∫0 (v 2 sin 60° − gt ) dt, y = v 2t sin 60 ° −

1 2 gt . 2

(3)

Setting v y = 0 in Eq. (2), we obtain the time at which the package reaches its highest point: t =

v 2 sin 60 ° . g

Substituting this result into Eq. (3) gives the maximum height: y =

v 2 2 sin 2 60 ° . 2g

Setting y = 100 ft solving, we obtain v 2 = 92.7 ft/s. Then substituting this value into Eq. (1) and solving gives k = 329 lb/ft. k = 329 lb/ft.

(2)

vy =

186

dy = v 2 sin 60 ° − gt, dt

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Problem 15.69 Suppose an object has a string or cable with constant tension T attached as shown. The force exerted on the object can be expressed in terms of polar coordinates as F = −Te r . Show that the work done on the object as it moves along an arbitrary plane path from a radial position r1 to a radial position r2 is U 12 = −T (r2 − r1 ).

Solution: U = ∫

The work done on the object is

F ⋅ d s.

Suppose that the arbitrary path is defined by d r = (dre r + rd θe θ ), and the work done is U = ∫

r2 r1

= −∫

F ⋅ d r = ∫ −T (e r ⋅ e r ) dr + ∫ r2

T (re r ⋅ e θ )r d θ

T dr = −T (r2 − r1 )

since e r ⋅ e s = 0 by definition.

Problem 15.70 The 2-kg collar is initially at rest at position 1. A constant 100-N force is applied to the string, causing the collar to slide up the smooth vertical bar. What is the velocity of the collar when it reaches position 2 ? (See Problem 15.69.)

200 mm 2

Solution:

The constant force on the end of the string acts through a distance s = 0.5 2 + 0.2 2 − 0.2 = 0.3385 m. The work done by the constant force is U F = Fs = 33.85 N-m. The work done by gravity on the collar is

500 mm 100 N

U gravity = ∫ (−mg) ds = −mgh = −(2)(9.81)(0.5) 0

= −9.81 N-m. From the principle of work and energy: 1 U F + U gravity = mv 2 , 2

2(U F + U gravity ) = 4.90 m/s. m

from which v =

Problem 15.71 The 8-kg collar starts from rest at position 1. The tension in the string is 200 N, and the y-axis points upward. If friction is negligible, what is the magnitude of the velocity of the collar when it reaches position 2? y

2 (4, 4, 2) m

(1, 1, 0) m

The work done by the weight of the collar as it moves from position 1 to position 2 is U weight = −mg (4 m − 1 m). The lengths of the spring at positions 1 and 2 are L1 =

(6 m − 1 m) 2 + (2 m − 1 m) 2 + (1 m − 0) 2 = 5.20 m,

L2 =

(6 m − 4 m) 2 + (2 m − 4 m) 2 + (1 m − 2 m) 2 = 3 m.

The work done by the string as the collar moves from position 1 to position 2 is U string = (200 N)( L1 − L 2 ).

(6, 2, 1) m 1

Solution:

Applying the principle of work and energy,

200 N

U weight + U spring = −mg (4 m − 1 m) + (200 N)( L1 − L 2 ) 1 mv 2 − 0, 2 2 and solving, we obtain v 2 = 7.14 m/s. =

x z

BandF_6e_ISM_C15_Pt1.indd 187

7.14 m/s.

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Problem 15.72 As the F/A-18 lands at 210 ft/s, the cable from A to B engages the airplane’s arresting hook at C. The arresting mechanism maintains the tension in the cable at a constant value, bringing the 26,000-lb airplane to rest at a distance of 72 ft. What is the tension in the cable? (See Problem 15.69.) Solution: U = −2T ( (72 ft) 2 + (33 ft) 2 − 33 ft) =

(

)

1 26,000 lb (0 − [210 ft/s] 2 ) 2 32.2 ft/s 2

T = 193,000 lb.

72 ft

C A

B 66 ft

Problem 15.73 If the airplane in Problem 15.72 lands at 240 ft/s, what distance does it roll before the arresting system brings it to rest? Solution: U = −2(193,000 lb)( s 2 + (33 ft) 2 − 33 ft) =

(

)

1 26,000 lb (0 − [240 ft/s] 2 ) 2 32.2 ft/s 2

Solving we find

s = 87.3 ft.

72 ft

C A

B 66 ft

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Problem 15.74 A spacecraft 320 km above the surface of the Earth is moving at escape velocity v esc = 10,900 m/s. What is its distance from the center of the Earth when its velocity is 50 percent of its initial value? The radius of the Earth is 6370 km. (See Example 15.7.) vesc

1 1 1 1 U 12 = mgR E2  −  = mv 22 − mv 12  r2 r1  2 2 −1  v 2 − v 12 1 r2 =  2 +  2  2 gR E r1 

 (5450 m/s) 2 − (10,900 m/s) 2  −1 1 r2 =  +  2 2  2(9.81 m/s )(6,370,000 m) 6,690,000 m  r2 = 26,600 km.

320 km

Problem 15.75 A piece of ejecta is thrown up by the impact of a meteor on the Moon. When it is 1000 km above the Moon’s surface, the magnitude of its velocity (relative to a nonrotating reference frame with its origin at the center of the Moon) is 200 m/s. What is the magnitude of its velocity just before it strikes the Moon’s surface? The acceleration due to gravity at the surface of the Moon is 1.62 m/s 2 . The Moon’s radius is 1738 km. 1000 km

Solution:

200 m/s

Solution:

The work done on the ejecta as it falls from 1000 km is RM RM R2 (−Wejecta ) ds = ∫ U ejecta = ∫ −mg M M ds RM + h RM + h s2

(

Solution:

Denote the work to be done by the rocket by U rocket. Denote R park = 6700 km, Rgeo = 42222 km. The work done by the satellite's weight as it moves from the parking orbit to the geosynchronous orbit is

U transfer = ∫

Rgeo Rpark

F ds = ∫

Rgeo Rpark

(−mg Rs ds ) 2 E 2

)

R M2  R M

 h , =  mg M = mg M R M  s  RM + h RM + h  U ejecta = 1.028 m × 10 6 N-m. From the principle of work and energy, at the Moon’s surface: m m U ejecta =  v 2  −  v 2   2  surface  2  RM + h from which

Problem 15.76 A satellite in a circular orbit of radius r around the Earth has velocity v = gR E 2 /r , where R E = 6370 km is the radius of the Earth. Suppose you are designing a rocket to transfer a 900-kg communication satellite from a circular parking orbit with 6700-km radius to a circular geosynchronous orbit with 42,222-km radius. How much work must the rocket do on the satellite?

The kinetic energy at h = 1000 km is

 mv 2  = 2 × 10 4 N-m.  2  RM + h

v surface =

2(1.028 × 10 6 + 2 × 10 4 ) = 1448 m/s.

U transfer = −4.5 × 10 9 N-m. From the principle of work and energy: 1 1 U transfer + U rocket =  mv 2  . −  mv 2  2  geo 2  park from which 1 1 U rocket =  mv 2  −  mv 2  − U transfer . 2  geo 2  park Noting m  gR E2   1 mv 2  =  = 4.24 × 10 9 N-m,  2  2  Rgeo  geo m R2   1 mv 2  =  g E  = 2.67 × 10 10 N-m,  2  2  R park  park from which U rocket = 2.25 × 10 10 N-m.

 1 R 2  Rgeo 1   = mgR E2  − , =  mg E  s  Rpark R park    Rgeo

Problem 15.77 The force exerted on a charged particle by a magnetic field is F = q v × B, where q and v are the charge and velocity of the particle and B is the magnetic field vector. Suppose that other forces on the particle are negligible. Use the principle of

BandF_6e_ISM_C15_Pt1.indd 189

work and energy to show that the magnitude of the particle’s velocity is constant. Solution:

The force vector F is given by a cross product involving v. This means that the force vector is ALWAYS perpendicular to the velocity vector. Hence, the force field does no work on the charged particle - it only changes the direction of its motion. Hence, if work is zero, the change in kinetic energy is also zero and the velocity of the charged particle is constant.

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Problem 15.78 The 10-lb box is released from rest at position 1 and slides down the smooth inclined surface to position 2. (a) If the datum is placed at the level of the floor as shown, what is the sum of the kinetic and potential energies of the box when it is in position 1? (b) What is the sum of the kinetic and potential energies of the box when it is in position 2? (c) Use conservation of energy to determine the magnitude of the box’s velocity when it is in position 2.

2 308

5 ft 2 ft Datum

Solution: (a)

T1 + V1 = 0 + (10 lb)(5 ft) = 50 ft-lb.

(b) T1 + V1 = T2 + V2 = 50 ft-lb. (c) 50 ft-1b =

(

)

10 lb 1 v 2 + (10 lb)(2 ft) ⇒ v 2 = 13.9 ft/s. 2 32.2 ft/s 2 2

Problem 15.79 The 0.45-kg soccer ball is 1 m above the ground when it is kicked upward at 12 m/s. Use conservation of energy to determine the magnitude of the ball’s velocity when it is 4 m above the ground. Obtain the answer by placing the datum (a) at the level of the ball’s initial position and (b) at ground level. 12 m/s

Solution:

x Datum 1m

1m x Datum

12 m/s (a) Datum

(b)

We use a cartesian coordinate system with the y axis upward. In case (a) we place the origin at the ball’s initial position, and in case (b) we place it at ground level. In both cases, we can write the potential energy of the ball as V = mgy. Datum

(a)

In case (a), conservation of energy states that 1 1 mv 2 + V1 = mv 2 2 + V2 : 2 1 2 1 1 (1) mv 1 2 + mgy1 = mv 2 2 + mgy 2 , 2 2 1 1 (12 m/s) 2 + 0 = v 2 2 + (9.81 m/s 2 )(3 m). 2 2 In Eq. (1) we have divided out the mass, which appeared in each term. In case (b), conservation of energy states that 1 1 mv 2 + V1 = mv 2 2 + V2 : 2 1 2 1 1 2 (2) mv + mgy1 = mv 2 2 + mgy 2 , 2 1 2 1 1 (12 m/s) 2 + (9.81 m/s 2 )(1 m) = v 2 2 + (9.81 m/s 2 )(4 m). 2 2 Comparing Eq. (1) to Eq. (2), they are equivalent. We will obtain the same equation for any height of datum we choose. Solving, we obtain v 2 = 9.23 m/s. (a), (b) 9.23 m/s.

190

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Problem 15.80 The 20-kg package moves to the right at 3 m/s on the smooth surface. The spring constant is k = 15,600 N/m. (a) What is the sum of the kinetic and potential energies of the system consisting of the package and spring? (b) Use conservation of energy to determine the velocity of the package when it has moved 80 mm to the right after contacting the spring.

Solution: (a) Only the spring does work on the package, and the spring force is conservative. The spring is unstretched initially, so from Eq. (15.24), its initial potential energy is zero. That means that the sum of the kinetic and potential energies in the beginning and throughout the motion of the package is equal to the initial kinetic energy of the package: 1 2 mv 2 1 = (20 kg)(3 m/s) 2 2 = 90 N-m.

T +V =

3 m/s

(b) Let s denote the displacement of the package to the right relative to its position when it comes into contact with the spring. The potential energy of the spring is

V =

1 2 ks . 2

Conservation of energy states that 1 1 mv 2 + V1 = mv 2 2 + V2 : 2 1 2 1 1 1 mv 1 2 + 0 = mv 2 2 + k s 2 , 2 2 2 1 1 (20 kg)(3 m/s) 2 + 0 = (20 kg) v 2 2 2 2 1 + (15,600 N/m)(0.08 m) 2 . 2

(1)

Solving, we obtain v 2 = 2.00 m/s. (a) 90 N-m. (b) 2.00 m/s.

Problem 15.81 The 0.4-kg collar starts from rest at position 1 and slides down the smooth rigid wire. The y-axis points upward. Use conservation of energy to determine the magnitude of the velocity of the collar when it reaches point 2.

Solution:

Assume gravity acts in the −y direction and that y = 0 1 is the datum. By conservation of energy, mv 2 + V = constant where 2 V = mgy. Thus, 1 2 1 mv + mgy1 = mv 22 + mgy 2 2 1 2 m = 0.4 kg, g = 9.81 m/s 2 , v 1 = 0, y 2 = 0, and y1 = 5 m. Thus

0 + (0.4)(9.81)(5) =

(5, 5, 2) m

1 (0.4)v 22 + 0 2

v 2 = 9.90 m/s.

0.4 kg

(5, 5, 2) m

x 2 z

0.4 kg

(3, 0, 2) m

x 2

(3, 0, 2) m

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Problem 15.82 At the instant shown, the 20-kg mass is moving downward at 1.6 m/s. Let d be the downward displacement of the mass relative to its present position. Use conservation of energy to determine the magnitude of the velocity of the 20-kg mass when d = 1 m.

Solution: 4 kg

m1 = 4 kg

20 kg

m 2 = 20 kg v 1 = 1.6 m/s g = 9.81 m/s 2

d = 1m 1.6 m s

Energy for the system is conserved

(

) (

1.6 m/s

)

1 1 1 1 m v2 + 0 + m v 2 + 0 = m1v 22 + m 2v 22 2 1 1 2 2 1 2 2 4 kg

+ m1 g ( d ) − m 2 g ( d )

(m1 + m 2 )v 12 = (m1 + m 2 )v 22 + 2(m1 − m 2 ) gd

d B

State (2)

VA 5 VB 5 0

Solution: m = 2 kg L = 1m Use conservation of energy State 1 θ = 0; State 2,0 < θ < 180 ° Datum: θ = 0, v 1 = 0, g = 9.81 m/s 2 1 2 1 mv + mg(0) = mv 22 + mg(−L sin θ) 2 1 2 KE =

Datum State (1)

Substituting known values and solving v 2 = 3.95 m/s.

Problem 15.83 The mass of the ball is m = 2 kg, and the string’s length is L = 1 m. The ball is released from rest in position 1 and swings to position 2, where θ = 40 °. (a) Use conservation of energy to determine the magnitude of the ball’s velocity at position 2. (b) Draw graphs of the kinetic energy, the potential energy, and the total energy for values of θ from zero to 180 °.

20 kg

1 2 mv 2 2

V = −mgL sin θ for all θ. Total energy is always zero

(datum value). u

(a) Evaluating at θ = 40 °, v 2 = 3.55 m/s.

192

KE, PE, and (PE + KE) N-m

20 15 10 5 0 –5 –10 –15 –20

Kinetic and Potential Energy vs a KE positive, PE negative, (KE + PE) zero KE PE TOT

60 80 100 120 Theta (degrees)

140

160

180

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Problem 15.84 The mass of the ball is m = 2 kg and the string’s length is L = 1 m. The ball is released from rest in position 1. When the string is vertical, it hits the fixed peg shown. (a) Use conservation of energy to determine the minimum angle θ necessary for the ball to swing to position 2. (b) If the ball is released at the minimum angle θ determined in part (a), what is the tension in the string just before and just after it hits the peg?

1 L 2

Solution:

Energy is conserved. v 1 = v 2 = 0 Use θ = 90 ° as

the datum. (a)

1 2 1 L mv + mg(−L cos θ1 ) = mv 22 − mg 2 1 2 2

mV 3 L

L 0 − mgL cos θ1 = 0 − mg 2 cos θ1 =

1 2

θ = 60 °. mg

(b) Use conservation of energy to determine velocity at the lowest point, (state 3) (v 1 ≡ 0) 1 2 1 mv − mgL cos60 ° = mv 32 − mgL 2 1 2 1 2 mv 3 = mgL − mgL /2 2 m2 v 32 = gL = 9.81 2 s v 3 = 3.13 m/s at θ = 0 °.

mV 3 (L/2)

Before striking the peg T1 − mg = mv 32 /L T1 = (2)(9.81) + (2)(9.81)/(1) T1 = 39.2 N. After striking the peg.

T − mg = mv 32 /(L /2) T = (2)(9.81) + 2[(2)(9.81)/1] T = 58.9 N.

Problem 15.85 A small pellet of mass m = 0.2 kg starts from rest at position 1 and slides down the smooth surface of the cylinder to position 2. The radius R = 0.8 m. Use conservation of energy to determine the magnitude of the pellet’s velocity at position 2 if θ = 45 °. 1

Solution:

Work is done on the pellet only by its weight. We can express the potential energy in terms of the pellet’s height above the horizontal surface:

V = mgR cos θ. Conservation of energy states that T1 + V1 = T2 + V2 : 1 0 + mgR cos 20 ° = mv 2 2 + mgR cos θ. 2 Setting θ = 45 ° and solving, we obtain v 2 = 1.91 m/s.

208 2 R

BandF_6e_ISM_C15_Pt1.indd 193

1.91 m/s.

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Problem 15.86 A small pellet of mass m = 0.2 kg starts from rest at position 1 and slides down the smooth surface of the cylinder to position 2. The radius R = 0.8 m. Use conservation of energy to determine the magnitude of the pellet’s velocity as a function of θ and draw a graph of it for 20 ° ≤ θ ≤ 40 °. 1

Work is done on the pellet only by its weight. We can express the potential energy in terms of the pellet’s height above the horizontal surface:

V = mgR cos θ. Conservation of energy states that T1 + V1 = T2 + V2 : 1 0 + mgR cos 20 ° = mv 2 2 + mgR cos θ. 2 We obtain v2 =

208 2 R

Solution:

2 gR (cos 20 ° − cos θ).

The graph is shown:

m 1.8 1.6 1.4

v2, m/s

1.2 1 0.8 0.6 0.4 0.2 0 20

u, degrees

Problem 15.87 The bar is smooth. The 10-kg slider at point A is given a downward velocity of 12 m/s. (a) What is the magnitude of the slider’s velocity at point C? (b) At the instant the slider passes point B, what is the magnitude of the normal force exerted on it by the bar?

1m C

Solution:

Only its weight does work on the slider. Let the datum be placed at the level of point A. In terms of a coordinate system with its origin at A and the y axis upward, the potential energy is V = mgy.

10 kg

A 1m

(a) Applying conservation of energy to points A and C, T1 + V1 = T2 + V2 : 1 1 m(12 m/s) 2 + 0 = mv 2 2 + mg(2 m). 2 2

We obtain v 2 = 10.2 m/s. (b) Applying conservation of energy to points A and B, T1 + V1 = T2 + V2 : 1 1 m(12 m/s) 2 + 0 = mv 2 2 + mg(−1 m), 2 2 We obtain v 2 = 12.8 m/s. Now consider the free-body of the slider when it is at B:

v2 : R (12.8 m/s) 2 N − (10 kg) g = (10 kg) . 1m

∑ Fn = N − mg = ma n = m

This yields N = 1730 N. Answer:

et mg

194

Newton’s second law in the normal direction is

(a) 12.8 m/s. (b) 1.73 kN.

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Problem 15.88 The bar is smooth. Determine the minimum downward velocity at point A that would cause the 10-kg slider to reach point D.

10 kg

The slider needs to be given a velocity large enough to reach the high point between C and D. We can assume its velocity at that point is zero, and use conservation of energy to determine the necessary initial velocity at A.

Only its weight does work on the slider. Let the datum be placed at the level of point A. In terms of a coordinate system with its origin at A and the y axis upward, the potential energy is V = mgy. Conservation of energy states that

1m C

Solution:

T1 + V1 = T2 + V2 : 1 mv 2 + 0 = 0 + mg(3 m). 2 1 This yields v 1 = 7.67 m/s. The velocity at point A must be greater than 7.67 m/s.

Greater than 7.67 m/s.

Problem 15.89 At the instant shown, the spring is unstretched and the 20-kg mass is moving downward at 2 m/s. The spring constant is k = 150 N/m. Use conservation of energy to determine the magnitude of its velocity when the 20-kg mass has moved downward 0.5 m.

Solution:

We treat the two masses, the spring, and the cable and pulley as a system. Let s denote the downward displacement of the 20-kg mass. When it moves downward, it loses potential energy and the 4-kg mass gains potential energy, so we write the total potential energy of the system as V = −(20 kg) gs + (4 kg) gs + = −(16 kg) gs +

1 2 ks 2

1 2 ks . 2

The kinetic energy of the system is 1 1 (20 kg)v 2 + (4 kg)v 2 2 2 1 2 = (24 kg)v , 2

T =

where v is the magnitude of the velocity of the masses. Conservation of energy states that

4 kg

20 kg

T1 + V1 = T2 + V2 : 1 1 (24 kg)(2 m/s) 2 + 0 = (24 kg)v 2 2 − (16 kg) g(0.5 m) 2 2 1 + (150 N/m)(0.5 m) 2 . 2 We obtain v 2 = 3.00 m/s. 3.00 m/s.

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Problem 15.90 A rock climber with a mass of 80 kg has a rope of length h = 2 m attached to a fixed piton for protection. If he falls, the worst-case scenario (other than the piton failing) is that he falls a distance 2h before coming to the end of the rope. When he does, the rope behaves like a linear spring with constant k = 28 kN/m. If this happens, use conservation of energy to determine the maximum force exerted on him by the rope.

Solution:

We solve the problem in two parts. The first part is from the moment he falls until he comes to the end of the rope. The only force acting on him is his weight. We place a coordinate system with its origin at his level when he reaches the end of the rope: y Initial position velocity is zero

2h Final position of first part of solution

x Datum

Writing his potential energy as V = mgy, conservation of energy states that T1 + V1 = T2 + V2 : 0 + mg(2h) =

1 mv 2 + 0. 2 2

Solving, his velocity when he reaches the end of the rope is h

v2 = =

4 gh 4(9.81 m/s 2 )(2 m)

= 8.86 m/s. The second part of the solution is his motion from the time he comes to the end of the rope until he stops. During this part, the rope acts on him like a linear spring. We use the same coordinate system: y

Initial position velocity is 8.86 m/s

k x Datum

Final position velocity is zero

1 For this part the potential energy is V = mgy + ky 2 . Conservation of 2 energy is T1 + V1 = T2 + V2 : 1 1 m(8.86 m/s) 2 + 0 = 0 + mgy 2 + ky 2 2 . 2 2 Solving this quadratic equation, we obtain y 2 = −0.502 m. The rope is stretched 0.502 m, so the maximum force it exerts on him is k(0.502 m) = 14,100 N (3160 lb). 14.1 kN.

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Problem 15.91 The collar A slides on the smooth horizontal bar. The spring constant k = 40 lb/ft. The weights are W A = 30 lb and W B = 60 lb. At the instant shown, the spring is unstretched and B is moving downward at 4 ft/s. Use conservation of energy to determine the velocity of B when it has moved downward 2 ft from its current position. (See Example 15.9.)

Solution:

We treat the slider, block, spring, and the cable and pulley as a system. Let s denote the downward displacement of the block B relative to its initial position, which is also the stretch of the spring. The total potential energy of the system is 1 V = −W B s + ks 2 . 2 Conservation of energy states that

T1 + V1 = T2 + V2 : 1 WA 1 WB 1 WA 2 1 WB 2 1 v + v − W B s + ks 2 . (4 ft/s) 2 + (4 ft/s) 2 + 0 = 2 g 2 g 2 g 2 2 g 2 2 Setting s = 2 ft and solving, we obtain v 2 = 6.68 ft/s. 6.68 ft/s.

Problem 15.92 The spring constant k = 700 N/m. The masses m A = 14 kg and m B = 18 kg. The horizontal bar is smooth. At the instant shown, the spring is unstretched and the mass B is moving downward at 1 m/s. How fast is B moving when it has moved downward 0.2 m from its present position?

0.3 m k 0.15 m

Solution: δ =

The unstretched length of the spring is

(0.3 m) 2 + (0.15 m) 2

= 0.335 m.

When B has moved 0.2 m, the length of the spring is ∆ =

(0.5 m) 2 + (0.15 m) 2 = 0.522 m.

Conservation of energy is 1 1 1 (m + m B )v 12 = (m A + m B )v 22 − m B gh + k (∆ − δ ) 2 2 A 2 2 2m B gh − k (∆ − δ ) 2 2 v 2 = v1 + m A + mB v2 =

(1 m/s) 2 +

2(18 kg)(9.81 m/s 2 )(0.2 m) − (700 N/m)(0.187 m) 2 32 kg

v 2 = 1.56 m/s.

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Problem 15.93 The semicircular bar is smooth. The unstretched length of the spring is 10 in. The 5-lb collar at A is given a downward velocity of 6 ft/s, and when it reaches B the magnitude of its velocity is 15 ft/s. Determine the spring constant k.

δA =

The stretch distances for the spring at A and B are

10 7 ft ) − ( ft ) = ft = 0.583 ft ( 17 12 12 12 2

7 10 ft ) + ( ft ) − ft = 0.184 ft ( 10 12 12 12

δB =

Conservation of energy gives

2 in

5 in

Solution:

A k

1 2 1 1 1 mv + kδ A2 + mgh = mv B2 + kδ B2 2 A 2 2 2 k =

m [v B2 − v A2 − 2 gh] δ A2 − δ B2

k =

− (6 ft/s) − 2(32.2 ft/s )(1 ft)   ( 32.25 lbft/s )  (15 ft/s)(0.583  ft) − (0.184 ft)

1 ft

k = 63.1 lb/ft. B

Problem 15.94 The mass of the ball is 1 kg, the spring constant k = 260 N/m, and the unstretched length of the spring is 0.1 m. When the system is released from rest in the position shown, with the string vertical, the spring contracts, pulling the mass to the right. Use conservation of energy to determine the magnitude of the velocity of the mass when the string and spring are parallel.

Solution:

The angle α = arctan (0.25 m/0.45 m) = 29.1 °. If we place the datum at the upper support, the height of the ball in position 1 is h1 = −0.3 m and the height in position 2 is h 2 = −(0.3 m) cos α. The initial length of the spring is L1 =

(0.15 m) 2 + (0.25 m) 2 = 0.292 m.

When the spring and string are parallel, their combined length is L =

(0.45 m) 2 + (0.25 m) 2 = 0.515 m,

so the length of the spring when they are parallel is L 2 = L − 0.3 m = 0.215 m. Applying conservation of energy, T1 + V1 = T2 + V2 : 0.3 m

1 1 1 0 + mgh1 + k ( L1 − 0.1 m) 2 = mv 2 2 + mgh 2 + k ( L 2 − 0.1 m) 2 . 2 2 2 Solving, we obtain v 2 = 2.32 m/s.

0.15 m

2.32 m/s.

Datum 0.25 m a

0.45 m

0.3 m

2 1

0.25 m

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Problem 15.95 The mass of the ball is 1 kg, the spring constant k = 260 N/m, and the unstretched length of the spring is 0.1 m. When the system is released from rest in the position shown, with the string vertical, the spring contracts, pulling the mass to the right. Determine the tension in the string when the string and spring are parallel.

Solution: Datum

0.3 m

0.45 m

2 1

0.3 m

0.15 m

0.25 m

(0.15 m) 2 + (0.25 m) 2 = 0.292 m.

When the spring and string are parallel, their combined length is 0.25 m

L =

(0.45 m) 2 + (0.25 m) 2 = 0.515 m,

so the length of the spring when they are parallel is L 2 = L − 0.3 m = 0.215 m. Applying conservation of energy, T1 + V1 = T2 + V2 : 1 1 1 0 + mgh1 + k ( L1 − 0.1 m) 2 = mv 2 2 + mgh 2 + k ( L 2 − 0.1 m) 2 . 2 2 2 Solving, we obtain v 2 = 2.32 m/s. Now we draw the free-body diagram of the ball when it’s in position 2,

a 0.3 m en

T a

et F

where the spring force F = k ( L 2 − 0.1 m). Newton’s second law in the normal direction is ∑ Fn = T − F − mg cos α = ma n = m

v22 . (0.3 m)

This gives T = 56.3 N. 56.3 N.

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Problem 15.96* In terms of polar coordinates, the force exerted on the object by the nonlinear spring is F = −[k (r − r0 ) + q(r − r0 ) 3 ]e r , where k and q are constants and r0 is the unstretched length of the spring. Determine the potential energy of the spring in terms of its stretch S = r − r0 .

Solution:

We must find a function of position V such that

dV = −F ⋅ d r = [k (r − r0 ) + q(r − r0 ) 3 ]e r ⋅ d r. (1) From Eq. (15.22), in polar coordinates d r = dre r + rd θe θ . (2) Because e r and e θ are perpendicular unit vectors, the dot product of two vectors a and b expressed in terms of polar coordinates is a ⋅ b = a r br + a θ bθ . Therefore from Eqs. (1) and (2) we have dV = −F ⋅ d r = [k (r − r0 ) + q(r − r0 ) 3 ] dr . (3) Noting that

S = r − r0 , u

dS = dr , Equation (3) becomes dV = (kS + qS 3 )dS , (4) or dV = kS + qS 3 . dS Integrating this equation, we obtain V =

1 2 1 kS + qS 4 + C , 2 4

where C is a constant. This expression satisfies Eq. (4) for any value of C, so we can choose C = 0. V =

Problem 15.97 The 20-kg cylinder is released at the position shown and falls onto the linear spring (k = 3000 N/m). Use conservation of energy to determine how far down the cylinder moves after contacting the spring.

1 2 1 kS + qS 4 . 2 4

Solution:

Choose the base of the cylinder as a datum. The potential energy of the piston at rest is V1 = mg(3.5) = 686.7 N-m. The conservation of energy condition after the spring has compressed to the point that the piston velocity is zero is mgh + 12 k (h − 1.5) 2 = mg(3.5), where h is the height above the datum. From which h 2 + 2bh + c = 0, where

( 32 − mgk )

b = − and

c = 2.25 −

7 mg . k

The solution is h = −b ± b 2 − c = 1.95 m, n = 0.919 m. The value h = 1.95 m has no physical meaning, since it is above the spring. The downward compression of the spring is S = 1.5 − 0.919 = 0.581 m.

1.5 m

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Problem 15.98 The 20-kg cylinder is released at the position shown and falls onto the nonlinear spring. In terms of the stretch S of the spring, its potential energy is V = 12 kS 2 + 14 qS 4 , where k = 3000 N/m and q = 4000 N/m 3 . What is the velocity of the cylinder when the spring has been compressed 0.5 m?

Solution: Note that S = 1.5 − h where h is the height above the datum, from which h = 1.5 − S . Use the solution to Problem 15.97. The conservation of energy condition when the spring is being compressed is 12 mv 2 + Vspring + mg(1.5 − S ) = mg(3.5), from which v =

7.0 g − 2 g(1.5 − S ) − 2

1.5 m

Vspring . m

The potential energy in the spring is Vspring = 12 (3000)(0.5 2 ) + 1 4 4 (4000)(0.5 ) = + 437.5 N-m. Substitute numerical values to obtain v = 2.30 m/s.

Problem 15.99 The ball’s position is described by the polar coordinates r and θ. The string exerts a force of constant magnitude T on it. Show that the potential energy associated with this force is V = Tr.

Solution:

We must find a function of position V such that

dV = −F ⋅ d r = Te r ⋅ d r. (1) From Eq. (15.22), in polar coordinates d r = dre r + rd θe θ . (2) Because e r and e θ are perpendicular unit vectors, the dot product of two vectors a and b expressed in terms of polar coordinates is a ⋅ b = a r br + a θ bθ .

Therefore from Eqs. (1) and (2) we have dV = −F ⋅ d r = Tdr , (3)

or dV = T. dr Integrating this equation, we obtain V = Tr + C, T

where C is a constant. This expression satisfies Eq. (3) for any value of C, so we can choose C = 0. V = Tr.

Problem 15.100 The system is at rest in the position shown, with the 12-lb collar A resting on the spring (k = 20 lb/ft), when a constant 30-lb force is applied to the cable. What is the velocity of the collar when it has risen 1 ft? (See Problem 15.99.)

30 lb

3 ft

A 2 ft k

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15.100 (Continued) Solution:

Choose the rest position as the datum. At rest, the compression of the spring is S1 =

−W = −0.6 ft. k

When the collar rises 1 ft the stretch is S 2 = S1 + 1 = 0.4 ft When the collar rises 1 ft the constant force on the cable has acted through a distance s =

32 + 22 −

(3 − 1) 2 + 2 2 = 0.7771 ft.

The work done on the system is U s = 12 k (S12 − S 22 ) − mg(1) + Fs. From the conservation of energy U s = 12 mv 2 from which v =

k 2 2 Fs (S − S 22 ) − 2 g + = 8.45 ft/s. m 1 m

Problem 15.101 A 1-kg disk slides on a smooth horizontal table and is attached to a string that passes through a hole in the table. A constant force T = 10 N is exerted on the string. At the instant shown, r = 1 m and the velocity of the disk in terms of polar coordinates is v = 6e θ (m/s). Use conservation of energy to determine the magnitude of the velocity of the disk when r = 2 m. (See Problem 15.99.)

Solution:

From Problem 15.99, the potential energy associated with the force exerted by the string is V = Tr. That is the only force that does work on the disk. Applying conservation of energy, 1 1 mv 2 + Tr1 = mv 2 2 + Tr2 : 2 1 2 1 1 (1 kg)(6 m/s) 2 + (10 N)(1 m) = (1 kg)v 2 2 + (10 N)(2 m). 2 2

Solving, we obtain v 2 = 4 m/s. 4 m/s. r

Problem 15.102* A 1-kg disk slides on a smooth horizontal table. A string attached to the disk passes through a hole in the table. A 1-kg suspended mass is attached to the string. At the instant shown, r = 2 m and the velocity of the disk in terms of polar coordinates is v = 8e θ (m/s). Use conservation of energy to determine the magnitude of the velocity of the disk when r = 3 m. (Because the disk undergoes central-force motion, the product of its radial position and transverse velocity, rv θ , is constant.)

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15.102 (Continued) Solution:

We treat the two masses and the string as a system. Let the mass of the disk be denoted by m = 1 kg and let the suspended mass be denoted by M = 1 kg. If we place the datum at the level of the suspended mass when r = 0, its height equals r and the potential energy associated with its weight is V = Mgr. The disk has radial and transverse components of velocity, v = v r e r + v θ e θ . The velocity of the suspended mass equals the radial velocity of the disk. The kinetic energy of the system is T =

1 1 m[(v r ) 2 + (v θ ) 2 ] + M (v r ) 2 . 2 2

We also have the given relation r1 (v θ ) 1 = r2 (v θ ) 2 : (2 m)(8 m/s) = (3 m)(v θ ) 2 . From this equation (v θ ) 2 = 5.33 m/s, so the only unknown in the conservation of energy equation is (v r ) 2 . Solving it, we obtain (v r ) 2 = 2.82 m/s. The magnitude of the disk’s velocity is (v r ) 2 2 + (v θ ) 2 2 =

(2.82 m/s) 2 + (5.33 m/s) 2

= 6.03 m/s. 6.03 m/s.

Applying conservation of energy to the system, 1 1 1 1 m[(v r ) 1 2 + (v θ ) 1 2 ] + M (v r ) 1 2 + Mgr1 = m[(v r ) 2 2 + (v θ ) 2 2 ] + M (v r ) 2 2 + Mgr2 : 2 2 2 2 1 1 1 2 2 2 2 [0 + (8 m/s) ] + 0 + (9.81 m/s )(2 m) = [(v r ) 2 + (v θ ) 2 ] + (v r ) 2 2 + (9.81 m/s 2 )(3 m). 2 2 2 (In the second line we divided out the equal masses.)

Problem 15.103 A satellite initially is inserted into orbit at a distance r0 = 8800 km from the center of the Earth. When it is at a distance r = 18, 000 km from the center of the Earth, the magnitude of its velocity is v = 7000 m/s. Use conservation of energy to determine its initial velocity v 0 . The radius of the Earth is 6370 km. (See Example 15.10.)

Solution: v0

mgR E2 mgR E2 1 2 1 mv − = mv 2 − r0 r 2 0 2 v0 =

1 1 v 2 + 2 gR E2  −   r0 r

v0 =

(7000 m/s) 2 + 2(9.81 m/s 2 )(6.37 × 10 6 m) 2

1 ( 8.8 ×110 m − 18 × 10 ) m 6

v 0 = 9760 m/s.

Problem 15.104 Astronomers detect an asteroid 100, 000 km from the Earth moving at 2 km/s relative to the center of the Earth. Suppose the asteroid strikes the Earth. Use conservation of energy to determine the magnitude of its velocity as it enters the atmosphere. (You can neglect the thickness of the atmosphere in comparison to the Earth’s 6370-km radius.)

Solution: Use the solution to Problem 15.103. The potential energy at a distance r is V = −

mgR E2 . r

The conservation of energy condition at r0 :

mgR E2 mgR E2 1 2 1 . mv − = mv 2 − 2 0 r0 2 r

Solve: v =

1 1 v 02 + 2 gR E2  − . r r0 

Substitute: r0 = 1 × 10 8 m, v 0 = 2 × 10 3 m/s, and r = 6.37 × 10 6 m. The velocity at the radius of the earth is v = 11 km/s.

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Problem 15.105* A satellite is in the elliptic Earth orbit shown. Its velocity in terms of polar coordinates when it is at the perigee A is v = 8640 e θ (m/s). Determine the velocity of the satellite in terms of polar coordinates when it is at point B.

Solution:

We have

r A = 8000 km = 8 × 10 6 m rB =

13,900 2 + 8000 2 km = 1.60 × 10 7 m.

Energy and angular momentum are conserved. Therefore mgR E2 mgR E2 1 2 1 2 + v2 ) − mv A − = m(v Br , r Av A = rBv Bθ Bθ 2 rA 2 rB

Solving we have 13,900 km A

v Bθ =

rA 8 × 10 6 m v = (8640 m/s) = 4310 m/s, rB A 1.60 × 10 7 m

v Br =

1 1 v 2A − v B2 θ + 2 gR E2  −  = 2480 m/s.  rB rA 

v = (2480 e r + 4310 e θ ) m/s.

16,000 km

8000 km

Problem 15.106 Use conservation of energy to determine the magnitude of the velocity of the satellite in Problem 15.105 at the apogee C. Using your result, confirm numerically that the velocities at perigee and apogee satisfy the relation r Av A = rC v C .

Solution: From Problem 15.105, r A = 8000 km, rC = 24000 km, v A = 8640 m/s, g = 9.81 m/s 2 , R E = 6370 km. From conservation of energy, mgR E2 mgR E2 1 2 1 mv − = mv c2 − 2 A rA 2 rc Factor m out of the equation, convert all distances to meters, and solve for v c . Solving, v C = 2880 m/s

Does r Av A = rCv C 13,900 km

Substituting the known values, we have r Av A = rCv C = 6.91 × 10 10 m 2 /s.

16,000 km

8000 km

Problem 15.107 The Voyager and Galileo spacecraft observed volcanic plumes, believed to consist of condensed sulfur or sulfur dioxide gas, above the surface of the Jovian satellite Io. The plume observed above a volcano named Prometheus was estimated to extend 50 km above the surface. The acceleration due to gravity at the surface is 1.80 m/s 2 . Using conservation of energy and neglecting the variation of gravity with height, determine the velocity at which a solid particle would have to be ejected to reach 50 km above Io’s surface.

204

Solution:

Conservation of energy yields: T1 + V1 = T2 + V2 : Using 1 the forms for a constant gravity field, we get mv 12 + 0 = 0 + mgy 2 2 1 2 Evaluating, we get v 1 = (1.8)(50,000), or v 1 = 424 m/s. 2

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Problem 15.108 Solve Problem 15.107 using conservation of energy and accounting for the variation of gravity with height. The radius of Io is 1815 km.

Solution: Conservation of energy yields: T1 + V1 = T2 + V2 : Only the form of potential energy changes from that used in Problem 15.107. Here we get mgR I2 mgR I2 1 2 mv 1 − = 0− . 2 RI rI Evaluating, (1.8)(1,815,000) 2 (1.8)(1,815,000) 2 1 2 v − = − . 2 1 1,815,000 1,815,000 + 50,000 or v 1 = 419 m/s.

Problem 15.109* What is the relationship between Eq. (15.21), which is the gravitational potential energy neglecting the variation of the gravitational force with height, and Eq. (15.23), which accounts for the variation? Express the distance from the center of the Earth as r = R E + y, where R E is the Earth’s radius and y is the height above the surface, so that Eq. (15.23) can be written as mgR E V = − . y 1+ RE

Solution: V = −

mgR E2 r

Define y /R E = ε = −

V = V |ε=0 +

mgR E2 mgR E mgR E = − = − y RE + y 1+ε 1+ RE

dV ε+ dε ε = 0

V = −mgR E +

mgR E y ε = −mgR E + mgR E RE (1 + ε) 2 ε = 0

= −mgR E + mgy QED.

By expanding this equation as a Taylor series in terms of y /R E and assuming that y /R E 1, show that you obtain a potential energy equivalent to Eq. (15.21). Problem 15.110 The potential energy associated with a force F acting on an object is V = 2 x 3 − 4 y 2 N-m, where x and y are in meters. (a) What are the cartesian components of F? (b) Determine the curl of F. (c) If an object subjected to F moves from point 1 to point 2 along path A, the resulting work done by F is U 12 = 2 N-m. What is the work done by F if the object moves from point 1 to point 2 along path B? Solution:

2 (1, 1) m

(a) The derivatives ∂V = 6x 2 , ∂x

∂V = − 8 y, ∂y

∂V = 0, ∂z

so from Eq. (15.28), the force is F = −6 x 2 i + 8 yj (N). (b) From Eq. (15.30), the i, j and k components of the curl are ∂ Fy ∂ Fz − = 0, ∂y ∂z

−

( ∂∂Fx − ∂∂Fz ) = 0, ∂∂Fx − ∂∂Fy = 0. z

The curl is zero. (We already knew that because F has a potential energy and so is conservative.) (c) F is conservative, so the work done is path-independent. The work is U 12 = 2 N-m. (a) F = −6 x 2 i + 8 yj (N). (b) Zero. (c) 2 N-m.

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Problem 15.111 A force F = 2 yi − x 2 j (N) acts on an object. (a) Show that F is not conservative. (b) Determine the work done by F as the object moves from point 1 to point 2 along path A. (c) Determine the work done if the object moves along path B.

2 (1, 1) m

Solution: (a) From Eq. (15.30), the i, j and k components of the curl are

(

∂ Fy ∂ Fz ∂ Fz ∂ Fx − = 0, − − ∂y ∂z ∂x ∂z

)

∂ Fy ∂ Fx = 0, − = −2 x − 2. ∂x ∂y

The curl is not zero. The force F is not conservative. (b) Consider the vertical part of path A. Because x = 0 on that part of the path, the component of F parallel to the path Fy = 0, so the work done is zero. On the horizontal part of path A, y = 1 m, so the component of F parallel to the path is Fx = 2 N. The work done is U A = (2 N)(1 m) = 2 N-m. (c) Consider the horizontal part of path B. Because y = 0 on that part of the path, the component of F parallel to the path Fx = 0, so the work done is zero. On the vertical part of path B, x = 1 m, so the component of F parallel to the path is Fx = −1 N. The work done is U B = (−1 N)(1 m) = −1 N-m. (a) The curl is not zero. (b) 2 N-m. (c) −1 N-m.

Problem 15.112 In terms of polar coordinates, the potential energy associated with the force F exerted on an object by a nonlinear spring is V =

Solution:

The force is given by

( ∂∂r e + 1r ∂∂θ e )( 12 k(r − r ) + 14 q(r − r ) )

F = −∇V = −

1 1 k (r − r0 ) 2 + q(r − r0 ) 4 , 2 4

= −[k (r − r0 ) + q(r − r0 ) 3 ]e r .

where k and q are constants and r0 is the unstretched length of the spring. Determine F in terms of polar coordinates. (See Practice Example 15.11.) Problem 15.113 In terms of polar coordinates, the force exerted on an object by a nonlinear spring is F = −[k (r − r0 ) + q(r − r0 ) 3 ]e r , where k and q are constants and r0 is the unstretched length of the spring. Use Eq. (15.31) to show that F is conservative. (See Practice Example 15.11.)

Solution: A necessary and sufficient condition that F be conservative is ∇ × F = 0.  re θ er  1  ∂ ∂ ∇×F =  ∂r ∂θ r  −[k (r − r ) + q(r − r ) 3 ] 0 0 0  =

206

e z  ∂  ∂ z  0 

1 [0 e r − 0re θ + 0 e z ] = 0. F is conservative. r

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Problem 15.114 The potential energy associated with a force F acting on an object is given in polar coordinates by V = −r sin 2 θ + r cos 2 θ ft-lb. (a) Determine F in terms of polar coordinates. (b) If the object moves from point 1 to point 2 along the circular path shown, how much work is done by F?

Solution: (a) The derivatives ∂V = − sin 2 θ + cos 2 θ, ∂r

∂V = −4r sin θ cos θ, ∂θ

∂V = 0, ∂z

so from Eq. (15.29), the force is F = (sin 2 θ − cos 2 θ) e r + 4 sin θ cos θ e θ (lb). (b) Because F is derived from a potential energy, we know it is conservative. Therefore we can calculate the work done when the object moves from 1 along the x axis to the origin, then from the origin along the y axis to 2.

y 2

When the object is on the x axis, θ = 0 and F = −e r = −i (lb). This constant force is direction toward the left, and as the object moves from 1 to the origin the work done is 1 ft x

U x axis = (1 ft)(1 lb) = 1 ft-lb. When the object is on the y axis, θ = 90 ° and F = e r = j(lb). As the object moves from the origin to 2 the work done is U y axis = (1 ft)(1 lb) = 1 ft-lb. We see that the work done as the object moves from 1 to 2 along any path is 2 ft-lb. (a) F = (sin 2 θ − cos 2 θ)e r + 4 sin θ cos θe θ (lb). (b) 2 ft-lb.

Problem 15.115 In terms of polar coordinates, the force exerted on an object of mass m by the gravity of a hypothetical two-dimensional planet is F = −(mg T R T /r )e r , where g T is the acceleration due to gravity at the surface, R T is the radius of the planet, and r is the distance of the object from the center of the planet. (a) Determine the potential energy associated with this gravitational force. (b) If the object is given a velocity v 0 at a distance r0 , what is its velocity v as a function of r?

r0 RT

Solution: (a) The potential is V = −∫ F ⋅ d r + C = ∫

( mgr R )e ⋅ (e dr ) + C T

= mg T RT ln(r ) + C , where C is the constant of integration. Choose r = RT as the datum, from which C = −mg T RT ln(RT ), and  r  V = mg T RT ln  .  RT  (Note: Alternatively, the choice of r = 1 length-unit as the datum, from which C = mg M RT ln(1), yields r V = mg T RT ln = mg T RT ln(r ).) 1

( )

(b) From conservation of energy, r   r  1 2 1 mv + mg T RT ln   = mv 02 + mg T RT ln  0 .  RT   RT  2 2 Solve for the velocity v =

v 02 + 2 gT ln

( rr ). 0

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Problem 15.116 By substituting Eqs. (15.27) into Eq. (15.30), confirm that ∇ × F = 0 if F is conservative. Solution:

Eq. 15.30 is

 i j k     ∂ ∂ ∂    ∂ 2V ∂ 2 V  + ∇ × F =  ∂x ∂y ∂ z  = i −    ∂ y ∂z y ∂ z  ∂   ∂V V V ∂ ∂   − − −  ∂x ∂y ∂ z    ∂ 2V ∂ 2V ∂ 2V ∂ 2 V  − j − + k  −  = 0.  ∂x ∂y ∂ x ∂z ∂ x ∂z ∂ x ∂ y 

(

)

Thus, F is conservative.

Problem 15.117 Determine whether the following forces are conservative: (a) F = (3 x 3 − 4 xy) i − 2 x 2 j;

(b) The i, j and k components of the curl are

(

∂ Fy ∂ Fz ∂ Fz ∂ Fx = 0, − − − ∂y ∂z ∂x ∂z ∂ Fy ∂ Fx − = 3 x 2 y − (−x ). ∂x ∂y

(b) F = ( x 2 − xy ) i + x 3 yj; (c) F = (2 xy 2 − y 3 ) i + (2 x 2 y − 3 xy 2 ) j + zk.

) = 0,

The force is not conservative. (c) The i, j and k components of the curl are

Solution:

(

∂ Fy ∂ Fz ∂ Fz ∂ Fx − = 0, − − ∂y ∂z ∂x ∂z ∂ Fy ∂ Fx − = −4 x − (−4 x ) = 0. ∂x ∂y

) = 0,

The force is conservative.

Problem 15.118 The driver of a 3000-lb car moving at 40 mi/h applies an increasing force on the brake pedal. The magnitude of the resulting frictional force exerted on the car by the road is f = 250 + 6s lb, where s is the car’s horizontal position (in feet) relative to its position when the brakes were applied. Assuming that the car’s tires do not slip, determine the distance required for the car to stop (a) by using Newton’s second law and (b) by using the principle of work and energy.

)

∂ Fy ∂ Fz ∂ Fz ∂ Fx − = 0, − − = 0, ∂y ∂z ∂x ∂z ∂ Fy ∂ Fx − = 4 xy − 3 y 2 − (4 xy − 3 y 2 ) = 0. ∂x ∂y

(a) From Eq. (15.30), the i, j and k components of the curl are

(a) and (c) are, (b) is not.

Solution: (a) Newton’s second law:  W  dv  g  dt = − f , where f is the force on the car in opposition to the motion. Use the chain rule:  W  dv  g v ds = − f = −(250 + 6s). Integrate and rearrange:

( 2Wg )(250s + 3s ) + C. 5280 At s = 0, v (0) = 40 ( ) = 58.67 ft/s, 3600

v2 = −

from which C = (58.67 2 ) = v 12 . The velocity is

( 2Wg )(250s + 3s ) + v (ft/s) .

v2 = −

2 1

At v = 0, s 2 + 2bs + c = 0, where Wv 2 125 = 41.67, c = − 1 = −53493.6. 6g 3 The solution: s = −b ± b 2 − c = 193.3 ft, = −276.7 ft, from which s = 193 ft. b =

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15.118 (Continued) (b) Principle of work and energy: The energy of the car when the brakes are first applied is 1  W  2 v = 1.605 ft-lb. 2  g  1 s

U = ∫ f ds = −∫ (250 + 6s) ds

1 W  1 W  ⋅  v 22 = ⋅  v 12 − (250 s + 3s 2 ). 2 g  2 g  When the car comes to a stop, v 2 = 0, from which

The work done is s

Rearrange:

= − (250 s + 3s 2 ).

From the principle of work and energy, after the brakes are applied, 1 W  1 W  U =  v 22 −  v 12 . 2 g  2 g 

Problem 15.119 As a 28,000-lb airplane takes off from rest at position s = 0, the total horizontal force acting on it given as a function of s in feet is F = T (1 − cs), where T = 26,000 lb and c = 0.0002 ft −1 . It requires 1000 ft of runway to take off. Use the principle of work and energy to determine the magnitude of the airplane’s velocity at takeoff.

1 W  .0 = ⋅  v 12 − (250 s + 3s 2 ). 2 g  Reduce: s 2 + 2bs + c = 0, where Wv 2 125 = 41.67, c = − 1 = − 53493.6. 3 6g

b =

The solution s = −b ± which s = 193 ft.

Solution: U 12 = ∫

Using the given expression, the work done on the airplane is

1000 ft 0

b 2 − c = 193.3 ft, − 276.6 ft, from

Te −cs ds

T ft = − [ e −cs ]1000 0 c T = [ 1 − e −c(1000 ft) ] c = 2.47E7 ft-lb. Then

1 2 mv : 2 1 28,000 lb 2 v , 2.47E7 ft-lb = 2 32.2 ft/s 2 U 12 =

(

)

which yields v = 239 ft/s. 239 ft/s.

Problem 15.120 An astronaut in a small rocket vehicle (combined mass = 1020 kg) is hovering 100 m above the surface of the Moon when she discovers that she is nearly out of fuel and can exert the thrust necessary to cause the vehicle to hover for only 7 more seconds. She quickly considers two strategies for getting to the surface: (a) Fall 20 m, turn on the thrust for 7 s, and then fall the rest of the way. (b) Fall 30 m, turn on the thrust for 7 s, and then fall the rest of the way. Which strategy gives her the best chance of surviving? The acceleration due to gravity is 1.62 m/s 2 .

Initial position (Velocity is zero) H1 (Falling)

100 m

H2 (Thrust on)

H3 (Falling)

Solution:

Their combined weight, which is the thrust T necessary to hover, is mg M = (1020 kg)(1.62 m/s 2 ) = 1650 N.

In the first phase, she falls a given distance H 1 . Equating the work done by the weight to the kinetic energy at the end of the first phase, 1 mg M H 1 = mv 1 2 , 2 the velocity at the end of the first phase is v1 =

2g M H 1 .

In the second phase, the thrust is turned on for a time T = 7 s. The thrust is equal to the weight, so the sum of the forces is zero and the acceleration is zero. The velocity at the end of the second phase is still v 1 and the distance fallen is H 2 = v 1T.

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The distance fallen in the third phase is H 3 = 100 m − H 1 − H 2 . Equating the work done by the weight during the third phase to the change in the kinetic energy, 1 1 mg M H 3 = mv 2 2 − mv 1 2 , 2 2 we can determine the velocity of impact on the surface v 2 . Substituting the values, in case (a) we obtain v 2 = 11.9 m/s and in case (b) we obtain v 2 = 10.0 m/s. Case (b) is slightly more advantageous. She should choose (b), impact velocity 10.0 m/s, instead of (a), impact velocity 11.9 m/s.

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Problem 15.121 The coefficients of friction between the 20-kg crate and the inclined surface are µ s = 0.24 and µ k = 0.22. If the crate starts from rest and the horizontal force F = 200 N, what is the magnitude of the velocity of the crate when it has moved 2 m? Solution:

308

Will the box slip? The normal force is

N = mg cos30 ° + F sin 30 ° = (20 kg)(9.81 m/s 2 ) cos30 ° + (200 N)sin 30 ° = 270 N. The friction force required for equilibrium is

308

f = F cos30 ° − mg sin 30 ° = (200 N) cos30 ° − (20 kg)(9.81 m/s 2 )sin 30 °

= 75.1 N. The maximum friction force that can be supported is µ s N = 64.8 N, so the box will slip and f = µ k N.

F N

The force directed up the plane is F cos30° − mg sin 30 ° − µ k N = (200 N) cos30 ° − (20 kg)(9.81 m/s 2 )sin 30 ° −(0.22)(270 N) = 15.7 N. Equating the work done as it moves 2 m up the plane to the change in its kinetic energy, (15.7 N)(2 m) =

1 (20 kg)v 2 2 − 0, 2

we obtain v 2 = 1.77 m/s. 1.77 m/s.

Problem 15.122 The coefficients of friction between the 20-kg crate and the inclined surface are µ s = 0.24 and µ k = 0.22. If the crate starts from rest and the horizontal force F = 40 N, what is the magnitude of the velocity of the crate when it has moved 2 m? Solution:

Will the box move up or down the plane? Suppose it moves down the plane:

308

mg F

mkN N

The normal force is N = mg cos30 ° + F sin 30 ° = (20 kg)(9.81 m/s 2 ) cos30 ° + (40 N)sin 30 ° = 190 N. The force directed down the plane is mg sin 30 ° − µ k N − F cos30 ° = (20 kg)(9.81 m/s 2 )sin 30 ° −0.22(170 N) − (40 N) cos30 ° = 21.7 N. We see that the box will move down the plane. Equating the work done as it moves 2 m down the plane to the change in its kinetic energy, 1 (21.7 N)(2 m) = (20 kg)v 2 2 − 0, 2 we obtain v 2 = 2.08 m/s. 2.08 m/s.

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Problem 15.123 The Union Pacific Big Boy locomotive, now a museum exhibit, weighs 1.19 million lb, and the tractive effort (tangential force) of its drive wheels was 135, 000 lb. If you neglect other tangential forces, what distance was required for the train to accelerate from zero to 60 mi/h?

Solution:

The potential associated with the force is

V = −∫ F ds = −Fs. 0

The energy at rest is zero. The energy at v = 60 mi/h = 88 ft/s is 0 =

1  W  2 v + V , 2  g 

from which s =

Problem 15.124 The 400-lb container A starts from rest at position s = 0. The horizontal force (in pounds) exerted on the container by the hydraulic piston is given as a function of the position s in feet by F = 30 + 6s. What is the magnitude of the container’s velocity when it has reached the position s = 2 ft? A

Solution:

The mass of the container is W 400 lb m = = = 12.4 slugs. g 32.2 ft/s 2 The work done as the container moves from its initial position to s = 2 ft is U 12 = ∫

s2 s1

∑ Ft ds

= ∫ F ds = ∫

1  W  2 v = 1061 ft. 2  gF 

0 2 ft 0

(30 + 6s) ds

= [ 30 s + 3s 2 ]20 ft = 72 ft-lb. Equating the work to the change in kinetic energy of the container, 1 mv 2 − 0, 2 2 we obtain v 2 = 3.40 ft/s.

U 12 =

3.40 ft/s.

Problem 15.125 The force exerted on a car by a prototype crash barrier as the barrier crushes is F = −(2000 s + 8000 s 3 ) N, where s is the distance in meters from the point of initial contact. A 2200-kg car traveling 100 km/h hits the barrier and comes to rest. Determine the maximum deceleration to which the passengers are subjected.

Solution:

The decelerating force on the car increases as the distance s increases. To determine the maximum force, we need to determine the value of s at which the car comes to rest.

The car’s initial velocity in m/s is 1000 m 1h 100 km/h = 27.8 m/s. 1 km 3600 s

(

)(

)

The work done by the barrier as a function of s is s

U 12 = ∫ −(2000 s + 8000 s 3 ) ds 0

= −[ 1000 s 2 + 2000 s 4 ]s0 = −(1000 s 2 + 2000 s 4 ). Equating this to the change in the car’s kinetic energy, 1 −(1000 s 2 + 2000 s 4 ) = 0 − mv 1 2 2 1 = − (2200 kg)(27.8 m/s) 2 . 2 This is a quadratic equation in terms of s 2 . Solving it, we obtain s = 4.51 m. The magnitude of the maximum force exerted on the car is F = 2000 s + 8000 s 3 = 2000(4.51 m) + 8000(4.51 m) 3 = 7.44E5 N. The maximum deceleration is F 7.44E5 N = m 2200 kg = 338 m/s 2 (34.5 g’s). 338 m/s 2 (34.5 g’s).

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Problem 15.126 In a preliminary design for a mail-sorting machine, parcels moving at 2 ft/s slide down a smooth ramp and are brought to rest by a linear spring. What is the largest value the spring constant can have if you do not want the 15-lb parcel to be subjected to a maximum deceleration greater than 12 g’s (12 times the acceleration due to gravity)? 2 ft/s

3 ft

Solution: The 12g requirement places a limit on the magnitude of the force the spring exerts on the parcel. Let s denote the maximum displacement of the parcel after it contacts the spring, which is also the amount the spring is compressed. Denote the weight of the parcel by W = 15 lb. From Newton’s second law, we have the constraint ∑ F = ma: (1) W  ks =  12 g = 12W . g Next, we will determine the parcel’s velocity after it slides down the ramp. Setting the work done by its weight equal to its change in kinetic energy, U 12 = W (3 ft) =

1  W  2 1 W  v −   (2 ft/s) 2 , 2  g  2 2 g 

we obtain v 2 = 4.11 ft/s. Now we apply conservation of energy to the motion of the parcel from its first contact with the spring until it comes to rest: T1 + V1 = T2 + V2 : (2) 1W 1 (4.11 ft/s) 2 + 0 = 0 + k s 2 . 2 g 2 We can solve Eqs. (1) and (2) for k and s, obtaining k = 353 lb/ft, s = 0.510 ft. A stronger spring would stop the parcel faster, so k must not be greater than this value. 353 lb/ft.

Problem 15.127 When the 1-kg collar is in position 1, the tension in the spring is 50 N, and the unstretched length of the spring is 260 mm. If the collar is pulled to position 2 and released from rest, what is its velocity when it returns to position 1?

300 mm

Solution:

The stretched length of the spring in position 1 is S1 = 0.3 − 0.26 = 0.04 m. The stretched length of the spring in position 2 is S 2 = 0.3 2 + 0.6 2 − 0.26 = 0.411 m. The spring constant is 50 k = = 1250 N/m. S1 1 The potential energy of the spring in position 2 is kS 22 . The potential 2 1 energy of the spring in position 1 is kS12 . The energy in the collar at 2 1 1 1 position 1 is kS 22 = mv 12 + kS12 , from which 2 2 2 v1 =

k 2 (S − S12 ) = 14.46 m/s. m 2

1 600 mm

300 mm

2 600 mm

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Problem 15.128 When the 1-kg collar is in position 1, the tension in the spring is 100 N, and when the collar is in position 2, the tension in the spring is 400 N. (a) What is the spring constant k? (b) If the collar is given a velocity of 15 m/s at position 1, what is the magnitude of its velocity just before it reaches position 2?

Solution: (a) Assume that the dimensions defining locations 1 and 2 remain the same, and that the unstretched length of the spring changes from that given in Problem 15.141. The stretched length of the spring in position 1 is S1 = 0.3 − S 0 , and in position 2 is S 2 = 0.3 2 + 0.6 2 − S 0 . The two conditions: 400 100 , 0.3 − S 0 = . k k Subtract the second from the first, from which k = 809 N/m. Substitute and solve: S 0 = 0.176 m, and S1 = 0.124 m, S 2 = 0.494 m. 0.6 2 + 0.3 2 − S 0 =

(b) The energy at the onset of motion at position 1 is 300 mm

1 2 1 mv + kS12 . 2 1 2 At position 2:

1 2 1 1 1 mv + kS12 = mv 22 + kS 22 , 2 1 2 2 2 from which

1 600 mm

Problem 15.129 The 50-lb weight is released from rest with the two springs (k A = 40 lb/ft, k B = 20 lb/ft) unstretched. (a) How far does the weight fall before rebounding? (b) What maximum velocity does it attain?

v2 =

v 12 +

k 2 (S − S 22 ) = 6.29 m/s. m 1

Solution: Suppose the connected springs are subjected to a force F. In terms of the resulting stretches of the two springs, we have F = k AS A , F = k BS B . The total resulting stretch is F F S = S A + SB = + kA kB

 1 1  = F  + ,  kA k B  From this argument, we see that the two connected springs are equivalent to a single spring with constant

k =

1 . 1 1 + kA kB

Let s denote the downward displacement of the weight relative to its initial position. Placing the datum at the initial position and applying conservation of energy, we have T1 + V1 = T2 + V2 : 0+0 =

(1) 1  W  2 1 v − Ws + ks 2 . 2  g  2 2

(a) By setting v 2 = 0 in Eq. (1), we can solve for the maximum displacement, obtaining s = 7.5 ft. (b) From Eq. (1), the condition for v 2 to be a maximum is

(

)

d 1 Ws − ks 2 = W − ks = 0. ds 2 This makes sense, the weight will accelerate until the force exerted by the springs equals the weight. Substituting s = W /k into Eq. (1), we obtain v 2 = 11.0 ft/s. (a) 7.5 ft. (b) 11.0 ft/s.

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Problem 15.130 The piston and the load it supports are accelerated upward by the gas in the cylinder. The total weight of the piston and load is 1000 lb. The cylinder wall exerts a constant 50-lb frictional force on the piston as it rises. The net force exerted on the piston by pressure is ( p 2 − p atm ) A, where p is the pressure of the gas, p atm = 2117 lb/ft 2 is atmospheric pressure, and A = 1 ft 2 is the cross-sectional area of the piston. Assume that the product of p and the volume of the cylinder is constant. When s = 1 ft, the piston is stationary and p = 5000 lb/ft 2 . What is the velocity of the piston when s = 2 ft?

Solution: At the rest position, p 0 As = p 0V = K, where V = 1 ft 3 , from which K = p 0 . Denote the datum: s 0 = 1 ft. The potential energy of the piston due to the gas pressure after motion begins is s

Vgas = −∫ F ds = −∫ ( p − p atm )A ds s

= p atm A(s − s 0 ) − ∫ pA ds. s0

From which Vgas = p atm A(s − s 0 ) − K ∫

s ds s0

 s  = p atm A(s − s 0 ) − K ln  .  s0 

The potential energy due to gravity is s

Vgravity = −∫ (−W ) ds = W (s − s 0 ). s0

The work done by the friction is s

U friction = ∫ (− f ) ds = − f (s − s 0 ), where f = 50 lb. s0

From the principle of work and energy: Piston

U friction =

Gas s

1  W  2 v + Vgas + Vgravity 2  g 

Rearrange: 1  W  2 v = U friction − Vgas − Vgravity . At s = 2ft, 2  g  1  W  2 v = −(−1348.7) − (1000) − 50 = 298.7 ft-lb, 2  g  2(298.7) g from which v = = 4.39 ft/s. W

Problem 15.131 When the 59,000-kg test rocket’s engine burns out at an altitude of 3 km, it is traveling 6 km/s at an angle of 60 ° relative to the horizontal. Neglect the variation in gravity with altitude. (a) If you neglect aerodynamic forces, what is the magnitude of the velocity of the rocket when it reaches an altitude of 8 km ? (b) If the actual velocity of the rocket when it reaches an altitude of 5 km is 5.8 km/s, how much work was done by aerodynamic forces as the rocket moved from 3 km to 6 km altitude?

Solution: (a) Equating the work done by the rocket’s weight to the change in its kinetic energy, 1 1 U 12 = −mg (8000 m − 3000 m) = mv 2 2 − m (6000 m/s) 2 , 2 2 We obtain v 2 = 5990 m/s. (b) Let the work done by aerodynamic forces be denoted by U a . Equating the work done by the rocket’s weight and aerodynamic forces to the change in its kinetic energy, 1 U 12 = −mg(8000 m − 3000 m) + U a = m (5800 m/s) 2 2 1 2 − m(6000 m/s) , 2 gives U a = −6.67E10 N-m. (a) 5990 m/s. (b) −6.67E10 N-m.

Source: Courtesy of Alexyz3d/Shutterstock.

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Problem 15.132 The 12-kg collar A is at rest in the position shown at t = 0 and is subjected to the tangential force F = 24 − 12t 2 N for 1.5 s. Neglecting friction, what maximum height h does the collar reach?

Solution:

Choose the datum at the initial point. The strategy is to determine the velocity at the end of the 1.5 s and then to use work and energy methods to find the height h. From Newton’s second law: m

dv = F = 24 − 12t 2 . dt

Integrating: v =

[Note: The displacement during this time must not exceed 2 m. Integrate the velocity: s =

( )

1 1.5 1 (24 − 12t 2 ) dt = [24t − 4t 3 ]1.5 = 1.875 m/s. 0 m ∫0 m

( m1 ) ∫ (24t − 4t ) dt 1.5

( m1 )[12t − t ] 2

4 1.5 0

= 1.82 m < 2 m,

so the collar is still at the datum level at the end of 1.5 s.] The energy condition as the collar moves up the bar is 1 2 1 mv = mv 2 + mgh. 2 0 2 At the maximum height h, the velocity is zero, from which h =

Problem 15.133 Suppose that, in designing a loop for a roller coaster’s track, you establish as a safety criterion that at the top of the loop the normal force exerted on a passenger by the roller coaster should equal 10 percent of the passenger’s weight. (That is, the passenger’s “effective weight” pressing him down into his seat is 10 percent of his actual weight.) The roller coaster is moving at 62 ft/s when it enters the loop. What is the necessary instantaneous radius of curvature ρ of the track at the top of the loop?

v 02 = 0.179 m. 2g

Solution:

The energy at the top of the loop is

1 2 1 2 mv = mv top + mgh, 2 0 2 where v 0 = 62 ft/s, h = 50 ft, and g = 32.2 ft/s 2 , from which v top = v 02 − 2 gh = 25 ft/s. From Newton’s second law: 2   v top m   = (1.1) mg,  ρ 

from which v2 ρ = top = 17.7 ft. 1.1 g

r 50 ft

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Problem 15.134 A 180-lb student runs at 15 ft/s, grabs a rope, and swings out over a lake. He releases the rope when his velocity is zero. (a) What is the angle θ when he releases the rope? (b) What is the tension in the rope just before he releases it? (c) What is the maximum tension in the rope?

30 ft

Solution: (a) The energy condition after the seizure of the rope is 1 2 1 mv = mv 2 + mgL (1 − cos θ), 2 0 2 where v 0 = 15 ft/s, L = 30 ft. When the velocity is zero, v 02 = 2 gL (1 − cos θ), from which cos θ = 1 −

v 02 = 0.883, θ = 27.9 °. 2 gL

(b) From the energy equation v 2 = v 02 − 2 gL (1 − cos θ). From Newton’s second law, (W /g)(v 2 /L ) = T − W cos θ, from which

( )

W  v2 T =   + W cos θ = 159.0 lb. g L (c) The maximum tension occurs at the angle for which dT = 0 = −2W sin θ − W sin θ, dθ from which θ = 0, from which  v2 Tmax = W  0 + 1  = 222 lb.   gL

Problem 15.135 If the student in Problem 15.134 releases the rope when θ = 25 °, what maximum height does he reach relative to his position when he grabs the rope?

Solution:

Use the solution to Problem 15.134. [The height when he releases the rope is h1 = L(1 − cos 25 °) = 2.81 ft. ] Before he releases the rope, the total energy is 1  W  2 1 W  v − WL =  v 2 − WL cos θ. 2  g  0 2 g  Substitute v 0 = 15 ft/s, θ = 25 ° and solve: v = 6.63 ft/s. The horizontal component of velocity is v cos θ = 6.01 ft/s. From conservation of energy: W (2.81) +

216

30 ft

1 1 m (6.63 2 ) = Wh + m(6.01 2 ) 2 2

from which h = 2.93 ft.

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Problem 15.136 A boy takes a running start and jumps on his sled at position 1. He leaves the surface at position 2 and lands in deep snow at a distance b = 30 ft. How fast was he moving at position 1?

15 ft

Solution: Let v 2 be the magnitude of his velocity at 2. We can integrate the constant acceleration to analyze his motion through the air. Let t = 0 be the instant he is at 2. We have dv x ax = = 0, (1) dt dx vx = = v 2 cos35 °, (2) dt

x = (v 2 cos35 °)t, (3) dv y ay = = −g, (4) dt dy vy = = −gt + v 2 sin 35 °, (5) dt 1 y = − gt 2 + (v 2 sin 35 °)t. (6) 2 From Eq. (3), the time at which he lands is t =

b . v 2 cos35 °

Substituting this time into Eq. (6) and setting y = −5 ft, we obtain the equation 2 1  b b  + (v sin 35°)   −5 ft = − g  2  v cos35° , 2  v 2 cos35 °  2 which we can solve for v 2 : v2 =

b 2g . 2 10 cos 35 ° + 2b sin 35 ° cos35 °

Problem 15.137 A boy takes a running start and jumps on his sled at position 1. He leaves the ground at position 2 and lands in deep snow at a distance b. Draw a graph of b as a function of his velocity at position 1 from zero to 20 ft/s.

358

5 ft

x 5 ft

Datum b

This gives v 2 = 28.8 ft/s. Now we can analyze his motion from 1 to 2. Placing the datum as shown in the figure, conservation of energy states that T1 + V1 = T2 + V2 : 1 1 2 mv + mg(15 ft) = mv 2 2 + mg(5 ft). 2 1 2 Solving gives v 1 = 13.7 ft/s. 13.7 ft/s.

15 ft

358

5 ft

Solution: Let v 2 be the magnitude of his velocity at 2. We can integrate the constant acceleration to analyze his motion through the air. Let t = 0 be the instant he is at 2. We have dv x = 0, (1) dt dx vx = = v 2 cos35 °, (2) dt

ax =

x = (v 2 cos35 °)t, (3) dv y = −g, (4) dt dy vy = = −gt + v 2 sin 35 °, (5) dt ay =

x 5 ft

Datum b

1 y = − gt 2 + (v 2 sin 35 °)t. (6) 2 From Eq. (3), the time at which he lands is t =

b . v 2 cos35 °

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15.137 (Continued) Substituting this time into Eq. (6) and setting y = −5 ft, we obtain the equation

For a given value of v 1 , we can solve Eq. (8) for v 2 and then solve Eq. (7) for b. The graph is shown:

  2  b b 1  −5 ft = − g  ,  + (v 2 sin 35 °)   v 2 cos35 °   2 v 2 cos35 ° 

40 38

which we can write as a quadratic equation in b :

36 (7)

Now we analyze the motion from 1 to 2. Placing the datum as shown in the figure, conservation of energy states that T1 + V1 = T2 + V2 : 1 1 mv 2 + mg (15 ft) = mv 2 2 + mg (5 ft). 2 1 2

34 32 b, ft

  2 g  b − (tan 35°) b − 5 = 0.  2v 2 2 cos 2 35 ° 

30 28

(8)

26 24 22 20

v1, ft/s

Problem 15.138 The 1-kg collar A is attached to the linear spring (k = 500 N/m) by a string. The collar starts from rest in the position shown, and the initial tension in the string is 100 N. What distance does the collar slide up the smooth bar?

Solution:

The deflection of the spring is 100 S = = 0.2 m. k 1 The potential energy of the spring is Vspring = kS 2 . The energy condi2 1 tion after the collar starts sliding is Vspring = mv 2 + mgh. At the max2 imum height, the velocity is zero, from which h =

Vspring k S 2 = 1.02 m. = mg 2mg

k A

Problem 15.139 The masses m A = 40 kg and m B = 60 kg. The collar A slides on the smooth horizontal bar. The system is released from rest. Use conservation of energy to determine the velocity of the collar A when it has moved 0.5 m to the right.

Solution:

Placing the datum for B at its initial position, conservation of energy gives T1 + V1 = T2 + V2 : Evaluating, we get 1 1 0 + 0 = (40)v 2 + (60)v 2 − (60)(9.81)(0.5) or v = 2.43 m/s. 2 2

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Problem 15.140 The spring constant is k = 850 N/m, m A = 40 kg, and m B = 60 kg. The collar A slides on the smooth horizontal bar. The system is released from rest in the position shown with the spring unstretched. Use conservation of energy to determine the velocity of the collar A when it has moved 0.5 m to the right.

Solution:

Let s denote the displacement of collar A relative to its initial position, which is also the stretch of the spring. Let the datum for the potential energy of B be placed at the level of B's initial position. The initial length of the cable from its attachment at A to the pulley is L1 =

The length when A has moved a distance s is L2 =

(0.9 m) 2 + (0.4 m) 2 .

(0.9 m − s) 2 + (0.4 m) 2 .

The distance the box B moves downward from its initial position when A has moved a distance s is L1 − L 2 . The velocity of B when A has moved a distance s is

d (L − L 2 ) dt 1 d = − (0.9 m − s) 2 + (0.4 m) 2 (1) dt 1 1 ds − = − [ (0.9 m − s) 2 + (0.4 m) 2 ] 2 2(0.9 m − s)(−1) 2 dt = Bv A ,

vB = 0.4 m

where we have defined

0.9 m

B =

(0.9 m − s) (0.9 m − s) 2 + (0.4 m) 2

and we have noted that ds /dt = v A . Conservation of energy states that T1 + V1 = T2 + V2 : (2) 1 1 1 0 + 0 = m Av A 2 + m Bv B 2 + ks 2 − m B g ( L1 − L 2 ). 2 2 2 By substituting Eq. (1) into Eq. (2), we can solve for v A : vA =

2m B g ( L1 − L 2 ) − ks 2 . (m A + m B B 2 )

When s = 0.5 m we obtain v A = 2.00 m/s. 2.00 m/s.

Problem 15.141 The y-axis is vertical and the curved bar is smooth. If the magnitude of the velocity of the 4-lb slider is 6 ft/s at position 1, what is the magnitude of its velocity when it reaches position 2?

Solution:

Choose the datum at position 2. At position 2, the energy

condition is 1  W  2 1 W  v + Wh =  v 22 , 2  g  1 2 g  where h = 2, from which

v2 =

v 12 + 2 gh =

6 2 + 2 g(2) = 12.83 ft/s.

2 ft 2

4 ft

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Problem 15.142 In Problem 15.141, determine the magnitude of the velocity of the slider when it reaches position 2 if it is subjected to the additional force F = 3 xi − 2 j (lb) during its motion. y

Solution: 0

U = ∫ F ⋅ d r = ∫ (−2) dy + ∫ 3 x dx 4 3 = [−2 y ] 02 +  x 2  = 4 + 24 = 28 ft-lb.  2 0

From the solution to Problem 15.141, the energy condition at position 2 is 1  W  2 1 W  v + Wh + U =  v 22 , 2  g  1 2 g 

from which v2 =

2 ft 2

v 12 + 2 gh +

2 g(28) = W

6 2 + 2 g (2) +

2 g(28) 4

= 24.8 ft/s. x

4 ft

Problem 15.143 Suppose that an object of mass m is beneath the surface of the Earth. In terms of a polar coordinate system with its origin at the Earth’s center, the gravitational force on the object is −(mgr /R E )e r , where R E is the radius of the Earth. Show that the potential energy associated with the gravitational force is V = mgr 2 /2 R E .

Problem 15.144 It has been pointed out that if tunnels could be drilled straight through the Earth between points on the surface, trains could travel between those points using gravitational force for acceleration and deceleration. (The effects of friction and aerodynamic drag could be minimized by evacuating the tunnels and using magnetically levitated trains.) Suppose that such a train travels from the North Pole to a point on the equator. Disregard the Earth’s rotation. Determine the magnitude of the velocity of the train (a) when it arrives at the equator and (b) when it is halfway from the North Pole to the equator. The radius of the Earth is R E = 3960 mi. (See Problem 15.143.)

Solution:

By definition, the potential associated with a force F is

V = −∫ F ⋅ d r. If d r = e r dr + re θ d θ, then  mgr   mgr  V = −∫  −  e ⋅ e r dr − ∫  − e ⋅ e r dθ  R E  r  R E  r θ  mgr 2   mgr  = −∫  − .  dr =   2 R E   R E 

Solution: Vgravity

The potential associated with gravity is

mgr 2 = . 2RE

With an initial velocity at the North Pole of zero, from conservation of energy, at any point in the path  mgr 2   mgr 2  1 2  2 R  = mv +  R . 2 2 E E NP (a) At the equator, the conservation of energy condition reduces to

( mg2R ) = 12 mv E

( mg2R ), E

from which v EQ = 0. (b) At the midway point, r = R E sin 45 ° = tion of energy

RE , and from conserva2

( mg2R ) = 12 mv + ( mg4R ), E

from which v M =

220

2 EQ

2 M

gR E = 18300 ft/s = 12500 mi/h. 2

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Problem 15.145 In Problem 15.123, what is the maximum power transferred to the locomotive during its acceleration?

Solution:

Problem 15.146 Just before it lifts off, a 10, 500-kg airplane is traveling at 60 ft/s. The total horizontal force exerted by the plane’s engines is 189 kN, and the plane is accelerating at 15 m/s 2 . (a) How much power is being transferred to the plane by its engines? (b) What is the total power being transferred to the plane?

Solution:

From Problem 15.123, the drive wheel traction force F = 135,000 lb is a constant, and the final velocity is v = 60 mi/h = 88 ft/s. The power transferred is P = Fv , and since the force is a constant, by inspection the maximum power transfer occurs at the maximum velocity, from which P = Fv = (135000)(88) = 11.88 × 10 6 ft-lb/sec = 21,600 hp.

(a) The power being transferred by its engines is P = Fv = (189 × 10 3 )(60) = 1.134 × 10 7 Joule/s = 11.3 MW. (b) Part of the thrust of the engines is accelerating the airplane: From Newton’s second law, dv m = T = (10.5 × 10 3 )(15) = 157.5 kN. dt The difference (189 − 157.5) = 31.5 kN is being exerted to overcome friction and aerodynamic losses. (c) The total power being transferred to the plane is Pt = (157.5 × 10 3 )(60) = 9.45 MW.

Problem 15.147 The “Paris Gun” used by Germany in World War I had a range of 120 km, a 37.5-m barrel, and a muzzle velocity of 1550 m/s and fired a 120-kg shell. (a) If you assume the shell’s acceleration to be constant, what maximum power was transferred to the shell as it traveled along the barrel? (b) What average power was transferred to the shell?

dv From Newton’s second law, m = F, from which, for a dt constant acceleration,

Solution: v =

( mF )t + C.

At t = 0, v = 0, from which C = 0. The position is s =

F 2 t + C. 2m

At t = 0, s = 0, from which C = 0. At s = 37.5 m, v = 1550 m/s, from which F = 3.844 × 10 6 N and t = 4.84 × 10 −2 s is the time spent in the barrel. The power is P = Fv , and since F is a constant and v varies monotonically with time, the maximum power transfer occurs just before the muzzle exit: P = F(1550) = 5.96 × 10 9 joule/s = 5.96 GW. (b) From Eq. (15.18) the average power transfer is 1

Pave = 2

mv 22 − 12 mv 12 = 2.98 × 10 9 W = 2.98 GW. t

Source: Courtesy of Sueddeutsche Zeitung Photo / Alamy Stock Photo.

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Chapter 16 Problem 16.1 The 20-kg crate is stationary at time t = 0. It is subjected to a horizontal force given as a function of time (in newtons) by F = 10 + 2t 2 . (a) Determine the magnitude of the linear impulse exerted on the crate from t = 0 to t = 4 s. (b) Use the principle of impulse and momentum to determine how fast the crate is moving at t = 4 s.

Solution: (a) The linear impulse is t2

∫t ΣFt dt = ∫0 (10 + 2t 2 ) dt 1

2 4s =  10t + t 3   3 0 = 82.7 N-s. (b) The principle of impulse and momentum states that t2

∫t ΣFt dt = mv 2 − mv 1:

82.7 N-s = (20 kg)v 2 − 0, which gives v 2 = 4.13 m/s. (a) 82.7 N-s. (b) 4.13 m/s.

Problem 16.2 The 40-kg crate is released from rest at time t = 0. The coefficient of kinetic friction between the crate and the inclined surface is µ k = 0.18. (a) Determine the magnitude of the linear impulse exerted on the crate from t = 0 to t = 4 s. (b) Use the principle of impulse and momentum to determine how fast the crate is moving at t = 4 s.

Solution:

The free-body diagram of the crate is shown: 308

mg mkN N (a) There is no acceleration normal to the surface, so the normal force is N = mg cos30 °. The linear impulse is t2

∫t ΣFt dt = ∫0 (mg sin 30° − µ k N ) dt 1

= (mg sin 30 ° − µ k N )(4 s) = 540 N-s. 308

(b) The principle of impulse and momentum states that t2

∫t ΣFt dt = mv 2 − mv 1: 1

540 kg-m/s = (40 kg)v 2 − 0, which gives v 2 = 13.5 m/s. (a) 540 kg-m/s. (b) 13.5 m/s.

Problem 16.3 The 1360-kg French Gazelle helicopter starts from rest on the runway. The pilot advances the throttle in such a way that the vertical thrust of its engine (in kN) is given as a function of time in seconds by T = 14.0 + 0.12t + 0.09t 2 . Use the principle of impulse and momentum to determine how fast the helicopter is moving upward at t = 4 s.

Source: Courtesy of Simon Belcher/Alamy Stock Photo.

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16.3 (Continued) Solution:

The helicopter’s weight is mg = (1360 kg)(9.81 m/s 2 ) = 13,300 N, so the total upward force on it is

14.0 − 13.3 + 0.12t + 0.09t 2 kN. Applying impulse and momentum to its upward motion, 1000 ∫

4s 0

(14.0 − 13.3 + 0.12t + 0.09t 2 ) dt = mv 2 − mv 1:

1000 [(14.0 − 13.3)t + 0.06t 2 + 0.03t 3 ] 40 s = (1360 kg)v 2 . This gives v 2 = 4.05 m/s. 4.05 m/s.

Problem 16.4 The 1.40E8-kg cargo ship starts from rest. The total force exerted on it by its engines and hydrodynamics drag (in newtons) is given as a function of time in seconds by ΣFt = 935,000 − 0.65t 2 . Use the principle of impulse and momentum to determine how fast the ship is moving in 15 minutes.

Solution:

Applying impulse and momentum, t2

∫t ΣFt dt = mv 2 − mv 1: 1

(15)(60) s

∫0

(935,000 − 0.65t 2 ) dt = (1.40E8 kg)v 2 − 0,

 935,000t − 0.65 t 3    3

(15)(60) s 0

= (1.40E8 kg)v 2 ,

Source: Courtesy of DifferR/Shutterstock.

which gives v 2 = 4.88 m/s (17.6 km/h). 4.88 m/s.

Problem 16.5 The combined mass of the motorcycle and rider is 180 kg. The coefficient of kinetic friction between the tires and the dirt road is µ k = 0.65. The rider starts from rest and spins the rear (drive) wheel. The normal force between the rear wheel and the road is 1200 N. Use the principle of impulse and momentum to estimate the velocity reached by the motorcycle in 2 s.

Solution:

The only horizontal force, the friction force on the rear wheel, is µ k (1200 N). Equating the impulse due to the constant horizontal force to the change in linear momentum, µ k (1200 N)(2 s) = m(v 2 − 0),

the velocity is 8.67 m/s. 8.67 m/s.

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Source: Courtesy of Arthur Bargan/Shutterstock.

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Problem 16.6 A bioengineer models the horizontal component of the force in Newtons generated by the wings of the 0.2-kg gull by an equation of the form F = F0 (1 + sin ωt ) N, where F0 and ω are constants. From video measurements of it taking off, he estimates that ω = 18 rad/s and determines that the bird requires 1.42 s to take off and is moving at 6.1 m/s when it does. Use the principle of impulse and momentum to determine the constant F0 .

Solution: t

∫0 F0 (1 + sin ωt ) dt = mv

(

F0 t − F0 =

t 1 1 cos ωt = F0 t + [1 − cos ωt ] = mv ω ω 0

t +

)

(

)

(0.2 kg)(6.1 m/s) mv = 1 1 (1 − cos[(18)(1.42)]) (1 − cos ωt ) (1.42 s) + 18 rad/s ω

F0 = 0.856 N.

Source: Courtesy of Janet Griffin/Shutterstock.

Problem 16.7 The 20-kg object is stationary at t = 0. The total force acting on it is ΣF = 60 i + 80 j + 40 k (N). Use the principle of impulse and momentum to determine the magnitude of the object’s velocity at t = 4 s.

Solution: t2

The linear impulse is 4s

∫t ΣF dt = ∫0 [60 i + 80 j + 40 k (N)] dt 1

= 240 i + 320 j + 160 k (kg-m/s). Applying impulse and momentum, t2

∫t ΣF dt = mv 2 − mv 1:

240 i + 320 j + 160 k (kg-m/s) = (20 kg)v 2 − 0, the velocity at t = 4 s is v 2 = 12 i + 16 j + 8k (m/s). Its magnitude is v 2 = 21.5 m/s.

21.5 m/s. x z

Problem 16.8 At time t = 0, the velocity of the 20-kg object is v = 3i − 2 j + 4 k (m/s). The total force acting on it is given as a function of time in seconds by ΣF = (60 + 6t ) i + 80 j + (40 + 8t )k (N). Use the principle of impulse and momentum to determine the magnitude of its velocity at t = 4 s.

x z

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16.8 (Continued) Solution: t2

The linear impulse is 4s

∫t ΣF dt = ∫0 [ (60 + 6t )i + 80 j + (40 + 8t )k (N) ]dt 1

= [ (60t + 3t 2 ) i + 80t j + (40t + 4t 2 )k ]40 s = 288 i + 320 j + 224 k (kg-m/s). Applying impulse and momentum, t2

∫t ΣF dt = mv 2 − mv 1: 1

288 i + 320 j + 224 k (kg-m/s) = (20 kg)v 2 − (20 kg)[ 3i − 2 j + 4 k (m/s) ], the velocity at t = 4 s is v 2 = 17.4 i + 14 j + 15.2k (m/s). Its magnitude is v 2 = 27.0 m/s. 27.0 m/s.

Problem 16.9 At time t = 0, the velocity of the 20-kg object is v = 3i − 2 j + 4 k (m/s). At t = 6 s, its velocity is v = 26 i + 22 j + 32k (m/s). What is the magnitude of the average total force acting on the object during the interval of time from t = 0 to t = 6 s?

Solution:

From Eq. (16.2), the average force is

1 ( mv 2 − mv 1 ) ΣFav = t 2 − t1 20 kg [ ( 26 i + 22 j + 32k ) − ( 3i − 2 j + 4 k ) ] m/s = 6s−0 =76.7 i + 80 j + 93.3k (N).

x z

Its magnitude is ΣFav = 145 N. 145 N.

Problem 16.10 The 5-lb collar at A is initially at rest. At t = 0, a force given in terms of time in seconds by F = (0.1 i + 0.3 j − 0.4 k)e −0.2t (lb)

is exerted on the collar, causing it to slide along the smooth horizontal bar. Use the principle of impulse and momentum to determine the magnitude of the collar’s velocity at t = 2 s.

Solution:

The collar’s mass is m = (5 lb)/(32.2 ft/s 2 ) = 0.155 slug.

Only the x component of the force affects the velocity. Applying impulse and momentum, t2

∫t Fx dt = mv x 2 − mv x1: 1

∫0 0.1e −0.4t dt = mv x 2 − 0. Evaluating the integral, we obtain v x2 = 1.06 ft/s. 1.06 ft/s.

F A x 4 ft

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Problem 16.11 The y -axis is vertical and the curved bar is smooth. The 5-kg slider starts from rest in position 1 and requires 4 s to slide to position 2. What is the magnitude of the average tangential force exerted on the slider as it moves from position 1 to position 2?

Solution:

We can use work and energy to determine the magnitude of the velocity at position 2, then use impulse and momentum to determine the average tangential force. Only the slider’s weight does work on it. Applying work and energy, we have 1 1 mv 2 − mv 1 2 : 2 2 2 1 mg(1 m) = mv 2 2 − 0, 2 U 12 =

y 1

obtaining v 2 = 4.43 m/s. Then applying impulse and momentum, (t 2 − t 1 ) Fav = mv 2 − mv 1: (4 s − 0) Fav = (4.43 m/s)(5 kg) − 0,

Gives Fav = 5.54 N. 2

5.54 N.

Problem 16.12 During the first 5 s of the 60,000-lb Gulfstream’s takeoff roll, the pilot increases the engine’s thrust at a constant rate from zero to its full thrust of 15,000 lb. (a) What impulse does the thrust exert on the airplane during the 5 s? (b) If you neglect other forces and assume that the thrust stays constant after 5 s, what total time is required for the airplane to accelerate from rest to its takeoff speed of 220 ft/s?

Solution: (a) The force exerted by the thrust during the first five seconds is T =

lb ( 15,000 )t. 5s

The linear impulse is

∫t ΣF dt = ∫0 ( t2

)

15, 000 lb t dt 5s

lb  1  ( 15, 000 )  2 t  5s 2

5s 0

= 37, 500 lb-s. (b) Applying impulse and momentum to the first five seconds, t2

∫t ΣF dt = mv 2 − mv 1: 1

37,500 lb-s =

60,000 lb ( 32.2 )v − 0, ft/s 2

the airplane’s velocity at t = 5 s is v 2 = 20.1 ft/s. Applying impulse and momentum to the motion after the first five seconds, t2

∫t ΣF dt = mv 2 − mv 1: 1

Source: Courtesy of Karolis Kavolelis/Shutterstock.

∫5 s(15,000 lb )dt = ( 32.2 ft/s 2 )(220 ft/s − 20.1 ft/s), t

60,000 lb

we obtain t = 29.8 s. (a) 37,500 slug/ft/s. (b) 29.8 s.

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Problem 16.13 The 10-kg box starts from rest on the smooth surface and is subjected to the horizontal force described in the graph. Use the principle of impulse and momentum to determine how fast the box is moving at t = 12 s.

Solution:

The simple shape of the force history suggests using geometry to calculate the impulse. Calculating the area of the triangle, rectangle, and triangle, the impulse from t = 0 to t = 12 s is 1 1 (4 s)(50 N) + (4 s)(50 N) + (4 s)(50 N) 2 2 = 400 N-s.

impulse =

Applying impulse and momentum to the box’s motion from t = 0 until t = 12 s,

impulse = mv 2 − mv 1: 400 N-s = (10 kg)v 2 − 0, the velocity of the box at t = 12 s is v 2 = 40 m/s.

F (N)

40 m/s.

50 N

t (s)

Problem 16.14 The 10-kg box starts from rest and is subjected to the horizontal force described in the graph. The coefficients of friction between the box and the surface are µ s = µ k = 0.2. Determine how fast the box is moving at t = 12 s.

F (N) 50 N

Solution: 0

t (s)

mg f

We can do so by calculating the area under F from t = 0 to t = 4 s, then subtract the area from t = 0 to t = t s = 1.57 s:

The normal force N = mg. The box does not move until the force F is large enough to cause it to slip, that is, when

impulse of F from t s to 4 s =

F = µ s N = 0.2(10 kg)(9.81 m/s 2 ) = 19.6 N.

Using this result, the total impulse of F from t = t s until t = 12 s is

Let t s denote the time at which the force reaches this value. Using the proportionality

impulse of F =

ts 19.6 N = 4s 50 N

1 1 (4 s)(50 N) − (1.57 s)(19.6 N). 2 2

1 1 (4 s)(50 N) − (1.57 s)( 19.6 N ) 2 2 1 + (4 s)(50 N) + (4 s)(50 N) = 385 N-s. 2

The impulse of the constant negative friction force from t = t s until t = 12 s is

gives t s = 1.57 s. After the box slips, the friction force remains f = µ k N = 0.2(10 kg) (9.81 m/s 2 ) = 19.6 N. The fact that the friction force is constant and the graph of the force F is simple suggests using geometry to determine the impulse of the forces on the box. The only slight complication is determining the impulse due to F from the time t s at which the box slips until t = 4 s:

impulse of friction = −(12 s − t s )µ k N = −(12 s − 1.57 s)0.2(10 kg)(9.81 m/s 2 ) = −205 N-s. Applying impulse and momentum, impulse of F + impulse of friction = mv 2 − mv 1:

385 kg-m/s −205 kg-m/s = (10 kg)v 2 −0,

50 N

the velocity of the box at t = 12 s is v 2 = 18.0 m/s.

19.6 N 0 ts

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18.0 m/s.

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Problem 16.15 The mass of the crate is 100 kg. The coefficient of kinetic friction between it and the sloping dock is µ k = 0.4. The crate starts from rest and the winch exerts a tension T = 1400 N. (a) What total impulse is applied to the crate during the first second of motion? (b) How fast is the crate moving after 1 s?

Solution: 308

T mg N mkN

(a) There is no acceleration normal to the surface, so the normal force is N = mg cos30 ° = (100 kg)(9.81 m/s 2 ) cos30° = 850 N. The total impulse is t2

∫t ΣF dt = ∫0 (T − mg sin 30° − µ k N ) dt 1

= (T − mg sin 30 ° − µ k N )(1 s) = [1400 N − (100 kg)(9.81 m/s 2 )sin 30 °

308

− 0.4(850 N)](1 s) = 570 kg-m/s. (b) Applying impulse and momentum, t2

∫t ΣF dt = mv 2 − mv 1: 1

570 kg-m/s = (100 kg)v 2 − 0, the velocity is v 2 = 5.70 m/s. (a) 570 kg-m/s. (b) 5.70 m/s.

Problem 16.16 The mass of the crate is 100 kg. The coefficient of kinetic friction between it and the sloping dock is µ k = 0.4. The crate starts from rest at time t = 0 and the winch exerts a tension given as a function of time in seconds T = 1400 + 600t N. (a) What total impulse is applied to the crate during the first second of motion? (b) How fast is the crate moving after 1 s? 308

Solution: 308

T mg

(b) Applying impulse and momentum,

∫t ΣF dt = mv 2 − mv 1: 1

mkN

870 kg-m/s = (100 kg)v 2 − 0,

(a) There is no acceleration normal to the surface, so the normal force is

the velocity is v 2 = 8.70 m/s. (a) 870 kg-m/s. (b) 8.70 m/s.

N = mg cos30 ° =

(100 kg)(9.81 m/s 2 ) cos30°

= 850 N. The total impulse is t2

∫t ΣF dt = ∫0 (T − mg sin 30° − µ k N ) dt 1

= ∫

1s 0

[ (1400 + 600t N) − (100 kg)(9.81 m/s 2 )sin 30 ° − 0.4(850 N) ] dt

= [ 1400 t + 300t 2 − (100 kg)(9.81 m/s 2 )t sin 30 ° − 0.4(850 N)t ]10 s = 870 N-s.

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Problem 16.17 In an assembly-line process, the 20-kg package A starts from rest and slides down the smooth ramp. Suppose that you want to design the hydraulic device B to exert a constant force of magnitude F on the package and bring it to rest in 0.15 s. What is the required force F ?

Solution:

We can use work and energy to determine how fast the package is moving when it comes into contact with B. The vertical distance the package moves is h = (2 m)sin 30 ° = 1 m, so equating the work done by its weight to the change in kinetic energy of the package,

After the package comes into contact with B, the force acting on it in the direction down the inclined plane is mg sin 30 ° − F . We apply impulse and momentum from the time t = 0 it comes into contact with B to t = 0.15 s when it comes to rest:

1 1 mgh = mv 2 2 − mv 1 2 : 2 2 (20 kg)(9.81 m/s 2 )(1 m) =

308

1 (20 kg)v 2 2 − 0, 2

∫t ΣF dt = mv 2 − mv 1: 1

we obtain v 2 = 4.43 m/s.

0.2 s

∫0

308

Solving gives F = 689 N. F

(mg sin30 ° − F ) dt = 0 − mv 1,

[(20 kg)(9.81 m/s 2 )sin30 ° − F ](0.15 s) = −(20 kg)(4.43 m/s).

689 N.

Problem 16.18 The 15-kg package A starts from rest and slides down the smooth ramp. If the hydraulic device B exerts a force of magnitude F = 300(1 + 5t ) N on the package, where t is in seconds measured from the time of first contact, what time after contact is required to bring the package to rest?

B 308

Solution:

After the package comes into contact with B, the force acting on it in the direction down the inclined plane is mg sin 30 ° − F = mg sin 30 ° − 300 − 1500t. We apply impulse and momentum from the time t = 0 it comes into contact with B to the time it comes to rest: t2

∫t ΣF dt = mv 2 − mv 1 :

we obtain v 2 = 4.43 m/s.

∫0 (mg sin 30° − 300 − 1500t ) dt = 0 − mv 1,

308

[(15 kg)(9.81 m/s 2 )sin 30 ° − 300]t 2 − 750t 2 2 = −(15 kg)(4.43 m/s).

Solving this quadratic equation, we obtain t 2 = 0.183 s. N

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0.183 s.

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Problem 16.19 In a cathode-ray tube, an electron (mass = 9.11 × 10 −31 kg) is projected at O with velocity v = (2.2 × 10 7 ) i (m/s). While the electron is between the charged plates, the electric field generated by the plates subjects it to a force F = −eEj. The charge on the electron is e = 1.6 × 10 −19 C (coulombs), and the electric field strength is E = 15sin(ωt ) kN/C, where the frequency ω = 2 × 10 9 s −1 . (a) What impulse does the electric field exert on the electron while it is between the plates? (b) What is the velocity of the electron as it leaves the region between the plates?

y 2 2 2 2 2 2 2 2

1 1 1 1 1 1 1 1 30 mm

Solution:

The x component of the velocity is unchanged. The 0.03 time spent between the plates is t = = 1.36 × 10 −9 s. 2.2 × 10 7

(a) The impulse is t2

∫t F dt = ∫0 (−eE ) dt = ∫0 −e(15 ×10 3 )(sin ωt ) dt 1

=  

1.36×10 (15 × 10 3 )e cos ωt  0 ω

−9

∫ t F dt = −2.3 × 10 −24 N-s. 1

The y component of the velocity is vy =

−2.3 × 10 −24 = −2.52 × 10 6 m/s. 9.11 × 10 −31

(b) The velocity on emerging from the plates is v = 22 × 10 6 i − 2.5 × 10 6 j m/s.

Problem 16.20 The two weights are released from rest at time t = 0. The coefficient of kinetic friction between the horizontal surface and the 5-lb weight is µ k = 0.4. Use the principle of impulse and momentum to determine the magnitude of the velocity of the 10-lb weight at t = 1 s. Strategy: Apply the principle to each weight individually.

Solution: Impulse = (10 lb)(1 s) − 0.4(5)(1 s) = 8 lb-s 8 lb-s =

15 lb ( 32.2 ) v ⇒ v = 17.2 ft/s. ft/s 2

5 lb

10 lb

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Problem 16.21 The two crates are released from rest. Their masses are m A = 60 kg and m B = 40 kg , and the coefficient of kinetic friction between crate A and the inclined surface is µ k = 0.2. What is the magnitude of the velocity of the crates after 1 s?

Solution:

The free-body diagrams are shown: 208

mAg T

mkN

mBg The normal force is N = m A g cos 20 °

208

= (60 kg)(9.81 m/s 2 ) cos 20 ° = 553 N. B

We apply impulse and momentum to crate A, t2

∫t ΣF dt = mv 2 − mv 1: 1

∫0 (T + m Ag sin 20° − µ k N ) dt = mv 2 − 0, [T + (60 kg)(9.81 m/s 2 )sin 20 ° − 0.2(553 N)](1 s) = (60 kg)v 2 , and also to crate B, t2

∫t ΣF dt = mv 2 − mv 1: 1

∫0 (−T + m B g) dt = m Bv 2 − 0, [−T + (40 kg)(9.81 m/s 2 )](1 s) = (40 kg)v 2 . We have two equations in terms of T and v 2 . Solving, we obtain T = 199 N, v 2 = 4.83 m/s. 4.83 m/s.

Problem 16.22 The two crates are released from rest. Their masses are m A = 20 kg, m B = 100 kg, and the surfaces are smooth. The angle θ = 15 °. What is the magnitude of the velocity of crate A after 1 s? Strategy: Apply the principle of impulse and momentum to each crate individually.

Solution:

Consider the individual free-body diagrams of the crates: T

NA B

mBg NB

A 158

mAg 158

Let v denote the velocity of crate A up the surface and the velocity of B down it. Impulse and momentum for crate A is (t 2 − t 1 )ΣFA = m Av 2 − m Av 1: (1 s)(T − m A g sin15 °) = m Av 2 − 0, A

and impulse and momentum for crate B is (t 2 − t 1 )ΣFB = m Bv 2 − m Bv 1:

(1 s)(m B g sin15 ° − T ) = m Bv 2 − 0.

We have two equations in terms of T and v 2 . Solving them, we obtain T = 84.6 N, v 2 = 1.69 m/s. 1.69 m/s.

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Problem 16.23 The two crates are released from rest. Their masses are m A = 20 kg, m B = 100 kg. The coefficient of kinetic friction between the contacting surfaces is µ k = 0.1. The angle θ = 15 °. What is the magnitude of the velocity of crate A after 1 s? A

Solution:

Consider the individual free-body diagrams of the

crates: mkNA T

u NA

mBg

mkNB

158

mAg

158

NA mkNA

We can see that the normal forces are N A = m A g cos15 ° and N B = N A + m B g cos15 °. Let v denote the velocity of crate A up the surface and the velocity of B down it. Impulse and momentum for crate A is (t 2 − t 1 )ΣFA = m Av 2 − m Av 1: (1 s)(T − m A g sin15 ° − µ k N A ) = m Av 2 − 0, and impulse and momentum for crate B is (t 2 − t 1 )ΣFB = m Bv 2 − m Bv 1: (1 s)(m B g sin15 ° − T − µ k N A − µ k N B ) = m Bv 2 − 0. Because the normal forces are known, we have two equations in terms of T and v 2 . Solving them, we obtain T = 78.3 N, v 2 = 0.429 m/s. 0.429 m/s.

Problem 16.24 At t = 0, a 20-kg projectile is given an initial velocity v 0 = 40 m/s at an initial angle θ 0 = 70 ° above the horizontal. Use the principle of impulse and momentum to determine the cartesian components of the projectile’s velocity at t = 3 s.

Solution:

The initial components of the velocity are v x = v 0 cos θ 0 , v y = v 0 sin θ 0 . The force acting on a projectile is

u0 x

F = −mg j. We write the principle of impulse and momentum in the form t2

∫t F dt = mv 2 − mv 1 : 1

∫0 −mg j dt = mv 2 − m(v 0 cos θ0 i + v 0 sin θ0 j), −mg j(3 s) = mv 2 − m(v 0 cos θ 0 i + v 0 sin θ 0 j). We obtain v 2 = v 0 cos θ 0 i + [v 0 sin θ 0 − (3 s) g] j = (40 m/s) cos(70 °) i + [(40 m/s)sin(70 °) − (3 s)(9.81 m/s 2 )] j = 13.7 i + 8.16 j (m/s). 13.7 i + 8.16 j (m/s).

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Problem 16.25 At time t = 0, the soccer player kicks the stationary 0.43-kg ball to a teammate. The ball reaches a maximum height of 1 m at a horizontal distance of 3 m from where he kicked it. The duration of the kick is 0.05 s. Neglecting aerodynamic drag, what is the magnitude of the average force exerted on the ball by the kick?

Solution: Let v x0 and v y0 be the horizontal and vertical components of the ball’s velocity following the kick. Applying work and energy to the ball’s vertical motion from just after the kick to its maximum height, 1 1 mv 2 − mv 1 2 : 2 2 2 1 −mg(1 m) = 0 − mv y 0 2 , 2 U 12 =

we obtain v y0 = 4.43 m/s. The ball’s vertical velocity as a function of time is v y = v y0 − gt. Setting v y = 0, we find that the time when the ball reaches its maximum height is t = 0.452 s. The horizontal position as a function of time is x = v x0 t. Setting x = 3 m and t = 0.467 s, we obtain v x0 = 6.64 m/s. Let Fx , Fy be the components of the average force exerted by the kick. Applying impulse and momentum in the x direction, Fx (t 2 − t 1 ) = mv x 2 − mv x1: Fx (0.05 s − 0) = (0.43 kg)(6.64 m/s) − 0, we obtain Fx = 57.1 N, and applying impulse and momentum in the y direction,

Source: Courtesy of Nirat.pix/Shutterstock.

( Fy − mg)(t 2 − t 1 ) = mv y 2 − mv y1: [ Fy − (0.43 kg)(9.81 m/s 2 )](0.05 s − 0) = (0.43 kg)(4.43 m/s) − 0, we obtain Fy = 42.3 N. The magnitude of the average force is Fx 2 + Fy 2 = 71.1 N. 71.1 N.

Problem 16.26* An object of mass m = 2 kg slides with constant velocity v 0 = 4 m/s on a smooth horizontal table (seen from above in the figure). The object is connected by a string of length L = 1 m to a fixed point O and is in the position shown, with the string parallel to the x-axis, at t = 0. Determine the x and y components of the force exerted on the object by the string as functions of time, then apply the principle of impulse and momentum to determine the velocity vector of the mass at t = 6 s.

Solution: y

v0 F

x The normal component of the acceleration of the object is v 0 2 /L, so the force exerted on it by the string is F = mv 0 2 /L. The string’s angular velocity is v 0 /L, so angle θ shown as a function of time is

y v0 O

v0 t L The components of the force as functions of time are θ =

Fx = −F cos θ = − x

mv 0 2 v cos 0 t L L

= −32 cos 4t N, Fy = −F sin θ = −

mv 0 2 v sin 0 t L L

= −32 sin4t N. We apply impulse and momentum in the x direction: t2

∫t Fx dt = mv x 2 − mv x1: 1

−∫

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6s 0

(1)

32cos4t dt = 2v x 2 ,

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16.26 (Continued) and apply it in the y direction: t2

∫t Fy dt = mv y 2 − mv y1: 1

−∫

6s 0

(2)

32sin 4t dt = 2v y 2 .

Evaluating the integrals in Eqs. (1) and (2), we obtain v x2 = 3.62 m/s, v y2 = −2.30 m/s. 3.62 i − 2.30 j (m/s).

Problem 16.27 The rail gun, which uses an electromagnetic field to impart velocity to an object, accelerates a 2-kg projectile from zero to 2.5 km/s in 0.0018 s. What is the magnitude of the average force exerted on the projectile? Solution:

Applying impulse and momentum in the form of Eq.

(16.4), (t 2 − t 1 ) ΣFt av = mv 2 − mv 1: (0.0018 s) ΣFt av = (2 kg)(2500 m/s) − 0, we obtain ΣFt av = 2.78E6 N.

Source: Courtesy of Metamorworks/Shutterstock.

2.78 MN.

Problem 16.28 The mass of the boat and its passenger is 420 kg. At time t = 0, the boat is moving at 14 m/s and its motor is turned off. The magnitude of the hydrodynamic drag force on the boat (in newtons) is given as a function of time by 830(1 − 0.08t ). Determine how long it takes for the boat’s velocity to decrease to 5 m/s.

Solution: mv 1 + ∫

t2 t1

The principle of impulse and momentum gives F dt = mv 2 t

(420 kg) (14 m/s) − ∫ 830(1 − 0.08t ) dt = (420 kg)(5 m/s) 0

−830(t − 0.04t 2 ) = −3780

Source: Courtesy of Zdorov Kirill Vladimirovich/Shutterstock.

t 2 − 25t + 114 = 0 t =

25 −

25 2 − 4(1)(114) = 5.99 s. 2

t = 5.99 s.

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Problem 16.29 The motorcycle starts from rest at time t = 0 and travels along a circular track with 400-m radius. From t = 0 to t = 5 s, the component of the total force on the motorcycle tangential to its path is Ft = 2250 N. The combined mass of the motorcycle and rider is 250 kg. (a) Use the principle of impulse and momentum to determine the magnitude of the motorcycle’s velocity at t = 5 s. (b) What horizontal force is exerted on the motorcycle in the direction perpendicular to its path at t = 5 s?

Solution: (a) Applying impulse and momentum, t2

∫t ΣFt dt = mv 2 − mv 1: 1

∫0

(2250 N) dt = (250 kg)v 2 − 0,

we obtain v 2 = 45 m/s (162 km/h). (b)

en N

We denote the horizontal force normal to the path (exerted by friction on the bike’s tires) by N. Newton’s second law in the normal direction at t = 5 s is v22 : R (45 m/s) 2 N = (250 kg) 400 m = 1270 N.

ΣFn = ma n = m

Source: Courtesy of Teemu Tretjakov/Shutterstock.

Problem 16.30 The motorcycle starts from rest at time t = 0 and travels along a circular track with 400-m radius. From t = 0 to t = 5 s, the component of the total force on the motorcycle tangential to its path is given as a function of time in seconds by Ft = 1100 + 160t + 90t 2 N. The combined mass of the motorcycle and rider is 250 kg. (a) Use the principle of impulse and momentum to determine the magnitude of the motorcycle’s velocity at t = 5 s. (b) What horizontal force is exerted on the motorcycle in the direction perpendicular to its path at t = 5 s?

(a) 45 m/s. (b) 1270 N.

Solution: (a) Applying impulse and momentum, t2

∫t ΣFt dt = mv 2 − mv 1: 1

∫0 (1100 + 160t + 90t 2 N) dt = (250 kg)v 2 − 0, [ 1100t + 80t 2 + 30t 3 ] 50 s

= (250 kg)v 2 ,

we obtain v 2 = 45 m/s (162 km/h). (b)

en et

ΣFn = ma n = m

Source: Courtesy of Teemu Tretjakov/Shutterstock.

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(a) 45 m/s. (b) 1270 N.

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Problem 16.31 The titanium rotor of a Beckman Coulter ultracentrifuge used in biomedical research contains 2-g samples at a distance of 41.9 mm from the axis of rotation. The rotor reaches its maximum speed of 130,000 rpm in 12 min. (a) Determine the average tangential force exerted on a sample during the 12 min the rotor is accelerating. (b) When the rotor is at its maximum speed, what normal acceleration are samples subjected to?

Solution: (a) Using the principle of impulse and momentum we have 0 + Fave ∆t = mv mv Fave = ∆t =

(

)( 2πrevrad )( 160mins )(0.0419 m) 60 s (12 min)( ) min

(0.002 kg) 130, 000

rev min

Fave = 0.00158 N. (b)

an =

(r ω ) 2 v2 = = rω 2 ρ r

(

rev   2π rad   1 min  = (0.0419 m)  130, 000    min   rev   60 s 

)

a n = 7.77 × 10 6 m/s 2 .

Problem 16.32 From t = 0 to t = 10 s, the angle θ between the 4500-kg airplane’s path and the horizontal increases from 0 to 40 ° with constant angular velocity. The pilot maintains the thrust equal to the drag, so that the net tangential force on the plane is the tangential component of its weight. At t = 0, the magnitude of the plane’s velocity is 130 m/s. Use the principle of impulse and momentum to determine the magnitude of its velocity at t = 10 s.

Solution: u u mg

The angle θ varies from 0 to (π /180 °)40 ° = 0.698 radians in 10 s, so θ = 0.0698t rad. The tangential component of the plane’s weight is mg sin θ and is opposite to the direction of its motion. Applying impulse and momentum, we have t2

∫t ΣF dt = mv 2 − mv 1:

10 s

∫0

−mg sin0.0698t dt = mv 2 − m(130 m/s).

Evaluating the integral, we obtain v 2 = 97.1 m/s. 97.1 m/s.

Problem 16.33 A 0.0457-kg golf ball is struck by a club. The club is in contact with the ball for 0.0006 s. After being struck, the ball travels 60 mm in the first 0.001 s. What was the magnitude of the average impulsive force exerted by the club? (Neglect the impulse due to the ball’s weight.) Solution:

The magnitude of the ball’s velocity after being struck is

0.06 m = 60 m/s. 0.001 s Applying impulse and momentum in the form of Eq. (16.4), (t 2 − t 1 )ΣFt av = mv 2 − mv 1: (0.0006 s)ΣFt av = (0.0457 kg)(60 m/s) − 0, we obtain ΣFt av = 4570 N (1030 lb).

Source: Courtesy of Iaremenko Sergii/Shutterstock.

4570 N.

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Problem 16.34 In a test of an energy-absorbing bumper, a 2800-lb car is driven into a barrier at 5 mi/h. The duration of the impact is 0.4 s. When the car rebounds from the barrier, the magnitude of its velocity is 1.5 mi/h. (a) What is the magnitude of the average horizontal force exerted on the car during the impact? (b) What is the average deceleration of the car during the impact?

Solution:

The velocities are

(

)

(

)

88 ft/s 88 ft/s v 1 = 5 mi/h = 7.33 ft/s, v 2 = 1.5 mi/h = 2.2 ft/s. 60 mi/h 60 mi/h (a) Using the principle of impulse and momentum we have − mv 1 + Fave∆t = mv 2 Fave = m

(v 1 + v 2 ) 2800 lb = ∆t 32.2 ft/s 2

( 7.33 ft/s0.4+s2.2 ft/s )

Fave = 2070 lb. (b) The average deceleration of the car during impact is a =

v 2 − (−v 1 ) (7.33 ft/s + 2.2 ft/s) = ∆t 0.4 s

a = 23.8 ft/s 2 .

Problem 16.35 A bioengineer, using an instrumented dummy to test a protective mask for a hockey goalie, launches the 0.12-kg puck so that it strikes the mask moving horizontally at 50 m/s (which is slightly faster than the highest recorded velocity for a puck). From photographs of the impact, she estimates the duration to be 0.02 s and observes that the puck rebounds at 6 m/s. What average impulsive force was exerted on the mask by the puck? Solution:

Applying impulse and momentum in the form of Eq. (16.4),

(t 2 − t 1 )ΣFt av = mv 2 − mv 1: (0.02 s)ΣFt av = (0.12 kg)(50 m/s) − (0.12 kg)(−6 m/s), we obtain ΣFt av = 336 N (75.5 lb). 336 N.

Problem 16.36 A fragile object dropped onto a hard surface may break because it is subjected to a large impulsive force. Suppose you drop a 0.15-kg Rolex from 1 m above a concrete floor, the duration of the impact is 0.002 s, and the watch bounces 20 mm above the floor. (a) What average impulsive force is the watch subjected to? (b) What is the watch’s average deceleration during the impact?

which gives v 1 = 0.626 m/s. Now we apply impulse and momentum to the bounce, treating upward as positive: ΣFav (t 2 − t 1 ) = mv 2 − mv 1: ΣFav (0.002 s) = (0.15 kg)(0.626 m/s) − (0.15 kg)(−4.43 m/s). This gives ΣFav = 379 N (85.2 lb). (b) The average acceleration is ΣFav /m = 2530 m/s 2 (258 g ' s). (a) 379 N. (b) 2530 m/s 2 .

Solution: (a) We can apply work and energy to determine the watch’s velocity just before it hits the floor: 1 1 mv 2 − mv 1 2 : 2 2 2 1 mg(1 m) = mv 2 2 − 0, 2 U 12 =

obtaining v 2 = 4.43 m/s. In the same way, we can determine its upward velocity immediately after it bounces: 1 1 mv 2 − mv 1 2 : 2 2 2 1 2 −mg(0.02 m) = 0 − mv 1 , 2 U 12 =

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Problem 16.37 The 0.43-kg soccer ball is given a kick with a 0.08-s duration that accelerates it from rest to a velocity of 20 m/s at 40 ° above the horizontal. Neglecting aerodynamic drag, what is the magnitude of the average force exerted on the ball by the kick? Solution:

The horizontal and vertical components of the ball’s velocity following the kick are v x2 = (20 m/s) cos 40 ° = 15.3 m/s, v y2 = (20 m/s)sin 40 ° = 12.9 m/s. Let Fx , Fy denote the horizontal and vertical components of the average force exerted by the kick. Applying impulse and momentum in the x direction, Fx (t 2 − t 1 ) = mv x 2 − mv x1: Fx (0.08 s − 0) = (0.43 kg)(15.3 m/s) − 0, we obtain Fx = 82.3 N, and applying impulse and momentum in the y direction,

Source: Courtesy of Anthony Saint James/ Photodisc/Getty Images.

( Fy − mg)(t 2 − t 1 ) = mv y 2 − mv y1: [ Fy

− (0.43 kg)(9.81 m/s 2 )](0.08 s − 0) =

(0.43 kg)(12.9 m/s) − 0,

we obtain Fy = 73.3 N. The magnitude of the average force is Fx 2 + Fy 2 = 110 N. 110 N.

Problem 16.38 An entomologist measures the motion of a 3-g locust during its jump and determines that the insect accelerates from rest to 3.4 m/s in 25 ms (milliseconds). The angle of takeoff is 55 ° above the horizontal. What are the horizontal and vertical components of the average impulsive force exerted by the locust’s hind legs during the jump?

Solution:

The impulse is

∫ t F dt = Fave (t 2 − t1 ) = m(v 2 ) = m(3.4 cos 55°i + 3.4 sin 55° j), 1

from which Fave (2.5 × 10 −2 ) = (5.85 × 10 −3 ) i + (8.36 × 10 −3 ) j N -s. The average total force is 1 ((5.85 × 10 −3 ) i + (8.36 × 10 −3 ) j) Fave = 2.5 × 10 −2 = 0.234 i + 0.334 j N. The impulsive force is Fimp = Fave − (−mg j) = 0.234 i + 0.364 j N.

Problem 16.39 A 5-oz baseball is 3 ft above the ground when it is struck by a bat. The horizontal distance to the point where the ball strikes the ground is 180 ft. Photographic studies indicate that the ball was moving approximately horizontally at 100 ft/s before it was struck, the duration of the impact was 0.015 s, and the ball was traveling at 30 ° above the horizontal after it was struck. What was the magnitude of the average impulsive force exerted on the ball by the bat?

238

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16.39 (Continued) Solution:

We need to analyze the ball’s trajectory to determine its velocity just after it is struck.

Denoting the ball’s velocity after it is struck by v 0 and the elevation angle above the horizontal by θ 0 = 30 °, the initial components of its velocity are v x = v 0 cos θ 0 , v y = v 0 sin θ 0 . The components of its acceleration are a x = 0, a y = −g. We can integrate the acceleration with respect to time to determine the velocity and position as functions of time. In the x direction, ax = vx 0

(1)

dx vx = = v 0 cos30 °, dt x

∫0 dx = ∫0 v 0 cos30° dt, (2) x = (v 0 cos30 °)t,

(4)

If we set x = 180 ft in Eq. (2) and y = −3 ft in Eq. (4), we can solve them for t and v 0 , obtaining t = 2.58 s and v 0 = 80.7 ft/s. The ball’s mass is

We write the principle of impulse and momentum in the form given by Eq. (16.2): (0.015 s)ΣFav = (0.00970 slug)(v 0 cos θ 0 i + v 0 sin θ 0 j) − (0.00970 slug)[(−100 ft/s) i].

ΣFav = 110 i + 26.1 j (lb).

dv y ay = = −g, dt

The magnitude is ΣFav = 113 lb.

∫v sin 30 dv y = ∫0 −g dt,

(3)

vy =

1 2 gt . 2

Solving yields

and in the y direction,

y = (v 0 sin 30 °)t −

(t 2 − t 1 )ΣFav = mv 2 − mv 1:

5 lb W 16 m = = = 0.00970 slug. 32.2 ft/s 2 g

dv x = 0, dt

∫v cos 30 dv x = 0,

∫0 dy = ∫0 (v 0 sin 30° − gt ) dt,

113 lb.

dy = v 0 sin 30 ° − gt, dt

Problem 16.40 Paraphrasing the official rules of racquetball, a standard racquetball is 2 14 inches in diameter, weighs 1.4 ounces (16 ounces = 1 pound), and bounces between 68 and 72 inches from a 100-inch drop at a temperature between 70 and 74 degrees Fahrenheit. Suppose that a ball bounces 71 inches when it is dropped from a 100-inch height. If the duration of the impact is 0.08 s, what average force is exerted on the ball by the floor?

Solution:

The velocities before and after the impact are

v1 =

2(32.2 ft/s 2 )(100 in)

v2 =

2(32.2 ft/s 2 )(71 in)

( 121 ftin ) = 23.2 ft/s,

( 121 ftin ) = 19.5 ft/s,

Using the principle of impulse and momentum we have −mv 1 + Fave ∆t = mv 2 v + v2 Fave = m 1 ∆t 1.4 1 = lb 16 32.2 ft/s 2

(

)(

+ 19.5 ft/s) ) (23.2 ft/s0.08 s

Fave = 1.45 lb.

Problem 16.41 The masses m A = m B . The surface is smooth. At t = 0, A is stationary, the spring is unstretched, and B is given a velocity v 0 toward the right. (a) In the subsequent motion, what is the velocity of the common center of mass of A and B? (b) What are the velocities of A and B when the spring is unstretched? Strategy: To do part (b), think about the motions of the masses relative to their common center of mass.

Solution: (a) The velocity of the center of mass does not change because there are no external forces on the system vc =

m Av 0 + m B 0 v v = 0 vC = 0 . (m A + m B ) 2 2

(b) Looking at the system from a reference frame moving with the center of mass, we have an oscillatory system with either the masses moving towards the center or away from the center. Translating back to the ground reference system, this means

k mA

M16_BEDF2222_06_SE_C16_Part1.indd 239

Either

v A = v 0 (to the right), v B = 0

v A = 0, v B = v 0 (to the right).

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Problem 16.42 The masses m A = 40 kg and m B = 30 kg, and k = 400 N/m. The two masses are released from rest on the smooth surface with the spring stretched 1 m. What are the magnitudes of the velocities of the masses when the spring is unstretched?

Solution:

From the solution of Problem 16.41, (1) m Av A + m Bv B = 0: or, evaluating, 40v A + 30v B = 0. Energy is conserved, Thus, (2) 1 2 1 1 kS = m Av A 2 + m Bv B 2 . Evaluating, we get 2 2 2 1 1 1 (400)(1) 2 = (40)v A2 + (30)v B2 2 2 2

Solving Equations (1) and (2),

k mA

v A = 2.07 m/s, v B = 2.76 m/s.

Problem 16.43 A girl weighing 80 lb stands at rest on a 325-lb floating platform. She starts running at 10 ft/s relative to the platform and runs off the end. Neglect the horizontal force exerted on the platform by the water. (a) After she starts running, what is her velocity relative to the water? (b) While she is running, what is the velocity of the common center of mass of the girl and the platform relative to the water? (See Practice Example 16.5.) Solution: (a) Momentum is conserved. 0 = m gv g + m bv p , v g /p = v g − v p 0 = (80 lb)v g + (325 lb)v p , (10 ft/s) = v g − v p Solving we find vg =

(10 ft/s)(325 lb) = 8.02 ft/s. 405 lb

v g = 8.02 ft/s to the right. (b) Since momentum is conserved, the velocity of the center of mass is always zero. v cm = 0.

Problem 16.44 Two railroad cars with weights W A = 120,000 lb and W B = 70,000 lb collide and become coupled together. Car A is full, and car B is half full, of carbolic acid. When the cars collide, the acid in B sloshes back and forth violently. (a) Immediately after the impact, what is the velocity of the common center of mass of the two cars? (b) When the sloshing in B has subsided, what is the velocity of the two cars? 2 ft/s

240

Solution: (a)

(120,000 lb)(2 ft/s) + (70,000 lb)(1 ft/s) (120,000 lb + 70,000 lb) = 1.63 ft/s.

v center of mass =

(b) After the sloshing stops v cars = v center of mass = 1.63 ft/s.

1 ft/s

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Problem 16.45 The weights of the railroad cars are W A = 120,000 lb and W B = 70,000 lb. The railroad track has a constant slope of 0.2 degrees upward toward the right. If the cars are 6 ft apart at the instant shown, what is the velocity of their common center of mass immediately after they become coupled together?

2 ft/s

Solution: t =

1 ft/s

1 m/s

Time to couple (both accelerate at the same rate) is 6 ft = 6 s. (2 ft/s − 1 ft/s)

2 m/s

Impulse—momentum is now u 0.28

lb 70,000 lb ( 120,000 )(2 ft/s) + ( 32.2 )(1 ft/s) 32.2 ft/s ft/s 2

−(190,000 lb)sin 0.2°(6 s) =

lb ( 190,000 )v 32.2 ft/s 2

center of mass

v center of mass = 0.957 ft/s.

Problem 16.46 A 500-kg satellite (S ) traveling at 7 km/s is hit by a 2-kg meteor ( M ) traveling at 12 km/s in the direction shown. The meteor is embedded in the satellite by the impact. Determine the magnitude of the velocity of their common center of mass after the impact and the angle β between the path of the center of mass and the original path of the satellite.

7 km/s

b 458 M

Solution: y

Its magnitude is v = 6.94 km/s. The angle β is

v b vS

458

vy  β = arctan   vx  = arctan

km/s ( 0.0338 ) 6.94 km/s

= 0.279 °. vM

6.94 km/s, β = 0.279 °.

In terms of the coordinate system shown, the velocity of the satellite is v S = (7 km/s) i and the velocity of the meteor is v M = −(12 km/s) cos 45 °i + (12 km/s) cos 45 ° j. From Eq. (16.9), the velocity of the common center of mass after the impact is v = =

mS v S + m M v M mS + m M (500 kg)(7 km/s) i + (0.5 kg)[−(12 km/s) cos 45 °i + (12 km/s) cos 45 ° j] (500 kg) + (0.5 kg)

= 6.94 i + 0.0338 j (km/s).

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Problem 16.47 A 500-kg satellite (S ) traveling at 7 km/s is hit by a 2-kg meteor ( M ). The meteor is embedded in the satellite by the impact. (a) What would the magnitude of the velocity of the meteor need to be to cause the angle β between the original path of the satellite and the path of the center of mass of the combined satellite and meteor after the impact to be 0.5 ° ? (b) What is the magnitude of the velocity of the center of mass after the impact?

b 458 M

Solution:

Its components are

    mS mM v x =  v cos 45°, (1)  (7 km/s) −   m S + m M  M  m S + m M 

v b vS

  mM v y =  v cos 45 °. (2)  m S + m M  M

458

The angle β satisfies

tan β = =

(a) In terms of the coordinate system shown, the velocity of the satellite is

Let v M denote the magnitude of the velocity of the meteor before inpact. In terms of the coordinate system shown,

vy vx m M v M cos 45 ° . m S (7 km/s) − m M v M cos 45 °

Setting β = 0.5 °, we can solve this equation for the velocity of the meteor, obtaining v M = 21.4 km/s.

v S = (7 km/s) i.

(b) From Eqs. (1) and (2), we can solve for the magnitude of the velocity of the common center of mass, obtaining v = 6.91 km/s.

v M = −v M cos 45 °i + v M cos 45 ° j.

(a) 21.4 km/s. (b) 6.91 km/s.

From Eq. (16.9), the velocity of the common center of mass after the impact is v =

7 km/s

mS v S + m M v M mS + m M

    mS mM =   (7 km/s) i +    m S + m M   m S + m M  [−v M cos 45 °i + v M cos45 ° j].

Problem 16.48 On September 26, 2022, the DART (Double Asteroid Redirection Test) mission demonstrated the feasibility of altering the orbit of an asteroid by impacting it with a spacecraft. Dimorphos is a relatively small asteroid with a mass of 2.75E10 kg. Assume that the DART spacecraft, with a mass of 550 kg, strikes the asteroid moving in the direction opposite to the asteroid’s motion with a relative velocity of 6.6 km/s. What is the magnitude of the resulting change in the velocity of Dimorphos? Assume that the mass of the spacecraft remains with the asteroid. (In reality, the physics of the high-velocity impact is expected to result in some mass being thrown outward, which could affect the amount of momentum imparted to the asteroid. One of the objectives of the test was to evaluate this phenomenon.)

242

Source: Courtesy of Dpa picture alliance/Alamy Stock Photo.

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16.48 (Continued) Solution:

Let m A = 550 kg, m B = 2.75E10 kg. Applying conservation of energy in terms of a reference frame that is fixed with respect to the asteroid, m Av A + 0 = (m A + m B )v ′B : (550 kg)(6600 m/s) = (550 kg + 2.75E10 kg)v ′B ,

we obtain v ′B = 1.32E-4 m/s. 0.132 mm/s.

Problem 16.49* An 80-lb boy sitting in a stationary 15-lb wagon wants to simulate rocket propulsion by throwing bricks out of the wagon. Neglect horizontal forces on the wagon’s wheels. If the boy has three bricks weighing 5 lb each and throws them with a horizontal velocity of 2 ft/s relative to the wagon, determine the velocity he attains (a) if he throws the bricks all at once and (b) if he throws them one at a time.

Solution: (a) Let mW denote the combined mass of the wagon and the boy, and let m B denote the mass of one brick. If he throws the bricks all at once, conservation of momentum requires that mW v W + 3m Bv B = 0,

The velocity of the wagon relative to the brick is again 2 ft/s :

where v W and v B are the velocities of the wagon and bricks relative to the ground (positive toward the right). We are told that the velocity of the wagon relative to the bricks is 2 ft/s:

v W − v B = 2 ft/s,

v W = 0.0952 ft/s,

so we have two equations for the two velocities, and we obtain

He throws the third brick. Again changing frames of reference, conservation of momentum requires that

v W = 0.273 ft/s,

v B = −1.73 ft/s.

(b) When he throws the first brick, conservation of momentum requires that

so we again have two equations for the two velocities, and we obtain v B = −1.90 ft/s.

mW v W + m Bv B = 0. The velocity of the wagon relative to the brick is 2 ft/s,

(mW + 2m B )v W + m Bv B = 0.

v W − v B = 2 ft/s,

The velocity of the wagon relative to the brick is 2 ft/s:

and we obtain

v W − v B = 2 ft/s,

v W = 0.100 ft/s,

so we have two equations for the two velocities, and we obtain

Summing the three velocities, the final velocity of the wagon (and boy) relative to the ground is

v W = 0.0909 ft/s,

v B = −1.91 ft/s.

Now he throws the second brick. It makes things simpler if we now use the wagon before he throws the second brick as our reference frame, instead of the ground. We can determine the velocities of the wagon and brick as before, and do the same thing when he throws the third brick. We must then sum the three velocities of the wagon to determine the final velocity of the wagon relative to the ground. Conservation of momentum requires that

v B = −1.90 ft/s.

v W = 0.0909 + 0.0952 + 0.100 = 0.286 ft/s. (a) v W = 0.273 ft/s. (b) v W = 0.286 ft/s.

(mW + m B )v W + m Bv B = 0.

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Problem 16.50 The catapult, designed to throw a line to ships in distress, throws a 2-kg projectile. The mass of the catapult is 36 kg, and it rests on a smooth surface. If the velocity of the projectile relative to the Earth as it leaves the tube is 50 m/s at θ 0 = 30 ° relative to the horizontal, what is the resulting velocity of the catapult toward the left?

Solution: 0 = m cv c + m p (50 cos30 °). Solving, vc =

−(2)(50 cos30 °) = −2.41 m/s. 36

Problem 16.51 The catapult, which has a mass of 36 kg and throws a 2-kg projectile, rests on a smooth surface. The velocity of the projectile relative to the catapult as it leaves the tube is 50 m/s at θ 0 = 30 ° relative to the horizontal. What is the resulting velocity of the catapult toward the left?

Solution: 0 = m cv c + m pv px , where v px − v c = 50 cos30 °. Solving, v c = −2.28 m/s.

Problem 16.52 A bullet with a mass of 10 g is moving horizontally with velocity v and strikes a 30-kg block of wood. After the impact, the block and embedded bullet slide 12 mm across the floor. The coefficient of kinetic friction between the block and the floor is µ k = 0.4. Determine the velocity v .

Solution:

Denote the mass of the bullet by m S = 0.01 kg and the mass of the block by m B = 30 kg, and define m = m S + m B . We can use conservation of momentum to relate the velocity of the bullet to the velocity of the block and imbedded bullet after the bullet hits. Let v S be the velocity of the bullet before the impact and let v B be the velocity of the bullet and block afterward. We have m Sv S = mv B . (1)

v mkN

We can use work and energy to determine v B , then use Eq. (1) to determine v S . The normal force N = mg. Let s = 0.012 m denote the distance the block slides. Applying work and energy, U 12 = T2 − T1 : 1 −(µ k mg)s = 0 − mv B 2 , 2 we obtain v B = 0.307 m/s. Then Eq. (1) gives v S = 921 m/s. 921 m/s.

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Problem 16.53 A 0.12-ounce bullet hits a suspended 15-lb block of wood and becomes embedded in it. The angle through which the wires supporting the block rotate as a result of the impact is measured and determined to be 7 °. What was the bullet’s velocity? Solution: mS = mB =

2 ft

The masses of the bullet and block are

( 161 lboz ) = 0.000233 slug,

( 0.12 oz )

32.2 ft/s 2

15 lb = 0.466 slug. 32.2 ft/s 2

We can use conservation of momentum to relate the velocity of the bullet to the velocity of the block and imbedded bullet after the bullet hits. Let v S be the velocity of the bullet before the impact and let v B be the velocity of the bullet and block afterward. We have m Sv S = (m S + m B )v B . (1)

2 ft 78

Using the given angle of rotation of the wires, we can calculate how high the block rises after the bullet hits. We can use conservation of energy to determine v B , and then use Eq. (1) to determine v S . The distance the block rises is h = 2 ft − ( 2 ft ) cos 7 ° = 0.0149 ft. Applying conservation of energy, T1 + V1 = T2 + V2 : 1 (m + m B )v B 2 + 0 = 0 + (m S + m B ) gh, 2 S we obtain v B = 0.980 ft/s. Then Eq. (1) gives v S = 1960 ft/s. 1960 ft/s.

Problem 16.54 The overhead conveyor drops the 12-kg package A into the 1.6-kg carton B. The package is tacky and sticks to the bottom of the carton. If the coefficient of friction between the carton and the horizontal conveyor is µ k = 0.2, what distance does the carton slide after the impact?

268

1m /s

Solution: The horizontal velocity of the package ( A) relative to the carton ( B) is

0.2 m/s

v A = (1) cos 26 ° − 0.2 = 0.699 m/s. Let v be the velocity of the combined object relative to the belt. m Av A = (m A + m B )v . Solving, v =

m Av A m A + mB

(12)(0.699) 12 + 1.6

= 0.617 m/s. Use work and energy to determine the sliding distance d: −µ k (m A + m B ) gd = 0 − d =

1 (m + m B )v 2 , 2 A

(0.617) 2 v2 = 2µ k g 2(0.2)(9.81)

d = 0.0969 m.

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Problem 16.55 A 28,000-lb bus collides with a 3500-lb car. The velocity of the bus before the collision is v B = 20 i (ft/s) and the velocity of the car is v C = 44 j (ft/s). The two vehicles become entangled and remain together after the collision. The coefficient of kinetic friction between the vehicles’ tires and the road is µ k = 0.6. (a) What is the magnitude of the velocity of the common center of mass of the two vehicles immediately after the collision? (b) Determine the approximate final position of their common center of mass relative to its position when the collision occurred. (Assume that the tires skid, not roll, on the road, and that no obstacles are encountered.)

Solution: (a) The masses of the bus and car are mB =

28,000 lb = 870 slug, 32.2 ft/s 2

mC =

3500 lb = 109 slug. 32.2 ft/s 2

Let m = m B + mC = 978 slug. The velocity of their common center of mass is v = =

m B v B + mC v C m (870 slug)[20 i (ft/s)] + (109 slug)[44 j (ft/s)] 978 slug

= 17.8i + 4.89 j (ft/s). Its magnitude is v = 18.4 ft/s. (b) y

mkN

v a

x The angle α is vy  4.89 ft/s α = arctan   = arctan = 15.4 °. vx  17.8 ft/s

(

)

After the collision, the vehicles will slide some distance s until friction brings them to rest. The normal force between the vehicles and the road is N = 28,000 + 3500 = 31,500 lb. Applying work and energy, U 12 = T2 − T1: 1 −(µ k N )s = 0 − m v 2 , 2 we obtain s = 8.80 ft. The coordinates of the center of mass relative to its position when the collision occurred are x = s cos α = 8.48 ft, y = s sin α = 2.33 ft. (a) 18.4 ft/s. (b) x = 8.48 ft, y = 2.33 ft.

Problem 16.56 The velocity of the 220-kg astronaut A relative to the space station is 40 i + 15 j (mm/s). The velocity of the 310-kg structural member B relative to the station is −20 i + 15 j (mm/s). When they approach each other, the astronaut clings to the structural member. (a) What is the velocity of their common center of mass? (b) Determine the x and y coordinates of the point at which the astronaut comes into contact with the structural member. (c) What is the x coordinate of the point at which they contact the station?

246

6m A 9m

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16.56 (Continued) Solution: y

m Ax A + mBx B (220 kg)(0) + (310 kg)(9 m) = m A + mB 220 kg + 310 kg = 5.26 m.

Using their velocity in the y direction, the time required for them to move from their original positions to where they reach the station is

6000 mm = 400 s. 15 mm/s

(a) The velocity of their common center of mass is m Av A + m B v B m A + mB (220 kg)[40 i + 15 j (mm/s)] + (310 kg)[−20 i + 15 j (mm/s)] = 220 kg + 310 kg = 4.91 i + 15 j (mm/s).

v =

The distance their common center of mass moves in the x direction during that time is (4.91 mm/s)(400 s) = 1960 mm. Adding this distance to the initial x coordinate of their center of mass, the x coordinate of the point where they contact the station is

(b) In the x direction, the speed at which they are approaching each other before they come into contact is

5.26 m + 1.96 m = 7.23 m.

40 mm/s + 20 mm/s = 60 mm/s, (a) v = 4.91i + 15 j (mm/s).

so the time required for them to move from their original positions to where they come into contact is

(b) x = 6.00 m, y = 2.25 m. (c) x = 7.23 m.

9000 mm = 150 s. 60 mm/s Using the astronaut’s velocity, the x coordinate of the point where they come into contact is x contact = (40 mm/s)(150 s) = 6000 mm, and the y coordinate is y contact = (15 mm/s)(150 s) = 2250 mm.

Problem 16.57 The weights of the two objects are W A = 8 lb and W B = 12 lb. Object A is moving at v A = 5 ft/s and undergoes a direct central impact with the stationary object B. Determine the velocities of the objects after the impact if it is (a) perfectly plastic; (b) perfectly elastic.

Solution: mA =

WA , g

The masses of the objects are mB =

WB . g

In the impact, momentum is conserved: m Av A + m Bv B = m Av ′A + m Bv ′B. (1) From Eq. (16.16), the definition of the coefficient of restitution is

vA A

e = B

v ′B − v ′A . (2) vA − vB

The velocities of the objects before the impact are known, so if the coefficient of restitution is known, there are two equations for v ′A and v ′B . (a) In a perfectly plastic collision, e = 0. (Notice that this means the two objects have the same velocity after the impact.) Setting v A = 5 ft/s and v B = 0 and solving Eqs. (1) and (2), we obtain v ′A = v ′B = 2 ft/s. (b) In a perfectly elastic collision, e = 1. Setting v A = 5 ft/s and v B = 0 and solving Eqs. (1) and (2), we obtain v ′A = −1 ft/s, v ′B = 4 ft/s. (You should check this result by making sure the total kinetic energy is the same before and after the impact.) (a) A and B, 2 ft/s to the right. (b) A, 1 ft/s to the left, B, 4 ft/s to the right.

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Problem 16.58 The weights of the two objects are W A = 8 lb and W B = 12 lb . Object A is moving at v A = 5 ft/s and undergoes a direct central impact with the stationary object B. The coefficient of restitution is e = 0.2. Determine the velocities of the objects after the impact.

Solution: mA =

The masses of the objects are

WA , g

mB =

WB . g

In the impact, momentum is conserved: m Av A + m Bv B = m Av ′A + m Bv ′B . From Eq. (16.16), the definition of the coefficient of restitution is

vA A

e = B

v ′B − v ′A . vA − vB

The velocities of the objects before the impact are known, so because the coefficient of restitution is given, there are two equations for v ′A and v ′B . Setting v A = 5 ft/s and v B = 0 and solving, we obtain v ′A = 1.4 ft/s, v ′B = 2.4 ft/s. A, 1.4 ft/s to the right; B, 2.4 ft/s to the right.

Problem 16.59 The objects A and B with velocities v A = 20 m/s and v B = 4 m/s undergo a direct central impact. Their masses are m A = 8 kg and m B = 12 kg. After the impact, the object B is moving to the right at 16 m/s. What is the coefficient of restitution? Solution:

vA A

vB B

In the impact, momentum is conserved:

m Av A + m Bv B = m Av ′A + m Bv ′B . From Eq. (16.16), the definition of the coefficient of restitution is e =

v ′B − v ′A . vA − vB

The velocities of the objects before the impact are known, and the velocity of B after the impact is known. There are two equations for v ′B and e. Setting v A = 20 m/s, v B = 4 m/s, and v ′B = 16 m/s and solving, we obtain v ′A = 2 m/s, e = 0.875. e = 0.875.

Problem 16.60 The 10-kg mass A and the 14-kg mass B slide on the smooth horizontal bar with the velocities shown. The coefficient of restitution is e = 0.6. Determine the velocities of the masses after they collide. Solution:

3 m/s

2 m/s

In the impact, momentum is conserved:

m Av A + m Bv B = m Av ′A + m Bv ′B . From Eq. (16.16), the definition of the coefficient of restitution is e =

v ′B − v ′A . vA − vB

The velocities of the objects before the impact are known, so because the coefficient of restitution is given, there are two equations for v ′A and v ′B . Setting v A = 3 m/s and v B = −2 m/s and solving, we obtain v ′A = −1.67 m/s, v ′B = 1.33 m/s. A, 1.67 m/s to the left; B, 1.33 m/s to the right.

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Problem 16.61 In a study of the effects of an accident on simulated occupants, the 1900-lb car with velocity v A = 30 mi/h collides with the 2800-lb car with velocity v B = 20 mi/h. The coefficient of restitution of the impact is e = 0.15. What are the velocities of the cars immediately after the collision?

Solution:

Momentum is conserved, and the coefficient of restitu-

tion is used. m Av A + m Bv B = m Av ′A + m Bv ′B ,

e(v A − v B ) = v ′B − v ′A

(1900 lb) (30 mi/h) + (2800 lb)(−20 mi/h) = (1900 lb)v ′A + (2800 lb)v ′B (0.15)([30 mi/h] − [−20 mi/h]) = v ′B − v ′A Solving we find v ′A = −4.26 mi/h,

v ′B = 3.24 mi/h.

Converting into ft/s we find v ′A = 6.24 ft/s to the left, v ′B = 4.76 ft/s to the right.

Problem 16.62 In a study of the effects of an accident on simulated occupants, the 1900-lb car with velocity v A = 30 mi/h collides with the 2800-lb car with velocity v B = 20 mi/h. The coefficient of restitution of the impact is e = 0.15. The duration of the collision is 0.22 s. Determine the magnitude of the average acceleration to which the occupants of each car are subjected.

Solution:

The velocities before the collision are converted into ft/s. The velocities after the collision were calculated in the preceding problem. The velocities are v A = 44 ft/s, v B = 29.3 ft/s, v ′A = − 6.24 ft/s, v ′B = 4.76 fts. The average accelerations are aA =

(−6.24 ft/s) − (44 ft/s) ∆v = = 228 ft/s 2 ∆t 0.22 s

aB =

(4.76 ft/s) − (−29.3 ft/s) ∆v = = 155 ft/s 2 ∆t 0.22 s

a A = 228 ft/s 2 (7.09 g′s), a B = 155 ft/s 2 (4.81g′s).

Problem 16.63 The balls are of equal mass m. Balls B and C are connected by an unstretched spring and are stationary. Ball A moves toward ball B with velocity v A . The impact of A with B is perfectly elastic (e = 1). (a) What is the velocity of the common center of mass of balls B and C immediately after the impact? (b) What is the velocity of the common center of mass of B and C at time t after the impact? Solution:

Consider the impact of balls A and B. From the equations

vA k A

mv A = mv ′A + mv ′B , e = 1=

v ′B − v ′A , vA

we obtain v ′A = 0, v ′B = v A . (a) The position of the center of mass is x =

mx B + mx C x + xC = B , m+ m 2

so dx 1 = dt 2

( dxdt + dxdt ). B

(b) With no external forces, dx 1 = const. = v A . dt 2

Immediately after the impact dx B /dt = v A and dx C /dt = 0, so dx 1 = v A. dt 2

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Problem 16.64* In Problem 16.63, what is the maximum compressive force in the spring as a result of the impact?

vA k A

Solution:

See the solution of Problem 16.63. Just after the collision of A and B, B is moving to the right with velocity v A , C is stationary, and the center of mass D of B and C is moving to the right

vB9 = vA

vD9 = ½ vA

vC9 = 0

with velocity 12 v A (Fig. a). Consider the motion in terms of a reference frame that is moving with D (Fig. b). Relative to this reference frame, B is moving to the right with velocity 12 v A and C is moving to the left with velocity 12 v A . There total kinetic energy is

(a) y

2 2 1 1 1 1 1 m v A + m v A = mv 2A . 2 2 2 2 4

(

)

(

)

When the spring has brought B and C to rest relative to D, their kinetic energy has been transformed into potential energy of the spring. This is when the compressive force in the spring is greatest. Setting 1 2 1 mv = kS 2 , we find that the compression of the spring is 4 A 2 S = −v A

C x (b)

½ vA

m . 2k

Therefore the maximum compressive force is k S = vA

mk . 2

Problem 16.65* The balls are of equal mass m. Balls B and C are connected by an unstretched spring and are stationary. Ball A moves toward ball B with velocity v A . The impact of A with B is perfectly elastic (e = 1). Suppose that you interpret this as an impact between ball A and an “object” D consisting of the connected balls B and C. (a) What is the coefficient of restitution of the impact between A and D ? (b) If you consider the total energy after the impact to be the sum of the kinetic energies 12 m(v ′A ) 2 + 12 (2m)(v ′D ) 2 , where v ′D is the velocity of the center of mass of D after the impact, how much energy is “lost” as a result of the impact? (c) How much energy is actually lost as a result of the impact? (This problem is an interesting model for one of the mechanisms of energy loss in impacts between objects. The energy “loss” calculated in part (b) is transformed into “internal energy”—the vibrational motions of B and C relative to their common center of mass.)

Solution: See the solution of Problem 16.135. Just after the impact of A and B, A is stationary and the center of mass D of B and C is moving with velocity 12 v A . A

vA9 = 0

vD9 = ½ vA

(a) The coefficient of restitution is e =

v ′D − v ′A vA

1 v −0 1 2 A = . 2 vA

(b) The energy before the impact is 12 mv 2A . The energy after is 1 1 1 m(v ′A ) 2 + (2m)(v ′D ) 2 = mv 2A . 2 2 4 The energy “lost” is 14 mv 2A . (c) No energy is actually lost. The total kinetic energy of A, B, and C just after the impact is 12 mv 2A .

vA k A

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Problem 16.66 Suppose that you investigate an accident in which a 3400-lb car A struck a parked 1960-lb car B. All four of car B’s wheels were locked, and skid marks indicate that it slid 20 ft after the impact. If you estimate the coefficient of kinetic friction between B’s tires and the road to be µ k = 0.8 and the coefficient of restitution of the impact to be e = 0.2, what was A’s velocity v A just before the impact? (Assume that only one impact occurred.)

Solution: We can use work-energy to find the velocity of car B just after the impact. Then we use conservation of momentum and the coefficient of restitution to solve for the velocity of A. In general terms we have 1 m v ′ 2 − µ k m B gd = 0 ⇒ v ′B = 2µ k gd 2 B B m Av A = m Av ′A + m Bv ′B ,

ev A = v ′B − v ′A

Putting in the numbers we have v ′B =

2(0.8)(32.2 ft/s 2 )(20 ft) = 32.1 ft/s,

(3400 lb)v A = (3400 lb)v ′A + (1960 lb) (32.1 ft/s) (0.2)v A = (32.1 ft/s) − v ′A

vA A

Solving the last two equations simultaneously we have v ′A = 23.7 ft/s,

v A = 42.2 ft/s (28.7 mi/h).

Problem 16.67 When the player releases the ball from rest at a height of 5 ft above the floor, it bounces to a height of 3.5 ft. If he throws the ball downward, releasing it at 3 ft above the floor, how fast would he need to throw it so that it would bounce to a height of 12 ft?

Solution:

We’ll analyze the release part of the problem first, then the throw part. Let W and m denote the weight and mass of the ball. First we apply work and energy as the ball falls from a height of 5 ft to the floor.

5 ft

v1 5 0 v2

5 ft

1 1 mv 2 − mv 1 2 : 2 2 2 1 2 mg(5 ft) = mv 2 − 0. 2 U 12 =

The ball’s downward velocity just before it hits the floor is v 2 = 17.9 ft/s. Now we do the same thing as the ball rises to a height of 3.5 ft. (We don’t know the ball’s velocity after it bounces. It clearly wasn’t a perfectly elastic bounce or it would have bounced to a height of 5 ft.)

e =

v2 5 0 v1

3.5 ft

1 1 mv 2 − mv 1 2 : 2 2 2 1 −mg(3.5 ft) = 0 − mv 1 2 . 2 U 12 =

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The ball’s upward velocity just after it hits the floor is v 1 = 15.0 ft/s. With this information we can determine the coefficient of restitution of the bounce. Applying the definition of the coefficient of restitution to the bounce (with the stationary floor treated as object B and with upward velocity considered positive), v ′B − v ′A 0 − 15.0 ft/s = , vA − vB −17.9 ft/s − 0

we obtain e = 0.837. Now we’re going to assume that the same coefficient of restitution applies to the throwing part of the problem. That is, the coefficient of restitution is independent of velocity. That isn’t entirely obvious, but we do that throughout this chapter, and in this case it’s the best we can do. First let’s determine what velocity would be needed after the ball bounces to reach a height of 12 ft. Applying work and energy,

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16.67 (Continued) v2 5 0

12 ft

v1 1 1 mv 2 − mv 1 2 : 2 2 2 1 2 −mg(12 ft) = 0 − mv 1 , 2 U 12 =

we obtain v 1 = 27.8 ft/s. That means the downward velocity the ball must have just before it bounces is v 1 /e = (27.8 ft/s)/0.837 = 33.2 ft/s. Finally, we apply work and energy to his downward throw. Let v 0 be the downward velocity with which he throws the ball:

v0 v2

3 ft

1 1 mv 2 − mv 1 2 : 2 2 2 1 1 mg(3 ft) = m(33.2 ft/s) 2 − mv 0 2 . 2 2 U 12 =

Solving, we obtain v 0 = 30.2 ft/s. 30.2 ft/s.

Problem 16.68 The 0.45-kg soccer ball is 1 m above the ground and stationary when it is kicked straight up at 12 m/s. If the coefficient of restitution between the ball and ground is e = 0.8, how high above the ground does the ball bounce on its first bounce?

Solution: We can apply conservation of energy to the ball’s motion from just after it is kicked to just before it hits the ground. Placing the datum at ground level, we have T1 + V1 = T2 + V2 : 1 1 2 m(12 m/s) + mg(1 m) = mv 2 2 + 0. 2 2 Solving, the ball’s velocity just before it bounces is v 2 = 12.8 m/s. Using the coefficient of restitution, the ball’s upward velocity just after it bounces is (12.8 m/s)e = 10.2 m/s.

Source: Courtesy of West_photo/Shutterstock.

Now we apply conservation of energy to the ball’s motion from just after it bounces until it reaches its maximum height: T1 + V1 = T2 + V2 : 1 m(10.2 m/s) 2 + 0 = 0 + mgh. 2 This gives h = 5.34 m. 5.34 m.

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Problem 16.69 The 0.45-kg soccer ball is stationary just before it is kicked upward at 12 m/s. If the impact lasts 0.02 s, what average force is exerted on the ball by the player’s foot?

Solution:

(Neglect gravity during the impact) Details of kick (all

in y direction)

FAV

Source: Courtesy of West_photo/Shutterstock.

v0 = 0 v 1 = 12 m/s m = 0.45 kg j : ∫ FAV dt = FAV ∆t = mv 1 − mv 0 FAV ∆t = mv 1 = ( 0.45 )( 12 ) = 5.40 N-s ∆t = 0.02 s, Solving FAV = 270 N

Problem 16.70 By making measurements directly from the photograph of the bouncing golf ball, estimate the coefficient of restitution.

Solution:

Let h1 denote the height of the first bounce (the distance from the surface to the bottom of the ball) and h 2 the height of the second bounce. Considering only the vertical motion, we will apply conservation of energy to the ball’s motion from the top of the first bounce to just before it hits the surface. Placing the datum at the surface, we have T1 + V1 = T2 + V2 : 1 0 + mgh1 = mv 2 2 + 0. 2

Solving, the velocity just before the ball bounces is 2 gh1 , so the velocity just after it bounces is e 2 gh1 . Now we apply conservation of energy to the ball’s motion from just after it bounces to the height of the second bounce. T1 + V1 = T2 + V2 : 1 m(e 2 gh1 ) 2 + 0 = 0 + mgh 2 . 2 From this equation we obtain the relation e =

h2 . h1

Measuring from the photograph, we estimate that h1 = 41.5 mm and h 2 = 24.5 mm, which gives a coefficient of restitution e = 0.768. e = 0.768.

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Problem 16.71 If you throw the golf ball in Problem 16.70 horizontally at 2 ft/s and release it 4 ft above the surface, what is the distance between the first two bounces?

Solution:

The ball has no horizontal acceleration, so the horizontal component of its velocity will be constant, 2 ft/s. We can apply conservation of energy to determine the vertical component of its velocity just before the first bounce: T1 + V1 = T2 + V2 : 1 mv 2 + 0, 2 2 1 0 + m(32.2 ft/s 2 )(4 ft) = mv 2 2 + 0. 2 0 + mgh1 =

Solving, the downward component of velocity just before the ball bounces is v 2 = 16.0 ft/s, so the upward component of velocity just after it bounces is (see Problem 16.70) ev 2 = (0.768)(16.0 ft/s) = 12.3 ft/s. The ball’s upward velocity as a function as a function of time after it bounces is 12.3 ft/s − (32.2 ft/s 2 )t,

Source: Courtesy of Canbedone/Shutterstock.

so the time it takes to go from the first bounce to the second bounce is t = 2

12.3 ft/s ( 32.2 ) ft/s 2

= 0.766 s. The distance between bounces is the product of this time and the horizontal component of velocity: (0.766 s)(2 ft/s) = 1.53 ft. 1.53 ft.

Problem 16.72 In a forging operation, the 100-lb weight is lifted into position 1 and released from rest. It falls and strikes a workpiece in position 2. If the weight is moving at 15 ft/s immediately before the impact and the coefficient of restitution is e = 0.3, what is the velocity of the weight immediately after the impact? Solution: The strategy is to treat the system as an in-line impact on a rigid, immovable surface. From Eq. (16.16) with B’s velocity equal to zero: v ′A = −ev A , from which v ′A = −0.3(−15) = 4.5 ft/s.

16 in 2 k

Workpiece 12 in

Problem 16.73 The 100-lb weight is released from rest in position 1. The spring constant is k = 120 lb/ft, and the springs are unstretched in position 2. If the coefficient of restitution of the impact of the weight with the workpiece in position 2 is e = 0.6, what is the magnitude of the velocity of the weight immediately after the impact?

16 in 2 k

Workpiece 12 in

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16.73 (Continued) Solution:

Work-energy will be used to find the velocity just before the collision. Then the coefficient of restitution will give the velocity after impact. 2

( 12 kx ) + mgh = 12 mv 2

v =

k 2 x + 2 gh m

   120 lb/ft  20 − 12 2 16  2 ft + 2(32.2 ft/s 2 ) ft = 10.96 ft/s   100 lb   12 12     32.2 ft/s 2  

(

)

(

)

v ′ = ev = (0.6)(10.96 ft/s) = 6.58 ft/s. v′ = 6.58 ft/s.

Problem 16.74* A bioengineer studying helmet design uses an experimental apparatus that launches a 2.4-kg helmet containing a 2-kg model of the human head against a rigid surface at 6 m/s. The head suspended within the helmet is not immediately affected by the impact of the helmet with the surface and continues to move to the right at 6 m/s, so the head then undergoes an impact with the helmet. If the coefficient of restitution of the helmet’s impact with the surface is 0.85 and the coefficient of restitution of the subsequent impact of the head with the helmet is 0.15, what is the velocity of the head after its initial impact with the helmet? Solution:

6 m/s

The helmet’s rebound velocity is

v helmet = (0.85)(6 m/s) = 5.1 m/s The collision of the helmet and head (2 kg)(6 m/s) + (2.4 kg)(−5.1 m/s) = (2 kg)v head ′ + (2.4 kg)v helmet ′ 0.15(6 − [−5.1]) m/s = v helmet ′ − v head ′ Solving we find v head ′ = −0.963 m/s.

Problem 16.75* (a) If the duration of the impact of the head with the helmet in Problem 16.74 is 0.004 s, what is the magnitude of the average force exerted on the head by the impact? (b) Suppose that the simulated head alone strikes the rigid surface at 6 m/s, the coefficient of restitution is 0.5, and the duration of the impact is 0.0002 s. What is the magnitude of the average force exerted on the head by the impact?

Solution:

See the solution to Problem 16.74

(a) (2 kg)(6 m/s) − Fave (0.004 s) = (2 kg)(−0.963 m/s) ⇒ Fave = 3.48 kN. (b) The velocity of the head after the collision is v = 0.5(6 m/s) = 3 m/s (2 kg)(6 m/s) − Fave (0.0002 s) = (2 kg)(−3 m/s)

6 m/s

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⇒ Fave = 90 kN.

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Problem 16.76 Two small balls, each of 1-lb weight, hang from strings of length L = 3 ft. The left ball is released from rest with θ = 35 °. The coefficient of restitution of the impact is e = 0.9. Through what maximum angle does the right ball swing?

Solution:

Using work-energy and conservation of momentum we have 1 mgL (1 − cos θ) = mv A 2 ⇒ v A = 2 gL (1 − cos θ) 2 mv A = mv ′A + mv ′B  1+ e vA  ⇒ v ′B = 2 ev A = v ′B − v ′A   v′ 2  1 mv ′ 2 = mgL (1 − cos φ) ⇒ φ = cos −1  1 − B   2 B 2 gL 

Putting in the numbers we find L u

vA =

2(32.2 ft/s 2 )(3 ft)(1 − cos35 °) = 5.91 ft/s,

1.9 (5.91 ft/s) = 5.61 ft/s, 2 [5.61 ft/s] 2 φ = cos −1 1 − = 33.2 °. 2[32.2 ft/s 2 ][3 ft]

v ′B =

(

)

φ = 33.2 °.

Problem 16.77 In Example 16.7, if the Apollo commandservice module approaches the Soyuz spacecraft with velocity 0.25 i + 0.04 j + 0.01 k (m/s) and the docking is successful, what is the velocity of the center of mass of the combined vehicles afterward?

Solution:

Momentum is conserved so

m Av A = (m A + m S )v comb ⇒ v comb = v comb =

mA v m A + mS A

18 (0.25i + 0.04 j + 0.01k) m/s. 18 + 6.6

v comb = (0.183i + 0.0293 j + 0.00732k) m/s.

Problem 16.78 The 3-kg object A and 8-kg object B undergo an oblique central impact. The coefficient of restitution is e = 0.8. Before the impact, v A = 10 i + 4 j + 8k (m/s) and v B = −2 i − 6 j + 5k (m/s). What are the velocities of A and B after the impact?

Solution:

Tangent to the impact plane the velocities do not change. In the impact plane we have (3 kg)(10 m/s) + (8 kg)(−2 m/s) = (3 kg) v Ax ′ + (8 kg) v Bx ′ 0.8(10 − [−2]) m/s = v Bx ′ − v Ax ′

Solving we find v Ax ′ = −5.71 m/s, v Bx ′ = 3.89 m/s y

Thus v A ′ = (−5.71i + 4 j + 8k ) m/s, v B ′ = (3.89 i − 6 j + 5k) m/s.

B x

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Problem 16.79 A baseball bat (shown with the bat’s axis perpendicular to the page) strikes a thrown baseball. Before their impact, the velocity of the baseball is v b = 132(cos 45 °i + cos 45 ° j) (ft/s) and the velocity of the bat is v B = 60(− cos 45 °i − cos 45 ° j) (ft/s). Neglect the change in the velocity of the bat due to the direct central impact. The coefficient of restitution is e = 0.02. What is the ball’s velocity after the impact? Assume that the baseball and the bat are moving horizontally. Does the batter achieve a potential hit or a foul ball?

vb baseball vB

Bat

Solution:

Tangent to the impact plane, the velocity does not change. Since we are neglecting the change in velocity of the bat, then 0.2([132 cos 45 °] − [−60 cos 45 °]) ft/s = (−60 cos 45 °) ft/s − v ballx ′

Solving we find v ball x ′ = −69.6 ft/s Thus v ball ′ = (−69.6 i + 132cos 45 ° j) ft/s = (−69.6 i + 93.3 j) ft/s The ball is foul.

Problem 16.80 The cue gives the cue ball A a velocity parallel to the y-axis. The cue ball hits the eight ball B and knocks it straight into the corner pocket. If the magnitude of the velocity of the cue ball just before the impact is 2 m/s and the coefficient of restitution is e = 1, what are the velocity vectors of the two balls just after the impact? (The balls are of equal mass.)

Solution:

Denote the line from the 8-ball to the corner pocket by BP. This is an oblique central impact about BP. Resolve the cue ball velocity into components parallel and normal to BP. For a 45° angle, the unit vec1 tor parallel to BP is e BP = (−i + j), and the unit vector normal to BP 2 1 is e BPn = ( i + j). Resolve the cue ball velocity before impact into 2 components: v A = v AP e BP + v APn e BPn . The magnitudes v AP and v APn are determined from v AP e BP 2 + v APn e BPn 2 = v A = 2 m/s

458

and the condition of equality imposed by the 45° angle, from which v AP = v APn = 2 m/s. The cue ball velocity after impact is v′A = v ′AP e BP + v APn e BPn , (since the component of v A that is at right angles to BP will be unchanged by the impact). The velocity of the 8-ball after impact is v′BP = v ′BP e BP . The unknowns are the magnitudes v′BP and v′AP . These are determined from the conservation of linear momentum along BP and the coefficient of restitution.

m Av AP = m Av ′AP + m Bv ′BP , and

v ′BP − v ′AP . v AP

For m A = m B , these have the solution v ′AP = 0, v ′BP = v AP , from which x

v′A = v APn e BPn = i + j (m/s) and v′B = v AP e BP = −i + j (m/s).

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Problem 16.81 In Problem 16.80, what are the velocity vectors of the two balls just after the impact if the coefficient of restitution is e = 0.9? y

Solution:

Use the results of the solution to Problem 16.80, where the problem is solved as an oblique central impact about the line from the 8-ball to the corner pocket. Denote the line from the 8-ball to the corner 1 pocket by BP. The unit vector parallel to BP is e BP = (−i + j), and 2 1 the unit vector normal to BP is e BPn = ( i + j). Resolve the cue ball 2 velocity before impact into components: v A = v AP e BP + v APn e BPn ,

458

where, from Problem 16.80, v AP = v APn = 2 m/s. The velocity of the 8-ball after impact is v′BP = v ′BP e BP . The unknowns are the magnitudes v′BP and v′AP . These are determined from the conservation of linear momentum along BP and the coefficient of restitution.

m Av AP = m Av ′AP + m Bv ′BP ,

and e =

v ′BP − v ′AP . v AP

For m A = m B , these have the solution 1 v ′AP = (1 − e)v AP = 0.05v AP , 2 and 1 v ′BP = (1 + e)v AP = 0.95v AP . 2 The result:

( )

v′A = v AP e BP + v APne BPn = (−0.05i + 0.05 j + i + j) = 0.95i + 1.05 j m/s v ′B = 0.95(−i + j) (m/s).

Problem 16.82 If the coefficient of restitution is the same for both impacts, show that the cue ball’s path after two banks is parallel to its original path.

Solution:

The strategy is to treat the two banks as two successive oblique central impacts. Denote the path from the cue ball to the first bank impact as CP1, the path from the first impact to the second as CP2, and the final path after the second bank as CP3. The cue ball velocity along CP1 is v A1 = v A1x i + v A1y j, and the angle is v  α = tan −1  Ax1 .  y Ay1  The component v A1y j is unchanged by the impact. The x component after the first impact is v A 2 x = −ev A1x , from which the velocity of the cue ball along path CP2 is v A 2 = −ev A1x i + v A1y j. The angle is  −ev A1x  . β = tan −1   v A1y  The x component of the velocity along path CP2 is unchanged after the second impact, and the y component after the second impact is v A3 y = −ev A1y . The velocity along the path CP3 is v A3 = −ev A1x i − ev A1y j, and the angle is  −ev A1x   = α. γ = tan −1   −ev A1y  The sides of the table at the two banks are at right angles; the angles α = γ show that the paths CP1 and CP3 are parallel.

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Problem 16.83 The velocity of the 170-g hockey puck is v P = 10 i − 4 j (m/s). If you neglect the change in the velocity v S = v S j of the stick resulting from the impact, and if the coefficient of restitution is e = 0.6, what should v S be to send the puck toward the goal?

Solution:

Directing the puck toward the goal requires that

(v′P ) x 10 m/s = tan 20 °. = (v′P ) y (v′P ) y We see that the y component of the puck’s velocity after it is struck must be (v′P ) y = 27.5 m/s.

Applying the definition of the coefficient of restitution to the y components of the velocities, we have

Direction of goal 208

vP x

e =

(v′S ) y − (v′P ) y (v P ) y − (v S ) y

v S − 27.5 m/s . −4 m/s − v S

Setting e = 0.6 and solving, we obtain v S = 15.7 m/s. v S = 15.7 m/s.

Problem 16.84 In Problem 16.83, if the stick responds to the impact the way an object with the same mass as the puck would and the coefficient of restitution is e = 0.6, what should v S be to send the puck toward the goal? y

Solution:

(v′P ) x 10 m/s = tan 20 °. = (v′P ) y (v′P ) y We see that the y component of the puck’s velocity after it is struck must be (v′P ) y = 27.5 m/s.

Direction of goal

Applying the definition of the coefficient of restitution to the y components of the velocities, we have

208

e =

(v′S ) y − (v′P ) y (v P ) y − (v S ) y

v ′S − 27.5 m/s . −4 m/s − v S

Directing the puck toward the goal requires that

We also apply conservation of linear momentum in the y direction: m(v S ) y + m(v P ) y = m(v′S ) y + m(v′P ) y : mv S + m(−4 m/s) = mv ′S + m(27.5 m/s). We have two equations for v S and v′S . Solving, we obtain v S = 35.3 m/s, v ′S = 3.87 m/s. v S = 35.3 m/s.

Problem 16.85 At the instant shown (t 1 = 0), the position of the 2-kg object’s center of mass is r = 6 i + 4 j + 2k (m) and its velocity is v = −16 i + 8 j − 12k (m/s). No external forces act on the object. What is the object’s angular momentum about the origin O at t 2 = 1 s?

y r O

Solution: H O = (6 i + 4 j + 2k) m × (2 kg)(−16 i + 8 j − 12k) m/s

H O = (−128i + 80 j + 224 k) kg-m 2 /s.

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Problem 16.86 The total external force on the 2-kg object is given as a function of time by ΣF = 2ti + 4 j (N). At time t 1 = 0, the object’s position and velocity are r = 0 and v = 0. (a) Use Newton’s second law to determine the object’s velocity v and position r as functions of time. (b) By integrating r × ΣF with respect to time from t 1 = 0 to t 2 = 6 s, determine the angular impulse about O exerted on the object during this interval of time. (c) Use the results of part (a) to determine the change in the object’s angular momentum from t 1 = 0 to t 2 = 6 s.

Solution: (a) F = (2ti + 4 j)N = (2 kg)a, a = (ti + 2 j) m/s 2 v =

( t2 i + 2tj ) m/s, r = ( t6 i + t j ) m. 2

(b) Angular Impulse = ∫

6s 0

M O dt

∫0 r × F dt = ∫0  ( 6 i + t 2 j ) m  × [(2ti + 4 j)N] dt 6s

 t3



∫0 ( − 3 k ) N-mdt 6s

4t 3

Angular Impulse = (−432k) kg-m 2 /s. y

∆H O = ∫

(c)

6s 0

M O dt = (−432k) kg-m 2 /s.

r O

Problem 16.87 A satellite is in the elliptic Earth orbit shown. Its velocity at perigee A is 8640 m/s. The radius of the Earth is 6370 km. (a) Use conservation of angular momentum to determine the magnitude of the satellite’s velocity at apogee C. (b) Use conservation of energy to determine the magnitude of the velocity at C. (See Example 16.9.)

13,900 km

Solution: (a) r Av A = rCv C = H 0 (8000)(8640) = (24000)v C v C = 2880 m/s (b)

mgR E2 mgR E2 1 2 1 mv A − = mv C2 − 2 rA 2 rC

16,000 km

8000 km

v2 v 2A gR E2 gR E2 − = C − 2 rA 2 rC where v A = 8640 m/s, g = 9.81 m/s, R E = 6370000 m, r A = 8,000,000 m, rC = 24,000,000 m. Solving for v C , v C = 2880 m/s.

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Problem 16.88 For the satellite in Problem 16.87, determine the magnitudes of the radial velocity v r and transverse velocity v θ at B. (See Example 16.9.)

13,900 km

Solution:

Use conservation of energy to find the velocity magnitude at B. Then use conservation of angular momentum to determine the components. mgR E2 mgR E2 1 1 mV A2 − = mVB2 − 2 rA 2 rB

where r A = 8 × 10 6 m, v A = 8640 m/s rB =

(8 × 10 6 ) 2 + (13.9 × 10 6 ) 2

rB = 16 × 10 6 m, R E = 6.370 × 10 6 m Solving, we get v B = 4990 m/s

16,000 km

8000 km

From conservation of angular momentum

8000 km

r Av A = rBv θ Solving, v θ = 4320 m/s Finally v r =

v B2 − v θ2 = 2500 m/s.

Problem 16.89 The bar rotates in the horizontal plane about a smooth pin at the origin. The 2-kg sleeve A slides on the smooth bar, and the mass of the bar is negligible in comparison to the mass of the sleeve. The spring constant is k = 40 N/m, and the spring is unstretched when r = 0. At t = 0, the radial position of the sleeve is r = 0.2 m and the angular velocity of the bar is ω 0 = 6 rad/s. What is the angular velocity of the bar when r = 0.25 m?

v0 k A

Solution:

The line of action of the force exerted on the sleeve by the spring passes through the origin, so this is central-force motion. That means the product of the radial position and the transverse component of the velocity rv θ = r 2 ω is constant, r1 2 ω 1 = r2 2 ω 2 : (0.2 m) 2 (6 rad/s) = (0.3 m) 2 ω 2 .

This gives ω 2 = 3.84 rad/s. 3.84 rad/s.

Problem 16.90 The bar rotates in the horizontal plane about a smooth pin at the origin. The 2-kg sleeve A slides on the smooth bar, and the mass of the bar is negligible in comparison to the mass of the sleeve. The spring constant is k = 30 N/m, and the spring is unstretched when r = 0. At t = 0, the radial component of the velocity of the sleeve is v r = 0, the radial position of the sleeve is r = 0.2 m, and the angular velocity of the bar is 6 rad/s. What are the radial and transverse components of the velocity of the sleeve when r = 0.3 m? Strategy: Apply both conservation of angular momentum and conservation of energy.

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16.90 (Continued) Solution:

When r = 0.2 m, the radial and transverse components of the velocity of the sleeve are (v r ) 1 = 0, (v θ ) 1 = r1ω 1 = (0.2 m)(6 rad/s) = 1.2 m/s.

(0.3 m) 2 ω 2 .

This gives ω 2 = 2.67 rad/s, and the transverse component of the velocity when r = 0.3 m is (v θ ) 2 = r2 ω 2 = (0.3 m)(2.67 rad/s) = 0.8 m/s. We can also apply conservation of energy to the motion. In terms of the radial and transverse components of velocity, the kinetic energy of the sleeve is T =

The compression of the spring is equal to the radial position r, so the potential energy is V =

1 2 kr . 2

We write conservation of energy as T1 + V1 = T2 + V2 : 1 1 1 m[(v r ) 1 2 + (v θ ) 1 2 ] + kr1 2 = T2 + kr2 2 , 2 2 2 1 1 1 (2 kg)[0 + (1.2 m/s) 2 ] + (30 N/m)(0.2 m) 1 2 = T2 + (30 N/m) 2 2 2 (0.3 m) 2 , which yields T2 = 0.69 N-m. Then using the expression for the kinetic energy, 1 m[(v r ) 2 2 + (v θ ) 2 2 ] : 2 1 0.69 N-m = (2 kg)[(v r ) 2 2 + (0.8 m/s) 2 ], 2 T2 =

we obtain (v r ) 2 = 0.224 m/s. v r = 0.224 m/s, v θ = 0.8 m/s.

1 m(v r 2 + v θ 2 ). 2

Problem 16.91* A 2-kg disk slides on a smooth horizontal table and is connected to an elastic cord whose tension is T = 6r N, where r is the radial position of the disk in meters. If the disk is at r = 1 m and is given an initial velocity of 4 m/s in the transverse direction, what are the magnitudes of the radial and transverse components of its velocity when r = 2 m ?

Solution:

The strategy is to (a) use the principle of conservation of angular momentum to find the transverse velocity and (b) use the conservation of energy to find the radial velocity. The angular momentum the instant after t = 0 is (r × mv) o = H 0 e z = (mrv θ ) o e z , from which H 0 = 8 kg-m 2 /s. In the absence of external transverse forces, the angular momentum impulse vanishes: t2

∫t (r × ∑ F) dt = 0 = H 2 − H 1, 1

so that H 1 = H 2 , that is, the angular momentum is constant. At

r = 2, v θ =

H0 8 = = 2 m/s. mr 4

From conservation of energy: 1 2 2 mv r0

+ 12 mv θ2o + 12 kS o2 = 12 mv r2 + 12 mv θ2 + 12 kS 2 .

Solve: vr =

v r20 + v θ2o − v θ2 +

( mk )(S − S ). 2 o

Substitute numerical values: Noting m = 2 kg, v r0 = 0, v θo = 4 m/s, k = 6 N/m, r = 2 m, v θ = 2 m/s, S o = 1 m, S = 2 m from which vr =

Problem 16.92* In Problem 16.91, determine the maximum value of r reached by the disk.

3 m/s . The velocity is v = 1.732e r + 2e θ (m/s).

Solution: The maximum value is the stretch of the cord when v r = 0. From the solution to Problem 16.91, v r2 = v r20 + v θ2o − v θ2 +

( mk )(S − S ) = 0, 2 o

H0 H m/s, v r20 = 0, v θo = 0 m/s, S o = r0 = 1 m, S = rm, mr m(1) and H 0 = 8 kg-m 2 /s. Substitute and reduce:

wherev θ = r

v r2 = 0 =

( Hm )(1 − r1 ) + ( mk )(1 − r ). 2 0 2

Denote x = r 2 and reduce x 2 + 2bx + c = 0, where

262

quadratic

canonical

form

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16.92 (Continued)

( 12 )( Hkm + 1) = −3.167, c = Hkm = 5.333. 2 0

b = −

2 0

2 = −b ± Solve r1,2

b 2 − c = 5.333, = 1, from which the greatest

positive root is rmax = 2.31 m. [Check: This value is confirmed by a graph of the value of f (r ) =

( Hm )(1 − r1 ) + ( mk )(1 − r ) 2 0 2

to find the zero crossing.]

Problem 16.93 A 1-kg disk slides on a smooth horizontal table and is attached to a string that passes through a hole in the table. (a) If the mass moves in a circular path of constant radius r = 1 m with a velocity of 2 m/s, what is the tension T ? (b) Starting from the initial condition described in part (a), the tension T is increased in such a way that the string is pulled through the hole at a constant rate until r = 0.5 m. Determine the value of T as a function of r while this is taking place.

Solution:

(a) Circular motion T = −mv 2 /re r

T = (1)(2) 2 /1 = 4 N (b) By conservation of angular momentum, r0v 0 r

mr0v 0 = mrv T , ∴ v T =

T m

and from Newton’s second law T = mv T2 /r = m T = (1)

( r rv ) /r 0 0

( (1)(2) ) /r = 4/r N r 3

Problem 16.94 In Problem 16.93, how much work is done on the mass in pulling the string through the hole as described in part (b)? Solution:

The work done is

0.5

(− mrr r ) e ⋅ dre

U 12 = ∫

2 2 0 0 3

1  0.5 = −(1)(1) 2 (2) 2   −2r 2   1 1 = −4  − +   2(0.5) 2 2  = −mr02v 02 ∫

0.5 dr

U 12 = −4[−1.5] = 6 N-m.

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Problem 16.95 Two gravity research satellites (m A = 250 kg, m B = 50 kg) are tethered by a cable. The satellites and cable rotate with angular velocity ω 0 = 0.25 revolution per minute. Ground controllers order satellite A to slowly unreel 6 m of additional cable. What is the angular velocity afterward?

v0 A

Solution:

12 m

The satellite may be rotating in (a) a vertical plane, or (b) in the horizontal plane, or (c) in some intermediate plane. The strategy is to determine the angular velocity for the three possibilities.

Case (a) Assume that the system rotates in the x -y plane, with y positive upward. Choose the origin of the coordinates at the center of mass of the system. The distance along the cable from the center of mass to A is 12m B = 2 m, from which the distance to B is 10 m. m A + mB Assume that both satellites lie on the x axis at t = 0. The radius position of satellite A is r A = −2( i cos ω 0 t + j sin ω 0 t ), and the radius position of satellite B is rB = 10( i cos ω 0 t + j sin ω 0 t ). The acceleration due to gravity is W = −

mgR E2 j = −mg′j, r2

from which W A = −m A g′j and WB = −m B g′j (or, alternatively). The angular momentum impulse is t2

∫t (r × ∑ F) dt =∫t (r A × WA + rB × WB ) dt. Carry out the indicated operations:  i j k   r A × W A =  −2(cos ω 0t ) −2(sin ω 0t ) 0    0 0  −m A g′  = 2m A g′ cos ω 0tk. rB × WB

 i j k     =  10(cos ω 0t ) 10(sin ω 0t )    −m B g′ 0 0   = −10 m B g′ cos ω 0tk.

Substitute into the angular momentum impulse: t2

∫t (r × ∑ F) dt = 0 = H 2 − H 1, 1

from which H 1 = H 2 ; that is, the angular momentum is conserved. From Newton’s second law, the center of mass remains unchanged as the cable is slowly reeled out.

in magnitude, where ω 0 = 0.026 rad/s. After 6 meters is reeled out, the distance along the cable from the center of mass to A is m B (6 + 12) = 3 m, m A + mB from which the distance to B is 15 m. The new angular velocity when the 6 m is reeled out is ω =

H = 0.0116 rad/s = 0.1111 rpm. 3 2 m A + 15 2 m B

Case (b): Assume that the system rotates in the x -z plane, with y positive upward. As above, choose the origin of the coordinates at the center of mass of the system. Assume that both satellites lie on the x axis at t = 0. The radius position of satellite A is r A = −2( i cos ω 0 t + k sin ω 0 t ), and the radius position of satellite B is rB = 10( i cos ω 0 t + k sin ω 0 t ). The force due to gravity is W A = −m A g′j and WB = −m B g′j. The angular momentum impulse is t2

∫t (r × ∑ F) dt = ∫t (r A × WA + rB × WB ) dt. Carry out the indicated operations: r A × W A = 2m A g′ cos ω 0ti, rB × WB = −10 m B g′ cos ω 0ti. Substitute into the angular momentum impulse: t2

∫t (r × ∑ F) dt = 0 = H 2 − H 1, 1

from which H 1 = H 2 ; that is, the angular momentum is conserved. By a repeat of the argument given in Case (a), the new angular velocity is ω = 0.0116 rad/s = 0.111 rpm. Case (c): Since the angular momentum is conserved, a repeat of the above for any orientation of the system relative to the gravity vector leads to the same result.

A repeat of the argument above for any additional length of cable leads to the same result, namely, the angular momentum is constant, from which the angular momentum is conserved as the cable is reeled out. The angular momentum of the original system is r A × m A v + rB × m B v = 4 m A ω 0 k + 100 m B ω 0 k = 157.1 kg-m 2 /s,

264

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Problem 16.96 The astronaut moves in the x−y plane at the end of a 10-m tether attached to a large space station at O. The total mass of the astronaut and his equipment is 120 kg. (a) What is the astronaut’s angular momentum about O before the tether becomes taut? (b) What is the magnitude of the component of his velocity perpendicular to the tether immediately after the tether becomes taut?

Solution: (a) The angular momentum by definition is  i j k    (r × mv) =  0 6 0  = −6(2)(120)k  (120)2 0 0    = −1440 k (kg-m 2 /s). (b) From conservation of angular momentum,

H = −1440 k = −(10)(120)v k, from which

2i (m >s)

v =

1440 = 1.2 m/s. (10)(120)

y 6m

v 10 m

Problem 16.97 The astronaut moves in the x−y plane at the end of a 10-m tether attached to a large space station at O. The total mass of the astronaut and his equipment is 120 kg. The coefficient of restitution of the “impact” that occurs when he comes to the end of the tether is e = 0.8. What are the x and y components of his velocity immediately after the tether becomes taut?

Solution:

From the solution of Problem 16.96, his velocity perpendicular to the tether is 1.2 m/s.

Before the tether becomes taught, his component of velocity parallel to the tether is v r = 2 cos36.9 ° = 1.6 m/s. After it becomes taught, v ′r = −ev r = −(0.8)(1.6) = −1.28 m/s.

The x and y components of his velocity are 2i (m >s)

v ′x = v ′r cos 36.9 ° + (1.2)sin 36.9 ° = −0.304 m/s and v ′y = v ′r sin 36.9 ° − (1.2)cos36.9 ° = −1.728 m/s.

2 m/s

10 m

u 5 36.98

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1.2 m/s

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Problem 16.98 A ball suspended from a string that goes through a hole in the ceiling at O moves with velocity v A in a horizontal circular path of radius r A . The string is then drawn through the hole until the ball moves with velocity v B in a horizontal circular path of radius rB . Use the principle of angular impulse and momentum to show that r Av A = rBv B . Strategy: Let e be a unit vector that is perpendicular to the ceiling. Although this is not a central-force problem—the ball’s weight does not point toward O— you can show that e ⋅ (r × ΣF) = 0, so that e ⋅ H O is conserved.

Solution:

Assume that the motion is in the x – y plane, and that the ball lies on the positive x axis at t = 0. The radius vector r A = r A ( i cos ω At + j sin ω At ),

where ω A is the angular velocity of the ball in the path. The velocity is v A = − jr A ω A sin ω At + jr A ω A cos ω At. The angular momentum per unit mass about the axis normal to the ceiling is

(

 i j k   r × mv =  r A cos ω At r A sin ω At 0  = k(r A2 ω A ). m   ω ω ω ω − r sin t r cos t 0   A A A A A A

)

Define the unit vector parallel to this angular momentum vector, e = k. From the principle of angular impulse and momentum, the external forces do not act to change this angular momentum. This is shown as follows: The external force is the weight, W = −mg k. The momentum impulse is

∫t (r × ∑ F) dt = ∫t (r A × W) dt. Carry out the operation rB

 i j k   r A × W =  r A cos ω At r A sin ω At 0  = r Amg cos ω Atj,   − mg 0 0   from which t2

∫t jrAmg cos ωt dt = − j(rAω Amg)(sin ω At 2 − sin ω At1 ) 1

= H 2 − H 1. Since this has no component parallel to the unit vector e = k, the angular momentum along the axis normal to the ceiling is unaffected by the weight, that is, the projection of the angular momentum impulse due to the external forces on the unit vector normal to the ceiling is zero e ⋅ H 2 = e ⋅ H 1 = 0, hence the angular momentum normal to the ceiling is conserved. This result holds true for any length of string, hence (r × v) A = kr A2 ω A = (r × v) B = krB2 ω B , from which r A2 ω A = rB2 ω B . Since v A = r A ω A , v B = rB ω B , the result can be expressed v Ar A = v BrB .

Problem 16.99 The Cheverton fire-fighting and rescue boat can pump 3.8 kg/s of water from each of its two pumps at a velocity of 44 m/s. If both pumps point in the same direction, what total force do they exert on the boat?

266

Solution:

The magnitude of the total force is

2(3.8 kg/s)(44 m/s) = 334 N.

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Problem 16.100 The mass flow rate of water through the fire truck’s nozzle is 2 slugs/s. Determine the magnitude of the horizontal force exerted on the truck by the flow of the water if the nozzle is pointed upward at an angle θ = 25 ° with respect to the horizontal. (Assume that the motion of the water can be analyzed as if it were a projectile.)

Solution:

We first need to analyze the trajectory of the water to estimate its velocity as it leaves the nozzle: y

x,y

We have a projectile problem in which we know the initial elevation angle θ and the coordinates x, y of a point on the path. We need to determine the initial velocity v 0 .

u 20 ft 12 ft 35 ft

The components of the acceleration are a x = 0, a y = −g. We can integrate these with respect to time in the usual way to determine x and y as functions of time: x = (v 0 cos θ)t, 1 y = (v 0 sin θ)t − gt 2 . 2 We can solve the first equation for the time and substitute it into the second one to obtain an equation for the initial velocity: v0 =

gx 2 . 2( x sin θ cos θ − y cos 2 θ)

Setting θ = 25 °, x = 35 ft, and y = 8 ft, we obtain v 0 = 53.7 ft/s. The magnitude of the force exerted on the truck by the water is dm f v = (2 slugs/s)(53.7 ft/s) dt 0 = 107 lb. The horizontal component is (107 lb) cos 25 ° = 97.4 lb. 97.4 lb.

Problem 16.101 The front-end loader moves at a constant speed of 2 mi/h scooping up iron ore. The constant horizontal force exerted on the loader by the road is 400 lb. What weight of iron ore is scooped up in 3 s?

Solution: dm f  dm f   88 ft/s  400 lb =  ⇒ = 136.4 lb-s/ft (2 mph)  60 mph   dt  dt In 3 s m f = (136.4 lb-s/ft)(3 s) = 409 lb-s 2 /ft W = m f g = (409 lb-s 2 /ft)(32.2 ft/s 2 ) = 13,200 lb.

Problem 16.102 The snowblower moves at 1 m/s and scoops up 750 kg/s of snow. Determine the force exerted by the entering flow of snow. Solution:

The mass flow rate is

 dm f   dt  = 750 kg/s. The velocity is v f = 1 m/s. The force exerted by the entering flow of snow is  dm f  F =   v = 750(1) = 750 N.  dt  f

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Problem 16.103 The snowblower scoops up 750 kg/s of snow. It blows the snow out the side at 45 ° above the horizontal from a port 2 m above the ground and the snow lands 20 m away. What horizontal force is exerted on the blower by the departing flow of snow?

Solution:

The strategy is to use the solution of Newton’s second law to determine the exit velocity. dv y From Newton’s second law (ignoring drag) m f = −m f g, from dt which v y = −gt + v f sin 45 ° (m/s), v x = v f cos 45 ° (m/s), and g y = − t 2 + (v f sin 45 °) t + 2 m, 2 x = (v f cos 45 °)t. 2 + 2bt imp + c = 0, where At y = 0, the time of impact is t imp

b = −

v f sin 45 ° 4 ,c = − . g g

The Solution: t imp = −b ± =

b2 − c

vf   1 ± 2g 

8g  . v 2f 

Substitute:  v f  v f  x = 20 =   1±  2  2g 

8g  . v 2f 

This equation is solved by iteration using TK Solver Plus to yield v f = 13.36 m/s. The horizontal force exerted on the blower is  dm f  F =   v = 750(13.36)cos45° = 7082.7 kN.  dt  f

Problem 16.104 A nozzle ejects a stream of water horizontally at 40 m/s with a mass flow rate of 30 kg/s, and the stream is deflected in the horizontal plane by a plate. Determine the force exerted on the plate by the stream in cases (a), (b), and (c). (See Example 16.12.) y

Solution:

Apply the strategy used in Example 16.7. The exit velocity is v fe = v 0 ( i cos θ + j sin θ) where θ is the total angle of deflection of the stream, and v 0 is the magnitude of the stream velocity. The inlet velocity is v fi = v 0 i. The force on the plate exerted by the exit stream is in a direction opposite the stream flow, whereas the force exerted on the plate by the inlet stream is in the direction of stream flow. The sum of the forces exerted on the plate (see Eq. (16.25)) is

∑ F = (dm f /dt )(v fi − v fe ), from which

458

∑ F = (dm f /dt )v 0 (i(1 − cos θ) − j sin θ). The mass flow is (dm f /dt ) = 30 kg/s and the stream velocity is v 0 = 40 m/s. For Case(a), θ = 45 °,

x (a)

∑ F = (30)(40)(0.2929i − 0.7071j) = 351.5i − 848.5 j (N).

The force is downward to the right. For Case (b) θ = 90 °,

∑ F = (30)(40)(i − j) = 1200 i − 1200 j (N). x

x (b)

268

(c)

For Case (c) θ = 180 °,

∑ F = (30)(40)(2i) = 2400 i (N).

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Problem 16.105 A stream of water with velocity 80i (m/s) and a mass flow of 12 kg/s strikes a turbine blade moving with constant velocity 20i (m/s). The angle α = 65 °. What is the magnitude of the force exerted on the blade by the water?

y a

20 m/s 80 m/s

Solution:

Let us assume that the reference frame is moving with the blade. The velocities of the stream relative to the blade are shown.

y 60 m/s

658

60 m/s

The water leaving the blade exerts a force on it opposite to the direction of the flow of magnitude

The water entering the blade exerts a force on it in the direction of the flow of magnitude dm f v = (12 kg/s)(60 m/s) dt f = 720 N.

dm f v = (12 kg/s)(60 m/s) dt f = 720 N. Summing the forces, the total force exerted on the blade is 720 i − 720 cos65 °i − 720sin65 ° j (N) = 416 i − 653 j (N). Its magnitude is 774 N. 774 N.

Problem 16.106 At the instant shown, the nozzle A of the lawn sprinkler is located at (0.1, 0, 0) m. Water exits each nozzle at 8 m/s relative to the nozzle with a mass flow rate of 0.22 kg/s. At the instant shown, the flow relative to the nozzle at A is in the direction of the unit vector e =

1 1 1 i− j+ k. 3 3 3

Determine the total moment about the z-axis exerted on the sprinkler by the flows from all four nozzles. Solution: e =

dm f 1 ( i + j + k), r = 0.1i, v = (8 m/s)e, = 0.22 kg/s dt 3

 dm f  M = 4  − (r × v) = (0.406 N-m)( j + k)  dt  The moment about the z -axis is M z = M • k = 0.406 N-m.

Problem 16.107* A 50 kg/s flow of gravel exits the chute at 4 m/s and falls 2 m onto a conveyer moving at 0.3 m/s. Determine the components of the force exerted on the conveyer by the impact of the gravel if θ = 0.

458 y 2m

0.3 m/s u x

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16.107* (Continued) Solution:

Let v 0 = 4 m/s denote the velocity of the gravel as it leaves the chute. The components of velocity are

v x = v 0 cos 45 ° = 2.83 m/s,

v x = 2.83 m/s − (0.3 m/s) cos θ = 2.53 m/s, v y = −6.87 m/s + (0.3 m/s)sin θ

v y = −v 0 cos 45 ° = −2.83 m/s. The horizontal component will be constant as the gravel falls. We can use conservation of energy to determine the vertical component when the gravel reaches the conveyer. We place the datum at that level. T1 + V1 = T2 + V2 :

= −6.87 m/s. Using these values, the components of force exerted on the conveyer by the gravel are Fx =

1 1 mv 2 + mgh = m(v ′y ) 2 + 0, 2 y 2

dm f v dt x

= (50 kg/s)(2.53 m/s) = 126 N,

v ′y = − v y 2 + 2 gh = − (2.83 m/s) 2 + 2(9.81 m/s 2 )(2 m)

Fy =

= −6.87 m/s. Now let us assume that the frame of reference is moving with the conveyer. In terms of that frame, the components of velocity of the gravel as it reaches the conveyer are

Problem 16.108* A 50 kg/s flow of gravel exits the chute at 4 m/s and falls 2 m onto a conveyer moving at 0.3 m/s. Determine the components of the force exerted on the conveyer by the impact of the gravel if θ = 30 °.

dm f v dt y

= (50 kg/s)(−6.87 m/s) = −344 N. 126 i − 344 j (N).

T1 + V1 = T2 + V2 : 1 2 1 mv + mgh = m(v ′y ) 2 + 0, 2 y 2 v ′y = − v y2 + 2 gh = − (2.83 m/s) 2 + 2(9.81 m/s 2 )(2 m) = −6.87 m/s.

458

Now let us assume that the frame of reference is moving with the conveyer. In terms of that frame, the components of velocity of the gravel as it reaches the conveyer are

v x = 2.83 m/s − (0.3 m/s) cos θ

= 2.57 m/s, v y = −6.87 m/s + (0.3 m/s)sin θ = −6.72 m/s.

0.3 m/s u x

Using these values, the components of force exerted on the conveyer by the gravel are Fx =

Solution: Let v 0 = 4 m/s denote the velocity of the gravel as it leaves the chute. The components of velocity are

= (50 kg/s)(2.57 m/s) = 44.6 N,

v x = v 0 cos 45 ° = 2.83 m/s, v y = −v 0 cos 45 ° = −2.83 m/s.

dm f v dt x

Fy =

The horizontal component will be constant as the gravel falls. We can use conservation of energy to determine the vertical component when the gravel reaches the conveyer. We place the datum at that level.

dm f v dt y

= (50 kg/s)(−6.72 m/s) = −336 N. 128i − 336 j (N).

270

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Problem 16.109 Suppose that you are designing a toy car that will be propelled by water that squirts from an internal tank at 10 ft/s relative to the car. The total weight of the car and its water “fuel” is to be 2 lb. If you want the car to achieve a maximum speed of 12 ft/s, what part of the total weight must be water? (See Example 16.11.) Solution:

208

See Example 16.10

12 ft/s − 0 = (10 ft/s) cos 20 ° ln

( 2Wlb ) ⇒ W = 0.558

The water must be W water = (2 lb) − (0.558 lb) = 1.44 lb.

Problem 16.110 The rocket consists of a 1000-kg payload and a 9000-kg booster. Eighty percent of the booster’s mass is fuel, and its exhaust velocity is 1200 m/s. If the rocket starts from rest and external forces are neglected, what velocity will it attain? (See Example 16.11.)

Booster

Payload

Solution:   10,000 kg v = (1200 m/s) ln  = 1530 m/s.  1,000 kg + 0.2[9,000 kg] 

Problem 16.111* The rocket consists of a 1000-kg payload and a booster. The booster has two stages whose total mass is 9000 kg. Eighty percent of the mass of each stage is fuel, and the exhaust velocity of each stage is 1200 m/s. When the fuel of stage 1 is expended, it is discarded and the motor of stage 2 is ignited. Assume that the rocket starts from rest and neglect external forces. Determine the velocity attained by the rocket if the masses of the stages are m1 = 6000 kg and m 2 = 3000 kg. Compare your result to the answer to Problem 16.110.

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Solution: First burn,

 Full mass = 10,000 kg   Empty mass = 4,000 kg + 0.2(6,000 kg) = 5,200 kg 

 Full mass = 4,000 kg Second burn   Empty mass = 1,000 kg + 0.2(300 kg) = 1,600 kg   10,000 kg   4,000 kg  v = (12,000 m/s) ln   + (12,000 m/s) ln    5200 kg   1600 kg  = 1880 m/s Much faster using stages.

Payload

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Problem 16.112 A rocket of initial mass m 0 takes off straight up. Its exhaust velocity v f and the mass flow rate of its engine m f = dm f /dt are constant. Show that, during the initial part of the flight, when aerodynamic drag is negligible, the rocket’s upward velocity as a function of time is m0 v = v f ln − gt. m 0 − m f t

(

)

Solution:

Start by adding a gravity term to the second equation in Example 16.10. (Newton’s second law). We get dv

∑ Fx = dt f v f − mg = m dtx . Substitute dm f dm = − dt dt as in the example and divide through by m to get

( m1 )dmdt − g = ddtv .

−v f

Integrate with respect to time gives the relation −v f ∫

t 0

dt ( m1 ) dm dt

t dv

−∫ g dt = ∫

dt.

Simplifying the integrals and setting appropriate limits when we change the variable of integration, we get −v f ∫

m dm m0

− g ∫ dt = ∫ d v x .

Integrating and evaluating at the limits of integration, we get v = v f ln(m 0 /m) − gt. Recalling that m = m 0 − m f t and substituting this in, we get the desired result.

Problem 16.113 The mass of the rocket sled in Practice Example 16.10 is 440 kg. Assuming that the only significant force acting on the sled in the direction of its motion is the force exerted by the flow of water entering it, what distance is required for the sled to decelerate from 300 m/s to 100 m/s? Fy

Solution:

From Example 16.9, the force in the x direction (the only force that affects the speed) is Fx = −ρ Av 2

From Newton’s Second law ma x = −ρ Av 2 dv x ρA 2 v = − ds m 100 dv s f ρA ∫300 v = − m ∫0 ds ρA ln(v ) 100 s 300 = − m f

( )

ax = v

( )

where ρ = 1000 kg/m 3 , A = 0.01 m 2 , m = 440 kg.

Solving s f = 48.3 m. x

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Problem 16.114* You grasp the end of a stationary chain that weighs 2 lb/ft and jerk it straight up off the floor at a constant speed v 0 = 3 ft/s. (a) Determine the upward force F you must exert as a function of the height s in feet. (b) How much work do you do in lifting the top of the chain to s = 4 ft? [Let m denote the mass of the suspended part of the chain. It can be shown by an analysis similar to that leading to Eq. (16.27) that the bottom of the suspended chain is subjected to a downward force of magnitude (dm /dt )v 0 .]

Solution: (a) Let γ 0 = 2 lb/ft denote the weight of the chain per unit length. The mass of the chain per unit length is γ 0 /g. The acceleration of the suspended chain is zero, so the sum of the forces on it must be zero. They include the force F, the weight of the suspended chain γ 0 s, and the given force acting at the bottom. The mass of the suspended part of the chain is (γ 0 /g)s, so

so the force necessary to lift it is γ  F = γ 0 s +  0 v 0 2  g  = (2 lb/ft)s +

2 lb/ft ( 32.2 )(3 ft/s) ft/s 2

= 2s + 0.559 lb.

 γ  ds γ  dm =  0  =  0 v 0 .  g  dt  g  dt

(b) The work done is

The force exerted on the bottom of the suspended chain as it rises is γ  dm v =  0 v 0 2 .  g  dt 0

U 12 = ∫

4 ft 0

(2s + 0.559 lb) ds

= 18.2 ft-lb.

The sum of the forces on the suspended chain is

(a) F = 2s + 0.559 lb. (b) 18.2 ft-lb.

γ  F − γ 0 s −  0 v 0 2 = 0,  g 

Problem 16.115* You grasp the end of a stationary chain that weighs 2 lb/ft and lift it straight up off the floor at a constant acceleration a 0 = 3 ft/s 2 . (a) Determine the upward force F you must exert as a function of the height s. (b) How much work do you do in lifting the top of the chain to s = 4 ft? (See Problem 16.114.)

Solution: (a) Let γ 0 = 2 lb/ft denote the weight of the chain per unit length. The mass of the chain per unit length is γ 0 /g. The forces acting on the rising chain include the force F, the weight of the suspended chain γ 0 s, and the force given in Problem 16.114 acting at the bottom. The mass of the suspended part of the chain is (γ 0 /g)s, so  γ  ds γ  dm =  0  =  0 v .  g  dt  g  dt The downward force exerted on the bottom of the suspended chain as it rises is

γ  dm v =  0 v 2 .  g  dt Applying Newton’s second law, γ  γ  F − γ 0 s −  0 v 2 =  0  sa 0 ,  g   g  so the force necessary to lift it is γ  γ  F = γ 0 s +  0 v 2 +  0  sa 0 . (1)  g   g  Because the acceleration is constant, we can determine the velocity of the chain as a function of s. The velocity and position as functions of time are v = a 0t, 1 s = a 0t 2 . 2

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16.115* (Continued) Solving the first equation for t and substituting it into the second one, we obtain v =

2a 0 s .

Substituting this into Eq. (1), the force as a function of s is γ  F = γ 0 s + 3 0  a 0 s  g  3a 0   = γ 0  1 + s  g  3( 3 ft/s 2 )  s = (2 lb/ft)  1 +  32.2 ft/s 2  = 2.56s lb. (b) The work done is U 12 = ∫

4 ft 0

(2.56s lb) ds

= 20.5 ft-lb. (a) F = 2.56s lb. (b) 20.5 ft-lb.

Problem 16.116* It has been suggested that a heavy chain could be used to gradually stop an airplane that rolls past the end of the runway. A hook attached to the end of the chain engages the plane’s nose wheel, and the plane drags an increasing length of the chain as it rolls. Let m be the airplane’s mass and v 0 its initial velocity, and let ρ L be the mass per unit length of the chain. Neglecting friction and aerodynamic drag, what is the airplane’s velocity as a function of s?

Solution:

Assume that the chain is laid out lengthwise along the runway, such that the aircraft hook seizes the nearest end as the aircraft proceeds down the runway. As the distance s increases, the length of chain s being dragged is (see figure). The mass of the chain being dragged is 2 ρ Ls . The mass “flow” of the chain is 2 d

( ρ2L ) = ρ L . s

From Newton’s second law,

( ρ2L + m ) ddtv = − ρ L2 . v2

Use the chain rule and integrate:

( ρ2L + m ) ddsv = − ρ2L , dvv = − 2 ρ Lρ L+ m ds, (2 ) ρL + C. ln (v ) = −ln ( m + 2 ) v

For v = v 0 at s = 0, C = ln (mv 0 ), and v =

mv 0 . ρ Ls 2

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Problem 16.117* In Problem 16.116, the frictional force exerted on the chain by the ground would actually dominate other forces as the distance s increases. If the coefficient of kinetic friction between the chain and the ground is µ k and you neglect all forces except the frictional force, what is the airplane’s velocity as a function of s?

Solution:

Assume that the chain layout is configured as shown in Problem 16.116. From Problem 16.116, the weight of the chain being ρ Lgs dragged at distance s is . From Newton’s second law 2 µ ρ Lgs ρ L s dv m+ = − k , 2 dt 2

(

)

where only the friction force is considered. Use the chain rule and integrate:

( m + ρ2L )v ddsv = − µ ρ2L , v dv = −µ g 2 m ρ+L ρ L ds, ( 2 ) s

from which

(

−2µ k g ρ s ρ Ls v2 = − m ln m + L m+ 2 ρL 2 m s

)) + C.

For v = v 0 at s = 0, C =

2µ k gm v 02 + (1 − ln (m)), 2 ρL

from which, after reduction

(

)

ρ s   2m v 2 = v 02 − 2 gµ k  s − ln 1 + L .  ρL 2m 

Problem 16.118 A turbojet engine is being operated on a test stand. The mass flow rate of air entering the compressor is 130 kg/s, and the mass flow rate of fuel is 0.25 kg/s. The effective velocity of air entering the compressor is negligible, and the exhaust velocity is 500 m/s. What is the thrust of the engine? (See Example 16.13.)

Solution: In terms of the flow rate of inlet air dm c /dt = 130 kg/s, the fuel flow rate dm f /dt = 0.25 kg/s, and the exhaust velocity v e = 500 m/s, the engine’s thrust is T =

( dmdt + dmdt )v c

= (130 kg/s + 0.25 kg/s)(500 m/s) = 65,100 N (14,600 lb). 65.1 kN.

Problem 16.119 An airplane is moving at 900 km/h. The mass flow rate of air entering the compressor of one of its engines is 130 kg/s, and the mass flow rate of fuel is 0.25 kg/s. The effective velocity of air entering the compressor is equal to the plane’s speed, and the exhaust velocity relative to the plane is 500 m/s. What is the net thrust of the engine? (See Example 16.13.)

Solution: 900 km/h

The airplane’s velocity in meters per second is

m 1h ( 1000 )( 3600 ) = 250 m/s. 1 km s

In terms of the flow rate of inlet air dm c /dt = 130 kg/s, the fuel flow rate dm f /dt = 0.25 kg/s, the velocity of the air entering the engine v i = 250 m/s, and the exhaust velocity v e = 500 m/s, the engine’s thrust is T =

( dmdt + dmdt )v − dmdt v c

= (130 kg/s + 0.25 kg/s)(500 m/s) − (130 kg/s)(250 m/s) = 32,600 N (7330 lb). 32.6 kN.

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Problem 16.120 A turbojet engine’s thrust reverser causes the exhaust to exit the engine at 20 ° from the engine centerline. The mass flow rate of air entering the compressor is 130 kg/s, and the air enters at 70 m/s. The mass flow rate of fuel is 0.25 kg/s, and the exhaust velocity relative to the plane is 500 m/s. What net braking force does the engine exert on the airplane?

Solution:

In terms of the flow rate of inlet air dm c /dt = 130 kg/s and the velocity of the air entering the engine v i = 70 m/s, the entering air exerts an effective braking force

dm c v = (130 kg/s)(70 m/s) dt i = 9100 N. The braking force exerted by the engine’s exhaust is

( dmdt + dmdt )v cos 20° = (130 kg/s + 0.25 kg/s)(500 m/s) cos 20° c

= 61,200 N. The net braking force is the sum of these, 70,300 N (15,800 lb). 208

70.3 kN.

208

Problem 16.121 The total external force on a 10-kg object is constant and equal to 90 i − 60 j + 20 k (N). At t = 2 s, the object’s velocity is −8i + 6 j (m/s). (a) What impulse is applied to the object from t = 2 s to t = 4 s? (b) What is the object’s velocity at t = 4 s?

Solution: (a) The impulse is t2

∫t F dt = 90(4 − 2)i − 60(4 − 2) j + 20(4 − 2)k 1

= 180 i − 120 j + 40 k (N-s). (b) The velocity is mv 2 − mv 1 = ∫

t2 t1

F dt,

from which v 2 = v1 + +

Problem 16.122 The total external force on an object is F = 10ti + 60 j (lb). At t = 0, the object’s velocity is v = 20 j (ft/s). At t = 12 s, the x component of its velocity is 48 ft/s. (a) What impulse is applied to the object from t = 0 to t = 6 s? (b) What is the object’s velocity at t = 6 s?

) i + ( 6 − 120 )j ( m1 ) ∫ F dt = ( −8 + 180 10 10 t2

40 ( 10 )k = 10i − 6 j + 4k (m/s).

Solution: (a) The impulse is t2

∫t F dt = [5t 2 ]60 i + [60t ]60 j = 180 i + 360 j (lb-s) 1

(b) The mass of the object is found from the x component of the velocity at 12 s. t2

m =

∫t F dt 1

48 − 0

5(12 2 ) = 15 slug. 48

The velocity at 6 s is mv 2 − mv 1 = ∫

t2 t1

F dt,

from which v 2 = 20 j +

276

180 360 i+ j = 12 i + 44 j (ft/s). 15 15

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Problem 16.123 An aircraft arresting system is used to stop airplanes whose braking to systems fail. The system stops a 47,500-kg airplane moving at 80 m/s in 9.15 s. (a) What impulse is applied to the airplane during the 9.15 s? (b) What is the average deceleration to which the passengers are subjected?

Solution: (a) The impulse is t2

∫t F dt = mv 2 − mv 1 = 47500(80) = 3.8 ×10 6 N-s. 1

(b) The average force is t2

∫t F dt

Fave =

t 2 − t1

3.8 × 10 6 = 4.153 × 10 5 N. 9.15

From Newton’s second law

( ddtv )

Problem 16.124 The 1895 Austrian 150-mm howitzer had a 1.94-m barrel, possessed a muzzle velocity of 300 m/s, and fired a 38-kg shell. If the shell took 0.013 s to travel the length of the barrel, what average force was exerted on the shell?

ave

Solution:

Fave = 8.743 m/s 2 . 47500

The average force is

Fave =

∫t F dt 1

t 2 − t1

(38)(300) = 877,000 N. 0.013

Problem 16.125 An athlete throws a shot weighing 16 lb. When he releases it, the shot is 7 ft above the ground and the angle between its velocity vector and the horizontal is 40 °. The time required for him to accelerate the shot from rest to release is 1.2 s. It travels a horizontal distance of 60 ft from the point where he releases it. Use the principle of impulse and momentum to determine the magnitude of the average force he exerts on the shot. Solution:

The mass of the shot is

W 16 lb = = 0.497 slugs. g 32.2 ft/s 2

m =

To apply Eq. (16.2), we need to know the shot’s velocity when it is released. We can determine that because the horizontal distance it travels after release is given.

v0 408

7 ft

Source: Courtesy of Wavebreakmedia/ Shutterstock.

60 ft Let t = 0 be the time at which the shot is released, and let v 0 be its initial velocity. The initial components of the velocity are v x = v 0 cos 40 °, v y = v 0 sin 40 ° and the components of the acceleration are a x = 0, a y = −g. We can integrate the acceleration with respect to time to determine the components of velocity and position as functions of time. Placing the origin of coordinates at the point of release, the results are dx = v 0 cos 40 °, dt x = (v 0 cos 40 °)t,

vx =

(1)

v y = v 0 sin 40 ° − gt, y = (v 0 sin 40 °)t −

1 2 gt . 2

(2)

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16.125 (Continued) Solving Eq. (1) for the time and substituting it into Eq. (1), we obtain an equation for the trajectory: y =

(t 2 − t 1 )(Fav − W j) = mv 2 − mv 1 : (1.2 s − 0)(Fav − W j) = m[(41.5 ft/s)cos 40° i + (41.5 ft/s)sin 40° j] − 0 .

gx 2 x sin 40 ° . − 2 cos 40 ° 2v 0 cos 2 40 °

We obtain the average force

Setting x = 60 ft and y = −7 ft, we can solve this equation for the initial velocity, obtaining v 0 = 41.5 ft/s. We can now apply Eq. (16.2). The forces acting on the shot while it is thrown are its weight and the average force Fav applied by the athlete:

Problem 16.126 The 6000-lb pickup truck A moving at 40 ft/s collides with the 4000-lb car B moving at 30 ft/s. (a) What is the magnitude of the velocity of their common center of mass after the impact? (b) Treat the collision as a perfectly plastic impact. How much kinetic energy is lost?

Fav = 13.2 i + 27.0 j (lb) with magnitude Fav = 30.1 lb. 30.1 lb.

Solution: (a) From the conservation of linear momentum,  W A + W B   W A   WB  v x ,  g v A +  g v B cos θ =  g  W A + W B   W B  v y .  g v B sin θ =  g Substitute numerical values, with θ = 30 ° to obtain v x = 34.39 ft/s, v y = 6 ft/s, from which v = v x2 + v y2 = 34.9 ft/s (b) The kinetic energy before the collision minus the kinetic energy after the collision is the loss in kinetic energy:

1  W A  2 1 W  1  W + W B  2  v +  B  v B2 −  A  v 2  g  A 2 g 2 g

308

= 15,7000 ft-lb. B

Problem 16.127 Two hockey players (m A = 80 kg, m B = 90 kg) converging on the puck at x = 0, y = 0 become entangled and fall. Before the collision, v A = 9 i + 4 j (m/s) and v B = −3i + 6 j (m/s). If the coefficient of kinetic friction between the players and the ice is µ k = 0.1, what is their approximate position when they stop sliding?

Solution:

Let m = m A + m B = 170 kg. The velocity of their common center of mass is m Av A + m B v B m (80 kg)[9i + 4 j (m/s)] + (90 kg)[−3 i + 6 j (m/s)] = 170 kg = 2.65 i + 5.06 j (m/s).

v =

Let its magnitude be denoted by v 1 = 5.71 m/s. The angle between their direction of motion and the x axis is vy  5.06 m/s θ = arctan   = arctan = 62.4 °. vx  2.65 m/s

(

)

As they slide, friction will slow their motion: y

The normal force is N = mg.

278

U 12 = T2 − T1 : 1 1 −( µ k mg)s = mv 2 2 − mv 1 2 , 2 2 1 −0.1m(9.81 m/s 2 )s = 0 − m(5.71 m/s) 2 . 2 Solving, we obtain s = 16.6 m. The coordinates of the point where their center of mass comes to rest are

mkN

We can apply work and energy to determine the distance s that they slide:

x = s cos θ = 7.70 m, y = s sin θ = 14.7 m. 7.70 i + 14.7 j (m).

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Problem 16.128 The cannon weighed 400 lb, fired a cannonball weighing 10 lb, and had a muzzle velocity of 200 ft/s. For the 10 ° elevation angle shown, determine (a) the velocity of the cannon after it was fired and (b) the distance the cannonball traveled. (Neglect drag.)

Solution:

Assume that the height of the cannon mouth above the ground is negligible. (a) Relative to a reference frame moving with the cannon, the cannonball’s velocity is 200 ft/s at. The conservation of linear momentum condition is −mCv C + m B (v B cos10 ° − v C ) = 0, from which vC =

m Bv B cos10 ° = 4.804 ft/s. m B + mC

108 The velocity of the cannon ball relative to the ground is

v B = (200 cos10 ° − v C ) i + 200 sin10 ° j = 192.16 i + 34.73 j ft/s, from which v 0 x = 192.16 ft/s, v 0 y = 34.73 ft/s. The maximum height is (from the conservation of energy for free fall) h max = v 02y /(2 g) = 18.75 ft, 18.75 ft, where the height of the cannon mouth above the ground is negligible. The time of flight (from the solution of Newton’s second law for free fall) is twice the time required to fall from the maximum height, t flight = 2

2h max = 2.16 s. g

From which the range is x impact = v 0 x t flight = 415 ft since v 0 x is constant during the flight.

108

Problem 16.129 A 1-kg ball moving horizontally at 12 m/s strikes a 10-kg block. The coefficient of restitution of the impact is e = 0.6, and the coefficient of kinetic friction between the block and the inclined surface is µ k = 0.4. What distance does the block slide before stopping?

Solution:

First we analyze the impact between the ball (b) and the block (B). The component of the ball’s velocity parallel to the inclined surface is v b = (12) cos 25 ° = 10.9 m/s. Solving the equations m bv b = m bv ′b + m Bv ′B , e =

v ′B − v ′b , vb

we obtain v′B = 1.58 m/s. We use work and energy to determine the distance d the block slides: 1 (−m B g sin 25 ° − µ k m B g cos 25 °)d = 0 − m B (v ′B ) 2 . 2 Solving yields d = 0.162 m.

v 258

Problem 16.130 A Peace Corps volunteer designs the simple device shown for drilling water wells in remote areas. A 70-kg “hammer,” such as a section of log or a steel drum partially filled with concrete, is hoisted to h = 1 m and allowed to drop onto a protective cap on the section of pipe being pushed into the ground. The combined mass of the cap and section of pipe is 20 kg. Assume that the coefficient of restitution is nearly zero. (a) What is the velocity of the cap and pipe immediately after the impact? (b) If the pipe moves 30 mm downward when the hammer is dropped, what resistive force was exerted on the pipe by the ground? (Assume that the resistive force is constant during the motion of the pipe.)

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16.130 (Continued) Solution:

(b) The work done by the resistive force exerted on the pipe by the cap s is ∫ −F ds = −Fs N-m, where s = 0.03 m. The work and ener0

gy principle for the hammer, pipe and cap is −Fs + (m H + m P ) gs 1 = (m H + m P )v 2 , from which 2 (m H + m P )(v 2 + 2 gs) (90)(3.45 2 + 2(9.81)(0.03)) F = = 2s 2(0.03) = 18,700 N.

(a) The conservation of momentum principle for the hammer, pipe and cap is m H v H + m pv p = (m H + m p )v , where v H , v p are the velocity of the hammer and pipe, respectively, before impact, and v is the velocity of their combined center of mass immediately after impact. The velocity of the hammer before impact (from the conservation of energy for a free fall from height h ) is v H = 2 gh = 2(9.81)(1) = 4.43 m/s. Since v P = 0, v =

mH 70 v = (4.43) = 3.45 m/s. mH + mP H 90

Problem 16.131 A tugboat (mass = 40,000 kg) and a barge (mass = 160,000 kg) are stationary with a slack hawser connecting them. The tugboat accelerates to 2 knots (1 knot = 1852 m/h) before the hawser becomes taut. Determine the velocities of the tugboat and the barge just after the hawser becomes taut (a) if the “impact” is perfectly plastic (e = 0) and (b) if the “impact” is perfectly elastic (e = 1). Neglect the forces exerted by the water and the tugboat’s engines.

(

)

m The tugboat’s initial velocity is v T = 2 1852 h 1h = 1.03 m/s. The equations governing the “impact” are 3600 s mT v T = mT v ′T + m Bv ′B ,

Solution:

(

)

e =

v ′B − v ′T . vT

(a) With e = 0, we obtain v ′T = v ′B = 0.206 m/s (0.4 knots). (b) With e = 1, we obtain v ′T = −0.617 m/s (− 1.2 knots), v ′B = 0.412 m/s (0.8 knots).

Problem 16.132 In Problem 16.131, determine the magnitude of the impulsive force exerted on the tugboat in the two cases if the duration of the “impact” is 4 s. Neglect the forces exerted by the water and the tugboat’s engines during this period.

280

Solution:

Since the tugboat is stationary at t = t 1, the linear impulse is

∫t F dt = Fave (t 2 − t1 ) = m Bv ′B , 1

from which Fave = m Bv ′B /4 = (4 × 10 4 )v ′B . For Case (a) v′B = 0.206 m/s, from which Fave = 8230 N . Case (b). v′B = 0.412 m/s, Fave = 16,500 N.

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Problem 16.133 The 10-kg mass A is moving at 5 m/s when it is 1 m from the stationary 10-kg mass B. The coefficient of kinetic friction between the floor and the two masses is µ k = 0.6, and the coefficient of restitution of the impact is e = 0.5. Determine how far B moves from its initial position as a result of the impact.

Solution: −µ k mg(1) =

1 2 1 mv − m(5) 2 . 2 A 2

Solving, v A = 3.64 m/s. Now analyze the impact: mv A = mv ′A + mv ′B , e =

5 m/s

Use work and energy to determine A’s velocity just before

impact:

v ′B − v ′A . vA

Solving these two equations, we obtain v′B = 2.73 m/s. We use work and energy to determine how far B slides: −µ k mgd = 0 −

1 m(v ′B ) 2 . 2

Solving, d = 0.632 m.

Problem 16.134 The kinetic coefficients of friction between the 5-kg crates A and B and the inclined surface are 0.1 and 0.4, respectively. The coefficient of restitution between the crates is e = 0.8. If the crates are released from rest in the positions shown, what are the magnitudes of their velocities immediately after they collide?

v A = a At = (8.01)(0.369) = 2.95 m/s v B = a Bt = (6.53)(0.369) = 2.41 m/s Conservation of linear momentum yields m Av A + m Bv B = m Av ′A + m Bv ′B : (5)(2.95) + (5)(2.41) = (5)v ′A + (5)v ′B

(1)

The coefficient of restitution is e =

v ′B − v ′A : vA − vB

Evaluating, we have

0.8 =

v ′B − v ′A 2.95 − 2.41

(2).

Solving equations (1) and (2), v ′A = 2.46 m/s, v ′B = 2.90 m/s. 0.1 m B

y 0.1 NA

608

mAg

aA x

Solution:

The free body diagrams of A and B are shown. From the diagramof A, wehave N A = m A g cos60 ° = (5)(9.81) cos60 ° = 24.5 N

∑ Fx = m A g sin60° − 0.1 N A = m Aa A . Solving, (5) (9.81)sin60 ° − 0.1(24.5) = (5)a A , a A = 8.01 m/s 2 . The 1 velocity of A is v A = a At and its position is x A = a At 2 . From 2 the free body diagram of B, we have N B = N A = 24.5 N and ∑ Fx = m A g sin60° − 0.4 N B = m Ba B . Solving we have (5)(9.81) sin 60 ° − 0.4(24.5) = (5)a B , or a B = 6.53 m/s 2 . The velocity of B is v B = a Bt and its position is x B = a Bt 2 /2. To find the time of impact, set x A = x B + 0.1:

1 2 2 a At

y 0.4 NB

mBg

aB x

= 12 a Bt 2 + 0.1

Solving for t t =

2(0.1) = a A − aB

2(0.1) = 0.369 s. 8.01 − 6.53

The velocities at impact are

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Problem 16.135 Solve Problem 16.134 if crate A has a velocity of 0.2 m/s down the inclined surface and crate B is at rest when the crates are in the positions shown.

Solution:

From the solution of Problem 16.134, A’s acceleration is a A = 8.01 m/s 2 so A’s velocity is v

∫0.2 dv = ∫0 a A dtv A = 0.2 + a At

1 a t 2 . From the solution of 2 A Problem 16.136, the acceleration, velocity, and position of B are

and its position is x A = 0.2t +

a B = 6.53 m/s 2 v B = a Bt x B = 12 a Bt 2 , At impact, x A = x B + 0.1; 0.2t + 12 a At 2 = 12 a Bt 2 + 0.1. Solving for t, we obtain t = 0.257 s so,

v A = 0.2 + (8.01)(0.257) = 2.26 m/s; v B = (6.53)(0.257) = 1.68 m/s m Av A + m Bv B = m Av ′A + m Bv ′B : 0.1 m B

(5)(2.26) + (5)(1.68) = (5)v ′A + (5)v ′B

(1)

v ′B − v ′A v ′B − v ′A : 0.8 = 2.26 − 1.68 vA − vB

(2)

e =

Solving Equations (1) and (2), v ′A = 1.74 m/s v ′B = 2.20 m/s. 608

Problem 16.136 A small object starts from rest at A and slides down the smooth ramp. It is moving horizontally when it hits the wall at the right after its first bounce. What was the coefficient of restitution? A

We can integrate the acceleration to determine the velocity and position (relative to the coordinate system shown) as functions of the time after the object leaves the ramp. Because the horizontal acceleration is zero, the horizontal component of the velocity is constant. v x = v 2 x = const., (1) x = v 2 xt, v y = v 2 y − gt, (2) y = v 2 yt −

608

3 ft

1 2 gt . (3) 2

Setting y = −1 ft. in Eq. (1), we can solve for the time at which the object reaches the floor, obtaining t = 0.699 s. Substituting this time into Eq. (1), the horizontal distance the object moves before it bounces is

1 ft

5 ft

b = v 2 x t = (5.67 ft/s)(0.699 s) = 3.97 ft. And from Eq. (2), the vertical component of velocity just before it bounces is v y = v 2 y − gt

Solution: y h1

= 9.83 ft/s − (32.2 ft/s 2 )(0.699 s) = −12.7 ft/s.

v2 608

The upward component of velocity just after it bounces is therefore ev 3 y = e(12.7 ft/s).

x Datum

b 5 ft

We can apply conservation of energy to determine the object’s velocity as it leaves the ramp. Placing the datum at the floor, we have T1 + V1 = T2 + V2 : 1 1 2 mv + mgh1 = mv 2 2 + mgh 2 , 2 1 2 1 2 0 + m(32.2 ft/s )(3 ft) = mv 2 2 + m(32.2 ft/s 2 )(1 ft). 2 This gives v 2 = 11.3 ft/s. The ball’s components of velocity at that point are v 2 x = (11.3 ft/s) cos60 ° = 5.67 ft/s,

The ball’s upward velocity as a function as a function of time after it bounces is e(12.7 ft/s) − (32.2 ft/s 2 )t. Setting this equal to zero, the time from when it bounces until it is moving horizontally is t =

( 12.7 ft/s ) e

32.2 ft/s 2

We want this time multiplied by the horizontal velocity to equal 6 ft − b, (5.67 ft/s)t = (5.67 ft/s)

(12.7 ft/s)e = 5 ft − b, 32.2 ft/s 2

from which we obtain e = 0.461. e = 0.461.

v 2 y = (11.3 ft/s)sin 60 ° = 9.83 ft/s.

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Problem 16.137 The cue gives the cue ball A a velocity of magnitude 3 m/s. The angle β = 0 and the coefficient of restitution of the impact of the cue ball and the eight ball B is e = 1. If the magnitude of the eight ball’s velocity after the impact is 0.9 m/s, what was the coefficient of restitution of the cue ball’s impact with the cushion? (The balls are of equal mass.)

b 30 8 A

Solution:

The strategy is to treat the two impacts as oblique central impacts. Denote the paths of the cue ball as P1 before the bank impact, P2 after the bank impact, and P3 after the impact with the 8-ball. The velocity of the cue ball is v AP1 = 3( i cos30 ° + j sin 30 °) = 2.6 i + 1.5 j (m/s).

The x component is unchanged by the bank impact. The y component after impact is v AP 2 y = −ev AP1y = −1.5e, from which the velocity of the cue ball after the bank impact is v AP2 = 2.6 i − 1.5ej. At impact with the 8-ball, the x component is unchanged. The y component after impact is obtained from the conservation of linear momentum and the coefficient of restitution. The two equations are m Av AP 2 y = m Av AP 3 y + m Bv BP 3 y and

v BP 3 y − v AP 3 y . v AP 2 y

For m A = m B , these equations have the solution v AP 3 y = 0 and v BP 3 y = v AP 2 y , from which the velocities of the cue ball and the 8-ball after the second impact are v AP3 = 2.6 i (m/s), and v BP3 = −1.5ej (m/s). The magnitude of the 8-ball velocity is v BP3 = 0.9 m/s, from which e =

Problem 16.138 What is the solution to Problem 16.137 if the angle β = 10 ° ?

0.9 = 0.6. 1.5

Solution:

Use the results of the solution to Problem 16.137. The strategy is to treat the second collision as an oblique central impact about the line P when β = 10 °. The unit vector parallel to the line P is e P = ( i sin10 ° − j cos10 °) = 0.1736 i − 0.9848 j.

b 30 8 A

The vector normal to the line P is e Pn = 0.9848i + 0.1736 j. The projection of the velocity v AP 2 = v AP 3e P + v AP 3 n e Pn . From the solution to Problem 16.13, v AP2 = 2.6 i − 1.5ej, from which the two simultaneous equations for the new components: 2.6 = 0.173 v AP 3 + 0.9848v AP 3 n , and −1.5e = −0.9848v AP 3 + 0.1736v AP 3 n . Solve: v AP3 = 0.4512 + 1.477e, v AP 3 n = 2.561 − 0.2605e. The component of the velocity normal to the line P is unchanged by impact. The change in the component parallel to P is found from the conservation of linear momentum and the coefficient of restitution: m Av AP 3 = m Av ′AP 3 + m Bv ′BP 3 , and

v ′BP 3 − v ′AP 3 . v AP 3

For m A = m B , these equations have the solution v′AP3 = 0 and v ′BP 3 = v AP 3 . From the value of v′BP3 = 0.9 m/s, 0.9 = 0.4512 + 1.477e, from which e =

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0.9 − 0.4512 = 0.304. 1.477

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Problem 16.139 What is the solution to Problem 16.137 if the angle β = 15 ° and the coefficient of restitution of the impact between the two balls is e = 0.9?

Solution:

Use the solution to Problem 16.137. The strategy is to treat the second collision as an oblique central impact about P when β = 15 °. The unit vector parallel to the line P is e P = ( i sin15 ° − j cos15 °) = 0.2588i − 0.9659 j. The vector normal to the line P is

b 30 8 A

e Pn = 0.9659 i + 0.2588 j. The projection of the velocity v AP 2 = v AP 3e P + v AP 3 n e Pn . From the solution to Problem 16.139, v AP2 = 2.6 i − 1.5ej, from which the two simultaneous equations for the new components: 2.6 = 0.2588v AP3 + 0.9659v AP 3 n , and −1.5e = −0.9659v AP 3 + 0.2588v AP 3 n . Solve: v AP 3 = 0.6724 + 1.449e, v AP 3 n = 2.51 − 0.3882e. The component of the velocity normal to the line P is unchanged by impact. The velocity of the 8-ball after impact is found from the conservation of linear momentum and the coefficient of restitution: m Av AP 3 = m Av ′AP 3 + m Bv ′BP 3 , v ′BP 3 − v ′AP 3 , v AP 3 where e B = 0.9. For m A = m B , these equations have the solution and

eB =

v ′BP 3 =

( 12 )(1 + e )v B

AP 3

= 0.95v AP 3 .

From the value of v′BP3 = 0.9 m/s, 0.9 = 0.95(0.6724 + 1.449e), from which e = 0.189.

Problem 16.140 A ball is given a horizontal velocity of 3 m/s at 2 m above the smooth floor. Determine the distance D between the ball’s first and second bounces if the coefficient of restitution is e = 0.6. 3 m/s

Solution:

The ball has no horizontal acceleration, so the horizontal component of its velocity will be constant, 3 m/s. We can apply conservation of energy to determine the vertical component of its velocity just before the first bounce: T1 + V1 = T2 + V2 : 1 0 + mgh1 = mv 2 2 + 0, 2 1 2 0 + m(9.81 m/s )(2 m) = mv 2 2 + 0. 2 Solving, the downward component of velocity just before the ball bounces is v 2 = 6.26 m/s, so the upward component of velocity just after it bounces is ev 2 = (0.6)(6.26 m/s) = 3.76 m/s.

The ball’s upward velocity as a function as a function of time after it bounces is 3.76 m/s − (9.81 m/s 2 )t, so the time it takes to go from the first bounce to the second bounce is t = 2

3.76 m/s ( 9.81 ) m/s 2

= 0.766 s. The distance between bounces is the product of this time and the horizontal component of velocity: (0.766 s)(3 m/s) = 2.30 m. 2.30 m.

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Problem 16.141* A basketball dropped on the floor from a height of 4 ft rebounds to a height of 3 ft. In the layup shot shown, the magnitude of the ball’s velocity is 5 ft/s, and the angles between its velocity vector and the positive coordinate axes are θ x = 42 °, θ y = 68 °, and θ z = 124 ° just before it hits the backboard. What are the magnitude of its velocity and the angles between its velocity vector and the positive coordinate axes just after the ball hits the backboard?

Solution:

Using work and energy to determine the dropped ball’s velocity just before and just after it hits the floor, 1 2 mv , 2 1 mg(3 ft) = m(v ′) 2 , 2 mg(4 ft) =

we obtain v = 16.05 ft/s and v′ = −13.90 ft/s. The coefficient of restitution is e = −

v′ = 0.866. v

The ball’s velocity just before it hits the backboard is

v = 5(cos 42 °i + cos68 ° j + cos124 °k) = 3.72 i + 1.87 j − 2.80 k (ft/s). The z component of velocity just after the impact is v ′z = −ev z = −(0.866)(−2.80) = 2.42 ft/s. x

The ball’s velocity after impact is v′ = 3.72 i + 1.87 j + 2.42k (ft/s). Its magnitude is v′ = 4.81 ft/s, and

( 3.72 ) = 39.5°, 4.81 1.87 θ = arccos( ) = 67.1°, 4.81 2.42 θ = arccos( ) = 59.8°. 4.81 θ x = arccos

Problem 16.142* In Problem 16.141, the basketball’s diameter is 9.5 in, the coordinates of the center of the basket rim are x = 0, y = 0, z = 12 in, and the backboard lies in the x–y plane. Determine the x and y coordinates of the point where the ball must hit the backboard so that the center of the ball passes through the center of the basket rim.

Solution:

See the solution of Problem 16.141. The ball’s velocity after impact in in./s is v′ = 44.6 i + 22.5 j + 29.1k (in./s)

From impact to the center of the rim, the distance the center of the ball moves in the z direction is 12 − 12 (9.5) = 7.25 in. The time required is t =

7.25 in. = 0.250 s. 29.1 in./s

Setting

x = 0 = x 0 + v xo t = x 0 + (44.6)(0.250) and y = 0 = y 0 + v y 0 t −

1 2 gt 2

1 (32.2)(12)(0.250) 2 , 2 we obtain x 0 = −11.13 in., y 0 = 6.42 in. = y 0 + (22.5)(0.250) −

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Problem 16.143 A satellite at r0 = 10,000 mi from the center of the Earth is given an initial velocity v 0 = 20,000 ft/s in the direction shown. Determine the magnitude of the transverse component of the satellite’s velocity when r = 20,000 mi. (The radius of the Earth is 3960 mi.)

Solution:

By definition,

H 0 = r × mv = mr0v 0 sin 45 ° = m(10, 000)(5280)(20, 000)sin 45 ° H 0 = 7.46 m × 10 11 slug-ft 2 /s. The gravitational force F = −

mgR E2 er, r2

where e r is a unit vector parallel to the radius vector r. Since (r × e r ) = 0, it follows that the angular momentum impulse is t2

∫t ( r × ∑ F ) dt = 0 = H 1 − H 0 ,

v0 u

458

from which H 0 = H 1 , and the angular momentum is conserved. Thus the angular momentum at the distance 20,000 mi is H 0 = mr1v θ1 sin 90 °. From which 7.467 m × 10 11 = 7071 ft/s 1.056 m × 10 8

v θ1 =

is the magnitude of the transverse velocity.

Problem 16.144 In Problem 16.143, determine the magnitudes of the radial and transverse components of the satellite’s velocity when r = 15,000 mi.

Solution:

From the solution to Problem 16.143, the constant angular momentum of the satellite is H 0 = 7.467 m × 10 11 slug-ft 2 /s. The transverse velocity at r = 15,000 mi is H = 8430 ft/s m(15,000)(5280)

vθ =

From conservation of energy:

( 12 mv ) 2 0

mgR E2 mgR E2 1 2 = mv − , r0 2 1 r =15,000 r1

(

)

from which

r u

r = 10,000

−

1 1 v 12 = v 02 + 2 gR E2  −  r r0 

458

= (20, 000) 2 + 2(32.17)(3960) 2

1 ( 15,1000 − 10000 )(5280)

= 2.224 × 10 8 (ft/s) 2 . The radial velocity is vr =

Problem 16.145 The snow is 2 ft deep and weighs 20 lb/ft 3 , the snowplow is 8 ft wide, and the truck travels at 5 mi/h. What force does the snow exert on the truck?

v 12 − v θ2 = 11,550 ft/s.

Solution: v = 5

The velocity of the truck is

5280 ( 3600 ) = 7.33 ft/s.

The mass flow is the product of the depth of the snow, the width of the plow, the mass density of the snow, and the velocity of the truck, from which  dm f   20   dt  = (2) g  (8)(7.33) = 72.84 slug/s. The force is  dm f  F =   v = 534 lb.  dt 

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Problem 16.146 An empty 55-lb drum, 3 ft in diameter, stands on a set of scales. Water begins pouring into the drum at 1200 lb/min from 8 ft above the bottom of the drum. The weight density of water is approximately 62.4 lb/ft 3 . What do the scales read 40 s after the water starts pouring?

Solution:

The mass flow rate is

 dm f   1200  1  dt  =  g  60 = 0.6217 slug/s

( )

After 40 seconds, the volume of water in the drum is volume =

( )

1200 40 = 12.82 ft 3 . 62.4 60

The height of water in the drum is h water =

8 ft

volume = 1.814 ft. π (1.5 2 )

The velocity of the water stream at point of impact is (from the conservation of energy for free fall) v = 2 g(8 − h water ) = 19.95 ft/s. The scale reading is  dm f  W = (volume)(62.4) + 55 +    dt  v = 12.82(62.4) + 55 + 0.6216(19.95) = 867 lb.

Problem 16.147 The ski boat’s jet propulsive system draws water in at A and expels it at B at 80 ft/s relative to the boat. Assume that the water drawn in enters with no horizontal velocity relative to the surrounding water. The maximum mass flow rate of water through the engine is 2.5 slugs/s. Hydrodynamic drag exerts a force on the boat of magnitude 1.5v lb, where v is the boat’s velocity in feet per second. Neglecting aerodynamic drag, what is the ski boat’s maximum velocity?

Solution:

Use the boat’s “rest” frame of reference. The force exerted by the inlet mass flow on the boat is  dm f  Finlet = −  v = −2.5 v.  dt 

The force exerted by the exiting mass flow on the boat is  dm f  Fexit =   (80) = 2.5(80)  dt  The hydrodynamic drag is Fdrag = −1.5 v. At top speed the sum of the forces vanishes: ∑ F = Finlet + Fexit + Fdrag = 0 = −2.5v + 2.5(80) − 1.5v = −4.0v + 2.5(80) = 0, from which v =

2.5(80) = 50 ft/s. 4.0

Source: Courtesy of Zdorov Kirill Vladimirovich/Shutterstock.

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Problem 16.148 The ski boat in Problem 16.147 weighs 2800 lb. The mass flow rate of water through the engine is 2.5 slugs/s, and the craft starts from rest at t = 0. Determine the boat’s velocity at (a) t = 20 s and (b) t = 60 s.

Solution:

Use the solution to Problem 16.147: the sum of the forces

on the boat is ∑ F = −4v + 2.5(80) lb. From Newton’s second law W dv g dt

( ) = −4.0v + 200.

Separate variables and integrate: 4g dv = dt, W 50 − v from which B

ln(50 − v ) = −

Source: Courtesy of Zdorov Kirill Vladimirovich / Shutterstock.

4g t + C. W

At t = 0, v = 0, from which C = ln (50), from which

(

ln 1 −

4g v = − t. W 50

)

Invert: 4g  − t v (t ) = 50  1 − e W .  

Substitute numerical values: W = 2800 lb, g = 32.17 ft/s 2 .

Problem 16.149* A crate of mass m slides across a smooth floor pulling a chain from a stationary pile. The mass per unit length of the chain is ρ L . If the velocity of the crate is v 0 when s = 0, what is its velocity as a function of s ? s

(a) At t = 20 s,

v = 30.1 ft/s.

(b) At t = 60 s,

v = 46.8 ft/s.

Solution:

The “mass flow” of the chain is

 dm f   dt  = ρ Lv . The force exerted by the “mass flow” is F = ρ Lv 2 . From Newton’s second law (ρ L s + m)

( ddtv ) = −ρ v . L

Use the chain rule: (ρ L s + m)v

dv = −ρ Lv 2 . ds

Separate variables and integrate:  m  ln(v ) = −ln  s +  + C,  ρ L  from which  mv  C = ln  0 .  ρL  Reduce and solve: v =

288

mv 0 . (ρ L s + m)

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Chapter 17 Problem 17.1 The angle θ is given as a function of time in seconds by θ = 0.1t 2 rad. At t = 3 s, determine the angular velocity of the bar and the magnitude of the velocity of point A.

Solution:

The angular velocity of the bar is

dθ = 0.2t dt = 0.2(3 s)

ω =

= 0.6 rad/s. From Eq. (17.2), the velocity of point A is A

v = ωr = (0.6 rad/s)(2 ft) = 1.2 ft/s.

0.6 rad/s, 1.2 ft/s.

Problem 17.2 The angle θ (in radians) is given as a function of time by θ = 0.2πt 2 . At t = 4 s, determine the magnitudes of (a) the velocity of point A and (b) the tangential and normal components of acceleration of point A. Solution:

We have dθ dω = 0.4 πt, α = = 0.4 π. dt dt

θ = 0.2πt 2 , ω = Then

(a) v = r ω = (2)(0.4 π )(4) = 10.1 ft/s. (b)

v = 10.1 ft/s.

a n = r ω 2 = (2)[(0.4 π ) (4)] 2 = 50.5 ft/s 2 ,

a n = 50.5 ft/s 2 ,

a t = rα = (2)(0.4 π) = 2.51 ft/s 2 .

a t = 2.51 ft/s 2 .

Problem 17.3 The mass A starts from rest at t = 0 and falls with a constant acceleration of 8 m/s 2 . When the mass has fallen one meter, determine the magnitudes of (a) the angular velocity of the pulley and (b) the tangential and normal components of acceleration of a point at the outer edge of the pulley. Solution:

We have

100 mm

a = 8 m/s 2 , v = 2as = 2(8 m/s 2 )(1 m) = 4 m/s, 4 m/s v ω = = = 40 rad/s, 0.1 m r 8 m/s 2 a α = = = 80 rad/s 2 . 0.1 m r (a)

ω = 40 rad/s.

(b) a t = rα = (0.1 m)(80 rad/s 2 ) = 8 m/s 2 , an =

rω 2

(0.1 m)(40 rad/s) 2

= 160 m/s 2 .

a t = 8 m/s 2 , a n = 160 m/s 2 .

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Problem 17.4 At the instant shown, the left disk has an angular velocity of 4 rad/s counterclockwise and an angular acceleration of 2 rad/s 2 clockwise. Assume there is no relative motion between the two disks at their point of contact. (a) What are the angular velocity and angular acceleration of the right disk? (b) What are the magnitudes of the velocity and acceleration of point A?

Solution: (a) We can see that the angular velocity ω of the right disk will be clockwise. Because the velocities of the disks are equal at their point of contact, we have (1 m)(4 rad/s) = (2.5 m) ω, giving ω = 1.6 rad/s clockwise. In the same way, we can see that the angular acceleration α of the right disk will be counterclockwise, and we have (1 m)(2 rad/s 2 ) = (2.5 m) α, giving α = 0.8 rad/s 2 counterclockwise. (b) The velocity of point A is

4 rad/s 2 rad/s2

v = (2 m) ω

A 2.5 m

= (2 m)(1.6 rad/s) = 3.2 m/s.

The tangential component of the acceleration of point A (the component tangent to its circular path) is a t = (2 m) α = (2 m)(0.8 rad/s 2 ) = 1.6 m/s 2 . The normal component is an = =

v2 r (3.2 m/s) 2 (2 m)

= 5.12 m/s 2 . The magnitude of the acceleration of A is a t 2 + a n 2 = 5.36 m/s 2 . (a) 1.6 rad/s clockwise, 0.8 rad/s 2 counterclockwise. (b) 3.2 m/s, 5.36 m/s 2 .

Problem 17.5 The angular velocity of the left disk is given as a function of time by ω A = 13 + 3t 2 rad/s. Through what angle does the right disk turn from t = 0 to t = 3 s?

100 mm 100 mm vA

vB 200 mm

vC 200 mm

Solution: ω A = (13 + 3t 2 ) rad/s. r A ω A = rB 200 ω B and rB100 ω B = rC ω C . Thus, ω B = 0.5ω A and ω C = 0.5ω B = 0.25ω A θC = ∫

290

3s ω 0

dt = ∫

3s 0

(

(13 + 3t 2 ) 13t + t 3 dt = 4 4

) = 16.5 rad. 3s 0

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Problem 17.6 The gear setting of a bicycle is sometimes expressed in “gear inches,” which is the number of inches the bicycle travels when the pedals go through one revolution. If the bicycle has 27-in diameter wheels, what is the number of gear inches for the gear setting on the bicycle shown? (The pedals are attached to the 80-mm chainring, and the rear wheel is attached to the 30-mm cog.)

30 mm 80 mm

Solution:

When the 80-mm chainring goes through one revolution, the chain connecting it to the 30-mm cog moves one circumference of the chain ring, or 2π (80) mm. The number of revolutions this causes the 30-mm cog to rotate is the distance the chain divided by the circumference of the cog: 2π (80) mm = 2.67 revolutions. 2π (80) mm

When the rear wheel, connected to the cog, rotates through one revolution, the bicycle moves a distance equal to the circumference of the wheel, π(27 in). Therefore the number of gear inches for the bicycle areπ (27 in)(2.67 revolutions) = 226 in. 226 in.

Problem 17.7 The bicycle’s pedals are attached to the 80-mm chainring, and its 27-in diameter rear wheel is attached to the 30-mm cog. If the pedals rotate at one revolution per second, what is the bicycle’s velocity in km/h?

30 mm 80 mm

Solution:

The given angular velocity of the pedals is 2π rad/s. Let ω L and ω R be the angular velocities of the left and right gears, and rL and rR their radii. Due to the chain, the velocity (relative to the bicycle) of a point at the outer edge of the two gears must be equal:ω L rL = ω RrR . Therefore the angular velocity of the cog, and rear wheel, is ωL =

rR 80 mm ω = (2 π rad/s) = 16.8 rad/s. rL R 30 mm

The bicycle’s velocity is equal to the velocity of a point of a point at the edge of the wheel, v = (16.8 rad/s) = 226 in/s

( 12 27 in ) = 226 in/s

1 km ( 3600 s ) ( 39,370 in ) 1 h

= 20.7 km/h. 20.7 km/h.

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Problem 17.8 The disk is rotating about the origin with a constant clockwise angular velocity of 100 rpm. Determine the x and y components of velocity of points A and B (in in/s).

Solution: ω = 100 rpm

Working with point A rA =

( 2πrevrad ) ( 160mins ) = 10.5 rad/s.

(8 in ) 2 + (8 in ) 2 = 11.3 in,

θ A = tan −1

( 88 inin ) = 45°

v A = r Aω = (11.3 in)(10.5 rad/s) = 118 in/s v A = v A (cos θ A i + sin θ A j) = (118 in/s)(cos 45 °i + sin 45 ° j) v A = (83.8i + 83.8 j) in/s.

A 8 in 8 in

Working with point B B

16 in

rB = 16 in, v B = rB ω = (16 in)(10.5 rad/s) = 168 in/s v B = −(168 j) in/s.

Problem 17.9 The disk starts from rest in the orientation shown at time t = 0 has clockwise angular velocity given in terms of time in seconds by ω = 2t 3 rad/s. What are the magnitudes of the acceleration of points A and B (in in/s 2 ) at t = 2 s?

Solution:

The radial distance from the axis to point A is

rA =

+ (8 in) 2 = 11.3 in.

(8 in) 2

The angular velocity and angular acceleration at t = 2 s are ω = 2t 3 = 2(2 s) 3 = 16 rad/s, dω = 6t 2 =6(2 s) 2 = 24 rad/s 2 . dt The velocity and components of acceleration of A at t = 2 s are α =

(a t ) A = r Aα = (11.3 in)(24 rad/s 2 ) = 272 in/s 2 , (a n ) A = r A ω 2 = (11.3 in)(16 rad/s) 2 = 2900 in/s 2 . Its magnitude is aA =

A 8 in 8 in

= B

16 in

(a t ) A 2 + (a n ) A 2 (272 in/s 2 ) 2 + (2900 in/s 2 ) 2 = 2910 in/s 2 .

The components of acceleration of B at t = 2 s are (a t ) B = rB α = (16 in)(24 rad/s 2 ) = 384 in/s 2 , (a n ) B = rB ω 2 = (16 in)(16 rad/s) 2 = 4096 in/s 2 . Its magnitude is aB = =

(a t ) B 2 + (a n ) B 2 (384 in/s 2 ) 2 + (4096 in/s 2 ) 2 = 4110 in/s 2 .

a A = 2910 in/s 2 , a B = 4110 in/s 2 .

Problem 17.10 The outer radius of the Corvette’s tires is 15 in. At the instant shown, the car is moving in a straight line at 100 km/h and is accelerating at 3 m/s 2 . Its tires are rolling without slipping. The coordinate system shown is not rotating and its origin is fixed with respect to the center of the tire. What are the velocity and acceleration relative to the coordinate system of the point of the tire with coordinates x = 15 in, y = 0?

Source: Courtesy of Ditch Green/Alamy Stock Photo.

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17.10 (Continued) Solution:

The points A and B shown are points of the tire and so are rotating about its center. We know that, relative to the coordinate system, the velocity of point A, which is the point of the tire in contact with the ground at this instant, is equal to the car’s velocity v in the direction shown. Point B, which is the point we are concerned with, has the same velocity v in the direction shown, so the velocity of point B is

at an

v B = v j = 27.8 j (m/s). We also know that, relative to the coordinate system, point A has a tangential component of acceleration in the direction shown that is equal to the car’s acceleration, a t = 3 m/s 2 . Point B has the same tangential component in the direction shown. Both points have the same normal components of acceleration toward the center of the tire,

an A

an =

We see that the acceleration of point B is

The tire’s radius and the car’s velocity in SI units are R = 15 in

a B = −a n i + a t j

m ( 0.0254 ) = 0.381 m, 1 in

v = 100,000 m/h

v2 = 2030 m/s 2 . R

= −2030 i + 3 j (m/s 2 ).

1h ( 3600 ) = 27.8 m/s. s

v B = 27.8 j (m/s), a B = −2030 i + 3 j (m/s 2 ).

Problem 17.11 The bar has a counterclockwise angular velocity of 6 rad/s. What are the x and y components of the velocities of points A and B?

Solution: A y

vA uA A

0.4 m

vB rB

uB B

x 0.2 m

0.4 m

The radial distances from the origin to points A and B are rA =

(0.4 m) 2 + (0.4 m) 2 = 0.566 m,

rA =

(0.4 m) 2 + (0.2 m) 2 = 0.447 m.

The magnitudes of their velocities are v A = r A ω = 3.39 m/s, v B = rB ω = 2.68 m/s. The angles θ A = 45 °, θ B = arctan(0.4 m/0.2 m) = 63.4 °. The velocities of points A and B are v A = −v A sin θ A i − v A cos θ A j = −2.4 i − 2.4 j (m/s), v B = v B cos θ B i + v B sin θ B j = 1.2 i + 2.4 j (m/s). v A = −2.4 i − 2.4 j (m/s), v B = 1.2 i + 2.4 j (m/s).

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Problem 17.12 The bar has a counterclockwise angular velocity of 6 rad/s and a counterclockwise angular acceleration of 40 rad/s 2 . What are the x and y components of the acceleration of point A?

A y

0.4 m

Solution:

x A

0.2 m

aAt

aAn

0.4 m

The radial distance from the origin to point A is rA =

(0.4 m) 2 + (0.4 m) 2 = 0.566 m.

The magnitude of its velocity is v A = r A ω = 3.39 m/s. The tangential and normal components of its acceleration are a At = r Aα = 22.6 m/s 2 , a An = r A ω 2 = 20.4 m/s 2 . The angle θ A = 45 °. The acceleration of point A is a A = −a At sin θ A i − a At cos θ A j + a An cos θ A i − a An sin θ A j = −1.60 i − 30.4 j (m/s 2 ). a A = −1.60 i − 30.4 j (m/s 2 ).

Problem 17.13 A disk of radius R = 0.5 m rolls on a horizontal surface. The relationship between the horizontal distance x the center of the disk moves and the angle β through which the disk rotates is x = Rβ . Suppose that the center of the disk is moving to the right with a constant velocity of 2 m/s. (a) What is the disk’s angular velocity? (b) Relative to a nonrotating reference frame with its origin at the center of the disk, what are the magnitudes of the velocity and acceleration of a point on the edge of the disk?

b R

Solution: (a) x = Rβ ⇒ x = Rβ ⇒ v = Rω ⇒ ω =

(b)

294

v 2 m/s = = 4 rad/s. R 0.5 m

v = Rω = (0.5 m)(4 rad/s) = 2 m/s a = a n = Rω 2 = (0.5 m)(4 rad/s) 2 = 8 m/s 2 .

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Problem 17.14 The turbine rotates relative to the coordinate system at 30 rad/s about a fixed axis coincident with the x axis. (a) What is its angular velocity vector? (b) What is its angular velocity vector if it is rotating at 30 rad/s in the opposite direction?

30 rad/s

Solution: (a) The angular velocity vector is parallel to the x axis. Curving the fingers of the right hand about the x axis in the direction of rotation, the thumb points in the positive x direction. The angular velocity vector is 30 i (rad/s). (b) Curving the fingers of the right hand about the x axis in the opposite direction, the thumb points in the negative x direction. The angular velocity vector is −30 i (rad/s).

(a) 30 i (rad/s). (b) − 30 i (rad/s).

Problem 17.15 The rectangular plate swings in the x – y plane from arms of equal length. What is the angular velocity vector of (a) the rectangular plate and (b) the bar AB?

x 10 rad/s

Solution:

Denote the upper corners of the plate by B and B′, and denote the distance between these points (the length of the plate) by L. Denote the suspension points by A and A′, the distance separating them by L ′. By inspection, since the arms are of equal length, and since L = L ′, the figure AA′ B′B is a parallelogram. By definition, the opposite sides of a parallelogram remain parallel, and since the fixed side AA′ does not rotate, then BB′ cannot rotate, so that the plate does not rotate and

ω BB ′ = 0 . Similarly, by inspection the angular velocity of the bar AB is

A9 B9

ω AB = 10 k (rad/s), where by the right hand rule the direction is along the positive z axis (out of the paper).

Problem 17.16 The length of bar AB is 0.4 m. Its angular velocity is constant. What are the magnitudes of the velocity and acceleration of the center of the translating rectangular plate?

x 10 rad/s

Solution:

The velocity and acceleration of every point of the translating plate are the same. The magnitude of the velocity of point B due to its circular motion is

v = r AB ω = (0.4 m)(10 rad/s) = 4 m/s. Because the angular velocity of the bar is constant, point B has only a normal component of acceleration, with magnitude a n = r AB ω 2 = (0.4 m)(10 rad/s) 2 = 40 m/s 2 . Velocity is 4 m/s, acceleration is 40 m/s 2 .

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Problem 17.17 The disk of radius R = 0.4 m rolls on the horizontal surface. The angle β = 0.6t 2 rad/s. At t = 2 s, what are the distance x and the magnitude of the velocity of the center of the disk?

Solution:

The distance

x = Rβ = 0.6 Rt 2 , so at t = 2 s the distance x = 0.6(0.4 m)(2 s) 2 = 0.96 m. The magnitude of the velocity of the center equals

v = R

dβ dx = R = 1.2 Rt, dt dt

so at t = 2 s v = 1.2(0.4 m)(2 s) = 0.96 m/s.

x = 0.96 m, velocity is 0.96 m/s.

R x x

Problem 17.18 The rigid body rotates with angular velocity ω = 10 rad/s. The distance r A /B = 0.5 m. The velocity of point A at the instant shown is v A = 3i + 4 j (m/s). (a) Determine velocity of A relative to B by representing it as shown in Fig. 17.10b. (b) What is the velocity vector of B?

v A

rA/B

Solution: (a) Representing the velocity of A relative to B as in Fig. 17.10b, y

v B

vA/B A

rA/B

we see that v A /B = r A /B ω and v A /B = r A /B ω j = (0.5 m)(10 rad/s) j = 5 j (m/s). (b) From the equation v A = v B + v A /B : 3i + 4 j (m/s) = v B + 5 j (m/s), we see that v B = 3i − j (m/s). (a) v A /B = 5 j (m/s). (b) v B = 3i − j (m/s).

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Problem 17.19 The bar is rotating in the counterclockwise direction with angular velocity ω. The x component of the velocity of point B is 6 m/s. Determine the components of the velocity of point A.

Solution:

The position of B relative to the origin O is

rB /O = (0.4 m) i − (0.2 m) j. The angular velocity vector of the bar is ω = ωk. In terms of the angular velocity, the velocity of B is

v B /O = ω × r B /O =

i j k 0 0 ω 0.4 m −0.2 m 0

= 0.2ω i + 0.4 ω j (m/s).

We know that the x component 0.2ω = 6 m/s,

0.4 m

which tells us that ω = 30 rad/s.

The position of A relative to the origin O is x

0.4 m

r A /O = −(0.4 m) i + (0.4 m)j.

0.2 m

The velocity of A is

v A /O = ω × r A /O =

k i j 0 0 ω −0.4 m 0.4 m 0

= −0.4 ω i − 0.4 ω j = −12 i − 12 j (m/s). v A = − 12 i − 12 j (m/s).

Problem 17.20 The bar is rotating in the counterclockwise direction with angular velocity ω. The magnitude of the velocity of point A relative to point B is 6 m/s. Determine the velocity of point B.

Solution: r A /B =

y A

0.4 m v

(0.8 m) 2 + (0.6 m) 2 = 1 m x

6 m/s v = = 6 rad/s. 1m r A /B v B = ω × rB = (6 rad/s)k × (0.4 i − 0.2 j) m ω =

0.2 m

v B = (1.2 i + 2.4 j) m/s.

Problem 17.21 The bracket is rotating about point O with counterclockwise angular velocity ω. The magnitude of the velocity of point A relative to point B is 4 m/s. Determine ω.

0.4 m

488

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120 mm

0.4 m

180 mm

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17.21 (Continued) Solution:

The position vector of A relative to B is

so the angular velocity is

r A /B = (−0.18 − 0.12 cos 48 °) i + (0.12sin 48 °) j (m)

ω =

= −0.260 i + 0.0892 j (m) = x A /B i + y A /B j.

Alternatively, we can write the equation v A /B = ω × r A /B

Its magnitude is r A /B =

v A /B 4 m/s = = 14.5 rad/s. r A /B 0.275 m

x A /B 2 + y A /B 2 = 0.275 m.

We can use a graphical approach: y

i 0

j 0

x A /B

y A /B

k ω 0

= −ω y A /B i + ω x A /B j. The magnitude is

vA/B

rA/B

v A /B = B

Solving this equation for the angular velocity yields ω =

We see that the magnitude of the velocity of A relative to B is v A / B = ωr A / B ,

(−ω y A /B ) 2 + (ω x A /B ) 2 .

v A /B v = A /B = 14.5 rad/s. r A /B (−y A /B ) 2 + ( x A /B ) 2

14.5 rad/s.

Problem 17.22 The bracket is rotating with counterclockwise angular velocity ω. The velocity of point O is 4 i + 2 j (m/s). The magnitude of the velocity of B relative to O is 3 m/s. What is the velocity of A?

A y

120 mm

Solution:

Applying a graphical approach to the motion of B rel-

ative to O, 488

180 mm

vB/O v B

rB/O

we see that the angular velocity is ω =

v B /O 3 m/s = = 16.7 rad/s. rB / O 0.18 m

The position vector of A relative to O is r A /O = (−0.12cos48 °) i + (0.12sin 48 °) j (m) = −0.0803i + 0.0892 j (m) = x A /O i + y A /O j. The velocity of A is v A = v O + ω × r A /O = ( 4 m/s ) i + ( 2 m/s ) j +

i 0

j 0

x A /O

y A /O

k ω 0

= 2.51i + 0.662 j (m/s). v A = 2.51i + 0.662 j (m/s).

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Problem 17.23 If the angular velocity of the bar is constant, what are the x and y components of the velocity of point A 0.1 s after the instant shown?

Solution: ω =

The angular velocity is given by

dθ = 5 rad/s, dt

∫0 d θ = ∫0 5 dt, and θ = 5t rad.

At t = 0.1 s, θ = 0.5 rad = 28.6 °.

A 2m

308

v A = v 0 + ω × rA / O = 0 +

5 rad/s x

i j k 0 0 5 . 2 cos 28.6 ° 2 sin 28.6 ° 0

Hence, v A = −10 sin 28.6 °i + 10 cos 28.6 ° j = −4.78i + 8.78 j.

Problem 17.24 The disk is rotating in the counterclockwise direction at 40 rad/s. Determine the velocity vectors of A and B.

Solution:

The angular velocity vector of the disk is ω = 40 k rad/s. The vectors from the origin to A and B are A: r A /O = 0. 1j m, B: rB /O = [0.1cos(45 °) i − 0.1sin(45 °) j] (m) = 0.0707 i − 0.0707 j (m).

The velocity of A is

v A = v O + ω × r A /O

i j k 0 0 40 rad/s 0 0.1 m 0

= 0+ 100 mm x

= −4 i (m/s). The velocity of B is v B = v O + ω × r B /O

i j k 0 0 40 rad/s 0.0707 m −0.0707 m 0

B = 0+ 458

458

= 2.83i + 2.83 j (m/s). v A = −4 i (m/s), v B = 2.83i + 2.83 j (m/s).

Problem 17.25 If the magnitude of the velocity of A relative to C is 6 m/s, and the disk is rotating clockwise, what is the disk’s angular velocity vector?

Solution:

We write the clockwise angular velocity vector of the disk as ω = −ωk. The position vectors of A and C are A: r A /O = 0. 1j m,

C: rC /O = [−0.1cos(45 °) i − 0.1sin(45 °) j] (m) = − 0.0707 i − 0.0707 j (m).

The position vector of A relative to C is

r A /C = r A /O − rC /O = 0.0707 i + 0.1707 j (m). The velocity of A relative to C is

100 mm x

v A /C = ω × r A /C =

B 458

458

i j k 0 0 ω 0.0707 m 0.1707 m 0

= −0.1707 ω i + 0.0707 ω j (m/s). Its magnitude is v A /C =

(−0.1707 ω ) 2 + (0.0707 ω ) 2

= ω (−0.1707) 2 + (0.0707) 2 . Setting v A /C = 6 m/s and solving, we obtain ω = 32.5 rad/s. ω = 32.5 k (rad/s).

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Problem 17.26 The outer radius of the Corvette’s tires is 14 in. It is traveling at 80 mi/h. Assume that the tires roll, not skid, on the road surface. (a) What is the angular velocity of its wheels? (b) In terms of the Earth-fixed coordinate system shown, determine the velocity (in ft/s) of the point of the tire with coordinates (−14 in, 0, 0).

Source: Courtesy of Ditch Green/Alamy Stock Photo. Solution: v (a) ω = = r

ft/s ( 6088mi/h ) = 101 rad/s. ω = 101 rad/s. 14 ( 12 ft )

(80 mi/h)

(b) v P = v O + ω × r 80[88] 14  ft i vP = − ft/s i + (101 rad/s)k × −   12  60

(

)

(

)

v P = (−117 i − 117 j) ft/s.

Problem 17.27 The disk is rolling with clockwise angular velocity 4 rad/s. Determine the velocity of point A by applying Eq. (17.6) to points A and B.

Solution:

Applying the right-hand rule, the angular velocity vector of the disk is ω = −4 k (rad/s). We know that because the disk is rolling, the velocity of point B at the instant shown is zero. The position of A relative to B is

458

0.6 m C A

r A /B = (0.6 m) j. Equation (17.6) is

v A = v B + ω × r A /B = 0 +

i j k 0 0 −4 rad/s 0 0.6 m 0

= 2.4 i (m/s). v A =2.4 i (m/s).

Problem 17.28 The disk is rolling. At the instant shown, the x component of the velocity of point C is 2 m/s. What is the velocity of point D?

D 458

0.6 m C A

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17.28 (Continued) Solution:

In this case we don’t know the angular velocity vector of the disk, so we write it as ω = ωk. We know that because the disk is rolling, the velocity of point B at the instant shown is zero. The position of point C relative to B is

Now we can determine the velocity of D. The position of point D relative to B is

rC /B = −(0.6 m) i + (0.6 m) j.

The velocity of D in terms of the velocity of B is

Equation (17.6) for the velocity of C in terms of the velocity of B is

v D = v B + ω × r D /B

v C = v B + ω × rC /B = 0 +

i j k 0 0 ω −0.6 m 0.6 m 0

= −0.6 ω i − 0.6 ω j (m/s) . We know that the x component equals 2 m/s,

rD /B = −(0.6 m) cos 45 °i + 0.6 m(1 + cos 45 °) j.

= 0+

i j k 0 0 −3.33 rad/s −(0.6 m)cos45° 0.6 m(1 + cos45°) 0

= 3.41i + 1.41 j (m/s). v D = 3.41i + 1.41 j (m/s).

−0.6 ω = 2 m/s, So ω = −3.33 rad/s.

Problem 17.29 The bar is moving in the x–y plane and is rotating in the counterclockwise direction. The velocity of point A relative to the reference frame is v A = 12 i − 2 j (m/s). The magnitude of the velocity of point A relative to point B is 8 m/s. What is the velocity of point B relative to the reference frame?

B 2m

Solution: v 8 m/s ω = = = 4 rad/s, r 2m v B = v A + ω × r B /A

308 A

= (12 i − 2 j) m/s + (4 rad/s)k × (2 m)(cos30 °i + sin30 ° j) v B = (8i + 4.93 j) m/s.

Problem 17.30 Points A and B of the 4-m bar slide on the plane surfaces. At the instant shown, point A is moving downward at 2 m/s. What is the velocity vector of the midpoint G of the bar? Strategy: First apply Eq. (17. 6) to points A and B to determine the bar’s angular velocity. Then apply Eq. (17. 6) to points A and G.

y A

Solution:

The position vector of B relative to A is

rB /A = (4 m)cos 70 °i − (4 m)sin 70 ° j = 1.37 i − 3.76 j (m). The velocity of B in terms of the velocity of A is

708

v B = v A + ω × r B /A v B i = −(2 m/s) j +

i j k 0 0 ω . 1.37 m −3.76 m 0

Equating the j components on the left and right sides of this equation, we obtain the equation 0 = −2 m/s + (1.37 m)ω = 0, yielding the angular velocity ω = 1.46 rad/s.

The velocity of G in terms of the velocity of A is v G = v A + ω × rG /A = −( 2 m/s ) j +

i j k 0 0 1.46 rad/s 0.684 m −1.88 m 0

= 2.75i − j (m/s).

The position vector of G relative to A is rG /A =

1 r = 0.684 i − 1.88 j (m). 2 B /A

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Problem 17.31 Bar AB rotates at 6 rad/s in the clockwise direction. Determine the velocity (in in/s) of the slider C.

Solution: v B = v A + ω AB × rB /A v B = 0 − (6 rad/s)k × (4 i + 4 j) in v B = (24 i − 24 j) in/s.

v C = v B + ω BC × rC /B v C = (24 i − 24 j) + ω BC k × (10 i − 7 j)

4 in

v C = (24 + 7 ω BC ) i + (−24 + 10 ω BC ) j

6 rad/s

Slider C cannot move in the j direction, therefore

3 in

24 = 2.4 rad/s. 10 v C = (24 + 7(2.4)) i = (43.8 in/s) i ω BC =

C 4 in

Problem 17.32 If θ = 45 ° and the sleeve P is moving to the right at 2 m/s, what are the angular velocities of bars OQ and PQ?

Solution:

From the figure, v 0 = 0, v P = v P i = 2 i (m/s)

v Q = v 0 + ω OQ k × ( L cos θ i + L sin θ j)  i: v Q x = − ω OQ L sin θ   j: v Q y = ω OQ L cos θ v P = v Q + ω QP k × ( L cos θ i − L sin θ j)

1.2 m

(1) (2)

2 = v Q x + ω QP L sin θ

(3)

j: 0 = v Q y + ω QP L cos θ

(4)

1.2 m

v C = 43.8 in/s to the right.

10 in

Eqns (1)–(4) are 4 eqns in the 4 unknowns ω OQ , ω QP , v Q x , v Q y Solving, we get

v Q x = 1 m/s, v Q y = −1 m/s ω OQ = −1.18k (rad/s), ω QP = 1.18k (rad/s).

Problem 17.33 The Apache helicopter is in planar motion in the x–y plane. At the instant shown, the position of its center of mass G is x = 9 ft, y = 12 ft and its velocity is v G = 60 i + 20 j (ft/s). The position of point T where the tail rotor is mounted is x = −28 ft, y = 22 ft. The helicopter is rotating in the clockwise direction at eight degrees per second. What is the velocity vector of point T ?

Solution:

The clockwise angular velocity of the helicopter in rad/s is

 π rad  8 deg/s   = 0.140 rad/s.  180 deg  The position of T relative to G is rT /G = (−28 − 9)i + (22 − 12)j = −37 i + 10 j (ft). The velocity of T is v T = v G + ω × rT /G = 60 i + 20 j (ft/s) +

i j k 0 0 −0.140 rad/s −37 ft 10 ft 0

= 61.4 i + 25.2 j (ft/s).

v T = 61.4 i + 25.2 j (ft/s). G x

Source: Courtesy of Kojihirano/Shutterstock.

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Problem 17.34 At the instant shown, the collar C is sliding toward the right at 4 m/s. Determine the angular velocity of bar AB and the velocity of point D.

Solution: We know the velocity of point C and the direction of the velocity of point B. Let us apply Eq. (17.6) to points B and C: v C = v B + ω BCD × rC /B : ( 4 m/s ) i = v B j +

i j k 0 0 ω BCD , 0.24 m 0.48 m 0

( 4 m/s ) i = v B j − ω BCD ( 0.48 m ) i + ω BCD ( 0.24 m ) j.

0.32 m

Equating i and j components, we obtain the equations

4 m/s = −ω BCD (0.48 m), 0 = v B + ω BCD (0.24 m), Yielding ω BCD = −8.33 rad/s, v B = 2 m/s. 0.48 m

0.32 m

Knowing the velocity of B, we can see that the angular velocity of bar AB is counterclockwise and equal to ω AB = (2 m/s)/(0.32 m) = 6.25 rad/s. Now we write the velocity of point D in terms of the velocity of point B:

v D = v B + ω BCD × rD /B :

0.24 m 0.16 m

i j k 0 0 ω BCD 0.4 m 0.8 m 0

= (2 m/s) j +

= (2 m/s) j − ω BCD (0.8 m) i + ω BCD (0.4 m) j = 6.67 i − 1.33 j (m/s). ω AB = 6.25 rad/s counterclockwise, v D = 6.67 i − 1.33 j (m/s).

Problem 17.35 The crank AB has a counterclockwise angular velocity of 1000 rpm (revolutions per minute). At the instant shown, what are the angular velocity of the connecting rod BC and the velocity of the piston C ?

Solution:

The angular velocity of the crank is

ω AB = 1000 rpm

50 mm

2π rad ( 1 revolution )( 160mins ) = 105 rad/s.

The velocity of point B is i 0

v B = v A + ω AB × rB /A = 0 +

j 0

k ω AB

0.05 m 0.05 m

175 mm

50 mm

= −ω AB (0.05 m) i + ω AB (0.05 m) j. Using the fact that we know the direction of the velocity of point C, we express it in terms of the velocity of point B, v C = v B + ω BC × rC /B : v C i = −ω AB (0.05 m) i + ω AB (0.05 m) j +

i 0

j 0

0.175 m −0.05 m

k ω BC

obtaining the two equations v C = −ω AB (0.05 m) + ω BC (0.05 m), 0 = ω AB (0.05 m) + ω BC (0.175 m). Solving, we obtain ω BC = −30.0 rad/s, v C = −6.73 m/s. ω BC = 30.0 rad/s clockwise, v C = −6.73i (m/s).

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Problem 17.36 The velocity of the piston is v C = −30 i (m/s). When it has moved to the left far enough that the crank AB is vertical, what is the angular velocity of the crank AB? Solution:

The length of the crank AB is

50 mm

(0.05 m) 2 + (0.05 m) 2 = 0.0707 m.

The length of the connecting rod BC is (0.05 m) 2 + (0.175 m) 2 = 0.182 m. When the crank AB is vertical, the horizontal distance from A to C is

175 mm

(0.182 m) 2 − (0.0707 m) 2 = 0.168 m.

50 mm

We express the velocity of B in terms of the velocity of C: v B = v C + ω × r B /C

y i j k 0 0 ω . −0.684 m 0.0707 m 0

v B i = −(30 m/s) i +

B 0.0707 m

Equating j components in this equation yields the equation 0 = (−0.684 m)ω,

so the angular velocity of BC is zero. Then equating i components gives the velocity of B:

0.168 m

v B = −30 i (m/s). With this result we can determine the counterclockwise angular velocity of the crank. ω AB =

30 m/s = 424 rad/s. 0.0707 m

424 rad/s counterclockwise.

Problem 17.37 The bar BC is rotating in the clockwise direction at 60 rad/s. At the instant shown, what is the angular velocity of bar AB? 350 mm

Solution:

In terms of the counterclockwise angular velocity ω AB of bar AB, the velocity of point B is v B = −ω AB (0.2 m) i. Let us express the velocity of point C in terms of the velocity of point D: i 0

v C = v D + ω CD × rC /D = 0 +

j 0

−0.35 m 0.35 m

B 200 mm D

k ω CD

300 mm

350 mm

= −ω CD (0.35 m) i − ω CD (0.35 m) j. Now we express the velocity of point C in terms of the velocity of point B: v C = v B + ω BC × rC /B :

−ω CD (0.35 m) = ω BC (0.3 m).

−ω CD (0.35 m) i − ω CD (0.35 m) j = i 0

−ω AB (0.2 m) i +

j 0

0.3 m 0.15 m

304

obtaining the two equations −ω CD (0.35 m) = −ω AB (0.2 m) − ω BC (0.15 m), Setting ω BC = −60 rad/s and solving, we obtain ω AB = 135 rad/s,

k ω BC 0

ω CD = 51.4 rad/s. ω AB = 135 rad/s counterclockwise.

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Problem 17.38 The disk is rolling without slipping in the clockwise direction at 10 rad/s. Determine the angular velocities of bars AB and BC.

A 3m

Solution:

Using the known angular velocity of the disk, the veloc-

ity of C is

v C = (1 m)(10 rad/s) i = 10 i (m/s). Let us write the velocity of B in terms of the velocity of A, v B = v A + ω AB × rB /A = 0+

i j k 0 0 ω AB 1 m −2 m 0

= 2ω AB i + ω AB j,

and also write the velocity of B in terms of the velocity of C,

v B = v C + ω BC × rB /C = 10 i (m/s) +

i j k 0 0 ω BC −3 m 0 0

= 10 i − 3ω BC j.

By equating our two equations for the velocity of B, we obtain the two equations 2ω AB = 10, ω AB = −3ω BC . We see that ω AB = 5 rad/s, ω BC = −1.67 rad/s. AB : 5 rad/s counterclockwise. BC : 1.67 rad/s clockwise.

Problem 17.39* At the instant shown, the bar BC is horizontal and the x component of the velocity of the midpoint G of bar BC is 20 in/s. Determine the y component of the velocity of G and the angular velocity of bar AB.

Solution:

Let L CD be the length of bar CD. Because bar BC is horizontal at the instant shown, we can see that L CD sin 30 ° = L AB sin 45 °, which yields L CD = 14.1 in.

In terms of the (unknown) angular velocity ω AB of bar AB, the velocity of point B is v B = v A + ω AB × rB /A

y B

12 in G

= 0+

j 0

k ω AB

L AB cos45 ° L AB sin 45 °

(1)

= −ω AB L AB sin 45 ° i + ω AB L AB cos45 ° j.

10 in A

i 0

458

308

Now we express the velocity of point C in terms of the velocity of point D: v C = v D + ω CD × rC /D = 0+

i 0

j 0

−L CD cos30 ° L CD sin 30 °

k ω CD 0

(2)

= −ω CD L CD sin 30 °i − ω CD L CD cos30 ° j.

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17.39* (Continued) Then we express the velocity of point C in terms of the velocity of point B, using the results (1) and (2): v C = v B + ω BC × rC /B : − ω CD L CD sin 30 °i = −ω AB L AB sin 45 °i + ω AB L AB cos 45 ° j +

j k 0 ω BC 0 0

v Gx = −ω AB L AB sin 45 °, (5) v Gy = ω AB L AB cos 45 ° + ω BC

− ω CD L CD cos30 ° j i 0 L BC

The i and j components of this equation gives the two equations

1 L . (6) 2 BC

Setting v Gx = 20 in/s in Eq. (5), we obtain ω AB = −2.83 rad/s. Then solving Eqs. (3) and (4) yields ω BC = 4.55 rad/s, ω CD = −2.83 rad/s. From Eq. (6), v Gy = 7.32 in/s.

ω AB = 2.83 rad/s clockwise, v Gy = 7.32 in/s.

Equating i and j components, we obtain the equations −ω CD L CD sin 30 ° = −ω AB L AB sin 45 °, (3) −ω CD L CD cos30 ° = ω AB L AB cos 45 ° + ω BC L BC . (4) Finally, we express the velocity of point G in terms of the velocity of point B: v G = v B + ω BC × rG /B : v Gx i + v Gy j = −ω AB L AB sin 45 °i i 0

+ ω AB L AB cos 45 ° j +

1 L 2 BC

j k 0 ω BC 0

= −ω AB L AB sin 45 °i + ω AB L AB cos 45 ° j 1 + ω BC L BC j. 2

Problem 17.40 At the instant shown, point E is moving upward at 5 m/s. Determine the angular velocities of bars AB and BC.

Solution: Using the known velocity of point E, the counterclockwise angular velocity of bar CDE is

400 mm C

5 m/s ω CDE = = 7.14 rad/s. 0.7 m Therefore the velocity of point C is

700 mm

v C = −(7.14 rad/s)(0.4 m) j = −2.86 j (m/s).

400 mm

700 mm

In terms of the angular velocity of bar AB, the velocity of B is v B = −ω AB (0.4 m) i. Let us write the velocity of B in terms of the velocity of C: v B = v C + ω BC × rB /C : −ω AB (0.4 m) i = −2.86 j (m/s) +

i j k 0 0 ω BC . −0.7 m 0.4 m 0

This yields the two equations −ω AB (0.4 m) = −ω BC (0.4 m), 0 = −2.86 m/s − ω BC (0.7 m). We see that ω AB = ω BC = −4.08 rad/s. AB : 4.08 rad/s clockwise. BC : 4.08 rad/s clockwise.

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Problem 17.41 Bar AB rotates at 4 rad/s in the counterclockwise direction. Determine the velocity of point C.

C B

400 mm D

Solution: The angular velocity of bar AB is ω = 4k (rad/s). The radius vector AB is rB /A = 300 i + 600 j (mm). The velocity of point B is

600 mm

600 mm 300 mm

from which v B = −2400 i + 1200 j (mm/s). The vector radius from B to C is rC /B = 600 i + (900 − 600) j = 600 i + 300 j (mm). The velocity of point C is

The radius vector from C to D is rD /C = 200 i − 400 j (mm). The velocity of point D is  i j k   0 v D = v C +  0 ω BC   200 −400 0   = v C + 400 ω BC i + 200 ω BC j (mm/s). The radius vector from E to D is rD /E = −300 i + 500 j (mm). The velocity of point D is

Problem 17.42 The upper grip and jaw of the pliers ABC is stationary. The lower grip DEF is rotating at 0.2 rad/s in the clockwise direction. At the instant shown, what is the angular velocity of the lower jaw CFG ?

 i j k   0 4  , v B = ω AB × rB /A =  0  300 600 0   

 i j k   v C = v B +  0 0 ω BC   600 300 0   = (−2400 − 300 ω BC ) i + (1200 + 600 ω BC ) j (mm/s).

500 mm

200 mm

 i j k  0 0 ω DE v D = ω DE × rD /E =   −300 500 0  = −500 ω DE i − 300 ω DE j (mm/s).

v E = v B + ω BE × rE / B = 0 + ω BE k × (0.07 i − 0.03 j) m

300 mm

     

Equate the expressions for the velocity of point D; solve for v C , to obtain one of two expressions for the velocity of point C. Equate the two expressions for v C , and separate components: 0 = (−500 ω DE − 100 ω BC + 2400) i, 0 = (1200 + 300 ω DE + 800 ω BC ) j. Solve ω DE = 5.51 rad/s, ω BC = −3.57 rad/s. Substitute into the expression for the velocity of point C to obtain v C = −1330 i − 941 j (mm/s).

Stationary A

30 mm G E

Solution:

70 mm

30 mm 30 mm

= ω BE (0.03i + 0.07 j) m v F = v E + ω DEF × rF / E = ω BE (0.03i + 0.07 j) m + (−0.2 rad/s) k × (0.03 m) i = [(0.03 m)ω BE i + (−0.006 m/s + {0.07 m}ω BE ) j] v C = v F + ω CFG × rC / G = [(0.03 m)ω BE i + (−0.006 m/s + {0.07 m}ω BE ) j] + ω CFG k × (0.03 m) j = [(0.03 m)(ω BE − ω CFG ) i + (−0.006 m/s + {0.07 m}ω BE ) j] Since C is fixed we have (0.03 m)(ω BE − ω CFG ) = 0 ω BE = 0.0857 rad/s −0.006 m/s + { 0.07 m } ω BE = 0 ⇒ ω CFG = 0.0857 rad/s So we have ω CFG = 0.0857 rad/s CCW.

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Problem 17.43 The horizontal member ADE supporting the scoop is stationary. The link BD is rotating in the clockwise direction at 2 rad/s. What is the angular velocity of the scoop? Solution:

C 2 ft

1 ft 6 in E

D 5 ft

The velocity of B in terms of the velocity of D is

2 ft 6 in

1 ft

Scoop

v B = v D + ω BD k × rB /D = 0 − (2 rad/s)k × ( i + 2 j) = 4 i − 2 j (ft/s). We can express the velocity of C in terms of the velocities of both B and D to determine ω CE :

vBC

v C = v B + ω BC k × rC /B = (4 i − 2 j) + ω BC k × (2.5i − 0.5 j) = [(4 + 0.5ω BC ) i + (−2 + 2.5ω BC j)] (ft/s)

vCE

1 rad/s

and

v C = v E − ω CE k × rC /E = 0 − ω CE k × (1.5 j) = 1.5ω CE i (ft/s).

Equating these results for the velocity of C, we obtain x: 4 + 0.5ω BC = 1.5ω CE and y: −2 + 2.5ω BC = 0. Solving yields ω BC = 0.8 rad/s and ω CE = 2.93 rad/s. 2.93 rad/s clockwise.

Problem 17.44 The radius of the disk is 1.2 m, and the length of the bar AB is 2.5 m. The disk is rolling without slipping, and point B slides on the plane surface. Determine the angular velocity of the bar AB and the velocity of point B.

4 rad/s A

Solution:

The contact point, P, is stationary. We have r A /P = ( 1.2 i + 1.2 j ) m, and v A = v p + ωk × r A /P = 0 + 4 k × (1.2 i + 1.2 j) = (−4.8i + 4.8 j) (m/s).

We need the vector rB /A . An easy way to determine this is to analyze the triangle ABP. Note that angle BPA is 45 ° and r A /P = 1.2 2 + 1.2 2 = 1.70 m. If we create two right triangles by dropping a perpendicular from point A to the plane surface, at C, we have, on the right, a right triangle with one side of length 1.2 m and hypotenuse 2.5 m. The other side of this triangle, x, along the horizontal plane, is given by x 2 = 2.5 2 − 1.2 2. Solving, we get x = 2.19 ft. Thus,

y A O

x B

rB /P = (1.2 + 2.19) i = 3.39 i m

and rB /A = ( 2.19 i − 1.2 j ) m.

We need the angular velocity of bar AB. We note that v B = v B i because B slides on the horizontal surface. We have v B i = v A + ω AB × rB /A = v A + ω AB k × (2.19 i − 1.2 j) = [(−4.8 + 1.2ω AB ) i + (4.8 + 2.19ω AB ) j] (m/s) or, in component form, x: v B = −4.8 + 1.2ω AB and y: 0 = 4.8 + 2.08ω AB . Solving, we obtain v B = −7.43i (m/s) and ω AB = −2.19k (rad/s).

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Problem 17.45 A motor rotates the circular disk mounted at A, moving the saw back and forth. (The saw is supported by a horizontal slot so that point C moves horizontally.) The radius AB is 5 in, and the link BC is 15 in long. In the position shown, θ = 40 ° and the link BC is horizontal. If the angular velocity of the disk is two revolutions per second counterclockwise, what is the velocity of the saw?

Solution: Let the radius AB be denoted by r = 5 in. The coordinates of point B are x B = r cos θ, y B = r sin θ, and the y coordinate of C is r sin θ. The length of the link BC is L = 15 in. The x coordinate of C can be determined from the equation ( x B − x C ) 2 + ( y B − y C ) 2 = L2 . The counterclockwise angular velocity of the disk is ω = 2(2π ) rad/s. The magnitude of the velocity of B is ωr, and its components are v Bx = −ωr sin θ,

y B

u x

we obtain the two equations v C = v Bx − ω BC ( y C − y B ), 0 = v By + ω BC ( x C − x B ). Solving, we obtain ω BC = 3.21 rad/s, v C = −40.4 in/s. 40.4 in/s toward the left.

v By = ωr cos θ.

Expressing the velocity of C in terms of the velocity of B, v C = v B + ω BC × rC /B : v C i = v Bx i + v By j +

i 0 xC − x B

j 0 yC − y B

k ω BC , 0

Problem 17.46 In Problem 17.45, what is the velocity of the saw when the disk has rotated to θ = 60 ° ?

y B

Solution:

Let the radius AB be denoted by r = 5 in. The coordinates of point B are x B = r cos θ, y B = r sin θ, and the y coordinate of C is r sin 40 °. The length of the link BC is L = 15 in. The x coordinate of C can be determined from the equation

u A

( x B − x C ) 2 + ( y B − y C ) 2 = L2 . The counterclockwise angular velocity of the disk is ω = 2(2π ) rad/s. The magnitude of the velocity of B is ωr, and its components are v Bx = −ωr sin θ,

we obtain the two equations

v By = ωr cos θ.

Expressing the velocity of C in terms of the velocity of B, v C = v B + ω BC × rC /B : v C i = v Bx i + v By j +

i 0 xC − x B

j 0 yC − y B

k ω BC 0

v C = v Bx − ω BC ( y C − y B ), 0 = v By + ω BC ( x C − x B ). Solving, we obtain ω BC = 2.10 rad/s, v C = −52.1 in/s.

Problem 17.47 The disks roll without slipping on the plane surface. The angular velocity of the left disk is 2 rad/s in the clockwise direction. What is the magnitude of the velocity of the midpoint of the 3-ft rod?

52.1 in/s toward the left.

2 rad/s

3 ft

1 ft

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17.47 (Continued) Solution:

The angular velocity of the left disk is ω L = −2k (rad/s). Points P and Q are stationary.

r R /L =

8i + j (ft) .

vrod

2 rad/s

The distance from point L to the center of the right disk is d = 3 2 − 1 2 = 8 ft. The vector from L to R is

The velocity of point L is

v L = ω L k × ( i + j) = −2k (rad/s) × ( i + j)(ft) = 2 i − 2 j (ft/s). The velocity of point R can be written in terms of point L and also in terms of point Q. The results can be equated to determine the angular velocities of both the rod and the right-hand disk. The velocity of point R is v R = v L + ω rod × rR /L : v R = (2 i − 2 j)(ft/s) + ω rod k(rad/s) × ( 8i + 1 j)(ft) = [(2 − ω rod ) i + (−2 +

8ω rod ) j] (ft/s).

Now we can obtain the velocity of the midpoint M of the 3-ft rod. Writing it in terms of point L, v M = v L + ω rod ×

( 12 r ) : R /L

v M = (2 i − 2 j)(ft/s) + ω rod k(rad/s) × (0.5 8i + 0.5 j)(ft)

Also,

= [ (2 − 0.5ω rod ) i + (−2 + 0.5 8ω rod ) j ](ft/s).

v R = v Q + ω R × rR /Q = ω R k (rad/s) × 2 j (ft) = −2ω R i (ft/s).

From this expression we obtain v M = 1.65 i − j(ft/s). Its magnitude is

Equating these two expressions, we obtain the equations

v M = 1.93 ft/s.

x : − 2ω R = (2 − ω rod ) and y : 0 = −2 +

8ω rod .

Solving, we get ω R = −0.646k (rad/s) and ω rod = 0.707k (rad/s).

1.93 ft/s.

Problem 17.48 The disk rolls on the curved surface. The bar rotates at 10 rad/s in the counterclockwise direction. Determine the velocity of point A.

A 10 rad/s

40 mm

Solution:

The radius vector from the left point of attachment of the bar to the center of the disk is rbar = 120 i (mm). The velocity of the center of the disk is

 i j k    v O = ω bar × rbar = 10(120)  0 0 1  = 1200 j (mm/s).  1 0 0   

120 mm

The radius vector from the point of contact with the disk and the curved surface to the center of the disk is rO /P = −40 i (m). The velocity of the point of contact of the disk with the curved surface is zero, from which  i j k  0 ωO v O = ω O × rO /P =  0  −40 0 0 

   = −40 ω j. O   

Equate the two expressions for the velocity of the center of the disk and solve: ω O = −30 rad/s. The radius vector from the center of the disk to point A is r A /O = 40 j (mm). The velocity of point A is

y 10 rad/s

 i j k    v A = v O + ω O × r A /O = 1200 j − (30)(40)  0 0 1   0 1 0    = 1200 i + 1200 j (mm/s).

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Problem 17.49 If ω AB = 2 rad/s and ω BC = 4 rad/s, what is the velocity of point C, where the excavator’s bucket is attached? Solution:

The radius vector AB is

y B vBC

vAB

5.5 m

rB /A = 3i + (5.5 − 1.6) j = 3i + 3.9 j (m).

The velocity of point B is

1.6 m

 i j k    v B = ω AB × rB /A =  0 0 2  = −7.8i + 6 j (m/s).  3 3.9 0   

x 4m

2.3 m

The radius vector BC is rC /B = 2.3i + (5 − 5.5) j = 2.3i − 0.5 j (m). The velocity at point C is  i j k   v C = v B + ω BC × rC /B = −7.8i + 6 j +  0 0 −4   2.3 −0.5 0    = −9.8i − 3.2 j (m/s).

Problem 17.50 If ω AB = 2 rad/s, what clockwise angular velocity ω BC will cause the vertical component of the velocity of point C to be zero? What is the resulting velocity of point C ? Solution:

y B vBC

vAB

5.5 m

Use the solution to Problem 17.49. The velocity of A

point B is v B = −7.8i + 6 j (m/s).

1.6 m x

The velocity of point C is v C = v B + ω BC × rC /B

2.3 m

  i j k   0 = −7.8i + 6 j +  0 −ω BC  ,   2.3 −0.5 0   v C = (−7.8 − 0.5ω BC ) i + (6 − 2.3ω BC ) j (m/s). For the vertical component to be zero, ω BC =

6 = 2.61 rad/s clockwise. 2.3

The velocity of point C is v C = −9.1i (m/s).

Problem 17.51 The steering linkage of a car is shown. Member DE rotates about the fixed pin E. The right brake disk is rigidly attached to member DE. The tie rod CD is pinned at C and D. At the instant shown, the Pitman arm AB has a counterclockwise angular velocity of 1 rad/s. What is the angular velocity of the right brake disk?

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17.51 (Continued) Brake disks

A B

180 mm

100 mm E 220 mm

C 460 mm

Steering link

340 mm

200 mm 70 mm

Solution:

Note that the steering link moves in translation only. Thus v B = v C .

Point E is not moving. Equating the components of the velocity of point E to zero, we have

v C = v B = v A + ω AB × rB / A = 0 + (1k) × (−0.18 j) = 0.18i

0.18 + 0.08ω CD − 0.2ω DE = 0, 0.34 ω CD + 0.07ω DE = 0.

v D = v C + ω CD × rD / C = 0.18i + ω CD k × (0.34 i − 0.08 j)

Solving these equations simultaneously, we find that

= (0.18 + 0.08ω CD ) i + (0.34 ω CD ) j v E = v D + ω DE × rE / D

ω CD = −0.171 rad/s, ω DE = 0.832 rad/s. The angular velocity of the right brake disk is then

= (0.18 + 0.08ω CD ) i + (0.34 ω CD ) j + ω DE k × (0.07 i + 0.2 j) = (0.18 + 0.08ω CD − 0.2ω DE ) i + (0.34 ω CD + 0.07ω DE ) j

ω disk = ω DE = 0.832 rad/s counterclockwise.

Problem 17.52 An athlete exercises her arm by raising the mass m. The shoulder joint A is stationary. The distance AB is 300 mm, and the distance BC is 400 mm. At the instant shown, ω AB = 1 rad/s and ω BC = 2 rad/s. How fast is the mass m rising?

608

vAB

Solution:

The magnitude of the velocity of the point C parallel to the cable at C is also the magnitude of the velocity of the mass m. The radius vector AB is rB /A = 300 i (mm). The velocity of point B is  i j k  v B = ω AB × rB /A =  0 0 ω AB  300 0 0 

308

vBC

   = 300 j (mm/s).   

The radius vector BC is rC /B = 400( i cos60 ° + j sin 60 °) = 200 i + 346.4 j (mm). The velocity of point C is  i j k   v C = v B + ω BC × rC /B = 300 j +  0 0 2   200 346.4 0    = −692.8i + 700 j (mm/s). The unit vector parallel to the cable at C is e C = − i cos30 ° + j sin 30 ° = −0.866 i + 0.5 j. The component of the velocity parallel to the cable at C is v C ⋅ e C = 950 mm/s , which is the velocity of the mass m.

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Problem 17.53 An athlete exercises her arm by raising the mass m. The shoulder joint A is stationary. The distance AB is 310 mm, and the distance BC is 390 mm. At the instant shown, ω AB = 1.2 rad/s and the mass is rising at 1 m/s. Determine the angular velocity ω BC .

C 308

vBC

608

vAB

Solution:

The velocity of B is

v B = ω AB (0.310 m) = 0.372 j (m/s). The velocity of C in terms of the velocity of B is v C = v B + ω BC × rC /B i 0

= 0.372 j (m/s) +

j 0

k ω BC

(0.39 m) cos60 ° (0.39 m)sin 60 °

= −ω BC (0.39 m)sin 60°i + [ 0.372 (m/s) + ω BC (0.39 m) cos60° ] j. A unit vector parallel to the cable at C is e = − cos30 °i + sin 30 ° j. The rate at which the mass is rising is equal to 1 m/s = e ⋅ v C = (− cos30 °)[ −ω BC ( 0.39 m ) sin 60° ] + (sin 30°)[0.372 (m/s) + ω BC (0.39 m) cos60°]. Solving this equation, we obtain ω BC = 2.09 rad/s. ω BC = 2.09 rad/s.

Problem 17.54 Points B and C are in the x–y plane. The angular velocity vectors of the arms AB and BC are ω AB = −0.3k (rad/s) and ω BC = 0.5k (rad/s). What is the velocity of point C ?

Solution:

Point A is fixed so v A = 0. The velocity of C is v C = v A + v B /A + v C /B , where

408

v B /A = ω AB × rB /A

= [(−0.3k) × (0.76)(cos 40 °i + sin 40 ° j)] (m/s), or v B /A = (0.147 i − 175 j) (m/s). Also, we have

308

v C /B = ω BC × rC /B = [(0.5k) × (0.92)(cos30°i + sin 30° j)] (m/s) or v C /B = (0.230 i + 0.398 j) (m/s). Adding, we obtain v C = 0.377 i + 0.224 j (m/s).

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Problem 17.55 If the velocity of point C of the robotic arm is v C = −0.15i + 0.42 j (m/s), what are the angular velocities of arms AB and BC ?

Solution:

From the solution to Problem 17.54,

308

rB /A = 0.582 i + 0.489 j (m) 408

rC /B = 0.797 i − 0.460 j (m) v B = ω AB k × rB /A

(v A = 0)

v C = v B + ω BC k × rC /B We are given

v C = −0.15i + 0.42 j + 0 k (m/s). Thus, we know everything in the v C equation except ω AB and ω BC . v C = ω AB k × rB /A + ω BC k × rC /B This yields two scalar equations in two unknowns i and j components. Solving, we get ω AB = 0.476 k (rad/s), ω BC = 0.179k (rad/s).

Problem 17.56 The link AB of the robot’s arm is rotating at 2 rad/s in the counterclockwise direction, link BC is rotating at 4 rad/s in the clockwise direction, and link CD is rotating at 1 rad/s in the counterclockwise direction. What is the velocity of point D ?

250

mm 308

208

C A

Solution:

The angular velocities, in vector form, are ω AB = 2k (rad/s), ω BC = −4 k (rad/s), and ω CD = k (rad/s).

x 250 mm

Point A is fixed. so v A = 0. The velocity of D is v D = v A + v B /A + v C /B + v D /C where v B /A = ω AB × rB /A = [(2k) × (0.300)(cos30 °i + sin 30 ° j)] (m/s) = (−0.300 i + 0.520 j) (m /s), v C /B = ω BC × rC /B = [(−4 k) × (0.250)(cos 20 °i − sin 20 ° j)] (m/s) = (−0.342 i − .940 j) (m/s), and v D /C = ω CD × rD /C = [(1k) × (0.250) i] (m /s) = 0.250 j (m/s). Adding up the parts, we obtain v D = −0.642 i − 0.170 j (m/s).

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Problem 17.57 The person squeezes the grips of the shears, causing the angular velocities shown. What is the resulting angular velocity of the jaw BD? Solution: v D = v C + ω CD × rD /C = 0 − (0.12 rad/s)k × (0.025i + 0.018 j) m = (0.00216 i − 0.003 j) m/s v B = v D + ω BD × rB /D = (0.00216 i − 0.003 j) m/s + ω BD k × (−0.05i − 0.018 j) m = (0.00216 m/s + {0.018 m}ω BD ) i + (−0.003 m/s − {0.05 m}ω BD ) j From symmetry we know that C and B do not move vertically −0.003 m/s − {0.05 m}ω BD = 0 ⇒

0.12 rad/s

ω BD = −0.06 rad/s

ω BD = 0.06 rad/s CW. B

18 mm

C E

0.12 rad/s

25 mm 25 mm

Problem 17.58 Determine the velocity v W and the angular velocity of the small pulley.

50 mm

Solution:

Since the radius of the bottom pulley is not given, we cannot use Eq (17.6) (or the equivalent). The strategy is to use the fact (derived from elementary principles) that the velocity of the center of a pulley is the mean of the velocities of the extreme edges, where the edges lie on a line normal to the motion, taking into account the directions of the velocities at the extreme edges. The center rope from the bottom pulley to the upper pulley moves upward at a velocity of v W . Since the small pulley is fixed, the velocity of the center is zero, and the rope to the left moves downward at a velocity v W , from which the left edge of the bottom pulley is moving at a velocity v W downward. The right edge of the bottom pulley moves upward at a velocity of 0.6 m/s. The velocity of the center of the bottom pulley is the mean of the veloc0.6 − v W ities at the extreme edges, from which v W = . 2 0.6 Solve: v W = = 0.2 m/s. 3

0.6 m/s

100 mm

The angular velocity of the small pulley is ω =

vW 0.2 = = 4 rad/s. r 0.05

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Problem 17.59 Determine the velocity of the block and the angular velocity of the small pulley.

2 in

9 in/s

3 in

Solution:

Denote the velocity of the block by v B . The strategy is to determine the velocities of the extreme edges of a pulley by determining the velocity of the element of rope in contact with the pulley. The upper rope is fixed to the block, so that it moves to the right at the velocity of the block, from which the upper edge of the small pulley moves to the right at the velocity of the block. The fixed end of the rope at the bottom is stationary, so that the bottom edge of the large pulley is stationary. The center of the large pulley moves at the velocity of the block, from which the upper edge of the bottom pulley moves at twice the velocity of the block (since the velocity of the center is equal to the mean of the velocities of the extreme edges, one of which is stationary) from which the bottom edge of the small pulley moves at twice the velocity of the block. The center of the small pulley moves to the right at 9 in/s. The velocity of the center of the small pulley is the mean of the velocities at the extreme edges, from which 9 =

from which vB =

2 9 = 6 in/s. 3

The angular velocity of small pulley is given by

2v B + v B 3 = v B, 2 2

Problem 17.60 The device shown below is used in the semiconductor industry to polish silicon wafers. The wafers are placed on the faces of the carriers. The outer and inner rings are then rotated, causing the wafers to move and rotate against an abrasive surface. If the outer ring rotates in the clockwise direction at 7 rpm and the inner ring rotates in the counterclockwise direction at 12 rpm, what is the angular velocity of the carriers?

 i j k    9 i = 2v B i + ω × 2 j =  0 0 ω  = 2v B i − 2ω i,  0 2 0    12 − 9 from which ω = = 1.5 rad/s. 2

Outer ring 1.0 m 0.6 m

12 rpm Carriers (3) Inner ring

Solution:

The velocity of pt. B is v B = (1 m)ω 0 i = ω 0 i. The velocity of pt. A is v A = −(0.6 m)ω i i. Then v B = v A + ω C × rB /A : ω 0 i = −0.6ω i i +

i j k 0 0 ωC 0 0.4 0

7 rpm

. y

The i component of this equation is ω 0 = −0.6ω i − 0.4 ω C , −0.6ω i − ω 0 0.4 −0.6(12 rpm) − 7 rpm = 0.4 = −35.5 rpm.

so ω C =

C A vi

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Problem 17.61 Suppose that the outer ring rotates in the clockwise direction at 5 rpm and you want the centerpoints of the carriers to remain stationary during the polishing process. What is the necessary angular velocity of the inner ring?

Outer ring

Solution:

See the solution of Problem 17.60. The velocity of pt. B is v B = ω 0 i and the angular velocity of the carrier is ωC =

−0.6ω i − ω 0 . 0.4

We want the velocity of pt. C to be zero: v C = 0 = v B + ω C × rC /B = ω 0 i +

i j k 0 0 ωC 0 −0.2 0

From this equation we see that ω C = −5ω 0 . Therefore the velocity of pt. A is

1.0 m 0.6 m

v A = v C + ω C × r A /C = 0 + (−5ω 0 k) × (−0.2 j) = −ω 0 i

12 rpm Carriers (3)

We also know that v A = −(0.6 m)ω i i,

Inner ring

ωi =

ω0 5 rpm = = 8.33 rpm. 0.6 0.6

7 rpm

Problem 17.62 The ring gear is fixed, and the hub and planet gears are bonded together. The connecting rod is rotating in the counterclockwise direction at 30 rpm. Determine the magnitude of the velocity of point A.

Planet gear Hub gear 140 mm

Connecting rod

340 mm 240 mm

Solution:

Denote the contact points between the sun gear and the planet gear by the subscript SP and the point of contact between the hub gear and the ring gear by the subscript HR. The angular velocity of the connecting rod is ω CR = 30

rev ( min )2π ( rad ) 1 ( min )k = π k (rad/s). rev 60 sec

720 mm

Sun gear Ring gear

The vector from the center of the sun gear to the center of the hub gear is rHub /Sun = (0.720 − 0.140) j = 0.580 j (m). The velocity of the center of the gear hub is

v Hub = ω CR × rHub /Sun = π k (rad/s) × 0.580 j(m) = −1.822 i (m/s)., or

v HR = 0 = v Hub + ω Hub k × 0.140 j = (−1.822 − 0.140 ω Hub ) i . Solving, we obtain ω Hub = −13.014 (rad/s) or ω Hub = −13.014 k (rad/s)..

Hub vHub

This is also the angular velocity of the planet bear. The velocity of point A is then given by

vCR

v A = v Hub + ω Hub k × (0.340) j = −1.822 i + (−13.014 k) × (0.200 j) = 2.60 i (m/s). Sun

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Problem 17.63 The large gear is fixed. Bar AB has a counterclockwise angular velocity of 2 rad/s. What are the angular velocities of bars CD and DE ?

4 in

10 in

16 in D

B 4 in

Solution:

The strategy is to express vector velocity of point D in terms of the unknown angular velocities of CD and DE, and then to solve the resulting vector equations for the unknowns. The vector distance AB is rB /A = 14 j (in.) The linear velocity of point B is

10 in

2 rad/s E

 i j k    v B = ω × rB /A =  0 0 2  = −28i (in/s).  0 14 0    The lower edge of gear B is stationary. The radius vector from the lower edge to B is rB = 4 j (in.), The angular velocity of B is  i j k  v B = ω B × rB =  0 0 ω B  0 4 0 

   = −4 ω i (in/s), B   

The velocity of point D is also given by  i j k  0 ω DE v D = ω DE × rD /E =  0  −10 14 0  = −14 ω DE i − 10 ω DE j (in/s).

v from which ω B = − B = 7 rad/s. The vector distance from B to C 4 is rC /B = 4 i (in.). The velocity of point C is

Equate components:

 i j k    v C = v B + ω B × rC /B = −28i +  0 0 7   4 0 0    = −28i + 28 j (in/s).

Solve:

     

(−28 + 14 ω DE ) i = 0, (16ω CD + 28 + 10 ω DE ) j = 0. ω DE = 2 rad/s, ω CD = −3 rad/s.

The negative sign means a clockwise rotation.

The vector distance from C to D is rD /C = 16 i (in.), and from E to D is rD /E = −10 i + 14 j (in.). The linear velocity of point D is  i j k  v D = v C + ω CD × rD /C = −28i + 28 j +  0 0 ω CD  16 0 0  = −28i + (16ω CD + 28) j (in/s).

     

Problem 17.64 If the bar has a clockwise angular velocity of 10 rad/s and v A = 20 m/s, what are the coordinates of the instantaneous center of the bar, and what is the value of v B?

y vA

Solution:

Let the coordinates of the instantaneous center of the bar be denoted by ( x C , 0) and let w = −10 rad/s be its angular velocity. We can see that

A 1m

B 1m

v A = w (1 m − x C ), v B = w (2 m − x C ). Solving these two equations, we obtain x C = 3 m, v B = 10 m/s. ( x C , y C ) = (3, 0) m, v B = 10 m/s.

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Problem 17.65 If v A = 16 m/s and v B = 48 m/s, what are the coordinates of the instantaneous center of the bar and what is its angular velocity?

y vA

Solution:

Let the coordinates of the instantaneous center of the bar be denoted by ( x C , 0) and let ω be its counterclockwise angular velocity. We can see that

B 1m

v A = ω (1 m − x C ), v B = ω (2 m − x C ). Solving these two equations, we obtain x C = 0.5 m, ω = 32 rad/s. ( x C , y C ) = (0.5, 0) m, 32 rad/s counterclockwise.

Problem 17.66 The velocity of point O of the bat is v O = −4 i + 0.5 j (ft/s), and the bat is rotating about the z-axis with a counterclockwise angular velocity of 6 rad/s. What are the x and y coordinates of the bat’s instantaneous center?

Solution:

Let the coordinates of the instantaneous center be denoted by ( x C , y C ). The vector from C to O is rO /C = −x C i − y C j (ft),

and the velocity of point O is can be written

v O = −4 i + 0.5 j (ft/s) = ω k × rO /C =

i 0 −x C

j 0 −y C

k ω . 0

This yields the two equations

−4 = ω y C , 0.5 = −ω x C , giving x C = −0.0833 ft, y C = −0.667 ft. ( x, y ) = (−0.0833, − 0.667) ft.

Problem 17.67 Points A and B of the 2-m bar slide on the plane surfaces. The velocity of point B is v B = 6 i (m/s). Use the instantaneous center to determine the components of the velocity of the midpoint G of the bar at the instant shown.

708

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17.67 (Continued) y

Solution:

Let L = 2 m. Drawing lines from points A and B perpendicular to their directions of motion, we locate the instantaneous center C. The distance from C to point B is L cos 20 °, so the angular velocity of the bar about its instantaneous center is ω =

vB 6 m/s = = 3.19 rad/s. L cos 20 ° (2 m) cos 20 °

208

From the geometry, we can see that the distance from C to the midpoint G is L/2 = 1 m. The magnitude of the velocity of G is therefore

208 208

L v G = ω = 3.19 m/s. 2 The velocity of G is

v G = v G cos 20 °i − v G sin 20 ° j

= 3 i − 1.09 j (m/s). v G = 3 i − 1.09 j (m/s).

Problem 17.68 The bar is in two-dimensional motion in the x–y plane. The velocity of point A is v A = 12 i + 3 j (ft/s) and B is moving in the direction parallel to the bar. Determine the velocity of B (a) by using Eq. (17.6) and (b) by using the instantaneous center of the bar.

B 4 ft

308

Solution:

(a) The velocity of point B can be written as v B = v B cos30 °i + v B sin 30 ° j (ft/s). The position vector of B relative to A is rB / A = 4 cos30 °i + 4 sin 30 ° j (ft) = 3.46 i + 2 j (ft). The velocity of B in terms of A is

v B = ωk × rB /C = ωk × ((3.46 − x C ) i + (2 − y C ) j) = [−(2 − y C )ω i + (3.46 − x C )ω j].

v B = v A + ωk × rB /A = [(12 i + 3 j) + ωk × (3.46 i + 2 j) ] (ft/s).

In terms of components, we have

For v A :

x : v B cos30 ° = 12 − 2ω

x : 12 = ω y C y : 3 = −ω x C

y : v B sin 30 ° = 3 + 3.46ω Solving, we get v B = 11.9 (ft/s) and ω = 0.850 (rad/s). counterclockwise, so

For v B :

v B = 10.3i + 5.95 j (ft/s).

y : v B sin 30 ° = (3.46 − x C )ω

(b) Let ( x C , y C ) denote the coordinates of the instantaneous center. This means that v C = 0. The position vector from C to A is r A /C = (−x C i − y C j)( ft ), and the position vector from C to B is rB /C = (3.46 − x C ) i + (2 − y C ) j (ft).

x : v B cos30 ° = −(2 − y C )ω We have four equations in four unknowns. Solving, we get ( x C , y C ) = (−3.52, 14.1) (ft) and v B = 11.9 (ft/s), so v B = 10.3i + 5.95 j (ft/s).

The velocity of B can be written v B = v B cos30 ° i + v B sin 30 °j (ft/s), and the velocities of A and B in terms of C are v A = (12 i + 3 j) (ft/s) = ωk × (−x C i − y C j) = ω y C i − ω x C j and

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Problem 17.69 Point A of the bar is moving at 5 m/s in the direction of the unit vector 0.966i − 0.259j, and the bar is rotating in the counterclockwise direction at 4 rad/s. (a) What is the velocity of point B? (b) What are the coordinates of the bar’s instantaneous center?

Solution: (a) The velocity of point B is v B = v A + ω k × r B /A = 5(0.966 i − 0.259 j) + 4 k × [2 cos(30 °) i + 2sin(30 °) j](m/s) = 0.830 i + 5.63 j (m/s). (b) Let the coordinates of the bar’s instantaneous center, C, be denoted by ( x C , y C ), and let the vector from C to A be denoted by r A /C = x A / C i + y A /C j.

We have v A = (5 m/s)(0.966 i − 0.259 j) = (4.83i − 1.30 j) (m/s).

We also know that v A = v C + ω × r A /C , where v C = 0.

Using these two expressions, we have v A = (4.83i − 1.30 j) (m/s) = 4 k × ( x A /C i + y A /C j)

308 A

= (−4 y A /C i + 4 x A /C j) (m/s). In component form, x : 4.83 = −4 y A /C and y : −1.30 = 4 x A /C . Hence ( x C , y C ) = (0.324, 1.21) (m). (a) v B = 0.830 i + 5.63 j (m/s). (b) ( x C , y C ) = (0.324, 1.21) m.

Problem 17.70 Bar AB rotates with a counterclockwise angular velocity of 10 rad/s. At the instant shown, what are the angular velocities of bars BC and CD? (See Practice Example 17.5.)

2m 10 rad/s

Solution:

The location of the instantaneous center for BC is shown, along with the relevant distances. Using the concept of the instantaneous centers we have

A 2m

v B = (10)(2) = ω BC (4) ω BC = 5 rad/s v C = (4.472)(5) = 2.236(ω CD ) ω CD = 10 rad/s We determine the directions by inspection ω BC = 5 rad/s clockwise. ω CD = 10 rad/s counterclockwise.

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Problem 17.71 Use instantaneous centers to determine the horizontal velocity of B.

Solution:

The instantaneous center of OA lies at O, by definition, since O is the point of zero velocity, and the velocity at point A is parallel to the x-axis:  i j k  v A = ω OA × r A /O =  0 0 −ω OA  0 6 0 

1 rad/s

A 6 in

   = 6 i (in/s).   

A line perpendicular to this motion is parallel to the y axis. The point B is constrained to move on the x axis, and a line perpendicular to this motion is also parallel to the y axis. These two lines will not intersect at any finite distance from the origin, hence at the instant shown the instantaneous center of bar AB is at infinity and the angular velocity of bar AB is zero. At the instant shown, the bar AB translates only, from which the horizontal velocity of B is the horizontal velocity at A: v B = v A = 6 i (in/s).

O 12 in

Problem 17.72 When the mechanism in Problem 17.71 is in the position shown here, use instantaneous centers to determine the horizontal velocity of B. 1 rad/s

Solution:

The strategy is to determine the intersection of lines perpendicular to the motions at A and B. The velocity of A is parallel to the bar AB. A line perpendicular to the motion at A will be parallel to the bar OA. From the dimensions given in Problem 17.71, the length of bar AB is r AB = 6 2 + 12 2 = 13.42 in. Consider the triangle OAB. The interior angle at B is  6  β = tan −1   = 24.1 °,  r AB 

B O

and the interior angle at O is θ = 90 ° − β = 65.9 °. The unit vector parallel to the handle OA is e OA = i cos θ + j sin θ, and a point on the line is L OA = L OA e OA , where L OA is the magnitude of the distance of the point from the origin. A line perpendicular to the motion at B is parallel to the y axis. At the intersection of the two lines L OA cos θ =

r AB , cos β

from which L OA = 36 in. The coordinates of the instantaneous center are (14.7, 32.9) (in.). Check: From geometry, the triangle OAB and the triangle formed by the intersecting lines and the base are similar, and thus the interior angles are known for the larger triangle. From the law of sines

A B

L OA r r AB = OB = = 36 in., sin 90 ° sin β sin β cos β and the coordinates follow immediately from L OA = L OA e OA . check. The vector distance from O to A is r A /O = 6( i cos θ + j sin θ) = 2.450 i + 5.478 j (in.). The angular velocity of the bar AB is determined from the known linear velocity at A.  i j k   0 0 v A = ω OA × r A /O =  −1   2.450 5.477 0    = 5.48i − 2.45 j (in/s). The vector from the instantaneous center to point A is r A /C = rOA − rC = 6e OA − (14.7 i + 32.86 j) = −12.25i − 27.39 j (in.)

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17.72 (Continued) The velocity at point A is  i j k  0 0 v A = ω AB × r A /C =  ω AB  −12.25 −27.39 0  = ω AB (27.39 i − 12.25 j) (ft/s).

     

Equate the two expressions for the velocity at point A and separate components, 5.48i = 27.39ω AB , − 2.45 j = −12.25ω AB j (one of these conditions is superfluous) and solve to obtain ω AB = 0.2 rad/s, counterclockwise.

[Check: The distance OA is 6 in. The magnitude of the velocity at A is ω OA (6) = (1)(6) = 6 in/s. The distance to the instantaneous center from O is 14.7 2 + 32.9 2 = 36 in., and from C to A is (36 − 6) = 30 in. from which 30 ω AB = 6 in/s, from which ω AB = 0.2 rad/s. check.]. The vector from the instantaneous center to point B is rB /C = rB − rC = 14.7 i − (14.7 i + 32.86 j = −32.86 j) (in.) The velocity at point B is  i j k   0 0.2  = 6.57 i (in/s). v B = ω AB × rB /C =  0  0 −32.86 0   

Problem 17.73 The angle θ = 45 °, and bar OQ is rotating in the counterclockwise direction at 0.2 rad/s. Use instantaneous centers to determine the velocity of the sleeve P. Solution:

2 ft

The velocity of Q is

v Q = 2ω 0Q = 2(0.2) = 0.4 ft/s.

2 ft

u O P

Therefore −−

ω PQ =

vQ 0.4 = = 0.2 rad/s 2 ft 2

(clockwise) and v P = 2 2ω PQ = 0.566 ft/s (v P is to the left).

Instantaneous center of bar PQ

vPQ Q

2√2 ft

vOQ O

Problem 17.74 At the instant shown, the disk is rolling in the clockwise direction at 10 rad/s. What are the angular velocities of bars AB and BC ? Solution:

The instantaneous center of the disk is at O. Let rOC be the distance from O to C, so the magnitude of the velocity of C is v C = (10 rad/s)rOC . Using this value, the components of the velocity of C are

0.2 m A 0.4 m

v C = (10 rad/s)rOC cos 45 °i + (10 rad/s)rOC cos 45 ° j

0.2 m

= (10 rad/s)(0.2 m) i + (10 rad/s)(0.2 m) j = 2 i + 2 j (m/s). The instantaneous center of bar AB is at A. Let ω AB denote the clockwise angular velocity of bar AB, and let r AB be its length. Then the magnitude of the velocity of B is ω ABr AB , and in terms of the angle α its components are v B = ω ABr AB sin αi + ω ABr AB cos α j = ω AB (r AB sin α) i + ω AB (r AB cos α) j

a C

a A

= ω AB (0.2 m) i + ω AB (0.4 m) j.

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17.74 (Continued) Finally, we write the velocity of C in terms of the velocity of B: v C = v B + ω BC × rC /B : 2 i + 2 j (m/s) = ω AB (0.2 m) i + ω AB (0.4 m) j +

i 0

j k 0 ω BC

0.6 m 0

This yields the two equations 2 m/s = ω AB (0.2 m), 2 m/s = ω AB (0.4 m) + ω BC (0.6 m). We obtain ω AB = 10 rad/s, ω BC = −3.33 rad/s. AB : 10 rad/s clockwise. BC : 3.33 rad/s clockwise.

Problem 17.75 Bar AB rotates at 6 rad/s in the clockwise direction. Use instantaneous centers to determine the angular velocity of bar BC.

4 in A

Solution:

Choose a coordinate system with origin at A and y axis vertical. Let C ′ denote the instantaneous center. The instantaneous center for bar AB is the point A, by definition, since A is the point of zero velocity. The vector AB is rB /A = 4 i + 4 j (in.). The velocity at B is

6 rad/s

3 in C 4 in

 i j k    v B = ω AB × rB /A =  0 0 −6  = 24 i − 24 j (in/s).  4 4 0   

10 in

The unit vector parallel to AB is also the unit vector perpendicular to the velocity at B,

C′

1 ( i + j). 2

e AB =

The vector location of a point on a line perpendicular to the velocity at B is L AB = L AB e AB , where L AB is the magnitude of the distance from point A to the point on the line. The vector location of a point on a perpendicular to the velocity at C is L C = (14 i + yj) where y is the y-coordinate of the point referenced to an origin at A. When the two lines intersect,

A C

L AB i = 14 i, 2 and y =

L AB = 14 2

from which L AB = 19.8 in., and the coordinates of the instantaneous center are (14, 14) (in.). [Check: The line AC ′ is the hypotenuse of a right triangle with a base of 14 in. and interior angles of 45 °, from which the coordinates of C ′ are (14, 14) in. check.]. The angular velocity of bar BC is determined from the known velocity at B. The vector from the instantaneous center to point B is rB /C = rB − rC = 4 i + 4 j − 14 i − 14 j = −10 i − 10 j.

324

The velocity of point B is  i j k  0 v B = ω BC × rB /C =  0 ω BC  −10 −10 0 

     

= ω BC (10i − 10 j )(in/s). Equate the two expressions for the velocity: 24 = 10 ω BC , from which ω BC = 2.4 rad/s ,

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Problem 17.76 The crank AB is rotating in the clockwise direction at 2000 rpm (revolutions per minute). (a) At the instant shown, what are the coordinates of the instantaneous center of the connecting rod BC ? (b) Use instantaneous centers to determine the angular velocity of the connecting rod BC at the instant shown.

B 50 mm

Solution: ω AB = 2000 rpm

min ( 2πrevrad )( 60 ) = 209 rad/s s

(a) The instantaneous center of BC is located at point Q

175 mm

Q = (225, 225) mm.

50 mm

(b) v B = ω AB ( AB) = ω BC (QB) ω BC =

AB 50 2 mm ω = (209 rad/s) = 59.8 rad/s QB AB (225 − 50) 2 mm

v C = ω BC (QC ) = (59.8 rad/s)(0.225 m) = 13.5 m/s v C = (13.5 m/s) i.

225 mm B

Problem 17.77 The disks roll on the plane surface. The left disk rotates at 2 rad/s in the clockwise direction. Use instantaneous centers to determine the angular velocities of the bar and the right disk.

225 mm

2 rad/s

3 ft

1 ft

Solution:

Choose a coordinate system with the origin at the point of contact of the left disk with the surface, and the x axis parallel to the plane surface. Denote the point of attachment of the bar to the left disk by A, and the point of attachment to the right disk by B. The instantaneous center of the left disk is the point of contact with the surface. The vector distance from the point of contact to the point A is r A /P = i + j (ft). The velocity of point A is

 i j k    v A = ω LD × r A /P =  0 0 −2  = 2 i − 2 j (ft/s).  1 1 0    The point on a line perpendicular to the velocity at A is L A = L A ( i + j), where L A is the distance of the point from the origin. The point B is at the top of the right disk, and the velocity is constrained to be parallel to the x axis. A point on a line perpendicular to the velocity at B is L B = (1 + 3cos θ) i + yj (ft), where θ = sin −1

B A

( 13 ) = 19.5°.

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17.77 (Continued) At the intersection of these two lines L A = 1 + 3cos θ = 3.83 ft, and the coordinates of the instantaneous center of the bar are (3.83, 3.83) (ft). The angular velocity of the bar is determined from the known velocity of point A. The vector from the instantaneous center to point A is

The vector from the instantaneous center to point B is

r A /C = r A − rC = i + j − 3.83i − 3.83 j = −2.83i − 2.83 j (ft).

  i j k   0 0.7071  = 1.294 i (ft/s). v B = ω AB × rB /A =  0   0 −1.83 0  

The velocity of point A is  i j k  0 0 v A = ω AB × r A /C =  ω AB  −2.83 −2.83 0 

     

The velocity of point B is

Using the fixed center at point of contact:  i j k  v B = ω RD × rB /P =  0 0 ω RD  0 2 0 

= ω AB (2.83i − 2.83 j) (ft/s). Equate the two expressions and solve: ω AB =

rB /C = rB − rC = (1 + 3cos θ) i + 2 j − 3.83i − 3.83 j = −1.83 j.

   = −2ω i (ft/s). RD   

Equate the two expressions for v B and solve:

2 = 0.7071 (rad/s) counterclockwise. 2.83

ω RD = −0.647 rad/s, clockwise.

Problem 17.78 At the instant shown, bar CD is rotating at 10 rad/s in the counterclockwise direction. What are the angular velocities of bars AB and BC ?

350 mm

Solution:

200 mm

A 300 mm

C B

350 mm

vC 458

The distance from D to C is rCD = (0.35 m)/ sin 45 ° = 0.495 m, so the magnitude of the velocity of C is v C = (10 rad/s)rCD = 4.95 m/s. The distance from C to the instantaneous center I of bar BC is rCI = (0.3 m)/ sin 45 ° = 0.424 m, so the clockwise angular velocity of bar BC is v ω BC = C = 11.7 rad/s. rCI The distance from A to I is 0.65 m, so the distance from B to I is rBI = 0.45 m and the magnitude of the velocity of B is v B = ω BC rBI = 5.25 m/s. Now the clockwise angular velocity of bar AB is ω AB =

vB = 26.3 rad/s. r AB

AB : 26.3 rad/s clockwise. BC : 11.7 rad/s clockwise.

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Problem 17.79 The horizontal member ADE supporting the scoop is stationary. The link BD is rotating in the clockwise direction at 1 rad/s. Use instantaneous centers to determine the angular velocity of the scoop.

5 ft

The triangle BDH is key to the solution of the problem. The angle θ between the horizontal and the line from D to B is

( 12 ftft ) = 63.4°.

The angle DEH is a right angle. We can see that 1 ft 6 in E

1 ft

v B = (2.24 ft)(1 rad/s) = 2.24 ft/s.

C 2 ft

The distance from D to B is BD = 1 2 + 2 2 = 2.24 ft. The magnitude of the velocity of point B is

θ = arctan

Solution:

2 ft 6 in

Scoop

tan θ =

HE , 3.5 ft

so HE = 7 ft and HC = 5.5 ft. The side HD is given by HD = 3.5 ft/ cos θ = 7.83 ft. Thus, HB = HD − BD = 5.59 ft. Also note that since H is the instantaneous center for BC, and we know v B = v B = 2.24 ft/s, we can find ω BC = v B /HB = 0.4 rad/s counterclockwise. Next, the magnitude of the velocity of point C can be determined from v C = HC ω BC = (5.5 ft)(0.4 rad/s) = 2.20 ft/s. Now, having the velocity of C, we find the clockwise angular velocity of the scoop as ω CE =

vC 2.20 ft/s = = 1.47 rad/s. CE 1.5 ft

1.47 rad/s clockwise.

Instantaneous center of bar BC

vBC

rBH

rCH B vBD 5 1rad/s D

Problem 17.80 The disk is in planar motion. The directions of the velocities of points A and B are shown. The velocity of point A is v A = 2 m/s. (a) What are the coordinates of the disk’s instantaneous center? (b) Determine the velocity v B and the disk’s angular velocity.

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vC vCE

3.5 ft

vB 63.48

308

708 B (0.5, 0.4)m x

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17.80 (Continued)

Solution:

Let the coordinates of the disk’s instantaneous center C be denoted by ( x C , y C ). Recall that v C = 0. Thus the velocities of A and B can be written as

C r C/A

v A, B = v C + ω × r A, B / C = ω × r A, B / C , v A,B = ωk × [( x A,B − x C ) i + ( y A,B − y C ) j].

708

r C/B B

For point A, we have

(0.5, 0.4) m

v A = v A cos30 °i + v A cos30 ° j, x : v A cos30 ° = −ω ( y A − y C ) = ω y C

y : v A sin 30 ° = ω ( x A − x C ) = −ω x C

308

For point B, x : v B cos 70 ° = −ω ( y B − y C ) = −ω (0.4 − y C ) y : v B sin 70 ° = ω ( x B − x C ) = ω (0.5 − x C ) The unknowns are ω, x C , y C , and v B . ( x C , y C ) = (−0.425, 0.737) m,

Solving,

obtain

ω = 2.35 rad/s counterclockwise, and v B = 2.31 m/s. (a) ( x C , y C ) = (−0.425, 0.737) m. (b) v B = 2.31 m/s, ω = 2.35 rad/s counterclockwise.

Problem 17.81 The rigid body rotates about the z-axis with counterclockwise angular velocity ω = 5 rad/s and counterclockwise angular acceleration α = 4 rad/s 2 . The distance r A /B = 1.2 m. (a) What are the rigid body’s angular velocity and angular acceleration vectors? (b) Determine the acceleration of point A relative to point B, first by using Eq. (17.9) and then by using Eq. (17.10).

Solution:

Note that rB /A = 1.2 i m.

(a) Applying the right-hand rule, the thumb points in the positive z-axis direction, and the angular velocity and angular acceleration vectors are ω = 5k rad/s and α = 4 k rad/s 2 . (b) To apply Eq. (17.9), we first evaluate the term

ω × (ω × r A /B ) = ω × y

i j k 0 0 5 rad/s 0 6 m/s 0

i j k 0 0 5 rad/s 1.2 m 0 0

= ω × [ 6 j (m/s) ]

= −30 i (m/s 2 ).

Then A

a A /B = α × r A /B + ω × (ω × r A /B )

rA/B =

0 0 4 rad/s 2 1.2 m 0 0

−30 i (m/s 2 )

= −30 i + 4.8 j (m/s 2 ). Applying Eq. (17.10), a A /B = α × r A /B − ω 2 r A /B i =

0 0 4 rad/s 2 1.2 m 0 0

− (5 rad/s) 2 [1.2 i (m)]

= −30 i + 4.8 j (m/s 2 ). (a) ω = 5k rad/s, α = 4 k rad/s 2 . (b) a A /B = −30 i + 4.8 j (m/s 2 ).

328

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Problem 17.82 The bar is rotating with a counterclockwise angular velocity of 6 rad/s and a counterclockwise angular acceleration of 40 rad/s 2 . Use Eq. (17.10) to determine the acceleration of point A. y

Solution:

Let L = 2 m, ω = 6 rad/s, and α = 40 rad/s 2 . The position vector of point A relative to the origin O is r A /O = L cos30 °i + L sin 30 ° j. The acceleration of point A is a A = a O + α × r A /O − ω 2 r A /O = 0+

6 rad/s 40 rad/s

i j k 0 0 α L cos30 ° L sin 30 ° 0

− ω 2 ( L cos30 °i + L sin 30 ° j)

= [−αL sin 30 ° − ω 2 L cos30 °]i + [αL cos30° − ω 2 L sin 30°] j = −102 i + 33.3 j (m/s 2 ). a A = −102 i + 33.3 j (m/s 2 ).

308

Problem 17.83 The bar rotates with a counterclockwise angular velocity of 20 rad/s and a counterclockwise angular acceleration of 6 rad/s 2 . (a) By applying Eq. (17.10) to point A and the fixed point O, determine the acceleration of A. (b) By using the result of part (a) and applying Eq. (17.10) to points A and B, determine the acceleration of B.

Solution: (a) a A = a 0 + α × r A /O − ω 2 r A /O where ω = 20 k rad/s α = 6k rad/s 2 r A /O = 1i, and a A = 0 a A = O + 6k × 1i − 400(1i)

a A = −400 i + 6 j (m/s 2 ). (b) a B = a A + α × rB /A − ω 2 rB /A where 20 rad/s 6 rad/s2 A

rB /A = 1i

B x

a B = −400 i + 6 j + 6k × 1i − 400(1i) a B = −800 i + 12 j (m/s 2 ).

Problem 17.84 The Apache helicopter is in planar motion in the x–y plane. At the instant shown, the position of its center of mass G is x = 9 ft, y = 12 ft, its velocity is v G = 60 i + 20 j (ft/s), and its acceleration is a G = 4 i + 2 j (ft/s 2 ). The position of point T where the tail rotor is mounted is x = −28 ft, y = 22 ft. The helicopter is rotating in the clockwise direction at 8 per second, and its angular acceleration is 5 per second squared counterclockwise. What is the acceleration vector of point T ?

Solution:

The counterclockwise angular velocity of the helicopter in

rad/s is  π rad  ω = −8 deg/s   = −0.140 rad/s.  180 deg  The counterclockwise angular acceleration is  π rad  α = 5 deg/s 2   = 0.0873 rad/s 2 .  180 deg  The position of T relative to G is rT /G = (−28 − 9)i + (22 − 12)j = −37 i + 10 j (ft) = xi + yj, where we define x = −37 ft, y = 10 ft.

The acceleration of T is

a T = a G + α × rT /G − ω 2 rT /G G

= 4 i + 2 j (ft/s 2 ) +

Source: Courtesy of Kojihirano/Shutterstock.

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i j k 0 0 α x y 0

− ω 2 ( xi + yj)

= (4 ft/s − αy − ω 2 x ) i + (2 ft/s + αx − ω 2 y ) j = 3.85i − 1.42 j (ft/s 2 ). a T = 3.85i − 1.42 j (ft/s 2 ).

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Problem 17.85 Point A of the rolling disk is moving toward the right and accelerating toward the right. The magnitude of the velocity of point C is 2 m/s, and the magnitude of the acceleration of point C is 14 m/s 2 . Determine the accelerations of points B and D. (See Practice Example 17.6.)

D 458

Solution:

300 mm C

First the velocity analysis

v C = v A + ω × rC /A = r ω i − ωk × (−ri) = r ω i + r ω j vc =

(r ω ) 2 + (r ω ) 2 =

2r ω ⇒ ω =

v = 2r

2 m/s = 4.71 rad/s. 2(0.3 m)

Now the acceleration analysis a C = a A + α × rC /A − ω 2 rC /A = rαi − αk × (−ri) − ω 2 (−ri) = (rα + ω 2r ) i + (αr ) j aC =

(αr + ω 2r ) 2 + (αr ) 2 = r (α + ω 2 ) 2 + α 2 (14 m/s 2 ) 2 = (0.3 m) 2 [(α + [4.71] 2 ) 2 + α 2 ]

Solving this quadratic equation for α we find α = 20.0 rad/s 2 . Now a B = r ω 2 j = (0.3 m)(4.71 rad/s) 2 j

a B = 6.67 j (m/s 2 )

a D = a A + α × r D /A − ω 2 r D /A = rαi − αk × (−r cos 45 °i + 4 sin 45 ° j) − ω 2 (−r cos 45 °i + 4 sin 45 ° j) = (r[1 + sin 45 °]α + r ω 2 cos 45 °) i + (rα cos 45 ° − r ω 2 sin 45 °) j Putting in the numbers we find

a D = 14.9 i − 0.480 j (m/s 2 ).

Problem 17.86 The disk rolls on the circular surface with a constant clockwise angular velocity of 1 rad/s. What are the accelerations of points A and B? Strategy: Begin by determining the acceleration of the center of the disk. Notice that the center moves in a circular path and the magnitude of its velocity is constant.

Solution: vB = 0 v 0 = v B + ωk × rO /B = (−1k) × (0.4 j) v 0 = 0.4 i m/s Point O moves in a circle at constant speed. The acceleration of O is a 0 = −v 02 /( R + r ) j = (−0.16)/(1.2 + 0.4) j

a 0 = −0.1 j (m/s 2 ). A

a B = a 0 − ω 2 rB /O = −0.1 j − (1) 2 (−0.4) j a B = 0.3 j (m/s 2 ).

0.4 m x

a A = a 0 − ω 2 r A /O = −0.1 j − (1) 2 (0.4) j a A = −0.5 j (m/s 2 ).

1.2 m

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Problem 17.87 The length of the bar is L = 4 ft and the angle θ = 30 °. The bar’s angular velocity is ω = 1.8 rad/s and its angular acceleration is α = 6 rad/s 2 . The endpoints of the bar slide on the plane surfaces. Determine the acceleration of the midpoint G. Strategy: Begin by applying Eq. (17.10) to the endpoints of the bar to determine their accelerations.

Solution:

Denote the top end of the bar B and the bottom end A. From geometry, the position vectors of B and G relative to A are L L rB /A = −L sin θ i + L cos θ j and rG /A = − sin θ i + cos θ j. 2 2

We know that a A = a Ax i and a B = a By j . We also note that v A = v Ax i and v B = a By j since A and B slide on the horizontal and vertical surfaces respectively, For the motion of point B relative to point A, we have a B = a A + α × r B /A − ω 2 r B /A :

a By j = a Ax i +

i 0 −L sin θ

j k 0 α L cos θ 0

− ω 2 (−L sin θ i + L cos θ j).

This yields the two equations 0 = a Ax − αL cos θ + ω 2 L sin θ, a By = −αL sin θ − ω 2 L cos θ. Solving, we obtain a Ax = 14.3 ft/s 2 , a By = −23.2 ft/s 2 . For the motion of point G relative to point A, we have

a G = a A + α × rG /A − ω 2 rG /A :

i 0

G a Gx i + a Gy j = a Ax i +

v a

j 0

k α

1 1 − L sin θ L cos θ 0 2 2 1 1 − ω 2 − L sin θ i + L cos θ j , 2 2 which yields the two equations

(

)

1 1 a Gx = a Ax − α L cos θ + ω 2 L sin θ, 2 2 1 1 2 a Gy = −α L sin θ − ω L cos θ. 2 2 Solving yields a Gx = 7.15 ft/s 2 , a Gy = −11.6 ft/s 2 . a G = 7.15 i − 11.6 j (ft/s 2 ).

Problem 17.88 The angular velocity and angular acceleration of bar AB are ω AB = 2 rad/s and α AB = 10 rad/s. The dimensions of the rectangular plate are 12 in × 24 in. What are the angular velocity and angular acceleration of the rectangular plate?

Solution: D B

A 20 in

rC/D

rA/B rC/A

458

12 in

458

B vAB aAB

458 C

The geometry of our problem is defined by three position vectors: r A /B = (1 ft)(− cos 45 °i − sin 45 ° j), rC /D = (1.67 ft)(− cos 45 °i − sin 45 ° j), and rC /A = 2 i (ft). The velocities of A and C are given by v A = ω AB × r A /B = ω AB k × (− cos 45 °i − sin 45 ° j) (ft/s) = ω AB (sin 45 °i − cos 45 ° j)) (ft/s), v C = ω CD × rC /D = ω CD k × (1.67 ft)(− cos 45 °i − sin 45 ° j) = 1.67ω CD (sin 45 °i − cos 45 ° j)) (ft/s).

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17.88 (Continued) The velocity of C can also be written v C = v A + ω AC × rC /A = v A + ω AC k × (2 ft) i = v A + 2ω AC j (ft/s). Equating the two expressions for v C , we get x : 1.67ω CD (sin 45 °) = ω AB (sin 45 °), y : 1.67ω CD (− cos 45 °) = ω AB (− cos 45 °) + 2ω AC . Solving these equations simultaneously, we get ω CD = 1.20 rad/s counterclockwise and ω AC = ω plate = 0. We can find the angular acceleration of the plate in the same way. The accelerations of points A and C are given by a A = ω AB × (ω AB × r A /B ) + α AB × r A /B = −ω 2AB r A /B + α AB × r A /B , 2 r a C = ω CD × (ω CD × rC /D ) + α CD × rC /D = −ω CD C /D + α CD × rC /D ,

and the acceleration of C can also be written a C = a A + ω AC × (ω AC × rC /A ) + α AC × rC / A = a A + α AC × rC /A , since ω AC = 0. Equating the two expressions for a C , we get 2 r −ω CD C /D + α CD × rC /D = a A + α AC × rC /A .

Substituting known values and equating components, we have 2 cos 45 ° + α 2 x : (1.67 ft)(ω CD CD sin 45 °) = ω AB cos 45 ° + α AB sin 45 °, 2 sin 45 ° − α 2 y : (1.67 ft)(ω CD CD cos 45 °) = ω AB sin 45 ° − α AB cos 45 + (2 ft)α AC .

Substituting the values of the angular velocities and solving, we get αCD = 6.96 (rad/s 2 ), α AC = −1.13 (rad/s 2 ). Angular velocity is zero, angular acceleration is 1.13 rad/s 2 clockwise.

Problem 17.89 The ring gear is stationary, and the sun gear has an angular acceleration of 10 rad/s 2 in the counterclockwise direction. Determine the angular acceleration of the planet gears.

Ring gear

7 in 34 in 20 in

Solution:

The strategy is to use the tangential acceleration at the point of contact of the sun and planet gears, together with the constraint that the point of contact of the planet gear and ring gear is stationary, to determine the angular acceleration of the planet gear. The tangential acceleration of the sun gear at the point of contact with the top planet gear is

Planet gears (3) Sun gear

 i j k    a ST = α × rS =  0 0 10  = −200 i (in/s 2 ).  0 20 0    This is also the tangential acceleration of the planet gear at the point of contact. At the contact with the ring gear, the planet gears are stationary, hence the angular acceleration of the planet gear satisfies  i j k  0 α P (−2rP ) =  0 αP  0 −14 0 

   = −200 i   

a ST RS

from which αP = −

332

200 = −14.29 (rad/s 2 ) (clockwise). 14

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Problem 17.90 Bar CD is rotating in the clockwise direction at 12 rad/s and has a counterclockwise angular acceleration of 240 rad/s 2 . What are the angular velocity and angular acceleration of bar AB?

y C

Solution:

The position vector of C relative to D is rC /D = −(2 m)i + (2 m) j. The velocity of C in terms of D is

v C = v D + ω CD × rC /D = 0 +

i j k 0 0 ω CD −2 m 2 m 0

D 2m

= −ω CD (2 m) i − ω CD (2 m) j. The position vector of B relative to A is rB /A = (2 m)j. The velocity of B in terms of A is v B = v A + ω AB × rB /A = 0 +

i j k 0 0 ω AB 0 2m 0

= 0+

The position vector of C relative to B is rC /B = (2 m)i. The velocity of C in terms of B is i j k 0 0 ω BC 2m 0 0

= −ω AB (2 m)i + ω BC (2 m)j. Equating the two expressions for the velocity of C, we obtain the two equations −ω CD = −ω AB , −ω CD = ω BC . We see that ω AB = ω CD = −12 rad/s, ω BC = −ω CD = 12 rad/s. To evaluate the acceleration, we follow the same steps we used to evaluate the velocity. The acceleration of C in terms of D is a C = a D + α CD × rC /D − ω CD 2 rC /D = 0+

The acceleration of B in terms of A is a B = a A + α AB × rB /A − ω AB 2 rB /A

= −ω AB (2 m)i.

v C = v B + ω BC × rC /B = −ω AB (2 m)i +

i j k 0 0 αCD −2 m 2 m 0

i j k 0 0 α AB 0 2m 0

− ω AB 2 (2 m) j

= −α AB (2 m)i − ω AB 2 (2 m)j. The acceleration of C in terms of B is a C = a B + α BC × rC /B − ω BC 2 rC /B = −α AB (2 m)i − ω AB 2 (2 m)j +

i j k 0 0 α BC 2m 0 0

− ω BC 2 (2 m)i

= [−α AB (2 m) − ω BC 2 (2 m)]i + [α BC (2 m) − ω AB 2 (2 m)] j. Equating the two expressions for the acceleration of C, we obtain the two equations −αCD + ω CD 2 = −α AB − ω BC 2 , −αCD − ω CD 2 = α BC − ω AB 2 . From these equations we obtain α AB = −48 rad/s 2, α BC = −240 rad/s 2 . Angular velocity is 12 rad/s clockwise,

− ω CD 2 [−(2 m)i + (2 m)j]

angular acceleration is 240 rad/s 2 clockwise.

= [−αCD (2 m) + ω CD 2 (2 m)]i + [−αCD (2 m) − ω CD 2 (2 m)] j.

Problem 17.91 The 1-m-diameter disk rolls, and point B of the 1-m-long bar slides, on the plane surface. Determine the angular acceleration of the bar and the acceleration of point B.

Solution: Choose a coordinate system with the origin at O, the center of the disk, with x axis parallel to the horizontal surface. The point P of contact with the surface is stationary, from which

10 rad/s2

 i j k  v P = 0 = v O + ω O × −R = v O +  0 0 ωO  0 −0.5 0 

4 rad/s

from which v O = −2 i (m/s). The velocity at A is

M17_BEDF2222_06_SE_C17_Part1.indd 333

   = v + 2 i, O   

 i j k   v A = v O + ω O × r A /O = −2 i +  0 0 ω O  = −2 i + 2 j (m/s)  0.5 0 0    The motion at point B is constrained to be parallel to the x axis. The line perpendicular to the velocity of B is parallel to the y axis. The line perpendicular to the velocity at A forms an angle at 45 ° with the x axis. From geometry, the line from A to the fixed center is the hypotenuse of a right triangle with base cos30 ° = 0.866 and interior angles 45 °. The coordinates of the fixed center are (0.5 + 0.866, 0.866) = (1.366, 0.866) in.

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17.91 (Continued) The vector from the instantaneous center to the point A is r A /C = r A − rC = −0.866 i − 0.866 j (m). The angular velocity of the bar AB is obtained from  i j k  v A = ω AB × r A /C =  0 0 ω AB  −0.866 −0.866 0  = 0.866ω AB i − 0.866ω AB j (m/s),

     

The vector B /A is rB /A = rB − r A = (0.5 + cos θ) i − 0.5 j − 0.5i = 0.866 i − 0.5 j (m). The acceleration of point B is a B = a A + α AB × r A /B − ω 2AB r A /B

from which ω AB = −

     

 i j k  0 0 α AB = −13i + 5 j +   − cos θ sin θ 0  − ω 2AB (−i cos θ + j sin θ).

2 = −2.31 (rad/s). 0.866

The acceleration of the center of the rolling disk is a O = −αRi = −10(0.5) i = −5i (m/s 2 ). The acceleration of point A is

The constraint on B insures that the acceleration of B will be parallel to the x axis. Separate components:

a A = a O + αO × r A /O − ω O2 r A /O

a B = −13 + 0.5α AB − ω 2AB (0.866), 0 = 5 + 0.866α AB + 0.5ω 2AB .

 i j k   = −5i +  0 0 α  − 16(0.5) i  0.5 0 0    = −13i + 5 j (m/s 2 ).

Solve: α AB = −8.85 (rad/s 2 ) , where the negative sign means a clockwise rotation. a B = −22.04 i (m/s 2 ).

Problem 17.92 The angle θ = 40 °. The sleeve P is moving to the right with a constant velocity of 3 m/s. Find the angular velocities and angular accelerations of bars OQ and PQ.

1.2 m

Solution:

1.2 m

rQ/O

rQ/P Solving, we obtain ω OQ = −1.94 rad/s, ω PQ = 1.94 rad/s.

The position vector of Q relative to O is rQ /O = (1.2 m) cos θ i + (1.2 m)sin θj. The velocity of Q in terms of O is

To evaluate the acceleration, we follow the same steps we used to evaluate the velocity. The acceleration of Q in terms of O is a Q = a O + α OQ × rQ /O − ω OQ 2 rQ /O

v Q = v O + ω OQ × rQ /O = 0+

i 0

j 0

− ω OQ

The position vector of Q relative to P is rQ /P = −(1.2 m) cos θ i + (1.2 m)sin θ j. The velocity of Q in terms of P is

−(1.2 m) cos θ (1.2 m)sin θ

2 [(1.2 m) cos θ i + (1.2 m)sin θ j]

+ (1.2 m)[αOQ cos θ − ω OQ 2 sin θ] j. The acceleration of Q in terms of P is a Q = a P + α PQ × rQ /P − ω PQ 2 rQ /P

v Q = v P + ω PQ × rQ /P = v Pi +

k αOQ

= (1.2 m)[−αOQ sin θ − ω OQ 2 cos θ]i

= −ω OQ (1.2 m)sin θ i + ω OQ (1.2 m) cos θ j.

j 0

(1.2 m) cos θ (1.2 m)sin θ

k ω OQ

(1.2 m) cos θ (1.2 m)sin θ

i 0

= 0+

k ω PQ 0

= [v P − ω PQ (1.2 m)sin θ]i − ω PQ (1.2 m) cos θ j. Equating the two expressions for the velocity of Q, we obtain the two equations

= 0+

i 0

j 0

−(1.2 m) cos θ (1.2 m)sin θ

k α PQ 0

− ω PQ 2 [−(1.2 m) cos θ i + (1.2 m)sin θ j] = (1.2 m)[−α PQ sin θ + ω PQ 2 cos θ]i + (1.2 m)[−α PQ cos θ − ω PQ 2 sin θ] j.

−ω OQ (1.2 m)sin θ = v P − ω PQ (1.2 m)sin θ, ω OQ = −ω PQ .

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17.92 (Continued) Equating the two expressions for the acceleration of Q, we obtain the two equations −αOQ sin θ − ω OQ 2 cos θ = −α PQ sin θ + ω PQ 2 cos θ, αOQ cos θ − ω OQ 2 sin θ = −α PQ cos θ − ω PQ 2 sin θ. Solving, we obtain αOQ = −4.51 rad/s 2 , α PQ = 4.51 rad/s 2 . OQ : 4.51 rad/s 2 clockwise. PQ : 4.51 rad/s 2 counterclockwise.

Problem 17.93 If θ = 50 ° and bar OQ has a constant clockwise angular velocity of 1 rad/s, what is the acceleration of sleeve P ?

1.2 m

Solution: u

ω oQ = −1k rad/s, α oQ = 0, a 0 = 0 a Q = a 0 + α oQ × rQ /o − ω 2 rQ /o

a Q = 0 + 0 − (1) 2 (1.2 cos 50 °i + 1.2sin 50 ° j) a Q = −0.771i − 0.919 j m/s 2 v Q = v 0 + ω oQ × rQ /o

2 r a p = a Q + αQp k × r p /Q − ω Qp p /Q

where

v0 = 0

= (−1k) × [(1.2 cos 50°) i + (1.2sin 50) j]

a p = a pi

= .919 i − 0.771 j (m/s).

r p /Q = 1.2 cos 50 °i − 1.2sin 50 ° j

v P = v Q + ω QP × rP /Q

2 (1.2) cos 50 ° i: a p = −0.771 + 1.2αQp sin 50 ° − ω Qp

(1)

2 (1.2)sin 50 ° j: 0 = −0.919 + 1.2αQp cos 50 ° + ω Qp

(2)

= v Q + ω QP k × (1.2 cos 50 °i − 1.2sin 50 ° j). i : v P = 0.919 + 1.2ω QP sin 50 °

We have two eqns in three unknowns a p , αQp , ω Qp .

j : O = −0.771 + 1.2ω QP cos 50 °

We need another eqn. To get it, we use the velocity relationships and determine ω QP . Note v p = v p i.

Solving, v P = 1.839 m/s, ω QP = 1 rad/s. Now going back to eqns. (1) and (2), we solve to get a P = −1.54 m/s 2 αQP = 0, (to the left).

Problem 17.94 The angle θ = 70 °. Bar OQ has a clockwise angular velocity of 8 rad/s and a clockwise angular acceleration of 20 rad/s 2 . Find (a) the angular acceleration of the bar PQ and (b) the acceleration of point P.

Solution:

400 mm

200 mm

We can determine the angle β by using the law of sines:

sin β sin θ = , 0.2 m 0.4 m obtaining β = 28.0 °.

The position vector of Q relative to O is rQ /O = (0.2 m) cos θ i + (0.2 m)sin θ j. The velocity of Q in terms of O is

rQ/O

v Q = v O + ω OQ × rQ /O = 0 +

i 0

j 0

k ω OQ

(0.2 m) cos θ (0.2 m)sin θ

Q rQ/P u

= −ω OQ (0.2 m)sin θi + ω OQ (0.2 m) cos θ j.

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17.94 (Continued) The position vector of Q relative to P is rQ /P = −(0.4 m) cos β i + (0.4 m)sin β j. The velocity of Q in terms of P is

The acceleration of Q in terms of P is a Q = a P + α PQ × rQ /P − ω PQ 2 rQ /P

v Q = v P + ω PQ × rQ /P i 0

= v Pi +

j 0

−(0.4 m) cos β (0.4 m)sin β

j 0

k α PQ

−(0.4 m) cos β (0.4 m)sin β

− ω PQ 2 [−(0.4 m) cos β i + (0.4 m)sin β j]

= [v P − ω PQ (0.4 m)sin β ]i − ω PQ (0.4 m) cos β j. Equating the two expressions for the velocity of Q, we obtain the two equations

i 0

= aPi +

k ω PQ

= {a P + (0.4 m)[−α PQ sin β + ω PQ 2 cos β ]} i + (0.4 m)[−α PQ cos β − ω PQ 2 sin β ] j.

−ω OQ (0.2 m)sin θ = v P − ω PQ (0.4 m)sin β ,

Equating the two expressions for the acceleration of Q, we obtain the two equations

ω OQ (0.2 m) cos θ = −ω PQ (0.4 m) cos β .

(0.2 m)[−αOQ sin θ − ω OQ 2 cos θ] = a P + (0.4 m) [−α PQ sin β + ω PQ 2 cos β ],

Setting ω OQ = −8 rad/s and solving, we obtain ω PQ = 1.55 rad/s, v P =1.79 m/s. To evaluate the acceleration, we follow the same steps we used to evaluate the velocity. The acceleration of Q in terms of O is a Q = a O + α OQ × rQ /O − ω OQ 2 rQ /O i 0

= 0+

j 0

(0.2 m) cos θ (0.2 m)sin θ − ω OQ

k αOQ

(0.2 m)[αOQ cos θ − ω OQ

2 sin θ ] =

(0.4 m)

[−α PQ cos β − ω PQ 2 sin β ]. Setting αOQ = −20 rad/s 2 and solving, we obtain α PQ = 36.7 rad/s 2 , a P = 5.42 m/s 2 . (a) 36.7 rad/s 2 counterclockwise. (b) 5.42 m/s 2 .

2 [(0.2 m) cos θ i + (0.2 m)sin θ j]

= (0.2 m)[−αOQ sin θ − ω OQ 2 cos θ ]i + (0.2 m)[αOQ cos θ − ω OQ 2 sin θ ] j.

Problem 17.95 At the instant shown, the piston’s velocity and acceleration are v C = −14 i (m/s) and a C = −2200 i (m/s 2 ). What is the angular acceleration of the crank AB? Solution:

The velocity analysis:

v B = V A + ω AB × rB / A

= 0 + ω AB k × (0.05i + 0.05 j)

50 mm

= (−0.05ω AB i + 0.05ω AB j)

v C = v B + ω BC × rC / B = (−0.05ω AB i + 0.05ω AB j) + ω BC k × (0.175i − 0.05 j) = (−0.05ω AB + 0.05ω AB ) i + (0.05ω AB + 0.175ω BC ) j = −14 i 175 mm

Equating components and solving we find that ω AB = 218 rad/s, ω BC = −62.2 rad/s. The acceleration analysis:

50 mm

a B = a A + α AB × rB / A − ω AB 2 rB / A = 0 + α AB k × (0.05i + 0.05 j) − (218) 2 (0.05i + 0.05 j) = (−0.05α AB − 2370) i + (0.05α AB − 2370) j 2 r a C = a B + α BC × rC / B − ω BC C/B

= (−0.05α AB − 2370) i + (0.05α AB − 2370) j + α BC k × (0.175i − 0.05 j) − (−62.2) 2 (0.175i − 0.05 j) = (−0.05α AB − 2370 + 0.05α BC − 678) i

Equating components and solving, we find α AB = −3530 rad/s 2 , α BC = 13,500 rad/s 2 Thus α AB = 3530 rad/s 2 clockwise.

+ (0.05α AB − 2370 + 0.175α BC + 194) j = −2200 i

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Problem 17.96 Bar CD has a clockwise angular velocity of 2 rad/s and a clockwise angular acceleration of 6 rad/s 2 . Determine the angular velocity ω AB and the angular acceleration α AB . Solution:

The position vector of C relative to rC /D = −2 i + j (m). The velocity of C in terms of D is v C = v D + ω CD × rC /D = 0 +

2m B

A vAB aAB

is C

i j k 0 0 ω CD (1) −2 m 1 m 0

1m D

= −ω CD i − 2ω CD j (m/s). The position vector of B relative to C is rB /C = i + j (m). Using the result (1), the velocity of B in terms of C is

v B = v C + ω BC × rB /C = −ω CD i − 2ω CD j (m/s) +

i j k 0 0 ω BC 1m 1m 0

Using this result, the acceleration of B in terms of C is a B = a C + α BC × rB / C − ω BC 2 rB / C = (−αCD + 2ω CD 2 ) i + (−2αCD − ω CD 2 ) j

= (−ω CD − ω BC ) i + (−2ω CD + ω BC ) j (m/s). The velocity of B can also be expressed in terms of the angular velocity of bar AB: v B = 2ω AB j (m/s). Equating the two expressions for the velocity of B, we obtain the two equations

i j k 0 0 α BC 1m 1m 0

− ω BC 2 ( i + j)

= (−αCD + 2ω CD 2 − α BC − ω BC 2 ) i + (−2αCD − ω CD 2 + α BC − ω BC 2 ) j

−ω CD − ω BC = 0, −2ω CD + ω BC = 2ω AB .

The acceleration of B can also be expressed in terms of the motion of bar AB:

Setting ω CD = −2 rad/s and solving, we obtain ω AB = 3 rad/s, ω BC = 2 rad/s.

Equating the two expressions for the acceleration of B, we obtain the two equations

The acceleration of C in terms of D is

−αCD + 2ω CD 2 − α BC − ω BC 2 = −2ω AB 2 ,

a C = a D + α CD × rC /D − ω CD 2 rC /D

−2αCD − ω CD 2 + α BC − ω BC 2 = 2α AB .

= 0+

a B = 2α AB j − 2ω AB 2 i.

i j k 0 0 αCD 0 −2 m 1 m

− ω CD

Setting αCD = −6 rad/s 2 and solving, we obtain α AB = 16 rad/s 2 , α BC = 28 rad/s 2 .

2 [−(2 m) i + (1 m) j]

ω AB = 3 rad/s, α AB = 16 rad/s 2 .

= (−αCD + 2ω CD 2 ) i + (−2αCD − ω CD 2 ) j (m/s 2 ).

Problem 17.97 The angular velocity and angular acceleration of bar AB are ω AB = 2 rad/s and α AB = 8 rad/s 2 . What is the acceleration of point D?

Solution:

First we must do a velocity analysis to find the angular velocity of BCD v B = v A + ω AB × rB /A = 0 + (2 rad/s)k × (0.32 m) i = (0.64 m/s) j v C = v B + ω BCD × rC /B = (0.64 m/s) j + ω BCD k × (0.24 i + 0.48 j)m

= (−0.48 m)ω BCD i + (0.64 m/s + { 0.24 m } ω BCD ) j

Since C cannot move in the j direction we know 0.64 m/s + {0.24 m}ω BCD = 0 ⇒ ω BCD = −2.67 rad/s

0.32 m C

Now we do the acceleration analysis a B = a A + α AB × rB / A − ω AB 2 rB / A = 0 + (8 rad/s 2 )k × (0.32 m) i − (2 rad/s) 2 (0.32 m) i = (−1.28i + 2.56 j) m/s 2

0.48 m

a C = a B + α BC × rC / B − ω BC 2 rC / B

vAB aAB B

= (−1.28i + 2.56 j) m/s 2 + α BCD k × (0.24 i + 0.48 j) m x

− (−2.67 rad/s) 2 (0.24 i + 0.48 j) m = (−2.99 m/s 2 − {0.48 m}α BCD ) i

0.32 m

0.24 m 0.16 m

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+ (−0.853 m/s 2 + {0.24 m}α BCD ) j

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17.97 (Continued) Since C cannot move in the j direction we know −0.853 m/s 2 + {0.24 m}α BCD = 0 ⇒ α BCD = 3.56 rad/s 2 Now we can find the acceleration of point D a D = a B + α BCD × rD /B − ω BCD 2 rD /B = (−1.28i + 2.56 j) m/s 2 + (3.56 rad/s 2 )k × (0.4 i + 0.8 j) m − (−2.67 rad/s) 2 (0.4 i + 0.8 j) m a D = (−.697 i − 1.71 j) m/s 2 .

Problem 17.98 The angular velocity ω AB = 6 rad/s. If the acceleration of the slider C is zero at the instant shown, what is the angular acceleration α AB?

B 4 in

Solution:

The velocity analysis: 3 in

v B = v A + ω AB × rB /A

aAB vAB

= 0 + (−6)k × (4 i + 4 j)

= (24 i − 24 j) v C = v B + ω BC + rC /B

4 in

10 in

= (24 i − 24 j) + ω BC k × (10 i − 7 j) = (24 + 7ω BC ) i + (24 + 10 ω BC ) j Since C cannot move in the j direction, we set the j component to zero and find that

The acceleration of C is zero. Equating both components to zero and solving, we find that

ω BC = −2.4 rad/s.

α AB = 19.0 rad/s 2 , α BC = 18.0 rad/s 2

The acceleration analysis:

Thus

α AB = 19.0 rad/s 2 clockwise.

a B = a A + α AB × rB / A − ω AB 2 rB / A = 0 − α AB k × (4 i + 4 j) − (6) 2 (4 i + 4 j) = (4α AB − 144) i + (−4α AB − 144) j a C = a B + α BC × rC / B − ω BC 2 rC / B = (4α AB − 144) i + (−4α AB − 144) j + α BC k × (10 i − 7 j) − (−2.4) 2 (10 i − 7 j) = (4α AB − 144 + 7α BC − 57.6) i + (−4α AB − 144 + 10α BC + 40.3) j

Problem 17.99 The angular velocity and angular acceleration of bar AB are ω AB = 5 rad/s and α AB = 10 rad/s 2 . Determine the angular acceleration of bar BC. 0.2 m

338

aAB

vAB

0.4 m

0.2 m

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17.99 (Continued) Solution:

Do a velocity analysis first to find all of the angular velocities. Let point E be the point on the wheel that is in contact with the ground. v B = v A + ω AB × rB / A = 0 + (5 rad/s)k × (−0.4 i + 0.2 j) m = (−i − 2 j) m/s

2 r a C = a B + α BC × rC / B − ω BC C/B

= (8i − 9 j) m/s 2 + α BC k × (0.6 m) i − (1.67 rad/s) 2 (0.6 m) i = (6.33 m/s 2 ) i + (−9 m/s 2 + {0.6 m}α BC ) j a D = a C + α wheel × rD / C − ω wheel 2 rD / C = (6.33 m/s 2 ) i + (−9 m/s 2 + {0.6 m}α BC ) j

v C = v B + ω BC × rC / B = (−i − 2 j) m/s + ω BC k × (0.6 m) i

+ α wheel k × (0.2 m) i − (5 rad/s) 2 (0.2 m) i

= (−1 m/s) i + (−2 m/s + {0.6 m}ω BC ) j

= (1.33 m/s 2 ) i + (−9 m/s 2 + {0.6 m}α BC

v E = v C + ω wheel × rE / C = (−1 m/s) i + (−2 m/s + {0.6 m}ω BC ) j + ω wheel k × (0.2 i − 0.2 j) m = (−1 m/s + {0.2 m}ω wheel ) i + (−2 m/s + {0.6 m}ω BC + {0.2 m}ω wheel ) j

+ {0.2 m}α wheel ) j Finally we work down to point E a E = a D + α wheel × rE / D − ω wheel 2 rE / D = (1.33 m/s 2 ) i + (−9 m/s 2 + {0.6 m}α BC + {0.2 m}α wheel ) j

Point E is the instantaneous center of the wheel. Therefore −1 m/s + { 0.2 m } ω wheel = 0 −2 m/s + { 0.6 m } ω BC + { 0.2 m } ω wheel = 0 ⇒

+ α wheel k × (−0.2 m) j − (5 rad/s) 2 (−0.2 m) j

}

= (1.33 m/s 2 + {0.2 m}α wheel ) i + (−4 m/s 2 + {0.6 m}α BC + {0.2 m}α wheel ) j

ω BC = 1.67 rad/s ω wheel = 5 rad/s

Now we do the acceleration analysis a B = a A + α AB × rB / A − ω AB 2 rB / A = 0 + (10 rad/s 2 )k × (−0.4 i + 0.2 j) m − (5 rad/s) 2 (−0.4 i + 0.2 j) m = (8i − 9 j) m/s 2

D moves horizontally therefore a D ⋅ j = 0 The wheel does not slip therefore a E ⋅ i = 0 We have −9 m/s 2 + { 0.6 m } α BC + { 0.2 m } α wheel = 0   

1.33 m/s 2 + { 0.6 m } α wheel = 0 ⇒

α wheel = −6.67 rad/s 2 α BC = 17.2 rad/s 2

α BC = 17.2 rad/s 2 CCW.

Problem 17.100 At the instant shown, bar AB is rotating at 10 rad/s in the counterclockwise direction and has a counterclockwise angular acceleration of 20 rad/s 2 . The disk rolls on the circular surface. Determine the angular accelerations of bar BC and the disk. Solution:

A 3m

The velocity analysis:

v B = v A + ω AB × rB /A = 0 + (10 k) × (1i − 2 j) = 20 i + 10 j

C B

v C = v B + ω BC × rC /B = (20 i + 10 j) + ω BC k × (3i) = (20) i + (10 + 3ω BC ) j v C = ω disk × r = ω disk k × (1 j) = −ω disk (1) i Equating the components of these two expressions for v C and solving, we find ω BC = −3.33 rad/s, ω disk = −20 rad/s, v C = 20 m/s. The acceleration analysis (note that point C is moving in a circle of radius 2 m ): a B = a A + α AB × rB / A − ω AB 2 r = 0 + (20 k) × (1i − 2 j) − (10) 2 (1i − 2 j) = (−60 i + 220 j) a C = a B + α BC × rC / B − ω BC 2 rC / B = (−60 i + 220 j) + α BC k × (3i) − (−3.33) 2 (3i) = (−93.3) i + (220 + 3α BC ) j a C = α disk × r +

v C2 (20) 2 j = α disk k × (1 j) + j = −α disk (1) i + 200 j 2m 2

Equating the components of these two expressions for a C and solving, we find α BC = −6.67 rad/s 2 , α disk = 93.3 rad/s 2 . α BC = 6.67 rad/s 2 clockwise, α disk = 93.3 rad/s 2 counterclockwise.

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Problem 17.101 If ω AB = 2 rad/s, α AB = 2 rad/s 2 , ω BC = 1 rad/s, and α BC = 4 rad/s 2 , what is the acceleration of point C where the scoop of the excavator is attached?

y B vBC

vAB a AB

aBC

5.5 m

Solution:

The vector locations of points A, B, C are

r A = 4 i + 1.6 j (m),

1.6 m

rB = 7 i + 5.5 j (m).

rC = 9.3i + 5 j (m).

The vectors

2.3 m

rB /A = rB − r A = 3i + 3.9 j (m), rC /B = rC − rB = 2.3i − 0.5 j (m).

The acceleration of point C in terms of the acceleration at point B is

The acceleration of point B is

2 (r a C = a B + α BC × rC /B − ω BC C /B )

a B = α AB × rB /A − ω 2AB rB /A .  i j k   a B =  0 0 2  − (2 2 )(3.0 i + 3.9 j),  3 3.9 0    a B = +2(−3.9 i + 3 j) − (4)(3i + 3.9 j)

 i j k   = −19.8i − 9.6 j +  0 − 0 4  − 1 2 (2.3i − 0.5 j),   2.3 −0.5 0  a C = −19.8i − 9.6 j − 2 i − 9.2 j − 2.3i + 0.5 j = −24.1i − 18.3 j (m/s 2 ).

= −19.8i − 9.6 j(m/s 2 ).

Problem 17.102 If the velocity of point C of the excavator is v C = 4 i (m/s) and is constant, what are ω AB , α AB , ω BC , and α BC ?

y B vBC

vAB a AB

Solution:

aBC

5.5 m

The strategy is to determine the angular velocities ω AB , ω BC from the known velocity at point C, and the angular velocities α AB , α BC from the data that the linear acceleration at point C is constant.

5m A 1.6 m

The angular velocities: The vector locations of points A, B, C are

r A = 4 i + 1.6 j (m),

2.3 m

rB = 7 i + 5.5 j (m), rC = 9.3i + 5 j (m). Solve: ω AB = −0.8787 rad/s, ω BC = −1.146 rad/s.

The vectors rB /A = rB − r A = 3i + 3.9 j (m),

The angular accelerations: The acceleration of point B is

rC /B = rC − rB = 2.3i − 0.5 j (m).

a B = a AB × rB /A − ω 2AB rB /A

The velocity of point B is

 i j k    =  0 0 α AB  − (ω 2AB )(3i + 3.9 j),  3 3.9 0    a B = −3.9α AB i + 3α AB j − 3ω 2AB i − 3.9ω 2AB j (m/s 2 ).

 i j k  v B = ω AB × rB /A =  0 0 ω AB  3 3.9 0 

   = −3.9ω i + 3ω j. AB AB   

The velocity of C in terms of the velocity of B v C = v B + ω BC × rC /B   i j k   = −3.9ω AB i + 3ω AB j +  0 0 −ω BC  ,   2.3 −0.5 0   v C = −3.9ω AB i + 3ω AB j − 0.5ω BC i − 2.3ω BC j (m/s).

The acceleration of C in terms of the acceleration of B is 2 r a C = a B + a BC × rC /B − ω BC C /B

 i j k  = a B +  0 −α BC 0  2.3 −0.5 0 

  2 (2.3i − 0.5 j)  − ω BC   

a C = (−3.9α AB − 3ω 2AB ) i + (3α AB − 3.9ω 2AB ) j

Substitute v C = 4 i (m/s), and separate components:

2 ) i + (− 2.3α 2 + (−0.5α BC − 2.3ω BC BC + 0.5ω BC ) j.

4 = −3.9ω AB − 0.5ω BC , 0 = 3ω AB − 2.3ω BC .

340

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17.102 (Continued) Substitute a C = 0 from the conditions of the problem, and separate components: 2 , 0 = −3.9α AB − 0.5α BC − 3ω 2AB − 2.3ω BC 2 . 0 = 3α AB − 2.3α BC − 3.9ω 2AB + 0.5ω BC

Solve: α BC = −2.406 rad/s 2 , α AB = −1.06 rad/s 2 .

Problem 17.103 The steering linkage of a car is shown. Member DE rotates about the fixed pin E. The right brake disk is rigidly attached to member DE. The tie rod CD is pinned at C and D. At the instant shown, the Pitman arm AB has a counterclockwise angular velocity of 1 rad/s and a clockwise angular acceleration of 2 rad/s 2 . What is the angular acceleration of the right brake disk? Brake disks

A B

180 mm

100 mm E 220 mm

C Steering link

Solution:

Note that the steering link translates, but does not rotate. The velocity analysis: v C = v B = v A + ω AB × rB / A = 0 + (1k) × (−0.18 j) = 0.18i v D = v C + ω CD × rD / C = 0.18i + ω CD k × (0.34 i − 0.08 j) = (0.18 + 0.08ω CD ) i + (0.34 ω CD ) j v E = v D + ω DE × rE / D = (0.18 + 0.08ω CD ) i + (0.34 ω CD ) j + ω DE k × (0.07 i + 0.2 j) = (0.18 + 0.08ω CD − 0.2ω DE ) i + (0.34 ω CD + 0.07ω DE ) j

Since point E is fixed, we can set both components to zero and solve. We find ω CD = −0.171 rad/s,

ω DE = 0.832 rad/s.

The acceleration analysis:

460 mm

340 mm

200 mm 70 mm

2r a E = a D + α DE × rE /D − ω DEE /D

= (−0.370 + 0.08αCD ) i + (0.182 + 0.34αCD ) j + α DE k × (0.07 i + 0.2 j) − (0.832) 2 (0.07 i + 0.2 j) = (−.418 + 0.08αCD − 0.2α DE ) i + (0.0436 + 0.34αCD + 0.07α DE ) j Since point E cannot move, we set both components of a E to zero and solve. We find: αCD = 0.278 rad/s 2 ,

α DE = −1.98 rad/s 2 .

The angular acceleration of the right brake is the same as the angular acceleration of DE. α right brake = 1.98 rad/s 2 clockwise.

a C = a B = a A + α AB × rC / B − ω 2ABB r /A = 0 + (−2k) × (−0.18 j) − (1) 2 (−0.18 j) = (−0.36 i + 0.18 j) a D = a C + αCD × rD / C − ω CD2 rD / C = (−0.36 i + 0.18 j) + αCD k × (0.34 i − 0.08 j) − (−0.171) 2 × (0.34 i − 0.08 j) = (−0.370 + 0.08αCD ) i + (0.182 + 0.34αCD ) j

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Problem 17.104 Bar AB has a constant clockwise angular velocity of 10 rad/s. Determine the magnitudes of the velocity and acceleration of point E.

Solution:

400 mm

In terms of the angular velocity of bar AB, ω AB = −10 rad/s, the velocity of point B is v B = −ω AB (0.4 m) i. The position vector of C relative to B is rC /B = 0.7 i − 0.4 j (m). The velocity of C in terms of B is

v C = v B + ω BC × rC / B = −ω AB (0.4 m) i +

400 mm

700 mm i j k 0 0 ω BC 0.7 m −0.4 m 0

A 700 mm

(1)

= [−ω AB (0.4 m) + ω BC (0.4 m)]i + ω BC (0.7 m)j (m/s). In terms of the angular velocity of bar CDE, the velocity of point C is v C = −ω CDE (0.4 m) j. Equating the two expressions for the velocity of C, we obtain the two equations −ω AB (0.4 m) + ω BC (0.4 m) = 0, ω BC (0.7 m) = −ω CDE (0.4 m).

In terms of the motion of bar CDE, the acceleration of C is a C = ω CDE 2 (0.4 m) i − αCDE (0.4 m)j. Equating the two expressions for the acceleration of C, we obtain the two equations α BC (0.4 m) − ω BC 2 (0.7 m) = ω CDE 2 (0.4 m),

Setting ω AB = −10 rad/s and solving, we obtain ω BC = −10 rad/s, ω CDE = 17.5 rad/s. The magnitude of the velocity of point E is

−40 + α BC (0.7 m) + ω BC 2 (0.4 m) = −αCDE (0.4 m).

v E = (0.7 m) ω CDE = (0.7 m)(17.5 rad/s) = 12.3 m/s.

Solving, we obtain α BC = 481 rad/s 2 , αCDE = −842 rad/s 2 . The acceleration of point E is

The acceleration of point B is

a E = −ω CDE 2 (0.7 m)i + αCDE (0.7 m)j = −214 i − 590 j (m/s 2 ).

a B = −ω AB 2 (0.4 m) j = −40 j (m/s 2 ).

Its magnitude is a E = 627 m/s 2 .

The acceleration of C in terms of B is

v E = 12.3 m/s, a E = 627 m/s 2 .

a C = a B + α BC × rC / B − ω BC 2 rC / B i j k 0 0 α BC 0.7 m −0.4 m 0

= −40 j (m/s 2 ) +

− ω BC 2 [(0.7 m) i + (−0.4 m) j] = [α BC (0.4 m) − ω BC 2 (0.7 m)]i + [−40 + α BC (0.7 m) + ω BC 2 (0.4 m)] j (m/s 2 ).

Problem 17.105 Bar CD has a counterclockwise angular velocity of 6 rad/s and a counterclockwise angular acceleration of 180 rad/s 2 . Determine the angular velocity ω AB and the angular acceleration α AB .

Solution: y rB/C

B rB/A

rC/D

The position vector of C relative to D is rC /D = −0.35i + 0.35 j (m). The velocity of C in terms of D is

350 mm 200 mm

aAB vAB

v C = v D + ω CD × rC / D = 0 +

i 0

j 0

−0.35 m 0.35 m

k ω CD 0

(1)

= −ω CD (0.35) i − ω CD (0.35) j (m/s). 300 mm

342

350 mm

The position vector of B relative to C is rB /C = −0.3i − 0.15 j (m). Using the result (1), the velocity of B in terms of C is

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17.105 (Continued) v B = v C + ω BC × rB / C

Using this result, the acceleration of B in terms of C is

= −ω CD (0.35) i − ω CD (0.35) j (m/s) i 0

j 0

a B = a C + α BC × rB / C − ω BC 2 rB / C

k ω BC

−0.3 m −0.15 m

= [−αCD (0.35) + ω CD 2 (0.35)]i + [−αCD (0.35) − ω CD 2 (0.35)] j

i 0

= [−ω CD (0.35) + ω BC (0.15)] i + [−ω CD (0.35)

j 0

k α BC

−0.3 m −0.15 m

− ω BC (0.3)] j (m/s).

− ω BC 2 [(−0.3 m) i + (−0.15 m) j]

= [−αCD (0.35) + ω CD 2 (0.35) + α BC (0.15) + ω BC 2 (0.3)]i +

The velocity of B can also be expressed in terms of the angular velocity of bar AB:

[−αCD (0.35) − ω CD 2 (0.35) − α BC (0.3) + ω BC 2 (0.15)] j (m/s 2 ).

v B = −ω AB (0.2) i (m/s).

The acceleration of B can also be expressed in terms of the motion of bar AB:

Equating the two expressions for the velocity of B, we obtain the two equations

a B = −α AB (0.2) i − ω AB 2 (0.2) j (m/s 2 ).

−ω CD (0.35) + ω BC (0.15) = −ω AB (0.2),

Equating the two expressions for the acceleration of B, we obtain the two equations

−ω CD (0.35) − ω BC (0.3) = 0. Setting ω CD = 6 rad/s and solving, we obtain ω AB = 15.8 rad/s, ω BC = −7 rad/s.

2r a C = a D + α CD × rC / D − ω CDC /D

i 0

j 0

k αCD

−0.35 m 0.35 m

−αCD (0.35) − ω CD 2 (0.35) − α BC (0.3) + ω BC 2 (0.15) = −ω AB 2 (0.2). Setting αCD = 180 rad/s 2 and solving, we obtain α AB = 225 rad/s 2 , α BC = −62.1 rad/s 2 .

The acceleration of C in terms of D is

= 0+

−αCD (0.35) + ω CD 2 (0.35) + α BC (0.15) + ω BC 2 (0.3) = −α AB (0.2),

ω AB = 15.8 rad/s, α AB = 225 rad/s 2 .

2 [−(0.35 m) i + (0.35 m) j] − ω CD 2 (0.35)]i = [−αCD (0.35) + ω CD 2 (0.35)] j (m/s 2 ). + [−αCD (0.35) − ω CD

Problem 17.106 If ω AB = 4 rad/s counterclockwise and α AB = 12 rad/s 2 counterclockwise, what is the acceleration of point C?

C B

400 mm D

Solution:

The velocity of B is

600 mm

500 mm

v B = v A + ω AB × rB / A = O+

i 0

j 0

k ω AB

0.3 0.6

A 600 mm 300 mm

= −0.6ω AB i + 0.3ω AB j.

200 mm

300 mm

The velocity of D is v D = v B + ω BD × rD / B = −0.6ω AB i + 0.3ω AB j +

Solving these two eqns with ω AB = 4 rad/s, we obtain

i j k 0 0 ω BD . 0.8 −0.1 0

(1)

v D = v E + ω DE × rD / E = O +

a B = a A + α AB × rB / A − ω 2AB rB / A j 0

k ω DE

−0.3 0.5

(2)

= O+

i 0

j 0

k α AB

0.3 0.6

− ω 2AB (0.3i + 0.6 j)

= (−0.6α AB − 0.3ω 2AB ) i + (0.3α AB − 0.6ω 2AB ) j.

Equating i and j components in Eqns. (1) and (2), we obtain −0.6ω AB + 0.1ω BD = −0.5ω DE ,

(3)

0.3ω AB + 0.8ω BD = −0.3ω DE .

(4)

M17_BEDF2222_06_SE_C17_Part2.indd 343

ω DE = 5.51 rad/s.

The acceleration of B is

We can also express the velocity of D as i 0

ω BD = −3.57 rad/s,

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17.106 (Continued) The acceleration of C is

The acceleration of D is 2 r a D = a B + α BD × rD / B − ω BD D/B

2 r a C = a B + α BD × rC / B − ω BD C/B

= (−0.6α AB − 0.3ω 2AB ) i + (0.3α AB − 0.6ω 2AB ) j +

i j k 0 0 α BD 0.8 −0.1 0

(5)

= (−0.6α AB − 0.3ω 2AB ) i + (0.3α AB − 0.6ω 2AB ) j

2 (0.8i − 0.1 j). − ω BD

i 0

j 0

k α BD

0.6 0.3

2 (0.6 i + 0.3 j). − ω BD

(9)

a C = −7.78i − 33.5 j (m/s 2 ).

We can also express the acceleration of D as 2 r a D = a E + α DE × rD / E − ω DE D/E

= O+

i 0

j 0

k α DE

−0.3 0.5

2 (− 0.3i + 0.5 j). − ω DE

(6)

Equating i and j components in Eqns. (5) and (6), we obtain 2 − 0.6α AB − 0.3ω 2AB + 0.1α BD − 0.8ω BD 2 , = −0.5α DE + 0.3ω DE 2 0.3α AB − 0.6ω 2AB + 0.8α BD + 0.1ω BD 2 . = −0.3α DE − 0.5ω DE

(7)

(8)

Solving these two eqns with α AB = 12 rad/s 2 , we obtain α BD = −39.5 rad/s 2 .

Problem 17.107 The angular velocities and angular accelerations of the grips of the shears are shown. What is the resulting angular acceleration of the jaw BD? Solution:

From 17.57 we know that ω BD = −0.06 rad/s

a D = a C + αCD × rD / C − ω CD 2 rD / C = 0 − (0.08 rad/s 2 )k × (0.025i + 0.018 j) m − (0.12 rad/s) 2 (0.025i + 0.018 j) m = (0.00108i − 0.00226 j) m/s 2 a B = a D + α BD × rB / D − ω BD 2 rB / D = (0.00108i − 0.00226 j) m/s 2 + α BD k × (−0.05i − 0.018 j) m − (−0.06 rad/s) 2 (−0.05i − 0.018 j) m = (0.00126 m/s 2 + {0.018 m}α BD ) i 0.12 rad/s

+ (−0.00219 m/s 2 − {0.05 m}α BD ) j From symmetry, B cannot accelerate in the j direction. Therefore − 0.00219 m/s 2 − {0.05 m}α BD = 0 ⇒ α BD = −0.0439 rad/s 2

D B

18 mm

C E

α BD = 0.0439 rad/s 2 CW. 25 mm 25 mm

344

0.08 rad/s2

0.12 rad/s

0.08 rad/s2

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Solution:

The constraint that the arm CD remain vertical means that the angular velocity of arm CD is zero. This implies that arm CD translates only, and in a translating, non-rotating element the velocity and acceleration at any point is the same, and the velocity and acceleration of arm CD is the velocity and acceleration of point C. The vectors:

y 300 m

B 158

30 0m m

Problem 17.108 If arm AB has a constant clockwise angular velocity of 0.8 rad/s, arm BC has a constant clockwise angular velocity of 0.2 rad/s, and arm CD remains vertical, what is the acceleration of part D?

170 mm 508

rB /A = 300( i cos 50 ° + j sin 50°) = 192.8i + 229.8 j (mm). rC /B = 300( i cos15° − j sin15°) = 289.78i − 77.6 j (mm). The acceleration of point B is a B = α AB × r A /B − ω 2AB r A /B = −ω 2AB (192.8i + 229.8 j) (mm/s 2 ), since α AB = 0. a B = −123.4 i − 147.1 j (mm/s). The acceleration of C in terms of the acceleration of B is 2 r a C = a B + α BC × rC /B − ω BC C /B 2 (289.8i − 77.6 j), = −123.4 i − 147.1 j − ω BC

since α BC = 0.a C = −135i − 144 j (mm/s 2 ). translating:

Since

a D = a C = −135i − 144 j (mm/s 2 ).

y 300 m

B 0m

158 C

Problem 17.109 If arm AB has a constant clockwise angular velocity of 0.8 rad/s and you want D to have zero velocity and acceleration, what are the necessary angular velocities and angular accelerations of arms BC and CD? Solution:

Except for numerical values, the solution follows the same strategy as the solution strategy for Problem 17.105. The vectors: rB /A = 300( i cos 50 ° + j sin 50°) = 192.8i + 229.8 j (mm). rC /B = 300( i cos15° − j sin15°) = 289.78i − 77.6 j (mm),

170 mm 508 A

x D

rC /D = 170 j (mm). The velocity of point B is  i j k   v B = ω AB × rB / A =  0 0 −0.8   192.8 229.8 0    = 183.8i − 154.3 j (mm/s). The velocity of C in terms of the velocity of B is  i j k  v C = v B + ω BC × rC / B = v B +  0 0 ω BC  289.8 −77.6 0  v C = 183.9 i − 154.3 j + ω BC (77.6 i + 289.8 j)(mm/s).

  .   

The velocity of C in terms of the velocity of point D:  i j k  v C = ω CD × rC /D =  0 0 ω CD  0 170 0 

   = −170 ω i (mm/s). CD   

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17.109 (Continued) Equate the expressions for v C and separate components:

The acceleration of point C in terms of the acceleration of point D:

183.9 + 77.6ω BC = −170 ω CD ,

 i j k  aC = =  0 0 αCD  0 170 0  2 a C = −170αCD i − 170 ω CD j (mm/s 2 ). 2 r αCD × rC / D − ω CD C/D

− 154.3 + 289.8ω BC = 0. Solve: ω BC = 0.532 rad/s, ω CD = 1.325 rad/s. Get the angular accelerations. The acceleration of point B is a B = α AB × rB / A − ω 2AB rB / A = −ω 2AB (192.8i + 229.8 j) = −123.4 i − 147.1 j (mm/s 2 ).

   − ω 2 (170 j). CD   

Equate the expressions and separate components: 2 −123.4 + 77.6α BC − 289.8ω BC = −170αCD , 2 2 . −147.1 + 289.8α BC + 77.6ω BC = −170 ω CD

The acceleration of point C in terms of the acceleration of point B: 2 r 2 a C = a B + α BC × rC / B − ω BC C / B = a B − ω BC rC / B . 2 i a C = −123.4 i − 147.1 j + 77.6α BC i + 289.8α BC j − 289.8ω BC 2 j (mm/s 2 ). + 77.6ω BC

Solve: α BC = −0.598 rad/s 2 , αCD = 1.482 rad/s 2 , where the negative sign means a clockwise angular acceleration.

Problem 17.110 If you want arm CD to remain vertical and you want part D to have velocity v D = 1.0 i (m/s) and zero acceleration, what are the necessary angular velocities and angular accelerations of arms AB and BC?

y 300 m

B 0m m

158

Solution:

The constraint that CD remain vertical with zero acceleration means that every point on arm CD is translating, without rotation, at a velocity of 1 m/s. This means that the velocity of point C is v C = 1.0 i (m/s), and the acceleration of point C is zero. The vectors:

170 mm 508 A

rB /A = 300( i cos 50 ° + j sin 50°) = 192.8i + 229.8 j (mm).

rC /B = 300( i cos15° − j sin15°) = 289.78i − 77.6 j (mm). The angular velocities of AB and BC: The velocity of point B is  i j k  v B = ω AB × rB / A =  0 0 ω AB  192.8 229.8 0  = ω AB (−229.8i + 192.8 j) (mm/s).

     

The velocity of C in terms of the velocity of B is

The acceleration of C in terms of the acceleration of B is

 i j k   v C = v B + ω BC × rC / B = v B +  0 0 ω BC  .  289.8 −77.6 0   v C = −229.8ω AB i + 192.8ω AB j + ω BC (77.6 i + 289.8 j) (mm/s)

2 r a C = a B + α BC × rC / B − ω BC C/B

The velocity of C is known, v C = 1000 i (mm/s). Equate the expressions for v C and separate components: 1000 = −229.8ω AB + 77.6ω BC , 0 = 192.8ω AB + 289.8ω BC . Solve: ω AB = −3.55 rad/s, ω BC = 2.36 rad/s,

 i j k   2 (289.8i − 77.6 j),  α BC  − ω BC 0 0 = aB +    289.8 −77.6 0   2 (289.8i − 77.6 j) (mm/s 2 ). a C = a B + α BC (77.6 i + 289.8 j) − ω BC The acceleration of point C is known to be zero. Substitute this value for a C , and separate components: 2 = 0, −229.8α AB − 192.8ω 2AB + 77.6α BC − 289.8ω BC 2 192.8α AB − 229.8ω 2AB + 289.8α BC + 77.6ω BC = 0.

where the negative sign means a clockwise angular velocity. The accelerations of AB and BC: The acceleration of point B is  i j k  a B = α AB × rB / A − ω 2AB rB / A =  0 0 α AB  192.8 229.8 0  − ω 2AB (192.8i + 229.8 j (mm/s 2 ).

     

Solve: α AB = −12.1 rad/s 2 , α BC = 16.5 rad/s 2 , where the negative sign means a clockwise angular acceleration.

a B = α AB (−229.8i + 192.8 j) − ω 2AB (192.8i + 229.8 j) (mm/s 2 )

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Problem 17.111 Link AB of the robot’s arm is rotating with a constant counterclockwise angular velocity of 2 rad/s, and link BC is rotating with a constant clockwise angular velocity of 3 rad/s. Link CD is rotating at 4 rad/s in the counterclockwise direction and has a counterclockwise angular acceleration of 6 rad/s 2 . What is the acceleration of point D?

Solution: The acceleration of B is a B = a A + α AB × rB /A − ω 2ABrB /A . Evaluating, we get a B = 0 + 0 − (2) 2 (0.3cos30 °i + 0.3sin30 ° j) = −1.039 i − 0.600 j (m/s 2 ). 2 r The acceleration of C is a C = a B + α BC × rC /B − ω BC C /B . Evaluating, we get

a C = −1.039 i − 0.600 j − (3) 2 (0.25cos 20 °i − 0.25sin 20 ° j) = −3.154 i + 0.170 j (m/s 2 ).

y 250

308

2 r The acceleration of D is a D = a C + αCD × rD /C − ω CD D /C . Evaluating, we get

208

a D = −3.154 i + 0.170 j +

i j k 0 0 6 0.25 0 0

− (4) 2 (0.25i)

= −7.154 i + 1.67 j (m/s 2 ).

250 mm

Problem 17.112 The upper grip and jaw of the pliers ABC is stationary. The lower grip DEF is rotating in the clockwise direction with a constant angular velocity of 0.2 rad/s. At the instant shown, what is the angular acceleration of the lower jaw CFG ?

Solution:

From 17.42 we know that ω BE = ω CFG = 0.0857 rad/s.

a E = a B + α BE × rE /B − ω AB 2 rE /B = 0 + α BE k × (0.07 i − 0.03 j) m − (0.0857 rad/s) 2 (0.07 i − 0.03 j) m = (−0.000514 m/s 2 + { 0.03 m } α BE ) i + (0.0002204 m/s 2 + { 0.07 m } α BE ) j

Stationary A

2r a F = a E + α EF × rF /E − ω EFF /E

= (− 0.000514 m/s 2 + {0.03 m}α BE ) i

30 mm G E D

+ (0.0002204 m/s 2 + {0.07 m}α BE ) j + 0 − (0.0857 rad/s) 2 (0.03 m) i

= (−0.00171 m/s 2 + {0.03 m}α BE ) i 70 mm

+ (0.0002204 m/s 2 + {0.07 m}α BE ) j

30 mm 30 mm

2r a C = a F + αCFG × rC /F − ω CFG C /F

= (− 0.00171 m/s 2 + {0.03 m}α BE ) i + (0.0002204 m/s 2 + {0.07 m}α BE ) j + αCFG k × (0.03 m) j − (0.0857 rad/s) 2 (0.03 m) j = (−0.00171 m/s 2 + {0.03 m}[α BE − αCFG ]) i + (0.07 m)α BE j Since C is fixed we have −0.00171 m/s 2 + {0.03 m}[α BE − αCFG ] = 0    (0.07 m)α BE = 0  ⇒

α BE = 0 αCFG = − 0.0571 rad/s 2 α BE = 0

αCFG = 0.0571 rad/s 2 CW.

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Problem 17.113 The horizontal member ADE supporting the scoop is stationary. If link BD has a clockwise angular velocity of 1 rad/s and a counterclockwise angular acceleration of 2 rad/s 2 , what is the angular acceleration of the scoop?

2 ft A

1 ft 6 in E

D 5 ft

Solution:

1 ft

2 ft 6 in

Scoop

The velocity of B is

v B = v D + ω BD × rB /D = 0 +

i j k 0 0 −1 1 2 0

We can also express a C as 2 r a C = a E + αCE × rC /E − ω CE C /E = 0 + (α CE k )

× (1.5 j) − (−1.467) 2 (1.5 j)

= 2 i − j (ft/s).

= −1.5αCE i − 3.23 j.

The velocity of C is

v C = v B + ω BC × rC /B = 2 i − j + 0 +

i 0

j 0

k ω BC

2.5 −.5

(1)

Equating i and j components in Equations (3) and (4), we get −5 + 0.5α BC − (0.4) 2 (2.5) = −1.5αCE , and 2.5α BC + (0.4) 2 (0.5) = −3.23.

We can also express v c as

Solving, we obtain

v C = v E + ω CE × rC /E = 0 + (ω CE k) × (1.5 j) = −1.5ω CE i. (2)

α BC = −1.32 rad/s 2

Equating i and j components in Equations (1) and (2) we get 2 + 0.5ω BC = −1.5ω CE , and −1 + 2.5ω BC = 0. Solving, we obtain ω BC = 0.400 rad/s and ω CE = −1.467 rad/s.

αCE = 4.04 rad/s 2 . y

The acceleration of B is 2 r a B = a D + α BD × rB /D − ω BD B /D ,

or a B = 0 +

(4)

i j k 0 0 2 1 2 0

1 rad/s

B vBC aBC

2 rad/s2

− (1) 2 ( i + 2 j)

C vCE aCE E

= −5i (ft/s 2 ). The acceleration of C is 2 r a C = a B + α BC × rC /B − ω BC C /B

a C = − 5i +

i 0

j 0

k α BC

2.5 −0.5

− (0.4) 2 (2.5i − 0.5 j).

(3)

Problem 17.114 The ring gear is fixed, and the hub and planet gears are bonded together. The connecting rod has a counterclockwise angular acceleration of 10 rad/s 2 . Determine the angular accelerations of the planet and sun gears. Solution:

Planet gear Hub gear Connecting rod

The x components of the accelerations of pts B and

140 mm

340 mm 240 mm

C are a Bx = 0,

720 mm

a Cx = −(10 rad/s 2 )(0.58 m)

Sun gear

= −5.8 m/s 2 . Let α P and α S be the angular accelerations of the planet and sun gears.

Ring gear

a B = a C + α P × rB /C − ω P2 rB /C = aC +

i j k 0 0 αP 0 0.14 0

− ω P2 (0.14 j).

The i component of this equation is. 0 = −5.8 − 0.14α P .

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17.114 (Continued) y

We obtain αP

= −41.4 rad/s 2 .

a D = a C + α P × rD /C − ω P2 rD /C

Also,

= aC +

i j k 0 0 −41.4 0 −0.34 0

B − ω P2 (−0.34 j).

C D

The i component of this equation is a Dx

= −5.8 − (41.4)(0.34) = −19.9 m/s 2 .

Therefore αS =

19.9 = 82.9 rad/s 2 . 0.24

Problem 17.115 The connecting rod has a counterclockwise angular velocity of 4 rad/s and a clockwise angular acceleration of 12 rad/s 2 . Determine the magnitude of the acceleration of point A.

Planet gear Hub gear 140 mm

Connecting rod

Solution:

See the solution of Problem 17.114. The velocities of pts B and C are

340 mm 240 mm

v B = O, v C = −(4)(0.58) i = −2.32 i (m/s). Let ω P and ω S be the angular velocities of the planet and sun gears.

720 mm

v B = v C + ω P × r B /C :

Sun gear

O = −2.32 i + (ω P k) × (0.14 j)

Ring gear

= (−2.32 − 0.14 ω P ) i. We see that ω P = −16.6 rad/s. Also, v D = v C + ω P × r D /C

The acceleration of C is

= −2.32 i + (−16.6k) × (−0.34 j)

a C = a E + (−12k) × rC /E − (4) 2 rC /E

= −7.95i (m/s), So

ωS =

7.95 = 33.1 rad/s. 0.24

= 0+

i j k −12 0 0 0 0.58 0

− (4) 2 (0.58 j)

The x components of the accelerations of pts B and C are = 6.96 i − 9.28 j (m/s 2 ).

a Bx = 0,

Then the acceleration of A is

a Cx = (12 rad/s 2 )(0.58 m)

a A = a C + α P × r A /C − ω P2 r A /C

= 6.96 m/s 2 . a B = a C + α P × rB /C − ω P2 rB /C = aC +

i j k αP 0 0 0 0.14 0

− ω P2 (0.14 j).

The i component is

= 6.96 i − 9.28 j +

i j k 0 0 49.7 0 0.34 0

− (−16.6) 2 (0.34 j)

= −9.94 i − 102.65 j (m/s 2 ). a A = 103 m/s 2 .

0 = 6.96 − 0.14α P , so α P = 49.7 rad/s 2 .

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Problem 17.116 The large gear is fixed. The angular velocity and angular acceleration of bar AB are ω AB = 2 rad/s and α AB = 4 rad/s 2 . Determine the angular accelerations of bars CD and DE.

4 in B

   = −4 ω i (in/s). B   

10 in A

vAB aAB

The acceleration of point D in terms of the acceleration of point C is 2 (16 i) a D = a C + αCD × 16 i − ω CD

υB = 7 rad/s. 4 The velocity of point C is

 i j k   2 (16 i), = a C +  0 0 αCD  − ω CD   16 0 0   a D = −396 i + (16αCD + 56) j (in/s 2 ).

ωB = −

v C = v B + ω B × rBC  i j k    = −28i +  0 0 7  = −28i + 28 j (in/s).  4 0 0   

The acceleration of point D in terms of the acceleration of point E is

The velocity of point D is      

The velocity of point D is also given by  i j k  0 ω DE v D = ω DE × rED =  0  −10 14 0  = −14 ω DE i − 10 ω DE j (in/s).

 i j k    a C = a B + α BC × 4 i − ω B2 (4 i) = −56 i +  0 0 14  − 49(4 i)  4 0 0    = −252 i + 56 j (in/s 2 ).

Equate the velocities v B to obtain the angular velocity of B:

 i j k  v D = v C + ω CD × rCD = −28i + 28 j +  0 0 ω CD  16 0 0  = −28i + (16ω CD + 28) j (in/s).

from which α BC = 14 rad/s 2 . The acceleration of point C in terms of the acceleration of point B is

The lower edge of gear B is stationary. The velocity of B is also  i j k  v B = ω B × rB =  0 0 ω B  0 4 0 

10 in

4 in

Solution: The strategy is to express vector velocity of point D in terms of the unknown angular velocities and accelerations of CD and DE, and then to solve the resulting vector equations for the unknowns. The angular velocities ω CD and ω DE . (See solution to Problem 17.51). The linear velocity of point B is  i j k    v B = ω AB × r AB =  0 0 2  = −28i (in/s).  0 14 0   

16 in

 i j k   2 (− 10 i + 14 j) a D =  0 0 α DE  − ω DE   −10 14 0   = (40 − 14α DE ) i − (10α DE + 56) j (in/s 2 ) Equate the expressions for a D and separate components: − 396 = 40 − 14α DE , 16αCD + 56 = −10α DE − 56. Solve:

     

α DE = 31.1 rad/s 2 , αCD = −26.5 rad/s 2 ,

where the negative sign means a clockwise angular acceleration.

Equate and separate components: (−28 + 14 ω DE ) i = 0, (16ω CD + 28 + 10 ω DE ) j = 0. Solve:

ω DE = 2 rad/s, ω CD = −3 rad/s.

The negative sign means a clockwise rotation. The angular accelerations. The tangential acceleration of point B is  i j k    a B = α AB × rB /A =  0 0 4  = −56 i (in/s 2 ).  0 14 0    The tangential acceleration at the point of contact between the gears A and B is zero, from which  i j k  a B = α BC × 4 j =  0 0 α BC  0 4 0 

350

   = −4α i (in/s 2 ), BC   

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Problem 17.117 The bar is rotating in the counterclockwise direction at 4 rad/s. The coordinate system shown is fixed with respect to the bar. At the instant shown, the velocity and acceleration of the sleeve A relative to the bar are v = 3 ft/s and a = 0. (a) Determine the velocity of A at the instant shown in terms of polar coordinates using Eq. (13.57). (b) Determine the velocity of A at the instant shown using Eq. (17.11).

y a v B

A 2 ft

Solution: (a) From Eq. (13.57), the velocity of the sleeve is dr e + rω e θ dt r = (3 ft/s)e r + (2 ft)(4 rad/s)e θ

v =

= 3 e r + 8 e θ (ft/s). (b) Equation (17.11) for the sleeve is v A = v B + v A rel + ω × r A /B = 0 + (3 ft/s) i +

i j k 0 0 4 rad/s 0 2 ft 0

= (3 ft/s) i + (4 rad/s)(2 ft) j = 3 i + 8 j (ft/s). (We could have written Eq. (17.11) in terms of polar components, but it’s easy to see the equivalence of the two results.) (a) v = 3 e r + 8 e θ ( ft/s ), (b) v A = 3 i + 8 j (ft/s).

Problem 17.118 The bar is rotating in the counterclockwise direction at 4 rad/s and has a counterclockwise angular acceleration of 12 rad/s 2 . The coordinate system shown is fixed with respect to the bar. At the instant shown, the velocity and acceleration of the sleeve A relative to the bar are v = 3 ft/s and a = 6 ft/s 2 . (a) Determine the acceleration of A at the instant shown in terms of polar coordinates using Eqs. (13.58) and (13.59). (b) Determine the acceleration of A at the instant shown using Eq. (17.15).

Solution: (a) From Eqs. (13.58) and (13.59), the acceleration of the sleeve is

(

)

 d 2r dθ 2  d 2θ dr d θ a =  2 −r e e + r 2 + 2  dt dt  r dt dt dt θ

( )

= [(6 ft/s 2 ) − (2 ft)(4 rad/s) 2 ]e r + [(2 ft)(12 rad/s 2 ) + 2(3 ft/s)(4 rad/s)]e θ = −26 e r + 48 e θ (ft/s 2 ). (b) Equation (17.15) for the sleeve is a A = a B + a A rel + 2ω × v A rel + α × r A /B − ω 2 r A /B

y = 0 + (6 ft/s 2 ) i + 2

a v B

A 2 ft

i +

i j k 0 0 4 rad/s 3 ft/s 0 0

0 0 12 rad/s 2 2 ft 0 0

− (4 rad/s) 2 (2 ft) i

= −26 i + 48 j (ft/s 2 ). (We could have written Eq. (17.15) in terms of polar components, but the equivalence of the two results is clear.) (a) a = −26 e r + 48 e θ (ft/s 2 ), (b) a = −26 i + 48 j (ft/s 2 ).

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Problem 17.119 Sleeve C slides at 1 m/s relative to bar BD. Use the body-fixed coordinate system shown to determine the velocity of C. y B

4 rad/s

1 m/s

The velocity of point B is

 i j k   0 2  = −1200( i − j) (mm/s). v B = ω AB × rB /A =  0  600 600 0    Use Eq. (17.11). The velocity of sleeve C is

v C = v B + v Arel + ω BD × rC /B .  i j k   v C = −1200 i + 1200 j + 1000 i +  0 0 4  .  400 0 0    v C = −200 i + 2800 j (mm/s) .

400 mm 600 mm

Solution:

2 rad/s A 600 mm

Problem 17.120 The angular accelerations of the two bars are zero and sleeve C slides at a constant velocity of 1 m/s relative to bar BD. What is the acceleration of sleeve C? 4 rad/s

1 m/s

a B = −ω 2AB rB /A = −4(600 i + 600 j) = −2400 i − 2400 j (mm/s 2 ).

2 r a C = a B + a Crel + 2ω BD × v Crel + α BD × rC /B − ω BD C /B .

400 mm 600 mm

From Problem 17.119, ω AB = 2 rad/s, ω BC = 4 rad/s. The acceleration of point B is

Use Eq. (17.15). The acceleration of C is

y B

Solution:

   − ω 2 (400 i), BD   

 i j k  0 ω BD a C = −2400 i − 2400 j + 2  0  1000 0 0  a C = −8800 i + 5600 j (mm/s 2 ).

2 rad/s A 600 mm

Problem 17.121 Bar BD has a constant counterclockwise angular velocity of 4 rad/s. At the instant shown, determine the angular velocity of bar AC and the velocity of the pin relative to the slot.

Solution: y C

rC/A

rC/B

C A

4 in

Assume that the coordinate system is fixed with respect to bar BC. The velocity of C relative to A is v C = v A + ω AC × rC /A

B = 0+ 7 in

352

i 0

j 0

k ω AC

7 in 4 in

= −ω AC (4 in) i + ω AC (7 in) j.

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17.121 (Continued) Equation (17.11) for the velocity of C relative to B is

Equating the two expressions, we obtain the equations

v C = v B + v C rel + ω BC × rC /B

−ω AC (4 in) = −(16 in/s),

= 0 + v C rel j +

ω AC (7 in) = v C rel ,

i j k 0 0 4 rad/s 0 0 4 in

Yielding ω AC = 4 rad/s, v C rel = 28 in/s.

= −(16 in/s) i + v C rel j.

4 rad/s counterclockwise, 28 in/s upward.

Problem 17.122 The velocity of pin C relative to the slot is 21 in/s upward and is decreasing at 42 in/s 2 . What are the angular velocity and angular acceleration of bar AC?

Solution:

See the solution of Problem 17.121. Solving Eqs. (3), (4), (7), and (8) with v Crel = 21 in/s and a C rel = −42 in/s 2 , we obtain

4 in

ω AC = 3 rad/s,

α AC = −6 rad/s 2 . 7 in

Problem 17.123 What should the angular velocity and acceleration of bar AC be if you want the angular velocity and acceleration of bar BD to be 4 rad/s counterclockwise and 24 rad/s 2 counterclockwise, respectively?

4 in

Solution:

See the solution of Problem 17.121. Solving Eqs. (3), (4), (7), and (8) with ω BD = 4 rad/s 2 and α BD = 24 rad/s 2 , we obtain A

ω AC = 4 rad/s, α AC =

52 rad/s 2 . 7 in

Problem 17.124 Bar AB has an angular velocity of 4 rad/s in the clockwise direction. What is the velocity of pin B relative to the slot?

60 mm

Solution:

The velocity of point B is

 j k  i 0 −ω AB v B = ω AB × rB /A =  0  115 60 0 

   = 240 i − 460 j (mm/s).   

80 mm

The velocity of point B is also determined from bar CB v B = v Brel + ω CB × (35i + 60 j),  i j k    v B = v Brel +  0 0 ω CB   35 60 0   v B = v Brel i − 60 ω CB i + 35ω CB j (mm/s).

35 mm

vBrel

Equate like terms: 240 = v Brel − 60 ω CB , − 460 = 35ω CB from which ω BC = −13.14 rad/s, v Brel = −548.6 mm/s.

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Problem 17.125 Bar AB has an angular velocity of 4 rad/s in the clockwise direction and an angular acceleration of 10 rad/s 2 in the counterclockwise direction. What is the acceleration of pin B relative to the slot?

60 mm

The acceleration of pin B in terms of bar BC is 2 r a B = a Brel i + 2ω BC × v Brel + α BC × rB / C − ω BC B/C,

  i j k    a B = a Brel i + 2  0 0 − 13.14     −548.6 0 0

80 mm

a B = α AB × rB /A − ω 2AB rB /A  i j k   0 10  − (16)(115i + 60 j) (mm/s 2 ), =  0  115 60 0    a B = −600 i + 1150 j − 1840 i − 960 j = −2440 i + 190 j (mm/s 2 ).

Solution: Use the solution to Problem 17.124, from which ω BC = −13.14 rad/s, v Brel = −548.6 mm/s. The angular acceleration and the relative acceleration. The acceleration of point B is

 i j k   +  0 0 α BC  − (13.14 2 )(35i + 60 j).  35 60 0   a B = a Brel i + 14,419.5 j + 35α BC j − 60α BC i

35 mm

− 6045.7 i − 10,364.1 j. Equate expressions for a B and separate components: −2440 = a Brel − 60α BC − 6045.7, 190 = 14,419.6 + 35α BC − 10364.1 . Solve: a Brel = −3021i (mm/s 2 ) , α BC = −110.4 rad/s 2 .

Problem 17.126 The hydraulic actuator BC of the crane is extending (increasing in length) at a rate of 0.2 m/s. At the instant shown, what is the angular velocity of the crane’s boom AD? Strategy: Use Eq. (17.8) to write the velocity of point C in terms of the velocity of point A, and use Eq. (17.11)to write the velocity of point C in terms of the velocity of point B. Then equate your two expressions for the velocity of point C. D

Solution:

Using Bar ACD

v C = v A + ω AD × rC /A = 0 + ω AD k × (3i + 1.4 j) m = ω AD (−1.4 i + 3 j) m Using cylinder BC v C = v B + v Brel + ω BC × rC /B = 0 + (0.2 m/s)

i + 2.4 j ( 1.2 ) + ω k × (1.2i + 2.4 j) m 1.2 + 2.4 2

= (0.0894 m/s − { 2.4 m } ω BC ) i + (0.1789 m/s + {1.2 m } ω BC ) j Equating the components of the two expressions we have (−1.4 m)ω AD = 0.0894 m/s − (2.4 m)ω BC  ω BC = 0.0940 rad/s ⇒  (3 m)ω AD = 0.1789 m/s + (1.2 m)ω BC ω AD = 0.0972 rad/s  ω AD = 0.0972 rad/s CCW.

C A

2.4 m 1m

1.8 m

354

1.2 m

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Problem 17.127 The hydraulic actuator BC of the crane is extending (increasing in length) at a constant rate of 0.2 m/s. At the instant shown, what is the angular acceleration of the crane’s boom AD?

Solution:

Use the angular velocities from 17.126

Bar ACD a C = a A + α ACD × rC /A − ω ACD 2 rC /A = 0 + α ACD k × (3i + 1.4 j) m − ω ACD 2 (3i + 1.4 j) m = (−{1.4 m}α ACD − {3 m}ω ACD 2 ) i + ({3 m}α ACD

− {1.4 m}ω ACD 2 ) j Cylinder BC a C = a B + a Brel + α BC × rC / B − ω BC 2 rC / B + 2ω BC × v Brel = 0 + 0 + α BC k × (1.2 i + 2.4 j) m − ω BC 2 (1.2 i + 2.4 j) m

+ 2ω BC k × (0.2 m/s)

= (−{2.4 m}α BC − {1.79 m/s}ω BC − {1.2 m}ω BC 2 ) i

2.4 m

i + 2.4 j ( 1.2 ) 1.2 + 2.4

+ ({1.2 m}α BC + {0.894 m/s}ω BC − {2.4 m}ω BC 2 ) j

Equating the two expressions for the acceleration of C we have −{1.4 m}α AD − {3 m}ω 2AD= −{2.4 m}α BC

1.8 m

1.2 m

2 − {1.79 m/s}ω BC − {1.2 m}ω BC

{3 m}α AD − {1.4 m}ω AD 2 = {1.2 m}α BC + { 0.894 m/s } ω BC − { 2.4 m } ω BC 2 Solving (using the angular velocities from 17.126) we find α BC = −0.0624 rad/s, α AD = 0.000397 rad/s α AD = 0.000397 rad/s CCW.

Problem 17.128 The angular velocity ω AC = 5 ° per second. Determine the angular velocity of the hydraulic actuator BC and the rate at which the actuator is extending. C

Solution:

The point C effectively slides in a slot in the arm BC. The angular velocity of ω AC = 5

aAC

π ( 180 ) = 0.0873 rad/s.

2.4 m

vAC

The velocity of point C with respect to arm AC is  i j k  0 ω AC v C = ω AC × rC /A =  0  2.6 2.4 0  = −0.2094 i + 0.2269 j (m/s),

     

The unit vector parallel to the actuator BC is e =

1.4 m

1.2 m

1.2 i + 2.4 j = 0.4472 i + 0.8944 j. 1.2 2 + 2.4 2

vCrel y

The velocity of point C in terms of the velocity of the actuator is v C = v Crel e + ω BC × rC /B .  i j k   v C = v Crel (0.4472 i + 0.8944 j) +  0 0 ω BC   1.2 2.4 0   v C = v Crel (0.4472 i + 0.8944 j) + ω BC (−2.4 i + 1.2 j).

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58/s

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17.128 (Continued) Equate like terms in the two expressions: −0.2094 = 0.4472v Crel − 2.4 ω BC , 0.2269 = 0.8944v Crel + 1.2ω BC . ω BC = 0.1076 rad/s = 6.17 deg/s, v Crel = 0.109 (m/s), which is also the velocity of extension of the actuator.

Problem 17.129 The angular velocity ω AC = 5 ° per second and the angular acceleration α AC = −2 ° per second squared. Determine the angular acceleration of the hydraulic actuator BC and the rate of change of the actuator’s rate of extension.

Solution:

Use the solution to Problem 17.128 for the velocities:

ω BC = 0.1076 rad/s, ω AC = 0.0873 rad/s v Crel = 0.1093 (m/s). The angular acceleration α AC = −2

π ( 180 ) = −0.03491 rad/s . 2

The acceleration of point C is a C = α AC × rC /A − ω 2AC rC /A  i j k    0 α AC  − ω 2AC (2.6 i + 2.4 j), =  0  2.6 2.4 0   a C = α AC (−2.4 i + 2.6 j) − ω 2AC (2.6 i + 2.4 j)

= 0.064 i − 0.109 j (m/s 2 ).

aAC

2.4 m

vAC

The acceleration of point C in terms of the hydraulic actuator is 2 r a C = a Crel e + 2ω BC × v Crel + α BC × rC / B − ω BC C / B,

1.4 m

1.2 m

 i j k   0 0 ω BC  a C = a Crel e + 2    0   0.4472v Crel 0.8944v Crel    i j k   2 (1.2 i + 2.4 j) +  0 0 α BC  − ω BC   1.2 2.4 0   a C = a Crel (0.4472 i + 0.8944 j) + 2ω BC (−0.0977 i + 0.0489 j) 2 (1.2 i + 2.4 j). + α BC (−2.4 i + 1.2 j) − ω BC

Equate like terms in the two expressions for a C . 0.0640 = 0.4472a Crel − 0.0139 − 2.4α BC − 0.0210, −0.1090 = 0.8944 a Crel − 0.0278 + 1.2α BC + 0.0105. Solve:

a Crel = −0.0378 (m/s 2 ) ,

which is the rate of change of the rate of extension of the actuator, and α BC = −0.0483 (rad/s 2 ) = −2.77 deg/s 2 .

356

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Problem 17.130 The sleeve at A slides upward at a constant velocity of 10 m/s. Bar AC slides through the sleeve at B. Determine the angular velocity of bar AC and the velocity at which the bar slides relative to the sleeve at B. (See Example 17.9.) A 1m

Solution:

The velocity of the sleeve at A is given to be v A = 10 j (m/s). The unit vector parallel to the bar (toward A) is

308

e = 1(cos30 °i + sin 30 ° j) = 0.866 i + 0.5 j. B

Choose a coordinate system with origin at B that rotates with the bar. The velocity at A is C

v A = v B + v Arel e + ω AC × r A /B  i j k  = 0 + v Arel e +  0 0 ω AB  0.866 0.5 0 

10 m/s

     

vA vArel

v A = (0.866 i + 0.5 j)v Arel + ω AC (−0.5i + 0.866 j) (m/s).

The given velocity is v A = 10 j (m/s). Equate like components in the two expressions for v A :

0 = 0.866v Arel − 0.5ω AC ,

10 = 0.5v Arel + 0.866ω AC . Solve:

ω AC = 8.66 rad/s (counterclockwise). v Arel = 5 m/s from B toward A.

Problem 17.131 The sleeve at A slides upward at a constant velocity of 10 m/s. Determine the angular acceleration of bar AC and the rate of change of the velocity at which the bar slides relative to the sleeve at B. (See Example 17.9.)

Solution:

Use the solution of Problem 17.130:

e = 0.866 i + 0.5 j ω AB = 8.66 rad/s, v Arel = 5 m/s. The acceleration of the sleeve at A is given to be zero. The acceleration in terms of the motion of the arm is a A = 0 = a Arel e + 2ω AB × v Arel e + α AB × r A /B − ω 2AB r A /B .

10 m/s

308

 i j k   0 0 ω AB  a A = 0 = a Arel e + 2v Arel   0.866 0.5 0      i j k   +  0 0 α AB  − ω 2AB (0.866 i + 0.5 j)  0.866 0.5 0    0 = (0.866 i + 0.5 j)a Arel − 43.3i + 75 j + α AB (− 0.5 i + 0.866 j) − 64.95 i − 37.5 j.

B Separate components: C

0 = 0.866 a Arel − 43.3 − 0.5 α AB − 64.95, 0 = 0.5a Arel + 75 + 0.866α AB − 37.5. Solve:

a Arel = 75 (m/s 2 ) (toward A). α AB = −86.6 rad/s 2 , (clockwise).

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Problem 17.132 Block A slides up the inclined surface at 2 ft/s. Determine the angular velocity of bar AC and the velocity of point C.

Solution:

The velocity at A is given to be

v A = 2(−i cos 20 ° + j sin 20 °) = −1.879 i + 0.6840 j (ft/s).

2 ft 6 in

From geometry, the coordinates of point C are

(

( )) = (7,3.89) (ft).

7 7,2.5 4.5

208 4 ft 6 in

The unit vector parallel to the bar (toward A) is

2 ft 6 in

e = (7 2 + 3.89 2 ) −1/ 2 (−7 i − 3.89 j) = −0.8742 i − 0.4856 j. The velocity at A in terms of the motion of the bar is  i j k    v A = v Arel e + ω AB × r A /B = v Arel e +  0 0 ω AC  ,  −4.5 −2.5 0   v A = −0.8742v Arel i − 0.4856v Arel j + 2.5ω AC i − 4.5ω AC j (ft/s).

Equate the two expressions for v A and separate components: −1.879 = −0.8742v Arel + 2.5ω AC ,

0.6840 = −0.4856v Arel − 4.5ω AC . Solve: v Arel = 1.311 ft/s,

A vArel

ω AC = −0.293 rad/s (clockwise). Noting that v A = 2 ft/s, the velocity at point C is   i j k    v C = v A (−0.8742 i − 0.4856 j) +  0 −0.293  , 0   2.5 3.89 − 2.5 0   v C = −0.738i − 1.37 j (ft/s).

Problem 17.133 Block A slides up the inclined surface at a constant velocity of 2 ft/s. Determine the angular acceleration of bar AC and the acceleration of point C.

Solution:

The velocities: The velocity at A is given to be

v A = 2(−i cos 20 ° + j sin 20 °) = −1.879 i + 0.6840 j (ft/s). From geometry, the coordinates of point C are 7 ( 7,2.5 ( 4.5 )) = (7,3.89) (ft).

The unit vector parallel to the bar (toward A) is

e = B

The velocity at A in terms of the motion of the bar is 2 ft 6 in

A x

208 4 ft 6 in

−7 i − 3.89 j = −0.8742 i − 0.4856 j. 7 2 + 3.89 2

2 ft 6 in

 i j k   v A = v Arel e + ω AC × r A /B = v Arel e +  0 0 ω AC  ,  −4.5 −2.5 0   v A = −0.8742v Arel i − 0.4856v Arel j + 2.5ω AC i − 4.5ω AC j (ft/s). Equate the two expressions and separate components: −1.879 = −0.8742v Arel + 2.5ω AC , 0.6842 = −0.4856v Brel − 4.5ω AC . Solve: v Arel = 1.311 ft/s, ω AC = −0.293 rad/s (clockwise).

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17.133 (Continued) The accelerations: The acceleration of block A is given to be zero. In terms of the bar AC, the acceleration of A is

Solve:

α AC = 0.1494 rad/s 2 (counterclockwise).

a A = 0 = a Arel e + 2ω AC × v Arel e + α AC × r A / B = ω 2AC r A / B .  i j k   0 = a Arel e + 2ω ACv Arel  0 0 1   −0.8742 −0.4856 0      i j k   +  0 0 α AC  − ω 2AC (−4.5i − 2.5 j).  −4.5 −2.5 0   0 = a Arel e + 2ω ACv Arel (−e y i + e x j) + α AC (2.5i − 4.5 j) − ω 2AC (−4.5i − 2.5 j).

a Arel = 0.4433 (ft/s 2 ) (toward A).

The acceleration of point C is a C = a Arel e + 2ω AC × v Arel + α AC × rC /B − ω 2AC rC /B  i j k   a C = a Arel e + 2ω ACv Arel  0 0 1   ex ey 0     i  j k   +  0 0 α AC  − ω 2AC (2.5i + (3.89 − 2.5) j).  2.5 3.89 − 2.5 0  

Separate components to obtain: Substitute numerical values: a C = −1.184 i + 0.711 j (ft/s 2 )

0 = −0.8742a Arel − 0.3736 + 2.5α AC + 0.3875, 0 = −0.4856a Arel + 0.6742 − 4.5α AC + 0.2153.

Problem 17.134 The angular velocity of the scoop is 1 rad/s clockwise. Determine the rate at which the hydraulic actuator AB is extending.

B 2 ft A

Solution:

5 ft

C is

2 ft 6 in

1 ft

Scoop

Equate the expressions and separate components:

Point B is constrained to move normally to the arm DB: The unit vector parallel to DB is e DB =

1 ft 6 in E

The point B slides in the arm AB. The velocity of point

 i j k    v C = ω scoop × (1.5 j) =  0 0 1  = 1.5i (ft/s).  0 15 0   

1i + 2 j = 0.4472 i + 0.8944 j. 12 + 2 2

1.364 = 0.9487v Brel − 2ω AB , −0.6818 = 0.3162v Brel + 6ω AB . Solve: ω AB = −0.1704 rad/s, v Brel = 1.078 ft/s which is the rate of extension of the actuator.

The unit vector normal to e DB is e NDB = 0.8944 i − 0.4472 j, from which the velocity of C in terms of BC is

v C = v B e NBD + ω BC × rC /B  i j k  = v B (0.8944 i − 0.4472 j) +  0 0 ω BC  2.5 −0.5 0 

  .   

vBrel

v C = v B (0.8944 i − 0.4472 j) + ω BC (0.5i + 2.5 j). Equate terms in v C , 1.5 = 0.8944v B + 0.5ω BC , O = −0.4472v B + 2.5ω BC . Solve: ω BC = 0.2727 rad/s, v B = 1.525 ft/s, from which v B = v B e NDB = 1.364 i − 0.6818 j (ft/s). The unit vector parallel to the arm AB is e AB =

6i + 2 j = 0.9487 i + 0.3162 j. 62 + 22

Choose a coordinate system with origin at A rotating with arm AB. The velocity of point B is v B = v Brel e AB + ω AB × rB /A  i j k  = v Brel (0.9487 i + 0.3162 j) +  0 0 ω AB  6 2 0 

  .   

v B = v Brel (0.9487 i + 0.3162 j) + ω AB (−2 i + 6 j).

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Problem 17.135 The angular velocity of the scoop is 1 rad/s clockwise and its angular acceleration is zero. Determine the rate of change of the rate at which the hydraulic actuator AB is extending.

2 ft A

rB /A = rB − r A = 6 i + 2 j (ft), rB /D = rB − rD = 1i + 2 j (ft). Assume that the scoop rotates at 1 rad/s about point E. The acceleration of point C is 2 a C = α Scoop × 15 j − ω scoop (1.5 j) = −1.5 j (ft/s 2 ),

since α scoop = 0. The vector from C to B is rB /C = rB − rC = −2.5i + 0.5 j (ft). The acceleration of point B in terms of point C is 2 r a B = a C + α BC × rB /C − ω BC B /C

 i j k  = 1.5 j +  0 0 α BC  −2.5 +0.5 0  from which

   − ω 2 (−2.5i + 0.5 j), BC   

2 ) i − (1.5 + 2.5α 2 a B = −(0.5α BC − 2.5ω BC BC + 0.5ω BC ) j.

The acceleration of B in terms of D is  i j k   2 ( i + 2 j).  = a D +  0 0 α BD  − ω BD  1 2  0   The acceleration of point D is zero, from which a B = −(2α BD + 2 ) i + (α 2 ω BD BD − 2ω BD ) j. Equate like terms in the two expressions for

Problem 17.136 Suppose that the curved bar in Example 17.10 rotates with a counterclockwise angular velocity of 2 rad/s. (a) What is the angular velocity of bar AB? (b) What is the velocity of block B relative to the slot?

The angle defining the position of B in the circular slot is

(

5 ft

1 ft

2 ft 6 in

Scoop

2 ) = −(2α 2 a B , − (0.5α BC − 2.5ω BC BD + ω BD ), − (1.5 + 2.5α BC + 2 ) = (α 2 ). From the solution to Problem 17.134, 0.5ω BC − 2 ω BD BD ω BC = 0.2727 rad/s, and v B = 1.525 ft/s. The velocity of point B is normal to the link BD, from which

ω BD =

vB = 0.6818 rad/s. 12 + 2 2

Substitute and solve for the angular accelerations: α BC = −0.1026 rad/s 2 , α BD = −0.3511 rad/s 2 . From which the acceleration of point B is 2 ) i + (α 2 a B = −(2α BD + ω BD BD − 2ω BD ) j

= 0.2372 i − 1.281 j (ft/s 2 ). The acceleration of point B in terms of the arm AB is a B = a Brel e B /A + 2ω AB × v Brel e B /A + α AB × rB /A − ω 2AB rB /A . From the solution to Problem 17.134: e B /A = 0.9487 i + 0.3162 j, v Brel ] = 1.078 ft/s, ω AB = −0.1705 rad/s. From which a B = a Brel (0.9487 i + 0.3162 j) + 0.1162 i − 0.3487 j + α AB (−2 i + 6 j) − 0.1743i − 0.0581 j.

2 r a B = a D + α BD × rB /D − ω BD B /D

Solution:

1 ft 6 in

Solution:

Choose a coordinate system with the origin at D and the x axis parallel to ADE. The vector locations of points A, B, C , and E are r A = −5i ft, rB = 1i + 2 j ft, rC = 3.5i + 1.5 j ft, rE = 3.5i ft. The vector AB is

Equate the accelerations of point B and separate components: 0.2371 = 0.9487a Brel − 2α AB − 0.0581, −1.281 = 0.3162a Brel + 6α AB − 0.4068. Solve: a Brel = 0.0038 ft/s 2 , which is the rate of change of the rate at which the actuator is extending.

The velocity of B in terms of BC is v B = v Brel e B + ω BC × rB /C  i j k    = v Brel (−0.7 i + 0.714 j) +  0 0 ω BC  ,  −142.9 350 0   v B = v Brel (−0.7 i + 0.714 j) + (−700 i − 285.8 j) (mm/s) Equate the expressions for the velocity of B and separate components:

)

350 β = sin −1 = 44.4 °. 500

−350 ω AB = −0.7v Brel − 700,

The vectors are

−857ω AB = 0.714v Brel − 285.8.

rB /A = (500 + 500 cos β ) i + 350 j = 857 i + 350 j (mm).

Solve:

rB /C = (−500 + 500 cos β ) i + 350 j (mm).

(a)

ω AB = −2 rad/s (clockwise).

The unit vector tangent to the slot at B is given by

(b)

v Brel = −2000 mm/s (toward C ).

e B = − sin β i + cos β j = −0.7 i + 0.714 j. The velocity of B in terms of AB is  i j k  0 ω AB v B = ω AB × rB /A =  0  857 350 0  = ω AB (−350 i + 857 j) (mm/s).

360

B      

v A

500 mm

350 mm

500 mm 1000 mm

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Problem 17.137 Suppose that the curved bar in Example 17.10 has a clockwise angular velocity of 4 rad/s and a counterclockwise angular acceleration of 10 rad/s 2 . What is the angular acceleration of bar AB?

Equate the expressions for the velocity of B and separate components: −350 ω AB = −0.7v Brel + 1400, 857ω AB = 0.714v Brel + 571.6. Solve: ω AB = 4 rad/s (counterclockwise). v Brel = 4000 mm/s (away from C). Get the accelerations: The acceleration of point B in terms of the AB is a B = α AB × rB /A − ω 2AB rB /A

Solution:

Use the solution to Problem 17.118 with new data.

Get the velocities: The angle defining the position of B in the circular slot is β = sin −1

( 350 ) = 44.4°. 500

  i j k   0 α AB  − ω 2AB (857 i + 350 j), =  0   857 350 0   a B = α AB (−350 i + 857 j) − 138713i − 5665 j(mm/s 2 ). The acceleration in terms of the arm BC is 2 r a B = a Brel + 2ω BC × v Brel e B + α BC × rB /C − ω BC B /C .

The vectors

Expanding term by term:

rB /A = (500 + 500 cos β ) i + 350 j = 857 i + 350 j (mm). a Brel = a Brel e B −

rB /C = (−500 + 500 cos β ) i + 350 j (mm).

( v500 ) e 2 Brel

= a Brel (−0.7 i + 0.7141 j) − 22,852.6 i − 22,400 j.

The unit vector tangent to the slot at B e B = − sin β i + cos β j = −0.7 i + 0.714 j.

Other terms:

The component normal to the slot at B is

2ω BC × v Brel e B = 22852 i + 22,400 j, α BC × rB /C = −3500 i − 1429.3 j,

e NB = cos β i + sin β j = 0.7141i + 0.7 j.

2 r −ω BC B /C = 2286.8i − 5600 j.

The velocity of B in terms of AB

Collect terms:

 i j k  0 ω AB v B = ω AB × rB /A =  0  857 350 0  = ω AB (−350 i + 857 j) (mm/s).

     

a B = a Brel (−0.7 i + 0.7141 j) − 22852.6 i − 22400 j + 22852.6 i + 22400 j − 3500 i − 1429.3 j + 2286.9 i − 5600 j a B = a Brel (−0.7 i + 0.7141 j) − 1213i − 7029.3 j.

The velocity of B in terms of BC is

Equate the two expressions for the acceleration of B to obtain the two equations:

v B = v Bre e B + ω BC × rB /C

−350α AB − 13,871 = −0.7a Brel − 1213.1,

  i j k    = v Brel (−0.7 i + 0.714 j) +  0 0 −ω BC  ,   −142.9 350 0   v B = v Brel (−0.7 i + 0.714 j) + (1400 i + 571.6 j) (mm/s).

857α AB − 5665 = 0.7141a Brel − 7029.3. Solve: a Brel = 29180 (mm/s 2 ), α AB = 22.65 rad/s 2 (counterclockwise).

Problem 17.138* The disk rolls on the plane surface with a counterclockwise angular velocity of 10 rad/s. Bar AB slides on the surface of the disk at A. Determine the angular velocity of bar AB. 10 rad/s

Choose a coordinate system with the origin at the point of contact between the disk and the plane surface, with the x axis parallel to the plane surface. Let A be the point of the bar in contact with the disk. The vector location of point A on the disk is r A = i cos 45 ° + j(1 + sin 45 °) = 0.707 i + 1.707 j (ft).

The unit vector parallel to the radius of the disk is

e A = cos 45 °i + sin 45 ° j = 0.707 i + 0.707 j.

2 ft

The unit vector tangent to the surface of the disk at A is

458 C

Solution:

1 ft

e NA = i sin 45 ° − j cos 45 ° = 0.707 i − 0.707 j. The angle formed by the bar AB with the horizontal is sin 45 ° β = sin −1 = 20.7 °. 2 The velocity of point A in terms of the motion of bar AB is

(

)

 i j k  v A = ω AB × r A /B =  0 ω AB 0  −2 cos β 2sin β 0  v A = ω AB (−0.707 i − 1.871 j) (ft/s).

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17.138* (Continued) The velocity of point A in terms of the point of the disk in contact with the plane surface is

v A = v Arel e NA + ω disk × r A

 i j k  = v Arel (0.707 i − 0.707 j) +  0 0 ω disk  0.707 1.707 0  v A = v Arel (0.707 i − 0.707 j) + (−17.07 i + 7.07 j).

  ,   

vArel

Equate the expressions and separate components: x

−0.707ω AB = 0.707v Arel − 17.07, −1.871ω AB = −0.707v Arel + 7.07. Solve: v Arel = 20.3 ft/s, ω AB = 3.88 rad/s (counterclockwise).

Problem 17.139* The disk rolls on the plane surface with a constant counterclockwise angular velocity of 10 rad/s. Determine the angular acceleration of bar AB.

10 rad/s

2 ft 458 C

Solution:

Use the results of the solution to Problem 17.138. Choose a coordinate system with the origin at the point of contact between the disk and the plane surface, with the x axis parallel to the plane surface. The vector location of point A on the disk is

1 ft

r A = i cos 45 ° + j(1 + sin 45 °) = 0.707 i + 1.707 j (ft). The unit vector tangent to the surface of the disk at A is e NA = i sin 45 ° − j cos 45 ° = 0.707 i − 0.707 j. The angle formed by the bar AB with the horizontal is β = sin −1 (sin 45 °/2) = 20.7 °. Get the velocities: The velocity of point A in terms of the motion of bar AB is  i j k  v A = ω AB × r A /B =  ω AB 0 0  −2cos β 2sin β 0  = ω AB (−0.707 i − 1.871 j) (ft/s).

     

y A

The acceleration of the center of the disk is zero. The velocity of point A in terms of the center of the disk is v A = v Arel e NA + ω disk × r A  i j k  = v Arel (0.707 i − 0.707 j) +  0 0 ω disk  0.707 1.707 0  v A = v Arel (0.707 i − 0.707 j) + (−17.07 i + 7.07 j).

  ,   

v 2Arel R

B aArel

Equate the expressions and separate components: −0.707ω AB = 0.707v Arel − 17.07, − 1.871ω AB = −0.707v Arel + 7.07. Solve: v Arel = 20.3 ft/s, ω AB = 3.88 rad/s (counterclockwise).

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17.139* (Continued) Get the accelerations: The acceleration of point A in terms of the arm AB is a A = α AB × r A /B − ω 2AB r A /B  i j k   0 0 =  α AB  + 28.15i − 10.64 j (ft/s 2 ),  −1.87 0.707 0    a A = α AB (−0.707 i − 1.87 j) + 28.15i − 10.64 j (ft/s 2 ).

since the acceleration of the disk is zero. 2 −ω disk r A /C = −70.7 i − 70.7 j.

Collect terms and separate components to obtain: −0.707α AB + 28.15 = 0.707a Arel − 290.3 + 286.6 − 70.7, −1.87α AB − 10.64 = −0.707a Arel − 290.3 + 286.6 − 70.7. Solve:

The acceleration of point A in terms of the disk is

a Arel = 80.6 ft/s 2 ,

2 a A = a Arel + 2ω disk × v Arel e NA + α disk × r A /C − ω disk r A /C .

α AB = 64.6 rad/s 2 (counterclockwise).

Expanding term by term: The acceleration a Arel is composed of a tangential component and a radial component: a Arel = a Arel e NA −

( v 1 )e 2 Arel

= a Arel (0.707 i − 0.707 j) − 290.3i − 29 0.3 j. 2ω disk × v Arel e NB = 286.6 i + 286.6 j, α disk × r A = 0,

Problem 17.140* Bar BC rotates with a counterclockwise angular velocity of 2 rad/s. A pin at B slides in a circular slot in the rectangular plate. Determine the angular velocity of the plate and the velocity at which the pin slides relative to the circular slot.

Solution:

Choose a coordinate system with the origin O at the lower left pin and the x axis parallel to the plane surface. The unit vector parallel to AB is e AB = i. The unit vector tangent to the slot at B is e NAB = j. The velocity of the pin in terms of the motion of BC is v B = ω BC × rB /C .  i j k  0 ω BC v B =  0  −60 30 0 

   = 2(−30 i − 60 j) = −60 i − 120 j (mm/s).   

The velocity of the pin in terms of the plate is A

B 40 mm

30 mm

 i j k  v B = v Brel j + ω AB × rB /A =  0 0 ω AB  40 30 0  = v Brel j + ω AB (−30 i + 40 j) (mm/s).

     

Equate the expressions and separate components to obtain −60 = −30 ω AB , −120 = v Brel + 40 ω AB .

60 mm

40 mm

Solve: v Brel = −200 j mm/s, ω AB = 2 rad/s (counterclockwise).

Problem 17.141* Bar BC rotates with a constant counterclockwise angular velocity of 2 rad/s. Determine the angular acceleration of the plate.

Solution:

Choose the same coordinate system as in Problem 17.140. Get the velocities: The unit vector parallel to AB is e AB = i. The unit vector tangent to the slot at B is e NAB = j. The velocity of the pin in terms of the motion of BC is v B = ω BC × rB /C

A 30 mm

 i j k  0 ω BC =  0  −60 30 0 

B 40 mm

= (−60 i − 120 j) (mm/s). C

40 mm

The velocity of the pin in terms of the plate is

60 mm

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17.141* (Continued)  i j k  v B = v Brel j +  0 0 ω AB  40 30 0  = v Brel e NAB + ω AB × rB /O

     

= v Brel j + ω AB (−30 i + 40 j) (mm/s).

Expand term by term: a Brel = a Brel e NAB −

ω AB = 2 rad/s (counterclockwise). Get the accelerations: The acceleration of the pin in terms of the arm BC is 2 r a B = α BC × rB /C − ω BC B /C

= 0 − 4(−60 i + 30 j) = 240 i − 120 j (mm/s 2 ). The acceleration of the pin in terms of the plate AB is

2ω AB × v Brel e NAB = 800 i (mm/s 2 ).  i j k   α AB × rB / O =  0 0 α AB   40 30 0   = α BA (−30 i + 40 j) (mm/s 2 ),

−60 = −30 ω AB , −120 = v Brel + 40 ω AB .

v Brel = −200 j mm/s,

2 Brel

= a Brel j − 1000 i (mm/s 2 ),

Equate the expressions and separate components to obtain

Solve:

( v40 )e

−ω 2AB (40 i + 30 j) = −160 i − 120 j (mm/s 2 ). Collect terms and separate components to obtain: 240 = −1000 + 800 − 30α BA − 160, −120 = a Brel + 40α BA − 120. Solve: a Brel = 800 mm/s 2 (upward), α AB = −20 rad/s 2 , (clockwise).

a B = a Brel + 2ω AB × v Brel e NAB + α AB × rB /O − ω 2AB rB /O .

Problem 17.142* By taking the derivative of Eq. (17.11) with respect to time and using Eq. (17.12), derive Eq. (17.13). Solution:

Eq (17.11) is

v A = v B + v Arel + ω × r A /B . Eq (17.12) is v Arel =

( dxdt ) i + ( dydt ) j + k( dzdt )k.

Assume that the coordinate system is body fixed and that B is a point on the rigid body, ( A is not necessarily a point on the rigid body), such that r A = rB + r A /B , where r A /B = xi + yj + zk, and x, y, z are the coordinates of A in body fixed coordinates. Take the derivative of both sides of Eq (17.11): dv A dv B d v Arel dr dω = + + × r A /B + ω × A /B . dt dt dt dt dt

Substitute into the derivative: d v Arel = a Arel + ω × v Arel . dt Noting ω ×

(

d r A /B dy dx dz i+ j+ k = ω× dt dt dt dt

)

( dtdi + y dtdj + z ddtk )

+ω× x

= ω × v Arel + ω × (ω × r A /B ). Collect and combine terms: the derivative of Eq (17.11) is a A = a B + a rel + 2ω × v Arel + α × r A /B + ω × (ω × r A /B ) , which is Eq (17.13).

By definition, dv A dv B dω = a A, = a B , and = α. dt dt dt The derivative: d v Arel d 2y dy d j d 2x d 2z dx d i dz d k = i+ j+ 2k+ + + . dt dt 2 dt 2 dt dt dt dt dt dt dt Using the fact that the derivative of a unit vector represents a rotation of the unit vector, dj di dk = ω × i, = ω × j, = ω × k. dt dt dt

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Problem 17.143 In Practice Example 17.11, suppose that the merry-go-round has counterclockwise angular velocity ω and counterclockwise angular acceleration α. The person A is standing still on the ground. Determine her acceleration relative to your reference frame at the instant shown.

Solution:

First the velocity analysis

v A = v B + v Arel + ω × r A /B 0 = 0 + v Arel + ωk × ( Ri) v Arel = −ω Rj Now the acceleration analysis a A = a B + a Arel + a × r A /B − ω 2 r A /B + 2ω × v Arel 0 = 0 + a Arel + αk × ( Ri) − ω 2 ( Ri) + 2(ωk) × (−ω Ri) a Arel = −αRj + ω 2 Ri + 2ω 2 Ri a Arel = −ω 2 Ri − αRj.

Problem 17.144 The x – y coordinate system is body fixed with respect to the bar. The angle θ (in radians) is given as a function of time by θ = 0.2 + 0.04 t 2 . The x coordinate of the sleeve A (in feet) is given as a function of time by x = 1 + 0.03t 3 . Use Eq. (17.16) to determine the velocity of the sleeve at t = 4 s relative to a nonrotating reference frame with its origin at B. (Although you are determining the velocity of A relative to a nonrotating reference frame, your answer will be expressed in components in terms of the body-fixed reference frame.)

Solution:

We have

θ = 0.2 + 0.04t 2 , θ = 0.08t x = 1 + 0.03t 3 , x = 0.09t 2 At the instant t = 4 s, θ = 0.32, x = 2.92, x = 1.44 Thus v A = v B + v Arel + ω × r A /B = 0 + x i + θ k × ( xi) = x i + xθ j = (1.44) i + (2.92)(0.32) j v A = (1.44 i + 0.934 j) ft/s.

x y A

Problem 17.145 The metal plate is attached to a fixed ball-and-socket support at O. The pin A slides in a slot in the plate. At the instant shown, x A = 1 m, dx A /dt = 2 m/s, and d 2 x A /dt 2 = 0, and the plate’s angular velocity and angular acceleration are ω = 2k (rad/s) and α = 0. What are the x, y, and z components of the velocity and acceleration of A relative to a nonrotating reference frame with its origin at O?

Solution:

The velocity is v A = v O + v Arel + ω × r A /O . The relative velocity is v Arel =

where

( dxdt ) i + ( dydt ) j + ( dzdt )k,

dy dx d dx dz = 2 m/s, = = 1 m/s, = 0, 0.25 x 2 = 0.5 x dt dt dt dt dt

and r A /O = xi + yj + zj = i + 0.25 j + 0, from which v A = 2 i + j + ω (k × ( i + 0.25 j)) = 2 i + j + 2(−0.25i + j) = 1.5i + 3 j (m/s).

The acceleration is a A = a O + a Arel + 2ω × v Arel + α × r A /O + ω × (ω × r A /O ). A

Noting a Arel =

y 5 0.25x2 m

( ddt x ) i + ( ddt y ) j + ( ddt z )k, 2

where

( ddt x ) = 0, ddt y = dtd 0.25x = 0.5( dxdt ) = 2, ( ddt z ) = 0. 2

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17.145 (Continued) Substitute: a A = 2 j + 2ω (k × (2 i + j)) + ω 2 (k × (k × ( i + 0.25 j)))  i  i j k  j      =  0 0 4  + 4 k ×  0 0  1 0.25  2 1 0      i j  a A = 2 j − 4 i + 4 ω j + 4  0 0  −0.25 1 

k  1  0   k  1  0  

= −8i + 9 j (m/s 2 ).

Problem 17.146 The pin A slides in a slot in the plate. Suppose that at the instant shown, x A = 1 m, dx A /dt = −3 m/s, and the plate’s angular velocity and angular acceleration are ω = −4 j + 2k (rad/s) and α = 3i − 6 j (rad/s 2 ). What are the x, y, and z components of the velocity and acceleration of A relative to a nonrotating reference frame that is stationary with respect to O?

Solution:

The velocity is v A = v O + v Arel + ω × r A /O . The relative velocity is v Arel =

( dxdt ) i + ( dydt ) j + ( dzdt )k,

dy dx d dx dz = − 3 m/s, = 0.25 x 2 = 0.5 x = − 15 m/s, = 0, dt dt dt dt dt and r A /O = xi + yj + zj = i + 0.25 j + 0, from which v A = −3i − where

1.5 j + ω × ( i + 0.25 j).  i j k   v A = −3i − 1.5 j +  0 −4 2  = −3i − 1.5 j − 0.5i + 2 j + 4 k  1 0.25 0    = −3.5i + 0.5 j + 4 k (m/s).

The acceleration is a A = a O + a Arel + 2ω × v Arel + α × r A /O + ω × (ω × r A /O ). Noting a Arel =

y 5 0.25x2 m

( ddt x ) i + ( ddt y ) j + ( ddt z )k, 2

where

( ddt x ) = 4 m/s , 2

( )

d 2y d2 dx 2 d 2x = 2 0.25 x 2 = 0.5 + 0.5 x = 6.5 (m/s 2 ), dt 2 dt dt dt 2

( )

( ddt z ) = 0, α = 3i − 6 j (rad/s ), v 2

Arel

= −3i − 1.5 j,

and from above: ω × r A /O = −0.5i + 2 j + 4 k. Substitute:  i  i j k  j k    a A = 4 i + 6.5 j + 2  0 −4 2  +  3 −6 0   1 0.25 0   −3 −1.5 0        i j k   +  0 −4 2  .  −0.5 2 4    a A = 4 i + 6.5 j + 2(3i + 6 j − 12k) + (6.75k) + (−20 i − j − 2k) a A = −10 i − 6.5 j − 19.25k (m/s 2 ).

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Problem 17.147 The coordinate system is fixed relative to the ship B. At the instant shown, the ship is sailing north at 5 m/s relative to the Earth, and its angular velocity is 0.26 rad/s counterclockwise. Using radar, it is determined that the position of the airplane is 1080 i + 1220 j + 6300 k (m) and its velocity relative to the ship’s coordinate system is 870 i − 45 j − 21k (m/s). What is the airplane’s velocity relative to the Earth? (See Example 17.12.)

y A

Solution:

v A = v B + v Arel + ω × r A /B = 5 j + (870 i − 45 j − 21k) + (0.26k) × (1080 i + 1220 j + 6300 k) = (870 i − 40 j − 21k) +

i j k 0 0 0.26 1080 1220 6300

v A = (553i + 241 j − 21k) m/s.

Problem 17.148 Car B is approaching the intersection at 10 m/s. Its navigation system determines that the car’s angular velocity is 0.8 rad/s in the clockwise direction. In terms of its car-fixed reference frame, the navigation system measures the position of car A to be x = 25 m, y = 0, and determines its velocity to be v Arel = −23.2 i + 12.8 j (m/s). What is the magnitude of the velocity of car A relative to an earth-fixed reference frame? That is, how fast is car A approaching the intersection? Solution:

x B

Applying Eq. (17.16),

v A = v B + v Arel + ω × r A /B = (10 m/s) i − (23.2 m/s) i + (12.8 m/s) j +

i j k 0 0 −0.8 rad/s 0 25 m 0

= −13.2 i − 7.20 j (m/s). Its magnitude is v A = 15.0 m/s. 15.0 m/s.

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Problem 17.149 The train on the circular track is traveling at a constant speed of 50 ft/s in the direction shown. The train on the straight track is traveling at 20 ft/s in the direction shown and is increasing its speed at 2 ft/s 2 . Determine the velocity of passenger A that passenger B observes relative to the given coordinate system, which is fixed to the car in which B is riding.

0f t

500 ft B

50 = 0.1 rad/s. 500 The velocity of A is v A = v B + v Arel + ω × r A /B .

Solution:

The angular velocity of B is ω =

50 ft/s

20 ft/s

At the instant shown, v A = −20 j (ft/s), v B = +50 j (ft/s), and r A /B = 500 i (ft), from which  i j k    v Arel = −20 j − 50 j −  0 0 0.1  = −20 j − 50 j − 50 j,  500 0 0    v Arel = −120 j (ft/s).

Problem 17.150 In Problem 17.149, determine the acceleration of passenger A that passenger B observes relative to the coordinate system fixed to the car in which B is riding.

Solution:

Use the solution to Problem 17.149:

v Arel = −120 j (ft/s), ω =

50 = 0.1 rad/s, 500

r A /B = 500 i (ft). The acceleration of A is a A = −2 j (ft/s), The acceleration of B is

a B = −500(ω 2 ) i = −5i ft/s 2 ,

and α = 0, from which a A = a B + a Arel + 2ω × v Arel + ω × (ω × r A /B ).

500 ft B

50 ft/s

368

Rearrange: a Arel = a A − a B − 2ω × v Arel − ω × (ω × r A / B ).

20 ft/s

  i  i j k  j k       2    a Rel = −2 j + 5i − 2  0 0 ω  − ω  k ×  0 0 1  .   500 0 0    0 −120 0          i j k   1  , a Rel = −2 j + 5i − 2(120 ω ) i − ω 2  0 0  0 500 0    a Arel = −14 i − 2 j (ft/s 2 ) .

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Problem 17.151 The satellite A is in a circular polar orbit (a circular orbit that intersects the Earth’s axis of rotation). The radius of the orbit is R, and the magnitude of the satellite’s velocity relative to a nonrotating reference frame with its origin at the center of the Earth is v A . At the instant shown, the satellite is above the equator. An observer B on the Earth directly below the satellite measures its motion using the Earth-fixed coordinate system shown. What are the velocity and acceleration of the satellite relative to B’s Earth-fixed coordinate system? The radius of the Earth is R E and the angular velocity of the Earth is ω E . (See Example 17.13.)

Solution:

From the sketch, in the coordinate system shown, the location of the satellite in this system is r A = ( R − R E ) i, from which r A /B = r A − 0 = ( R − R E ) i. The angular velocity of the observer is ω E = −ω E k. The velocity of the observer is v B = −ω E R E k. The velocity of the satellite is v A = v A j. The relative velocity is v Arel = v A − v B − ω E × r A /B , j k  1 0  0 0  

 i  0 v Arel = v A j + ω E R E k − (ω E )   R − RE  = v A j + Rω E k . From Eqs (17.26) and (17.27)

a Arel = a A − a B − 2ω E × v Arel − α × r A /B − ω E × (ω E × r A /B ).

The accelerations: vA

( vR ) i, a = −ω R i, α = 0, from which

a A = ω 2A Ri = − x

2 A

2 E

 i j k  v 2A 0 + iω E2 R E − 2  0 ω E R  0 v R ω  A E   i j k   − ω E ×  ω E 0  . 0  R− R 0 0   E

( )

a Arel = −

     

a Arel = ω 2A Ri + ω E2 R E i  i j  − 2ω E2 Ri −  0 ω E  0 0 

 k   0  −ω E ( R − R E ) 

( vR ) i + ω R i − 2ω Ri + ω (R − R )i 2 A

= −

2 E

(( vR ) + ω R ) i.

a Arel = −

Problem 17.152 A car A at north latitude L drives north on a north-south highway with constant speed v. The Earth’s radius is R E , and the Earth’s angular velocity is ω E . (The Earth’s angular velocity vector points north.) The coordinate system is Earth fixed, and the x-axis passes through the car’s position at the instant shown. Determine the car’s velocity and acceleration (a) relative to the Earth-fixed coordinate system and (b) relative to a nonrotating reference frame with its origin at the center of the Earth.

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2 E

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17.152 (Continued) Solution:

(a) In earth fixed coords,

v rel = v j,

a rel = −v 2 /R E i. (motion in a circle)

x A

(b) v A = v Arel + ω E × r A /B + v B (v B = 0) L

= vj + (ω E sin Li + ω E cos Lj) × R E i v A = v j − ω E R E cos Lk .

a A = a B + a Arel + 2ω E × v Arel + α × r A /B + ω E × (ω E × r A /B )

where ω E = ω E sin Li + ω E cos Lj and r A /B = R E i aA = 0 −

v2 i + 2v ω E sin Lk RE

+ (ω E sin Li + ω E cos Lj) × (−ω E R E cos Lk)  v2  a A = − + ω E2 R E cos 2 L  i  RE  + (ω E2 R E sin L cos L ) j + 2v ω E sin Lk .

Problem 17.153 The airplane B conducts flight tests of a missile. At the instant shown, the airplane is traveling at 200 m/s relative to the Earth in a circular path of 2000-m radius in the horizontal plane. The coordinate system is fixed relative to the airplane. The x-axis is tangent to the plane’s path and points forward. The y-axis points out the plane’s right side, and the z-axis points out the bottom of the plane. The plane’s bank angle (the inclination of the z-axis from the vertical) is constant and equal to 20 °. Relative to the airplane’s coordinate system, the pilot measures the missile’s position and velocity and determines them to be r A /B = 1000 i (m) and v A /B = 100.0 i + 94.0 j + 34.2k (m/s). (a) What are the x, y, and z components of the airplane’s angular velocity vector? (b) What are the x, y, and z components of the missile’s velocity relative to the Earth?

Solution: (a) The bank angle is a rotation about the x axis; assume that the rotation is counterclockwise, so that the z axis is rotated toward the positive y axis. The magnitude of the angular velocity is ω =

200 = 0.1 rad/s. 2000

In terms of airplane fixed coordinates, ω = 0.1( i sin 20 ° − j cos20 °) (rad/s). ω = 0.03242 j − 0.0940 k rad/s. (b) The velocity of the airplane in earth fixed coordinates is v A = v B + v Arel + ω × r A /B = 200 i + 100 i + 94.0 j + 34.2k   i j k   +  0 0.0342 −0.0940    1000 0 0   v A = 300 i + 94.0 j + 34.2k − 94.0 j − 34.2k = 300 i (m/s).

2000 m

370

208

x A

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Problem 17.154 To conduct experiments related to long-term spaceflight, engineers construct a laboratory on Earth that rotates about the vertical axis at B with a constant angular velocity ω of one revolution every 6 s. They establish a laboratory-fixed coordinate system with its origin at B and the z-axis pointing upward. An engineer holds an object stationary relative to the laboratory at point A, 3 m from the axis of rotation, and releases it. At the instant he drops the object, determine its acceleration relative to the laboratory-fixed coordinate system, (a) assuming that the laboratory-fixed coordinate system is inertial and (b) not assuming that the laboratory-fixed coordinate system is inertial, but assuming that an Earthfixed coordinate system with its origin at B is inertial. (See Example 17.14.)

Solution:

(a) If the laboratory system is inertial, Newton’s second law is F = ma. The only force is the force of gravity; so that as the object free falls the acceleration is −gk = −9.81k (m/s 2 ) . If the earth fixed system is inertial, the acceleration observed is the centripetal acceleration and the acceleration of gravity: ω × (ω × rB ) − g, where the angular velocity is the angular velocity of the coordinate system relative to the inertial frame. j k   

2 



 i 



 3 0 0  

( 26π )  k ×  0 0 1   − 9.81k

ω × (ω × r A /B ) − g = − =

( π3 ) i − 9.81k 2

= 3.29 i − 9.81k (m/s 2 ).

y v

3m y

A x

Problem 17.155 The disk rotates in the horizontal plane about a fixed shaft at the origin with constant angular velocity ω = 10 rad/s. The 2-kg slider A moves in a smooth slot in the disk. The spring is unstretched when x = 0 and its constant is k = 400 N/m. Determine the acceleration of A relative to the body-fixed coordinate system when x = 0.4 m. Strategy: Use Eq. (17.25) to express Newton’s second law for the slider in terms of the body-fixed coordinate system.

Solution: T = −kx = (−400)(0.4) T = −160 i Newtons

∑ F = −160 i = ma Arel + m[ a B + 2 ω × v rel 0

+ α × r A /B + ω × (ω × r A /B )] where ω = −10 k, r A /B = 0.4 i, m = 2 − 160 i = 2a Arel − 80 i a Arel = −40 i (m/s 2 ).

y y

T A

x x

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Problem 17.156* Engineers conduct flight tests of a rocket at 30 ° north latitude. They measure the rocket’s motion using an Earth-fixed coordinate system with the x-axis pointing upward and the y-axis directed northward. At a particular instant, the mass of the rocket is 4000 kg, the velocity of the rocket relative to the engineers’ coordinate system is 2000 i + 2000 j (m/s), and the sum of the forces exerted on the rocket by its thrust, weight, and aerodynamic forces is 400 i + 400 j (N). Determine the rocket’s acceleration relative to the engineers’ coordinate system (a) assuming that their Earthfixed coordinate system is inertial and (b) not assuming that their Earth-fixed coordinate system is inertial.

Solution:

N y x 308

Use Eq. (17.22):

 

i j k  ω × (−R E ω E cos λ)k = sin λ cos λ 0   0 0 −1    = (−R E ω E2 cos 2 λ) i + ( R E ω E2 cos λ sin λ) j. R E ω E2 cos λ 

∑ F − m[a B + 2ω × v Arel + α × r A /B + ω × (ω × r A /B )] = ma Arel . (a) If the earth fixed coordinate system is assumed to be inertial, this reduces to ∑ F = ma Arel , from which a Arel =

1 1 ∑ F = 4000 (400 i + 400 j) m

Collect terms, a Arel = +( R E ω E2 cos 2 λ) i − ( R E ω E2 cos λ sin λ) j − 4000 ω E (sin λ − cos λ)k + 0.1i + 0.1 j.

= 0.1i + 0.1 j (m/s 2 ). (b) If the earth fixed system is not assumed to be inertial, a B = −R E ω E2 cos 2 λ i + R E ω E2 cos λ sin λ j, the angular velocity of the rotating coordinate system is ω = ω E sin λi + ω E cos λ j (rad/s). The relative velocity in the earth fixed system is v Arel = 2000 i + 2000 j (m/s), and r A /B = R E i (m). 2ω × v Arel

 i j k    = 2ω E  sin λ cos λ 0   2000 2000 0    = 4000 ω E (sin λ − cos λ)k

Substitute values: R E = 6336 × 10 3 m, ω E = 0.73 × 10 −4 rad/s, λ = 30 °, a Arel = 0.125i + 0.0854 j + 0.1069k (m/s 2 ). Note: The last two terms in the parenthetic expression for a A in

∑ F − m[a B + 2ω × v A rel + α × r A/B + ω × (ω × r A/B )] = ma Arel can be neglected without significant change in the answers.

 i j k   ω × (ω × r A /B ) = ω ×  ω E sin λ ω E cos λ 0   RE 0 0   = ω × (−R E ω E cos λ)k

Problem 17.157* Consider a point A on the surface of the Earth at north latitude L. The radius of the Earth is R E , and the Earth’s angular velocity is ω E . A plumb bob suspended just above the ground at point A will hang at a small angle β relative to the vertical because of the Earth’s rotation. Show that β is related to the latitude by ω 2 R sin L cos L tan β = E E 2 . g − ω E R E cos 2 L

N x

L B

Strategy: Using the Earth-fixed coordinate system shown, express Newton’s second law in the form given by Eq. (17.22).

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17.157* (Continued) Use Eq. (17.25). ∑ F − m[a B + 2ω × v Arel + α × r A /B + ω × (ω × r A /B )] = ma Arel . The bob is stationary, so that v Arel = 0. The origin of the coordinate system is stationary, so that a B = 0. The external force is the weight of the bob ∑ F = mg. The relative acceleration is the apparent acceleration due to gravity, ma Arel = mg Apparent . Substitute:

Solution:

g Apparent = g − ω × (ω × r A /B )  i j k   = g i − ω ×  ω E sin L ω E cos L 0   RE 0 0   g Apparent = g i − ω × (−R E ω E cos L )k

The vertical component of the apparent acceleration due to gravity is g vertical = g − R E ω E2 cos 2 L. The horizontal component of the apparent acceleration due to gravity is g horizontal = R E ω E2 cos L sin L. From equation of angular motion, the moments about the bob suspension are M vertical = (λ sin β )mg vertical and M horizontal = (λ cos β )mg horizontal , where λ is the length of the bob, amd m is the mass of the bob. In equilibrium, M vertical = M horizontal , from which g vertical sin β = g horizontal cos β . Substitute and rearrange: tan β =

g horizontal R E ω E2 cos L sin L = . g vertical g − R E ω E2 cos 2 L

 i j k   = g i + R E ω E2 cos L  sin L cos L 0   0 −1  0   g Apparent = g i − ( R E ω E2 cos 2 L ) i + ( R E ω E2 cos L sin L ) j.

b C

m Fg

Problem 17.158* Suppose that a space station is in orbit around the Earth and two astronauts on the station toss a ball back and forth. They observe that the ball appears to travel between them in a straight line at constant velocity. (a) Write Newton’s second law for the ball as it travels between the astronauts in terms of a nonrotating coordinate system with its origin fixed to the station. What is the term ΣF? Use the equation you wrote to explain the behavior of the ball observed by the astronauts. (b) Write Newton’s second law for the ball as it travels between the astronauts in terms of a nonrotating coordinate system with its origin fixed to the center of the Earth. What is the term ΣF? Explain the difference between this equation and the one you obtained in part (a).

Solution:

An earth-centered, non-rotating coordinate system can be treated as inertial for analyzing the motions of objects near the earth (See Section 17.2.) Let O be the reference point of this reference frame, and let B be the origin of the non-rotating reference frame fixed to the space station, and let A denote the ball. The orbiting station and its contents and the station-fixed non-rotating frame are in free fall about the earth (they accelerate relative to the earth due to the earth’s gravitational attraction), so that the forces on the ball in the fixed reference frame exclude the earth’s gravitational attraction. Let g B be the station’s acceleration, and let g A be the ball’s acceleration relative to the earth due to the earth’s gravitational attraction. Let ∑ F be the sum of all forces on the ball, not including the earth’s gravitational attraction. Newton’s second law for the ball of mass m is ∑ F + mg A = ma A = m(a B + a A /B ) = mg B + ma A /B . Since the ball is within a space station whose dimensions are small compared to the distance from the earth, g A is equal to g B within a close approximation, from which ∑ F = ma A /B . The sum of the forces on the ball not including the force exerted by the earth’s gravitational attraction equals the mass times the ball’s acceleration relative to a reference frame fixed with respect to the station. As the astronauts toss the ball back and forth, the only other force on it is aerodynamic drag. Neglecting aerodynamic drag, a A /B = 0, from which the ball will travel in a straight line with constant velocity. (b) Relative to the earth-centered non-rotating reference frame, Newton’s second law for the ball is ∑ F = ma A where ∑ F is the sum of all forces on the ball, including aerodynamic drag and the force due to the earth’s gravitational attraction. Neglect drag, from which a A = g A ; the ball’s acceleration is its acceleration due to the earth’s gravitational attraction, because in this case we are determining the ball’s motion relative to the earth. Note: An obvious unstated assumption is that the time of flight of the ball as it is tossed between the astronauts is much less than the period of an orbit. Thus the very small acceleration differences g A − g B will have a negligible effect on the path of the ball over the short time interval.

B gB

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A/B

A gA

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Problem 17.159 If θ = 60 ° and bar OQ rotates in the counterclockwise direction at 5 rad/s, what is the angular velocity of bar PQ?

Solution:

By applying the law of sines, β = 25.7 °. The velocity

of Q is

Q 400 mm

200 mm u O

v Q = v 0 + ω OQ × rQ /O or vQ = 0 +

i j k 0 0 5 0.2 cos60 ° 0.2sin 60 ° 0

= − sin 60°i + cos60° j.

y Q

The velocity of P is

vPQ 5 rad/s

v p i = v Q + ω PQ × rP /Q i 0

= −sin60 °i + cos60 ° j +

j 0

0.4 cos β −0.4sin β

608

k ω PQ .

P x

O P

Equating i and j components v P = − sin 60 ° + 0.4 ω PQ sin β , and 0 = cos60 ° + 0.4 ω PQ cos β . Solving, we obtain v P = −1.11 m/s and ω PQ = −1.39 rad/s.

Problem 17.160 If θ = 55 ° and the sleeve P is moving to the left at 2 m/s, what are the angular velocities of bars OQ and PQ?

Q 400 mm

200 mm u

Solution:

By applying the law of sines, β = 24.2 ° The velocity

of Q is

v Q = v 0 + ω OQ × rQ /O = 0+

i 0

j 0

k ω OQ

0.2cos 55 ° 0.2sin 55 °

(1)

= −0.2ω OQ sin 55 °i + 0.2ω OQ cos55° j We can also express v Q as v Q = v P + ω PQ × rQ /P = −2 i +

i 0

j 0

−0.4 cos β 0.4sin β

k ω PQ . 0

Equating i and j components in Equations (1) and (2), we get −0.2ω OQ sin 55 ° = −2 − 0.4 ω PQ sin β , and 0.2ω OQ cos55 ° = −0.4 ω PQ cos β . Solving, we obtain ω OQ = 9.29 rad/s ω PQ = −2.92 rad/s.

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Problem 17.161 Determine the vertical velocity v H of the hook and the angular velocity of the small pulley.

Solution:

The upper pulley is fixed so that it cannot move, from which the upward velocity of the rope on the right is equal to the downward velocity on the left, and the upward velocity of the rope on the right of the lower pulley is 120 mm/s. The small pulley is fixed so that it does not move. The upward velocity on the right of the small pulley is v H mm/s, from which the downward velocity on the left is v H mm/s. The upward velocity of the center of the bottom pulley is the mean of the difference of the velocities on the right and left, from which vH = Solve,

120 mm/s

120 − v H . 2 v H = 40 mm/s.

The angular velocity of the small pulley is 40 mm

ω =

vH 40 = = 1 rad/s. R 40

Problem 17.162 If the crankshaft AB is turning in the counterclockwise direction at 2000 rpm, what is the velocity of the piston?

Solution: The angle of the crank with the vertical is 45°. The angular velocity of the crankshaft is ω = 2000

( 260π ) = 209.44 rad/s.

The vector location of point B (the main rod bearing) rB = 2(−i sin 45 ° + j cos 45 °) = 1.414(−i + j) in. The velocity of point B (the main rod bearing) is

 i j k   v B = ω × rB /A = 1.414 ω  0 0 1   −1 1 0    = −296.2( i + j) (in/s).

5 in

From the law if sines the interior angle between the connecting rod and 2 5 the vertical at the piston is obtained from = , from which sin θ sin 45 ° 2sin 45 ° θ = sin −1 = 16.43 °. 5

(

458 B

)

The location of the piston is rC = (2sin 45 ° + 5cos θ) j = 6.21 j (in.). The vector rB /C = rB − rC = −1.414 i − 4.796 j (in.). The piston is constrained to move along the y axis. In terms of the connecting rod the velocity of the point B is  i j k  v B = v C + ω BC × rB /C = v C j +  0 0 ω BC  1.414 −4.796 0  = v C j + 4.796ω BC i − 1.414 ω BC j (in/s).

     

Equate expressions for v B and separate components: −296.2 = 4.796ω BC , −296.2 = v C − 1.414 ω BC . Solve:

v C = −383.5 j (in/s) = −32 j (ft/s) , ω BC = −61.8 rad/s.

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Problem 17.163 If the piston is moving with velocity v C = 20 j (ft/s), what are the angular velocities of the crankshaft AB and the connecting rod BC? y

Solution:

Use the solution to Problem 17.162. The vector location of point B (the main rod bearing) rB = 1.414(−i + j) in. From the law if sines the interior angle between the connecting rod and the vertical at the piston is θ = sin −1

( 2sin545° ) = 16.43°.

The location of the piston is rC = (2 sin 45 ° + 5 cos θ) j = 6.21 j (in.). The piston is constrained to move along the y axis. In terms of the connecting rod the velocity of the point B is C

 i j k  v B = v C + ω BC × rB /C = 240 j +  0 0 ω BC  −1.414 −4.796 0  = + − vB 240 j 4.796ω BC i 1.414 ω BC j (in/s).

5 in

     

In terms of the crank angular velocity, the velocity of point B is 458 B x

 i j k   v B = ω AB × rB /A = 1.414 ω AB  0 0 1   −1 1 0    = −1.414 ω AB ( i + j) (in/s). Equate expressions and separate components: 4.796ω BC = −1.414 ω BC = −1.414 ω AB . Solve: ω BC = 38.65 rad/s (counterclockwise). ω AB = 131.1 rad/s = −12515 rpm (clockwise).

Problem 17.164 If the piston is moving with velocity v C = 20 j (ft/s) and its acceleration is zero, what are the angular accelerations of the crankshaft AB and the connecting rod BC?

Use the solution to Problem 17.163. rB /A = 1.414(−i + j in., ω AB = −131.1 rad/s, rB /C = rB − rC = −1.414 i − 4.796 j (in.), ω BC = 38.65 rad/s. For point B, a B = a A + α AB × rB /A − ω 2AB rB /A

 i j k    = 1.414α AB  0 0 1  − 1.414 ω 2AB (−i + j),  −1 1 0    a B = −1.414α AB ( i + j) + 24291( i − j) (in/s 2 ).

In terms of the angular velocity of the connecting rod,

2 r a B = a C + α BC × rB /C − ω BC B /C ,

5 in

 i j k   2 (− 1.414 i − 4.795 j) (in/s 2 ), a B = α BC  0 0 1  − ω BC  −1.414 −4.796 0    a B = 4.796α BC i − 1.414α BC j + 2112.3i + 71630 j (in/s 2 ).

458 B

Equate expressions and separate components: A

−1.414α AB + 24291 = 4.796α BC + 2112.3, −1414α AB − 24291 = 1.414α BC + 7163.

Solution:

Solve:

α AB = −13,605 rad/s 2 (clockwise). α BC = 8636.5 rad/s 2 (counterclockwise).

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Problem 17.165 Bar AB rotates at 6 rad/s in the counterclockwise direction. Use instantaneous centers to determine the angular velocity of bar BCD and the velocity of point D.

8 in C

Solution: The strategy is to determine the angular velocity of bar BC from the instantaneous center; using the constraint on the motion of C. The vector rB /A = rB − r A = (8i + 4 j) − 4 j = 8i (in.). The velocity of point B is v B = ω AB × rB /A = ω AB (k × 8i) = 48 j (in/s). The velocity of point B is normal to the x axis, and the velocity of C is parallel to the x axis. The instantaneous center of bar BC has the coordinates (14, 0). The vector

12 in 6 rad/s A

rB /IC = rB − rIC = (8i + 4 j) − (14 i − 4 j) = −6 i (in.).

   = −6ω j = 48 j, BC   

ω BC = −

48 = −8 rad/s. 6

The velocity of point C is  i j k  v C = ω BC × rC /IC =  0 0 ω BC  0 12 0 

   = 96 i (in/s).   

x 6 in

4 in

The velocity of point D is normal to the unit vector parallel to BCD 6 i + 12 j = 0.4472 i + 0.8944 j. 6 2 + 12 12

e =

from which

8 in

The velocity of point B is  i j k  v B = ω BC × rB /IC =  0 0 ω BC  −6 0 0 

The intersection of the projection of this unit vector with the projection of the unit vector normal to velocity of C is occurs at point C, from which the coordinates of the instantaneous center for the part of the bar CD are (14, 12). The instantaneous center is translating at velocity v C , from which the velocity of point D is v D = v C + ω BC × rD /ICD = 96 i − 8(k × (4 i + 8 j)) = 160 i − 32 j (in/s).

Problem 17.166 Bar AB rotates with a constant angular velocity of 6 rad/s in the counterclockwise direction. Determine the acceleration of point D.

D 8 in C

Solution:

Use the solution to Problem 17.165. The accelerations are determined from the angular velocity, the known accelerations of B, and the constraint on the motion of C. The vector rB /A = rB − r A = 8i (in.). The acceleration of point B is a B = α AB × rB /A

− ω 2r

A /B

12 in

0 − 36(8i) = −288i (in/s 2 ).

From the solution to Problem 17.165, ω BC = −8 rad/s. (clockwise).

6 rad/s A

rC /B = rC − v B = (14 i + 12 j) − (8i) = 6 i + 12 j (in.). The vector rB /C = −rC /B . The acceleration of point B in terms of the angular acceleration of point C is

8 in

6 in

4 in

2 r a B = a C + α BC × rBC − ω BC B /C

 i j k    = aCi +  0 α BC  − 64(−6 i − 12 j). 0  −6 −12 0   a B = a C i + 12α BC i − 6α BC j + 384 i + 768 j. Equate the expressions and separate components: −288 = a C + 12α BC + 384, 0 = −6α BC + 768.

Solve α BC = 128 rad/s 2 (counterclockwise), a C = −2208 in/s 2 . The acceleration of point D is 2 r a D = a C + α BC × rD /C − ω BC D /C

 i j k  = −2208i +  0 0 α BC  4 8 0 

a D = −2208i + (128)(−8i + 4 j) − (64)(4 i + 8 j) = −3490 i (in/s 2 )

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   − ω 2 (4 i + 8 j). BC    .

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Problem 17.167 Point C is moving to the right at 20 in/s. What is the velocity of the midpoint G of bar BC? y

Solution: v B = v A + w AB × rB /A = 0 +

i j k 0 0 ω AB . 4 4 0

Also,

4 in

A 3 in

C 4 in

i j k 0 0 ω BC −10 7 0

v B = v C + w BC × rB /C = 20 i +

10 in

Equating i and j components in these two expressions, −4 ω AB = 20 − 7ω BC , 4 ω AB = −10 ω BC , and solving, we obtain ω AB = −2.94 rad/s, ω BC = 1.18 rad/s. Then the velocity of G is i 0

v G = v C + ω BC × rG /C = 20 i +

j 0

k ω BC

−5 3.5

= 15.88i − 5.88 j (in/s).

Problem 17.168 Point C is moving to the right with a constant velocity of 20 in/s. What is the acceleration of the midpoint G of bar BC?

4 in

Solution:

See the solution of Problem 17.167. 3 in

a B = a A + α AB × rB /A − ω 2AB rB /A = O+

i j k 0 0 α AB 4 4 0

C − ω 2AB (4 i + 4 j).

4 in

2 r Also, a B = a C + α BC × rB /C − ω BC B /C

= O+

i j k 0 0 α BC 0 −10 7

= O+

Problem 17.169 If the velocity of point C is v C = 1.0 i (in/s), what are the angular velocity vectors of arms AB and BC? B

j 0

k α BC

−5 3.5

2 (− 5i + 3.5 j) − ω BC

3 in C 10 in

Solution:

Use the solution to Problem 17.167: The velocity of the point B is determined from the known velocity of point C and the known velocity of C:  i j k   v B = v C + ω BC × rB /C = 1.0 i +  0 0 ω BC   −10 7 0   = 1.0 i − 7ω BC i − 10 ω BC j. The angular velocity of bar AB is determined from the velocity of B.

4 in

i 0

= −8.18i − 26.4 j (in/s 2 ).

2 , 4α AB − 4 ω 2AB = −10α BC − 7ω BC

10 in

2 r Then a G = a C + α BC × rG /C − ω BC G /C

2 (− 10 i + 7 j). − ω BC

2 , −4α AB − 4 ω 2AB = −7α BC + 10 ω BC

4 in

and solving yields α AB = −4.56 rad/s 2 , α BC = 4.32 rad/s 2 .

Equating i and j components in these two expressions,

 i j k  v B = ω AB × rB /A =  0 0 ω AB  4 4 0 

   = −4ω ( i − j) (in/s) AB   

Equate expressions, separate components, 1.0 − 7ω BC = −4 ω AB , − 10 ω BC = 4 ω AB . Solve: ω AB = −0.147 rad/s, ω BC = 0.0588 rad/s, from which ω AB = −0.147k (rad/s), ω BC = 0.0588k (rad/s).

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Problem 17.170 Points B and C are in the x – y plane. The angular velocity vectors of arms AB and BC are ω AB = −0.5k (rad/s) and ω BC = −2.0 k (rad/s). Determine the velocity of point C.

Solution:

The vector

rB /A = 760( i cos15 ° − j sin15 °) = 734.1i − 196.7 j (mm). The vector rC /B = 900( i cos 50 ° + j sin 50 °) = 578.5i + 689.4 j (mm).

The velocity of point B is

760

 i j k   v B = ω AB × rB /A = −0.5  0 0 1   734.1 −196.7 0    v B = −98.35i − 367.1 j (mm/s).

The velocity of point C v C = v B + ω BC × rC /B m

158

508

90 0m

= −98.35i − 367.1 j + (2)(k × (578.5i + 689.4 j)), v C = −98.35i − 367.1 j − 1378.9 i + 1157.0 j

= −1477.2 i + 790 j (mm/s).

Problem 17.171 If the velocity of point C is v C = 1.0 i (m/s), what are the angular velocity vectors of arms AB and BC?

Solution:

Use the solution to Problem 17.170.

The vector rB /A = 760( i cos15 ° − j sin15 °) = 734.1i − 196.7 j (mm). The vector

rC /B = 900( i cos 50 ° + j sin 50 °) = 578.5i + 689.4 j (mm). The velocity of point B is 760

 i j k  0 0 v B = ω AB × rB /A =  ω AB  734.1 −196.7 0 

     

= 196.7ω AB i + 734.1ω AB j (mm/s).

508

0m m

The velocity of point C is 158

v C = v B + ω BC × rC /B  i j k  = 196.7ω AB i + 734.1ω AB j +  0 0 ω BC  578.5 687.4 0 

  ,   

1000 i = 196.7ω AB i + 734.1ω AB j − 687.4ω BC i + 578.5ω BC j (mm/s). Separate components: 1000 = 196.7ω AB − 687.4 ω BC , 0 = 734.1ω AB + 578.5ω BC . Solve: ω AB = 0.933k (rad/s), ω BC = −1.184 k (rad/s).

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Problem 17.172 The angular velocity vectors of arms AB and BC are ω AB = −0.5k (rad/s) and ω BC = 2.0 k (rad/s), and their angular acceleration vectors are α AB = 1.0 k (rad/s 2 ) and α BC = 1.0 k (rad/s 2 ). What is the acceleration of point C?

Solution:

Use the solution to Problem 17.170.

The vector rB /A = 760( i cos15 ° − j sin15 °) = 734.1i − 196.7 j (mm). The vector rC /B = 900( i cos 50 ° + j sin 50 °) = 578.5i + 689.4 j (mm).

The acceleration of point B is

760

 i j k   a B = α AB × rB /A − ω 2AB rB /A  0 0 1   734.1 −196.7 0   

− (0.5 2 )(734.1i − 196.7 j) (mm/s 2 ). a B = 196.7 i + 734.1 j − 183.5i + 49.7 j = 13.2 i + 783.28 j (mm/s 2 ).

The acceleration of point C is

158

508

2 r a C = a B + α BC × rC /B − ω BC C /B

 i j k   = a B +  0 0 1  − (2 2 )(578.5i + 689.4 j)  578.5 689.4 0    a C = 13.2 i + 783.3 j − 689.4 i + 578.5 j − 2314 i − 2757.8 j (mm/s 2 ). a C = −2990 i − 1396 j (mm/s 2 ).

 i j k  0 0 v B = ω AB × rB /A =  ω AB  734.1 −196.7 0 

     

158

= 196.7ω AB i + 734.1ω AB j (mm/s).

Use the solution to Problem 17.171. The vector rB /A = 760( i cos15 ° − j sin15 °) = 734.1i − 196.7 j (mm). The vector rC /B = 900( i cos50 ° + j sin 50°) = 578.5i + 689.4 j (mm). The velocity of point B is

508

760

Solution:

Problem 17.173 The velocity of point C is v C = 1.0 i (m/s) and a C = 0. What are the angular velocity and angular acceleration vectors of arm BC?

The velocity of point C is v C = v B + ω BC × rC /B  i j k  = 196.7ω AB i + 734.1ω AB j +  0 0 ω BC  578.5 689.4 0 

     

1000 i = 196.7ω AB i + 734.1ω AB j − 689.4 ω BC i + 578.5ω BC j (mm/s). Separate components: 1000 = 196.7ω AB − 689.4 ω BC , 0 = 734.1ω AB + 578.5ω BC .

The acceleration of point B is  i j k  a B = α AB × rB /A − ω 2AB rB /A =  α AB 0 0  734.1 −196.7 0 

Solve: ω AB = 0.933k (rad/s), ω BC = −1.184 k (rad/s).

     

− (ω 2AB )(734.1i − 196.7 j) (mm/s) 2 . a B = 196.7α AB i + 734.1α AB j − 639.0 i + 171.3 j (mm/s 2 ).

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17.173 (Continued) The acceleration of point C is 2 r a C = a B + α BC × rC /B − ω BC C /B

 i j k  = a B +  0 α BC 0  578.5 689.4 0 

   − (ω 2 ) (578.5i + 689.4) BC   

a C = 0 = α AB (196.7 i + 734.1 j) + α BC (−689.4 i + 578.5 j) − 811.0 i − 966.5 j − 639.0 i + 171.3 j (mm/s 2 ) Separate components: 196.7α AB − 689.4α BC − 811.3 − 639.0 = 0, 734.1α AB + 578.5α BC − 966.5 + 171.3 = 0. Solve: α AB = 2.24 rad/s 2 , α BC = −1.465 (rad/s 2 ).

Problem 17.174 The crank AB has a constant clockwise angular velocity of 200 rpm. What are the velocity and acceleration of the piston P?

B 2 in

A P

Solution: 200 rpm = 200(2π )/60 = 20.9 rad/s. v B = v A + ω AB × rB /A = 0 +

2 in i j k 0 0 −20.9 . 2 2 0

v B = v P + ω BP × rB /P = v P i +

Also,

i 0

Also,

j k 0 ω BP .

−6 2

6 in

2 r a B = a P + α BP × rB /P − ω BP B /P

= aPi +

i 0

j k 0 α BP

−6 2

2 (− 6 i + 2 j). − ω BP

Equating i and j components,

Equating i and j components in these two expressions,

2 , −2(20.9) 2 = a P − 2α BP + 6ω BP

−(−20.9)(2) = v P − 2ω BP , (−20.9)(2) = −6ω BP ,

2 , −2(20.9) 2 = −6α BP − 2ω BP

we obtain ω BP = 6.98 rad/s and v P = 55.9 in/s.

and solving, we obtain a P = −910 in/s 2 .

a B = a A + α AB × rB /A − ω 2AB rB /A = 0 + 0 − (−20.9) 2 (2 i + 2 j).

Problem 17.175 Bar AB has a counterclockwise angular velocity of 10 rad/s and a clockwise angular acceleration of 20 rad/s 2 . Determine the angular acceleration of bar BC and the acceleration of point C.

Solution:

Choose a coordinate system with the origin at the left end of the horizontal rod and the x axis parallel to the horizontal rod. The strategy is to determine the angular velocity of bar BC from the instantaneous center; using the angular velocity and the constraint on the motion of C, the accelerations are determined. The vector rB /A = rB − r A = (8i + 4 j) − 4 j = 8i (in.).

y 8 in

The velocity of point B is

6 in

 i j k    v B = ω AB × rB /A =  0 0 10  = 80 j (in/s).  8 0 0   

10 rad/s A

B 20 rad/s2

The acceleration of point B is

4 in C

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 i j k     a B = α AB × rB /A A /B =  0 0 − 20  − 100(8i)   8 0 0   2 = −800 i − 160 j (in/s ). − ω 2r

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17.175 (Continued) The velocity of point B is normal to the x axis, and the velocity of C is parallel to the x axis. The line perpendicular to the velocity at B is parallel to the x-axis, and the line perpendicular to the velocity at C is parallel to the y axis. The intercept is at (14, 4), which is the instantaneous center of bar BC. Denote the instantaneous center by C ′′.

The acceleration of point C is 2 r a C = a B + α BC × rC /B − ω BC C /B

The vector rB /C ′′ = rB − rC ′′ = (8i + 4 j) − (14 i − 4 j) = −6 i (in.). The velocity of point B is  i j k  v B = ω BC × rB /IC =  0 0 ω BC  −6 0 0  from which ω BC = −

   − 1066.7 i + 711.1 j (in/s 2 ).   

 j k  i  = −800 i − 160 j +  0 0 α BC  6 −4 0 

The acceleration of point C is constrained to be parallel to the x axis. Separate components:

   = −6ω j = 80 j, BC   

a C = −800 + 4α BC − 1066.7, 0 = −160 + 6α BC + 711.1. Solve:

80 = −13.33 rad/s. 6

a C = −2234 i (in/s 2 ), α BC = −91.9 rad/s 2 (clockwise).

The vector rC /B = rC − rB = (14 i) − (8i + 4 j) = 6 i − 4 j (in.).

Problem 17.176 The angular velocity of arm AC is 1 rad/s counterclockwise. What is the angular velocity of the scoop?

1m D C 1m

Solution:

Choose a coordinate system with the origin at A and the y axis vertical. The vector locations of B, C and D are rB = 0.6 i (m), rC = −0.15i + 0.6 j (m), rD = (1 − 0.15) i + 1 j = 0.85i + j (m), from which rD /C = rD − rC = 1i + 0.4 j (m), and rD /B = rD − rB = 0.25i + j (m). The velocity of point C is  i j k  0 0 ω AC v C = ω AC × rC /A =   −0.15 0.6 0  = −0.6 i − 0.15 j (m/s).

     

0.6 m

0.15 m

The velocity of D in terms of the velocity of C is  i j k  v D = v C + ω CD × rD /C = −0.6 i − 0.15 j +  0 0 ω CD  1 0.4 0  = −0.6 i − 0.15 j + ω CD (−0.4 i + j).

     

0.6 m

 i j k  v D = ω DB × rD /B =  0 0 ω DB  0.25 1 0 

Scoop

   = ω (−i + 0.25 j). DB   

Equate expressions and separate components: −0.6 − 0.4 ω CD = −ω DB , − 0.15 + ω CD = 0.25ω DB . Solve:

The velocity of point D in terms of the angular velocity of the scoop is

ω CD = 0.333 rad/s, ω DB = 0.733 rad/s (counterclockwise).

Problem 17.177 The angular velocity of arm AC is 2 rad/s counterclockwise and its angular acceleration is 4 rad/s 2 clockwise. What is the angular acceleration of the scoop?

Solution:

 i j k   v C = ω AC × rC /A =  0 0 ω AC  = −1.2 i − 0.3 j (m/s).  −0.15 0.6 0   The velocity of D in terms of the velocity of C is

D C 1m 0.6 m B

Use the solution to Problem 17.176. Choose a coordinate system with the origin at A and the y axis vertical. The vector locations of B, C and D are rB = 0.6 i (m), rC = −0.15i + 0.6 j (m), rD = (1 − 0.15) i + 1 j = 0.85i + j (m), from which rD /C = rD − rC = 1i + 0.4 j (m), and rD /B = rD − rB = 0.25i + j (m). The velocity of point C is

 i j k  v D = v C + ω CD × rD /C = −1.2 i − 0.3 j +  0 0 ω CD  1 0.4 0 

     

= −1.2 i − 0.3 j + ω CD (−0.4 i + j). 0.15 m

382

0.6 m

Scoop

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17.177 (Continued) The velocity of point D in terms of the angular velocity of the scoop is  i j k    v D = ω DB × rD /B =  0 0 ω DB  = ω DB (−i + 0.25 j).  0.25 1 0   Equate expressions and separate components:

The acceleration of point D in terms of the angular acceleration of point B is 2 r a D = α BD × rD /B − ω BD D /B

 i j k  =  0 0 αCD  0.25 1 0 

−1.2 − 0.4 ω CD = −ω DB , − 0.3 + ω CD = 0.25ω DB .

   − ω 2 (0.25i + j). BD   

a D = α BD (−i + 0.25 j) − 0.538i − 2.15 j.

Solve: ω CD = 0.667 rad/s, ω DB = 1.47 rad/s. The angular acceleration of the point C is

Equate expressions for and separate components:

a C = α AC × rC /A − ω 2AC rC /A

−0.4αCD + 2.56 = −α BD − 0.538,

 i j k  0 0 α AC =   −0.15 0.6 0 

   − ω 2 (−0.15i + 0.6 j), AC   

αCD − 1.98 = 0.25α BD − 2.15. Solve: αCD = −1.052 rad/s 2 , α BD = −3.51 rad/s 2 ,

a C = 2.4 i + 0.6 j + 0.6 i − 2.4 j = 3i − 1.8 j (m/s 2 ). The acceleration of point D in terms of the acceleration of point C is

where the negative sign means a clockwise acceleration.

2 r a D = a C + αCD × rD /C − ω CD D /C

 i j k  = 3i − 1.8 j +  0 0 αCD  1 0.4 0 

   − ω 2 ( i + 0.4 j). CD   

a C = αCD (−0.4 i + j) + 2.56 i − 1.98 j (m/s 2 ).

Problem 17.178 If you want to program the robot so that, at the instant shown, the velocity of point D is v D = 0.2 i + 0.8 j (m/s) and the angular velocity of arm CD is 0.3 rad/s counterclockwise, what are the necessary angular velocities of arms AB and BC?

Solution:

The position vectors are:

rB /A = 300( i cos30 ° + j sin30 °) = 259.8i + 150 j (mm), rC /B = 250( i cos20 ° − j sin 20°) = 234.9 i − 85.5 j (mm), rC /D = −250 i (mm). The velocity of the point B is

250

m 00

308

 i j k  v B = ω AB × rB /A =  0 0 ω AB  259.8 150 0 

208

v B = −150 ω AB i + 259.8ω AB j.

     

x 250 mm

The velocity of point C in terms of the velocity of B is  i j k   v C = v B + ω BC × rC /B = v B +  0 0 ω BC   234.9 −85.5 0   = −150 ω AB i + 259.8ω AB j + 85.5ω BC i + 234.9ω BC j (mm/s). The velocity of point C in terms of the velocity of point D is  i j k   v C = v D + ω CD × rC /D = 200 i + 800 j + 0.3  0 0 1   −250 0 0    = 200 i + 725 j (mm/s). Equate the expressions for v C and separate components: −150 ω AB + 85.5ω BC = 200, and 259.8ω AB + 234.9ω BC = 725. Solve: ω AB = 0.261 rad/s, ω BC = 2.80 rad/s.

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Problem 17.179 The ring gear is stationary, and the sun gear rotates at 120 rpm in the counterclockwise direction. Determine the angular velocity of the planet gears and the magnitude of the velocity of their centerpoints.

Ring gear

7 in 34 in 20 in

Solution:

Denote the point O be the center of the sun gear, point S to be the point of contact between the upper planet gear and the sun gear, point P be the center of the upper planet gear, and point C be the point of contact between the upper planet gear and the ring gear. The angular velocity of the sun gear is ωS =

Planet gears (3) Sun gear

120(2π ) = 4 π rad/s, 60

from which ω S = 4 πk (rad/s). At the point of contact between the sun gear and the upper planet gear the velocities are equal. The vectors are: from center of sun gear to S is rP /S = 20 j (in.), and from center of planet gear to S is rS /P = −7 j (in.). The velocities are: v S /O = v O + ω S × (20 j) = 0 + ω S (20)(k × j)  i j k    v S /O = 20 ω S  0 0 1  = −20(ω S )  0 1 0    i = −251.3i (in/s).

y C

From equality of the velocities, v S /P = v S /O = −251.3i (in/s). The point of contact C between the upper planet gear and the ring gear is stationary, from which

v S /P = −251.3i = v C + ω P × rC /S  i j k  = 0 +  0 0 ωP  −14 0 0 

   = 14 ω i = −251.3i P   

from which ω P = 17.95 rad/s. The velocity of the centerpoint of the top most planet gear is v P = v S /P + ω P × rP /S = −251.3i + (−17.95)(−7)(k × j)  i j k    = −251.3i + 125.65  0 0 1   0 1 0    v P = −125.7 i (in./s) The magnitude is v PO = 125.7 in./s. By symmetry, the magnitudes of the velocities of the centerpoints of the other planetary gears is the same.

Problem 17.180 Arm AB is rotating at 10 rad/s in the clockwise direction. Determine the angular velocity of arm BC and the velocity at which the arm slides relative to the sleeve at C.

B 1.8

308 A

C 2m

384

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17.180 (Continued) Solution:

The position vectors are

The velocity of B in terms of the velocity of C is

rB /A = 1.8( i cos30 ° + j sin 30 °) = 1.56 i + 0.9 j (m). rB /C = rB /A − 2 i = −0.441i + 0.9 j (m). The velocity of point B is  j k   i v B = ω AB × rB /A =  0 0 −10  = 9 i − 15.6 j (m/s)  1.56 0.9 0   

Equate the expressions for v B and separate components: 9 = 0.4401v Crel − 0.9ω BC , and −15.6 = −0.8979v Crel − 0.441ω BC .

The unit vector from B to C is e BC =

 i j k    v B = v rel + ω BC × rB /C = v rel +  0 0 ω BC  ,  −0.441 0.9 0   v B = 0.4401v Crel i − 0.8979v Crel j − 0.9ω BC i − 0.441ω BC j (m/s).

Solve:

−rB /C = 0.4401i − 0.8979 j. r B /C

The relative velocity is parallel to this vector: v Crel = v Crel e BC = v Crel (0.4401i − 0.8979 j) (m/s)

v Crel = 17.96 m/s (toward C). ω BC = −1.22 rad/s (clockwise)

Problem 17.181 Arm AB is rotating with an angular velocity of 10 rad/s and an angular acceleration of 20 rad/s 2 , both in the clockwise direction. Determine the angular acceleration of arm BC. Solution:

B 1.8

308

Use the solution to 17.180. The vector A

rB /A = 1.8( i cos30 ° + j sin 30 °) = 1.56 i + 0.9 j (m). rB /C = rB /A − 2 i = −0.441i + 0.9 j (m).

C 2m

The angular velocity: ω BC = −1.22 rad/s, and the relative velocity is v Crel = 17.96 m/s. The unit vector parallel to bar BC is e = 0.4401i − 0.8979 j

Collecting terms,

The acceleration of point B is

a B = a Crel (0.4401i − 0.8979 j) − α BC (0.9 i + 0.4411 j)

aB =

− 38.6 i − 20.6 j (m/s 2 ).

α AB × rB /A − ω 2AB rB /A

 i j k   0 α AB  − ω 2AB (1.56 i + 0.9 j), =  0  1.56 0.9 0    a B = 18i − 31.2 j − 155.9 i − 90 j = −137.9 i − 121.2 j (m/s 2 ). The acceleration of point B in terms of the acceleration of bar BC is a B = a Crel + 2ω BC × v Crel

Equate the two expressions for a B and separate components: −137.9 = 0.4401a Crel − 0.9α BC − 38.6, and − 121.2 = −0.8979a Crel − 0.4411α BC − 20.6. Solve:

a Crel = 46.6 m/s 2 (toward C)

α BC = 133.1 rad/s 2 .

2 r + α BC × rB /C − ω BC B /C .

Expanding term by term: a Crel = a Crel (0.4401 i − 0.8979 j) (m/s 2 ),  i j k   0 0 1  2ω BC × v Crel = 2v Crel ω BC   0.440 −0.8979 0  = −39.26 i − 19.25 j (m/s 2 ),  i  0 α BC × rB / C =   −0.4411 

j 0 0.9

k α BC 0

     

= α BC (−0.9 i − 0.4411 j) (m/s 2 ) − ω 2BC (−0.4411i + 0.9 j) = 0.6539 i − 1.334 j (m/s 2 ).

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Problem 17.182 Arm AB is rotating with a constant counterclockwise angular velocity of 10 rad/s. Determine the vertical velocity and acceleration of the rack R of the rack-and-pinion gear.

B C 12 in

10 in 6 in D

Solution:

The vectors:

rB /A = 6 i + 12 j (in.).

rC /B = 16 i − 2 j (in.). 6 in

The velocity of point B is

16 in

6 in

 i j k    v B = ω AB × rB /A =  0 0 10  = −120 i + 60 j (in/s).  6 12 0    The velocity of point C in terms of the velocity of point B is  i j k  v C = v B + ω BC × rC /B = v B +  0 0 ω BC  16 −2 0  = −120 i + 60 j + ω BC (2 i + 16 j) (in/s)

     

The velocity of point C in terms of the velocity of the gear arm CD is  i j k  0 ω CD v C = ω CD × rC /D =  0  −6 10 0  = −10 ω CD i − 6ω CD j (in/s).

     

Equate the two expressions for v C and separate components: −120 + 2ω BC = −10 ω CD , 60 + 16ω BC = −6ω CD . Solve:

ω BC = −8.92 rad/s, ω CD = 13.78 rad/s,

where the negative sign means a clockwise rotation. The velocity of the rack is  i j k  v R = ω CD × rR /D =  0 0 ω CD  6 0 0  v R = 82.7 j (in/s) = 6.89 j (ft/s).

   = 6ω j, CD   

The acceleration of point C in terms of the gear arm is 2 r a C = αCD × rC /D − ω CD C /D

 i j k  0 αCD =  0  −6 10 0 

   − ω 2 (−6 i + 10 j) (in/s 2 ), CD   

a C = −10αCD i − 6αCD j + 1140 i − 1900 j (in/s 2 ). Equate expressions for a C and separate components: 2α BC − 1873 = −10αCD + 1140, 16α BC − 1041 = −6αCD − 1900. Solve:

αCD = 337.3 rad/s 2 , and α BC = −180.2 rad/s 2 .

The acceleration of the rack R is the tangential component of the acceleration of the gear at the point of contact with the rack:   i j k   a R = αCD × rR /D =  0 0 αCD  = 6αCD j (in/s 2 ).   6 0 0   a R = 2024 j (in/s 2 ) = 169 j (ft/s 2 ).

The angular acceleration of point B is a B = −ω 2AB rB /A = −100(6 i + 12 j) = −600 i − 1200 j (in/s 2 ). The acceleration of point C is

aCD D

2 r a C = a B + α BC × rC /B − ω BC C /B ,

 i j k  a C = a B +  0 0 α BC  16 −2 0 

vCD

r R/D

   − ω 2 (16 i − 2 j) BC   

2 (16 i − 2 j). = a B + 2α BC i + 16α BC j − ω BC

Noting 2 (16 i − 2 j) = − 600 i − 1200 j − 1272.7 i + 159.1 j a B − ω BC

= −1872.7 i − 1040.9 j, from which a C = +α BC (2 i + 16 j) − 1873i − 1041 j (in/s 2 )

386

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Problem 17.183 The rack R of the rack-and-pinion gear is moving upward with a constant velocity of 10 ft/s. What are the angular velocity and angular acceleration of bar BC?

B C 12 in

10 in 6 in D

Solution: The constant velocity of the rack R implies that the angular acceleration of the gear is zero, and the angular velocity of the 120 gear is ω CD = = 20 rad/s. The velocity of point C in terms of 6 the gear angular velocity is  i j k    v C = ω CD × rC /D =  0 0 20  = −200 i − 120 j (in/s).  −6 10 0    The velocity of point B in terms of the velocity of point C is  i j k  v B = v C + ω BC × rB /C = v C +  0 0 ω BC  −16 2 0 

  ,   

R 6 in

16 in

6 in

The acceleration of point B in terms of the acceleration of C is 2 r a B = a C + α BC × rB /C − ω BC B /C

 i j k  0 α BC = a C +  0  −16 2 0 

   − ω 2 (−16 i + 2 j). BC   

a B = α BC (−2 i − 16 j) + 2400 i − 4000 j + 2680 i − 335 j.

v B = −200 i − 120 j − 2ω BC i − 16ω BC j (in/s).

a B = −2α BC i − 16α BC j + 5080 i − 433.5 j (in/s 2 ).

The velocity of point B in terms of the angular velocity of the arm AB is

The acceleration of point B in terms of the angular acceleration of arm AB is

 i j k  v B = ω AB × rB /A =  0 0 ω AB  6 12 0  = −12ω AB i + 6ω AB j (in/s).

     

a B = α AB × rB /A − ω 2AB rB /A = α AB (k × (6 i + 12 j)) − ω 2AB (6 i + 12 j) (in/s 2 ) a B = α AB (−12 i + 6 j) − 1263.2 i − 2526.4 j (in/s 2 ). Equate the expressions for a B and separate components:

Equate the expressions for v B and separate components

−2α BC + 5080 = −12α AB − 1263.2,

−200 − 2ω BC = −12ω AB , − 120 − 16ω BC = 6ω AB .

−16α BC − 4335 = 6α AB − 2526.4.

Solve: ω AB = 14.5 rad/s, ω BC = −12.94 rad/s , where the negative sign means a clockwise rotation. The angular acceleration of the point C in terms of the angular velocity of the gear is

Solve: α AB = −515.2 rad/s 2 ,

α BC = 80.17 rad/s 2 .

2 r 2 a C = αCD × rC /D − ω CD C /D = 0 − ω CD (− 6 i + 10 j)

= 2400 i − 4000 j (in/s 2 ).

Problem 17.184 Bar AB has a constant counterclockwise angular velocity of 2 rad/s. The 1-kg collar C slides on the smooth horizontal bar. At the instant shown, what is the tension in the cable BC?

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17.184 (Continued) Solution: v B = v A + ω AB × rB /A = 0 +

i j k 0 0 2 1 2 0

= −4 i + 2 j (m/s).

v C = v B + ω BC × rC /B : i j k 0 0 ω BC . 2 −2 0

v C i = −4 i + 2 j +

From the i and j components of this equation, v C = −4 + 2ω BC , 0 = 2 + 2ω BC , we obtain ω BC = −1 rad/s. a B = a A + α AB × rB /A − ω 2AB rB /A = 0 + 0 − (2) 2 ( i + 2 j) = −4 i − 8 j (m/s 2 ). 2 r a C = a B + α BC × rC /B − ω BC C /B :

a C i = −4 i − 8 j +

i j k 0 0 α BC 2 −2 0

− (−1) 2 (2 i − 2 j).

From the i and j components of this equation, a C = −4 + 2α BC − 2, 0 = −8 + 2α BC + 2, we obtain a C = 0. The force exerted on the collar at this instant is zero, so TBC = 0.

Problem 17.185 An athlete exercises her arm by raising the 8-kg mass m. The shoulder joint A is stationary. The distance AB is 300 mm, the distance BC is 400 mm, and the distance from C to the pulley is 340 mm. The angular velocities ω AB = 1.5 rad/s and ω BC = 2 rad/s are constant. What is the tension in the cable?

C 308

vBC

608

vAB A

Solution: a B = −ω 2ABrB /A i = −(1.5) 2 (0.3) i = −0.675i (m/s 2 ). 2 r a C = a B − ω BC C /B

= −0.675i − (2) 2 (0.4 cos60 °i + 0.4sin60 ° j)

= −1.475i − 1.386 j (m/s 2 ).

308

a C ⋅ e = (−1.475)(− cos30 °) + (−1.386)(sin30 °) = 0.585 m/s 2 . This is the upward acceleration of the mass, so 608

T − mg = m(0.585), T = (8)(9.81 + 0.585)

= 83.2 N.

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Problem 17.186 The hydraulic actuator BC of the crane is extending (increasing in length) at a constant rate of 0.2 m/s. When the angle β = 35 °, what is the angular velocity of the crane’s boom AD?

Solution:

Using AD we have

v C = v A + ω AD × rC /A = 0 + ω AD k × (3 m)(cos35 °i + sin35 ° j) = (3 m)ω AD (−sin35 °i + cos35 ° j)

Using cylinder BC v C = v B + v Crel + ω BC × rC /B  [3cos35 ° − 2]i + [3sin35 °] j  = 0 + (0.2 m/s)   [3cos35 ° − 2] 2 + [3sin35 °] 2  + ω BC k × [3cos35 ° − 2]i + [3sin35°] j = (0.0514 m/s − {1.72 m } ω BC ) i + (0.193 m/s − { 0.457 m } ω BC ) j Equating components and solving we find ω BC = 0.133 rad/s, ω AD = 0.103 rad/s ω AD = 0.103 rad/s.

Problem 17.187 The coordinate system shown is fixed relative to the ship B. The ship uses its radar to measure the position of a stationary buoy A and determines it to be 400 i + 200 j (m). The ship also measures the velocity of the buoy relative to its body-fixed coordinate system and determines it to be 2 i − 8 j (m/s). What are the ship’s velocity and angular velocity relative to the Earth? (Assume that the ship’s velocity is in the direction of the y-axis.)

Solution: v A = 0 = v B + v Arel + ω × r A /B = v B j + 2i − 8 j +

i j k 0 0 ω . 400 200 0

Equating i and j components to zero, 0 = 2 − 200 ω 0 = v B − 8 + 400 ω we obtain ω = 0.01 rad/s and v B = 4 m/s.

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Chapter 18 Problem 18.1 A horizontal force F = 30 lb is applied to the 285-lb refrigerator as shown. Friction is negligible. (a) What is the magnitude of the refrigerator’s acceleration? (b) What normal forces are exerted on the refrigerator by the floor at A and B?

Solution:

Assume that the refrigerator rolls without tipping. We have the following equations of motion. F 5 30 lb 230 lb 60 in.

F 28 in.

A 60 in

∑ Fx : (30 lb) = ( 32.2 ft/s 2 )a 230 lb

28 in A

B 14 in

14 in

∑ Fy : A + B − 230 lb = 0 ∑ M G : −(30 lb)(32 in.) − A(14 in.) + B(14 in.) = 0 Solving we find (a)

a = 4.2 ft/s 2 .

(b)

A = 80.7 lb, B = 149.3 lb. Since A > 0 and B > 0 then our assumption is correct.

Problem 18.2 Solve Problem 18.1 if the coefficient of kinetic friction at A and B is µ k = 0.1.

Solution:

Assume sliding without tipping F 5 30 lb 230 lb 60 in.

28 in. mA 60 in

mB N

28 in A

B 14 in

14 in

∑ Fx : (30 lb) − (0.1)( A + B) = ( 32.2 ft/s 2 ) a 230 lb

∑ Fy : A + B − 230 lb = 0 ∑ M G : −(30 lb)(32 in.) − A(14 in.) + B(14 in.) − (0.1)( A + B)(28 in.) = 0 Solving, we find

390

(a)

a = 0.98 ft/s 2 .

(b)

A = 57.7 lb, B = 172 lb.

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Problem 18.3 The largest horizontal force F the man can exert on the 285-lb refrigerator without causing it to tip over is the force that causes the normal force at A to equal zero. Determine that force and the resulting acceleration of the refrigerator if the kinematic coefficient of friction between the supports and the floor is µ k = 0.2.

Solution: F

60 in 285 lb

mkB

28 in

14 in

F The mass of the refrigerator is W 285 lb m = = = 8.85 slugs. g 32.2 ft/s 2

Applying Newton’s second law in the horizontal direction,

60 in

∑ Fx = 20 lb − µ k B = (8.85 slugs)a x . (1) 28 in A

B 14 in

14 in

The sum of the moments about the center of mass is

∑ M center of mass = (14 in)B − (28 in)µ k B − (60 in − 28 in) F = 0. (2) Because there is no acceleration in the y direction, we have

∑ Fy = B − 285 lb = 0. Solving Eqs. (1) and (2) with B = 285 lb and µ k = 0.2, we obtain F = 74.8 lb, a x = 2.01 ft/s 2 . F = 74.8 lb, acceleration is 2.26 ft/s 2 .

Problem 18.4 The Boeing 747 begins its takeoff run at time t = 0. The normal forces exerted on its tires at A and B are N A = 150 kN and N B = 2800 kN, respectively. If you assume that these forces are constant and neglect horizontal forces other than the thrust T, how fast is the airplane moving at t = 5 s? (See Practice Example 18.1.)

T A

3m B 26 m

Solution: Let m denote the airplane’s mass. Newton’s second law in the horizontal direction is

∑ Fx = T = ma x . There is no acceleration in the vertical direction, so we have

∑ Fy = N A + N B − mg = 0. The sum of the moments about the center of mass is

∑ M center of mass = (2 m) N B − (24 m) N A − (2 m)T = 0. Solving these equations with the given reactions at A and B, we obtain a x = 2.33 m/s 2 , T = 1000 kN, and m = 301,000 kg. If the given reactions remain constant, the plane’s acceleration remains constant and its velocity after 5 seconds is (5 s)a x = 16.6 m/s. 16.6 m/s.

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Problem 18.5 The crane moves to the right with constant acceleration, and the 800-kg load moves without swinging. (a) What is the acceleration of the crane and load? (b) What are the tensions in the cables attached at A and B? Solution:

From Newton’s second law: Fx = 800 a N. The sum of the forces on the load:

∑ Fx = FA sin 5° + FB sin 5° − 800a = 0. ∑ Fy = FA cos5° + FB cos5° − 800 g = 0.

The sum of the moments about the center of mass:

∑ M CM = −1.5FA cos 5° + 1.5FB cos 5° − FA sin 5 ° − FB sin 5 ° = 0.

1.5 m

Solve these three simultaneous equations: a = 0.858 m/s 2 , 58

FA = 3709 N ,

FB = 4169 N.

1.0 m mg 1.5 m 1.5 m

Problem 18.6 The total weight of the go-cart and driver is 240 lb. The location of their combined center of mass is shown. The rear drive wheels together exert a 24-lb horizontal force on the track. Neglect the horizontal forces exerted on the front wheels. (a) What is the magnitude of the go-cart’s acceleration? (b) What normal forces are exerted on the tires at A and B?

Solution:

240 lb

24 lb

∑ Fx : (24 lb) = ( 32.2 ft/s 2 )a 240 lb

15 in 6 in

4 in

B 16 in

392

60 in

∑ Fy : N A + N B − (240 lb) = 0 ∑ M G : − N A (16 in.) + N B (44 in.) + (24 lb)(15 in.) = 0 Solving we find (a)

a = 3.22 ft/s 2 .

(b)

N A = 182 lb, N B = 58 lb.

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Problem 18.7 The combined mass of the bicycle and rider is 65 kg. The location of their common center of mass is shown. The dimensions are h = 0.95 m, b = 0.57 m, c = 0.43 m. If the normal force exerted by the road on the bicycle’s rear tire at the instant shown is N B = 380 N, what is the bicycle’s acceleration? Solution:

Let m and F denote their combined mass and the horizontal force exerted by the road, respectively. Newton’s second law in the horizontal direction is

∑ Fx = F = ma x . There is no acceleration in the vertical direction, so we have

∑ Fy = N A + N B − mg = 0. A

The sum of the moments about the center of mass is

B b

∑ M center of mass = cN B − bN A − hF = 0.

Solving these equations with the given mass and reaction at B, we obtain a x = 0.268 m/s 2 , F = 17.4 N, and N A = 258 N. 0.268 m/s 2 .

Problem 18.8 The moment of inertia of the disk about O is I = 20 kg-m 2 . At t = 0, the stationary disk is subjected to a constant 50 N-m torque. (a) What is the magnitude of the resulting angular acceleration of the disk? (b) How fast is the disk rotating (in rpm) at t = 4 s?

Solution: (a) M = I α ⇒ α =

M 50 N-m = = 2.5 rad/s 2 . I 20 kg-m 2

α = 2.5 rad/s 2 . (b) The angular velocity is given by ω = αt = (2.5 rad/s 2 )(4 s) = 10 rad/s

50 N-m

( 21πrev )( 60 s ) rad 1 min

= 95.5 rpm. ω = 95.5 rpm.

Problem 18.9 The 20-lb bar is at rest on a smooth table. The figure shows the bar viewed from above. The bar’s moment of inertia about its center of mass is I = 0.828 slug-ft 2 . At time t = 0, a force F = 4 j (lb) is applied at A. At that instant, what is the acceleration of point A? y

Solution: y G

A F

rA/G x

The mass of the bar is m = A

2 ft

20 lb W = = 0.621 slugs. g 32.2 ft/s 2

Newton’s second law tells us the acceleration of the center of mass,

∑ Fy = ma y : 4 lb = (0.621 slugs)a y , x giving a y = 6.44 ft/s 2 .

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18.9 (Continued) The force exerts a clockwise moment about the center of mass, resulting in a clockwise angular acceleration (2 ft)(4 lb) M = = 9.66 rad/s 2 . I 0.828 slug-ft 2 Now we can write the acceleration of point A in terms of the acceleration of the center of mass: a A = a G + α × r A /G − ω 2 r A /G i = (6.44 ft/s 2 ) j +

0 0 −9.66 rad/s 2 −2 ft 0 0

−0

= 25.8 j (ft/s 2 ). a A = 25.8 j (ft/s 2 ).

Problem 18.10 The 20-lb bar is on a smooth horizontal table. The figure shows the bar viewed from above. The bar’s moment of inertia about its center of mass is I = 0.828 slug-ft 2 . At time t = 0, the bar is rotating at 1 rad/s in the counterclockwise direction and a force F = 4 j (lb) is applied at A. At that instant, what is the acceleration of point B?

Solution: y G

A F

rA/G x

The mass of the bar is

m =

W 20 lb = = 0.621 slugs. g 32.2 ft/s 2

Newton’s second law tells us the acceleration of the center of mass, A

2 ft

ΣFy = ma y : 4 lb = (0.621 slugs)a y , giving a y = 6.44 ft/s 2 . x The force exerts a clockwise moment about the center of mass, resulting in a clockwise angular acceleration (2 ft)(4 lb) M = = 9.66 rad/s 2 . I 0.828 slug-ft 2 Now we can write the acceleration of point B in terms of the acceleration of the center of mass: a B = a G + α × r B /G − ω 2 r B /G i = (6.44 ft/s 2 ) j +

0 0 −9.66 rad/s 2 2 ft 0 0

− (1 rad/s) 2 (2 ft) i

= −2 i − 12.9 j (ft/s 2 ). a B = −2 i − 12.9 j (ft/s 2 ).

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Problem 18.11 The SpaceX capsule is not rotating at time t = 0. Point A has coordinates x = 1.4 m, y = 1.7 m. The capsule’s moment of inertia about its center of mass is I = 32,200 kg-m 2 . Its attitude control thruster exerts a force T = 400 N for 2 s. How long does it take the capsule to rotate 10 ° relative to its initial orientation?

Solution: T

208

1.7 m

1.4 m

Ten degrees is 10 °

( π180rad ) = 0.175 rad.

The counterclockwise moment of the force about the center of mass is M = (1.7 m)T cos 20 ° − (1.4 m)T sin 20° = 447 N-m. The counterclockwise angular acceleration while the thruster is firing is

Source: Courtesy of NASA Photo/Alamy Stock Photo.

α =

(447 N-m) M = = 0.0139 rad/s 2 . I 32,200 kg-m 2

The angular velocity and displacement during that period are

y T A

ω = αt,

208

θ =

1 2 αt . 2

At t = 2 s, their values are G x

ω 2 seconds = 0.0278 rad/s,

θ 2 seconds = 0.0278 rad (1.59 °).

After the thruster has fired, the capsule will continue to rotate with constant angular velocity. Letting t now denote the time after the first two seconds, we want the time at which the capsule has rotated ten degrees: θ 2 seconds + ω 2 seconds t = 0.0175 rad (ten degrees in radians). Solving, we obtain t = 5.28 s. Adding 2 second, the time required is 7.28 s. 7.28 s.

Problem 18.12 The moment of inertia of the helicopter’s rotor is 420 slug-ft 2 . The rotor starts from rest. At t = 0, the pilot begins advancing the throttle so that the torque exerted on the rotor by the engine (in ft-lb) is given as a function of time in seconds by T = 200t. (a) How long does it take the rotor to turn ten revolutions? (b) What is the rotor’s angular velocity (in rpm) when it has turned ten revolutions?

Solution:

Find the angular acceleration

T = Iα ⇒ α =

T 200t = = 0.476t I 420

Now answer the kinematics questions α = 0.476t, ω = 0.238t 2 , θ = 0.0794t 3 . (a) When it has turned 10 revolutions, (10 rev)

( 2πrevrad ) = 0.0794t

⇒

t = 9.25 s.

(b) The angular velocity is ω = 0.238(9.25) 2 = 20.4 rad/s

( 21πrev )( 60 s ) = 195 rpm rad 1 min

ω = 195 rpm.

Source: Courtesy of Roman Babakin/Shutterstock.

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Problem 18.13 The moments of inertia of the pulleys are I A = 0.0025 kg-m 2 , I B = 0.045 kg-m 2 , and I C = 0.036 kg-m 2 . A 5 N-m counterclockwise couple is applied to pulley A. Determine the resulting counterclockwise angular accelerations of the three pulleys.

Solution:

The tension in each belt changes as it goes around each

pulley. The unknowns are ∆T AB , ∆TBC , α A , α B , αC . We will write three dynamic equations and two constraint equations

∑ M A : (5 N-m) − ∆T AB (0.1 m) = (0.0025 kg-m 2 )α A ∑ M B : ∆T AB (0.2 m) − ∆TBC (0.1 m) = (0.045 kg-m 2 )α B

100 mm

∑ M C : ∆TBC (0.2 m) = (0.036 kg-m 2 )αC

100 mm

(0.1 m)α A = (0.2 m)α B A 200 mm

200 mm

(0.1 m)α B = (0.2 m)αC . Solving, we find ∆T AB = 42.2 N, ∆TBC = 14.1 N, α A = 313 rad/s 2 , α B = 156 rad/s 2 , αC = 78.1 rad/s 2 .

Problem 18.14 The moment of inertia of the wind- tunnel fan is 7200 kg-m 2 . It starts from rest at time t = 0. The torque exerted on it by the engine is given as a function of time in seconds by T = 940 − 4t N-m. How long does it take for the fan to reach 100 rpm (revolutions per minute)? Solution:

The given angular velocity in rad/s is

100 rpm = 100 rpm

rad 1 min ( 21πrev )( 60 s ) = 10.5 rad/s.

From the angular equation of motion ∑ M = I α, the fan’s angular acceleration is α =

dω ∑ M = 940 − 4t N-m . = dt I 7200 kg-m 2

Source: Courtesy of Aviation C2017/Alamy Stock Photo.

Integrating, we have 1

∫0 d ω = 7200 ∫0 (940 − 4t ) dt: ω =

1 (940t − 2t 2 ). 7200

Solving this equation for t with ω = 10.5 rad/s, we obtain t = 103 s. 103 s.

Problem 18.15 The moment of inertia of the pulley about its axis is I = 0.005 kg-m 2 . If the 1-kg mass A is released from rest, how far does it fall in 0.5 s? Strategy: Draw individual free-body diagrams of the pulley and the mass.

100 mm

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18.15 (Continued) Solution:

The two free-body diagrams are shown.

The five unknowns are T , O x , O y , α, a. We can write four dynamic equations and one constraint equation, however, we only need to write two dynamic equations and the one constraint equation.

∑ M O : −T (0.1 m) = −(0.005 kg-m 2 )α, ∑ Fy : T − (1 kg)(9.81 m/s 2 ) = −(1 kg)a, a = (0.1 m)α. Solving we find T = 3.27 N, a = 6.54 m/s 2 , α = 65.4 rad/s 2 . Now from kinematics we know d =

1 2 1 at = (6.54 m/s 2 )(0.5 s) 2 2 2

d = 0.818 m.

Problem 18.16 The radius of the pulley is 125 mm and the moment of inertia about its axis is I = 0.05 kg-m 2 . If the system is released from rest, how far does the 20-kg mass fall in 0.5 s? What is the tension in the rope between the 20-kg mass and the pulley?

4 kg

Solution:

The free-body diagrams are shown.

20 kg

We have six unknowns T1 , T2 , O x , O y , a, α. We have five dynamic equations and one constraint equation available. We will use three dynamic equations and the one constraint equation

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18.16 (Continued)

∑ M O : (T1 − T2 )(0.125 m) = −(0.05 kg-m 2 )α, ∑ Fy1 : T1 − (4 kg)(9.81 m/s 2 ) = (4 kg)a, ∑ Fy 2 : T2 − (20 kg)(9.81 m/s 2 ) = −(20 kg)a, a = (0.125 m)α. Solving we find T1 = 62.3 N, T2 = 80.8 N, a = 5.77 m/s 2 , α = 46.2 rad/s 2 . From kinematics we find d =

1 2 1 at = (5.77 m/s 2 )(0.5 s) 2 = 0.721 m. 2 2

d = 0.721 m, T2 = 80.8 N.

Problem 18.17 The moment of inertia of the pulley is 0.3 slug-ft 2 . The horizontal surface is smooth. If the system is released from rest, how long does it take for the 4-lb weight to move 1 ft in each case? Solution:

2 lb

2 lb (a)

(b)

T In this case we have the equations of motion

N 2 lb Let a denote the acceleration of the 4-lb weight. The angular acceleration of the pulley is given in terms of the pulley radius by α = a /R. We have the equations of motion

∑ F = T = ( 32.2 ft/s 2 )a, 4 lb

∑ M = R(2 lb) − RT = I α =

a (0.3 slug-ft 2 ) . R

Solving, we obtain T = 0.188 lb, a s = 1 ft =

6 in

Case (a):

4 lb

∑ F = T1 = ( 32.2 ft/s 2 )a, 4 lb

∑ M = RT2 − RT1 = I α = ( 0.3 slug-ft 2 ) R , ∑ F = 2 lb − T2 = ( 32.2 ft/s 2 )a. 2 lb

Solving, we obtain T1 = 0.179 lb, T2 = 1.91 lb, a = 1.44 ft/s 2 . Setting s = 1 ft =

= 1.51 ft/s 2 . Setting

1 2 at 2

gives t = 1.18 s.

1 2 at 2

(a) 1.15 s. (b) 1.18 s.

gives t = 1.15 s. Case (b):

4 lb

N T2 T2

2 lb

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Problem 18.18 The 5-kg slender bar is released from rest in the horizontal position shown. Determine the bar’s counterclockwise angular acceleration (a) at the instant it is released, and (b) at the instant when it has rotated 45°.

1.2 m

Solution: (a) The free-body diagram is shown.

(a)

L 1 ∑ M O : mg 2 = 3 mL2α α =

3g 3(9.81 m/s 2 ) = = 12.3 rad/s 2 . 2L 2(1.2 m)

α = 12.3 rad/s 2 .

(b)

(b) The free-body diagram is shown. L

∑ M O : mg 2 cos 45° = 3 mL2α α =

3g 3(9.81 m/s 2 ) cos 45 ° = cos 45 ° 2L 2(1.2 m)

α = 8.67 rad/s 2 .

Problem 18.19 The 5-kg slender bar is released from rest in the horizontal position shown. At the instant when it has rotated 45°, its angular velocity is 4.16 rad/s. At that instant, determine the magnitude of the force exerted on the bar by the pin support. (See Example 18.5.) Solution:

1.2 m

First find the angular acceleration. L

∑ M O : mg 2 cos45° = 3 mL2α α =

3g 3(9.81 m/s 2 ) cos45° = 8.67 rad/s 2 cos45° = 2(1.2 m) 2L

Using kinematics we find the acceleration of the center of mass. a G = a O + α × rG /O − ω 2 rG /O a G = 0 + (8.67)k × (0.6)(− cos45°i − sin 45° j) −(4.16) 2 (0.6)(− cos45°i − sin 45° j) = (11.0 i + 3.66 j) m/s 2 . From Newton’s second law we have

∑ Fx : O x = ma x = (5 kg)(11.0 m/s 2 ) = 55.1 N ∑ Fy : O y − mg = ma y O y = m( g + a y ) = (5 kg)(9.18 m/s 2 + 3.66 m/s 2 ) = 67.4 N The magnitude of the force in the pin is now O =

O x2 + O y2 =

(55.1 N) 2 + (67.4 N) 2 = 87.0 N.

O = 87.0 N.

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Problem 18.20 The 5-kg slender bar is released from rest in the horizontal position shown. Determine the magnitude of its angular velocity when it has fallen to the vertical position. Strategy: Draw the free-body diagram of the bar when it has fallen through an arbitrary angle θ and apply the equation of angular motion to determine the bar’s angular acceleration as a function of θ. Then use the chain rule to write the angular acceleration as α =

Solution:

1.2 m

d ω dθ dω dω = = ω. dt d θ dt dθ

First find the angular acceleration. L

∑ M O : mg 2 cos θ = 3 mL2α ⇒ α = 2 L cos θ Using the hint we have α = ω

ω 90° 3g 3g dω cos θ ⇒ ∫ ω d ω = ∫ cos θ d θ = 0 0 2L 2L dθ

90° 3g 3g 1 2 sin θ  ω = = 0 2 2L 2L

ω =

3g = L

3(9.81 m/s 2 ) = 4.95 rad/s. (1.2 m)

ω = 4.95 rad/s.

Problem 18.21 The T-shaped object consists of two 4-kg slender bars of length L = 1 m welded together. The y-axis is vertical. The object is released from rest in the position shown. What is its angular acceleration at that instant? Solution: Let m = 4 kg. The moment of inertia of the horizontal bar about O is I horizontal =

1 mL2 = 0.333 kg-m 2 . 12

y L 2

x L 2

L 2

Applying the parallel-axis theorem, the moment of inertia of the vertical bar about O is I vertical =

1 L 2 mL2 + m = 1.33 kg-m 2 . 12 2

( )

The moment of inertia of the T-shaped object about O is I = I horizontal + I vertical = 1.67 kg-m 2 . When the object is released, the counterclockwise moment about O exerted by the weight of the vertical bar is M =

L mg = 19.6 N-m. 2

The object’s counterclockwise angular acceleration is α =

M = 11.8 rad/s 2 . I

11.8 rad/s 2 (counterclockwise).

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Problem 18.22 The T-shaped object consists of two 4-kg slender bars of length L = 1 m welded together. The y-axis is vertical. The object is released from rest in the position shown. What is its angular velocity when it has rotated 90 ° from its initial position? (See Problem 18.20.)

y L 2

x L 2

Solution: Let m = 4 kg. The moment of inertia of the horizontal bar about O is I horizontal =

L 2

1 mL2 = 0.333 kg-m 2 . 12

Applying the parallel-axis theorem, the moment of inertia of the vertical bar about O is I vertical =

The counterclockwise moment exerted about O by the weight of the left bar is

1 L 2 mL2 + m = 1.33 kg-m 2 . 12 2

( )

( L2 cos θ )mg.

The moment of inertia of the T-shaped object about O is

M =

I = I horizontal + I vertical = 1.67 kg-m 2 .

The object’s counterclockwise angular acceleration is

Consider the object when it has rotated through an angle θ:

α =

Lmg M = cos θ. (1) I 2I

Now we apply the chain rule: α =

dω d ω dθ dω = = ω. dt d θ dt dθ

With this expression and Eq. (1), we can separate variables and integrate. Lmg cos θ d θ, 2I ω Lmg θ ∫0 ω d ω = 2I ∫0 cos θ dθ, Lmg 1 2 ω = sin θ. 2 2I ω dω =

At θ = 90 ° we obtain ω = 4.85 rad/s. 4.85 rad/s (counterclockwise).

Problem 18.23 The length of the slender bar is l = 6 m and its mass is m = 40 kg. It is released from rest in the position shown. (a) If x = 1 m, what is the bar’s angular acceleration at the instant it is released? (b) What value of x results in the largest initial angular acceleration when the bar is released? What is the angular acceleration? (c) Draw a graph of the bar’s angular acceleration versus x for 0 ≤ x ≤ 3 m.

Solution: 1 (a) The moment of inertia about the fixed point is I = ml 2 + mx 2 . 12 The angular equation of motion is ΣM fixed point = mgx = I α, so we obtain α =

12 gx . l 2 + 12 x 2

Using the given numbers, we obtain m

(b) To find the value for x, we differentiate the equation for α and equate to zero. We have x l

 12 gx 12 g 288gx 2 dα d  − 2 = 0 = = 2 dx dx  (l 2 + 12 x 2 )  (l + 12 x 2 ) (l + 12 x 2 ) 2 Solving for x, we get x =

l = 12

corresponding angular acceleration is

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α = 2.45 rad/s 2 .

6 = 1.73 m, and the 12 α = 2.83 rad/s 2 .

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18.23 (Continued) (c) Plotting α as a function of x using the equation derived above, we have

alpha vs x for bar 3

alpha [rad/s^2)

2.5 2 1.5 1 .5 0

1.5 x [m]

2.5

Note that the maximum value forα is as calculated and at the appropriate value of x.

Problem 18.24 Model the arm ABC as a single rigid body. Its mass is 320 kg, and the moment of inertia about its center of mass is I = 360 kg-m 2 . Point A is stationary. If the hydraulic piston exerts a 14-kN force on the arm at B, what is the arm’s angular acceleration?

Solution:

I A = I G + md 2 = (360 kg-m 2 ) + (320 kg)([1.10 m] 2 + [1.80 m] 2 ) = 1780 kg-m 2 . The angle between the force at B and the horizontal is θ = tan −1

1.5 m ( 1.4 ) = 47.0°. m

The rotational equation of motion is now

1.80 m 1.40 m B

The moment of inertia about the fixed point A is

∑ M A : (14 kN)sin θ(1.4 m) − (14 kN) cos θ(0.8 m)

0.30 m

− (320 kg)(9.81 m/s 2 )(1.80 m) = (1780 kg-m 2 )α.

0.80 m x

A 0.70 m

Solving, we find α = 0.581 rad/s 2 . α = 0.581 rad/s 2 (counterclockwise).

Problem 18.25 The truck’s bed weighs 8000 lb and its moment of inertia about O is 33,000 slug-ft 2 . At the instant shown, the coordinates of the center of mass of the bed are (10, 12) ft and the coordinates of point B are (15, 11) ft. If the bed has a counterclockwise angular acceleration of 0.2 rad/s 2 , what is the magnitude of the force exerted on the bed at B by the hydraulic cylinder AB?

B 308

Solution:

The rotational equation of motion is

∑ M O : F sin 30°(15 ft) + F cos30°(11 ft) − (8000 lb)(10 ft)

A x

= (33,000 slug-ft 2 )(0.2 rad/s 2 ) Solving for F we find

402

F = 5090 lb.

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Problem 18.26 Arm BC has a mass of 12 kg and the moment of inertia about its center of mass is 3 kg-m 2 . Point B is stationary and arm BC has a constant counterclockwise angular velocity of 2 rad/s. At the instant shown, what are the couple and the components of force exerted on arm BC at B? y

Solution: Since the angular acceleration of arm BC is zero, the sum of the moments about the fixed point B must be zero. Let M B be the couple exerted by the support at B. Then  i j k   M B + rCM /B × mg = M B +  0.3cos 40 ° 0.3sin 40 ° 0  = 0.  0 −117.7 0    M B = 27.05k (N-m) is the couple exerted at B. From Newton’s second law: B x = ma x , B y − mg = ma y where a x , a y are the accelera-

0 30 m m

a = α × rCM /O − ω 2 rCM /O 408

tions of the center of mass. From kinematics:

700 mm

= −(2 2 )( i 0.3cos 40 ° + j0.3sin 40 °) = −0.919 i − 0.771 j (m/s 2 ), where the angular acceleration is zero from the problem statement. Substitute into Newton’s second law to obtain the reactions at B: B x = −11.0 N, B y = 108.5 N.

Problem 18.27 Arm BC has a mass of 12 kg and the moment of inertia about its center of mass is 3 kg-m 2 . At the instant shown, arm AB has a constant clockwise angular velocity of 2 rad/s and arm BC has a counterclockwise angular velocity of 2 rad/s and a clockwise angular acceleration of 4 rad/s 2 . What are the couple and the components of force exerted on arm BC at B?

0 30 m m

408 A

Solution:

Because the point B is accelerating, the equations of angular motion must be written about the center of mass of arm BC. The vector distances from A to B and B to G, respectively, are

700 mm

rB /A = rB − r A = 0.7 i, rG /B = 0.3 cos(40 °) i + 0.3 sin(40 °) j = 0.2298i + 0.1928 j (m). The acceleration of point B is 2 (0.7 i) (m/s 2 ). a B = α × rB /A − ω 2AB rB /A = −ω AB

The acceleration of the center of mass is 2 r a G = a B + α BC × rG /B − ω BC G /B

 i j k   0 0 a G = −2.8i +  −4  − 0.9193i − 0.7713 j  0.2298 0.1928 0   

 i j k    M G = M B + rB /G × B =  −0.2298 −0.1928 0   −35.37 97.43 0    = M B k − 29.21k (N-m). Note I = 3 kg-m 2 and α BC = −4 k (rad/s 2 ), from which M B = 29.21 + 3(−4) = 17.21 N-m.

= −2.948i − 1.691 j (m/s 2 ). From Newton’s second law, B x = ma Gx = (12)(−2.948) = −35.37 N , By

B y − mg = ma Gy , B y = (12)(−1.691) + (12)(9.81) = 97.43 N. From the equation of angular motion, M G = Iα BC . The moment about the center of mass is

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MB Bx

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Problem 18.28 The arm ADE of the excavator is stationary. The weight of the shovel is W = 300 lb. The moment of inertia of the shovel about E is 56 slug-ft 2 . At the instant shown, the shovel is stationary. The hydraulic cylinder AB exerts an 800-lb force at B that is directed from B toward A. At that instant, what is the shovel’s angular acceleration? (Neglect the masses of members AB, BC , and BD. That means you can treat them as two-force members in analyzing the forces acting on joint B.)

Solution: Let T AB = 800 lb be the force exerted at B by the hydraulic cylinder. We need to analyze the joint B to determine the force exerted on the shovel at C. y b

TAB

TBC g

TBD

The angles β = arctan

( 153 inin ) = 11.3°, γ = arctan( 154 inin ) = 14.9°

From the equilibrium equations

∑ Fx = −T AB + TBC cos β − TBD sin γ = 0, ∑ Fy = −TBC sin β − TBD cos γ = 0,

Hydraulic cylinder Shovel

15 in

3 in E

The force exerted on the shovel by the link BC is FC = −TBC cos β i + TBC sin β j

C 12 in

12 in 7 in 20 in

we obtain TBC = 775 lb, TBD = −157 lb.

= −759 i + 152 j (lb). W

The moment (in in-lb) about E exerted by this force is rC /E × FC =

i j k 7 in 12 in 0 −759 lb 152 lb 0

= 10,200 k (in-lb),

which is counterclockwise. Including the moment due to the weight of the shovel, the total moment about E is

∑ M = 10,200 in-lb − (27 in)(300 lb) = 2080 (in-lb), which is 173 ft-lb. ft-lb. Using this value, the counterclockwise angular acceleration of the shovel is α =

∑ M = 3.09 rad/s 2 . I

3.09 rad/s 2 (counterclockwise).

Problem 18.29 The arm ADE of the excavator is stationary. The weight of the shovel is W = 300 lb. The moment of inertia of the shovel about E is 56 slug-ft 2 . At the instant shown, the shovel is stationary. The hydraulic cylinder AB exerts an 800-lb force at B that is directed from B toward A. At that instant, what is the magnitude of the force exerted on the shovel at E? (See Problem 18.28.)

Hydraulic cylinder Shovel

15 in

3 in E

C 12 in

12 in 7 in 20 in

404

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18.29 (Continued) Solution:

Let T AB = 800 lb be the force exerted at B by the hydraulic cylinder. We need to analyze the joint B to determine the force exerted on the shovel at C.

The moment (in in-lb) about E exerted by this force is i j k 7 in 12 in 0 −759 lb 152 lb 0

rC /E × FC =

y b

TAB

TBC

which is counterclockwise. Including the moment due to the weight of the shovel, the total moment about E is

∑ M = 10,200 in-lb − (27 in)(300 lb) = 2080 (in-lb),

TBD

which is 173 ft-lb. ft-lb. Using this value, the counterclockwise angular acceleration of the shovel is

The angles

α =

β = arctan

(

)

3 in = 11.3 °, 15 in

= 10,200 k (in-lb),

γ = arctan

(

)

4 in = 14.9 ° 15 in

From the equilibrium equations

∑ M = 3.09 rad/s 2 . I

The angular velocity is zero, so the acceleration of the center of mass of the shovel at this instant is 27 ft ) α j = 6.95 j (ft/s ). ( 12

∑ Fx = −T AB + TBC cos β − TBD sin γ = 0,

a =

∑ Fy = −TBC sin β − TBD cos γ = 0,

Let FE denote the force exerted on the shovel at E. Writing Newton’s second law for the shovel,

we obtain TBC = 775 lb, TBD = −157 lb.

W 

The force exerted on the shovel by the link BC is

∑ F = FC + FE − Wj =  g  a,

FC = −TBC cos β i + TBC sin β j

we obtain FE = 759 i + 213 j (lb). Its magnitude is 789 lb.

= −759 i + 152 j (lb).

789 lb.

158

508

m m

760

90 0

Problem 18.30 Points B and C lie in the x – y plane. The y-axis is vertical. The center of mass of the 18-kg arm BC is at the midpoint of the line from B to C, and the moment of inertia of the arm about the axis through the center of mass that is parallel to the z-axis is 1.5 kg-m 2 . At the instant shown, the angular velocity and angular acceleration vectors of arm AB are ω AB = 0.6k (rad/s) and α AB = −0.3k (rad/s 2 ). The angular velocity and angular acceleration vectors of arm BC are ω BC = 0.4 k (rad/s) and α BC = 2k (rad/s 2 ). Determine the force and couple exerted on arm BC at B.

Solution:

The acceleration of point B is a B = a A + α AB × r A /B − ω 2AB r A /B or i j k −0.3 0 0 0 0.76cos15 ° −0.76sin15 °

aB =

− (0.6) 2 (0.76cos15 °i − 0.76sin15 ° j) = −0.323i − 0.149 j (m/s 2 ) The acceleration of the center of mass G of arm BC is 2 r a G = a B + α BC × rG /B − ω BC G /B

a B = −0.323i − 0.149 j By

i j k 0 0 2 0.45cos50 ° 0.45sin 50 ° 0

− (0.4) 2 (0.45cos50 °i + 0.45sin 50 ° j),

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Bx MB

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18.30 (Continued) or a G = −1.059 i + 0.374 j (m/s 2 ). The free body diagram of arm BC is: Newton’s second law is

The equation of angular motion is

∑ M G = I BC α BC : or

∑ F = B x i + (B y − mg) j = ma G :

(0.45sin 50 °) B x − (0.45cos 50 °) B y + M B = (1.5)(2)

Solving for M B , we obtain M B = 62.6 N-m.

B x i + [ B y − (18)(9.81)] j = 18(−1.059 i + 0.374 j). Solving, we obtain B x = −19.1 N, B y = 183.3 N. Hence, FB = −19.1i, +183.3j (N).

Problem 18.31 Points B and C lie in the x – y plane. The y-axis is vertical. The center of mass of the 18-kg arm BC is at the midpoint of the line from B to C, and the moment of inertia of the arm about the axis through the center of mass that is parallel to the z-axis is 1.5 kg-m 2 . At the instant shown, the angular velocity and angular acceleration vectors of arm AB are ω AB = 0.6k (rad/s) and α AB = −0.3k (rad/s 2 ). The angular velocity vector of arm BC is ω BC = 0.4 k (rad/s). If you want to program the robot so that the angular acceleration of arm BC is zero at this instant, what couple must be exerted on arm BC at B?

Solution:

From the solution of Problem 18.30, the acceleration of point B is a B = −0.323i − 0.149 j (m/s 2 ). If α BC = 0, the acceleration of the center of mass G of arm BC is 2 r a G = a B − ω BC G /B = − 0.323i − 0.149 j

− (0.4) 2 (0.45cos50 °i + 0.45sin 50 ° j) = −0.370 i − 0.205 j (m/s 2 ). From the free body diagram of arm BC in the solution of Problem 18.30. Newton’s second law is

∑ F = B x i + (B y − mg) j = ma G : B x i + [ B y − (18)(9.81)] j = 18(−0.370 i − 0.205 j). Solving, we obtain B x = −6.65 N, B y = 172.90 N. The equation of angular motion is

∑ M G = I BC α BC = 0: (0.45sin 50 °) B x − (0.45cos 50 °) B y + M B = 0.

406

508

158

Solving for M B , we obtain M B = 52.3 N-m.

760

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Problem 18.32 The radius of the 4-kg uniform disk is R = 120 mm. It rolls without slipping on the inclined surface. The angle θ = 35 °. If the disk is released from rest, how far down the plane does its center move in one second? (See Practice Example 18.3.) Solution:

The moment of inertia of the disk about its center is

1 1 I = mR 2 = (4 kg)(0.12 m) 2 = 0.0288 kg-m 2 . 2 2 Let a denote the acceleration of the center of the disk and α its clockwise angular acceleration. Because the disk is rolling, they are related by a = Rα. (1) The equation of angular motion of the disk is

∑ M = Rf = I α, (2) and Newton’s second law is

∑ F = mg sin θ − f = ma. (3) Solving Eqs. (1)–(3), we obtain f = 7.50 N, α = 31.3 rad/s 2 , a = 3.75 m/s 2 . The distance the center moves in one second is 1 1 s = at 2 = (3.75 m/s 2 )(1 s) 2 = 1.88 m. 2 2 1.88 m.

Problem 18.33 The radius of the 4-kg uniform disk is R = 120 mm. The coefficient of static friction between the inclined surface and the disk is µ s = 0.25. What is the maximum value of the angle θ for which the disk will roll on the inclined surface without slipping? Solution: Let a denote the acceleration of the center of the disk and α its clockwise angular acceleration. If we assume that the disk is rolling, they are related by

a = Rα. (1) The equation of angular motion of the disk is

∑ M = Rf = I α, (2) and Newton’s second law is

∑ F = mg sin θ − f = ma. (3) Substituting Eq. (1) into Eq. (3), we get mg sin θ − f = mRα. Then solving this equation together with Eq. (2) and noting that the moment of inertia of the uniform disk equals mR 2 /2, we obtain f =

1 mg sin θ. (4) 3

The largest friction force for which slipping will not occur is f = µ s N. Substituting this value into Eq. (4) and noting that the normal force N = mg cos θ, we obtain the equation tan θ = 3µ s . Setting µ s = 0.25, the maximum angle is θ = 36.9 °. θ = 36.9 °.

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Problem 18.34 A homogeneous circular disk (a) and a sphere (b), each of mass m = 10 kg and radius R = 0.1 m, are released from rest and roll down an inclined surface. The angle θ = 30 °. Determine the velocity of the centers of the rolling objects 2 s after they are released.

Solution:

y (b)

(a)

mg f

x N

In Case (a), I = mR 2 /2 and we have mg sin θ 2 = g sin 30 ° = 3.27 m/s 2 . 1 3 m 2 The velocity in 2 seconds is v = (2 s)a = 6.54 m/s. a =

Let α denote the clockwise angular acceleration and a the acceleration of the center down the plane. We have the equations

∑ Fx = mg sin θ − f = ma, (1) ∑ M center = Rf = Ia. (2) a = Rα

In Case (b), I = 2mR 2 /5 and we have a =

mg sin θ 5 = g sin 30 ° = 3.50 m/s 2 . 2 7 m 5

The velocity in 2 seconds is v = (2 s)a = 7.01 m/s.

Because there is rolling, a = Rα. Using this relation together with Eqs. (1) and (3), we obtain the expression

(a) 6.54 m/s. (b) 7.01 m/s.

mg sin θ . a = I m+ 2 R

Problem 18.35 The stepped disk weighs 40 lb and its moment of inertia is I = 0.2 slug-ft 2 . If the disk is released from rest, how long does it take its center to fall 3 ft? (Assume that the string remains vertical.) Solution:

The moment about the center of mass is M = −RT. From the equation of angular motion: −RT = I α , from which Iα T = − . From the free body diagram and Newton’s second law: R ∑ Fy = T − W = ma y , where a y is the acceleration of the center of mass. From kinematics: a y = −Rα. Substitute and solve: ay =

(

W . I + m 2 R

)

4 in 8 in

The time required to fall a distance D is t =

408

2D = ay

2 D ( I + R 2 m) . R 2W

For D = 3 ft, R =

4 W = 0.3333 ft, W = 40 lb, m = = 1.24 slug, 12 g

I =

t = 0.676 s.

0.2 slug -ft 2 ,

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Problem 18.36 The radius of the pulley is R = 100 mm and its moment of inertia is I = 0.1 kg-m 2 . The mass m = 5 kg. The spring constant is k = 135 N/m. The system is released from rest with the spring unstretched. At the instant when the mass has fallen 0.2 m, determine (a) the angular acceleration of the pulley, and (b) the tension in the rope between the mass and the pulley. Solution: The force in the spring is kx. There are five unknowns (O x , O y , T , a, α), four dynamic equations, and one constraint

m x

equation.

∑ M O : (kx) R − TR = −I α, ∑ Fy : T − mg = −ma, a = Rα Solving we find (a) α = =

R(mg − kx ) I + mR 2 (0.1 m)([5 kg][9.81 m/s 2 ] − [135 N/m][0.2 m]) 0.1 kg-m 2 + (5 kg)(0.1 m) 2

α = 14.7 rad/s 2 . (b) T = m( g − Rα) = (5 kg)(9.81 m/s 2 − [0.1 m][14.7 rad/s 2 ]) T = 41.7 N.

Problem 18.37 The radius of the pulley is R = 100 mm and its moment of inertia is I = 0.1 kg-m 2 . The mass m = 5 kg. The spring constant is k = 135 N/m. The system is released from rest with the spring unstretched. What maximum distance does the mass fall before rebounding? Strategy: Assume that the mass has fallen an arbitrary distance x. Write the equations of motion for the mass and the pulley and use them to determine the acceleration a of the mass as a function of x. Then apply the chain rule:

m x

dv dx dv dv = = v. dt dx dt dx

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18.37 (Continued) Solution:

The force in the spring is kx. There are five unknowns (O x , O y , T , a, α), four dynamic equations, and one constraint equation.

∑ M O : (kx) R − TR = −I α, ∑ Fy : T − mg = −ma, a = Rα Solving we find R 2 (mg − kx ) dv = v I + mR 2 dx

a =

x R 2 (mg − kx )

∫0 v dv = ∫0

I + mR 2

dx =

x R2 (mg − kx ) dx = 0 I + mR 2 ∫ 0

Thus mgx −

2 mg 1 2 kx = 0 ⇒ x = 0 or x = k 2

The maximum distance is x =

2 mg 2(5 kg)(9.81 m/s 2 ) = = 0.727 m x = 0.727 m. k 135 N/m

Problem 18.38 The mass of the disk is 45 kg, and its radius is R = 0.3 m. The spring constant is k = 600 N/m. The disk is rolled to the left until the spring is compressed 0.5 m and released from rest. (a) If you assume that the disk rolls, what is its angular acceleration at the instant it is released? (b) What is the minimum coefficient of static friction for which the disk will not slip when it is released?

Solution:

x 0 = −0.5

k = 600 N/m Fs

m = 45 kg

R = 0.3 m I0 =

1 mR 2 = 2.025 N-m 2 , Fs = kx 2

∑ Fx : − Fs − f = ma 0 x

∑ Fy : N − mg = 0 ∑ M 0 : − f R = I 0α Rolling implies a 0 x = −Rα

Four eqns, four unknowns (a 0 x , α, N , f ) (a) Solving f = 100 N, N = 441.5 N

We have, at x = −0.5 m

α = −14.81 rad/s 2 (clockwise)

− kx − f = ma 0 x

a 0 x = 4.44 m/s 2 .

N − mg = 0 − Rf = I 0 α

(b) for impending slip,

a 0 x = −Rα

f = µs N µ s = f/N = 100/441.5 µ s = 0.227.

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Problem 18.39 The disk weighs 12 lb and its radius is 6 in. It is stationary on the surface when the force F = 10 lb is applied. (a) If the disk rolls on the surface, what is the acceleration of its center? (b) What minimum coefficient of static friction is necessary for the disk to roll instead of slipping when the force is applied?

Solution:

There are five unknowns ( N , f , a, α, µ s ), three dynamic equations, one constraint equation, and one friction equation.

∑ Fx : F − f = ma, ∑ Fy : N − mg = 0, ∑ M G : − fr = −( 2 mr 2 )α, 1

a = rα, f = µs N. Solving, we find (a) a =

(b) µ s =

2(10 lb) 2F = 12 lb 3m 3 32.2 ft/s 2

(

)

= 8.94 ft/s 2 .

(10 lb) F = = 0.278 3mg 3(12 lb)

a = 17.9 ft/s 2 .

µ s = 0.278.

Problem 18.40 At t = 0, a sphere with mass m = 10 kg and radius R = 0.1 m is placed on a horizontal surface. Initially, the velocity of its center is zero and it has a clockwise angular velocity ω 0 = 40 rad/s. The coefficient of kinetic friction between the sphere and surface is µ k = 0.1. What maximum velocity will the center of the sphere attain? (You can see the analogy to spinning car tires and pool balls.)

The normal force N = mg, so the friction force is µ k N = µ k mg. Newton’s second law gives

∑ Fx = µ k mg = ma x .

Solution:

The velocity of the center of the disk toward the right as a function of time is therefore

v x = a x t = µ k gt. (2)

x mg N

mk N

Initially, and for some period of time, the sphere at its point of contact will be moving (slipping) relative to the surface. The resulting friction force causes the sphere to accelerate to the right and also exerts a counterclockwise moment about the its center, which causes its clockwise angular velocity to decrease. When the velocity v x of the center toward the right increases and the clockwise angular velocity ω decrease to the point where v x = Rω, (1) the sphere is rolling, the friction force is zero, and the center of the sphere stops accelerating. That is when the maximum velocity occurs.

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The moment of inertia of the sphere about its center is 2 I = mR 2 = 0.04 kg-m 2 . 5 Treating the clockwise direction as positive, the equation of angular motion is 2 ∑ M = −µ k mgR = I α = 5 mR 2α, so the disk’s clockwise angular velocity as a function of time is ω = ω 0 + αt = ω0 −

5µ k g (3) t. 2R

Substituting Eqs. (2) and (3) into the condition (1), we can solve for the time at which the disk stops slipping, which is the time at which the center stops accelerating: t =

2 Rω 0 = 1.16 s. 7µ k g

Substituting this time into Eq. (2), the maximum velocity attained by the center is 1.14 m/s.

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Problem 18.41 The angle θ = 30 °. At t = 0, a sphere with mass m = 10 kg and radius R = 0.1 m is placed on the inclined surface. Initially, it has clockwise angular velocity ω 0 = 10 rad/s and its center is moving at v 0 = 4 m/s up the surface. The coefficient of kinetic friction between the sphere and surface is µ k = 0.1. Where is the center of the sphere relative to its initial position when the sphere begins rolling without slipping? Solution:

The velocity of the center of the disk up the incline as a function of time is

x 308

v = v 0 + a x t. = v 0 + ( µ k cos θ − sin θ ) gt.

mk N

N Initially, and for some period of time, the disk at its point of contact will be moving (slipping) relative to the surface. The motion of the center of the disk will be determined by the combined effects of the resulting friction force and the component of the sphere’s weight down the incline. The friction force also exerts a counterclockwise moment about the center, which causes the clockwise angular velocity to decrease. We need to determine when the velocity and the angular velocity satisfy the condition for rolling,

2 mR 2 . 5

Treating the clockwise direction as positive, the equation of angular motion is 2

∑ M = −µ k NR = −µ k mg cos θ R = I α = 5 mR 2α, so the disk’s clockwise angular velocity as a function of time is ω = ω 0 + αt = ω0 −

The normal force is N = mg cos θ, so the friction force is µ k mg cos θ. Newton’s second law gives

∑ Fx = µ k mg cos θ − mg sin θ = ma x .

5µ k g cos θ (3) t. 2R

Substituting Eqs. (2) and (3) into the condition (1), we can solve for the time at which the disk stops slipping:

v = Rω, (1) and where the center of the disk is when that occurs.

(2)

The moment of inertia of the sphere about its center is I =

t =

(

Rω 0 − v 0 = 1.55 s. (4) 7 µ k cos θ − sin θ g 2

)

Let s denote the displacement of the center of the sphere up the inclined surface. We can integrate Eq. (2) to determine the displacement as a function of time: s = v 0t +

1 (µ cos θ − sin θ) gt 2 . 2 k

Substituting the result (4), we obtain s = 1.32 m. 1.32 m up the inclined surface.

Problem 18.42 The 120-kg cylindrical disk is at rest when the force F is applied to a cord wrapped around it. The static and kinetic coefficients of friction between the disk and the surface are 0.2. Determine the angular acceleration of the disk if (a) F = 500 N and (b) F = 1000 N. Strategy: First solve the problem by assuming that the disk does not slip, but rolls on the surface. Determine the frictional force, and find out whether it exceeds the product of the coefficient of friction and the normal force. If it does, you must rework the problem assuming that the disk slips.

412

F 300 mm

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18.42 (Continued) Solution:

Choose a coordinate system with the origin at the center of the disk in the at rest position, with the x axis parallel to the plane surface. The moment about the center of mass is M = −RF − Rf , from which −RF − Rf = I α. From which f =

 i j k    =  0 0 α  − ω 2 Rj = −Rαi − ω 2 Rj,  0 R 0   

F 500 = − = −167 N. 3 3

ax 6.67 = − = −22.22 rad/s 2 . R 0.3

(b) For F = 1000 N the acceleration is 4F 4000 = = 13.33 m/s 2 . 3m 300

The friction force is f = F − ma x = 1000 − 1333.3 = −333.3 N.

from which a y = 0 and a x = −Rα. Substitute for f and solve: 2F

α = −

ax =

a G = a x i = α × Rj − ω 2 Rj

( m + RI )

f = F − ma x = −

Note: −µ k W = −0.2 mg = −196.2 N, the disk does not slip. The angular velocity is

Iα −RF − I α = −F − . R R

From Newton’s second law: F − f = ma x , where a x is the acceleration of the center of mass. Assume that the disk rolls. At the point of contact a P = 0; from which 0 = a G + α × rP /G − ω 2 rP /G .

ax =

(a) For F = 500 N, the friction force is

The drum slips. The moment equation for slip is −RF + Rµ k gm = I α, from which α =

−RF + Rµ k gm 2µ k g 2F = − + = −53.6 rad/s 2 . I mR R

1 For a disk, the moment of inertia about the polar axis is I = mR 2 , 2 from which

4F 2000 ax = = = 6.67 m/s 2 . 3m 300

W f

Problem 18.43 The ring gear is fixed. The mass and moment of inertia of the sun gear are m S = 320 kg and I S = 40 kg-m 2 . The mass and moment of inertia of each planet gear are m P = 38 kg and I P = kg-m 2 . If a couple M = 200 N-m is applied to the sun gear, what is the latter’s angular acceleration?

Solution: M S = 200 N-m Sun Gear:

∑ M 0 : M S − 3RF = I S α S

Planet Gears:

∑ M c : Gr − Fr = I Pα P ∑ Ft : F + G = m P a ct

0.18 m

0.86 m

Ring gear

From kinematics a ct = −rα P 2α P rP = −Rα S We have 5 eqns in 5 unknowns. Solving,

α S = 3.95 rad/s 2 (counterclockwise).

0.50 m

F er

Planet gears (3) Sun gear

r C IP

et 3 small disks

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O Is

F F

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Problem 18.44 In Problem 18.43, what is the magnitude of the tangential force exerted on the sun gear by each planet gear at their points of contact when the 200 N-m couple is applied to the sun gear? Ring gear

0.18 m

Solution:

See the solution to Problem 18.43. Solving the 5 eqns in 5 unknowns yields α S = 3.95 rad/s 2 , G = 9.63 N, a Gt = 0.988 m/s 2 , α P = −5.49 rad/s 2 , F = 27.9 N.

0.86 m

F is the required value.

0.50 m

Planet gears (3) Sun gear

Problem 18.45 The 10-kg ladder is released from rest in the position shown. Model it as a slender bar and neglect friction. What are the magnitudes of the forces exerted on it by the floor and wall at the instant it is released?

258

Solution: y A FA u

G obtaining the scalar equations

mg B

FB Let m = 10 kg, L = 4 m, θ = 25 °. Denote the center of mass of the ladder by G. The moment of inertia of the ladder about its center of mass is I =

a B = αL cos θ, (4) 0 = a A + αL sin θ. (5) We also write the acceleration of the center of mass in terms of the acceleration of point A, a G = a A + α × rG /A − ω 2 rG /A :

1 mL2 = 13.3 kg-m 2 . 12

We have the equations of motion

a Gx i + a Gy j = a A j +

i 0

j 0

k α

− 0,

∑ Fx = FA = ma Gx , (1)

1 1 L sin θ − L cos θ 2 2

∑ Fy = FB − mg = ma Gy , (2)

obtaining the scalar equations 1 a Gx = αL cos θ, (6) 2 1 a Gy = a A + αL sin θ. (7) 2 Substituting the value of a A from Eq. (5) into Eq. (7) yields 1 a Gy = − αL sin θ. (8) 2 Equations (1)–(3), (6), and (8) provide 5 equations in terms of a Gx , a Gy , α, FA , FB . Solving them, we obtain a Gx = 2.82 m/s 2 , a Gy = −1.31 m/s 2 , α = 1.55 rad/s 2 , FA = 28.2 N, FB = 85.0 N.

∑ M G = FB 2 L sin θ − FA 2 L cos θ = I α. (3) We write the acceleration of point B in terms of the acceleration of point A, a B = a A + α × r B /A − ω 2 r B /A : aBi = a A j +

i j k 0 0 α L sin θ −L cos θ 0

− 0,

Floor: 85.0 N; wall: 28.2 N.

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Problem 18.46 At the instant shown, the 10-kg ladder is rotating in the counterclockwise direction at 1.0 rad/s. Model it as a slender bar and neglect friction. What are the magnitudes of the forces exerted on it by the floor and wall at that instant?

258

Solution: y A FA u

G We also write the acceleration of the center of mass in terms of the acceleration of point A,

mg B

a G = a A + α × rG /A − ω 2 rG /A :

FB Let m = 10 kg, L = 4 m, θ = 25 °. Denote the center of mass of the ladder by G. The moment of inertia of the ladder about its center of mass is I =

a Gx i + a Gy j = a A j +

1 mL2 = 13.3 kg-m 2 . 12

− ω2

We have the equations of motion

∑ Fx = FA = ma Gx , (1) ∑ Fy = FB − mg = ma Gy ,

(2)

1 1 ∑ M G = FB 2 L sin θ − FA 2 L cos θ = I α. (3) We write the acceleration of point B in terms of the acceleration of point A, a B = a A + α × r B /A aBi = a A j +

− ω 2r

B /A :

i j k 0 0 α L sin θ −L cos θ 0

− ω 2 ( L sin θ i − L cos θ j),

obtaining the scalar equations

i 0

j 0

k α

1 1 L sin θ − L cos θ 2 2

( 12 L sin θi − 12 L cos θ j ),

obtaining the scalar equations 1 1 a Gx = αL cos θ − ω 2 L sin θ, (6) 2 2 a Gy = a A +

1 1 αL sin θ + ω 2 L cos θ. (7) 2 2

Substituting the value of a A from Eq. (5) into Eq. (7) yields 1 1 a Gy = − αL sin θ − ω 2 L cos θ. (8) 2 2 Equations (1)–(3), (6), and (8) provide 5 equations in terms of a Gx , a Gy , α, FA , FB . Solving them, we obtain a Gx = 1.97 m/s 2 , a Gy = −3.13 m/s 2 , α = 1.55 rad/s 2 , FA = 19.7 N, FB = 66.8 N. Floor: 66.8 N; wall: 19.7 N.

a B = αL cos θ − ω 2 L sin θ, (4) 0 = a A + αL sin θ + ω 2 L cos θ. (5)

Problem 18.47 The 4-kg slender bar is released from rest in the position shown. Determine its angular acceleration at that instant if (a) the surface is rough and the bar does not slip, and (b) the surface is smooth.

608

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18.47 (Continued) Solution:

that information into the equation for the acceleration of point G in terms of the acceleration of point A:

a G = a A + α × rG /A − ω 2 rG /A :

G aG j = a Ai +

mg 608

j 0

k α

1 − L cos θ 2

1 L sin θ 2

− 0.

This yields the scalar equations

1 0 = a A − α L sin θ, 2 1 a G = −α L cos θ. 2

(a) Let θ = 60 °, m = 4 kg, L = 1 m. Let α denote the bar’s counterclockwise angular acceleration. If the bar doesn’t slip, it rotates about a fixed point, the bottom of the bar A. The bar’s moment of inertia about A is IA =

i 0

The equations of motion for the bar are

∑ Fy = N − mg = ma G ,

1 2 mL = 1.33 kg-m 2 . 3

∑ M G = N 2 L cos θ = I G α.

From the equation of angular motion 1

The bar’s moment of inertia about its center of mass is

∑ M A = mg 2 L cos60° = I Aα,

IG =

we obtain α = 7.36 rad/s 2 .

1 mL2 = 0.333 kg-m 2 . 12

We have 4 equations in terms ofa A , a G , α, N . Solving them, we obtain a A = 7.28 m/s 2 , a G = −4.20 m/s 2 , α = 16.8 rad/s 2 , N = 22.4 N.

(b) If the surface is smooth, the friction force f is zero and no horizontal forces act on the bar. Its center of mass G will fall straight down. Point A will not be fixed, but will move horizontally. We put

16.8 rad/s 2 (counterclockwise).

Problem 18.48 The 4-kg slender bar is released from rest in the position shown. The coefficient of static friction between the bar and the surface is µ s = 0.3, and the coefficient of kinetic friction is µ k = 0.15. Determine the bar’s angular acceleration at the instant it is released.

Solution:

The question is, will the bar slip when it is released? In most slip/no slip problems, the procedure is to first assume it doesn’t slip and solve for the friction force. If the friction force is smaller than the materials will support, we’re through. But if it is greater, we must rework the problem assuming slip.

y G mg 608

1m Let θ = 60 °, m = 4 kg, L = 1 m. Let α denote the bar’s counterclockwise angular acceleration. If the bar doesn’t slip, it rotates about a fixed point, the bottom of the bar A. The bar’s moment of inertia about A is IA = 608

1 2 mL = 1.33 kg-m 2 . 3

From the equation of angular motion 1

∑ M A = mg 2 L cos60° = I Aα, we obtain α = 7.36 rad/s 2 . This tells us the acceleration of the center of mass: 1 1 a G = − Lα sin θ i − Lα cos θ j 2 2 = −3.19 i − 1.84 j (m/s 2 ) = a Gx i + a Gy j.

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18.48 (Continued) From the equations of motion

This yields the scalar equations

∑ Fx = − f = ma Gx ,

1 a Gx = a A − α L sin θ, (2) 2 1 a Gy = −α L cos θ. (3) 2 The equations of motion for the bar are

∑ Fy = N − mg = ma Gy , we obtain f = 12.7 N, N = 31.9 N. The largest friction force that can be supported is µ s N = 9.56 N, so the bar slips. (You probably guessed that.) If the bar slips, we know that f = µ k N. (1) The one bit of kinematic information we have is that point A will move horizontally. We put that information into the equation for the acceleration of point G in terms of the acceleration of point A: a G = a A + α × rG /A − ω 2 rG /A :

a Gx i + a Gy j = a A i +

i 0

j 0

k α

1 − L cos θ 2

1 L sin θ 2

− 0.

Problem 18.49 The angle θ = 40 °. The identical slender carbon-fiber bars are 16 in long. Their weight is negligible, which means you can treat them as two-force members. The homogeneous aluminum plate weighs 10 lb, and its dimensions are 20 in by 6 in. The system is released from rest in the position shown. At that instant, what is the angular acceleration of the bars?

∑ Fx = − f = ma Gx , (4) ∑ Fy = N − mg = ma Gy , (5) 1

∑ M G = N 2 L cos θ − f 2 L sin θ = I G α. (6) The bar’s moment of inertia about its center of mass is 1 IG = mL2 = 0.333 kg-m 2 . 12 We have 6 equations in terms of a A , a Gx , a Gy , α, N , f . Solving them, we obtain a A = 5.12 m/s 2 , a Gx = −0.946 m/s 2 , a Gy = −3.50 m/s 2 , α = 14.0 rad/s 2 , N = 25.2 N, f = 3.78 N. 14.0 rad/s 2 (counterclockwise).

Solution: y B

A u

x WP b

The mass of the plate is mP =

WP 10 lb = = 0.311 slug. g 32.2 ft/s 2

Let α denote the counterclockwise angular acceleration of the bars and their length by L = 16/12 ft. The acceleration of the translating plate is a P = Lα sin θi − Lα cos θ j. The forces A and B are the axial forces exerted on the plate by the bars. We have the equations

∑ Fx = A cos θ + B cos θ = m P Lα sin θ, ∑ Fy = A sin θ + B sin θ − WP = −m P Lα cos θ, ∑ M center of mass = −bA sin θ − hA cos θ + bB sin θ − hB cos θ = 0. Solving, we obtain A = 2.06 lb, B = 4.36 lb, α = 18.5 rad/s 2 . 18.5 rad/s 2 (counterclockwise).

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Problem 18.50 The angle θ = 40 °. Each slender bar is 16 in long and weighs 5 lb. The homogeneous aluminum plate weighs 10 lb, and its dimensions are 20 in by 6 in. The system is released from rest in the position shown. At that instant, what is the angular acceleration of the bars?

Solution: y Ay

x Wp b

u u

WB Bx

The masses of the plate and each bar are mP =

WP 10 lb = = 0.311 slug, 32.2 ft/s 2 g

mB =

WB 5 lb = = 0.155 slug. g 32.2 ft/s 2

The moment of inertia of the bars about the point where they are supported in terms of the length L = 16/12 ft is IB =

1 m L2 = 0.0920 slug-ft 2 . 3 B

Let α denote the counterclockwise angular acceleration of the bars. The acceleration of the ends of the bars, which is the acceleration of the translating plate, is a P = Lα sin θi − Lα cos θ j. The forces A x , A y , B x , B y are the reactions between the plate and the bars. For the plate we have the equations

∑ Fx = Ax + B x = m P Lα sin θ, (1) ∑ Fy = Ay + B y − WP = −m P Lα cos θ, (2) ∑ M center of mass = −hAx − bAy − hB x + bB y = 0, (3) and for the bars we write the equations 1

∑ M pin = −L sin θ Ax + L cos θ Ay + 2 L cos θWB = I Bα, (4) 1

∑ M pin = −L sin θB x + L cos θB y + 2 L cos θWB = I Bα. (5) We have five equations in terms of A x , A y , B x , B y , α. Solving, we obtain A x = 1.78 lb, A y = 0.868 lb, B x = 3.76 lb, B y = 2.53 lb, α = 20.8 rad/s 2 . 20.8 rad/s 2 (counterclockwise).

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Problem 18.51 The mass of the suspended object A is 8 kg. The mass of the pulley is 5 kg, and its moment of inertia is 0.036 kg-m 2 . If the force T = 70 N, what is the magnitude of the acceleration of A? Solution:

Given

m A = 8 kg, m B = 5 kg, I B = 0.036 kg-m 2

120 mm

R = 0.12 m, g = 9.81 m/s 2 , T = 70 N The FBDs A

The dynamic equations

∑ FyB : T2 + T − m B g − B y = m B a By ∑ FyA : B y − m A g = m Aa Ay

∑ M B : −T2 R + TR = I Bα B Kinematic constraints

mBg

a By = a Ay , a By = Rα B Solving we find

a Ay = 0.805 m/s 2 .

We also have By

a By = 0.805 m/s 2 , α B = 6.70 rad/s, T2 = 68.0 N, B y = 84.9 N.

mAg

Problem 18.52 The suspended object A weighs 20 lb. The pulleys are identical, each weighing 10 lb and having moment of inertia 0.022 slug-ft 2 . If the force T = 15 lb, what is the magnitude of the acceleration of A?

4 in

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18.52 (Continued) Solution: g =

Given

32.2 ft/s 2 , W

mA =

= 20 lb, Wdisk = 10 lb, I =

Wdisk

0.022 slug-ft 2

W WA 4 , m disk = disk , R = ft, T = 15 lb g g 12

The FBDs T4

The dynamic equations

Wdisk

∑ Fy1 : T2 + T − T1 − m disk g = m disk a1

∑ Fy 2 : T4 + T1 − T3 − m disk g = m disk a 2 ∑ Fy 3 : T3 − m A g = m Aa A ∑ M 1 : TR − T2 R = I α1 ∑ M 2 : T1R − T4 R = I α 2

The kinematic constraints a1 = Rα1 , a 2 = Rα 2 , a1 = 2 Rα 2 , a A = a 2 Solving we find

a A = 3.16 ft/s 2 .

We also have a1 = 6.32 ft/s 2 , a 2 = 3.16 ft.s 2 , α1 = 19.0 rad/s 2 , α 2 = 9.48 rad/s 2 T1 = 16.8 lb, T2 = 13.7 lb, T3 = 22.0 lb, T4 = 16.2 lb.

Problem 18.53 The 2-kg slender bar and 5-kg block are released from rest in the position shown. If friction is negligible, what is the block’s acceleration at that instant? (See Example 18.6.) Solution:

L = 1 m, m = 2 kg

Assume directions for B x , B y , I G =

1m 558

M = 5 kg

1 m L2 12 B

∑ Fx : B x = ma G (1) x

∑ Fy : B y − mg = m B a G (2) y

∑ M G: (

)

(

)

L L cos θ B y + sin θ B x = I G α (3) 2 2

∑ Fx : −B x = Ma 0 x (4)

L cos u 2

∑ Fy : N − B y − Mg = 0 (5)

From kinematics, ω = 0 (initially) a 0 = a G + αk × r0/G where

G u

L L r0/G = cos θ i − sin θ j 2 2

 a 0 x = a Gx + (αL /2)sin θ   0 = a Gy + (αL /2) cos θ

420

(L / 2) sin u

From the diagram a 0 = a 0 x i

O (6) (7)

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18.53 (Continued) We know θ = 55 °, I G = 0.167 kg-m 2 , L = 1 m, m = 2 kg, M = 5 kg. We have 7 eqns in 7 unknowns

(a G x , a G y , a 0 x , α, B x , B y , N ), Solving, we get

B x = −5.77 N,

(opposite the assumed direction)

B y = 13.97 N, a G x = −2.88 m/s 2 , a G y = −2.83 m/s 2 α = 9.86 rad/s 2 , a0x

N = 63.0 N N

= 1.15 m/s 2 (to the right).

Problem 18.54 The 2-kg slender bar and 5-kg block are released from rest in the position shown. What minimum coefficient of static friction between the block and the horizontal surface would be necessary for the block not to move at the instant the system is released? (See Example 18.6.)

1m 558

Solution:

This solution is very similar to that of Problem 18.53. We add a friction force f = µ s N and set a 0 x = 0. L = 1m

m = 2 kg

M = 5 kg IG =

1 mL2 = 0.167 kg-m 2 12

∑ Fx : B x = ma G (1) x

∑ Fy : B y − mg = ma G (2) y

∑ M G: (

)

(

)

L L cos θ B y + sin θ B x = I G α (3) 2 2

(These are the same as in Problem 18.53) Note: In Prob 18.53, B x = −5.77 N (it was in the opposite direction to that assumed). This resulted in a 0 x to the right. Thus, friction must be to the left

Bx By

∑ Fx : −B x − µ s N = ma 0 x = 0 (4) ∑ Fy : N − B y − Mg = 0 (5)

From kinematics, a 0 = a G + α × r0/G = 0 O = a Gx + (αL/2) sin θ (6) O = a Gy + (αL/2) cosθ (7) Solving 7 eqns in 7 unknowns, we get

M Bx

B x = −6.91 N, B y = 14.78 N,

msN

a Gx = −3.46 m/s 2 , a Gy = −2.42 m/s 2 N = 63.8 N, α = 8.44 rad/s 2 µ s = 0.108.

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Problem 18.55 The angle θ = 35 °. The 3-kg slender bar has length L = 2 m. The 5-kg homogeneous disk has radius R = 0.5 m. The coefficient of static friction between the disk and the surface is µ s = 0.2, and the coefficient of kinetic friction is µ k = 0.15. The system is released from rest in the position shown. What is the angular acceleration of the disk at the instant of release?

L R

Solution: u

B y

Cy D

mg D

To obtain kinematic relationships, we can write the acceleration of the center of the bar B in terms of the acceleration of the center of the disk D: a B = a D + α B × r B /D − ω B 2 r B /D :

Mg N

We first assume the disk does not slip and solve the problem. If the resulting friction force is greater than the maximum value that can be supported, we must then solve the problem again assuming the disk slips. The forces C x and C y are the reactions the bar and disk exert on each other. The equations of motion for the bar are

i 0

j 0

k αB

1 − L sin θ 2

1 L cos θ 2

a Bx i + a By j = a D i +

− 0.

This yields the scalar equations 1 a Bx = a D − α B L cos θ, (8) 2

∑ Fx = −C x = ma Bx , (1)

1 a By = −α B L sin θ. (9) 2

∑ Fy = C y − mg = ma By , (2)

We have 9 equations in terms of the unknowns a Bx , a By , α B , a D , α D , C x , C y, N, f .

∑ M point B = −C x 2 L cos θ + C y 2 L sin θ = I Bα B . (3) The equations of motion for the disk are

∑ Fx = C x − f = Ma D , (4) ∑ Fy = −C y + N − Mg = 0, (5)

Solving them (using MATLAB), we obtain a Bx = −2.88 m/s 2 , a By = −2.83 m/s 2 , α B = 4.93 rad/s 2 , a D = 1.15 m/s 2 , α D = −2.31 rad/s 2 , C x = 8.65 N, C y = 20.9 N, N = 70.0 N, f = 2.88 N. Okay, does the disk slip? The maximum friction force that can be supported is f = µ k N = 14.0 N, so we’re through. 2.31 rad/s 2 (clockwise).

∑ M point D = −Rf = I Dα D . (6) If the disk rolls, we have the relation a D = −Rα D . (7)

Problem 18.56 The slender bar weighs 40 lb, and the crate weighs 80 lb. At the instant shown, the velocity of the crate is zero and it has an acceleration of 14 ft/s 2 toward the left. The horizontal surface is smooth. Determine the couple M and the tension in the rope.

6 ft

3 ft

422

6 ft

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18.56 (Continued) Solution: There are six unknowns ( M , T , N , O x , O y , α), five dynamic equations, and one constraint equation. We use the following subset of the dynamic equations.

∑ M O : M − (40 lb)(1.5 ft) − T cos 45 °(6 ft) − T sin 45 °(3 ft) =

(

)

1 40 lb (45 ft 2 )α, 3 32.2 ft/s 2

∑ Fx : − T cos 45° = −( 32.2 ft/s 2 )(14 ft/s 2 ) 80 lb

The constraint equation is derived from the triangle shown. We have L = 45 ft, d = 6 2 ft, θ = 63.4 °. x = L cos θ +

d 2 − L2 sin 2 θ

(

x = −L sin θ −

)

L2 cos θ sin θ θ d 2 − L2 sin 2 θ

Since the velocity x = 0, then we know that the angular velocity ω = θ = 0. Taking one more derivative and setting ω = 0, we find

(

x = −L sin θ −

)

L2 cos θ sin θ θ ⇒ −(14 ft/s 2 ) d 2 − L2 sin 2 θ

(

)

L2 cos θ sin θ α d 2 − L2 sin 2 θ Solving these equations, we find that = −L sin θ −

α = 1.56 rad/s 2 ,

M = 402 lb-ft, T = 49.2 lb.

Problem 18.57 The slender bar weighs 40 lb, and the crate weighs 80 lb. At the instant shown, the velocity of the crate is zero and it has an acceleration of 14 ft/s 2 toward the left. The coefficient of kinetic friction between the horizontal surface and the crate is µ k = 0.2. Determine the couple M and the tension in the rope. Solution:

There are seven unknowns ( M , T , N , O x , O y , α, f ) , five dynamic equations, one constraint equation, and one friction equation. We use the following subset of the dynamic equations.

6 ft

3 ft

∑ M O : M −(40 lb)(1.5 ft)

6 ft

−T cos 45°(6 ft) −T sin 45 °(3 ft) =

(

)

1 40 lb (45 ft 2 )α, 3 32.2 ft/s 2

∑ Fx : −T cos 45° + (0.2) N = −( 32.2 ft/s 2 )(14 ft/s 2 ) 80 lb

∑ Fy : T sin 45° + N − (80 lb) = 0. The constraint equation is derived from the triangle shown. We have L =

45 ft, d = 6 2 ft, θ = 63.4 °.

x = L cos θ +

(

x = −L sin θ −

d 2 − L2 sin 2 θ

)

L2 cos θ sin θ θ d 2 − L2 sin 2 θ

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18.57 (Continued) Since the velocity x = 0 , then we know that the angular velocity ω = θ = 0. Taking one more derivative and setting ω = 0, we find

(

L2 cos θ sin θ θ ⇒ − (14 ft/s 2 ) d 2 − L2 sin 2 θ

(

L2 cos θ sin θ α d 2 − L2 sin 2 θ

x = −L sin θ − = −L sin θ −

) )

Solving these equations, we find that α = 1.56 rad/s 2 , N = 168 N,

M = 470 lb-ft, T = 59.9 lb.

Problem 18.58 Bar AB is rotating with a constant clockwise angular velocity of 10 rad/s. The 8-kg slender bar BC slides on the horizontal surface. At the instant shown, determine the total force (including its weight) acting on bar BC and the total moment about its center of mass.

Solution:

We first perform a kinematic analysis to find the angular acceleration of bar BC and the acceleration of the center of mass of bar BC. First the velocity analysis: v B = v A + ω AB × rB /A = 0 + (−10 k) × (0.4 i + 0.4 j) = (−4 i + 4 j) v C = v B + ω BC × rC /B = (−4 i + 4 j) + ω BC k × (0.8i − 0.4 j) = (−4 + 0.4 ω BC ) i + (4 + 0.8 ω BC ) j

Since C stays in contact with the floor, we set the j component to zero ⇒ ω BC = −5 rad/s. Now the acceleration analysis.

a B = a A + α AB × rB /A − ω AB 2 rB /A = 0 + 0 − (10) 2 (0.4 i + 0.4 j) = (−40 i − 40 j)

0.4 m A

10 rad/s

0.4 m

a C = a B + α BC × rC /B − ω BC 2 rC /B = (−40 i − 40 j) + α BC k × (0.8i − 0.4 j) − (−5) 2 (0.8i − 0.4 j)

0.8 m

= (−60 + 0.4α BC ) i + (−30 + 08α BC ) j Since C stays in contact with the floor, we set the j component to zero ⇒ α BC = 37.5 rad/s 2 . Now we find the acceleration of the center of mass G of bar BC. a G = a B + α BC × rG /B − ω BC 2 rG /B = (−40 i − 40 j) + (37.5)k × (0.4 i − 0.2 j) − (−5) 2 (0.4 i − 0.2 j) = (−42.5i − 20 j) m/s 2 . The total force and moment cause the accelerations that we just calculated. Therefore F = ma G = (8 kg)(−42.5i − 20 j) m/s 2 = (−340 i − 160 j) N, M = Iα =

1 (8 kg)([0.8 m] 2 + [0.4 m] 2 )(37.5 rad/s 2 ) = 20 N-m. 12

F = (−340 i − 160 j) N,

424

M = 20 N-m (counterclockwise).

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Problem 18.59 The masses of the slender bars AB and BC are 10 kg and 12 kg, respectively. The angular velocities of the bars are zero at the instant shown and the horizontal force F = 150 N. The horizontal surface is smooth. Determine the angular accelerations of the bars.

0.4 m

Solution:

Given

m AB = 10 kg, m BC = 12 kg, g = 9.81 m/s 2 L AB = 0.4 m, L BC =

0.4 2 + 0.2 2 m, F = 150 N 0.4 m

0.2 m

The FBDs The dynamic equations L

∑ M A : −m AB g 2AB + B y L AB = 3 m AB L AB 2α AB ∑ FBCx : −B x − F = m BC a BCx

∑ FBCy : −B y − m BC g + N = m BC a BCy ∑ M BCG : (B x − F )(0.2 m) + (B y + N )(0.1 m) =

1 m L 2α 12 BC BC BC

Ax Ay

mAB g

The kinematic constraints a BCy = α AB L AB + α BC (0.1 m) F

mBC g

a BCx = α BC (0.2 m) α AB L AB + α BC (0.2 m) = 0 Solving we find

α AB = 20.6 rad/s 2 , α BC = −41.2 rad/s 2

α AB = 20.6 rad/s 2 CCW , α BC = 41.2 rad/s 2 CW . We also find a BCx = −8.23 m/s 2 , a BCy = 4.12 m/s 2 N = 244 N, B x = −51.2 N, B y = 76.5 N.

Problem 18.60 Let the total moment of inertia of the car’s two rear wheels and axle be I R , and let the total moment of inertia of the two front wheels be I F . The radius of the tires is R, and the total mass of the car, including the wheels, is m. If the car’s engine exerts a torque (couple) T on the rear wheels and the wheels do not slip, show that the car’s acceleration is a =

RT . R 2m + I R + I F

Strategy: Isolate the wheels and draw three free-body diagrams.

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18.60 (Continued) Solution: The free body diagrams are as shown: We shall write three equations of motion for each wheel and two equations of motion for the body of the car: We shall sum moments about the axles on each wheel.

bodies in the x direction, we get f R + f F = (m B + m R + m F )a = ma. (1) From the moment equations for the wheels, we get f F = −I F a /R 2 and f R = −I R a /R 2 + T /R. Substituting these into Eq. (1), we get a = RT /(mR 2 + I R + I F ) as required.

Rear Wheel:

mBg

∑ Fx = Fx + f R = m R a, ∑ Fy = N R − m R g − Fy = 0,

( )

∑ M Raxle = Rf R − T = I Rα = I R −

a R

Gx Fy

Front Wheel:

∑ Fx = G x + f F = m F a,

∑ Fy = N F − m F g − G y = 0,

mR g

∑ M Faxle = Rf F = I F α = I F ( − R ) a

mF g

Gx T

Car Body: fR

∑ Fx = −Fx − G x = m Ba,

∑ Fy = Fy + G y − m Bg = 0. Summing the y equations for all three bodies, we get N R + N F = (m B + m R + m F ) g = mg. Summing the equations for all three

Problem 18.61 The combined mass of the motorcycle and rider is 160 kg. Each 9-kg wheel has a 330-mm radius and a moment of inertia I = 0.8 kg-m 2 . The engine drives the rear wheel by exerting a couple on it. If the rear wheel exerts a 400-N horizontal force on the road and you do not neglect the horizontal force exerted on the road by the front wheel, determine (a) the motorcycle’s acceleration and (b) the normal forces exerted on the road by the rear and front wheels. (The location of the center of mass of the motorcycle, not including its wheels, is shown.)

723 mm

Solution:

In the free-body diagrams shown, m w = 9 kg and m = 160 − 18 = 142 kg. Let a be the motorcycle’s acceleration to the right and let α be the wheels’ clockwise angular acceleration. Note that a = 0.33α.

(1)

649 mm

1500 mm

Source: Courtesy of Arthur Bargan/Shutterstock.

Front Wheel:

∑ Fx = B x + f F = m ω a,

(2)

∑ Fy = B y + N F − m ω g = 0,

(3)

∑ M = − f F (0.33) = I α .

(4)

Rear Wheel:

426

∑ Fx = Ax + f R = m ω a,

(5)

∑ Fy = Ay + N R − m ω g = 0,

(6)

∑ M = M − f R (0.33) = I α.

(7)

Bx M

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18.61 (Continued) Motorcycle:

∑ Fx = − Ax − B x = ma,

(8)

∑ Fy = − Ay − B y − mg = 0,

(9)

Ay M

Bx m

wg fR

∑ M = −M + ( Ax + B x )(0.723 − 0.33) + B y (1.5 − 0.649) − A y (0.649) = 0.

(10)

wg fF

Solving Eqs. (1)–(10) with f R = 400 N, we obtain (a) a = 2.39 rad/s 2 and

(b) N R = 455 N, N F = 1115 N.

Problem 18.62 In Problem 18.61, if the front wheel lifts slightly off the road when the rider accelerates, determine (a) the motorcycle’s acceleration and (b) the torque exerted by the engine on the rear wheel. Solution:

See the solution of Problem 18.61. We set N F = 0 and replace Eq. (4) by f F = 0. Then solving Eqs. (1)–(10), we obtain

(a)

a = 9.34 m/s 2 ,

(b)

M = 516 N-m.

723 mm

649 mm

1500 mm

Source: Courtesy of Arthur Bargan/Shutterstock.

Problem 18.63 The moment of inertia of the vertical handle about O is 0.12 slug-ft 2 . The object B weighs 15 lb and rests on a smooth surface. The weight of the bar AB is negligible (which means that you can treat the bar as a two-force member). If the person exerts a 0.2-lb horizontal force on the handle 15 in above O, what is the resulting angular acceleration of the handle?

A 6 in

B O 12 in

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18.63 (Continued) Solution:

Let α be the clockwise angular acceleration of the handle. The acceleration of B is:

a B = a A + α AB × rB /A : i 0

a B i = (6/12)αi +

j 0

1 −0.5

15 in.

k α AB

6 in.

we see that α AB = 0 and a B = (6/12)α

(1).

The free body diagrams of the handle and object B are as shown. Note that β = arctan(6/12) = 26.6 °. Newton’s second law for the object B is C cos β = (15/32.2)a B ,

15 lb C

(2)

The equation of angular motion for the handle is (15/12) F − (6/12)C cos β = (0.12)α

(3).

Solving Equations (1)–(3) with F = 0.2 lb, we obtain α = 1.06 rad/s 2 .

aA = (6/12)

Problem 18.64 Model the bar the person is holding as a 4-lb, 18-in slender bar. The box at B weighs 8 lb and slides on the smooth horizontal surface. The weight of the carbon bar AB is negligible, which means you can treat it as a two-force member. The system is stationary when the person exerts a forward force F = 2 lb in the direction perpendicular to the bar, at 15 in from point O. At that instant, what is the acceleration of the box?

Solution: y

A 1 ft

0.5 ft b

First, some geometry. The x coordinate of B is x B = 1.12 ft. Then from the law of sines (Appendix A),

sin β sin 90 ° = , 1 ft 1.12 ft A

6 in

(0.5 ft) 2 + (1 ft) 2 =

sin γ sin 90 ° = , 0.5 ft 1.12 ft

the angles β = 63.4 °, γ = 26.6 °. The coordinates of A are x A = (0.5 ft) cos β = 0.224 ft, y A = (0.5 ft)sin β = 0.447 ft. The position vector of B relative to A is

12 in B

rB /A = ( x B − x A ) i + ( y B − y A ) j = 0.894 i − 0.447 j (ft) = x B /A i + y B /A j. The acceleration of point A is a A = a O + α OA × r A /O − ω OA 2 r A /O = 0+

i 0 xA

j 0 yA

k αOA 0

−0

= −αOA y A i + αOA x A j. We write the acceleration of point B in terms of the acceleration of point A, a B = a A + α AB × rB /A − ω AB 2 rB /A : a B i = −αOA y A i + αOA x A j +

428

i 0

j 0

x B /A

y B /A

k α AB 0

− 0,

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18.64 (Continued) which yields the scalar equations

The moment of inertia of the bar about O is

a B = −αOA y A − α AB y B /A , (1) 0 = αOA x A + α AB x B /A . (2)

2 1 18 m ft = 0.0932 slug-ft 2 . 3 OA 12

(

)

Let C denote the compressive force exerted by the link AB. For the bar we write the angular equation of motion

∑ M O = −( 12 ft ) F + (0.5 ft)C − ( 12 cos β ft )WOA = I αOA . (3)

F WOA

For the box we write Newton’s second law

∑ Fx = C cos γ = m B a B . (4)

We have four equations in terms of αOA , α AB , a B , C . Solving them, we obtain αOA = −22.5 rad/s 2 , α AB = 5.62 rad/s 2 , a B = 12.6 ft/s 2 , C = 3.49 lb.

Denote the weights of the bar OA and box B by WOA , W B . Their masses are W 4 lb mOA = OA = = 0.124 slug, 32.2 ft/s 2 g mB =

I =

12.6 ft/s 2 (toward the right).

WB 8 lb = = 0.248 slug. g 32.2 ft/s 2

Problem 18.65 Bars OQ and PQ each weigh 6 lb. The weight of the collar P and the friction between the collar and the horizontal bar are negligible. If the system is released from rest with θ = 45 °, what are the angular accelerations of the two bars? Solution: Let α OQ and α PQ be the clockwise angular acceleration of bar OQ and the counterclockwise angular acceleration of bar PQ. The acceleration of Q is a Q = a 0 + α 0Q × rQ / 0 =

i 0

j 0

2 ft

u O P

k −α OQ

2 cos 45 ° 2sin 45°

= 2α OQ sin 45° i − 2α OQ cos 45 ° j. y

The acceleration of P is

Q aOQ

a P = a Q + α PQ × rP /Q i 0

a P i = 2αOQ sin 45 °i − 2αOQ cos 45 ° j +

j 0

2 cos 45 ° 2sin 45 °

k α PQ .

458

Qy Qx

a P = 2αOQ sin 45 ° − 2α PQ sin 45 ° (1) 0 = −2αOQ cos 45 ° + 2α PQ cos 45 ° (2) The acceleration of the center of mass of bar PQ is a G = a Q + α PQ × rG /Q = 2α OQ sin 45 ° i i 0

j 0

cos 45 ° − sin 45 °

Equating i and j components,

−2αOQ cos 45 ° j +

aPQ

6 lb

Qy Qx 6 lb N

k α PQ . 0

Hence, a Gx = 2αOQ sin 45 ° + α PQ sin 45 ° (3)

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18.65 (Continued) a Gy = −2αOQ cos 45 ° + α PQ cos 45 ° (4) From the diagrams: The equation of angular motion of bar OQ is ∑ M 0 = I 0 αOQ : Q x (2sin 45 °) − Q y (2 cos 45 °) + 6 cos 45 ° =

1 (6/32.2)(2) 2 αOQ (5) 3

The equations of motion of bar PQ are

∑ Fx = −Q x = (6/32.2)a Gx (6) ∑ Fy = N − Q y − 6 = (6/32.2)a Gy (7) 1

∑ M = ( N + Q y + Q x )(cos 45°) = 12 (6/32.2)(2) 2 α PQ (8) Solving Equations (1)–(8), we obtain αOQ = α PQ = 6.83 rad/s 2 .

Problem 18.66 In Problem 18.65, what are the angular accelerations of the two bars if the collar P weighs 2 lb?

2 ft

Solution:

In the solution of Problem 18.65, the free body diagram of bar PQ has a horizontal component P to the left where P is the force exerted on the bar by the collar. Equations (6) and (8) become

∑ Fx = −Q x − P = (6/32.2)a Gx

2 ft

u O

∑ M = ( N − P + Q y + Q x )(cos 45°) = 12 (6/32.2)(2) 2 α PQ and the equation of motion for the collar is P = (2/32.2)a P solving equations (1–9), we obtain αOQ = α PQ = 4.88 rad/s 2 .

Problem 18.67 The 4-kg slender bar is pinned to 2-kg sliders at A and B. If friction is negligible and the system is released from rest in the position shown, what is the angular acceleration of the bar at that instant?

Solution: Express the acceleration of B in terms of the acceleration of A, a B = a A + α AB × rB /A : a B cos 45 °i − a B sin 45 ° j = −a A j +

i 0

j 0

k α AB ,

0.5 −1.2 or

a B cos 45 ° = 1.2α AB , (1)

and − a B sin 45 ° = −a A + 0.5α AB , (2)

We express the acceleration of G in terms of the acceleration of A, a G = a A + α AB × rG /A : a G = a Gx i + a Gy j = −a A j +

i 0

j 0

0.25 −0.6

1.2 m or

k α AB , 0

a Gx = 0.6α AB , (3)

and a Gy = −a A + 0.25α AB , (4) 458 B

The free body diagrams are as shown. The equations of motion are Slider A: N − A x = 0 (5)

0.5 m

and (2) (9.81) + A y = 2a A , (6) Slider B: P − [ B x + B y + (2)(9.81)]cos 45 ° = 0, (7)

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18.67 (Continued) and [(2)(9.81) − B x + B y ]cos 45 ° = 2a B , (8) Bar:

A x + B k = 4 a Gx (9)

and

A y + B y − (4)(9.81) = 4 a Gy (10)

A aA G

aAB

1 ( L /2)[( B x − A x ) cos β + ( B y − A y )sin β ] = (4) L2α AB (11) 12 where and

L =

x aB

(0.5) 2 + (1.2) 2 m

β arctan(0.5/1.2) = 22.6 °.

Solving Equations (1)–(11), we obtain α AB =

5.18 rad/s 2 .

B (4)(9.81)

By Bx

Ay Ax

N (2)(9.81) By

Bx (2)(9.81)

Problem 18.68* The weights of the uniform bars are W AB = 1 lb and W BC = 2 lb. The slider C weighs 6 lb, and the horizontal groove it slides in is smooth. The system is at rest in the position shown when the clockwise couple M = 2 ft-lb is applied to bar AB. At that instant, what is the acceleration of the slider C?

Solution:

The masses are

m AB =

W W AB = 0.0311 slug, m BC = BC = 0.0621 slug, g g

mC =

WC = 0.186 slug. g

The lengths of the bars are B 4 in A

(4 in) 2 + (4 in) 2 = 5.66 in = 0.471 ft,

L BC =

(10 in) 2 + (7 in) 2 = 12.2 in = 1.02 ft.

The moment of inertia of bar AB about the fixed point A and the moment of inertia of bar BC about its center of mass G are

3 in C 4 in

L AB =

10 in

IA =

1 m L 2 = 0.00230 slug-ft 2 , 3 AB AB

IG =

1 m L 2 = 0.00536 slug-ft 2 . 12 BC BC

Because there is no angular velocity, the acceleration of B in terms of the counterclockwise angular acceleration of bar AB is a B = α AB L AB (− cos 45 °i + cos 45 ° j). The acceleration of C in terms of the acceleration of B is a C = a B + α BC × rC /B :

a C i = α AB L AB (− cos 45 °i + cos 45 ° j) +

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i j k α BC 0 0 10 7 ft − ft 0 12 12

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18.68* (Continued) which yields the scalar equations 7 a C = −α AB L AB cos 45 ° + α , (1) 12 BC 10 0 = α AB L AB cos 45 ° + α . (2) 12 BC

we write the angular equation of motion:

Recall that G denotes the center of mass of bar BC. We can also write the acceleration of G in terms of the acceleration of B:

For bar BC and the slider,

∑ M A = I Aα AB : −M −

( 124 ft ) B + ( 124 ft ) B − ( 122 ft )W x

a G = a B + α BC × rG /B :

j 0

k α BC

5 3.5 ft − ft 12 12

= I Aα AB . (5)

C WBC

C Cx

a Gx i + a Gy j = α AB L AB (− cos 45 °i + cos 45 ° j) i 0

we write the equations of motion

giving the scalar equations

∑ Fx = −B x − C x = m BC a Gx , (6)

3.5 a Gx = −α AB L AB cos 45 ° + α , (3) 12 BC

∑ Fy = −B y − WBC − C y = m BC a Gy , (7)

a Gy = α AB L AB cos 45 ° +

5 α . (4) 12 BC

∑ M G = ( 12 ft ) B x + ( 12 ft ) B y − ( 12 ft )C x − ( 12 ft )C y 3.5

3.5

= I BC α BC ,

For bar AB,

(8)

∑ Fx = C x = mC aC . (9)

We have 9 equations in terms of the unknowns α AB , α BC , a Gx , a Gy , a C , B x , B y , C x , C y . Solving them (using MATLAB), we obtain α AB = −32.3 rad/s 2 , α BC = 12.9 rad/s 2 , a Gx = 14.6 ft/s 2 , a Gy = −5.39 ft/s 2 , a C = 18.3 ft/s 2 , B x = −4.32 lb, B y = 1.96 lb, C x = 3.42 lb, C y = −3.62 lb.

WAB

18.3 ft/s 2 (toward the right).

Problem 18.69* Bar AB rotates in the horizontal plane with a constant angular velocity of 10 rad/s in the counterclockwise direction. The masses of the slender bars BC and CD are 3 kg and 4.5 kg, respectively. Determine the x and y components of the forces exerted on bar BC by the pins at B and C at the instant shown.

Solution:

First let’s do the kinematics

Velocity v B = v A + ω AB × rB / A = 0 + (10 rad/s)k × (0.2 m) j = −(2 m/s) i v C = v B + ω BC × rC / B = −(2 m/s) i + ω BC k × (0.2 m) i = −(2 m/s) i + (0.2 m)ω BC j

v D = v C + ω CD × rD / C = −(2 m/s) i + (0.2 m)ω BC j + ω CD k × (0.2 m)( i − j)

= (−[2 m/s] + [0.2 m]ω CD ) i + (0.2 m)(ω BC + ω CD ) j Since D is pinned we find ω CD = 10 rad/s, ω BC = −10 rad/s

0.2 m

10 rad/s

Acceleration a B = a A + α AB × rB /A − ω AB 2 rB /A x

= 0 + 0 − (10 rad/s) 2 (0.2 m) j = −(20 m/s 2 ) j a C = a B + α BC × rC /B − ω BC 2 rC /B

0.2 m

= −(20 m/s 2 ) j + α BC k × (0.2 m) i − (−10 rad/s) 2 (0.2 m) i

0.2 m

= −(20 m/s 2 ) i + ([0.2 m]α BC − 20 m/s 2 ) j a D = a C + α CD × rD /C − ω CD 2 rD /C = −(20 m/s 2 ) i + ([0.2 m]α BC − 20 m/s 2 ) j + αCD k × (0.2 m)(i − j) − (10 rad/s) 2 (0.2 m)(i − j) = (−40 m/s 2 + [0.2 m]αCD )i + ([0.2 m][α BC + α BC ]) j

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18.69* (Continued) 0.2 m

Since D is pinned we find α BC = −200 rad/s 2 , αCD = 200 rad/s 2 Now find the accelerations of the center of mass G. B

a G = a B + α BC × rG1/B − ω BC 2 rG1/B

= −(20 m/s 2 ) j + (−200 rad/s 2 )k × (0.1 m) i − (−10 rad/s) 2 (0.1 m) i 0.2 m = (−10 i − 40 j) m/s 2 10 rad/s

The FBDs A

The dynamics

0.4 m

∑ FBCx : B x + C x = (3 kg)(−10 m/s 2 ) ∑ FBCy : B y + C y = (3 kg)(−40 m/s 2 )

1 ∑ M G1 : (C y − B y )(0.1 m) = 12 (3 kg)(0.2 m) 2 (−200 rad/s 2 )

∑ M D : C x (0.2 m) + C y (0.2 m) = 3 (4.5 kg)( 2[0.2 m]) 2 (200 rad/s 2 )

Solving we find Dx

B x = −220 N, B y = −50 N C x = 190 N, C y = −70 N.

Problem 18.70 The 2-kg bar rotates in the horizontal plane about the smooth pin. The 6-kg collar A slides on the smooth bar. At the instant shown, r = 12 m, ω = 0.4 rad/s, and the collar is sliding outward at 0.5 m/s relative to the bar. If you neglect the moment of inertia of the collar (that is, treat the collar as a particle), what is the bar’s angular acceleration? Strategy: Draw individual free-body diagrams of the bar and the collar, and write Newton’s second law for the collar in polar coordinates.

Solution:

Diagrams of the bar and collar showing the force they exert on each other in the horizontal plane are: the bar’s equation of angular motion is 1

∑ M 0 = I 0α: − Nr = 3 (2)(2) 2 α (1) In polar coordinates, Newton’s second law for the collar is

∑ F = ma: Ne θ = m  ( dt 2 − r ω 2 ) e r + ( rα + 2 dt ω ) e θ  . d 2r

Equating e θ components,

(

N = m rα + 2

)

dr ω = (6)[rα + 2(0.5)(0.4)] (2) dt

v Solving Equations (1) and (2) with r = 1.2 m gives α = −0.255 rad/s 2 .

N r

2m O

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Problem 18.71 In Problem 18.70, suppose that the moment of inertia of the collar about its center of mass is 0.2 kg-m 2 . Determine the angular acceleration of the bar and compare your answer with the answer to Problem 18.70.

Solution: Let C be the couple the collar and bar exert on each other: The bar’s equation of angular motion is 1

∑ M 0 = I 0α: − Nr − C = 3 (2)(2) 2 α

(1)

The collar’s equation of angular motion is

∑ M = I α:

C = 0.2α

(2)

From the solution of Problem 18.70, the e θ component of Newton’s second law for the collar is N = (6)[rα + 2(0.4)(0.4)]

(3)

Solving Equations (1)–(3) with r = 1.2 m gives α = −0.250 rad/s 2 .

Problem 18.72 The slender bar has mass m and length l. The axes L and L O are perpendicular to the bar. Use integration to determine the moment of inertia I about L and the moment of inertia I O about L O and show that IO = I +

L LO l

( 12 l ) m. 2

Solution: To determine the moment of inertia about L, we place the origin of a cartesian coordinate system at the center of mass with the x axis parallel to the bar:

To determine the moment of inertia about L O , we place the origin of the coordinate system at the end of the bar with the x axis parallel to the bar: y

y LO

L x

dm x

x dx

dx l

The moment of inertia is We define an element of the bar of length dx at a distance x from the origin. If ρ denotes the density (mass per unit volume) of the material and A the cross-sectional area of the bar, the mass of the bar is m = ρ Al and the mass of the element is dm = ρ Adx. The moment of inertia is I = ∫ r 2 dm = ∫ m

434

l/2

−l / 2

l l 1 1 1 I O = ∫ r 2 dm = ∫ x 2ρ Adx = ρ A  x 3  = ρ Al 3 = ml 2 . m 0 3 0 3 3

I =

1 1 ml 2 , I O = ml 2 . 12 3

l/2 1 1 1 x 2ρ Adx = ρ A  x 3  = ρ Al 3 = ml 2 .  3  −l / 2 12 12

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Problem 18.73 Two homogeneous slender bars, each of mass m and length l, are welded together to form the T-shaped object. Use integration to determine the moment of inertia of the object about the axis through point O that is perpendicular to the bars.

Solution:

Divide the object into two pieces, each corresponding to a slender bar of mass m; the first parallel to the y-axis, the second to the x-axis. By definition l

I = ∫ r 2 dm + ∫ r 2 dm. 0

For the first bar, the differential mass is dm = ρ A dr . Assume that the second bar is very slender, so that the mass is concentrated at a distance l from O. Thus dm = ρ A dx, where x lies between the limits − 2l ≤ x ≤ 2l . The distance to a differential dx is r = l 2 + x 2 . Thus the definition becomes 1

I = ρ A∫ r 2 dr + ρ A ∫ 2l (l 2 + x 2 ) dx

−

r3 l

x3  2 I = ρ A   + ρ A  l 2 x +   3 0  3  −1

Problem 18.74 The slender bar lies in the x – y plane. Its mass is 6 kg and the material is homogeneous. Use integration to determine its moment of inertia about the z-axis.

ml 2

(

)

1 1 17 2 ml . +1+ = 3 12 12

Solution: Iz = ∫

1m 0

+ ∫

The density is ρ =

6 kg = 2 kg/m 3m

ρ x 2 dx 2m

ρ[(1 m + s cos 50 °) 2 + (s sin 50 °) 2 ] ds

I z = 15.1 kg-m 2 .

y 2m

2m 508

x 508

Problem 18.75 The slender bar lies in the x – y plane. Its mass is 6 kg and the material is homogeneous. Use integration to determine its moment of inertia about the y-axis.

Solution: Iy = ∫

1m 0

+ ∫ y

The density is ρ =

6 kg = 2 kg/m 3m

ρ x 2 dx 2m

ρ[(1 m + s cos 50 °) 2 ] ds

I y = 12.0 kg-m 2 .

508

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Problem 18.76 The thin rectangular plate lies in the x – y plane and has uniform thickness t and mass m. Determine its moment of inertia about the y-axis and compare your result with the value given in Appendix C. Strategy: Use an element of mass of the plate in the form of a vertical strip of width dx.

Solution: y

We define an element of the plate of width dx at a distance x from the y axis. If ρ denotes the density (mass per unit volume) of the material and t the thickness of the plate, the mass of the plate is m = ρbht and the mass of the element is dm = ρht dx. The moment of inertia is b b 1 1 1 I y axis = ∫ r 2 dm = ∫ x 2ρht dx = ρht  x 3  = ρhtb 3 = mb 2 . m 0 3 0 3 3

1 mb 2 . 3

I y axis =

Problem 18.77 The thin rectangular plate lies in the x – y plane and has uniform thickness t and mass m. Determine its moment of inertia about the y ′-axis and compare your result with the value given in Appendix C. (See Problem 18.76.)

Solution: y9

y9 x9 dx9

We define an element of the plate of width dx at a distance x ′ from the y ′ axis. If ρ denotes the density (mass per unit volume) of the material and t the thickness of the plate, the mass of the plate is m = ρbht and the mass of the element is dm = ρht dx. The moment of inertia is I y ′ axis = ∫ r 2 dm = ∫ m

b = I y ′ axis =

Problem 18.78 The homogeneous thin plate is of uniform thickness and weighs 20 lb. Determine its moment of inertia about the y-axis.

b/2

−b / 2

b/2

1 ( x ′) 2ρht dx ′ = ρht  ( x ′) 3  3  −b / 2

1 1 ρhtb 3 = mb 2 . 12 12 1 mb 2 . 12

y y542

1 2 x ft 4

436

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18.78 (Continued) Solution:

The area is

The definition of the moment of inertia is

I = ∫ r 2 dm.

A = ∫

The distance from the y-axis is x, where x varies over the range m W −4 ≤ x ≤ 4. Let τ = = be the area mass density. The mass A gA W of an element y dx is dm = y dx. Substitute into the definition: gA I y-axis = =

−4

(

3 4

( 4 − x4 ) dx =  4 x − 12x  2

−4

= 21.333 ft 2

The moment of inertia about the y-axis is W 20 (3.2) = (3.2) = 1.99 slug-ft 2 . g 32.17

I ( y -axis) =

)

W 4 2 x2 x 4− dx gA ∫−4 4 W  4x 3 x 5  +4 W = [68.2667]. −   20  −4 gA  3 gA

Problem 18.79 Determine the moment of inertia of the 20-lb plate about the x-axis.

W The differential mass is dm = dy dx. The distance of gA a mass element from the x-axis is y, thus

Solution:

y y542

1 2 x ft 4

I =

4− W +4 4 dx y 2 dy gA ∫−4 ∫ 0

+4 W x2 3 4− dx ∫ 3gA −4 4

W  3 5 x7 4 64 x − 4 x 3 + x − 3gA  20 448  −4

W [234.057]. 3gA

(

From the solution to Problem 18.78, A = 21.333 ft 2 . Thus the moment of inertia about the x-axis is I x-axis =

Problem 18.80 The mass of the object is 10 kg. Its moment of inertia about L1 is 10 kg-m 2 . What is its moment of inertia about L 2 ? (The three axes lie in the same plane.)

)

W (234.057) W = (3.657) = 2.27 slug-ft 2 . 3g (21.333) g

Solution:

The strategy is to use the data to find the moment of inertia about L, from which the moment of inertia about L 2 can be determined. I L = −(0.6) 2 (10) + 10 = 6.4 kg-m 2 ,

from which I L 2 = (1.2) 2 (10) + 6.4 = 20.8 kg-m 2 .

0.6 m L

0.6 m L1

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Problem 18.81 An engineer gathering data for the design of a maneuvering unit determines that the astronaut’s center of mass is at x = 1.01 m, y = 0.16 m and that her moment of inertia about the z-axis is 105.6 kg-m 2 . The astronaut’s mass is 81.6 kg. What is her moment of inertia about the z ′ -axis through her center of mass?

x9 x

Solution:

The distance from the z ′ axis to the z axis is d = x 2 + y 2 = 1.02257 m. The moment of inertia about the z ′ axis is I z ′ -axis = −d 2 m + I z -axis = −(1.0457)(81.6) + 105.6 = 20.27 kg-m 2 .

Problem 18.82 The dimension l = 0.3 m. The masses of the horizontal and vertical bars are 4 kg and 2 kg, respectively. Use the parallel-axis theorem to determine the moment of inertia of the T-shaped object about the z-axis.

Solution: IL =

l x, x9

The moment of inertia of the left bar about the z axis is 2l

2 1 1 m (2l ) 2 +  (2l )  m L 2  12 L

= 0.48 kg-m 2 .

The moment of inertia of the right bar is IR = =

1 m l 2 + (2l ) 2 m R 12 R

l x

0.735 kg-m 2 . 1 (2l) 2

The moment of inertia of the object is I z = I L + I R = 1.22 kg-m 2 .

I z = 1.22 kg-m 2 .

Problem 18.83 The dimension l = 0.3 m. The masses of the horizontal and vertical bars are 4 kg and 2 kg, respectively. Use the parallel-axis theorem to determine the moment of inertia of the T-shaped object about the z ′-axis.

l x, x9

438

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18.83 (Continued) Solution:

1 l 3

x, x9

2 l 3

x The moment of inertia of the left bar about the z ' axis is

1 1 2 m L (2l ) 2 +  l  m L 3  12 = 0.160 kg-m 2 .

I 'L = The x coordinate of the center of mass of the object is x =

x L m L + x Rm R lm L + 2lm R 4 = = l = 0.4 m. mL + mR mL + mR 3

The moment of inertia of the right bar is 1 2 2 m l2 + l mR 12 R 3 2 = 0.0950 kg-m .

( )

I 'R =

The moment of inertia of the object is I z ' = I ' L + I ' R = 0.255 kg-m 2 . I z ' = 0.255 kg-m 2 .

Problem 18.84 The mass of the homogeneous slender bar is 30 kg. Determine its moment of inertia about the z-axis.

0.8 m

Solution: Iz =

The density is ρ =

30 kg = 10 kg/m 3m

x 2m

0.6 m

1 1 (10 kg)(1.0 m) 2 + (20 kg)(2 m) 2 3 12 + (20 kg)[(1.6 m) 2 + (0.8 m) 2 ]

I z = 74 kg-m 2 .

Problem 18.85 The mass of the homogeneous slender bar is 30 kg. Determine the moment of inertia of the bar about the z ′-axis through its center of mass. Solution:

First locate the center of mass

x =

(10 kg)(0.3 m) + (20 kg)(1.6 m) = 1.167 m 30 kg

y =

(10 kg)(0.4 m) + (20 kg)(0.8 m) = 0.667 m 30 kg

0.8 m

x 0.6 m

Using the answer to 18.100 I z ′ = (74 kg-m 2 ) − (30 kg)(1.167 2 + 0.667 2 )m 2 I z ′ = 19.8 kg-m 2 .

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Problem 18.86 The homogeneous slender bar weighs 5 lb. Determine its moment of inertia about the z-axis.

Solution:

The Bar’s mass is m = 5/32.2 slugs. Its length is L = L1 + L 2 + L 3 = 8 + 8 2 + 8 2 + π (4) = 31.9 in. The masses of the parts are therefore,

(

5 ) = 0.0390 slugs, )( 32.2

M1 =

L1 8 m = 31.9 L

M2 =

 2(64)  5 L2 = 0.0551 slugs, m =    31.9  32.2 L

M3 =

L3 4π m = 31.9 L

4 in

(

)

5 ) = 0.0612 slugs. )( 32.2

The center of mass of part 3 is located to the right of its center C a distance 2 R /π = 2(4)/π = 2.55 in. The moment of inertia of part 3 about C is x9

∫m r 2 dm = m3r 2 = (0.0612)(4) 2 = 0.979 slug-in 2 . 3

x 8 in

The moment of inertia of part 3 about the center of mass of part 3 is therefore I 3 = 0.979 − m 3 (2.55) 2 = 0.582 slug-in 2 . The moment of inertia of the bar about the z axis is 1 1 m L2 + m 2 L22 + I 3 + m 3 [(8 + 2.55) 2 + (4) 2 ] 3 1 1 3 = 11.6 slug-in 2 = 0.0803 slug-ft 2 .

I ( z axis) =

Problem 18.87 Determine the moment of inertia of the 5-lb bar about the z ′ -axis through its center of mass.

In the solution of Problem 18.86, it is shown that the moment of inertia of the bar about the z axis is I ( z axis) = 11.6 slug-in 2 . The x and y coordinates of the center of mass coincide with the centroid of the bar: x =

Solution:

x 1 L1 + x 2 L 2 + x 3 L 3 L1 + L 2 + L 3

2(4)  π (4) (4)(8) + (4) 8 2 + 8 2 +  8 +  π  = = 6.58 in, 8 + 8 2 + 8 2 + π (4) y =

4 in

= x9

y 1 L1 + y 2 L 2 + y 3 L 3 L1 + L 2 + L 3 0 + (4) 8 2 + 8 2 + π (4)(4) = 3.00 in. 8 + 8 2 + 8 2 + π (4)

The moment of inertia about the z′ axis is x

I ( z ′ axis ) = I ( z axis) − (x 2 + y 2 )

8 in

Problem 18.88 The rocket is used for atmospheric research. Its weight and its moment of inertia about the z-axis through its center of mass (including its fuel) are 10,000 lb and 10,200 slug-ft 2 , respectively. The rocket’s fuel weighs 6000 lb, its center of mass is located at x = −3 ft, y = 0, z = 0, and the moment of inertia of the fuel about the axis through the fuel’s center of mass parallel to the z-axis is 2200 slug-ft 2 . When the fuel is exhausted, what is the rocket’s moment of inertia about the axis through its new center of mass parallel to the z-axis?

440

5 ( 32.2 ) = 3.44 slug-in . 2

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18.88 (Continued) Solution:

Denote the moment of inertia of the empty rocket as I E about a center of mass x E , and the moment of inertia of the fuel as I F about a mass center x F . Using the parallel axis theorem, the moment of inertia of the filled rocket is

Using a value of g = 32 ft/s 2 , mF =

WF 6000 = = 186.34 slug, 32.2 g

I R = I E + x E2 m E + I F + x F2 m F ,

mE =

(W R − W F ) 10000 − 6000 = = 124.23 slug. g 32.2

about a mass center at the origin ( x R = 0). Solve:

186.335 ( 124.224 )(−3) = 4.5 ft

From which x E = −

I E = I R − x E 2m E − I F − x F 2m F .

The objective is to determine values for the terms on the right from the data given. Since the filled rocket has a mass center at the origin, the mass center of the empty rocket is found from

is the new location of the center of mass.

0 = mE x E + mF x F ,

I E = I R − x E2 m E − I F − x F2 m F

Substitute:

= 10200 − 2515.5 − 2200 − 1677.01

from which

= 3810 slug-ft 2 .

m  x E = − F  x F .  mE 

Problem 18.89 The mass of the homogeneous thin plate is 36 kg. Determine the moment of inertia of the plate about the x-axis.

Solution:

Divide the plate into two areas: the rectangle 0.4 m by 0.6 m on the left, and the rectangle 0.4 m by 0.3 m on the right. The m mass density is ρ = . A

The area is A = (0.4)(0.6) + (0.4)(0.3) = 0.36 m 2 ,

from which 0.4 m

0.4 m

ρ =

36 = 100 kg/m 2 . 0.36

The moment of inertia about the x-axis is

0.3 m

I x-axis = ρ 0.3 m

( 13 ) (0.4)(0.6 ) + ρ ( 13 ) (0.4)(0.3) = 3.24 kg-m . 3

Problem 18.90 Determine the moment of inertia of the 36-kg plate about the z-axis.

Solution:

The basic relation to use is I z -axis = I x -axis + I y -axis . The value of I x-axis is given in the solution of Problem 18.89. The moment of inertia about the y-axis using the same divisions as in Problem 8.89 and the parallel axis theorem is I y-axis = ρ

y 0.4 m

( 13 )(0.6)(0.4) + ρ ( 121 )(0.3)(0.4) 3

+ (0.6) 2 ρ (0.3)(0.4) = 5.76 kg-m 2 ,

0.4 m

from which 0.3 m

I z -axis = I x -axis + I y -axis = 3.24 + 5.76 = 9 kg-m 2 .

0.3 m x

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Problem 18.91 The mass of the homogeneous thin plate is 20 kg. Determine its moment of inertia about the x-axis.

Solution:

A = (0.2 m)(0.8 m) + (0.2 m)(0.4 m) + ρ =

Break the plate into the three regions shown.

1 (0.4 m)(0.6 m) = 0.36 m 2 2

20 kg = 55.6 kg/m 2 0.36 m 2

Using the integral tables we have

1000 mm

1 1 I x = (0.2 m)(0.8 m) 3 + (0.2 m)(0.4 m) 3 + (0.2 m)(0.4 m)(0.6 m) 2 3 12 400 mm

1 1 (0.6 m)(0.4 m) 3 + (0.6 m)(0.4 m)(0.667 m) 2 36 2

= 0.1184 m 4 I x − axis = (55.6 kg/m 2 )(0.1184 m 4 ) = 6.58 kg-m 2 .

400 mm

200 mm

1000

800 200 400 200

Problem 18.92 The mass of the homogeneous thin plate is 20 kg. Determine its moment of inertia about the y-axis.

Solution:

See the solution to 18.91

1 1 I y = (0.8 m)(0.2 m) 3 + (0.4 m)(0.2 m) 3 + (0.2 m)(0.4 m)(0.3 m) 2 3 12 +

1 1 (0.4 m)(0.6 m) 3 + (0.6 m)(0.4 m)(0.6 m) 2 36 2

= 0.0552 m 4 1000 mm

I y-axis = (55.6 kg/m 2 )(0.0552 m 4 ) = 3.07 kg-m 2 .

400 mm

200 mm

442

200 mm

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Problem 18.93 The thermal radiator (used to eliminate excess heat from a satellite) can be modeled as a homogeneous thin rectangular plate. The mass of the radiator is 5 slugs. Determine its moments of inertia about the x -, y - and z-axes.

Solution:

The area is A = 9(3) = 27ft 2 .

The mass density is ρ =

m 5 = = 0.1852 slugs/ft 2 . A 27

The moment of inertia about the centroid of the rectangle is y 3 ft

6 ft

I xc = ρ

( 121 )9(3 ) = 3.75 slug-ft ,

I yc = ρ

( 121 )3(9 ) = 33.75 slug-ft .

Use the parallel axis theorem:

3 ft

I x -axis = ρ A(2 + 1.5) 2 + I xc = 65 slug-ft 2 , I y -axis = ρ A(4.5 − 3) 2 + I yc = 45 slug-ft 2 .

2 ft x

I z -axis = I x -axis + I y -axis = 110 slug-ft 2 .

Problem 18.94 The mass of the homogeneous thin plate is 2 kg. Determine the moment of inertia of the plate about the axis through point O that is perpendicular to the plate.

80 mm 10 mm

Solution:

By determining the moments of inertia of the area about the x and y axes, we will determine the moments of inertia of the plate about the x and y axes, then sum them to obtain the moment of inertia about the z axis, which is I 0 .

30 mm

30 mm 130 mm

The areas are 1 (130)(80)mm 2 , 2 A2 = π (10) 2 mm 2 . A1 =

Using Appendix B, Ix =

1 1 (130)(80) 3 −  π (10) 4 + (30) 2 A2  4  12

= 5.26 × 10 6 mm 4 , 1 1 I y = (80)(130) 3 −  π (10) 4 + (100) 2 A2  4  4 = 40.79 × 10 6 mm 4 .

30 mm x

O 100 mm

Therefore m I = 2150 kg-mm 2 , A1 − A2 x m I ( y axis) = I = 16700 kg-mm 2 . A1 − A2 y I ( x axis) =

Then I ( z axis) = I ( x axis) + I ( y axis) = 18850 kg-mm 2 . I ( z axis) = 0.0188 kg-m 2 .

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Problem 18.95 The homogeneous cone is of mass m. Use the method described in Example 18.11 to determine its moment of inertia about the z-axis. Compare your result with the value given in Appendix C. Solution:

From Appendix C, the volume of the cone is

( 1/3 ) π R 2 h, so in terms of the density (mass per unit volume) ρ of the

material, the mass of the cone is

m = ( 1/3 ) πρ R 2 h. (1) We define a disk-shaped element of the cone that is at a position z and has thickness dz:

Expressing the mass of the element as the product of the density ρ and the volume of the element, we obtain

x z

( I z ) element =

dz z

R z. (2) h

Using the result from Appendix C for a thin disk, the moment of inertia of the disk element about the z axis is ( I z ) element =

where we used Eq. (2). To obtain the moment of inertia of the cone about the z axis, we integrate this expression from z = 0 to z = h: I z = ∫ ( I z ) element = ∫

The radius of the element is a linear function of z, r =

( )

1( R 4 1 ρπr 2 dz ) r 2 = πρ z dz, (3) 2 2 h

1 m r 2. 2 element

h πρ R 4 0

2h 4

z 4 dz =

πρ R 4 h . 10

Solving Eq. (1) for ρ and substituting it into this expression, we get Iz =

3 mR 2 . 10

Iz =

3 mR 2 . 10

Problem 18.96 The homogeneous cone is of mass m. Use the method described in Example 18.11 to determine its moment of inertia about the x-axis. Compare your result with the value given in Appendix C.

Solution:

From Appendix C, the volume of the cone is ( 1/3 ) π R 2 h, so in terms of the density (mass per unit volume) ρ of the material, the mass of the cone is h

m = ( 1/3 ) πρ R 2 h. (1) We define a disk-shaped element of the cone that is at a position z and has thickness dz:

Using the result from Appendix C for a thin disk and applying the parallelaxis theorem, the moment of inertia of the disk element about the x axis is ( I x ) element =

x z

( I x ) element =

z The radius of the element is a linear function of z,

444

1 m r 2 + m element z 2 . 4 element

Expressing the mass of the element as the product of the density ρ and the volume of the element, we obtain

r =

1 (ρπr 2 dz )r 2 + (ρπr 2 dz ) z 2 4

( )

R 4 R 2 1 z dz + ρπ z z 2 dz πρ h h 4

= πρ

( 4h + h ) z dz, R4

R z. (2) h

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18.96 (Continued) where we used Eq. (2). To obtain the moment of inertia of the cone about the z axis, we integrate this expression from z = 0 to z = h:

(

Iz = m

)

R4 R2 I x = ∫ ( I x ) element = ∫ πρ + 2 z 4 dz 4 0 h 4h h

= πρ

(

Solving Eq. (1) for ρ and substituting it into this expression, we get

( 203 R + 35 h ).

Iz = m

)

R 4h R 2h 3 + . 20 5

( 203 R + 35 h ).

Problem 18.97 The homogeneous truncated cone consists of titanium with density ρ = 4420 kg/m 3 . The dimension R = 50 mm. Use the method described in Example 18.11 to determine its moment of inertia about the z-axis.

x R

180 mm

Solution: Let R = 0.05 m, b = 0.18 m. We define a disk-shaped element of the cone that is at a position z and has thickness dz:

z 180 mm

x z

Expressing the mass of the element as the product of the density ρ and the volume of the element, we obtain

( I z ) element =

dz z

R z. (1) 2b

Using the result from Appendix C for a thin disk, the moment of inertia of the disk element about the z axis is ( I z ) element =

)

where we used Eq. (1). To obtain the moment of inertia of the truncated cone about the z axis, we integrate this expression from z = b to z = 2b:

The radius of the element is a linear function of z, r =

(

πρ R 4 1 1 R 4 (ρπr 2 dz )r 2 = πρ z dz = , (2) 2 2 2b 32b 4

I z = ∫ ( I z ) element = ∫

2 b πρ R 4 b

32b 4

z 4 dz =

31πρ R 4 b . 160

Substituting the values, we obtain I z = 0.00303 kg-m 2 . I z = 0.00303 kg-m 2 .

1 m r 2. 2 element

Problem 18.98 The homogeneous truncated cone consists of titanium with density ρ = 4420 kg/m 3 . The dimension R = 50 mm. Use the method described in Example 18.11 to determine its moment of inertia about the x-axis.

Solution: We define a disk-shaped element of the cone that is at a position z and has thickness dz: y

x y

R The radius of the element is a linear function of z, r =

180 mm

R z. (1) 2b

180 mm

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18.98 (Continued) Using the result from Appendix C for a thin disk and applying the parallel-axis theorem, the moment of inertia of the disk element about the x axis is ( I x ) element =

where we used Eq. (1). To obtain the moment of inertia of the cone about the z axis, we integrate this expression from z = b to z = 2b: I x = ∫ ( I x ) element = ∫

1 m r 2 + m element z 2 . 4 element

= πρ

Expressing the mass of the element as the product of the density ρ and the volume of the element, we obtain ( I x ) element = =

)

(

)

I x = 0.315 kg-m 2 .

( 64Rb + 4Rb ) z dz 4

5 2b

( 64Rb + 4Rb ) z5  4

( 64Rb + 4Rb )z dz

Substituting the values, we obtain I x = 0.315 kg-m 2 .

)

R 4 R 2 2 1 z dz + ρπ z z dz πρ 4 2b 2b

= πρ

πρ

31R 4 b 31R 2 b 3 = πρ + . 320 20

1 (ρπr 2 dz )r 2 + (ρπr 2 dz ) z 2 4

(

2b b

Problem 18.99 The homogeneous rectangular parallelepiped is of mass m. Determine its moments of inertia about the x -, y - and z-axes and compare your results with the values given in Appendix C.

a z

Solution:

Consider a rectangular slice normal to the x-axis of dimensions b by c and mass dm. The area density of this slice is dm The moment of inertia about the y axis of the centroid of ρ = . bc a thin plate is the product of the area density and the area moment of 1 1 2 inertia of the plate: dI y = ρ bc 3 , from which dI y = c dm. 12 12 By symmetry, the moment of inertia about the z axis is

( )

dI z =

b c

( )

( 121 )b dm.

from which

Since the labeling of the x -y - and z-axis is arbitrary,

Ix =

( 121 )(b + c )∫ dm = 12m (b + c ). 2

By symmetry, the argument can be repeated for each coordinate, to obtain

dI x = dI z + dI y , where the x-axis is normal to the area of the plate. Thus

( )

1 dI x = (b 2 + c 2 ) dm, 12

Iy =

Problem 18.100 The sphere-capped cone consists of material with density 7800 kg/m 3 . The radius R = 80 mm. Determine its moment of inertia about the x-axis.

Solution:

Given ρ = 7800 kg/m 3 , R = 0.08 m

m 2 (a + c 2 ) 12

Iz =

m 2 (b + a 2 ). 12

Using the tables we have Ix =

(

)

(

)

3 1 2 2 ρ π R 2 [4 R] R 2 + ρ πR 3 R 2 10 3 5 3

I x = 0.0535 kg-m 2 .

446

x 4R

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Problem 18.101 Determine the moment of inertia of the sphere-capped cone described in Problem 18.100 about the y-axis.

Solution:

The center of mass of a half-sphere is located a distance 3 R from the geometric center of the circle. 8

(

)(

2 − ρ πR 3 3

3R 2 2 + ρ πR 3 8 3

)

(

)(

3R 2 4R + 8

)

1 3 3 2 2 2 I y = ρ π R 2 [4 R] R + ρ πR 3 R 2 [4 R] 2 + 3 5 20 5 3

(

)( )

(

)

I y = 2.08 kg-m 2 .

Problem 18.102 The circular cylinder is made of aluminum (Al) with density 2700 kg/m 3 and iron (Fe) with density 7860 kg/m 3 . Determine its moment of inertia about the x ′-axis. (See Example 18.10.)

y y9 Al z

1 I x = [(2700 kg/m 2 )π (0.1 m) 2 (0.6 m)](0.1 m) 2 2 +

600 mm

Solution:

200 mm 600 mm x, x9

1 [(7860 kg/m 2 )π (0.1 m) 2 (0.6 m)](0.1 m) 2 2

I x = 0.995 kg-m 2 .

Problem 18.103 Determine the moment of inertia of the composite cylinder described in Problem 18.102 about the y ′-axis. (See Example 18.10.)

y y9 Al

Solution:

First locate the center of mass

600 mm

200 mm

[(2700 kg/m 3 )π (0.1 m) 2 (0.6 m)](0.3 m) x =

+ [(7860 kg/m 3 )π (0.1 m) 2 (0.6 m)](0.9 m) (2700 kg/m 3 )π (0.1 m) 2 (0.6 m) + (7860 kg/m 3 )π (0.1 m) 2 (0.6 m)

600 mm x, x9

x = 0.747 m 1 1 I y = [(2700 kg/m 3 )π (0.1 m) 2 (0.6 m)]  (0.6 m) 2 + (0.1 m) 2   12  4 + [(2700 kg/m 3 )π (0.1 m) 2 (0.6 m)]( x − 0.3 m) 2 1 1 + [(7680 kg/m 3 )π (0.1 m) 2 (0.6 m)]  (0.6 m) 2 + (0.1 m) 2   12  4 + [(7680 kg/m 3 )π (0.1 m) 2 (0.6 m)](0.9 m − x ) 2 I y = 20.1 kg-m 2 .

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Problem 18.104 The homogeneous machine part is made of aluminum alloy with mass density ρ = 2800 kg/m 3 . Determine the moment of inertia of the part about the z-axis.

y 20 mm x

40 mm 120 mm

40 mm

Solution:

We divide the machine part into the 3 parts shown: (The dimension into the page is 0.04 m ) The masses of the parts are m1 = (2800)(0.12)(0.08)(0.04) = 1.075 kg, 1 m 2 = (2800) π (0.04) 2 (0.04) = 0.281 kg, 2

m 3 = (2800)π (0.02) 2 (0.04) = 0.141 kg.

0.08 m

Using Appendix C and the parallel axis theorem the moment of inertia of part 1 about the z axis is I ( z axis )1 =

0.12 m

1 m [ (0.08) 2 + (0.12) 2 ] + m1 (0.06) 2 12 1

C 2 0.12 m

= 0.00573 kg-m 2 .

+ x – 0.04 m

0.12 m

x 0.02 m

The moment of inertia of part 2 about the axis through the center C that is parallel to the z axis is 1 2 2 m2 R

= 12 m 2 (0.04) 2

The distance along the x axis from C to the center of mass of part 2 is 4(0.04)/(3π ) = 0.0170 m. Therefore, the moment of inertia of part 2 about the z axis through its center of mass that is parallel to the axis is 1 2 2 2 m 2 (0.04) − m 2 (0.0170)

= 0.000144 kg-m 2 .

Using this result, the moment of inertia of part 2 about the z axis is I ( z axis)2 = 0.000144 + m 2 (0.12 + 0.017) 2 = 0.00544 kg-m 2 . The moment of inertia of the material that would occupy the hole 3 about the z axis is I ( z axis )3 = 12 m 3 (0.02) 2 + m 3 (0.12) 2 = 0.00205 kg-m 2 . Therefore, I ( z axis) = I ( z axis)1 + I ( z axis)2 − I ( z axis)3 = 0.00911 kg-m 2 .

Problem 18.105 Determine the moment of inertia of the machine part described in Problem 18.104 about the x-axis.

Solution:

We divide the machine part into the 3 parts shown in the solution to Problem 18.104. Using Appendix C and the parallel axis theorem, the moments of inertia of the parts about the x axis are: I ( x axis)1 =

= 0.0007168 kg-m 2

1 1 I ( x axis) 2 = m 2  (0.04) 2 + (0.04) 2   12  4

20 mm x

= 0.0001501 kg-m 2

1 1 I ( x axis) 3 = m 3  (0.04) 2 + (0.02) 2   12  4

40 mm 120 mm

1 m [(0.08) 2 + (0.04) 2 ] 12 1

40 mm

= 0.0000328 kg-m 2 . Therefore, I ( x axis) = I ( x axis)1 + I ( x axis) 2 − I ( x axis) 3 = 0.000834 kg-m 2 .

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Problem 18.106 The object shown consists of steel of density ρ = 7800 kg/m 3 . Determine its moment of inertia about the axis L O through point O.

20 mm

Solution:

Part (1): The semi-cylinder of radius R = 0.02 m, height h1 = 0.01 m. Part (2): The rectangular solid L = 0.1 m by h 2 = 0.01 m by w = 0.04 m. Part (3): The semi-cylinder of radius R = 0.02 m, h1 = 0.01 m Part (4): The cylinder of radius R = 0.02 m, height h = 0.03 m. Part (1)

O 100 mm Part (2) 10 mm

Divide the object into four parts:

m1 =

ρπ R 2 h1 = 0.049 kg, 2

I1 =

m1 R 2 = 4.9 × 10 −6 kg-m 2 , 4

m 2 = ρ wLh 2 = 0.312 kg, I 2 = (1/12)m 2 ( L2 + w 2 ) + m 2 ( L /2) 2

30 mm

= 0.00108 kg-m 2 .

LO Part (3)

m 3 = m1 = 0.049 kg, 2

( 43πR ) m + I + m ( L − 43πR )

I3 = −

= 0.00041179 kg-m 2 . Part (4)

m 4 = ρπ R 2 h = 0.294 kg, I 4 = ( 12 ) m 4 ( R 2 ) + m 4 L2 = 0.003 kg-m 2 .

The composite: I L 0 = I 1 + I 2 − I 3 + I 4 = 0.00367 kg-m 2 .

Problem 18.107 Determine the moment of inertia of the object described in Problem 18.106 about the axis through the center of mass of the object parallel to L O .

20 mm O 100 mm

Solution: x =

The center of mass is located relative to L 0 is given by 10 mm

( 43πR ) + m (0.05) − m ( 0.1 − 43πR ) + m (0.1)

m1 −

30 mm

m1 + m 2 − m 3 + m 4

= 0.066 m, I c = −x 2 m + I Lo = −0.00265 + 0.00367 = 0.00102 kg-m 2 .

Problem 18.108 The thick plate consists of steel of density ρ = 15 slug/ft 3 . Determine the moment of inertia of the plate about the z-axis.

Solution:

Divide the object into three parts: Part (1) the rectangle 8 in by 16 in, Parts (2) & (3) the cylindrical cut outs. Part (1):

m1 = ρ8(16)(4) = 4.444 slugs. I 1 = (1/12)m1 (16 2 + 8 2 ) = 118.52 slug-in 2 .

y Part (2):

4 in

2 in

z Part (3):

4 in 4 in

8 in

4 in

( 121 ) = 0.4363 slug, 3

m (2 2 ) I2 = 2 + m 2 (4 2 ) = 7.854 slug-in 2 . 2

2 in x

m 2 = ρπ (2 2 )(4)

4 in

m 3 = m 2 = 0.4363 slugs, I 3 = I 2 = 7.854 slug-in 2 .

The composite: I z -axis = I 1 − 2 I 2 = 102.81 slug-in 2 I z -axis = 0.715 slug-ft 2 .

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Problem 18.109 Determine the moment of inertia of the object described in Problem 18.108 about the x-axis. y

Solution:

Use the same divisions of the object as in Problem 18.108.

( 121 )m (8 + 4 ) = 29.63 slug-in ,

Part (1):

I 1x -axis =

Part (2):

I 2 x -axis = (1/12)m 2 (3(2 2 ) + 4 2 ) = 1.018 slug-in 2 .

The composite:

4 in

2 in

I x -axis = I 1x -axis − 2 I 2 x -axis = 27.59 slug in 2

2 in x

= 0.1916 slug-ft 2 .

4 in 4 in

4 in

8 in

4 in

Problem 18.110 The Rutan Long-EZ is at the beginning of its takeoff run. Its weight is 1000 lb, and the initial thrust T exerted by its engine is 300 lb. Assume that the thrust is horizontal, and neglect the tangential forces exerted on the wheels. (a) If the acceleration of the airplane remains constant, how long will it take to reach its takeoff speed of 80 mi/h? (b) Determine the normal force exerted on the forward landing gear at the beginning of the takeoff run.

Solution: a =

6 in

The acceleration under constant thrust is

300(32.2) T = = 9.66 ft/s 2 . m 1000

W T

The time required to reach 80 mph = 117.33 ft/s is t =

7 ft

1 ft

v 117.33 = = 12.1 s. a 9.66

The sum of the vertical forces: ∑ Fy = R + F − W = 0. The sum of the moments: ∑ M = 7 F − 0.5T − 1R = 0. Solve: R =

856.25 lb, F = 143.75 lb.

Problem 18.111 At the beginning of its takeoff run, the plane is accelerating at 10 ft/s 2 . Each of the 1-ft diameter rear wheels weighs 8 lb, and the moment of inertia about its axis is 0.05 slug-ft 2 . What horizontal force is exerted on each rolling rear wheel by the runway?

6 in

Solution:

1 ft

7 ft

a We have R f

∑ M center = Rf = I α: (0.5 ft) f = (0.05 slug-ft 2 )(20 rad/s 2 ), N

If the center of the wheel is accelerating to the right at 10 ft/s 2 , its clockwise angular acceleration is α =

450

yielding f = 2 lb. 2 lb.

10 ft/s 2 10 ft/s 2 = = 20 rad/s 2 . R 0.5 ft

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Problem 18.112 A 2-kg box is subjected to a 40-N horizontal force. Neglect friction. (a) If the box remains on the floor, what is its acceleration? (b) Determine the range of values of c for which the box will remain on the floor when the force is applied.

40 N c

Solution:

a =

100 mm

(a) From Newton’s second law, 40 = (2)a, from which 40 = 20 m/s 2 . 2

(b) The sum of forces: ∑ Fy = A + B − mg = 0. The sum of the moments about the center of mass: ∑ M = 0.1B − 0.1 A −

40 N

40 c = 0. Substitute the value of B from the first equation into the second equation and solve for c:

(0.1)mg − (0.2) A c = 40

The box leg at A will leave the floor as A ≤ 0, from which c ≤

(0.1)(2)(9.81) ≤ 0.0491 m 40

B 100 mm

100 mm

for values of A ≥ 0.

Problem 18.113 The slender 2-slug bar AB is 3 ft long. It is pinned to the cart at A and leans against it at B. (a) If the acceleration of the cart is a = 20 ft/s 2 , what normal force is exerted on the bar by the cart at B? (b) What is the largest acceleration a for which the bar will remain in contact with the surface at B?

Solution:

B a

608

Newton’s second law applied to the center of mass of

the bar yields −B + A x = ma Gx , A y − W = ma Gy , − Ay

θ θ ( L cos ) + (B + A )( L sin ) = I α, 2 2 x

where a Gx , a Gy are the accelerations of the center of mass. From kinematics, a G = a A + α × rG /A − ω 2AB rG /A = 20 i ft/s 2 where α = 0, ω AB = 0 so long as the bar is resting on the cart at B and is pinned at A. Substitute the kinematic relations to obtain three equations in three unknowns:

Ax Ay

−B + A x = ma, A y − W = 0, − Ay

θ θ ( L cos ) + (B + A )( L sin ) = 0. 2 2 x

W cot θ ma − . For W = mg = 64.34 lb, θ = 60 °, 2 2 m = 2 slug, and a = 20 ft/s 2 , B = −1.43 lb, from which the bar has moved away from the cart at point B. (b) The acceleration that produces a zero normal force is Solve: B =

a = g cot θ = 18.57 ft/s 2 .

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Problem 18.114 To determine a 4.5-kg tire’s moment of inertia, an engineer lets the tire roll down an inclined surface. If it takes the tire 3.5 s to start from rest and roll 3 m down the surface, what is the tire’s moment of inertia about its center of mass?

330 mm

158

Solution:

From Newton’s second law and the angular equation

of motion, a

mg sin15 ° − f = ma, R

Rf = I α. From these equations and the relation a = Rα, we obtain mg sin15 ° a = . m + I /R 2

mg f

(1)

We can determine the acceleration from

1 s = at 2 : 2 3 =

1 a(3.5) 2 , 2

obtaining a = 0.490 m/s 2 . Then from Eq. (1) we obtain I = 2.05 kg-m 2 .

Problem 18.115 The angle θ = 30 °. At t = 0, a sphere with mass m = 10 kg and radius R = 0.1 m is placed on the inclined surface. Initially, it has clockwise angular velocity ω 0 = 50 rad/s and its center is moving at v 0 = 8 m/s up the surface. The coefficient of kinetic friction between the sphere and surface is µ k = 0.1. What are the sphere’s angular velocity and the velocity of its center at t = 1 s?

The normal force is

Solution:

N = mg cos θ,

x 308

so the friction force is µ k mg cos θ. Newton’s second law gives

∑ Fx = µ k mg cos θ − mg sin θ = ma x . The velocity of the center of the disk up the incline as a function of time is

mk N

v = v 0 + a x t. = v 0 + (µ k cos θ − sin θ) gt.

At t = 1 s, this gives v = 3.94 m/s. Initially, and for some period of time, the disk at its point of contact will be moving (slipping) relative to the surface. The motion of the center of the disk will be determined by the combined effects of the resulting friction force and the component of the sphere’s weight down the incline. The friction force also exerts a counterclockwise moment about the center, which causes the clockwise angular velocity to decrease. We can use the equations of motion to determine the velocity and angular velocity at t = 1 s. Then we need to make sure the sphere is still slipping on the surface.

452

The moment of inertia of the sphere about its center is I =

2 mR 2 . 5

Treating the clockwise direction as positive, the equation of angular motion is 2

∑ M = −µ k NR = −µ k mg cos θ R = I α = 5 mR 2α,

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18.115 (Continued) so the disk’s clockwise angular velocity as a function of time is ω = ω 0 + αt = ω0 −

5µ k g cos θ t. 2R

At t = 1 s, this gives ω = 28.8 rad/s. We observe ω R = 2.88 m/s < v , so the sphere is still slipping at t = 1 s.

that

28.8 rad/s (clockwise), 3.94 m/s up the inclined surface.

Problem 18.116 Model the excavator’s arm ABC as a single rigid body. Its mass is 1200 kg, and the moment of inertia about its center of mass is I = 3600 kg-m 2 . The angular velocity of the arm is zero and its angular acceleration is 1 rad/s 2 counterclockwise. What force does the vertical hydraulic cylinder exert on the arm at B?

y C

B 2.4 m

Solution: d =

The distance from A to the center of mass is

3.0 m

(3.4) 2 + (3) 2 = 4.53 m.

The moment of inertia about A is

1.7 m

I A = I + d 2 m = 28,270 kg-m 2 . From the equation of angular motion: 1.7 B − 3.4 mg = I Aα. Substitute α = 1.0 rad/s 2 , to obtain B = 40,170 N. B

Problem 18.117 Model the excavator’s arm ABC as a single rigid body. Its mass is 1200 kg, and the moment of inertia about its center of mass is I = 3600 kg-m 2 . The angular velocity of the arm is 2 rad/s counterclockwise and its angular acceleration is 1 rad/s 2 counterclockwise. What are the components of the force exerted on the arm at A?

y C

B 2.4 m A

Solution:

3.0 m x

The acceleration of the center of mass is

 i j k   a G = α × rG /A − ω 2 rG /B =  0 0 α  − ω 2 (3.4 i + 3 j)  3.4 3 0   

1.7 m

= −16.6 i − 8.6 j m/s 2 . From Newton’s second law: A x = ma Gx = −19,900 N, A y + B − mg = ma Gy . From the solution to Problem 18.132, B = 40,170 N, from which A y = −38,720 N.

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Problem 18.118 To decrease the angle of elevation of the stationary 200-kg ladder, the gears that raised it are disengaged, and a fraction of a second later a second set of gears that lower it are engaged. At the instant the gears that raised the ladder are disengaged, what is the ladder’s angular acceleration and what are the components of force exerted on the ladder by its support at O? The moment of inertia of the ladder about O is I O = 14,000 kg-m 2 , and the coordinates of its center of mass at the instant the gears are disengaged are x = 3 m, y = 4 m.

Solution: α = −

The moment about O, −mg x = I oα, from which

(200)(9.81)(3) = −0.420 rad/s 2 . 14,000

The acceleration of the center of mass is

y (x , y)

 i j k    a G = α × rG /O − ω 2 rG /O =  0 0 α  = −4αi + 3α j  3 4 0    2 a G = 1.68i − 1.26 j (m/s ).

x Fx

From Newton’s second law: Fx = ma Gx = 336 N, Fy − mg = ma Gy , from which Fy = 1710 N.

Problem 18.119 The 12-lb horizontal bar is connected to the 8-lb disk by a smooth pin at A. The system is released from rest in the position shown. What are the angular accelerations of the bar and disk at that instant?

O 3 ft

Solution:

1 ft

The masses of the disk and bar are

mD =

WD 8 lb = = 0.248 slug, 32.2 ft/s 2 g

mB =

WB 12 lb = = 0.373 slug. 32.2 ft/s 2 g

Let L = 3 ft and R = 1 ft. In terms of the counterclockwise angular acceleration of the bar, the acceleration of point A is

y Ay O

a A = Lα B j. We write the acceleration of the center of mass G of the disk in terms of the acceleration of point A: a G = a A + α D × rG /A − ω D 2 rG /A : a Gx i + a Gy j = Lα B j +

i j k 0 0 αD R 0 0

− 0.

This yields the scalar equations a Gx = 0, a Gy = Lα B + Rα D . The moments of inertia of the disk about G and the bar about O are 1 1 I D = m D R 2 = 0.124 slug-ft 2 , I B = m B L2 = 1.12 slug-ft 2 . 2 3

454

For the disk we write the equations of motion

∑ Fx = − Ax = m D a Gx , ∑ Fy = − Ay − W D = m D a Gy , ∑ M G = RAy = I Dα D , and for the bar we write the equation 1

∑ M O = LAy − 2 LWB = I Bα B . Solving these equations, we obtain a Gx = 0, a Gy = − 35.4 ft/s 2 , α D = 6.44 rad/s 2 , α B = −14.0 rad/s 2 , A x = 0, A y = 0.800 lb. Bar: 14.0 rad/s 2 (clockwise). Disk: 6.44 rad/s 2 (counterclockwise).

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Problem 18.120 The slender 10-lb bar with length l = 3 ft is released from rest in the position shown (θ = 35 °). The floor and wall are smooth. What is the magnitude of the acceleration of the bar’s center of mass at the instant it is released?

Solution:

A FA

u u

mg B

obtaining the scalar equations

The bar’s mass is m =

10 lb = 0.311 slug. 32.2 ft/s 2

Denote the center of mass of the bar by G. The moment of inertia of the bar about its center of mass is I =

1 αL cos θ, (6) 2 1 a Gy = a A + αL sin θ. (7) 2 a Gx =

1 mL2 = 0.233 slug-ft 2 . 12

We have the equations of motion

∑ Fx = FA = ma Gx , (1) ∑ Fy = FB − mg = ma Gy , (2) 1 1 ∑ M G = FB 2 L sin θ − FA 2 L cos θ = I α. (3)

Substituting the value of a A from Eq. (5) into Eq. (7) yields 1 a Gy = − αL sin θ. (8) 2 Equations (1)-(3), (6), and (8) provide 5 equations in terms of a Gx , a Gy , α, FA , FB . Solving them, we obtain a Gx = 11.3 ft/s 2 , a Gy = −7.95 ft/s 2 , α = 9.23 rad/s 2 , FA = 3.52 lb, FB = 7.53 lb. The magnitude of the acceleration of the center of mass is a G = 13.9 ft/s 2 .

We write the acceleration of point B in terms of the acceleration of point A, a B = a A + α × r B /A − ω 2 r B /A : aBi = a A j +

i j k 0 0 α L sin θ −L cos θ 0

− 0,

obtaining the scalar equations a B = αL cos θ, (4) 0 = a A + αL sin θ. (5) We also write the acceleration of the center of mass in terms of the acceleration of point A, a G = a A + α × rG /A − ω 2 rG /A : a Gx i + a Gy j = a A j +

i 0

j 0

1 1 L sin θ − L cos θ 2 2

k α

− 0,

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Problem 18.121 The slender 5-lb bar with length l = 3 ft is released from rest in the position shown (θ = 35 °). The coefficient of kinetic friction between the bar and the floor and wall is µ k = 0.1. What is the bar’s angular acceleration at the instant it is released?

Solution:

mkFA

FA u

mg obtaining the scalar equations

mkFB

a B = αL cos θ, (4) 0 = a A + αL sin θ. (5)

The bar’s mass is

We also write the acceleration of the center of mass in terms of the acceleration of point A,

10 lb m = = 0.311 slug. 32.2 ft/s 2

a G = a A + α × rG /A − ω 2 rG /A :

Denote the center of mass of the bar by G. The moment of inertia of the bar about its center of mass is I =

a Gx i + a Gy j = a A j +

1 mL2 = 0.233 slug-ft 2 . 12

We have the equations of motion

i 0

1 1 L sin θ − L cos θ 2 2

∑ Fx = FA − µ k FB = ma Gx , (1)

obtaining the scalar equations

∑ Fy = FB + µ k FA − mg = ma Gy , (2)

a Gx =

1 −µ k FA L sin θ = I α. (3) 2

− 0,

a Gy = a A +

1 αL sin θ. (7) 2

Substituting the value of a A from Eq. (5) into Eq. (7) yields

We write the acceleration of point B in terms of the acceleration of point A,

1 a Gy = − αL sin θ. (8) 2

a B = a A + α × r B /A − ω 2 r B /A :

Equations (1)-(3), (6), and (8) provide 5 equations in terms of a Gx , a Gy , α, FA , FB . Solving them, we obtain a Gx = 8.03 ft/s 2 , a Gy = −5.62 ft/s 2 , α = 6.54 rad/s 2 , FA = 3.29 lb, FB = 7.92 lb.

aBi = a A j +

i j k 0 0 α L sin θ −L cos θ 0

− 0,

6.54 rad/s 2 (ounterclockwise).

Problem 18.122 Bar AB rotates with a constant angular velocity of 10 rad/s in the counterclockwise direction. The masses of the slender bars BC and CDE are 2 kg and 3.6 kg, respectively. The y-axis points upward. Determine the components of the forces exerted on bar BC by the pins at B and C at the instant shown.

y B 400 mm A

10 rad/s

700 mm

456

k α

1 αL cos θ, (6) 2

∑ M G = FB 2 L sin θ − µ k FB 2 L cos θ − FA 2 L cos θ

j 0

400 mm

700 mm

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18.122 (Continued) Solution:

The velocity of point B is

By Bx

 i j k   v B = ω AB × rB =  0 0 10   0 0.4 0    = −0.4(10) i = −4 i (m/s).

The velocity of point C is  i j k  v C = v B + ω BC × rC /B = −4 i +  0 0 ω BC  0.7 −0.4 0 

WBC Cx Cy

     

WCE

= −4 i + 0.4 ω BC i + 0.7ω BC j (m/s). From the constraint on the motion at point C , v C = v C j. Equate components: 0 = −4 + 0.4 ω BC , v C = 0.7ω BC , from which ω BC = 10 rad/s, v C = 7 m/s. The velocity at C in terms of the angular velocity ω CDE , v C = v D + ω CDE × rC /D  i j k  = 0 +  0 0 ω CDE  −0.4 0 0  from which ω CDE = −

   = −0.4 ω CDE j,   

7 = −17.5 rad/s. 0.4

The acceleration of point B is a B = −ω 2AB rB = −(10 2 )(0.4) j = −40 j (m/s 2 ). 2 r The acceleration at point C is a C = a B + α BC × rC /B − ω BC C /B .

 i j k   2 (0.7 i − 0.4 j) (m/s 2 ). 0 a C = −40 j +  0 α BC  − ω BC   0.7 −0.4 0   2 ) i + (− 40 + 0.7α 2 2 a C = +(0.4α BC − 0.7ω BC BC + 0.4 ω BC ) j (m/s ). The acceleration in terms of the acceleration at D is  i j k   2 0 αCDE  − ω CDE (−0.4 i) a C =  0   −0.4 0 0   2 = −0.4αCDE j + 0.4 ω CDE i.

Problem 18.123 At the instant shown, the arms of the robotic manipulator have constant counterclockwise angular velocities ω AB = −0.5 rad/s, ω BC = 2 rad/s, and ω CD = 4 rad/s. The mass of arm CD is 10 kg, and its center of mass is at its midpoint. At this instant, what force and couple are exerted on arm CD at C ?

Equate components and solve: α BC = 481.25 rad/s 2 , αCDE = −842.19 rad/s 2 . The acceleration of the center of mass of BC is  i j k  a G = −40 j +  0 0 α BC  0.35 −0.2 0 

from which a G = 61.25i + 148.44 j (m/s 2 ) The equations of motion: B x + C x = m BC a Gx , B y + C y − m BC g = m BC a Gy , where the accelerations a Gx , a Gy are known. The moment equation, 0.35C y + 0.2C x − 0.2 B x − 0.35B y = I BC α BC , where α BC , is known, and I BC =

( 121 )m L

where

ID =

2 BC

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= 0.1083 kg-m 2 , 0.4C y − 0.15mCE g = I D αCE ,

( 121 )m L CE

2 CE

+ (0.15) 2 mCE = 0.444 kg-m 2 ,

is the moment of inertia about the pivot point D, and 0.15 m is the distance between the point D and the center of mass of bar CDE. Solve these four equations in four unknowns by iteration: B x = −1959 N, B y = 1238 N, C x = 2081 N, C y = −922 N.

250

m 0m

308 A

   − ω 2 (0.35i − 0.2 j), BC   

208

x 250 mm

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18.123 (Continued) Solution:

Ay Ax

The relative vector locations of B, C , and D are

rB /A = 0.3( i cos30 ° + j sin 30 °) = 0.2598i + 0.150 j (m),

rC /B = 0.25( i cos 20 ° − j sin 20 °)

Ax mg B

= 0.2349 i − 0.08551 j (m),

308

rD /C = 0.25i (m). The acceleration of point B is a B = −ω 2AB rB /A = −(0.5 2 )(0.3cos30 °i + 0.3sin 30 ° j), a B = −0.065i − 0.0375 j (m/s 2 ).

For the arm CD the three equations of motion in three unknowns are C y − mCD g = mCD a Gy , C x = mCD a Gx , M − 0.125C y = 0,

The acceleration at point C is

which have the direct solution: 2 r 2 a C = a B − ω BC C /B = a B − ω BC (0.2349 i − 0.08551 j).

a C = −1.005i + 0.3045 j (m/s 2 ).

C y = 101.15 N, C x = −30.05 N.

The acceleration of the center of mass of CD is

M = 12.64 N-m,

2 (0.125i) (m/s 2 ), a G = a C − ω CD

where the negative sign means a direction opposite to that shown in the free body diagram.

from which a G = −3.005i + 0.3045 j (m/s 2 ).

Problem 18.124 Each bar is 1 m in length and has a mass of 4 kg. The inclined surface is smooth. If the system is released from rest in the position shown, what are the angular accelerations of the bars at that instant?

Solution:

For convenience, denote θ = 45 °, β = 30 °, and L = 1 m. The acceleration of point A is

 i  a A = αOA × r A /O =  0  L cos θ 

j 0

k αOA

L sin θ

458

308

  .   

a A = αOA (−iL sin θ + jL cos θ) (m/s 2 ). C

The acceleration of A is also given by

j 0

k α AB

L sin θ

D mg

  .   

a A = a B − iα AB L sin θ − jα AB L cos θ (m/s 2 ).

The acceleration of the center of mass of AB is

From the constraint on the motion at B, Equate the expressions for the acceleration of A to obtain the two equations:

a GAB = a A + α AB × rGAB /A

(1) −αOA L sin θ = a B cos β − α AB L sin θ, and (2) αOA L cos θ = a B sin β − α AB L cos θ.

458

Cx 125 mm

a A = a B + α AB × r A /B .  i  a A = a B +  0  −L cos θ 

    i j k     = aA +  0 0 α AB  ,     L cos θ L sin θ  − 0   2 2  Lα AB α AB L a GAB = a A + sin θ i + cos θ j (m/s 2 ), 2 2

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18.124 (Continued) from which

(7) − A y − mg + B cos β = ma GABy ,

(3) a GABx = −αOA L sin θ +

Lα AB sin θ (m/s 2 ), 2

(8) A y

Lα AB (4) a GABy = αOA L cos θ + cos θ. 2

−B

The equations of motion for the bars: for the pin supported left bar: (5) A y L cos θ − A x L sin θ − mg where I OA =

( L2 )cos θ = I α , OA

( L2 )cos θ sin β = I

where I GAB =

CAB α AB ,

( 121 )mL = ( 13 ) kg-m . 2

These eight equations in eight unknowns are solved by iteration: A x = −19.27 N, A y = 1.15 N, αOA = 0.425 rad/s 2 , α AB = −1.59 rad/s 2 , B = 45.43 N, a GABx = −0.8610 m/s 2 , a GABy = −0.2601 m/s 2 .

( mL3 ) = 43 kg-m . 2

( L2 )cos θ + A ( L2 )sin θ + B( L2 )sin θ cos β

The equations of motion for the right bar: (6) − A x − B sin β = ma GABx ,

Problem 18.125 Each bar is 1 m in length and has a mass of 4 kg. The inclined surface is smooth. If the system is released from rest in the position shown, what is the magnitude of the force exerted on bar OA by the support at O at that instant? O

Solution:

B 308

The acceleration of the center of mass of the bar OA is

 i   0 a GOA = αOA × rG /A = a A +   L cos θ   2  a GOA = −

458

j 0

k αOA

L sin θ 2

   ,    

L sin θ L cos θ αOA i + αOA j (m/s 2 ). 2 2

Fy mg

The equations of motion: Fx + A x = ma GOAx , Fy + A y − mg = ma GOAy .

Use the solution to Problem 18.140: θ = 45 °, α GA = A x = −19.27 N, m = 4 kg, from which Fx = 18.67 N, Fy = 38.69 N,

0.425 rad/s 2 ,

from which F =

Fx2 + Fy2 = 42.96 N.

Problem 18.126* The fixed ring gear lies in the horizontal plane. The hub and planet gears are bonded together. The mass and moment of inertia of the combined hub and planet gears are m HP = 130 kg and I HP = 130 kg-m 2 . The moment of inertia of the sun gear is I S = 60 kg-m 2 . The mass of the connecting rod is 5 kg, and it can be modeled as a slender bar. If a 1 kN-m counterclockwise couple is applied to the sun gear, what is the resulting angular acceleration of the bonded hub and planet gears?

Planet gear Hub gear Connecting rod

140 mm

340 mm 240 mm 720 mm

Sun gear Ring gear

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18.126* (Continued) Solution:

The moment equation for the sun gear is

Hub & Planet Gears

For the hub and planet gears:

Q R

(1) M − 0.24 F = I s α s . F

(2) (0.48)α HP = −0.24α s , M

(3) F − Q − R = m HP (0.14)(−α HP ),

Sun Gear

(4) (0.14)Q + 0.34 F − I HP (−α HP ).

Connecting Rod

For the connecting rod: These six equations in six unknowns are solved by iteration:

(5) (0.58) R = I CRαCR , 1 where I CR = m (0.58 2 ) = 0.561 kg-m 2 . 3 GR (6) (0.58)αCR = −(0.14)α HP .

( )

F = 1482.7 N, α s = 10.74 rad/s 2 , α HP = −5.37 rad/s 2 , Q = 1383.7 N, R = 1.25 N, αCR = 1.296 rad/s 2 .

Problem 18.127* The system is stationary at the instant shown. The net force exerted on the piston by the exploding fuel–air mixture and friction is 5 kN to the left. A clockwise couple M = 200 N-m acts on the crank AB. The moment of inertia of the crank about A is 0.0003 kg-m 2 . The mass of the connecting rod BC is 0.36 kg, and its center of mass is 40 mm from B on the line from B to C. The connecting rod’s moment of inertia about its center of mass is 0.0004 kg-m 2 . The mass of the piston is 4.6 kg. What is the piston’s acceleration? (Neglect the gravitational forces on the crank and connecting rod.)

125 m

B 408 A

Equate the two expressions for the acceleration of point B, note ω AB = ω BC = 0, and separate components:

Solution:

(1) −0.05α AB sin 40 ° = a C − 0.125α BC sin β ,

From the law of sines:

(2) 0.05α AB cos 40 ° = −0.125α BC cos β .

sin β sin 40 ° = , 0.05 0.125

The acceleration of the center of mass of the connecting rod is 2 r a GCR = a C + α BC × rGCR /C − ω BC GCR /C ,

from which β = 14.9 °. The vectors

 i j k    a GCR = a C i +  0 0 α BC   −0.085cos β 0.085sin β 0   2 (− 0.085cos β i + 0.085sin β j) (m/s 2 ), − ω BC

rB /A = 0.05( i cos 40 ° + j sin 40 °)rB /A = 0.0383i + 0.0321 j (m). rB /C = 0.125(−i cos β + j sin β ) (m). rB /C = −0.121i + 0.0321, (m).

from which

The acceleration of point B is

(3) a GCRx = a C − 0.085α BC sin β (m/s 2 ),

a B = α AB × rB /A − ω 2AB rB /A ,  i j k  a B =  0 0 α AB  0.0383 0.0321 0 

(4) a GCRy = −0.085α BC cos β (m/s 2 ).

     

The equations of motion for the crank: (5) B y (0.05cos 40 °) − B x (0.05sin 40 °) − M = I Aα AB

− ω 2AB (0.0383i + 0.0321 j) (m/s 2 ).

For the connecting rod:

The acceleration of point B in terms of the acceleration of point C is  i j k  a B = a C + α BC × rB /C = a C i +  α BC 0 0  −0.121 0.0321 0 

     

(6) −B x + C x = mCR a GCRx (7) −B y + C y = mCR a GCRy (8) C y (0.085cos β ) + C x (0.085sin β ) + B x (0.04 sin β ) + B y (0.04 cos β ) = I GCRα BC

2 (− 0.121i + 0.0321 j) (m/s 2 ). − ω BC

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18.127* (Continued) For the piston:

(9) −C x − 5000 = m P a C . These nine equations in nine unknowns are solved by iteration:

0.05

0.125 40

α AB = 1255.7 rad/s 2 , α BC = −398.2 rad/s 2 ,

Bx Cx

a GCRx = −44.45 m/s 2 , a GCRy = 32.71 m/s 2 , Cx

B y = 1254.6 N, B x = −4739.5 N,

5000 N

Bx Ay

C x = −4755.5 N, C y = 1266.3 , N a C = −53.15 m/s 2 .

Problem 18.128* If the crank AB described in Problem 18.127 has a counterclockwise angular velocity of 2000 rpm at the instant shown, what is the piston’s acceleration due to the couple M and the force exerted by the fuel–air mixture?

125 m

m 50

B 408 A

Solution:

The angular velocity of AB is

( )

2π ω AB = 2000 = 209.44 rad/s. 60 The angular velocity of the connecting rod BC is obtained from the expressions for the velocity at point B and the known value of ω AB :  i j k  v B = ω AB × rB /A =  0 0 ω AB  0.05cos 40 ° 0.05sin 40 ° 0 

  .   

v B = −0.05sin 40 °ω AB i + 0.05cos 40 °ω AB j (m/s).  i j k  v B = v C i +  ω BC 0 0  −0.125cos β 0.125sin β 0 

  ,   

v B = v C i − 0.125sin βω BC i − 0.125cos βω BC j (m/s). From the j component, 0.05cos 40 °ω AB = −0.125cos βω BC , from which ω BC = −66.4 rad/s. The nine equations in nine unknowns obtained in the solution to Problem 18.127 are (1) −0.05α AB sin 40 ° − 0.05ω 2AB cos 40°

(8) C y (0.085cos β ) + C x (0.085sin β ) + B x (0.04 sin β ) + B y (0.04 cos β ) = I GCRα BC . (9) −C x − 5000 = m P a C . These nine equations in nine unknowns are solved by iteration: α AB = −39,386.4 rad/s 2α BC = 22,985.9 rad/s 2 , a GCRx = −348.34 m/s 2 , a GCRy = −1984.5 m/s 2 , B y = 1626.7 N, B x = −3916.7 N, C x = −4042.1 N, C y = 912.25 N, a c = −208.25 (m/s 2 ).

2 cos β , = a C − 0.125α BC sin β + 0.125ω BC

(2) 0.05α AB cos 40 ° − 0.05ω 2AB sin 40 ° 2 sin β , = −0.125α BC cos β − 0.125ω BC 2 cos β (m/s 2 ), (3) a GCRx = a C − 0.085α BC sin β + 0.085ω BC 2 sin β (m/s 2 ), (4) a GCRy = −0.085α BC cos β − 0.085ω BC

(5) B y (0.05cos 40 °) − B x (0.05sin 40 °) − M = I Aα AB , (6) −B x + C x = mCR a GCRx , (7) −B y + C y = mCR a GCRy ,

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Chapter 19 Problem 19.1 Two identical homogeneous disks have mass m = 10 kg and radius R = 0.4 m. Disk (a) is not rotating and is moving at 5 m/s. Disk (b) is rotating about a fixed support at its center. Determine the rate of rotation of disk (b) that causes the disks to have the same kinetic energy. What is the kinetic energy in terms of N-m and ft-lb?

Solution:

The kinetic energy of disk A is

1 1 T A = mv 2 = (10 kg)(5 m/s) 2 = 125 N-m. 2 2 The moment of inertia of disk B about its center is I =

1 1 mR 2 = (10 kg)(0.4 m) 2 = 0.8 kg-m 2 . 2 2

The kinetic energy of disk B is TB =

1 2 1 I ω = (0.8 kg-m 2 ) ω 2 = (0.4 kg-m 2 ) ω 2 . 2 2

Equating the kinetic energies and solving for the angular velocity of disk B, we obtain ω = 17.7 rad/s (2.81 rev/s). The kinetic energy in ft-lb is T = 125 N-m = 125 N-m (a)

(b)

Problem 19.2 The moment of inertia of the rotor of a medical centrifuge used in medical research is I = 0.4 kg-m 2 . The rotor starts from rest and the motor exerts a constant torque of 0.6 N-m on it for 30 revolutions. (a) How much work does the motor do on the rotor? (b) Use work and energy to determine the rotor’s angular velocity (in rpm) when it has rotated through 30 revolutions.

1 lb ( 4.448 )( 1 ft ) = 92.2 ft-lb. N 0.3048 m

17.7 rad/s, T = 125 N-m = 92.2 ft-lb.

Solution: (a) The angle in radians through which the rotor rotates in 30 revolutions is θ = (30 revolutions)

π radians ( 12revolution ) = 188 rad.

The work done by the constant torque is U = (0.6 N-m)θ = (0.6 N-m)(188 rad) = 113 N-m. (b) Applying the principle of work and energy to the rotor, 1 2 Iω : 2 1 113 N-m = (0.4 kg-m 2 )ω 2 , 2 U =

we obtain ω = 23.8 rad/s (227 rpm). (a) 113 N-m. (b) 227 rpm.

Problem 19.3 The 20-kg disk is at rest when the constant 10 N-m counterclockwise couple is applied. Determine the disk’s angular velocity (in rpm) when it has rotated through four revolutions (a) by applying the equation of angular motion ΣM = I α, and (b) by applying the principle of work and energy.

462

10 N-m 0.25 m

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19.3 (Continued) Solution:

(b) Applying the principle of work energy

(a) First we find the angular acceleration

W =

∑ M = Iα 1 (20 kg)(0.25 m) 2 α ⇒ α = 16 rad/s 2 Next we will integrate the angular acceleration to find the angular velocity.

(10 N-m)(4 rev)

(10 N-m) =

ω ω dω ω2 = α ⇒ ∫ ω dω = ∫ α dθ ⇒ = αθ 0 0 dθ 2

ω =

2αθ =

2(16 rad/s 2 )(4 rev)

1 2 Iω 2

( 2πrevrad ) = 12  12 (20 kg)(0.25 m)  ω 2

ω = 28.4 rad/s

( 2πrevrad )( 60mins ) = 271 rpm

ω = 271 rpm.

rev ( 2πrevrad )( 60mins ) = 271 min

ω = 271 rpm.

Problem 19.4 A space station has a moment of inertia about a principal axis of 2.00E8 kg-m 2 . It is not rotating at t = 0 when its attitude control gyroscopes apply a constant 25 N-m torque about that axis until the station has rotated one degree. No torque is applied while the station rotates an additional 88 °, and then an opposite 25 N-m torque is applied until the station has rotated through a total angle of 90 °. (a) Use work and energy to determine the maximum angular velocity the station attains. (b) How long does it take to make the 90° attitude change?

Solution: (a) One degree in radians is π rad θ = (1 °) = 0.0175 rad. 180 ° The work done by the constant torque while the station rotates one degree is

(

)

U = M θ = (25 N-m)(0.0175 rad) = 0.436 N-m. Applying the principle of work and energy to the station, 1 2 Iω : 2 1 0.436 N-m = (2.00E8 kg-m 2 )ω 2 , 2 U =

we obtain ω = 6.61E-5 rad/s. This is the maximum angular velocity. (b) During the initial phase of the motion, the angular acceleration is α =

M 25 N-m = = 1.25E-7 rad/s 2 . I 2.00E8 kg-m 2

The angular velocity as a function of time is ω = αt. From our result for the maximum angular velocity, the time required for the station to rotate the first degree is t =

ω 6.61E-5 rad/s = = 528 s. α 1.25E-7 rad/s 2

The last, decelerating phase of the motion will take the same amount of time. We need to know how long the middle phase, when the station is rotating at the maximum angular velocity for 88 degrees, takes. We set θ = ωt:

(

)

π rad 88 ° = (6.61E-5 rad/s)t, 180 ° obtaining t = 23,300 s. The total time required is t = 2(528 s) + 23,300 s = 24,300 s (6.75 hours). (a) 6.61E-5 rad/s. (b) 24,300 s (6.75 hours).

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Problem 19.5 The helicopter’s rotor starts from rest. Suppose that its engine exerts a constant 1200 ft-lb couple on the rotor and aerodynamic drag is negligible. The rotor’s moment of inertia is I = 400 slug-ft 2 . (a) Use work and energy to determine the magnitude of the rotor’s angular velocity when it has rotated through 5 revolutions. (b) What average power is transferred to the rotor while it rotates through 5 revolutions?

Solution: (a) U = (1200 ft-lb)(10 π rad) =

1 (400 lb-s 2 -ft)ω 2 2

Source: Courtesy of HeliHead/Shutterstock.

ω = 13.7 rad/s. (b) To find the average power we need to know the time 1200 ft-lb = (400 lb-s 2 -ft) α α = 3 rad/s 2 , ω = (3 rad/s 2 )t, θ =

1 (3 rad/s 2 ) t 2 2

10 π rad =

1 (3 rad/s 2 )t 2 ⇒ t = 4.58 s 2

Power =

(1200 ft-lb)(10 π rad) U = 4.58 s t

= 8240 ft-lb/s

hp ( 5501 ft-lb/s ) = 15.0 hp.

Problem 19.6 The helicopter’s rotor starts from rest. The moment exerted on it (in N-m) is given as a function of the angle through which it has turned in radians by M = 6500 − 20θ. The rotor’s moment of inertia is I = 540 kg-m 2 . Determine the rotor’s angular velocity (in rpm) when it has turned through 10 revolutions.

Solution:

We will integrate to find the work.

W = ∫ Md θ = ∫ =

10(2 π ) 0

π) (6500 − 20 θ) d θ = [6500 θ − 10 θ 2 ]10(2 0

Source: Courtesy of HeliHead/Shutterstock.

3.69 × 10 5 N-m

Using the principle of work energy we can find the angular velocity. W =

1 2 Iω ⇒ ω = 2

ω = 37.0 rad/s

2W = I

2(3.69 × 10 5 N-m) = 37.0 rad/s 540 kg-m 2

( 21πrev )( 60 s ) = 353 rpm. rad min

ω = 353 rpm.

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Problem 19.7 The SpaceX cargo capsule is not rotating at time t = 0 when its attitude control thruster starts exerting a constant force T = 400 N. Point A has coordinates x = 1.4 m, y = 1.7 m. The capsule’s moment of inertia about its center of mass is I = 32,200 kg-m 2 . Use work and energy to determine the angle through which the capsule has rotated when its angular velocity reaches 4 °/s.

Solution: T

208 1.7 m

G 1.4 m

Four degrees per second is 4 deg/s

( π180rad° ) = 0.0698 rad/s.

The counterclockwise moment of the force about the center of mass is M = (1.7 m)T cos 20 ° − (1.4 m)T sin 20° = 447 N-m. Applying work and energy,

Source: Courtesy of NASA Photo/Alamy Stock Photo.

Mθ =

y T A

208

(447 N-m)θ =

1 2 Iω : 2 1 (32,200 kg-m 2 )(0.0698 rad/s) 2 , 2

we obtain θ = 0.175 rad (10.0 °).

Problem 19.8 The 6-kg bar is released from rest in the horizontal position 1 and falls to position 2. (a) Use conservation of energy to determine the magnitude of the bar’s angular velocity when it is in position 2. (b) What force is exerted on the bar by the pin at A when the bar is in position 2?

10.0 °.

Solution: (a) The bar’s moment of inertia about A is IA =

1 2 1 mL = (6 kg)(2 m) 2 = 8 kg-m 2 . 3 3

2m 1

x Datum

mg 2 Placing the datum at the level of position 1, the potential energy associated with the bar’s weight when it is at position 1 is V1 = 0, and L at position 2 it is V2 = −mg . Applying conservation of energy, 2 V1 + T1 = V2 + T2 : 2

0 + 0 = −mg

L 1 + I Aω 2 , 2 2

we obtain ω = 3.84 rad/s.

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19.8 (Continued) (b) Consider the free-body diagram of the bar when it is in position 2:

The moment about A is zero, so the bar’s angular acceleration is zero. The acceleration of the center of mass due to its circular motion is a =

ω2 j. 1 L 2

Applying Newton’s second law,

∑ Fx = Ax = ma x = 0, mg

∑ Fy = Ay − mg = ma y = m( L ), 2ω 2

we obtain A x = 0, A y = 147 N. (a) 3.84 rad/s. (b) 147j (N).

Problem 19.9 The 20-lb bar is released from rest in the horizontal position 1 and falls to position 2. In addition to the force exerted on it by its weight, it is subjected to a constant counterclockwise couple M = 30 ft-lb. Determine the bar’s counterclockwise angular velocity in position 2.

Solution:

T1 + V1 + U 12 = T2 + V2 0 + 0 + Mθ =

(

)

L 1 1 2 2 mL ω − mg sin θ 2 2 3 3g 6Mθ sin θ + mL2 L

⇒ ω = Thus

4 ft 1

A M

We will use the energy equation in the form

6(30 ft-lb)

ω =

408

(

40 ° π rad ) ( 180 3(32.2 ft/s ) ° sin (40 °) = 5.31 rad/s. + 2

)

20 lb-s (4 ft) 2 32.2 ft

4 ft

ω = 5.31 rad/s.

M 2

Problem 19.10 The dimensions R = 0.5 ft and L = 2 ft. The object consists of a 4-lb slender bar bonded to a 10-lb disk. It is released from rest in the horizontal position 1 and falls to position 2. Use conservation of energy to determine the magnitude of its angular velocity when it is in position 2. y

Solution:

The masses of the bar and disk are

mB =

WB 4 lb = = 0.124 slug, 32.2 ft/s 2 g

mD =

WD 10 lb = = 0.311 slug. g 32.2 ft/s 2

The bar’s moment of inertia about A is IB =

1 m L2 = 0.166 slug-ft 2 . 3 B

Using the parallel-axis theorem, the moment of inertia of the disk about A is

R 1 458

x ID =

1 m R 2 + ( L + R) 2 m D = 1.98 slug-ft 2 . 2 D

The moment of inertia of the bonded object about A is I A = I B + I D = 2.15 slug-ft 2 .

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19.10 (Continued) y

Placing the datum at the level of position 1, the potential energy in position 1 is V1 = 0, and in position 2 it is

1 A

Datum

( 12 L sin 45° )W − (L + R)sin 45°W = −20.5 ft-lb.

V2 = −

Applying conservation of energy, V1 + T1 = V2 + T2 :

0 + 0 = V2 +

1 I ω 2, 2 A

we obtain ω = 4.37 rad/s.

4.37 rad/s.

Problem 19.11* The dimensions R = 0.5 ft and L = 2 ft. The object consists of a 4-lb slender bar bonded to a 10-lb disk. It is released from rest in the horizontal position 1 and falls to position 2. What force is exerted on the object by the pin at A when it is in position 2?

y L R 1

458

Solution:

The masses of the bar and disk are

mB =

WB 4 lb = = 0.124 slug, 32.2 ft/s 2 g

mD =

WD 10 lb = = 0.311 slug. g 32.2 ft/s 2

The bar’s moment of inertia about A is IB =

1 m L2 = 0.166 slug-ft 2 . 3 B

Using the parallel-axis theorem, the moment of inertia of the disk about A is 1 I D = m D R 2 + ( L + R) 2 m D = 1.98 slug-ft 2 . 2 The moment of inertia of the bonded object about A is I A = I B + I D = 2.15 slug-ft 2 .

Applying conservation of energy, V1 + T1 = V2 + T2 : 0 + 0 = V2 +

we obtain ω = 4.37 rad/s.

The radial distance from A to the center of mass G of the combined object is 1

1 A

WD 2

1 I ω 2, 2 A

Datum

r = 2

Lm B + ( L + R)m D mB + mD

= 2.07 ft.

WB G

Placing the datum at the level of position 1, the potential energy in position 1 is V1 = 0, and in position 2 it is

( 12 L sin 45° )W − (L + R)sin 45°W = −20.5 ft-lb.

V2 = −

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19.11* (Continued) To determine the forces at A, we need to determine the acceleration of the center of mass G and apply Newton’s second law. At the instant shown, the angular equation of motion is ∑MA =

1 LW B cos 45 ° + ( L + R)W D cos 45 ° = I Aα, 2

yielding the counterclockwise angular acceleration α = 9.56 rad/s 2 . The polar components of the acceleration of the center of mass are a θ = r α = 19.8 ft/s 2 ,

a r = −r ω 2 = −39.6 ft/s 2 .

Writing Newton’s second law in the transverse and radial directions, we have

∑ Fθ = Ax cos 45° − Ay cos 45° + W D cos 45° + WB cos 45° = (m B + m D )a θ ,

∑ Fr = − Ax cos 45° − Ay cos 45° + W D cos 45° + WB cos 45° = (m B + m D )a r . Solving, we obtain A x = 18.3 lb, A y = 20.1 lb. A x = 18.3 lb, A y = 20.1 lb.

Problem 19.12 The mass of each box is 4 kg. The radius of the pulley is 120 mm and its moment of inertia is 0.032 kg-m 2 . The surfaces are smooth. If the system is released from rest, how fast are the boxes moving when the left box has moved 0.5 m to the right?

Solution: Let m = 4 kg, I = 0.0322 kg-m 2 , and R = 0.12 m. If v is the magnitude of the velocity of the boxes toward the right, the clockwise angular velocity of the pulley is ω = v /R. As the conservative system move from position 1 to position 2, the only change in its potential energy comes from the change in height of the right box. If we place the datum for the right box is at its initial position, so that V1 = 0, and the boxes move a distance d, the potential energy is V2 = −mgd sin 30 °. Applying conservation of energy, we have V1 + T1 = V2 + T2 :

308

0 + 0 = −mgd sin 30° + 2

( 12 mv ) + 12 I ( vR ) . 2

Solving, we obtain v 2 = 1.39 m/s. 1.39 m/s.

Problem 19.13 The mass of each box is 4 kg. The radius of the pulley is 120 mm and its moment of inertia is 0.032 kg-m 2 . The system is released from rest, and when the left box has moved 0.8 m, it is moving at 1 m/s. What is the kinetic coefficient of friction between the surface and the two boxes?

468

308

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19.13 (Continued) Solution:

Let m = 4 kg, I = 0.032 kg-m 2 , and R = 0.12 m. If v is the magnitude of the velocity of the boxes toward the right, the clockwise angular velocity of the pulley is ω = v /R.

The normal force on the left box is N L = mg. The normal force on the right box is N R = mg cos30 °. As the boxes move a distance d, work and energy for the left box is T L d − (µ k mg) d =

for the right box it is (mg sin30 °)d − T R d − (µ k mg cos30°)d =

1 2 mv , 2

and for the pulley it is TR d − TL d =

1 v 2 I . 2 R

( )

Setting d = 0.8 m and v = 1 m/s and solving, we obtain T L = 9.59 N, T R = 11.0 N, µ k = 0.181.

1 2 mv , 2

µ k = 0.181.

mkNL

TR mkNR

Problem 19.14 The 4-kg bar is released from rest in the horizontal position 1 and falls to position 2. The unstretched length of the spring is 0.4 m and the spring constant is k = 20 N/m. What is the magnitude of the bar’s angular velocity when it is in position 2?

In position 1 the spring is stretched a distance

d 1 = 0.6 m − 0.4 m = 0.2 m, while in position 2 the spring is stretched a distance d2 =

(1.6 m) 2 + (1 m) 2 − 2(1.6 m)(1 m)cos60° − 0.4 m = 1.0 m

Using conservation of energy we have

0.6 m

Solution:

T1 + V1 = T2 + V2 1

608

1 2 1 1 kd =  (4 kg)(1 m) 2  ω 2  2 1 2 3 −(4 kg)(9.81 m/s 2 )(0.5 m) sin60 ° +

1 2 kd 2 2

Solving we find ω = 3.33 rad/s. 2

Problem 19.15 The moments of inertia of two gears that can turn freely on their pin supports are I A = 0.002 kg-m 2 0.002 kg-m 2 and I B = 0.006 kg-m 2 . The gears are at rest when a constant couple M = 2 N-m is applied to gear B. Neglecting friction, use the principle of work and energy to determine the angular velocities of the gears when gear A has turned 100 revolutions.

90 mm A

60 mm

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19.15 (Continued) Solution: Let θ A and θ B denote the clockwise angular displacement of gear A and the counterclockwise angular displacement of gear B. We have the relations R Aθ A = R B θ B ,

R Aω A = R B ω B .

Applying work and energy, MθB =

1 1 I ω 2 + I Bω B 2 : 2 A A 2

2  R 1 1 R M  A θ A  = I Aω A 2 + I B  A ω A  .    RB  RB 2 2 When θ A = 100(2π ) rad, this equation yields ω A = 599 rad/s. Then ω B = ( R A /R B )ω A = 399 rad/s. A: 599 rad/s (clockwise). B: 399 rad/s (counterclockwise).

Problem 19.16 The moments of inertia of gears A and B are I A = 0.02 kg-m 2 and I B = 0.09 kg-m 2 . Gear A is connected to a torsional spring with constant k = 12 N-m/rad. The spring is initially unstretched. Gear B is rotated five revolutions counterclockwise and released from rest. What are the magnitudes of the angular velocities of the gears when the spring is unstretched?

Solution:

Let θ A and θ B denote the clockwise angular displacement of gear A and the counterclockwise angular displacement of gear B. We have the relations R Aθ A = R B θ B ,

R Aω A = R B ω B .

Using Eq. (19.17), conservation of energy is V1 + T1 = V2 + T2 : 1 1 1 kθ 2 + 0 = 0 + I A ω A 2 + I B ω B 2 2 2 2 A =

2 1 1 R I A ω A 2 + I B  A ω A  .  2 2  RB

(1)

When θ B = 5(2π ) rad, θ A = ( R B /R A )θ B = 44.9 rad and Eq. (1) yields ω A = 614 rad/s. Then ω B = ( R A /R B )ω A = 430 rad/s. A: 614 rad/s. B: 430 rad/s.

200 mm B

A 140 mm

Solution: All pulleys rotate in a positive direction: Problem 19.17 The moments of inertia of three pulleys that can turn freely on their pin supports are 0.1 ω C kg-m = ( 2). ω B , I A = 0.002 kg-m 2 , I B = 0.036 kg-m 2 , and I C = 0.032 0.2 0.032 kg-m 2 . The pulleys are stationary when a con0.1 ωB = ( )ω , stant couple M = 2 N-m is applied to pulley A. What 0.2 A is the angular velocity of pulley A when it has turned 10 from which revolutions? ωC = 100 mm

From the principle of work and energy: U = T2 − T1 , where T1 = 0 since the pulleys start from a stationary position. The work done is

100 mm M A 200 mm

470

0.1 ( 0.2 )ω.

200 mm

U = ∫

θA 0

M d θ = 2(2π )(10) = 40 π N-m

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19.17 (Continued) The kinetic energy is T2 = =

( 12 ) I ω + ( 12 ) I ω + ( 12 ) I ω 2 A

2 B

2 C

0.1 0.1 ( 12 )( I + ( 0.2 ) I + ( 0.2 ) I )ω , 2

2 A

from which T2 = 0.0065ω 2A , and U = 40 π = 0.0065ω 2A . Solve: ω A =

40 π = 139.0 rad/s. 0.0065

Problem 19.18 Model the arm ABC as a single rigid body. Its mass is 320 kg, and the moment of inertia about its center of mass is I = 440 kg-m 2 . Starting from rest with its center of mass 1.8 m above the ground (position 1), arm ABC is pushed upward by hydraulic actuators. When it is in the position shown (position 2), the arm has a counterclockwise angular velocity of 0.6 rad/s. How much work did the actuators do in moving the arm from position 1 to position 2?

1.80 m 1.40 m B

C 0.30 m

Solution: d =

The distance from the support A to the center of mass is

(1.8 m) 2 + (0.8 m + 0.3 m) 2 = 2.11 m.

Applying the parallel-axis theorem, the moment of inertia of the arm about A is I A = I + d 2 m = 1870 kg-m 2 . When the arm is in position 2, its kinetic energy is T2 =

1 1 I ω 2 = (1870 kg-m 2 )(0.6 rad/s) 2 = 337 N-m. 2 A 2

The height above the ground of the center of mass in position 2 is 2.25 + 0.80 + 0.30 = 3.35 m. The work done by the weight of the arm as it moves from position 1 to position 2 is U W = −mg(3.35 m − 1.8 m) = −4870 N-m.

0.80 m

Let U A denote the work done by the actuators. We have

A 0.70 m

U 12 = U A + U W = T2 − 0 : 2.25 m

U A − 4870 N-m = 337 N-m, obtaining U A = 5200 N-m (3840 ft-lb). 5200 N-m.

Problem 19.19 The mass of the circular disk is m = 30 kg and its radius is R = 0.2 m. The disk is stationary when a constant clockwise couple M = 20 N-m is applied to it, causing it to roll to the right. What is the velocity of the center of the disk when the center has moved a distance b = 5 m?

Solution: y

b M

As the center of the disk moves a distance b, the weight does no work because it is normal to the motion, and the normal and friction forces do no work because the point of the disk on which they act has zero velocity (see Example 19.1). The moment of inertia of the disk is I =

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1 mR 2 = 0.6 kg-m 2 . 2

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19.19 (Continued) We will treat clockwise angular motion as positive. The clockwise angle through which the disk rotates as the center moves a distance b is θ = b /R, so the work done by the constant couple is U 12 = M θ = M

b . R

Applying work and energy, U 12 = M

b 1 1 = mv 2 + I ω 2 , (1) R 2 2

where v is the velocity of the center. Solving this equation together with the rolling relation v = Rω, we obtain v = 4.71 m/s, ω = 23.6 rad/s. 4.71 m/s.

Problem 19.20 The mass of the circular disk is m = 40 kg and its radius is R = 0.2 m. The disk is stationary when a constant clockwise couple M = 20 N-m is applied to it, causing it to roll down the inclined surface. When the center of the disk has moved a distance b = 5 m, it is moving at 6 m/s. Use work and energy to determine the angle β .

Solution: y

mg x

b R

M As the center of the disk moves a distance b, the normal and friction forces do no work because the point of the disk on which they act has zero velocity (see Example 19.1). The disk’s weight does work U W = mgb sin β . b

The moment of inertia of the disk is I =

1 mR 2 = 0.8 kg-m 2 . 2

We will treat clockwise angular motion as positive. The clockwise angle through which the disk rotates as the center moves a distance b is θ = b /R, so the work done by the constant couple is U C = Mθ = M

b . R

Applying work and energy, U 12 = U W + U C = mgb sin β + M

b 1 1 = mv 2 + I ω 2 , (1) R 2 2

where v is the velocity of the center. Using the rolling relation ω = v /R and solving Eq. (1) for the angle, we obtain β = 17.2 °. β = 17.2 °.

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Problem 19.21 The mass of the stepped disk is 18 kg and its moment of inertia is 0.28 kg-m 2 . If the disk is released from rest, what is its angular velocity when the center of the disk has fallen 1 m?

Solution:

Let r = 0.1 m and h = 1 m. The force exerted by the suspending string does no work because the velocity of the disk at the point where the force is exerted is zero. The work done by the weight of the disk is

U 12 = mgh. Applying work and energy, U 12 = mgh =

1 2 1 mv + I ω 2 . 2 2

Solving this equation together with the relation v = r ω, we obtain v = 2.77 m/s, ω = 27.7 rad/s. 27.7 rad/s. 0.1 m

0.2 m

Problem 19.22 The 100-kg homogenous cylindrical disk ( R = 0.3 m) is at rest when the constant force F = 350 N is applied to a cord wrapped around it, causing the disk to roll. The angle β = 20 °. Use the principle of work and energy to determine the angular velocity of the disk when it has turned one revolution.

Solution: y

R F

mg x

b R

Let R = 0.3 m. If the disk turns one revolution, its center moves a distance b = 2π R.

The normal and friction forces do no work because the point of the disk on which they act has zero velocity (see Example 19.1). The disk’s weight does work U W = mgb sin β . The moment of inertia of the disk is I =

1 mR 2 = 0.8 kg-m 2 . 2

The velocity of the top of the disk where the force acts is twice the velocity v of the center (think about the instantaneous center), so the distance through which the force F moves is twice the distance the center moves, and the work done is U F = 2bF = 4 π RF . Applying work and energy, U 12 = U W + U F = mgb sin β + 4 π RF =

1 2 1 mv + I ω 2 . (1) 2 2

where v is the velocity of the center. Solving Eq. (1) together with the rolling relation v = Rω, we obtain v = 5.10 m/s, ω = 17.0 rad/s. 17.0 rad/s.

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Problem 19.23 The spring, with constant k = 5 lb/ft, is initially unstretched. The 2-slug homogeneous disk is rolled one revolution to the right and released from rest. If the disk rolls without slipping, what is the magnitude of the velocity of its center when the spring is unstretched?

Solution: I =

Let R = 1 ft. The moment of inertia of the disk is

1 mR 2 = 0.5 slug-ft 2 . 2

When the disk is rolled one revolution to the right, its center moves a distance b = 2π R = 6.28 ft, and the resulting potential energy of the spring is V1 =

1 2 kb . 2

Applying conservation of energy, k

V1 + T1 = V2 + T2 : 1 2 1 1 kb + 0 = 0 + mv 2 + I ω 2 . 2 2 2

1 ft

Solving this equation together with the relation v = r ω, we obtain v = 8.11 ft/s, ω = 8.11 rad/s. 8.11 ft/s.

Problem 19.24 The system is released from rest. The moment of inertia of the pulley is 0.03 slug-ft 2 . The slanted surface is smooth. Determine the magnitude of the velocity of the 10-lb weight when it has fallen 2 ft.

Solution:

Let W A = 5 lb, W B = 10 lb. Their masses are

WA 5 lb = = 0.155 slug, g 32.2 ft/s 2 W 10 lb mB = B = = 0.311 slug. 32.2 ft/s 2 g mA =

Let h = 2 ft. As the 10-lb weight falls, The work done on the two weighs by gravity is U 12 = −W A h sin 20 ° + W B h.

5 lb

Applying work and energy to the system,

6 in

U 12 = −W A h sin 20 ° + W B h =

208

1 1 1 m v 2 + m Bv 2 + I ω 2 . 2 A 2 2

Solving this equation together with the relation v = Rω, we obtain v = 7.52 ft/s, ω = 15.0 rad/s. 10 lb

Problem 19.25 The system is released from rest. The moment of inertia of the pulley is 0.03 slug-ft 2 . The coefficient of kinetic friction between the 5-lb weight and the slanted surface is µ k = 0.4. Determine the magnitude of the velocity of the 10-lb weight when it has fallen 2 ft.

7.52 ft/s.

Solution:

Let W A = 5 lb, W B = 10 lb. Their masses are

WA 5 lb = = 0.155 slug, g 32.2 ft/s 2 W 10 lb mB = B = = 0.311 slug. 32.2 ft/s 2 g

mA =

The normal force exerted on weight A by the surface is N A = W A cos 20 °. Let h = 2 ft. As weight B falls, The work done on the two weighs by gravity and friction is U 12 = −W A h sin 20 ° − µ kW A h cos 20 ° + W B h. Applying work and energy to the system,

5 lb

U 12 = −W Ah sin 20 ° − µ kW Ah cos 20 ° + W B h =

6 in 208

1 m v2 2 A

1 1 m v 2 + Iω 2. 2 B 2

Solving this equation together with the relation v = Rω, we obtain v = 6.62 ft/s, ω = 13.2 rad/s. 10 lb 6.62 ft/s.

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Problem 19.26 Each of the cart’s four wheels weighs 3 lb, has a radius of 5 in, and has moment of inertia I = 0.01 slug-ft 2 . The cart (not including its wheels) weighs 20 lb. The cart is stationary when the constant horizontal force F = 10 lb is applied. How fast is the cart going when it has moved 2 ft to the right?

Solution:

Use work energy

U 12 = Fd = (10 lb)(2 ft) = 20 ft-lb T1 = 0

(

)

1 20 lb-s 2 2 v 2 32.2 ft

T2 =

(

)

2  1 3 lb-s 2 2  1 v + 4  v + (0.01 slug-ft 2 )   2 32.2 ft 2 5/12 ft 

(

)

T1 + U 12 = T2 ⇒ v = 5.72 ft/s. F

Problem 19.27 The total moment of inertia of the car’s two rear wheels and axle is I R , and the total moment of inertia of the two front wheels is I F . The radius of the tires is R, and the total mass of the car, including the wheels, is m. The car is moving at velocity v 0 when the driver applies the brakes. If the car’s brakes exert a constant retarding couple M on each wheel and the tires do not slip, determine the car’s velocity as a function of the distance s from the point where the brakes are applied. (See Example 19.3.)

Solution: When the car rolls a distance s, the wheels roll through an angle s /R so the work done by the brakes is V12 = −4 M (s /R). Let m F be the total mass of the two front wheels, m R the mass of the rear wheels and axle, and m c the remainder of the car’s mass. When the car is moving at velocity v its total kinetic energy is T = =

1 1 1 1 m v 2 + m Rv 2 + I R (v /R) 2 + I F (v /R) 2 2 c 2 2 2 1 [m + ( I R + I F )/R 2 ] v 2 . 2

From the principle of work and energy, V12 = T2 − T1: − 4 M (s / R) =

1 1 [m + ( I R + I F )/R 2 ]v 2 − [m + ( I R + I F )/R 2 ]v 02 2 2

Solving for v, we get v =

v 02 − 8 Ms /[Rm + ( I R + I F )/R].

Problem 19.28 The slender bar has a mass of 6 kg. The spring (k = 10 N/m) is unstretched in the position shown. The bar is given a clockwise angular velocity of 30 °/s. What is its angular velocity in degrees per second when it has rotated through an angle of 20 °?

Solution: B A

Let m = 6 kg, L = 2 m, and L S = 1 m. The bar’s moment of inertia about the pin support is IO =

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19.28 (Continued) When the bar has rotated through an angle θ, the coordinates of point A in terms of the coordinate system shown are

x A = L sin θ, y A = L cos θ. The coordinates of end B of the spring are x B = −L S , y B = L. The stretch of the spring is S = =

V1 + T1 = V2 + T2 : 1 1 1 1 1 mgL + I O ω 1 2 + 0 = kS 2 2 + mgL cos 20° + I O ω 2 2 2 2 2 2 2 =

(x A − x B ) 2 + (y A − y B ) 2 − L S

2 1  k ( L sin15° + L S ) 2 + ( L cos 20° − L ) 2 − L S  2 

( L sin θ + L S ) 2 + ( L cos θ − L ) 2 − L S .

1 1 mgL cos 20° + I O ω 2 2 , 2 2

where ω 1 = 30 °(π /180 °) rad/s. Solving, we obtain ω 2 = 0.755 rad/s (43.2 deg/s).

With the datum at y = 0, the potential energy of the system is V =

Applying conservation of energy, we have

1 2 1 kS + mgL cos θ. 2 2

43.2 deg/s.

Problem 19.29 The radius of the pulley is R = 100 mm and its moment of inertia is I = 0.1 kg-m 2 . The mass m = 5 kg. The spring constant is k = 135 N/m. The system is released from rest with the spring unstretched. Determine how fast the mass is moving when it has fallen 0.5 m. Solution:

T1 = 0, V1 = 0, T2 =

v 1 1 (5 kg) v 2 + (0.1 kg-m 2 ) 2 2 0.1 m

(

V2 = −(5 kg)(9.81 m/s 2 )(0.5 m) +

)

1 (135 N/m)(0.5 m) 2 2

T1 + V1 = T2 + V2 ⇒ v = 1.01 m/s.

Problem 19.30 The masses of the bar and disk are 14 kg and 9 kg, respectively. They are welded together at A. Determine the angular velocity of the bar when it has rotated through an angle of (a) 45° and (b) 90 °.

A O 1.2 m

Solution: (a) Let m B = 14 kg, m D = 9 kg, L = 1.2 m, R = 0.3 m. The bar’s moment of inertia about O is IB =

Applying conservation of energy,

1 m L2 = 6.72 kg-m 2 . 3 B

V1 + T1 = V2 + T2 : (2) 1 0 + 0 = V2 + I O ω 2 . 2

Using the parallel-axis theorem, the moment of inertia of the disk about O is 1 I D = m D R 2 + L2 m D = 13.4 kg-m 2 . 2 The moment of inertia of the combined object about O is I O = I B + I D = 20.1 kg-m 2 .

0.3 m

Setting θ = 45 ° in Eq. (1) and solving Eq. (2), we obtain ω = 3.64 rad/s. (b) Setting θ = 90 ° in Eq. (1) and solving Eq. (2), we obtain ω = 4.33 rad/s.

Placing the datum at the level of position 1, the potential energy in position 1 is V1 = 0, and when the bar has rotated through an angle θ it is

(a) 3.64 rad/s. (b) 4.33 rad/s.

( 12 L sin θ )m g − (L sin θ)m g. (1)

V2 = −

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Problem 19.31 The masses of the bar and disk are 14 kg and 9 kg, respectively. The system is released from rest with the bar horizontal. Determine the angular velocity of the bar when it is vertical if the bar and disk are connected by a smooth pin at A.

Solution: Let m B = 14 kg, m D = 9 kg, L = 1.2 m, R = 0.3 m. The bar’s moment of inertia about O is IB =

1 m L2 = 6.72 kg-m 2 . 3 B

We can account for the disk not rotating by setting its moment of inertia about O equal to I D = L2 m D = 13.0 kg-m 2 . The moment of inertia of the combined object about O is I O = I B + I D = 20.0 kg-m 2 .

A O 1.2 m

Placing the datum at the level of position 1, the potential energy in position 1 is V1 = 0, and when the bar has rotated through an angle θ it is

0.3 m

( 12 L sin θ )m g − (L sin θ)m g. (1)

V2 = −

Applying conservation of energy, V1 + T1 = V2 + T2 : (2) 1 0 + 0 = V2 + I O ω 2 . 2 Setting θ = 45 ° in Eq. (1) and solving Eq. (2), we obtain ω = 4.38 rad/s. 4.38 rad/s.

Problem 19.32 The 40-kg crate is pulled up the inclined surface by the winch. The coefficient of kinetic friction between the crate and the surface is µ k = 0.4. The moment of inertia of the drum on which the cable is being wound is I A = 4 kg-m 2 . The crate starts from rest, and the couple exerted on the drum is M = 50 N-m. What is the crate’s velocity when it has moved 1 m?

0.15

Solution:

mkN

208

Let m = 40 kg, R = 0.15 m, and b = 1 m. The angle through which the drum turns as the crate moves a distance b is b /R. The normal force exerted on the crate by the surface is N = mg cos 20 °. The work done by the crate’s weight and friction when the crate moves a distance b is U C = −mgb sin 20 ° − µ k mgb cos 20 °. The work done by the winch is U W = Mb /R. Applying work and energy,

208 UC + UW =

1 2 1 mv + I A ω 2 . 2 2

We solve this equation together with the relation v = Rω, obtaining v = 0.689 m/s, ω = 4.59 rad/s. 0.689 m/s.

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Problem 19.33 The 40-kg crate is pulled up the inclined surface by the winch. The coefficient of kinetic friction between the crate and the surface is µ k = 0.4. The moment of inertia of the drum on which the cable is being wound is I A = 4 kg-m 2 . The crate starts from rest, and the couple exerted on the drum is given as a function of the angle through which it has rotated (in radians) by M = 50 − 0.3θ N-m. What is the crate’s velocity when it has moved 1 m?

0.15

Solution:

mkN

208

Let m = 40 kg, R = 0.15 m, and b = 1 m, and let M = 50 − 0.3θ N-m = c − d θ. The angle through which the drum turns as the crate moves a distance b is b /R. The normal force exerted on the crate by the surface is N = mg cos 20 °. The work done by the crate’s weight and friction when the crate moves a distance b is U C = −mgb sin 20 ° − µ k mgb cos 20 °. The work done by the winch is U W = ∫ M dθ = ∫

208

b /R 0

(c − d θ ) d θ

b /R 1 =  cθ − d θ 2   0 2

cb db 2 . − R 2R 2

Applying work and energy, UC + UW =

1 2 1 mv + I A ω 2 . 2 2

We solve this equation together with the relation v = Rω, obtaining v = 0.643 m/s, ω = 4.28 rad/s. 0.643 m/s.

Problem 19.34 The mass of the 2-m slender bar is 8 kg. A torsional spring exerts a counterclockwise couple kθ on the bar, where k = 40 N-m/rad and θ is in radians. The bar is released from rest with θ = 5 °. What is the magnitude of the bar’s angular velocity when θ = 60 °?

Solution:

We will use conservation of energy

T1 = 0 V1 = (8 kg)(9.81 m/s 2 )(1 m) cos 5 ° + T2 =

5 π rad ) )( 180

11 (8 kg) (2 m) 2  ω 2  2  3

V2 = (8 kg)(9.81 m/s 2 )(1 m) cos60 ° + u

(

1 N-m 40 2 rad

(

1 N-m 40 2 rad

60 π rad ) )( 180

T1 + V1 = T2 + V2 ⇒ ω = 1.79 rad/s.

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Problem 19.35 The mass of the suspended object A is 8 kg. The mass of the pulley is 5 kg, and its moment of inertia is 0.036 kg-m 2 . If the force T = 70 N is applied to the stationary system, what is the magnitude of the velocity of A when it has risen 0.2 m? Solution:

When the mass A rises 0.2 m, the end of the rope rises

0.4 m.

120 mm

T1 = 0, V1 = 0, T2 =

v 1 1 (13 kg) v 2 + (0.036 kg-m 2 ) 2 2 0.12 m

(

)

V2 = (13 kg)(9.81 m/s 2 )(0.2 m), U = (70 N)(0.4 m)

T1 + V1 + U = T2 + V2 ⇒ v = 0.568 m/s.

Problem 19.36 The mass of the pulley A is 6 kg, and its moment of inertia is 0.28 kg-m 2 . The mass of the pulley B is 2 kg, and its moment of inertia is 0.036 kg-m 2 . The system is released from rest. Use work and energy to determine how fast the 20-kg mass is moving when it has fallen 0.2 m.

Solution: B

v A

1 2v

0.2 m Let m A = 6 kg, I A = 0.28 kg-m 2 , m B = 2 kg, and I B = 0.036 kg-m 2 . We denote the masses of the two suspended weights by mC = 8 kg and m D = 20 kg.

0.3 m

20 kg 8 kg

Let the downward displacement of the object D be d = 0.2 m, and let v be its downward velocity. This is also the upward velocity of the rope on the left side of pulley B. Noting that the leftmost point of pulley A is its instantaneous center, we see that the center of pulley A will have velocity v /2. That means it will move upward a distance d /2. The clockwise angular velocity of pulley B will be ω B = v /rB , and the counterclockwise angular velocity of pulley A will be ω A = (v /2)/r A . The total work done on the system by gravity is U 12 = m D gd − (m A + mC ) g

( 12 d ).

Applying work and energy, U 12 =

1 1 1 1 2 1 (m + mC ) v + I Aω 2A + I B ω B2 + m Dv 2 . 2 A 2 2 2 2

( )

Solving, we obtain v = 1.42 m/s, ω A = 2.37 rad/s, ω B = 7.12 rad/s. 1.42 m/s.

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Problem 19.37 The 18-kg ladder is released from rest with θ = 10 °. The wall and floor are smooth. Modeling the ladder as a slender bar, use conservation of energy to determine the angular velocity of the bar when θ = 40 °.

Solution:

Choose the datum at floor level. The potential energy at the initial position is

V1 = mg

( L2 )cos10°.

At the final position, V2 = mg

( L2 )cos 40°.

The instantaneous center of rotation has the coordinates ( L sin θ, L cos θ), where θ = 40 ° at the final position. The distance of the center of rotation L from the bar center of mass is . The angular velocity about this center 2 2 is ω = v , where v is the velocity of the center of mass of the ladder. L The kinetic energy of the ladder is

( )

u 4m

T2 =

( 12 )mv + ( 12 )( mL )ω = ( mL6 )ω , 12 2

where v =

( L2 )ω has been used.

From the conservation of energy, V1 = V2 + T2 , from which mg = mg

( L2 )cos10°

( L2 )cos 40° + ( mL6 )ω . 2

Solve: ω = 1.269 rad/s.

Problem 19.38 The horizontal surface is smooth. The 4-kg slender bar is released from rest with θ = 50 °. What is the magnitude of its angular velocity when θ = 25 °?

Solution: u

G y

equating j components gives the relation 1 v G = − ω L cos θ. 2

A N

Let L = 2 m. The bar’s moment of inertia about its center of mass is 1 I = mL2 = 2.67 kg-m 2 . 12 There are no horizontal forces on the bar, which means its center of mass G will fall straight down. Writing the velocity of the bottom of the bar A in terms of the velocity of G,

(1)

The normal force N does no work on the bar. Placing the datum at the surface, the potential energy associated with its weight in terms of θ is V =

1 mgL sin θ. 2

Applying conservation of energy, V1 + T1 = V2 + T2 : 1 1 1 1 mgL sin 50 ° + 0 = mgL sin 25 ° + mv G 2 + I ω 2 . 2 2 2 2

(2)

Solving Eqs. (1) and (2), we obtain ω = 2.42 rad/s. 2.42 rad/s (counterclockwise).

v A = v G + ω × r A /G :

v Ai = v G j +

480

i 0

j 0

1 1 L cos θ − L sin θ 2 2

k ω , 0

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Problem 19.39 The mass and length of the bar are m = 4 kg and l = 1.2 m. The spring constant is k = 180 N/m. If the bar is released from rest in the position θ = 10 °, what is its angular velocity when it has fallen to θ = 20 °?

Solution:

If the spring is unstretched when θ = 0, the stretch of the

spring is S = l(1 − cos θ). The total potential energy is v = mg (l /2) cos θ +

1 2 kl (1 − cos θ) 2 . 2

From the solution of Problem 19.37, the bar's kinetic energy is k

1 2 2 ml ω . 6

T =

We apply conservation of energy. T1 + V1 = T2 + V2 : 1 2 kl (1 − cos10 °) 2 2 1 1 = ml 2 ω 22 + mg (l /2) cos 20 ° + kl 2 (1 − cos 20 °) 2 . 6 2

0 + mg (l /2) cos10 ° + l

Solving for ω 2 , we obtain ω 2 = 0.804 rad/s.

Problem 19.40 The 4-kg slender bar is pinned to a 2-kg slider at A and to a 4-kg homogeneous cylindrical disk at B. Neglect the friction force on the slider and assume that the disk rolls. If the system is released from rest with θ = 60 °, what is the bar’s angular velocity when θ = 0? (See Example 19.4.)

Choose the datum at θ = 0. The instantaneous center of the bar has the coordinates ( L cos θ, L sin θ) (see figure), and the distance from the center L of mass of the bar is , from which the angular velocity about the bar's 2 instantaneous center is v =

( L2 )ω,

where v is the velocity of the center of mass. The velocity of the slider is v A = ω L cos θ,

and the velocity of the disk is v B = ω L sin θ 1m

The potential energy of the system is V1 = m A gL sin θ1 + mg

( L2 )sin θ . 1

At the datum, V2 = 0. The kinetic energy is u

T2 = B

200 mm

( 12 )m v + ( 12 )mv + ( 12 )( mL )ω + ( 12 )m v 12 2 A A

2 B B

( 12 )( m 2R )( vR ) , B

where at the datum v A = ω L cos 0 ° = ω L,

Solution:

v B = ω L sin 0 ° = 0, v V

v = ω

( L2 ).

From the conservation of energy: V1 = T2 . Solve: ω = 4.52 rad/s.

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Problem 19.41* The sleeve P slides on the smooth horizontal bar. The mass of each bar is 4 kg and the mass of the sleeve P is 2 kg. If the system is released from rest with θ = 60 °, what is the magnitude of the velocity of the sleeve P when θ = 40 °?

1.2 m

1.2 m G

Q u 1.2 m

1.2 m

Point P is constrained to horizontal motion. We conclude that u

ω PQ = − ω OQ ≡ ω, v P = 2ω (1.2 m)sin θ P

We also need the velocity of point G v G = v Q + ω PQ × rG /Q

Solution:

= −ω (1.2 m)(− sin θ i + cos θ j) + ωk × (0.6 m)(cos θ i − sin θ j)

We have the following kinematics

v Q = ω ΟQ k × rQ /O

= [(1.8 m)ω sin θ]i − [(0.6 m)ω cos θ] j

= ω ΟQ k × (1.2 m)(cos θ i + sin θ j)

Now use the energy methods

= ω ΟQ (1.2 m)(− sin θ i + cos θ j)

T1 = 0, V1 = 2(4 kg)(9.81 m/s 2 )(0.6 m) sin 60 °;

v P = v Q + ω PQ × rP /Q

T2 =

= ω ΟQ (1.2 m)(− sin θ i + cos θ j) + ω PQ k × (1.2 m)(cos θ i − sin θ j) = [(ω PQ − ω OQ )(1.2 m)sin θ ]i + [(ω PQ + ω OQ )(1.2 m) cos θ] j

(

)

(

)

1 1 1 1 [4 kg][1.2 m] 2 ω 2 + [4 kg][1.2 m] 2 ω 2 2 3 2 12 +

1 (4 kg)([(1.8 m)ω sin 40 °] 2 + [(0.6 m) ω cos 40 °] 2 ) 2

1 (2 kg)(2ω [1.2 m]sin 40 °) 2 2

V2 = 2(4 kg)(9.81 m/s 2 )(0.6 m)sin 40°; Solving T1 + V1 = T2 + V2 ⇒ ω = 1.25 rad/s ⇒ v P = 1.94 m/s.

Problem 19.42* The masses of the slender bars AC and BD are 6 kg and 3 kg, respectively. The slider C is of negligible mass and the horizontal bar is smooth. The system is released from rest with θ = 45 °. What is the angular velocity of bar AC when θ = 25 °?

Solution: A

B A

vBD

D 1m

B 1m

vAC

u C

Let L BD = 1 m, L AC = 2 m, m BD = 3 kg, m AC = 6 kg. The moment of inertia of bar BD about D and the moment of inertia of bar AC about B are 1 1 I BD = m BD L BD 2 = 1 kg-m 2 , I AC = m L 2 = 2 kg-m 2 . 3 12 AC AC We observe that ω AC = ω BD . Because the bar BD rotates about the fixed point D, the magnitude of the velocity of B, which is the center of mass of bar AC, is v B = L BD ω BD .

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19.42* (Continued) In terms of the angle θ, the potential energy of the system is

Applying conservation of energy,

1 V = m BD gL BD sin θ + m AC gL BD sin θ. 2

V1 + T1 = V2 + T2 : 1 1 m gL sin 45 ° + m AC gL BD sin 45 ° + 0 = m BD gL BD sin 25 ° 2 BD BD 2 + m AC gL BD sin 25 °

The kinetic energies of the bars are TBD =

1 1 1 I ω 2 , T AC = m ACv B 2 + I AC ω AC 2 2 BD BD 2 2 1 1 = m AC ( L BD ω BD ) 2 + I AC ω BD 2 . 2 2

1 I ω 2 2 BD BD 1 + m AC ( L BD ω BD ) 2 2 1 + I AC ω BD 2 . 2 Solving, we obtain ω BD = ω AC = 2.16 rad/s. +

2.16 rad/s (counterclockwise).

Problem 19.43* The masses of the slender bars AB and BC are 6 kg and 4 kg, respectively. The horizontal surface is smooth. The system is released from rest in the position shown. What is the angular velocity of bar AB when it has rotated through an angle of 20 °? A

i 0

v C i = ω AB ( y B − 1 m) i − ω AB x B j +

j 0

k ω BC

L BC cos β −L BC sin β

obtaining the equation 0 = −ω AB x B + ω BC L BC cos β . (1)

We also write the velocity of the center of mass G of bar BC in terms of the velocity of B, 1 v G = v B + ω BC × rG /B : 2

1m C 2m

i 0

j 0

1 1 L cos β − L BC sin β 2 BC 2

k ω BC

obtaining the equations

A u

v Gx = ω AB ( y B − 1 m) + ω BC

B G b

v Gy = −ω AB x B + ω BC C

Let L AB = 2 m, L BC = 2 m. The moment of inertia of bar AB about A and the moment of inertia of bar BC about its center of mass G are 1 1 m L 2 = 8 kg-m 2 , I BC = m L 2 = 0.667 kg-m 2 . 3 AB AB 12 BC BC

When bar AB has rotated through an angle θ, the coordinates of B are x B = L AB cos θ, y B = 1 m − L AB sin θ. The angle β = arcsin( y B /L BC ), and the coordinates of C are x C = x B + L BC cos β , y C = 0. Let ω AB denote the clockwise angular velocity of bar AB, and ω BC the counterclockwise angular velocity of bar BC. The velocity of B is

v B = v A + ω AB × rB /A = 0 +

i 0 xB

j k −ω AB 0 yB − 1 m 0

= ω AB ( y B − 1 m) i − ω AB x B j. We write the velocity of C in terms of the velocity of B,

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v Gx i + v Gy j = ω AB ( y B − 1 m) i − ω AB x B j

Solution:

I AB =

v C = v B + ω BC × rC /B :

1 L sin β , (2) 2 BC

1 L cos β . (3) 2 BC

The kinetic energies of the bars are 1 1 1 T AB = I AB ω AB 2 , TBC = m BC (v Gx 2 + v Gy 2 ) + I BC ω BC 2 . 2 2 2 Placing the datum at y = 0, their potential energy is 1 1 V = m AB g  y B + (1 m − y B )  + m BC g y B   2 2 1 1 = m AB g(1 m + y B ) + m BC gy B . 2 2 Applying conservation of energy, V1 + T1 = V2 + T2 : 1 1 1 1 m AB g(2 m) + m BC g(1 m) + 0 = m AB g(1 m + y B ) + m BC gy B 2 2 2 2 1 + I AB ω AB 2 2 1 + m BC (v Gx 2 + v Gy 2 ) 2 1 + I BC ω BC 2 . 2 Solving this equation together with Eqs. (1)–(3), we obtain ω AB = 2.22 rad/s, ω AB = 3.02 rad/s. 2.22 rad/s (clockwise).

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Problem 19.44 Bar AB weighs 6 lb. The sliders A and B each weigh 2 lb, and the bars supporting them are smooth. The system is released from rest in the position shown. What is the velocity of the slider B when it has moved 4 in? y

The coordinates of A and B satisfy the equations (notice that x A is negative) yA 9 ( x B − x A ) 2 + y A 2 = L2 , = − . xA 4 When the bar is in position 2, x B2 = 12 in + 4 in = 1.33 ft, these two equations yield x A 2 = −0.156 ft, y A 2 = 0.350 ft. The position vector of A relative to B in position 2 is r A /B = (−x B 2 + x A 2 ) i + y A 2 j. We write the velocity of A in terms of the velocity of B, v A = v B + ω × r A /B :

i j 0 0 (−x B 2 + x A 2 ) y A 2

v A (e x i + e y j) = v B i + 9 in obtaining the equations

B x 4 in

e xv A = v B − ω y A2 , (1) e yv A = ω (−x B 2 + x A 2 ). (2)

12 in

We also write the velocity of the center of mass G of the bar in terms of the velocity of B,

Solution:

vG = vB + ω ×

v Gx i + v Gy j = v B i +

The masses are WA W 6 lb = B = = 0.186 slug, g g 32.2 ft/s 2 W 2 lb m AB = AB = = 0.0621 slug. 32.2 ft/s 2 g

m A = mB =

The length of the bar is L =

484

(4 in + 12 in) 2 + (9 in) 2 = 18.4 in = 1.53 ft.

1 r : 2 A /B i 0

k ω , 0

obtaining the equations 1 v Gx = v B − ω y A2 , (3) 2 1 v Gy = ω (−x B 2 + x A 2 ). (4) 2 Placing the datum at y = 0, the potential energies in positions 1 and 2 are 1 1 V1 = y A1W A + y A1W AB , V2 = y A 2W A + y A 2W AB . 2 2 The kinetic energy in position 2 is 1 1 1 1 m v 2 + m Av B 2 + m AB (v Gx 2 + v Gy 2 ) + I ω 2 . 2 A A 2 2 2 Applying work and energy, V1 = V2 + T2 , and using Eqs. (1)–(4), we obtain v A = 4.61 ft/s, v B = 2.86 ft/s, v Gx = 2.37 ft/s, T2 =

Let e be the unit vector that points down the slanted bar from A,

v Gy = −2.11 ft/s, ω = 2.83 rad/s.

(4 in) i − (9 in) j = 0.406 i − 0.914 j = e x i + e y j. (4 in) 2 + (9 in) 2

j 0

1 1 y (−x B 2 + x A 2 ) 2 A2 2

Its moment of inertia about its center of mass is 1 I = m L2 = 0.0363 slug-ft 2 . 12 AB

e =

k ω , 0

2.86 ft/s.

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Problem 19.45* Each bar has a mass of 8 kg and a length of 1 m. The spring constant is k = 100 N/m, and the spring is unstretched when θ = 0. If the system is released from rest with the bars vertical, what is the magnitude of the angular velocity of the bars when θ = 30 °? Solution:

The stretch of the spring is 2l (1 − cos θ), so the poten-

tial energy is 1 l 3l k[2l (1 − cos θ)] 2 + mg cos θ + mg cos θ 2 2 2 = 2kl 2 (1 − cos θ) 2 + 2mgl cos θ.

v =

The velocity of pt B is v B = v A + ω AB × rB /A = 0 +

i j k 0 0 ω −l sin θ l cos θ 0

= −ωl cos θ i − ωl sin θ j.

The velocity of pt. G is C

v G = v B + ω BC × rG /B

= −ωl cos θ i − ωl sin θ j +

i 0

j 0

k −ω

l sin θ 2

l cos θ 2

3l l = − ω cos θ i − ω sin θ j. 2 2

2l cos u 3l cos u 2

B v

From this expression,

1 v G 2 = (1 + 8sin 2 θ) l 2 ω 2 . 4 Applying conservation of energy to the initial and final positions,

l cos u 2 x

2mgl = 2kl 2 (1 − cos30 °) + 2mgl cos30 ° 1 1 2 2 1 1 + ml ω + ml 2 ω 2 2 3 2 12 1 1 + m(1 + 8sin 2 30 °) l 2 ω 2 . 2 4

(

)

(

)

Solving for ω, we obtain ω = 1.93 rad/s.

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Problem 19.46* The system starts from rest with the crank AB vertical. A constant couple M exerted on the crank causes it to rotate in the clockwise direction, compressing the gas in the cylinder. Let s be the displacement (in meters) of the piston to the right relative to its initial position. The net force toward the left exerted on the piston by atmospheric pressure and the gas in the cylinder is 350/(1 − 10 s) N. The moment of inertia of the crank about A is 0.0003 kg-m 2 . The mass of the connecting rod BC is 0.36 kg, and the center of mass of the rod is at its midpoint. The connecting rod’s moment of inertia about its center of mass is 0.0004 kg-m 2 . The mass of the piston is 4.6 kg. If the clockwise angular velocity of the crank AB is 200 rad/s when it has rotated 90° from its initial position, what is M? (Neglect the work done by the weights of the crank and connecting rod.)

125 m

vBC r C/B

r B/A

The distance s that the piston moves to the right is s = rB /Ax + rC /Bx −

(0.125) 2 − (0.05) 2 (5)

The velocity of B is vB = v A +

i 0

j 0

rB /Ax

rB /Ay

k −ω 0

= ωrB /Ay i − ωrB /Ax j.

The velocity of C is v C = v B + ω BC × rC /B :

v C i = ωrB /Ay i − ωrB /Ax j +

i 0

j 0

rc /Bx

rc /By

k ω BC 0

Equating i and j components, we can solve for v c and ω BC in terms of ω:

v C = [rB /Ay − rC /By (rB /Ax /rC /Bx )]ω (6) A

ω BC = (rB /Ax /rC /Bx )ω (7) The velocity of the center of mass G (the midpoint) of the connecting rod BC is 1 v G = v B + ω BC × rC /B = ωrB /Ay i − ωrB /Ax j 2

Solution:

As the crank rotates through an angle θ the work done by the couple is M θ. As the piston moves to the right a distance s, the work done on the piston by the gas is −∫

s 0

350 ds . 1 − 10 s

i 0

j 0

k ω BC

1 r 2 C /Bx

1 r 2 C /By

v C = (ωrB /Ay −

, or

1 1 ω r ) i − (ωrB /Ax − ω BC rC /Bx ) j (8) 2 BC C /By 2

∫0.1− s 35 z = 35ln (1 − 10s).

Let I A be the moment of inertia of the crank about A, I BC the moment of inertia of the connecting rod about its center of mass, m BC the mass of the connecting rod, and m p the mass of the piston. The principle of work and energy is: U 12 = T2 − T1:

The total work is U 12 = M θ + 35ln (1 − 10 s).

M θ + 35ln(1 − 10 s) =

Letting 1 − 10 s = 10 z, this is dz

0.1

From the given dimensions of the crank and connecting rod, the vector components are rB /Ax = 0.05sin θ, (1) rB /Ay = 0.05cos θ (2) rC /Bx =

1 1 1 2 I ω 2 + I BC ω BC + m BC v G 2 2 A 2 2 1 2 + m pv C (9) 2

Solving this equation for M and using Eqs. (1)–(8) with ω = 200 rad/s and θ = π /2 rad, we obtain M = 28.2 N-m.

(0.125) 2 − (0.05cos θ) 2 (3)

rC /By = −0.05cos θ (4)

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Problem 19.47* At t = 0, a sphere is placed on a horizontal surface with an initial clockwise angular velocity ω 0 . If it continues to slip on the surface, show that the work done on the sphere while its center moves a distance b to the right and the sphere rotates through a clockwise angle θ is

µ k mg ( b − Rθ ). Strategy: Modify the analysis that was used in Practice Example 19.2 to show that the work done by friction when the disk does not slip is zero.

Solution:

The normal force N = mg. In Example 19.1, the work done by a force F is expressed in the form t2

∫t F ⋅ v p dt, 1

where v p is the velocity of the point of application of the force. If the disk slips, the velocity of the point of the disk where the friction force acts is not zero. If ω is the clockwise angular velocity of the disk and the velocity of the center of the disk is vi, the velocity of the point where the friction force acts is v p = vi − Rωi.

x mg

mk N

Using this expression, the work done by the friction force F = µ k N i = µ k mgi is t2

∫t F ⋅ v p dt = ∫t µ k mg(v − Rω)dt, which we can write as µ k mg ∫

t2 t1

vdt − µ k mgR ∫

t2 t1

ωdt = µ k mg ∫ dx − µ k mgR ∫ d θ = µ k mg(b − Rθ).

Problem 19.48 The moment of inertia of the disk about O is 22 kg-m 2 . At t = 0, the stationary disk is subjected to a constant 50 N-m torque. (a) Determine the angular impulse exerted on the disk from t = 0 to t = 5 s. (b) What is the disk’s angular velocity at t = 5 s?

50 N-m

Solution: (a)

Angular Impulse = ∫

5s 0

∑ M dt = (50 N-m)(5 s)

= 250 N-m-s CCW (b) 250 N-m-s = (22 kg-m 2 )ω

⇒ ω = 11.4 rad/s (counter clockwise).

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Problem 19.49 The moment of inertia of the JT9D-20 jet engine’s rotating assembly is 190 kg-m 2 . The assembly starts from rest at t = 0. The engine’s starter exerts a couple on it that is given as a function of time in seconds by M = 1500(1 + 2e −0.2t ) N-m. (a) What is the magnitude of the angular impulse exerted on the assembly from t = 0 to t = 20 s? (b) What is the magnitude of the angular velocity of the assembly (in rpm) at t = 20 s? Solution: (a) The angular impulse is t2

20 s

∫t ∑ M dt = ∫0

1500(1 + 2e −0.2t )dt = 44,700 N-m-s.

(b) The angular velocity is ω =

angular impulse 44,700 N-m-s = = 235 rad/s (2250 rmp). I 190 kg-m 2

2250 rpm.

Problem 19.50 At time t = 0, the SpaceX capsule is not rotating and is stationary relative to a space station it is approaching. Point A has coordinates x = 1.4 m, y = 1.7 m. The capsule’s mass is 8800 kg, and the moment of inertia about its center of mass is I = 32,200 kg-m 2 . Its attitude control thruster exerts a constant force T = 400 N for a short interval of time ∆t, giving the capsule a counterclockwise angular velocity of 1 °/s. What is the resulting change in the velocity of the center of mass? (Disregard the change in the capsule’s orientation during the time the thruster is firing.) Solution:

The angular velocity induced by the thruster firing is

ω 2 = 1 deg/s

( π180rad° ) = 0.0175 rad/s.

T A

208

G x

Applying angular impulse and momentum, t2

∫t ∑ Mdt = H 2 − H 1: 1

∆t

∫0

T 208 1.7 m

G 1.4 m

[(1.7 m)T cos 20 ° − (1.4 m)T sin 20°]dt = I ω 2 − 0,

[(1.7 m)(400 N) cos 20 ° − (1.4 m)(400 N)sin 20 °]∆t = I (0.0175 rad/s), the interval of time is ∆t = 1.26 s. Then from linear impulse and momentum, t2

∫t ∑ F dt = mv 2 − mv 1: 1

∆t

∫0 (−T cos 20°i − T sin 20° j) dt = mv 2 − 0, (−T cos 20 °i − T sin 20 ° j) ∆t = mv 2 , we obtain v 2 = −0.0536 i − 0.0195 j (m/s). −0.0536 i − 0.0195 j (m/s).

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Problem 19.51 The combined mass of the astronaut and his equipment is 122 kg, and the moment of inertia about their center of mass is 45 kg-m 2 . The maneuvering unit exerts an impulsive force T of 0.2-s duration, giving him a counterclockwise angular velocity of 1 rpm. (a) What is the average magnitude of the impulsive force? (b) What is the magnitude of the resulting change in the velocity of the astronaut’s center of mass?

T 300 mm

Solution: (a) From the principle of moment impulse and angular momentum, t2

∫t ∑ M dt = I ω 2 − I ω1, 1

where ω 1 = 0 since the astronaut is initially stationary. The angular velocity is ω2 =

2π = 0.1047 rad/s 60

from which ∫

0.2 0

TR dt = I ω 2 ,

from which T (0.3)(0.2) = 45 (ω 2 ), T =

45ω 2 = 78.54 N. 0.06

(b) From the principle of impulse and linear momentum t2

∫t ∑ F dt = m(v 2 − v 1 ) 1

where v 1 = 0 since the astronaut is initially stationary. 0.2

∫0 T dt = mv 2 , from which v 2 =

T (0.2) = 0.129 m/s. m

Problem 19.52 A flywheel attached to an electric motor is initially at rest. At t = 0, the motor exerts a couple M = 200 e −0.1t N-m on the flywheel. The moment of inertia of the flywheel is 10 kg-m 2 . (a) What is the flywheel’s angular velocity at t = 10 s? (b) What maximum angular velocity will the flywheel attain? Solution: (a) From the principle of moment impulse and angular momentum, t2

∫t ∑ M dt = I ω 2 − I ω1, 1

where ω 1 = 0, since the motor starts from rest. 200

∫0 200e −0.1t dt = 0.1 [−e −0.1t ]100 = 2000[1 − e −1 ], = 1264.2 N-m-s. From which ω 2 =

1264.24 = 126 rad/s. 10

(b) An inspection of the angular impulse function shows that the angular velocity of the flywheel is an increasing monotone function of the time, so that the greatest value occurs as t → ∞. ω 2max = lim

200

t →∞ (10)(0.1)

[1 − e −0.1t ] → 200 rad/s.

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Problem 19.53 A main landing gear wheel of a Boeing 777 has a radius of 0.62 m and its moment of inertia is 24 kg-m 2 . After the plane lands at 75 m/s, the skid mark of the wheel’s tire is measured and determined to be 18 m in length. Determine the average friction force exerted on the wheel by the runway. Assume that the airplane’s velocity is constant during the time the tire skids (slips) on the runway. Solution:

The tire skids for t =

18 m = 0.24 s. 75 m/s

75 m/s When the skidding stops the tire is turning at the rate ω = = 0.62 m 121 rad/s Frt = I ω ⇒ F =

(24 kg-m 2 ) (121 rad/s) Iω = F = 19.5 kN. rt (0.62 m) (0.24 s)

Problem 19.54 The force a club exerts on a 0.045-kg golf ball is shown. The ball is 42 mm in diameter and can be modeled as a homogeneous sphere. The club is in contact with the ball for 0.0006 s, and the magnitude of the velocity of the ball’s center of mass after the ball is hit is 36 m/s. What is the magnitude of the ball’s angular velocity after it is hit? Solution:

Linear momentum:

Ft = mv ⇒ F =

Angular momentum Ftd = I ω ⇒ ω =

2.5 mm

mv t

(0.045 kg) (36 m/s) (0.0025 m) Ftd mv d = = I I 2/5 (0.045 kg) (0.021 m) 2

Solving ω = 510 rad/s.

Problem 19.55 Disk A initially has a counterclockwise angular velocity ω 0 = 50 rad/s. Disks B and C are initially stationary. Disk A is moved into contact with disk B. Determine the angular velocities of the three disks when they have stopped slipping relative to each other. The masses of the disks are m A = 2 kg, m B = 12 kg, mC = 6 kg. (Approximate the moments of inertia by using the value for a uniform disk.) Solution:

The moments of inertia are

(1)

∫0 −R A f AB dt = I Aω A − I Aω 0 , ∫t

∫0 −R B f AB dt + ∫0 R B f BC dt = I Bω B − 0,

490

A C

∫t ∑ M dt = H 2 − H 1: 1

(3)

∫0 RC f BC dt = I C ω C − 0. Solving Eqs. (1) and (3) for the terms ∫

tf 0

f AB dt and ∫

tf 0

f BC dt and sub-

stituting them into Eq. (2), we obtain the equation I IA I I ω = A ω A − B ω B + C ω C . (4) RA 0 RA RB RC

R A ω A = −R B ω B ,

∑ M dt = H 2 − H 1:

0.3 m

We also have the relations

for disk B, t2

0.4 m

0.2 m

and for disk C,

1 1 I A = m A R A 2 = 0.04 kg-m 2 , I B = m B R B 2 = 0.96 kg-m 2 , 2 2 1 2 2 I C = mC RC = 0.27 kg-m , 2 Let ω A , ω B , ω C be the final counterclockwise angular velocities. Let t f be the time at which slip ceases. For disk A we have

∫t ∑ M dt = H 2 − H 1:

(2)

R B ω B = −RC ω C . (5)

Solving Eqs. (4) and (5), we obtain A: 5 rad/s (counterclockwise). B: 2.5 rad/s (clockwise). C: 3.33 rad/s (counterclockwise).

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Problem 19.56 The radius of the 0.16-kg cue ball is 28 mm. The coefficient of sliding friction between the ball and table is µ k = 0.2. The ball can be modeled as a homogeneous sphere. The force exerted on the cue ball by the cue is horizontal. The ball is struck at h = 14 mm, giving it “back spin,” and the resulting velocity of its center of mass is 2 m/s. What are the velocity and angular velocity of the ball at t = 0.4 s? (Neglect friction in determining the ball’s initial velocity and angular velocity.)

From Appendix C, the ball’s moment of inertia is 2 2 mR 2 = (0.16 kg)(0.028 m) 2 = 5.02E-5 kg-m 2 . 5 5

I =

Applying angular impulse and momentum, t2

∫t ΣMdt = H 2 − H 1: 1

∫t −( R − h)F dt = I ω 2 − 0,

(2)

−( R − h) ∫

t2 t1

F dt = I ω 2 .

From Eqs. (1) and (2), we obtain ω 2 = −89.3 rad/s. We now know the initial velocity toward the left, v 1 = 2 m/s, and counterclockwise angular velocity, ω 1 = −89.3 rad/s, of the ball’s motion after it is struck.

R mg

N R

mkN

The normal force is N = mg = (0.16 kg)(9.81 m/s 2 ) = 1.57 N. The friction force µ k N = 0.2(1.57 N) = 0.314 N is constant. We can use impulse and momentum to determine the ball’s velocity and angular velocity as functions of time.

Linear momentum is t2

∫t ∑ F dt = mv 2 − mv1: 1

Solution:

(3)

∫ −µ k N dt = −µ k Nt = mv 2 − m(2 m/s), 0

and angular momentum is

R mg

∫t ∑ M dt = I ω 2 − I ω1: 1

∫t

t2 1

(4)

R µ k N dt = Rµ k Nt = I ω 2 − I (−89.3 rad/s).

Solving, we obtain v 2 = 1.22 m/s, ω 2 = −19.2 rad/s.

N Let v denote the ball’s velocity toward the left, and let ω be its counterclockwise angular velocity. Applying linear impulse and momentum,

1.22 m/s, 19.2 rad/s clockwise.

∫t ∑ F dt = mv 2 − mv 1: 1

∫t F dt = (0.16 kg)(2 m/s) − 0,

(1)

= 0.32 kg-m/s.

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Problem 19.57 The force exerted on the cue ball by the cue is horizontal. Determine the value of h for which the ball rolls without slipping. (Assume that the frictional force exerted on the ball by the table is negligible.) Solution:

From the principle of moment impulse and angular

momentum, t2

∫t (h − R) F dt = I (ω 2 − ω1 ), 1

where ω 1 = 0 since the ball is initially stationary. From the principle of impulse and linear momentum R

∫t F dt = m(v 2 − v 1 )

where v 1 = 0 since the ball is initially stationary. Since the ball rolls, v 2 = Rω 2 , from which the two equations: (h − R) F (t 2 − t 2 ) = I ω 2 , F (t 2 − t 1 ) = mRω 2 . F

The ball is a homogenous sphere, from which I =

2 mR 2 . 5

h–R

Substitute: (h − R)mRω 2 = Solve: h =

2mR 2 ω2. 5

( 75 ) R.

Problem 19.58 In Example 19.7, we neglected the moments of inertia of the two masses m about the axes through their centers of mass in calculating the total angular momentum of the person, platform, and masses. Suppose that the moment of inertia of each mass about the vertical axis through its center of mass is I M = 0.001 kg-m 2 . If the person’s angular velocity with her arms extended to r1 = 0.6 m is ω 1 = 1 revolution per second, what is her angular velocity ω 2 when she pulls the masses inward to r2 = 0.2 m? Compare your result to the answer obtained in Example 19.7.

Solution:

H O1 = ( I P + 2mr12 + 2 I M )ω 1

( revs )

= (0.4 kg-m 2 + 2[4 kg][0.6 m] 2 + 2[0.001 kg-m 2 ]) 1 H O 2 = ( I P + 2mr22 + 2 I M )ω 2

= (0.4 kg-m 2 + 2[4 kg][0.2 m] 2 + 2[0.001 kg-m 2 ])ω 2 H O1 = H O 2 ⇒

The final distance from A to the common center of mass is (0)(250) + (18)(50) = 3 m. 250 + 50 The total angular momentum about the center of mass is conserved. x =

The initial distance from A to the common center of

(0)(250) + (12)(50) = 2 m. 250 + 50

rev . s

rev . s rev Without the moments of inertia (Example 19.6) we found ω = 4.56 . s

12 m

mass is x0 =

ω = 4.55

If we include the moments of inertia of the weights then ω = 4.55

Problem 19.59 Two gravity research satellites (m A = 250 kg, I A = 350 kg-m 2 ; m B = 50 kg, I B = 16 kg-m 2 ) are tethered by a cable. The satellites and cable rotate with angular velocity ω 0 = 0.25 rpm. Ground controllers order satellite A to slowly unreel 6 m of additional cable. What is the angular velocity afterward? Solution:

Using the numbers from Example 19.6, we conserve angu-

lar momentum

or (2)(250)(2)ω 0 + 350 ω 0 + (10)(50)(10)ω 0 + 16ω 0 = (3)(250)(3)ω + 350 ω + (15)(50)(15)ω + 16ω. We obtain ω = 0.459ω 0 = 0.115 rpm.

x 0 m A ( x 0 ω 0 ) + I A ω 0 + (12 − x 0 )m B [(12 − x 0 )ω 0 ] + I B ω 0 = xm A ( x ω ) + I A ω + (18 − x )m B [(18 − x )ω ] + I B ω;

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Problem 19.60 The 2-kg bar rotates in the horizontal plane about the smooth pin. The 6-kg collar A slides on the smooth bar. Assume that the moment of inertia of the collar A about its center of mass is negligible; that is, treat the collar as a particle. At the instant shown, the angular velocity of the bar is ω 0 = 60 rpm and the distance from the pin to the collar is r = 1.8 m. Determine the bar’s angular velocity when r = 2.4 m. Solution:

v0 k A

Angular momentum is conserved

 1 (2 kg)(3 m) 2 + (6 kg)(1.8 m) 2  (60 rpm)  3  1 =  (2 kg)(3 m) 2 + (6 kg)(2.4 m) 2  ω 2 3  Solving we find ω 2 = 37.6 rpm.

Problem 19.61 The 2-kg bar rotates in the horizontal plane about the smooth pin. The 6-kg collar A slides on the smooth bar. The moment of inertia of the collar A about its center of mass is 0.2 kg-m 2 . At the instant shown, the angular velocity of the bar is ω 0 = 60 rpm and the distance from the pin to the collar is r = 1.8 m. Determine the bar’s angular velocity when r = 2.4 m and compare your answer to that of Problem 19.60. Solution:

v0 k A

Angular momentum is conserved

 1 (2 kg)(3 m) 2 + (6 kg)(1.8 m) 2 + (0.2 kg-m 2 )  (60 rpm)  3  1 =  (2 kg)(3 m) 2 + (6 kg)(2.4 m) 2 + (0.2 kg-m 2 )  ω 2 3  Solving we find ω 2 = 37.7 rpm.

Problem 19.62* The 2-kg bar rotates in the horizontal plane about the smooth pin. The 6-kg collar A slides on the smooth bar. The moment of inertia of the collar A about its center of mass is 0.2 kg-m 2 . The spring is unstretched when r = 0, and the spring constant is k = 10 N/m. At the instant shown, the angular velocity of the bar is ω 0 = 2 rad/s, the distance from the pin to the collar is r = 1.8 m, and the radial velocity of the collar is zero. Determine the radial velocity of the collar when r = 2.4 m. v0 k A

Solution:

Angular momentum is conserved

 1 (2 kg)(3 m) 2 + (6 kg)(1.8 m) 2 + (0.2 kg-m 2 )  (2 rad/s)  3  1  =  (2 kg)(3 m) 2 + (6 kg)(2.4 m) 2 + (0.2 kg-m 2 )  ω 2 3  Thus ω 2 = 1.258 rad/s Energy is conserved 11 (2 kg)(3 m) 2 + (6 kg)(1.8 m) 2 + (0.2 kg-m 2 )  (2 rad/s) 2  2  3 1 2 V1 = (10 N/m)(1.8 m) 2 1 1 T2 =  (2 kg)(3 m) 2 + (6 kg)(2.4 m) 2 + (0.2 kg-m 2 )  ω 22  23 1 + (6 kg)v r 2 2 1 V2 = (10 N/m)(2.4 m) 2 2 T1 =

Solving we find that v r = 1.46 m/s. r

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Problem 19.63 The circular bar is welded to the vertical shafts, which can rotate freely in bearings at A and B. Let I be the moment of inertia of the circular bar and shafts about the vertical axis. The circular bar has an initial angular velocity ω 0 , and the mass m is released in the position shown with no velocity relative to the bar. Determine the angular velocity of the circular bar as a function of the angle β between the vertical and the position of the mass. Neglect the moment of inertia of the mass about its center of mass; that is, treat the mass as a particle. Solution:

b m

Angular momentum about the vertical axis is conserved:

I ω 0 + ( Rsin β 0 )m( Rsin β 0 )ω 0 = I ω + ( Rsin β )m ( Rsin β )ω.

Solving for ω,  I + mR 2sin 2 β 0  ω =  ω .  I + mR 2sin 2 β  0

Problem 19.64 The 10-lb bar is released from rest in the 45° position shown. It falls and the end of the bar strikes the horizontal surface at P. The coefficient of restitution of the impact is e = 0.6. When the bar rebounds, through what angle relative to the horizontal will it rotate?

3 ft

Solution:

The mass of the bar is m = W /g = (10 lb)/(32.2 ft/s 2 ) = 0.311 slug. Its moment of inertia about the pin support is

1 2 1 mL = (0.311 slug)(3 ft) 2 = 0.932 slug-ft 2 . 3 3 First we use conservation of energy to find how fast the bar is rotating when it reaches the horizontal. Applying work and energy,

458 P

I =

U 12 = T2 − T1: W

( 12 L sin 45° ) = 12 I ω

2 − 0,

we obtain ω 2 = −4.77 rad/s. The bar’s downward velocity at P just before it hits the surface is (3 ft)(4.77 rad/s) = 14.3 ft/s. Applying the coefficient of restitution, the bar’s upward velocity at P just after the impact is e(14.3 ft/s) = 8.59 ft/s. Its counterclockwise angular

velocity is ω 1 = (8.59 ft/s)/L = 2.86 rad/s. Now we apply work and energy going the other way, U 12 = T2 − T1: 1 1 −W L sin θ = 0 − I ω 1 2 , 2 2

(

)

obtaining sin θ = 0.255, which gives θ = 14.7 °. 14.7 °.

Problem 19.65 The 10-lb bar is released from rest in the 45° position shown. It rebounds to a position 25° relative to the horizontal. What was the coefficient of restitution of the impact at P?

3 ft

Solution:

The mass of the bar is m = W /g = (10 lb)/(32.2 ft/s 2 ) = = 0.311 slug. Its moment of inertia about the pin support is

I =

1 2 1 mL = (0.311 slug)(3 ft) 2 = 0.932 slug-ft 2 . 3 3

First we use conservation of energy to find how fast the bar is rotating when it reaches the horizontal. Applying work and energy, U 12 = T2 − T1: 1 1 W L sin 45 ° = I ω 22 − 0, 2 2

(

458

)

we obtain ω 2 = −4.77 rad/s. The bar’s downward velocity at P just before it hits the surface is v 2 = (3 ft)(4.77 rad/s) = 14.3 ft/s. Applying the coefficient of restitution, the bar’s upward velocity at P just after the impact is e(14.3 ft/s) = v 1 . Its counterclockwise angular

velocity is ω 1 = v 1 /L = e(14.3 ft/s)/(3 ft). Now we apply work and energy going the other way, U 12 = T2 − T1: 1 1 −W L sin 25 ° = 0 − I ω 1 2 2 2 1 = − I [e(14.3 ft/s)/(3 ft)] 2 . 2 Solving this equation, we obtain e = 0.773.

(

)

e = 0.773.

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Problem 19.66 The 4-kg bar is released from rest in the horizontal position above the fixed projection at A. The distance b = 0.35 m. The impact of the bar with the projection is plastic; that is, the coefficient of restitution of the impact is e = 0. What is the bar’s angular velocity immediately after the impact?

0.2 m A

Solution:

First we do work energy to find the velocity of the bar just before impact. T1 + V1 = T2 + V2 1 0 + mgh = mv 22 + 0 ⇒ v 2 = 2 = 1.98 m/s

2 gh =

2 (9.81 m/s 2 )(0.2 m)

b Angular momentum is conserved about the impact point A. After the impact, the bar pivots about the fixed point A. mv 2 b = m =

12bv L + b )ω ⇒ ω = ( 12 L + 12b 2

12(0.35 m)(1.98 m/s) (1 m) 2 + 12(0.35 m) 2

ω 3 = 3.37 rad/s.

Problem 19.67 The 4-kg bar is released from rest in the horizontal position above the fixed projection at A. The coefficient of restitution of the impact is e = 0.6. What value of the distance b would cause the velocity of the bar’s center of mass to be zero immediately after the impact? What is the bar’s angular velocity immediately after the impact?

0.2 m A b

Solution:

First we do work energy to find the velocity of the bar just before impact. T1 + V1 = T2 + V2 1 0 + mgh = mv 22 + 0 ⇒ v 2 = 2 = 1.98 m/s

2 gh =

2(9.81 m/s 2 )(0.2 m)

Angular momentum is conserved about point A. The coefficient of restitution is used to relate the velocities of the impact point before and after the collision. 1 mv 2 b = mv 3b + mL2 ω 3 , ev 2 = ω 3b − v 3 , v 3 = 0, 12

Solving we have e 0.6 L = (1 m) = 0.224 m, 12 12 12v 2 b 12(1.98 m/s)(0.224 m) = = 5.32 rad/s. ω3 = L2 (1 m) 2 b =

b = 0.224 m, ω 3 = 5.32 rad/s.

Problem 19.68 The mass of the ship is 3.0E7 kg, and the moment of inertia of the vessel about its center of mass is 2.6E10 kg-m 2 . Wind causes the ship to drift sideways at 0.1 m/s and strike the stationary piling at P. When that occurs, crewmen secure the ship to the piling at that point. What is the ship’s angular velocity after the impact?

16 m 45 m

Solution:

Let m = 3.0E7 kg, I = 2.6E10 kg-m 2 . The angle β = arctan(16 m/45 m) = 19.6 °, and the radial distance from P to the center of mass when impact occurs is h/ cos β = 47.8 m. The ship’s counterclockwise angular momentum about P before the impact is equal to its value after the impact:

 h  h(mv ) =  (mv ) + I ω ′. (1)  cos β 

Before impact v

Because the ship is fixed at P after the impact, we also have the relation  h   cos β  ω ′ = v ′. (2) Solving Eqs. (1) and (2), we obtain v ′ = 0.0683 m/s, ω ′ = 0.00143 rad/s.

b v9

After impact

0.00143 rad/s.

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Problem 19.69 The mass of the ship is 3.0E7 kg and the moment of inertia of the vessel about its center of mass is 2.6E10 kg-m 2 . Wind causes the ship to drift sideways at 0.1 m/s and strike the stationary piling at P. The coefficient of restitution of the impact is e = 0.15. What is the ship’s angular velocity after the impact? Solution:

16 m 45 m

Let m = 3.0E7 kg, I = 2.6E10 kg-m 2 .

The ship’s velocity toward the piling before the impact is v = 0.1 m/s, and its upward velocity at point P immediately after the impact is v P′ = ev = 0.015 m/s. The ship’s counterclockwise angular momentum about P before the impact is equal to its value after the impact:

h G

h(mv ) = h(mv ′) + I ω ′. (1)

Before impact

We write the velocity of the center of mass G just after the impact in terms of the velocity at point P: i j k v G = −v ′ j = v′P + ω′ × rG /P = v P′ j + 0 0 ω′ , −h y G 0 obtaining the equation

P y v9p

G v9

−v ′ = v ′P − hω ′. (2)

After impact x

Solving Eqs. (1) and (2), we obtain v ′ = 0.0655 m/s, ω ′ = 0.00179 rad/s. 0.00179 rad/s (counterclockwise).

Problem 19.70 A 0.35-oz bullet moving horizontally at 2900 ft/s strikes a 5-lb suspended 2 × 4 near the lower end and is embedded in the wood. What is the angular velocity of the 2 × 4 immediately after the impact? Solution:

Let the masses of the bullet and plank be denoted by

( 161 lboz )( 32.21ft/s ) = 6.79E-4 slug, 1 M = 5 lb( ) = 0.155 slug. 32.2 ft/s m = 0.35 oz

3 ft

The moment of inertia of the plank about its center of mass is 1 I = ML2 = 0.116 slug-ft 2 . 12 Linear momentum in the horizontal direction is conserved in the impact: mv = mv ′ + MV ′. (1) Let O be the fixed point that is coincident with the center of mass G of the plank before the impact. We can determine the angular momentum of the bullet about O by calculating the “moment” of its linear momentum about O. Angular momentum about O is conserved in the impact:

( 12 L )mv = ( 12 L )mv ′ + I ω′. (2)

The bullet moves with the plank after the impact, so we can express its velocity in terms of the velocity of the center of mass of the plank and its angular velocity:

G O

= V ′i +

496

i 0

yielding the equation 1 v ′ = V ′ + Lω ′. (3) 2

G O

v ′i = V ′ i + ω ′ × r B / G j k 0 ω′ , 1 0 − L 0 2

Solving Eqs. (1)–(3), we obtain v′ = 49.9 ft/s, V ′ = 12.5 ft/s, ω′ = 24.9 rad/s. 24.9 rad/s (counterclockwise).

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Problem 19.71 The 2-kg sphere A is moving toward the right at 4 m/s when it strikes the end of the 5-kg slender bar B. Immediately after the impact, the sphere A is moving toward the right at 1 m/s. What is the angular velocity of the bar after the impact? Solution:

B 1m

Angular momentum is conserved about point O.

1 m Av A1 L = m Av A 2 L + m B L2 ω 2 3 3m A (v A1 − v A 2 ) ω2 = mBL 3(2 kg)(4 m/s − 1 m/s) = 3.6 m/s. ω2 = (5 kg)(1 m)

4 m/s

ω 2 = 3.6 m/s (counterclockwise).

Problem 19.72 The 2-kg sphere A is moving toward the right at 4 m/s when it strikes the end of the 5-kg slender bar B. The coefficient of restitution is e = 0.4. The duration of the impact is 0.002 seconds. Determine the magnitude of the average horizontal force exerted on the bar by the pin support as a result of the impact.

B 1m

Solution:

System angular momentum is conserved about point O. The coefficient of restitution is used to relate the relative velocities before and after the impact. We also use the linear impulse momentum equation for the ball and for the bar. A

1 m L2 ω 2 , ev A1 = ω 2 L − v A 2 , 3 B L m Av A1 − F ∆t = m Av A 2 , F ∆t + R∆t = m B ω 2 . 2 m Av A1 L = m Av A 2 L +

4 m/s

Solving we find R =

(1 + e) m A m Bv A1 1.4(2 kg)(5 kg)(4 m/s) = = 1270 N 2(3m A + m B )∆t 2(11 kg)(0.002 s)

R = 1.27 kN.

Problem 19.73 The 2-kg sphere A is moving toward the right at 10 m/s when it strikes the unconstrained 4-kg slender bar B. What is the angular velocity of the bar after the impact if the sphere adheres to the bar?

Solution: v9B

v9B B vA

v9A

Before impact After impact Total linear momentum is conserved: m Av A = m Av ′A + m Bv ′B . (1) Total angular momentum about the point of impact is conserved: 1 0 = I ω ′B − L m Bv ′B = 0. (2) 4 Because the sphere adheres to the bar, we can express its velocity in terms of the velocity of the center of mass of the bar and its angular velocity: 1 v ′A = v ′B + L ω ′B . (3) 4 Solving Eqs. (1)–(3), we obtain v ′A = 4.67 m/s, v ′A = 2.67 m/s, ω ′B = 8 rad/s.

( )

10 m/s

0.25 m

( )

8 rad/s (counterclockwise).

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Problem 19.74 The 2-kg sphere A is moving toward the right at 10 m/s when it strikes the unconstrained 4-kg slender bar B. The coefficient of restitution of the impact is e = 0.4. What are the velocity of the sphere and the angular velocity of the bar after the impact?

Solution:

v9B

P B

v9A

Before impact

After impact

v9p

Total linear momentum is conserved: 1m

m Av A = m Av ′A + m Bv ′B . (1) Total angular momentum about the point of impact is conserved:

0 = I ω ′B −

10 m/s

0.25 m

( 41 L )m v ′ = 0. (2) B B

Let P denote the point of the slender bar where the sphere strikes. The coefficient of restitution is e =

v ′P − v ′A . (3) vA − 0

The velocity of P after the impact can be expressed in terms of the velocity of the center of mass of the bar and its angular velocity: v ′P = v ′B +

( 14 L )ω′ . (4) B

Solving Eqs. (1)–(4), we obtain v ′A = 2.53 m/s, v ′P = 6.53 m/s, v ′B = 3.73 m/s, ω ′B = 11.2 rad/s. Velocity is 2.53 m/s, angular velocity is 11.2 rad/s (counterclockwise).

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Problem 19.75 A 0.01-kg particle A is moving toward the right at 100 m/s when it strikes the unconstrained 0.24-kg T-shaped object B. (The vertical and horizontal uniform bars have the same mass per unit length.) If the particle adheres on impact, what is the resulting angular velocity of the T-shaped object?

0.2 m A vA

Solution:

0.6 m

v9A v9B B

v9B

0.3 m

Let P denote the point of the object where the impact occurs. Total angular momentum about P is conserved:

C Before impact

After impact

Denote the mass of the object by m B = 0.24 kg. The mass of the vertical bar is mV = (0.6/0.9)(0.24 kg) = 0.16 kg, and the mass of the horizontal bar is m H = (0.3/0.9)(0.24 kg) = 0.08 kg. Let C denote the point where the two bars are attached. The sum of their moments of inertia about C is 1 1 IC = m (0.3 m) 2 + mv (0.6 m) 2 = 0.0198 kg-m 2 . 12 H 3 The vertical distance from the point where the two bars are attached to their common center of mass, denoted by B, is (0.3 m)mv + (0)m H h = = 0.2 m. mv + m H Applying the parallel-axis theorem, the moment of inertia of the object about its center of mass is I = IC

− h 2m

0.0102 kg-m 2 .

Total linear momentum is conserved in the impact:

0 = I ω ′B + (0.2 m)m Bv ′Bx = 0. (3) Because the particle adheres to the object, we can express its velocity after the impact occurs in terms of the velocity of the center of mass of the object and its angular velocity: v′A = v′B + ω ′B × r A / B : v ′Ax i + v ′Ay j = v ′Bx i + v ′By j +

i j k 0 0 ω ′B , 0 0.2 m 0

yielding the equations v ′Ax = v ′Bx − ω ′B (0.2 m), (4) v ′Ay = v ′By . (5) From Eqs. (2) and (5), we see that v ′Ay = v ′By = 0. Solving Eqs. (1), (3) and (4), we obtain v ′Ax = 7.48 m/s, v ′Bx = 3.85 m/s, ω ′B = −18.1 rad/s. 18.1 rad/s (clockwise).

m Av Ax = m Av ′Ax + m Bv ′Bx , (1) 0 = m Av ′Ay + m Bv ′By . (2)

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Problem 19.76* A 0.01-kg particle A is moving toward the right at 100 m/s when it strikes the unconstrained 0.24-kg T-shaped object B. (The vertical and horizontal uniform bars have the same mass per unit length.) The coefficient of restitution of the impact is e = 0.5. What is the resulting angular velocity of the T-shaped object?

0.2 m A vA

Solution:

0.6 m

B A

vAx

v9Ax

v9px v9B B

0.3 m

v9Bx h

Let P denote the point of the object where the impact occurs. Total angular momentum about P is conserved:

C Before impact

After impact

1 1 m (0.3 m) 2 + mV (0.6 m) 2 = 0.0198 kg-m 2 . 12 H 3

The vertical distance from the point where the two bars are attached to their common center of mass, denoted by B, is (0.3 m)mV + (0)m H h = = 0.2 m. mV + m H Applying the parallel-axis theorem, the moment of inertia of the object about its center of mass is I = I C − h 2 m B = 0.0102 kg-m 2 . Total linear momentum is conserved in the impact: m Av Ax = m Av ′Ax + m Bv ′Bx , (1) 0 = m Av ′Ay + m Bv ′By . (2)

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0 = I ω ′B + (0.2 m)m Bv ′Bx = 0. (3) We can express the velocity of P after the impact occurs in terms of the velocity of the center of mass of the object and its angular velocity: v′P = v′B + ω′B × rP / B : v ′Px i + v ′Py j = v ′Bx i + v ′By j +

i j k 0 0 ω ′B 0 0.2 m 0

yielding the equations v ′Px = v ′Bx − ω ′B (0.2 m), (4) v ′Py = v ′By . (5) The coefficient of restitution is v ′ − v ′Ax e = Px . (6) v Ax − 0 No vertical force is exerted on the particle A in the impact, so v ′Ay = 0, and then from Eqs. (2) and (5) we see that v ′By = v ′Py = 0. Solving Eqs. (1), (3), (4) and (6), we obtain v ′Ax = −38.8 m/s, v ′Bx = 5.78 m/s, v ′Px = 11.2 m/s, ω ′B = −27.2 rad/s. 27.2 rad/s (clockwise).

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Problem 19.77 A 10-lb slender bar of length l = 2 ft is released from rest in the horizontal position at a height h = 2 ft above a peg (Fig. a). A small hook at the end of the bar engages the peg, and the bar swings from the peg (Fig. b). What is the bar’s angular velocity immediately after it engages the peg? l

Solution:

Work energy is used to find the velocity just before impact.

T1 + V1 = T2 + V2 1 2 mv + 0 ⇒ v 2 = 2 gh 2 2 Angular momentum is conserved about the peg. 0 + mgh =

l 1 = ml 2 ω 3 , 2 3 3v 2 3 2 gh 3 2(32.2 ft/s 2 )(2 ft) = ω3 = = = 8.51 rad/s. 2l 2l 2(2 ft) mv 2

ω 3 = 8.51 rad/s. h

(a)

(b)

Problem 19.78 A 0.01-kg particle A is moving toward the right at 100 m/s when it strikes the unconstrained 40-kg sphere B. The radius of the sphere is 0.15 m. If the particle adheres on impact, what is the resulting angular velocity of the sphere? B

A vA

Because the particle adheres to the sphere, we can express its velocity after the impact occurs in terms of the velocity of the center of mass of the sphere and its angular velocity: v′A = v′B + ω ′B × r A / B : v ′Ax i + v ′Ay j = v ′Bx i + v ′By j +

i 0

j 0

−R sin 45 ° R sin 45 °

k ω ′B , 0

yielding the equations

458

v ′Ax = v ′Bx − ω ′B R sin 45 °, (4) v ′Ay = v ′By − ω ′B R sin 45 °. (5) Solving Eqs. (1)–(5), we obtain v ′Ax = 0.0562 m/s, v ′Ay = 0.0312 m/s, v ′Bx = 0.0250 m/s, v ′By = −7.80E-6 m/s, ω′B = −0.294 rad/s.

Solution:

0.2964 rad/s (clockwise). y A

458

x R

The moment of inertia of the sphere is I =

2 2 m R 2 = (40 kg)(0.15 m) 2 = 0.36 kg-m 2 . 5 B 5

Total linear momentum is conserved in the impact: m Av Ax = m Av ′Ax + m Bv ′Bx , (1) 0 = m Av ′Ay + m Bv ′By . (2) Let P denote the point of the sphere where the impact occurs. Total angular momentum about P is conserved: 0 = I ω ′B + ( R sin 45 °)m Bv ′Bx + ( R sin 45 °)m Bv ′By = 0. (3)

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Problem 19.79 The 1-slug disk rolls at velocity v = 10 ft/s toward a 6-in step. The wheel remains in contact with the step and does not slip while rolling up onto it. What is the wheel’s velocity once it is on the step?

Solution:

We apply conservation of angular momentum about 0 to analyze the impact with the step.

v1 18 in

R v

6 in ( R − h)mv 1 + I

( vR ) = Rmv + I ( vR ). (1) 1

Then we apply work and energy to the “climb” onto the step. 1 1 1 v 2 1 v 2 −mgh =  mv 32 + I 3  −  mv 22 + I 2  . (2)  2  2 2 R  2 R  1 Solving Eqs. (1) and (2) with v 1 = 10 ft/s and I = mR 2 , we obtain 2 v 3 = 6.25 ft/s.

( )

Problem 19.80 The 1-slug disk rolls toward a 6-in step. The wheel remains in contact with the step and does not slip while rolling up onto it. What is the minimum velocity v the disk must have in order to climb up onto the step?

( )

Solution:

See the solution of Problem 19.79 solving Eqs. (1) and (2) with v 3 = 0, we obtain v 1 = 5.96 ft/s.

18 in

6 in

Problem 19.81 The length of the bar is 1 m, and its mass is 4 kg. Just before the bar hits the floor, its angular velocity is ω = 0 and its center of mass is moving downward at 2 m/s. If the end of the bar adheres to the floor on impact, what is the bar’s angular velocity afterward?

708

Solution: Angular momentum about the point of impact is conserved: v9

G v9G

Solving Eqs. (1) and (2), we obtain v ′G = 0.513 rad/s, ω ′ = 1.03 rad/s. 1.03 rad/s (counterclockwise).

708

Before impact

( 12 L cos 70° )mv = ( 12 L )mv ′ + I ω′. (2)

After impact

Let L = 1 m, m = 4 kg, v G = 2 m/s. The bar’s moment of inertia about its center of mass is 1 I = mL2 = 0.333 kg-m 2 . 12 After impact, the bar will be rotating about its lower end. The velocity of its center of mass and its angular velocity are related by v ′G =

( 12 L )ω′. (1)

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Problem 19.82 The length of the bar is 1 m, and its mass is 4 kg. Just before the bar hits the smooth floor, its angular velocity is ω = 0 and its center of mass is moving downward at 2 m/s. The coefficient of restitution of the impact with the floor is e = 0.6. What is the bar’s angular velocity afterward?

Let L = 1 m, m = 4 kg, v G = 2 m/s. The bar’s moment of inertia about its center of mass is 1 I = mL2 = 0.333 kg-m 2 . 12 The smooth floor exerts no horizontal force on the bar, so its center of mass has no horizontal component of velocity after the impact. Angular moment about the point of impact is conserved: 1 1 L cos 70 ° mv G = − L cos 70 ° mv ′G + I ω ′. (1) 2 2

(

)

(

)

Let P denote the point of the bar that comes into contact with the floor. The velocity of P after the impact can be expressed in terms of the velocity of the center of mass and the angular velocity,

v′P = v′G + ω ′ × rP / G :

708

v ′Px i + v ′Py j = v ′G j +

Solution:

v9G

v9Py

708

Before impact

After impact

v9Px

i 0

j 0

k ω′ ,

1 1 L cos 70° − L sin 70 ° 2 2

yielding the equations 1 v ′Px = ω ′ L sin 70 °, (2) 2 1 v ′Py = v ′G + ω ′ L cos 70 °. (3) 2 The coefficient of restitution is v ′Py v ′Py e = − = . (4) v Py vG (Notice that we define v G and v ′G to be positive downward.) Solving Eqs. (1), (3) and (4), we obtain v ′G = 0.369 m/s, v ′Px = 1.20 m/s, ω′ = 4.86 rad/s. 4.86 rad/s (counterclockwise).

Problem 19.83 The length of the bar is 1 m, and its mass is 4 kg. Just before the bar hits the smooth floor, it has angular velocity ω and its center of mass is moving downward at 2 m/s. The coefficient of restitution of the impact with the floor is e = 0.6. What value of ω would cause the bar to have no angular velocity after the impact?

708

Solution: Let P denote the point of the bar that comes into contact with the floor. The velocity of P just before the impact can be expressed in terms of the velocity of the center of mass and the angular velocity, v P = v G + ω × rP / G :

v v9G

vG vPy

708

v9Py

vPx

Before impact

P After impact

Let L = 1 m, m = 4 kg, v G = 2 m/s. The bar’s moment of inertia about its center of mass is 1 I = mL2 = 0.333 kg-m 2 . 12 The smooth floor exerts no horizontal force on the bar, so the center of mass of the bar has no horizontal component of velocity after the impact. Angular moment about the point of impact is conserved: 1 1 L cos 70 ° mv G + I ω = − L cos 70 ° mv ′G . (1) 2 2

(

)

(

)

v Px i + v Py j = −v G j +

i 0

j 0

1 1 L cos 70 ° − L sin 70 ° 2 2

k ω , 0

yielding the equations 1 v Px = ω L sin 70 °, (2) 2 1 v Py = −v G + ω L cos 70 °. (3) 2 Because the bar is translating immediately after the impact, v ′Py = v ′G . (4) The coefficient of restitution is v ′Py v ′G , (5) e = − = 1 v Py v G − ω L cos 70 ° 2 where we substituted Eqs. (3) and (4). Solving Eqs. (1) and (5), we obtain v ′G = 2.05 m/s, ω = −8.32 rad/s. 8.32 rad/s (clockwise).

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Problem 19.84 During her parallel-bars routine, the velocity of the 90-lb gymnast’s center of mass is 4 i − 10 j (ft/s) and her angular velocity is zero just before she grasps the bar at A. In the position shown, her moment of inertia about her center of mass is 1.8 slug-ft 2 . If she stiffens her shoulders and legs so that she can be modeled as a rigid body, what is the velocity of her center of mass and her angular velocity just after she grasps the bar?

Solution:

Let v ′ and ω′ be her velocity and angular velocity after she grasps the bar. The angle θ = 20.0 ° and r = 23.4/12 = 1.95 ft. Conservation of angular momentum about A is k ⋅ (r × mv) = rmv ′ + I ω ′: k⋅

i j k x y 0 4 m −10 m 0

= rmv ′ + I ω ′,

− 10 mx − 4 my = rmv ′ + I ω ′. Solving this equation together with the relation v ′ = r ω ′, we obtain ω′ = 3.15 rad/s, v ′ = 6.14 ft/s. Her velocity is

v′ = v ′ cos θ i − v ′ sin θ j = 5.77 i − 2.10 j (ft/s).

(–8, –22) in

u v9 (x, y)

Problem 19.85 The 20-kg homogeneous rectangular plate is released from rest (Fig. a) and falls 200 mm before coming to the end of the string attached at the corner A (Fig. b). Assuming that the vertical component of the velocity of A is zero just after the plate reaches the end of the string, determine the angular velocity of the plate and the magnitude of the velocity of the corner B at that instant.

v9 y A v9

Just before

Just after

The velocity of A just after is v′A = v′G + ω′ × r A / G

A 200 mm

300 mm

= −v ′ j +

B 500 mm

The j component of v′A is zero:

B (a)

(b)

−v ′ + 0.25ω ′ = 0.

Solution:

We use work and energy to determine the plate's downward velocity just before the string becomes taut. 1 mg(0.2) = mv 2 . 2 Solving, v = 1.98 m/s. The plate’s moment of inertia is 1 (20)[ (0.3) 2 + (0.5) 2 ] = 0.567 kg-m 2 . 12 Angular momentum about A is conserved: I =

0.25(mv ) = 0.25(mv ′) + I ω ′.

(1)

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i j k −ω ′ . 0 0 −0.25 0.15 0

(2)

Solving Eqs. (1) and (2), we obtain v ′ = 1.36 m/s and ω′ = 5.45 rad/s. The velocity of B is v′B = v′G + ω′ × rB / G = −v ′ j +

i j k 0 0 −ω ′ 0.25 −0.15 0

v′B = −0.818i − 2.726 j (m/s).

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Problem 19.86* The two bars A and B are each 2 m in length, and each has a mass of 4 kg. In Fig. a, bar A has no angular velocity and is moving to the right at 1 m/s, and bar B is stationary. If the bars bond together on impact (Fig. b), what is their angular velocity ω′ after the impact?

B B

Solution:

Let m = 4 kg, L = 2 m, v A = 1 m/s. The simplest approach to this problem is to realize that after the impact there is, in effect, one bar with mass 2m, length 2L, moment of inertia I C = (1/12)(2m)(2 L ) 2 , and angular velocity ω′. From conservation of angular momentum about the point of impact,

A v9

( 12 L )mv = I ω′, A

1 m/s

we obtain ω′ = 0.375 rad/s. ω′ = 0.375 rad/s.

(a)

Problem 19.87* The two bars A and B are each 2 m in length, and each has a mass of 4 kg. In Fig. a, bar A has no angular velocity and is moving to the right at 1 m/s, and bar B is stationary. If the bars do not bond together and the coefficient of restitution of their impact is e = 0.6, what are the angular velocities of the bars afterward?

(b)

B B

Solution: A

v9B

A v9

v9B

1 m/s v9BP v9AP v9A

v9A

Before impact

After impact

Linear momentum is conserved during the impact: mv A = mv ′A + mv ′B . (1) Angular momentum of each bar about the contact point is conserved:

( 12 L )mv = mv ′ ( 12 L ) + I ω′ , (2) 1 0 = −( L ) mv ′ + I ω ′ . (3) 2 A

(a)

(b)

The velocity of the point of impact of each bar is related to the velocity of the bar’s center of mass and its angular velocity: v ′AP = v ′A −

( 12 L )ω′ , (5)

v ′BP = v ′B +

( 12 L )ω′ . (6)

Solving Eqs. (1)–(6), we obtain v ′A = 0.8 m/s, v ′B = 0.2 m/s, v ′AP = 0.2 m/s, v ′BP = 0.8 m/s, ω ′A = 0.6 rad/s, ω ′B = 0.6 rad/s. A and B, 0.6 rad/s (counterclockwise).

The coefficient of restitution is e =

506

v ′BP − v ′AP . (4) vA

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Problem 19.88* Two bars A and B are each 2 m in length, and each has a mass of 4 kg. In Fig. a, bar A has no angular velocity and is moving to the right at 1 m/s, and bar B is stationary. If the bars bond together on impact (Fig. b), what is their angular velocity ω′ after the impact?

Solution: Linear momentum is conserved: x − DIR: m Av A = m Av ′Ax + m Bv ′Bx , (1) 0 = m Av ′Ay + m Bv ′By . (2)

y − DIR:

Total angular momentum is conserved. Calculating it about 0, B

l l 0 = − m Av ′Ay + I A ω ′ − m Bv ′Bx + I B ω ′. (3) 2 2

We also have the kinematic relation v′B = v′A + ω′ × rB / A :

v ′Bx i + v ′By j = v ′Ax i + v ′Ay j + 1 m/s (a)

A (b)

i 0 l 2

j k 0 ω′ , l 0 2

l ω ′, (4) 2 l v ′By = v ′Ay + ω ′. (5) 2 v ′Bx = v ′Ax −

Solving Eqs. (1)–(5), we obtain ω′ = 0.3 rad/s.

v9By v9Bx v9Ay v9 vA Before impact

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v9Ax

After impact

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Problem 19.89* The horizontal velocity of the landing airplane is 60 m/s, its vertical velocity (rate of descent) is 2 m/s, and its angular velocity is zero. The mass of the airplane is 8000 kg, and the moment of inertia about its center of mass is 98,000 kg-m 2 . When the rear wheels touch the runway, they remain in contact with it. What is the plane’s angular velocity just after it touches down? (Neglect forces on the airplane other than the vertical component of the force exerted on the wheels by the runway.)

Applying linear impulse and momentum, mv ′V − mv V = −∫

t2 t1

F dt. (1)

Let b and h denote the horizonal and vertical distances from the center of mass to P. Applying angular impulse and momentum, I ω′ = b ∫

t2 t1

F dt. (2)

The velocity of P after touchdown can be expressed in terms of the velocity of the center of mass and the angular velocity: v′P = v′G + ω′ × rP / G : v ′Px i + v ′Py j = −v ′H i − vV′ j +

1.8 m

i j k 0 0 ω′ , b −h 0

yielding the equations v ′Px = −v ′H + hω ′, v ′Py = −vV′ + bω ′.

0.3 m

Because v ′Py = 0, we obtain the equation v ′V = bω ′. (3)

Solution:

Solving Eqs. (1)–(3), we obtain v ′V = 0.0146 m/s, ω ′ = 0.0486 rad/s. y

0.0486 rad/s (counterclockwise).

Before touchdown

v9 v9H

v9V

After touchdown

Let m = 8000 kg, I = 98,000 kg-m 2 , v H = 60 m/s, v V = 2 m/s. When it touches down, the plane is subjected to an upward vertical force F at the point of contact P. The resulting linear impulse causes the plane’s vertical velocity at P to become zero. Because there is no horizontal force on the plane, the horizontal velocity of its center of mass remains the same, v ′H = v H .

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Problem 19.90* The horizontal velocity of the landing airplane is 60 m/s, its vertical velocity (rate of descent) is 2 m/s, and its angular velocity is zero. The mass of the airplane is 8000 kg, and the moment of inertia about its center of mass is 98,000 kg-m 2 . The coefficient of restitution of the impact of the wheels with the runway is e = 0.4. What is the plane’s angular velocity immediately after it touches down? (Neglect forces on the airplane other than the vertical component of the force exerted on the wheels by the runway.)

Applying linear impulse and momentum, mv ′V − mv V = −∫

t2 t1

F dt. (1)

Let b and h denote the horizonal and vertical distances from the center of mass to P. Applying angular impulse and momentum, I ω′ = b ∫

t2 t1

F dt. (2)

1.8 m

yielding the equations v ′Px = −v ′H + hω ′, v ′Py = −vV′ + bω ′.

0.3 m

i j k 0 0 ω′ , b −h 0

(3)

The coefficient of restitution is e = −

Solution:

v ′Py v ′Py = . (4) v Py vV

Solving Eqs. (1)–(4), we obtain v ′V = −0.780 m/s, v ′Py = 0.8 m/s, ω′ = 0.0681 rad/s.

Before touchdown

0.0681 rad/s (counterclockwise).

v9 v9H

v9V

After touchdown

Let m = 8000 kg, I = 98,000 kg-m 2 , v H = 60 m/s, v v = 2 m/s. When it touches down, the plane is subjected to an upward vertical force F at the point of contact P. The resulting linear impulse causes the plane’s vertical velocity at P to become zero. Because there is no horizontal force on the plane, the horizontal velocity of its center of mass remains the same, v ′H = v H .

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Problem 19.91* While attempting to drive on an icy street for the first time, a student skids his 1260-kg car ( A) into the university president’s stationary 2700-kg Rolls-Royce Corniche ( B). The point of impact is P. Assume that the impacting surfaces are smooth and parallel to the y-axis and that the coefficient of restitution of the impact is e = 0.5. The moments of inertia of the cars about their centers of mass are I A = 2400 kg-m 2 and I B = 7600 kg-m 2 . Determine the angular velocities of the cars and the velocities of their centers of mass after the collision. (See Example 19.10.) y

The angular momentum of each car about the point of impact is conserved. −0.6m Av A = −0.6m Av ′A + I A ω ′A (2) 0 = 0.6m Bv ′B + I B ω ′B (3) The velocity of car A at the point of impact after the collision is v′AP = v′A + ω′A × rP / A = v ′A i +

1.7 −0.6

The corresponding equation for car B is v′BP = v′B + ω′B × rP / B = v ′B i +

i 0

j 0

k ω ′B ,

v ′BP = (v ′B − 0.6ω ′B ) i − 3.2ω ′B j

5 km/h 0.6 m 1.7 m

v′AP = ( v ′A + 0.6ω ′A ) i + 1.7ω ′A j.

−3.2 0.6 A

k ω ′A

P 0.6 m

The x components of the velocities at P are related by the coefficient of restitution:

e = 3.2 m

(v ′B − 0.6ω ′B ) − (v ′A + 0.6ω ′A ) (4) vA

Solving Eqs. (1)–(4), we obtain v ′A = 0.174 m/s, v ′B = 0.567 m/s

Solution: vA =

ω ′A = −0.383 rad/s, ω ′B = −0.12 rad/s.

Car A’s initial velocity is

5000 = 1.39 m/s 3600

The force of the impact is parallel to the x-axis so the cars are moving in the x direction after the collision. Linear momentum is conserved: m Av A = m Av ′A + m Bv ′B

(1)

Problem 19.92* The student in Problem 19.91 claimed that he was moving at 5 km/h prior to the collision, but police estimate that the center of mass of the Rolls-Royce was moving at 1.7 m/s after the collision. What was the student’s actual speed? (See Example 19.10.)

Solution:

Setting v ′B = 1.7 m/s in Eqs. (1)–(4) of the solution of Problem 19.91 and treating v A as an unknown, we obtain v A = 4.17 m/s = 15.0 km/h.

5 km/h 0.6 m 1.7 m

P 0.6 m

3.2 m

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Problem 19.93 Each slender bar is 48 in long and weighs 20 lb. Bar A is released in the horizontal position shown. The bars are smooth, and the coefficient of restitution of their impact is e = 0.8. Determine the angle through which B swings afterward.

v h

28 in v9B

v9 A B

h v9Ap b

Solution:

Choose a coordinate system with the x axis parallel to bar A in the initial position, and the y axis positive upward. The strategy is (a) from the principle of work and energy, determine the angular velocity of bar A the instant before impact with B; (b) from the definition of the coefficient of restitution, determine the value of e in terms of the angular velocities of the two bars at the instant after impact; (c) from the principle of angular impulse-momentum, determine the relation between the angular velocities of the two bars after impact; (d) from the principle of work and energy, determine the angle through which bar B swings. The angular velocity of bar A before impact: The angle of swing of bar A is 28 β = sin −1 = 35.69 °. 48 Denote the point of impact by P. The point of impact is

v9Bp

b F

( )

−h = −Lcosβ = −3.25 ft. The change in height of the center of mass of bar A is L h − cos β = − . 2 2 From the principle of work and energy, U = T2 − T1 , where T1 = 0, since the bar is released from rest. The work done by the weight of bar A is h 2

Wh . 0 2 1 The kinetic energy the bar is T2 = I ω2, 2 A A

U = ∫

−

− W dh =

from which ω A =

( )

Wh = IA

3g cos β = 4.43 rad/s, L

mL2 . 3 The angular velocities at impact: By definition, the coefficient of restitution is v ′ − v ′APx e = BPx . v APx − v BPx where I A =

Bar B is at rest initially, from which v BPx = 0, v ′BPx = v ′BP , and v ′APx = v ′AP cos β , v APx = v AP cos β , from which v ′ − v ′AP cos β e = BP . v AP cos β The angular velocities are related by v ′BP = hω ′B , v ′AP = L ω ′A , v AP = Lω A , from which e =

hω ′B − ( L cos β )ω ′A ω − ω ′A = B , ωA ( L cos β )ω A

from which (1) ω ′B − ω ′A = eω A

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The force reactions at P : From the principle of angular impulse momentum, t2

∫t Fh dt = Fh(t 2 − t1 ) = I B (ω ′B − 0), and 1

∫t −F ( L cos β ) dt = −F ( L cos β )(t 2 − t1 ) 1

= I A (ω ′A − ω A ). Divide the second equation by the first: ω′ − ω A −L cos β = −1 = A , h ω ′B from which (2) ω ′B + ω ′A = ω A . Solve (1) and (2): (1 − e) ω ′A = ω A, 2 (1 + e) ω ′B = ω A. 2 The principle of work and energy: From the principle of work and energy, U = T2 − T1 , where T2 = 0, since the bar comes to rest after rotating through an angle γ. The work done by the weight of bar B as its center of mass rotates through the angle γ is −

L cos γ

U = ∫ L 2 −

− W B dh = −W

( L2 ) (1 − cos γ ).

The kinetic energy is 1 1 T1 = I ω′ 2 = I (1 + e) 2 ω 2A 2 B B 8 B I (1 + e) 2 (3g cos β ) WL (1 + e) 2 cos β . = B = 8L 8 Substitute into U = −T1 to obtain

( )

cos γ = 1 −

( )

(1 + e) 2 cos β = 0.3421, 4

from which γ = 70 °.

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Problem 19.94* The Apollo CSM ( A) approaches the Soyuz Space Station ( B). The mass of the Apollo is m A = 18 Mg, and the moment of inertia about the axis through its center of mass parallel to the z-axis is I A = 114 Mg-m 2 . The mass of the Soyuz is m B = 6.6 Mg, and the moment of inertia about the axis through its center of mass parallel to the z-axis is I B = 70 Mg-m 2 . The Soyuz is stationary relative to the reference frame shown, and the CSM approaches with velocity v A = 0.21i + 0.05 j (m/s) and no angular velocity. What is the angular velocity of the attached spacecraft after docking?

Denote the vectors from P to the centers of mass by rP /GA = −7.3i (m), and rP /GB = +4.3i (m). The angular momentum about the origin is conserved: rP /GA × m A v GA = rP / GA × m A v′GA + I Aω ′A + rP / GB × m B v′GB . Denote the vector distance from the center of mass of the Apollo to the center of mass of the Soyuz by rB /A = 11.6 i (m). From kinematics, the instant after contact: v′GB = v′GA + ω′ × rB / A . Reduce:  i j k    v′GB = v′GA +  0 0 ω ′   11.6 0 0    = v ′GAx i + ( v ′GAy + 11.6ω ′ ) j,

7.3 m

from which v ′GBx = v ′GAx ,

(A)

v ′GBy = v ′GAy + 11.6ω ′ ⋅ rP / GA × m A v GA

(2)

 i j k   0 0  = (−0.365m A ) k, =  −7.3  0.21m A 0.05m A 0      i j k   rP / GA × m A v′GA =  −7.3 0 0   m Av ′GAx m Av ′GAy 0    = (−7.3m Av ′GAy )k.

x 4.3 m

 i j k    rP / GB × m B v′GB =  4.3 0 0   m Bv ′GAx m A (v ′GAy + 11.6ω ′) 0    = 4.3m B (v ′GAy + 11.6ω ′)k.

(B)

Solution:

The docking port is at the origin on the Soyuz, and the configuration the instant after contact is that the centers of mass of both spacecraft are aligned with the x axis. Denote the docking point of contact by P. (P is a point on each spacecraft, and by assumption, lies on the x-axis. ) The linear momentum is conserved:

Collect terms and substitute into conservation of angular momentum expression, and reduce: −0.365m A = (−7.3m A + 4.3m B ) v ′GAy + (49.9m B + I A + I B ) ω ′.

m A v GA = m A v′GA + m Bv ′GB ,

From (1) and (2),

from which m A (0.2) = m Av ′GAx + m Bv ′GBx ,

v ′GAy =

and (1) m A (0.05) = m Av ′GAy + m Bv ′GBy .

v9GAy

These two equations in two unknowns can be further reduced by substitution and algebraic reduction, but here they have been solved by iteration: ω′ = −0.003359 rad/s ,

v9GBy v9GAx

0.05m A − 11.6m B ω ′ . m A + mB

v9GBx

′ v GAy = 0.04704 m/s,

from which v9

512

11.6 m

′ ′ v GBy = v GAy + 11.6ω ′ = 0.00807 m/s.

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Problem 19.95 The moment of inertia of the pulley is 0.2 kg-m 2 . The system is released from rest. Use the principle of work and energy to determine the velocity of the 10-kg cylinder when it has fallen 1 m.

Solution:

150 mm

mLg

mR g

Choose a coordinate system with the y axis positive upward. Denote m L = 5 kg, m R = 10 kg, h R = −1 m, R = 0.15 m. From the principle of work and energy, U = T2 − T1 where T1 = 0 since the system is released from rest. The work done by the left hand weight is UL = ∫ 5 kg

10 kg

hL 0

−m L g dh = −m L gh L .

The work done by the right hand weight is UL = ∫

hR 0

−m R g dh = −m R gh R .

Since the pulley is one-to-one, h L = −h R , from which U = U L + U R = (m L − m R ) gh R . The kinetic energy is 1 1 1 T2 = I ω2 + m v2 + m v 2. 2 P 2 L L 2 R R

( )

Since the pulley is one-to-one, v L = −v R . From kinematics v ω = R, R from which 1 IP T2 = + m L + m R v R2 . 2 R2

( )(

)

Substitute and solve: vR =

(

Problem 19.96 The moment of inertia of the pulley is 0.2 kg-m 2 . The system is released from rest. Use momentum principles to determine the velocity of the 10-kg cylinder 1 s after the system is released.

2(m L − m R ) gh R IP + mL + mR R2

)

= 2.026… = 2.03 m/s.

150 mm

Solution:

Use the coordinate system and notations of Problem 19.95. From the principle of linear impulse-momentum for the left hand weight: t2

∫t (TL − m L g) dt = m L (v L 2 − v L1 ) = m Lv L 2 , 1

since v L1 = 0, from which (1)

T L (t 2 − t 1 ) = m L g(t 2 − t 1 ) + m Lv 2 .

5 kg

10 kg

For the right hand weight, t2

∫t (TR − m R g) dt = m Rv R 2 , 1

from which (2)

T R (t 2 − t 1 ) = m R g(t 2 − t 1 ) + m Rv R 2 . From the principle of angular impulse-momentum for the pulley: t2

∫t (TL − TR ) R dt = I P ω 2 ,

Substitute (1) and (2) into (3): IP ω . R 2 Since the pulley is one to one, v L 2 = −v R 2 . From kinematics: v ω 2 = R2 , R from which (m L − m R ) g(t 2 − t 1 ) + m Lv L 2 − m Rv R 2 =

from which (3)

I (T L − T R )(t 2 − t 1 ) = P ω 2 . R

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v R2 = −

(m R − m L ) g(t 2 − t 1 ) = −2.05 m/s. IP + mL + mR R2

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Problem 19.97 Arm BC has a mass of 12 kg, and the moment of inertia about its center of mass is 3 kg-m 2 . Point B is stationary. Arm BC is initially aligned with the (horizontal) x-axis with zero angular velocity, and a constant couple M applied at B causes the arm to rotate upward. When it is in the position shown, its counterclockwise angular velocity is 2 rad/s. Determine M. y

Solution:

Assume that the arm BC is initially stationary. Denote R = 0.3 m. From the principle of work and energy, U = T2 − T1 , where T1 = 0. The work done is

(

408

Problem 19.98 The cart is stationary when a constant force F is applied to it. What will the velocity of the cart be when it has rolled a distance b? The mass of the body of the cart is m c , and each of the four wheels has mass m, radius R, and moment of inertia, I.

)

( 12 )mv + ( 12 ) I ω . 2

( 12 )(mR + I )ω . 2

Substitute into U = T2 and solve: M =

(mR 2 + I BC )ω 2 + 2mgR sin θ = 44.2 N-m. 2θ

Solution:

From the principle of work and energy, U = T2 − T1 , where T1 = 0. The work done is b

U = ∫ F dx = Fb. 0

The kinetic energy is T2 =

−mg dh = M θ − mgR sin θ.

From kinematics, v = Rω, from which T2 =

R sin θ 0

The angle is π θ = 40 = 0.6981 rad. 180 The kinetic energy is T2 =

0 30 m m M

U = ∫ M d θ +∫

( 12 )m v + 4( 12 )mv + 4( 12 ) I ω . C

From kinematics, v = Rω, from which T2 =

( m2 + 2m + 2 RI )v . C

Substitute into U = T2 and solve: v =

514

2 Fb mC + 4m + 4

( RI )

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Problem 19.99 Each pulley has moment of inertia I = 0.003 kg-m 2 , and the mass of the belt is 0.2 kg. If a constant couple M = 4 N-m is applied to the bottom pulley, what will its angular velocity be when it has turned 10 revolutions?

Solution:

Assume that the system is initially stationary. From the principle of work U = T2 − T1 , where T1 = 0. The work done is θ

U = ∫ M d θ = M θ, 0

where the angle is θ = 10 ( 2π ) = 62.83 rad. The kinetic energy is 1 1 T2 = m v2 + 2 I ω 2. 2 belt 2 pulley

100 mm

( )

From kinematics, v = Rω, where R = 0.1 m, from which T2 =

( 12 )(R m 2

belt

+ 2 I pulley )ω 2 .

Substitute into U = T2 and solve: ω =

2Mθ

( R 2m belt + 2 I pulley )

= 250.7 rad/s.

Problem 19.100 The ring gear is fixed. The mass and moment of inertia of the sun gear are m S = 22 slug and I S = 4400 slug-ft 2 . The mass and moment of inertia of each planet gear are m P = 2.7 slug and I P = 65 slug-ft 2 . A couple M = 600 ft-lb is applied to the sun gear. Use work and energy to determine the angular velocity of the sun gear after it has turned 100 revolutions.

vS vS

Ring gear

7 in

34 in

M 20 in The velocity of the outer radius of the sun gear is v S = R S ω S . The velocity of the center of mass of the planet gears is the average velocity of the velocity of the sun gear contact and the ring gear contact, Planet gears (3) Sun gear

vP =

since v R = 0. The angular velocity of the planet gears is ωP =

Solution:

Denote the radius of planetary gear, R P = 7 in. = 0.5833 ft, the radius of sun gear R S = 20 in. = 1.667 ft, and angular velocities of the sun gear and planet gear by ω S , ω P . Assume that the system starts from rest. From the principle of work and energy U = T2 − T1 , where T1 = 0. The work done is θ

U = ∫ M d θ = M θ, 0

θ = (100)(2π ) = 200 π = 628.3 rad.

( )

Collect terms: R ω ω P =  S  S ,  RP  2 vP =

ωS =

(( 12 ) m v + ( 12 ) I ω ). 2 P P

RS ω . 2 S

2Mθ = 12.5 rad/s.   R S  2   3      I S +  m P R S2 + I P  4 R P   

2 P

M19_BEDF2222_06_SE_C19_Part2.indd 515

vP . RP

Substitute into U = T2 and solve:

where the angle is

The kinetic energy is 1 T2 = I ω2 + 3 2 S S

vS + vR v = S, 2 2

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Problem 19.101 The moments of inertia of gears A and B are I A = 0.014 slug-ft 2 and I B = 0.100 slug-ft 2 . Gear A is connected to a torsional spring with constant k = 0.2 ft-lb/rad. If the spring is unstretched and the surface supporting the 5-lb weight is removed, what is the velocity of the weight when it has fallen 3 in?

Solution: Denote W = 5 lb, s = 3 in = 0.25 ft is the distance the weight falls, rB = 10 in = 0.833 ft, r A = 6 in = 0.5 ft, rB′ = 3 in = 0.25 ft, are the radii of the gears and pulley. Choose a coordinate system with y positive upward. From the conservation of energy T + V = const. Choose the datum at the initial position, such that V1 = 0, T1 = 0, from which T2 + V2 = 0 at any position 2 . The gear B rotates in a negative direction and the gear A rotates in a positive direction. By inspection, s 0.25 θB = − ′ = − = −1 rad, rB 0.25 r  θ A = − B  θ B = 1.667 rad,  rA 

10 in

3 in

v = rB′ ω B = 0.25 ω B , 6 in

r  ω A = − B  ω B = 1.667 ω B .  rA 

B A

5 lb

The moment exerted by the spring is negative, from which the potential energy in the spring is θA θA 1 Vspring = −∫ M d θ =∫ kθ d θ = kθ A2 = 0.2778 ft-lb. 0 0 2 The force due to the weight is negative, from which the potential energy of the weight is Vweight = −∫

−s 0

(−W ) dy = −Ws = −1.25 ft-lb.

The kinetic energy of the system is T2 =

( 12 ) I ω + ( 12 ) I ω + ( 12 ) Wg v . A

2 A

2 B

Substitute: T2 = 1.1888v 2 , from which Vspring + V weight + 1.1888v 2 = 0. Solve v = 0.9043 ft/s (downward).

Problem 19.102 Consider the system in Problem 19.101. (a) What maximum distance does the 5-lb weight fall when the supporting surface is removed? (b) What maximum velocity does the weight achieve? Solution:

Use the solution to Problem 19.101:

Vspring + V weight + 1.1888v 2 = 0. Vspring = −∫

θA 0

Vweight = −∫

−s 0

M d θ =∫

θA 0

kθ d θ =

1 2 kθ = 4.444 s 2 , 2 A

6 in

10 in

B A

(−W ) dy = −Ws,

from which 4.4444 s 2 − 5s + 1.1888v 2 = 0. (a) The maximum travel occurs when v = 0, from which s max =

3 in

5 lb

5 = 1.13 ft. 4.4444

(where the other solution s max = 0 is meaningless here). (b) The maximum velocity occurs at dv 2 4.4444 5 = 0 = 2 s− = 0, ds 1.1888 1.1888 from which s v − max = 0.5625 ft. This is indeed a maximum, since d 2 (v 2 ) 4.444 = 2 > 0, ds 2 1.1888

(

)

and v max = 1.09 ft/s (downward).

516

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Problem 19.103 Each of the go-cart’s front wheels weighs 5 lb and has a moment of inertia of 0.01 slug-ft 2 . The two rear wheels and rear axle form a single rigid body weighing 40 lb and having a moment of inertia of 0.1 slug-ft 2 . The total weight of the rider and go-cart, including its wheels, is 240 lb. The go-cart starts from rest, its engine exerts a constant torque of 15 ft-lb on the rear axle, and its wheels do not slip. Neglecting friction and aerodynamic drag, how fast is the go-cart moving when it has traveled 50 ft?

Solution:

From the principle of work and energy: U = T2 − T1 , where T1 = 0, since the go-cart starts from rest. Denote the rear and front wheels by the subscripts A and B, respectively. The radius of the rear wheels r A = 6 in. = 0.5 ft. The radius of the front wheels is rB = 4 in. = 0.3333 ft. The rear wheels rotate through an angle 50 θA = = 100 rad, from which the work done is 0.5 U = ∫

θA 0

M d θ = M θ A = 15(100) ft-lb.

The kinetic energy is T2 =

( 12 ) Wg v + ( 12 ) I ω + 2( 12 ) I ω 2

2 A

2 B

(for two front wheels). The angular velocities are related to the v v gocart velocity by ω A = = 2v , ω B = = 3v , from which rA rB 2 T2 = 4.020v ft-lb. Substitute into U = T2 and solve: v = 19.32 ft/s.

6 in

4 in

B 60 in

15 in.

6 in. 16 in.

4 in. 60 in.

Problem 19.104 Determine the maximum power and the average power transmitted to the go-cart in Problem 19.103 by its engine. Solution:

The maximum power is Pmax = M ω Amax , where v max Mv max . From which Pmax = . Under constant torque, R R the acceleration of the go-cart is constant, from which the maximum velocity is the greatest value of the velocity, which will occur at the end of the travel. From the solution to Problem 19.103, v max = 19.32 ft/s, from which Mv max 15(19.32) = = 580 ft-lb/s. Pmax = rA 0.5 U The average power is Pave = . From the solution to Problem t 19.103, U = 1500 ft-lb. Under constant acceleration, v = at, and 1 2s 100 1 = = 5.177 s, s = at 2 , from which s = v t, and t = 2 v 19.32 2 from which U 1500 Pave = = = 290 ft-lb/s. t 5.177 ω Amax =

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6 in

4 in

B 60 in

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Problem 19.105 The system starts from rest with the 4-kg slender bar horizontal. The mass of the suspended cylinder is 10 kg. What is the angular velocity of the bar when it is in the position shown? Solution:

From the principle of work and energy: U = T2 − T1 , where T1 = 0 since the system starts from rest. The change in height of the cylindrical weight is found as follows: By inspection, the distance between the end of the bar and the pulley when the bar is in the horizontal position is d 1 = 2 2 + 3 2 = 3.61 m. The law of cosines is used to determine the distance between the end of the bar and the pulley in the position shown: d2 =

458

2 2 + 3 2 − 2(2)(3) cos 45 ° = 2.125 m,

from which h = d 1 − d 2 = 1.481 m. The work done by the cylindrical weight is U cylinder = ∫

−h 0

−m c g ds = m c gh = 145.3 N-m.

The work done by the weight of the bar is U bar = ∫

cos 45° 0

−m b g dh = −m b g cos 45 ° = −27.75 N-m,

from which U = U cylinder + U bar = 117.52 N-m. From the sketch, (which shows the final position) the component of velocity normal to the bar is v sin β , from which 2ω = v sin β . From the law of sines:  3  sin β =  sin 45 ° = 0.9984.  d2 

The kinetic energy is T2 =

458

( 12 )m v + ( 12 ) I ω , c

2 b

from which T2 = 22.732ω 2 , where v =

( sin2 β )ω and I = m(23 ) 2

has been used. Substitute into U = T2 and solve: ω = 2.274 rad/s.

Problem 19.106 The 0.1-kg slender bar and 0.2-kg cylindrical disk are released from rest with the bar horizontal. The disk rolls on the curved surface. What is the angular velocity of the bar when it is vertical?

40 mm

Solution: From the principle of work and energy, U = T2 − T1, where T1 = 0. Denote L = 0.12 m, R = 0.04 m, the angular velocity of the bar by ω B , the velocity of the disk center by v D , and the angular velocity of the disk by ω D . The work done is U = ∫

−

L 2

−m B g dh +∫

−L 0

( L2 )m g + Lm g.

−m D g dh =

120 mm

v From kinematics, v D = Lω B , and ω D = D . The kinetic energy is R 1 1 1 T2 = I B ω B2 + m Dv D2 + I D ω D2 . 2 2 2

( )

Substitute the kinematic relations to obtain T2 =

( 12 )( I + m L + I ( RL ) )ω , B

m B L2 , 3 2 m R ID = D , 2 1 from which T2 = 2

2 B

where I B =

( )( m3 + 3m2 ) L ω . B

2 B

Substitute into U = T2 and solve: ωB =

518

6 g(m B + 2m D ) = 11.1 rad/s. (2m B + 9m D ) L

L/2

L/2 vD

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Problem 19.107 The slender bar of mass m is released from rest in the vertical position and allowed to fall. Neglecting friction and assuming that it remains in contact with the floor and wall, determine the bar’s angular velocity as a function of θ.

C L 2 G

u v

The velocity of the center of mass is

Solution:

The strategy is

(a) to use the kinematic relations to determine the relation between the velocity of the center of mass and the angular velocity about the instantaneous center, and (b) the principle of work and energy to obtain the angular velocity of the bar. The kinematics: Denote the angular velocity of the bar about the instantaneous center by ω. The coordinates of the instantaneous center of rotation of the bar are ( L sin θ, L cos θ). The coordinates of the center of mass of the bar are

( L2 sin θ, L2 cos θ ).

   i j k   0 0 v G = ω × rG / C =  ω    L L  − sin θ − cos θ 0    2 2 ωL = ( i cos θ − j sin θ), 2 ωL from which v G = . 2 The principle of work and energy: U = T2 − T1 where T1 = 0. The work done by the weight of the bar is L U = mg (1 − cos θ). 2 The kinetic energy is

( )

T2 =

( 12 )mv + ( 12 ) I ω . 2 G

Substitute v G =

mL2 ωL mL2 ω 2 and I B = to obtain T2 = . 12 6 2

Substitute into U = T2 and solve: ω =

3g(1 − cos θ) . L

The vector distance from the instantaneous center to the center of mass is L rG /C = − ( i sin θ + j cos θ). 2

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Problem 19.108 The 4-kg slender bar is pinned to 2-kg sliders at A and B. If friction is negligible and the system starts from rest in the position shown, what is the bar’s angular velocity when the slider at A has fallen 0.5 m?

vA u2

458

( x B 2 , y B 2 ) = (0.5 + d , −(1.2 + d )) = (0.8138, −1.514). The vector distance from A to B is r B /A = ( x B 2 − x A 2 ) i + ( y B 2 − y A 2 ) j = 0.8138i − 1.014 j (m).

1.2 m

Check: rB /A = L = 1.3 m. check. The vector distance from A to the center of mass is rG /A = ( x G 2 − x A 2 ) i + ( y G 2 − y A 2 ) j = 0.4069 i − 0.5069 j (m). L Check: rG /A = = 0.65 m. check. 2 The kinematic relations: From kinematics, v B = v A + ω × rB / A . The slider A is constrained to move vertically, and the slider B moves at a 45 ° angle, from which v A = −v A j (m/s), and

458 B

0.5 m

Solution:

Choose a coordinate system with the origin at the initial position of A and the y axis positive upward. The strategy is

(a) to determine the distance that B has fallen and the center of mass of the bar has fallen when A falls 0.5 m, (b) use the coordinates of A, B, and the center of mass of the bar and the constraints on the motion of A and B to determine the kinematic relations, and

 i j k   0 0 v B = v A +  ω   0.8138 −1.014 0    = i(1.014 ω ) + j(−v A + 0.8138ω ), from which (1) v B =

1.014 ( cos )ω = 1.434ω, 45 °

and (2) v A = v B sin 45 ° + 0.8138ω = 1.828ω.

The displacement of B: Denote the length of the bar by L = 1.2 2 + 0.5 2 = 1.3 m. Denote the horizontal and vertical dis-

 i j k   0 0 v G = v A + ω × rG / A = −v A j +  ω   0.4069 −0.5069 0    = (0.5069ω ) i + (−v A + 0.4069ω ) j,

placements of B when A falls 0.5 m by d x and d y , which are in the ratio dy = tan 45 ° = 1, from which d x = d y = d. The vertical distance dx between A and B is reduced by the distance 0.5 m and increased by the distance d y , and the horizontal distance between A and B is increased by the distance d x , from which L2 = (1.2 − 0.5 + d y ) 2 + (0.5 + d x ) 2 . Substitute d x = d y = d and L = 1.3 m and reduce to obtain d 2 + 2bd + c = 0, where b = 0.6, and c = −0.475. Solve: d 1,2 = −b ± b 2 − c = 0.3138 m, or = −1.514 m, from which only the positive root is meaningful. The final position coordinates: The coordinates of the initial position of the center of mass of the bar are L L ( x G1 , y G1 ) = sin θ1 , − cos θ1 = (0.25, − 0.6) (m), 2 2 0.5 where θ1 = sin −1 = 22.61 ° L is the angle of the bar relative to the vertical. The coordinates of the final position of the center of mass of the bar are L L (x G 2 , y G 2 ) = sin θ 2 , − 0.5 − cos θ 2 = (0.4069, − 1.007), 2 2 0.5 + d where θ 2 = sin −1 = 38.75 °. L The vertical distance that the center of mass falls is h = y G 2 − y G1 = −0.4069 m. The coordinates of the final positions of A and B are, respectively ( x A 2 , y A 2 ) = (0, −0.5), and

(

520

v B = (v B cos45°) i − (v B sin45 °) j.

)

( )

(

)

The velocity of the center of mass of the bar is

v G = 0.5069ω i − 1.421ω j (m/s), from which (3) v G = 1.508ω The principle of work and energy: From the principle of work and energy, U = T2 − T1 , where T1 = 0. The work done is U = ∫

−d 1 0

−m B g dh +∫

−0.5 0

−m A g dh +∫

−h 0

−m bar g dh,

U = m B g(0.3138) + m A g(0.5) + m bar g(0.4069) = 31.93 N-m. The kinetic energy is T2 =

( 12 )m v + ( 12 )m v + ( 12 )m v + ( 12 ) I ω , 2 B B

2 A A

2 bar G

bar

m bar L2 , and (1), (2) and (3) to obtain T2 = 10.23ω 2 . 12 Substitute into U = T2 and solve:

substitute I bar =

ω =

31.93 = 1.77 rad/s. 10.23

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Problem 19.109 The homogeneous hemisphere of mass m is released from rest in the position shown. If it rolls on the horizontal surface, what is its angular velocity when its flat surface is horizontal?

2 The hemisphere’s moment of inertia about O is mR 2 , 5 so its moment of inertia about G is 2 2 3 83 I = mR 2 − R m = mR 2 . 5 8 320 The work done is

Solution:

( )

(

mg R −

3R 8

)

5 3 R = mgR. 8 8

Work and energy is 3 1 1 mgR = mv 2 + I ω 2 8 2 2 1  5R  2 1 83 = m mR 2 ω 2 . ω +  2  8 2 320

( )

(

)

ω =

5R 8

Solving for ω, we obtain 3R 8

15g . 13 R

Problem 19.110 The homogeneous hemisphere of mass m is released from rest in the position shown. It rolls on the horizontal surface. What normal force is exerted on the hemisphere by the horizontal surface at the instant the flat surface of the hemisphere is horizontal? Solution:

See the solution of Problem 19.109. The acceleration

of G is a G = a O + α × rG / O − ω 2 rG / O i j k 0 0 α 0 −h 0

= aO i +

3R 8

− ω 2 (−h j), O

so a Gy = ω 2 h. Therefore N − mg = ma Gy = mω 2 h. Using the result from the solution of Problem 19.109, N = mg + m

15g 3 R ( 13 )( ) = 1.433 mg. R 8

Problem 19.111 The slender bar rotates freely in the horizontal plane about a vertical shaft at O. The bar weighs 20 lb and its length is 6 ft. The slider A weighs 2 lb. If the bar’s angular velocity is ω = 10 rad/s and the radial component of the velocity of A is zero when r = 1 ft, what is the angular velocity of the bar when r = 4 ft? (The moment of inertia of A about its center of mass is negligible; that is, treat A as a particle.) v

Solution:

From the definition of angular momentum, only the radial position of the slider need be taken into account in applying the principle of the conservation of angular momentum; that is, the radial velocity of the slider at r = 4 ft does not change the angular momentum of the bar. From the conservation of angular momentum: I bar ω 1 + r12A

( Wg )ω = I ω + r ( Wg )ω . A

bar

2 2A

W  W  W I bar + r12A  A  = bar L2 + r12A  A  = 7.52 slug-ft 2  g   g  3g W  W  W I bar + r212 A  A  = bar L2 + r22A  A  = 8.46 slug-ft 2 ,  g   g  3g 7.52 ( 8.46 )ω = 8.90 rad/s. 1

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Substitute numerical values:

from which ω 2 = O

3R 5 h 8

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Problem 19.112 A satellite is deployed with angular velocity ω = 1 rad/s (Fig. a). Two internally stored antennas that span the diameter of the satellite are then extended, and the satellite’s angular velocity decreases to ω′ (Fig. b). By modeling the satellite as a 500-kg sphere of 1.2-m radius and each antenna as a 10-kg slender bar, determine ω′.

v 1.2 m v' (a)

Solution:

Assume (I) in configuration (a) the antennas are folded inward, each lying on a line passing so near the center of the satellite that the distance from the line to the center can be neglected; (II) when extended, the antennas are entirely external to the satellite. Denote R = 1.2 m, L = 2 R = 2.4 m. The moment of inertia of the antennas about the center of mass of the satellite in configuration ( a ) is mL2 I ant-folded = 2 = 9.6 kg-m 2 . 12 The moment of inertia of the antennas about the center of mass of the satellite in configuration ( b ) is

(

I ant-ext = 2

2.4 m

2.4 m (b)

)

( mL ) + 2( R + L2 ) m = 124.8 kg-m . 12 2

The moment of inertia of the satellite is 2 I sphere = m sphere R 2 = 288 kg-m 2 . 5 The angular momentum is conserved, I sphere ω + I ant-folded ω = I sphere ω ′ + I ant-ext ω ′, where I ant-folded , I ant-ext are for both antennas, from which 297.6ω = 412.8ω ′. Solve ω′ = 0.721 rad/s.

Problem 19.113 An engineer decides to control the angular velocity of a satellite by deploying small masses attached to cables. If the angular velocity of the satellite in configuration (a) is 4 rpm, determine the distance d in configuration (b) that will cause the angular velocity to be 1 rpm. The moment of inertia of the satellite is I = 500 kg-m 2 and each mass is 2 kg. (Assume that the cables and masses rotate with the same angular velocity as the satellite. Neglect the masses of the cables and the moments of inertia of the masses about their centers of mass.) Solution:

4 rpm

1 rpm

(a)

(b)

4 rpm

From the conservation of angular momentum,

522

d =

ω  ( I + 8m) 1  − I  ω2  = 19.8 m. 2m

1 rpm

2m 2m

( I + 2m(2 2 ))ω 1 = ( I + 2md 2 )ω 2 . Solve:

d (a)

d (b)

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Problem 19.114 The homogeneous cylindrical disk of mass m rolls on the horizontal surface with angular velocity ω. If the disk does not slip or leave the slanted surface when it comes into contact with it, what is the angular velocity ω′ of the disk immediately afterward?

R b

Solution:

The velocity of the center of mass of the disk is parallel to the surface before and after contact. The angular momentum about the point of contact is conserved mv R cos β + I ω = mv ′R + I ω ′. From kinematics, v = Rω and v ′ = Rω ′. Substitute into the angular moment condition to obtain:

v9 v

(mR 2 cos β + I )ω = (mR 2 + I )ω ′. Solve: ω ′ =

(2cos β + 1) ω. 3

Problem 19.115 The 10-lb bar falls from rest in the vertical position and hits the smooth projection at B. The coefficient of restitution of the impact is e = 0.6, the duration of the impact is 0.1 s, and b = 1 ft. Determine the average force exerted on the bar at B as a result of the impact.

3 ft

Solution: Choose a coordinate system with the origin at A and the x axis parallel to the plane surface, and y positive upward. The strategy is to (a) use the principle of work and energy to determine the velocity before impact, (b) the coefficient of restitution to determine the velocity after impact, (c) and the principle of angular impulse-momentum to determine the average force of impact.

B b

From the principle of work and energy, U = T2 − T1 , where L T1 = 0. The center of mass of the bar falls a distance h = . 2 L The work done by the weight of the bar is U = mg . The 2 1 mL2 kinetic energy is T2 = I ω 2 , where I = . Substitute into 2 3 3g U = T2 and solve: ω = − , where the negative sign on the L square root is chosen to be consistent with the choice of coordinates. By v ′ − v ′A , where v A , v ′A definition, the coefficient of restitution is e = B vA − vB are the velocities of the bar at a distance b from A before and after

( )

impact. Since the projection B is stationary before and after the impact, v B = v ′B = 0, from which v ′A = −ev A . From kinematics, v A = bω, and v ′A = bω ′, from which ω ′ = −eω. The principle of angular impulse-momentum about the point A is t 2 − t1 mL2 ∫0 bFB dt = (I ω ′ − I ω) = 3 (ω ′ − ω), where FB is the force exerted on the bar by the projection at B, from which mL2 bFB (t 2 − t 1 ) = − (1 + e)ω. 3 Solve: FB =

mL2 (1 + e) 3g = 84.63 lb. 3b(t 2 − t 1 ) L

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Problem 19.116 The 10-lb bar falls from rest in the vertical position and hits the smooth projection at B. The coefficient of restitution of the impact is e = 0.6 and the duration of the impact is 0.1 s. Determine the distance b for which the average force exerted on the bar by the support A as a result of the impact is zero. Solution:

3 ft

From the principle of linear impulse-momentum,

∫t ∑ F dt = m(v ′G − v G ),

where v G , v ′G are the velocities of the center of mass of the bar before and after impact, and ∑ F = FA + FB are the forces exerted on the L L bar at A and B. From kinematics, v ′G = ω ′, v G = ω. From 2 2 the solution to Problem 19.115, ω ′ = −eω, from which

( )

FA + FB = −

( )

mL (1 + e) ω. 2(t 2 − t 1 )

If the reaction at A is zero, then FB = −

mL (1 + e) ω. 2(t 2 − t 1 )

From the solution to Problem 19.115, FB = −

mL2 (1 + e) ω. 3b(t 2 − t 1 ) 2 L = 2 ft. 3

Substitute and solve: b =

Problem 19.117 The 1-kg sphere A is moving at 2 m/s when it strikes the end of the 2-kg stationary slender bar B. If the velocity of the sphere after the impact is 0.8 m/s to the right, what is the coefficient of restitution?

Solution:

Denote the distance of the point of impact P from the end of the bar by d = 0.4 m. The linear momentum is conserved: (1) m Av A = m Av ′A + m Bv ′CM , where v ′CM is the velocity of the center of mass of the bar after impact, and v A , v ′A are the velocities of the sphere before and after impact. By definition, the coefficient of restitution is v ′p − v ′A e = , where v p , v ′p are the velocities of the bar at the point vA − vp of impact. The point P is stationary before impact, from which (2)

2 m/s 400 mm

v ′p − v ′A = ev A . From kinematics,

(

)

L − d ω ′. 2 Substitute (2) into (3) to obtain (3) v ′CM = v ′P −

(4) v ′CM = v ′A + ev A −

( L2 − d )ω′.

Substitute (4) into (1) to obtain (5) ( m A − em B )v A = ( m A + m B )v ′A −

v9A

v9P

( L2 − d )m ω′. B

The angular momentum of the bar about the point of impact is conserved: L (6) 0 = − − d m Bv ′CM + I CM ω ′. 2 Substitute (3) into (6) to obtain,

(

(7)

)

( L2 − d )m ( v ′ + ev ) = ( I B

( L2 − d ) m )ω′, 2

m B L2 . 12 Solve the two equations (5) and (7) for the two unknowns to obtain: ω′ = 1.08 rad/s and e = 0.224. where I CM =

524

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Problem 19.118 The slender bar is released from rest in the position shown in Fig. a and falls a distance h = 1 ft. When the bar hits the floor, its tip is supported by a depression and remains on the floor (Fig. b). The length of the bar is 1 ft and its weight is 4 oz. What is angular velocity ω of the bar just after it hits the floor?

458

Solution: Choose a coordinate system with the x axis parallel to the surface, with the y axis positive upward. The strategy is to use the principle of work and energy to determine the velocity just before impact, and the conservation of angular momentum to determine the angular velocity after impact. From the principle of work and energy, U = T2 − T1 , where T1 = 0 since the bar is released from rest. The work done is 1 U = −mgh, where h = 1 ft. The kinetic energy is T2 = mv 2 . 2 Substitute into U = T2 − T1 and solve: v = −2 gh = 8.02 ft/s.

( )

The conservation of angular momentum about the point of impact is (r × mv) = (r × mv′) + I B ω ′. At the instant before the impact, the perpendicular distance from the point of impact to the center of mass velocity L vector is cos 45 °. After impact, the center of mass moves in an arc of 2 L radius about the point of impact so that the perpendicular distance 2 L . From the definition from the point of impact to the velocity vector is 2 of the cross product, for motion in the x, y plane, L ( r × mv ) ⋅ k = cos45 ° mv , 2 L ( r × mv ′ ) ⋅ k = mv ′, 2 and I Bω′ ⋅ k = I B ω ′. L From kinematics, v ′ = ω ′. Substitute to obtain: 2

( ) ( )

v h (a)

(b)

( )

( ( )

ω′ =

)

mv ( cos45 ° ) 3v 3 −gh = = = 8.51 rad/s. 2I mL 2L 2 2L + B 2 L

(

)

Problem 19.119 The slender bar is released from rest with θ = 45° and falls a distance h = 1 m onto the smooth floor. The length of the bar is 1 m and its mass is 2 kg. If the coefficient of restitution of the impact is e = 0.4, what is the angular velocity of the bar just after it hits the floor?

Solution:

Choose a coordinate system with the x axis parallel to the plane surface and y positive upward. The strategy is to (a) use the principle of work and energy to obtain the velocity the instant before impact, (b) use the definition of the coefficient of restitution to find the velocity just after impact, (c) get the angular velocity-velocity relations from kinematics and (d) use the principle of the conservation of angular momentum about the point of impact to determine the angular velocity. From the principle of work and energy, U = T2 − T1 , where T1 = 0. The center of mass of the bar also falls a distance h before impact. −h

The work done is U = ∫ −mg dh = mgh. The kinetic energy is 0 1 2 . Substitute into U = T and solve: v T2 = mv Gy 2 Gy = − 2 gh , 2 where the negative sign on the square root is chosen to be consistent with the choice of coordinates. Denote the point of impact on the bar by P. From the definition of the coefficient of restitution, v ′Py − v ′By e = . v By − v Py Since the floor is stationary before and after impact, v B = v ′B = 0, and (1) v ′Py = −ev Py = −ev Gy . From kinematics: v′P = v′G + L ω′ × rP / G , where rP / G = ( i cos θ − j sin θ), from which 2

( )

v9 vGy

v9Gy

 i j k  L  0 0 ω ′  2   cos sin 0 θ θ −   L = v ′Gy j + ( iω ′ sin θ + jω ′ cos θ). 2

(2) v′Py = v′Gy +

( )

Substitute (1) into (2) to obtain (3) v ′Gy = −ev Gy −

v9px

( L2 )ω′ cos θ.

The angular momentum is conserved about the point of impact:

( L2 )cos θ mv

(4) −

( L2 )cos θ mv ′ + I ω′.

= −

Substitute (3) into (4) to obtain −

(

)

.L L 2 m cos θ(1 + e)v G = I G + m cos 2 θ ω ′. 2 2

( )

Solve: ω′ = −

mL (1 + e) cos θ v Gy , L cos θ 2 2 IG + m 2

ω′ = − v9Gx

( )

(

) )

6(1 + e) cos θ (− 2 gh ) = 10.52 rad/s , (1 + 3cos 2 θ)

where I G =

mL2 has been used. 12

v9py

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Problem 19.120 The slender bar is released from rest and falls a distance h = 1 m onto the smooth floor. The length of the bar is 1 m and its mass is 2 kg. The coefficient of restitution of the impact is e = 0.4. Determine the angle θ for which the angular velocity of the bar after it hits the floor is a maximum. What is the maximum angular velocity? Solution:

From the solution to Problem 19.119,

6(1 + e) cos θ ω′ = 2 gh . (1 + 3cos 2 θ) Take the derivative: −6(1 + e)sin θ 6(1 + e)(6) cos 2 θ sin θ dω′ = 0 = 2 gh + 2 gh dθ (1 + 3cos 2 θ) (1 + 3cos 2 θ) 2 = 0, from which 3cos 2 θ max − 1 = 0, cos θ max =

1 , θ = 54.74 ° , 3 max

′ and ω max = 10.7 rad/s .

Problem 19.121 A nonrotating slender bar A moving with velocity v 0 strikes a stationary slender bar B. Each bar has mass m and length l. If the bars adhere when they collide, what is their angular velocity after the impact? Solution:

From the conservation of linear momentum, mv 0 = mv ′A + mv ′B . From the conservation of angular momentum about the mass center of A 0 = I Bω′A + I Bω′B + (r × mv′B )

l 2 v0

 i j k     L 0  , = ( I B ω ′A + I B ω ′B )k +  0 2    v ′B 0 0    L 0 = ( I B ω ′A + I B ω ′B )k − m v ′ k. 2 B

( ) )

Since the bars adhere, ω ′A = ω ′B , from which 2 I B ω ′ = m

( L2 )v ′ . B

From kinematics   i  v′B = v′A + ω′ × r AB = v ′A i +  0   0  L = v ′A − ω ′ i, 2 L from which v ′B = v ′A − ω ′. 2

(

j 0 L 2

 k  ω ′   0  

)

v9B v9A

( )

Substitute into the expression for conservation of linear momentum to obtain L v0 − ω′ 2 v ′B = . 2 Substitute into the expression for the conservation of angular momentum to obtain:

( )

ω′ =

( mL2 )v

mL2 4I B + 4

from which ω′ =

526

⋅ IB =

mL2 , 12

6v 0 . 7L

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Problem 19.122 An astronaut translates toward a nonrotating satellite at 1.0 i (m/s) relative to the satellite. Her mass is 136 kg, and the moment of inertia about the axis through her center of mass parallel to the z-axis is 45 kg-m 2 . The mass of the satellite is 450 kg and its moment of inertia about the z-axis is 675 kg-m 2 . At the instant the astronaut attaches to the satellite and begins moving with it, the position of her center of mass is (−1.8, −0.9, 0) m. The axis of rotation of the satellite after she attaches is parallel to the z-axis. What is their angular velocity? y

(3) 0.9m Av Ax = −1.8m Av ′Ay + 0.9m Av ′Ax + ( I A + I S )ω ′. From kinematics:  i j k   v′A = v′S + ω′ × r A / S = v S +  0 0 ω ′   −1.8 −0.9 0    = (v ′Sx + 0.9ω ′) i + (v ′Sy − 1.8ω ′) j from which (4) v ′Ax = v ′Sx + 0.9ω ′, (5) v ′Ay = v ′Sy − 1.8ω ′. With v Ax = 1.0 m/s, these are five equations in five unknowns. The number of equations can be reduced further, but here they are solved by iteration (using TK Solver Plus) to obtain: v ′Ax = 0.289 m/s, v ′Sx = 0.215 m/s, v ′Ay = −0.114 m/s, v ′Sy = 0.0344 m/s, v ′Sy = 0.0344 m/s, ω′ = 0.0822 rad/s.

x 1 m/s

v9 v9Sy

v9Ay

v9Sx v9Ax

Solution:

Choose a coordinate system with the origin at the center of mass of the satellite, and the y axis positive upward. The linear momentum is conserved: (1) m Av Ax = m Av ′Ax + m Sv ′Sx ,

(2) 0 = m Av ′Ay + m Sv ′Sy . The angular momentum about the center of mass of the satellite is conserved: r A / S × m A v A = r A / S × m A v′A + ( I A + I S )ω ′, where r A /S = −1.8i − 0.9 j( m ), and v A = 1.0 i(m/s).   i j k  i j k     −1.8 −0.9 0  =  −1.8 −0.9 0        m v′  m Av Ax 0 0    A Ax m Av ′Ay 0   + ( I A + I S )ω ′, from which

Problem 19.123 In Problem 19.122, suppose that the design parameters of the satellite’s control system require that the angular velocity of the satellite not exceed 0.02 rad/s. If the astronaut is moving parallel to the x-axis and the position of her center of mass when she attaches is (−1.8, −0.9, 0) m, what is the maximum relative velocity at which she should approach the satellite? Solution:

x 1 m/s

From the solution to Problem 19.122, the five equa-

tions are: (1) m Av Ax = m Av ′Ax + m Sv ′S , (2) 0 = m Av ′Ay + m Sv ′Sy (3) 0.9m Av Ax = −1.8m Av ′Ay + 0.9m Av ′Ax + ( I A + I S )ω ′ (4) v ′Ax = v ′Sx + 0.9ω ′, (5) v ′Ay = v ′Sy − 1.8ω ′.

With ω′ = 0.02 rad/s, these five equations have the solutions: v Ax = 0.243 m/s , v ′Ax = 0.070 m/s, v ′Sx = 0.052 m/s, v ′Ay = −0.028 m/s, v ′Sy = 0.008 m/s.

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Problem 19.124 A 2800-lb car skidding on ice strikes a concrete abutment at 3 mi/h. The car’s moment of inertia about its center of mass is 1800 slug-ft 2 . Assume that the impacting surfaces are smooth and parallel to the y-axis and that the coefficient of restitution of the impact is e = 0.8. What are the angular velocity of the car and the velocity of its center of mass after the impact? y

Solution: Let P be the point of impact on with the abutment. (P is located on the vehicle.) Denote the vector from P to the center of mass of the vehicle by r = a i + 2 j, where a is unknown. The velocity of the vehicle is 5280 v Gx = 3 = 4.4 ft/s. 3600 From the conservation of linear momentum in the y direction, mv Gy = mv ′Gy , from which v ′Gy = v Gy = 0. Similarly, v Ay = v ′Ay = 0. By definition, e =

v9Gx

v9Px

The conservation of angular momentum about P is r × mv G = r × mv′G + I ω′. From kinematics, v′G = v′P + ω′ × r. Reduce:

 i j k    v ′Gx = −(ev Gx + 2ω ′), r × mv G =  a 2 0   4.4 m 0 0    = −8.8mk.

from which

(

v9Py

 i j k    v′G = v ′Px i + ω′ × r =  0 0 ω ′   a 2 0    = (v ′Px − 2ω ′) i + aω ′ j = (−ev Gx − 2ω ′) i + aω ′ j,

3 mi/h 2 ft

v9Gy

 i j k   r × mv′G =  a 2 0   −(ev Gx + 2ω ′)m 0 0    = 2(ev Gx + 2ω ′)mk, I ω′ = I ω ′k.

)

v ′Px − v ′Ax . v Ax − v Px

Substitute into the expression for the conservation of angular momentum to obtain ω′ =

Assume that the abutment does not yield under the impact, v Ax = v ′Ax = 0, from which v ′Px = −ev Px = −ev Gx = −(0.8)(4.4) = −3.52 ft/s.

(

−15.84 I 22 + m

)

−15.84 = −0.641 rad/s. (4 + 20.68)

The velocity of the center of mass is v′G = v ′Gx i = −(ev Gx + 2ω ′) i = −2.24 i (m/s)

Problem 19.125 A 170-lb wide receiver jumps vertically to receive a pass and is stationary at the instant he catches the ball. At the same instant, he is hit at P by a 180-lb linebacker moving horizontally at 15 ft/s. The wide receiver’s moment of inertia about his center of mass is 7 slug-ft 2 . If you model the players as rigid bodies and assume that the coefficient of restitution is e = 0, what is the wide receiver’s angular velocity immediately after the impact? Solution:

14 in P

Denote the receiver by the subscript B, and the tacker 14 = 1.17 ft. The conservation of 12 linear momentum: (1) m Av A = m Av ′A + m Bv ′B . From the definition, v ′ − v ′A . Since v BP = 0, (2) ev A = v ′BP − v ′A . The angular e = BP v A − v BP momentum is conserved about point P: (3) 0 = −m B dv ′B + I B ω ′B . by the subscript A. Denote d =

( )

From kinematics: (4) v ′A = v ′B + d ω ′B . The Solution: For e = 0, from (1), mB v ′A = v A − v′ . mA B

( )

v9B

From (2) and (4) v ′A = v ′B + d ω ′B . From (3) v ′B =

( dmI )ω′. B

v9A

v9BP

Combine these last two equations and solve: ω′B =

528

vA = 4.445… = 4.45 rad/s. I B  m  1 + B  + d  dm B  mA 

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Chapter 20 Problem 20.1 The secondary reference frame has a constant angular velocity Ω = −2 i + 6 j + 4 k (rad/s) relative to an Earth-fixed reference frame. The rigid body has a constant angular velocity ω rel = 8i + 4 j + 2k (rad/s) relative to the secondary frame. A and B are points of the rigid body. B has a constant velocity v B = 20 i − 10 j + 15k (m/s) relative to the Earthfixed frame, and the position of A relative to B is r A /B = 3i + 2 j − 4 k (m). What are the velocity and acceleration of A relative to the Earth-fixed frame?

vrel

Solution:

The angular velocity of the rigid body relative to the Earth-

fixed frame is ω = Ω + ω rel = 6 i + 10 j + 6k (rad/s). The velocity of A relative to the Earth-fixed frame is v A = v B + ω × r A /B = 20 i − 10 j + 15k (m/s) +

= 20 i − 10 j + 15k − 52 i + 42 j − 18k (m/s) = −32 i + 32 j − 3k (m/s). Because the components of ω are constant, from Eq. (20.4) the angular acceleration of the rigid body is

α = Ω×ω =

A y

rA/B

i j k 6 rad/s 10 rad/s 6 rad/s 3m 2m −4 m

i j k −2 rad/s 6 rad/s 4 rad/s 6 rad/s 10 rad/s 6 rad/s

= −4 i + 36 j − 56k (rad/s 2 ). The acceleration of A relative to the Earth-fixed frame is a A = a B + α × r A /B + ω × (ω × r A /B )

O z

x Secondary reference frame

i = 0+

−4 rad/s 2 3m

36 rad/s 2 −56 rad/s 2 2m −4 m

i j k 6 rad/s 10 rad/s 6 rad/s −52 m/s 42 m/s −18 m/s

= −464 i − 388 j + 656k (m/s 2 ). v A = −32 i + 32 j − 3k (m/s), a A = −464 i − 388 j + 656k (m/s 2 ).

Problem 20.2 In Practice Example 20.1, suppose that the center of the tire moves at a constant speed of 5 m/s as the car turns. (As a result, when the angular velocity of the tire relative to an Earth-fixed reference frame is expressed in terms of components in the secondary reference frame, ω = ω x i + ω y j + ω z k, the components ω x , ω y , and ω z are constant.) What is the angular acceleration α of the tire relative to an Earth-fixed reference frame? Solution:

The angular acceleration is then α = Ω × ω = −(0.5 rad/s)k × (−13.9 j − 0.5k) rad/s α = −6.94 i rad/s 2 .

The angular velocity of the secondary coordinate system is

(5 m/s) v Ω = − k = − k = −(0.5 rad/s)k R (10 m) The angular velocity of the wheel with components in the secondary coordinate system is ω = Ω−

(5 m/s) v j = −(0.5 rad/s)k − j = (−13.9 j − 0.5k) rad/s r (0.36 m)

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Problem 20.3 Expressed in terms of the body-fixed coordinate system shown, the cube has a constant angular velocity ω = 4 i + 2 j − 2k (rad/s) relative to the primary reference frame. The center of mass has a constant velocity v G = 20 i + 14 j + 12k (m/s) relative to the primary frame. What are the velocity and acceleration of point A of the cube relative to the primary frame?

Solution:

The velocity of A relative to the primary frame is

v A = v G + ω × r A /G = 20 i + 14 j + 12k (m/s) +

i j k 4 rad/s 2 rad/s −2 rad/s 1m 1m 1m

= 20 i + 14 j + 12k + 4 i − 6 j + 2k (m/s) = 24 i + 8 j + 14 k (m/s).

Because ω is constant, the acceleration of A relative to the primary frame is a A = a G + α × r A /G + ω × (ω × r A /G )

= 0+ 0+

i j k 4 rad/s 2 rad/s −2 rad/s 4 m/s −6 m/s 2 m/s

= −8i − 16 j − 32k (m/s 2 ). z x

v A = 24 i + 8 j + 14 k (m/s), a A = −8i − 16 j − 32k (m/s 2 ).

Primary reference frame

Problem 20.4 The coordinate system shown is fixed with respect to the cube. The angular velocity of the cube relative to the primary reference frame, ω = −6.4 i + 8.2 j + 12k (rad/s), is constant. The acceleration of the center of mass G of the cube relative to the primary reference frame at the instant shown is a G = 136 i + 76 j − 48k (m/s 2 ). What is the acceleration of point A of the cube relative to the primary reference frame at the instant shown?

Solution:

The vector from G to A is

rG /A = ( i + j + k) m. The acceleration of point A is a A = a G + ω × (ω × rG /A ) = (136 i + 76 j − 48k) m/s 2 + (−6.4 i + 8.2 j + 12k) ×

i j k −6.4 8.2 12 m/s 2 1 1 1

Carrying out the vector algebra, we have y

a A = (−205i − 63.0 j − 135k) m/s 2 .

z x

530

Primary reference frame

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Problem 20.5 The secondary reference frame has a constant angular velocity Ω = −2 i + 4 j + 4 k (rad/s) relative to the primary reference frame. The cube has a constant angular velocity ω rel = 8i + 4 j + 2k (rad/s) relative to the secondary frame. The center of mass has a constant velocity v G = 20 i + 14 j + 12k (m/s) relative to the primary frame. What are the velocity and acceleration of point A of the cube relative to the primary frame?

Solution:

The angular velocity of the cube relative to the primary

frame is ω = Ω + ω rel = 6 i + 8 j + 6k (rad/s). The velocity of A relative to the primary frame is v A = v G + ω × r A /G = 20 i + 14 j + 12k (m/s) +

i j k 6 rad/s 8 rad/s 6 rad/s 1m 1m 1m

= 20 i + 14 j + 12k + 2 i − 2k (m/s) = 22 i + 14 j + 10 k (m/s).

Because ω is constant, from Eq. (20.4) the angular acceleration of the rigid body is α = Ω×ω =

i j k −2 rad/s 4 rad/s 4 rad/s 6 rad/s 8 rad/s 6 rad/s

= −8i + 36 j − 40 k (rad/s 2 ). z

The acceleration of A relative to the primary frame is x

a A = a G + α × r A /G + ω × (ω × r A /G ) i = 0+

Primary reference frame +

−8 rad/s 2 1m

36 rad/s 2 −40 rad/s 2 1m 1m

i j k 6 rad/s 8 rad/s 6 rad/s 2 m/s 0 −2 m/s

= 60 i − 8 j − 60 k (m/s 2 ). v A = 22 i + 14 j + 10 k (m/s), a A = 60 i − 8 j − 60 k (m/s 2 ).

Problem 20.6 Relative to an Earth-fixed reference frame, points A and B of the rigid parallelepiped are fixed and it rotates about the axis AB with an angular velocity of 30 rad/s. Determine the velocities of points C and D relative to the Earth-fixed reference frame. Solution:

B D

Given

(0.4 i + 0.2 j − 0.4 k) = (20 i + 10 j − 20 k) rad/s 0.6 rC /A = (0.2 m) j, rD /A = (0.4 i + 0.2 j) m ω = (30 rad/s)

v C = ω × rC /A = (4 i + 4 k) m/s, v D = ω × rD /A = (4 i − 8 j) m/s.

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30 rad/s 0.4 m

0.4 m

0.2 m

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Problem 20.7 Relative to an Earth-fixed reference frame, the vertical shaft rotates about its axis with constant angular velocity ω 0 = 4 rad/s. The secondary xyz coordinate system is fixed with respect to the shaft and its origin is stationary. The z-axis is horizontal, and the angle between the x-axis and the horizontal is β = 30 °. Relative to the secondary coordinate system, the disk (radius = 0.5 m) rotates with constant angular velocity ω d = 6 rad/s. (a) What is the velocity of point A of the disk relative to the Earth-fixed reference frame at the instant shown? (b) What is the disk’s angular acceleration relative to the Earth-fixed reference frame?

Solution: (a) Because the secondary reference frame is fixed relative to the rotating shaft, its angular velocity relative to the Earth-fixed frame is the angular velocity of the shaft: Ω = −ω 0 sin β i + ω 0 cos β j. The angular velocity of the disk relative to the secondary frame is ω rel = ω d j. We see that the angular velocity of the disk relative to the Earth-fixed frame is ω = Ω + ω rel = −ω 0 sin β i + (ω 0 cos β + ω d ) j. The velocity of point A is

v A = v O + ω × r A /O = 0 +

k 0 R

= (ω 0 cos β + ω d ) Ri + (ω 0 sin β ) Rj

i j −ω 0 sin β ω 0 cos β + ω d 0 0

= 4.73 Ri + j (m/s). (b) The components of ω are constant. From Eq. (20.4), the angular acceleration of the disk relative to the Earth-fixed frame is

α = Ω×ω =

i −ω 0 sin β

j ω 0 cos β

−ω 0 sin β ω 0 cos β + ω d

k 0 0

= [−(ω 0 sin β )(ω 0 cos β + ω d ) + (ω 0 cos β )(ω 0 sin β )]k

= −ω 0 ω d sin β k = −12k rad/s 2 . (a) v A = 4.73 Ri + j (m/s). (b) α = −12k rad/s 2 .

Problem 20.8 Relative to an Earth-fixed reference frame, the vertical shaft rotates about its axis with angular velocity ω 0 = 4 rad/s. The secondary xyz coordinate system is fixed with respect to the shaft and its origin is stationary. Relative to the secondary coordinate system, the disk (radius = 8 in) rotates with angular velocity ω d = 6 rad/s. At the instant shown, determine the velocity of point A (a) relative to the secondary reference frame, and (b) relative to the Earth-fixed reference frame.

A 458

Solution: (a) Because the secondary reference frame is fixed relative to the rotating shaft, its angular velocity relative to the Earth-fixed frame is the angular velocity of the shaft: Ω = ω 0 j.

The angular velocity of the disk relative to the secondary frame is ω rel = ω d i. We see that the angular velocity of the disk relative to the Earthfixed frame is ω = Ω + ω rel = ω d i + ω 0 j. Let O denote the origin of the secondary frame. The velocity of point A relative to the secondary frame is v A = v O + ω rel × r A /O = 0 +

i ωd 0

j 0

(b) The velocity of point A relative to the Earth-fixed frame is

k 0

R cos 45 ° R cos 45 °

= −ω d R cos 45 ° j + ω d R cos 45 °k = −33.9 j + 33.9k (in/s).

532

v A = v O + ω × r A /O = 0 +

i ωd 0

j ω0

k 0

R cos 45 ° R cos 45 °

= ω 0 R cos 45 °i − ω d R cos 45 ° j + ω d R cos 45 °k = 22.6 i − 33.9 j + 33.9k (in/s). (a) − 33.9 j + 33.9k (in/s). (b) 22.6 i − 33.9 j + 33.9k (in/s).

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Problem 20.9 Relative to an Earth-fixed reference frame, the vertical shaft rotates about its axis with constant angular velocity ω 0 = 6 rad/s. The secondary xyz coordinate system is fixed with respect to the shaft and its origin is stationary. Relative to the secondary coordinate system, the disk (radius = 8 in) rotates with constant angular velocity ω d = 12 rad/s. At the instant shown, what is the acceleration of point A relative to the Earthfixed reference frame?

The components of ω are constant. From Eq. (20.4), the angular acceleration of the disk relative to the Earth-fixed frame is j ω0 ω0

k 0 0

= −ω 0 ω d k. Let O denote the origin of the secondary frame. The cross product i ωd

ω × r A /O =

i 0 ωd

α = Ω×ω =

j ω0

k 0

R cos 45 ° R cos 45 °

= ω 0 R cos 45 ° i − ω d R cos 45 ° j + ω d R cos 45 ° k. The acceleration of A relative to the primary frame is vd

a A = a O + α × r A / O + ω × (ω × r A / O ) i 0

= 0+

458

j 0

k −ω 0 ω d

0 R cos 45 ° R cos 45 ° x

i ωd

j ω0

k 0

ω 0 R cos 45 ° −ω d R cos 45 ° ω d R cos 45 ° = ω 0 ω d R cos 45 °i + ω 0 ω d R cos 45 °i − ω d 2 R cos 45 ° j − (ω d 2 R cos 45 ° + ω 0 2 R cos 45 °)k v0

= 815 i − 815 j − 1020 k (in/s 2 ).

a A = 815i − 815 j − 1020 k (in/s 2 ).

Solution:

Because the secondary reference frame is fixed relative to the rotating shaft, its angular velocity relative to the Earth-fixed frame is the angular velocity of the shaft: Ω = ω 0 j.

The angular velocity of the disk relative to the secondary frame is ω rel = ω d i. We see that the angular velocity of the disk relative to the Earth-fixed frame is ω = Ω + ω rel = ω d i + ω 0 j.

Problem 20.10 The radius of the disk is R = 2 ft. It is perpendicular to the horizontal part of the shaft and rotates relative to it with constant angular velocity ω d = 36 rad/s. Relative to an Earth-fixed reference frame, the shaft rotates about the vertical axis with constant angular velocity ω 0 = 8 rad/s. (a) Determine the velocity relative to the Earth-fixed reference frame of point P, which is the uppermost point of the disk. (b) Determine the disk’s angular acceleration vector α relative to the Earth-fixed reference frame. (See Example 20.2.) Solution: (a)

y 3 ft P vd z v0 R

v P = (8 rad/s) j × (3 ft) i + [(36 i + 8 j) rad/s] × (2 ft) j = (48 ft/s)k.

(b)

α = (8 rad/s) j × [(36 i + 8 j) rad/s] = −(288 rad/s 2 )k.

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Problem 20.11 The vertical shaft supporting the disk antenna is rotating with a constant angular velocity ω 0 = 0.2 rad/s. The angle θ from the horizontal to the antenna’s axis is 30° at the instant shown and is increasing at a constant rate of 15° per second. The secondary xyz coordinate system shown is fixed with respect to the dish. (a) What is the dish’s angular velocity relative to an Earth-fixed reference frame? (b) Determine the velocity of the point of the antenna with coordinates (4, 0, 0) m relative to an Earth-fixed reference frame. Solution:

The relative angular velocity is

ω rel = (15 °/s)

( π180rad° ) = 12π rad/s

(a) ω = Ω + ω rel = (0.2sin 30 °i + 0.2 cos30 ° j) +

( 12π )k

ω = (0.1i + 0.173 j + 0.262k) rad/s. (b) v = ω × r = (0.1i + 0.173 j + 0.262k) × (4 i) v = (1.05 j − 0.693k) m/s.

Problem 20.12 The vertical shaft supporting the disk antenna is rotating with a constant angular velocity ω 0 = 0.2 rad/s. The angle θ from the horizontal to the antenna’s axis is 30° at the instant shown and is increasing at a constant rate of 15° per second. The secondary xyz coordinate system shown is fixed with respect to the dish. (a) What is the dish’s angular acceleration relative to an Earth-fixed reference frame? (b) Determine the acceleration of the point of the antenna with coordinates (4, 0, 0) m relative to an Earth-fixed reference frame. Solution:

The angular velocity is

ω rel = (15 °/s)

( π180rad° ) = 12π rad/s.

ω = Ω + ω rel = (0.2sin 30 °i + 0.2 cos30 ° j) +

( 12π )k

= (0.1i + 0.173 j + 0.262k) rad/s (a) The angular acceleration is α = Ω × ω rel = (0.2sin 30 °i + 0.2 cos30° j) ×

( 12π k )

α = (0.0453i − 0.0262 j) rad/s 2 . (b) The acceleration of the point a = α × r + ω × (ω × r ) = (0.0453i − 0.0262 j) × (4 i) + (0.1i + 0.173 j + 0.262k) × [(0.1i + 0.173 j + 0.262k) × (4 i)] a = (−0.394 i + 0.0693 j + 0.209k) m/s 2 .

534

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Problem 20.13 The radius of the circular disk is R = 0.2 m and b = 0.3 m. The disk rotates with angular velocity ω d = 10 rad/s relative to the horizontal bar. The horizontal bar rotates with angular velocity ω b = 6 rad/s relative to the vertical shaft, and the vertical shaft rotates with angular velocity ω 0 = 2 rad/s relative to an Earth-fixed reference frame. Assume that the secondary reference frame shown is fixed with respect to the horizontal bar. At the instant shown, what is the velocity of point P, which is the uppermost point of the disk, relative to the Earth-fixed reference frame?

Solution:

In solving this problem, it must be remembered that Eq. (20.1) requires that points A and B be points of the same rigid body. Let O be the origin of the secondary frame, and let C be the center of the disk. We will first apply Eq. (20.1) to points O and C to determine the velocity of point C, then apply it to points C and P. y

y b P

vd z v0 R

At the instant shown, the angular velocity of the horizontal bar and secondary reference frame relative to the Earth-fixed reference frame is Ω = ω 0 j + ω b k. The velocity of point C is v C = v O + Ω × rC /O = 0 +

i j 0 ω0

k ωb

= ω b b j − ω 0 b k. vb

The angular velocity of the disk relative to the secondary frame is ω rel = ω d i,

so the angular velocity of the disk relative to the Earth-fixed frame is

ω = Ω + ω rel = ω d i + ω 0 j + ω b k. x v0

The velocity of point P is v P = v C + ω × r P /C = ω bbj − ω 0 bk +

i ωd 0

j ω0 R

k ωb 0

= ω b b j − ω 0 b k − ω b Ri + ω d Rk = −1.2 i + 1.8 j + 1.4 k (m/s). v P = −1.2 i + 1.8 j + 1.4 k (m/s).

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Problem 20.14 The object in Fig. a is supported by bearings at A and B in Fig. b. The horizontal circular disk is welded to a vertical shaft that rotates with angular velocity ω 0 = 8 rad/s relative to an Earth-fixed reference frame. The horizontal bar rotates relative to the disk with angular velocity ω. At the instant shown, the segment of the bar ending in point C is vertical and a laser-doppler instrument indicates that the magnitude of the velocity of C relative to the Earth-fixed frame is 1.2 m/s. What is the angular velocity ω ? Solution:

The angular velocity of the circular disk and secondary reference frame relative to the Earth-fixed reference frame is Ω = ω 0 j.

so the angular velocity of ABC relative to the Earth-fixed frame is ω = Ω + ω rel = ω i + ω 0 j. Let O be the origin of the secondary frame. The velocity of C relative to the Earth-fixed frame is v C = v O + ω × rC /O = 0+

i j k ω ω0 0 0.1 m 0.1 m 0

= (0.1 m)(ω − ω 0 )k. Setting v C = 1.2 m/s, we obtain ω = 20 rad/s. ω = 20 rad/s.

The angular velocity of the member ABC relative to the secondary frame is ω rel = ω i, y

C 0.1 m A

0.2 m

0.1 m 0.4 m (a) v0

(b)

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Problem 20.15 The object in Fig. a is supported by bearings at A and B in Fig. b. The horizontal circular disk is welded to a vertical shaft that rotates with constant angular velocity ω 0 = 8 rad/s relative to an Earthfixed reference frame. The horizontal bar rotates relative to the disk with angular velocity ω. At the instant shown, the segment of the bar ending in point C is vertical and an accelerometer mounted at C indicates that the x component of the acceleration of C relative to the Earth-fixed frame is 16 m/s 2 . What is the angular velocity ω ?

Let O denote the origin of the secondary frame. The cross product i j k ω ω0 0 0.1 m 0.1 m 0

ω × rC /O =

= (0.1 m)(ω − ω 0 )k. The acceleration of C relative to the primary frame is a C = a O + α × rC /O + ω × (ω × rC /O ) = 0+

Solution:

The angular velocity of the circular disk and secondary reference frame relative to the Earth-fixed reference frame is

i j k −ω 0 ω 0 0 0.1 m 0.1 m 0

i j ω ω0 0 0

k 0 (0.1 m)(ω − ω 0 )

= [ω 0 ω (0.1 m) + (0.1 m)ω 0 (ω − ω 0 )]i + [−ω 0 ω (0.1 m)

Ω = ω 0 j.

−(0.1 m)ω (ω − ω 0 )] j.

The angular velocity of the member ABC relative to the secondary frame is ω rel = ω i, so the angular velocity of ABC relative to the Earth-fixed frame is

Setting the i component equal to 16 m/s 2 and solving, we obtain ω = 14 rad/s. ω = 14 rad/s.

ω = Ω + ω rel = ω i + ω 0 j. The components of ω are constant. From Eq. (20.4), the angular acceleration of the disk relative to the Earth-fixed frame is α = Ω×ω =

i j 0 ω0 ω ω0

k 0 0

= −ω 0 ωk.

C 0.1 m A

0.2 m

0.1 m 0.4 m (a) v0

(b)

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Problem 20.16 Relative to a primary reference frame, the gyroscope’s circular frame rotates about the vertical axis at 2 rad/s. The 60-mm diameter wheel rotates at 10 rad/s relative to the frame. Determine the velocities of points A and B relative to the primary reference frame.

Solution:

Let the secondary reference frame shown be fixed with respect to the gyroscope’s frame. The angular velocity of the SRF is Ω = 2 j (rad/s). The angular velocity of the wheel relative to the SRF is ω rel = 10 k (rad/s), so the wheel’s angular velocity is ω = Ω + ω rel = 2 j + 10 k (rad/s)

Let O denote the origin. The velocity of pt. A is y v A = v 0 + ω × r A /O = 0 +

2 rad/s Frame

60 mm

Wheel

= 80 i (mm/s).

The velocity of pt. B is v B = v 0 + ω × r B /O = 0+

i j k 0 2 10 30 cos 20 ° 30 sin 20 ° 0

= −102.6 i + 281.9 j − 56.4 k (mm/s).

208

i j k 0 2 10 0 0 40

10 rad/s

x 80 mm

Problem 20.17 Relative to a primary reference frame, the gyroscope’s circular frame rotates about the vertical axis with a constant angular velocity of 2 rad/s. The 60-mm diameter wheel rotates with a constant angular velocity of 10 rad/s relative to the frame. Determine the accelerations of points A and B relative to the primary reference frame.

Solution: See the solution of Problem 20.16. From Eq. (20.4), the wheel’s angular acceleration is α = Ω×ω =

= 0+

2 rad/s Frame Wheel

i j k 0 2 10 80 0 0

The acceleration of pt. B is a B = a 0 + α × rB /O + ω × (ω × rB /O ) = 0+

208 10 rad/s

+ x 80 mm

538

i j k 20 0 0 0 0 40

= −160 k (mm/s 2 ).

B A

= 20 i (rad/s 2 ).

The acceleration of pt. A is a A = a 0 + α × r A /O + ω × (ω × r A /O )

60 mm

i j k 0 2 0 0 2 10

i j k 20 0 0 30 cos 20 ° 30 sin 20 ° 0 i j k 0 2 10 −102.6 281.9 −56.4

= −2932 i − 1026 j + 410 k (mm/s 2 ).

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Problem 20.18 The point of the spinning top remains at a fixed point on the floor, which is the origin O of the secondary reference frame shown. The angle θ = 30 ° between the z-axis and the vertical is constant. The x-axis remains parallel to the floor and rotates about O with angular velocity ω 0 = 6.3 rad/s. The angular velocity of the top relative to the secondary frame is ω rel = 250 k (rad/s). At the present instant, determine the velocity relative to the Earth-fixed reference frame of the point P of the top with coordinates ( x, y, z ) = (0, −30, 50) mm. z

Solution:

The secondary reference frame rotates about the vertical axis with angular velocity ω 0 = 6.3 rad/s. Applying the right-hand rule, its angular velocity vector is Ω = ω 0 sin θ j + ω 0 cos θk. The angular velocity of the top relative to the secondary frame is ω rel = 250 k (rad/s) = ω rel k, so the angular velocity of the top relative to the Earth-fixed frame is ω = Ω + ω rel = ω 0 sin θ j + (ω 0 cos θ + ω rel )k. The velocity of P relative to the Earth-fixed frame is v P = v O + ω × r P /O = 0+

i 0

j ω 0 sin θ

0 −0.03 m

k ω 0 cos θ + ω rel 0.05 m

= [ω 0 sin θ(0.05 m) + (ω 0 cos θ + ω rel )(0.03 m)]i = 7.82 i (m/s). y

7.82 i (m/s).

O v0 x

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Problem 20.19 The point of the spinning top remains at a fixed point on the floor, which is the origin O of the secondary reference frame shown. The angle θ = 30 ° between the z-axis and the vertical is constant. The x-axis remains parallel to the floor and rotates about O with constant angular velocity ω 0 = 6.3 rad/s. The constant angular velocity of the top relative to the secondary frame is ω rel = 250 k (rad/s). (a) What is the top’s angular acceleration relative to the Earth-fixed reference frame? (b) Determine the acceleration relative to the Earth-fixed reference frame of the point of the top with coordinates ( x, y, z ) = (0, −30, 50) mm.

Solution: (a) The secondary reference frame rotates about the vertical axis with angular velocity ω 0 = 6.3 rad/s. Applying the right-hand rule, its angular velocity vector is Ω = ω 0 sin θ j + ω 0 cos θk. The angular velocity of the top relative to the secondary frame is ω rel = 250 k (rad/s) = ω rel k, so the angular velocity of the top relative to the Earth-fixed frame is ω = Ω + ω rel = ω 0 sin θ j + (ω 0 cos θ + ω rel )k. The components of ω are constant. From Eq. (20.4), the angular acceleration of the disk relative to the Earth-fixed frame is i j 0 ω 0 sin θ

α = Ω×ω =

k ω 0 cos θ

0 ω 0 sin θ ω 0 cos θ + ω rel

= [ω 0 sin θ(ω 0 cos θ + ω rel ) − ω 0 2 sin θ cos θ]i

= ω 0 ω rel sin θ i = 788 i (m/s 2 ). y

(b) The cross product ω × r P /O =

i 0

j ω 0 sin θ

k ω 0 cos θ + ω rel

0 −0.03 m

0.05 m

= [ω 0 sin θ(0.05 m) + (ω 0 cos θ + ω rel )(0.03 m)]i.

The acceleration of P relative to the primary frame is

v0 x

a P = a O + α × rP / O + ω × (ω × rP / O ) = 0+

i ω 0 ω rel sin θ 0

j 0

k 0

−0.03 m 0.05 m i 0

j k ω 0 sin θ ω 0 cos θ + ω rel

ω 0 sin θ(0.05 m) + (ω 0 cos θ + ω rel )(0.03 m)

= −ω 0 ω rel sin θ(0.05 m) j − ω 0 ω rel sin θ(0.03 m)k + (ω 0 cos θ + ω rel )[ω 0 sin θ(0.05 m) + (ω 0 cos θ + ω rel )(0.03 m)] j − ω 0 sin θ[ω 0 sin θ(0.05 m) + (ω 0 cos θ + ω rel )(0.03 m)]k = 1960 j − 48.3 k (m/s 2 ).

(a) 788 i (m/s 2 ). (b) 1960 j − 48.3k (m/s 2 ).

540

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Problem 20.20* The cone rolls on the horizontal surface, which is fixed with respect to an Earth-fixed reference frame. The x-axis of the secondary reference frame remains coincident with the cone’s axis, and the z-axis remains horizontal. As the cone rolls, the z-axis rotates in the horizontal plane with an angular velocity of 2 rad/s. (a) What is the angular velocity vector Ω of the secondary reference frame? (b) What is the angular velocity vector ω rel of the cone relative to the secondary reference frame? (See Example 20.3.) Strategy: To solve part (b), use the fact that the velocity relative to the Earth-fixed reference frame of points of the cone in contact with the surface is zero.

Solution: (a) The angle β = arctan(R /h) = arctan(0.2/0.4) = 26.6 °. The angular velocity of the secondary reference frame is Ω = ω 0 sin β i + ω 0 cos β j = 2(sin 26.6 °i + cos 26.6 ° j) = 0.894 i + 1.789 j (rad/s). (b) The cone’s angular velocity relative to the secondary reference frame can be written ω rel = ω rel i, so the cone’s angular velocity is ω = Ω + ω rel = (0.894 + ω rel ) i + 1.789 j (rad/s). To determine ω rel , we use the fact that the point P in contact with the surface has zero velocity: v P = v 0 + ω × r P /O = 0 +

i 0.894 + ω rel 0.4

j k 1.789 0 −0.2 0

= 0.

Solving, we obtain ω rel = −4.47 (rad/s), so ω rel = −4.47 i (rad/s). y 0.2 m

2 rad/s z

b 0.4 m

Problem 20.21* The cone rolls on the horizontal surface, which is fixed with respect to an Earth-fixed reference frame. The x-axis of the secondary reference frame remains coincident with the cone’s axis, and the z-axis remains horizontal. As the cone rolls, the z-axis rotates in the horizontal plane with an angular velocity of 2 rad/s. Determine the velocity relative to the Earth-fixed reference frame of the point of the base of the cone with coordinates x = 0.4 m, y = 0 and z = 0.2 m. (See Example 20.3.)

Solution:

See the solution of Problem 20.20. The cone’s angular

velocity is ω = Ω + ω rel = (0.894 i + 1.789 j) − 4.472 i = −3.578i + 1.789 j (rad/s) Let A denote the pt with coordinates (0.4, 0, 0.2) m. Its velocity is v A = v 0 + ω × r A /O = 0 +

i j k −3.578 1.789 0 0.4 0 0.2

= 0.358i + 0.716 j − 0.716k (m/s).

0.2 m

2 rad/s z 0.4 m

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Problem 20.22* The cone rolls on the horizontal surface, which is fixed with respect to an Earth-fixed reference frame. The x-axis of the secondary reference frame remains coincident with the cone’s axis, and the z-axis remains horizontal. As the cone rolls, the z-axis rotates in the horizontal plane with a constant angular velocity of 2 rad/s. Determine the acceleration relative to the Earthfixed reference frame of the point of the base of the cone with coordinates x = 0.4 m, y = 0, and z = 0.2 m. (See Example 20.3.)

Solution:

See the solutions of Problems 20.20 and 20.21. The cone’s angular acceleration is

α = Ω×ω =

i j k 0.894 1.789 0 −3.578 1.789 0

= 8.000 k (rad/s 2 ).

The acceleration of the point is a A = a 0 + α × r A /O + ω × ( ω × r A /O ) = 0+

i j k 0 0 8 0.4 0 0.2

i j k −3.578 1.789 0 0.358 0.716 −0.716

= −1.28i + 0.64 j − 3.20 k (m/s 2 ). 0.2 m

2 rad/s z 0.4 m

Problem 20.23* The radius and length of the cylinder are R = 0.2 m and L = 0.6 m. One end of the cylinder rolls on the floor while its center, the origin of the secondary reference frame, remains stationary. The angle β = 40 ° is constant. The z-axis remains coincident with the cylinder’s axis. As the cylinder rolls, the y-axis rotates in a horizontal plane with angular velocity ω 0 = 4 rad/s. (a) What is the angular velocity of the cylinder relative to the Earth-fixed reference frame? (b) What is the velocity relative to the Earth-fixed reference frame of the point of the cylinder with coordinates x = 0, y = R, z = L /2? Strategy: In determining the angular velocity of the cylinder, use the fact that the velocity relative to an Earthfixed frame of the point of the cylinder in contact with the floor is zero. x z

Solution: Æ x A

G mg

P mg

(a) The figure shows a side view of the cylinder. Because the y axis rotates in the horizontal plane with angular velocity ω 0 and the center of mass is stationary, the angular velocity vector of the secondary reference frame is

Ω = ω 0 cos β i + ω 0 sin β k. Relative to the secondary reference frame, the cylinder rotates about the z axis, so its angular velocity relative to the secondary frame is of the form ω rel = ω rel k.

v0 b

542

z b

The angular velocity of the cylinder relative to the Earth-fixed reference frame is ω = Ω + ω rel = ω 0 cos β i + (ω 0 sin β + ω rel )k, (1)

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20.23* (Continued) but we don’t know the value of ω rel . To determine it, we will use the fact that the velocity of the point of contact P of the cylinder with the surface is zero relative to the Earth-fixed reference frame: v P = v G + ω × r P /G :

v A = v G + ω × r A /G :

i j k ω 0 cos β 0 ω 0 sin β + ω rel 1 −R 0 − L 2

0 = 0+

(b) Let A denote the point of the cylinder with coordinates x = 0 y = R, z = L /2. Its velocity relative to the Earth-fixed reference frame is

= 0+

(

)

1 =  (ω 0 sin β + ω rel )(−R) − (ω 0 cos β ) − L  j.  2  Solving this equation for ω rel and substituting the result into Eq. (1), we obtain L ω = ω 0 cos β i + ω cos β k 2R 0 = 3.06 i + 4.60 k (rad/s).

Problem 20.24* The radius and length of the cylinder are R = 0.2 m and L = 0.6 m. One end of the cylinder rolls on the floor while its center, the origin of the secondary reference frame, remains stationary. The angle β = 40 ° is constant. The z-axis remains coincident with the cylinder’s axis. As the cylinder rolls, the y-axis rotates in a horizontal plane with constant angular velocity ω 0 = 4 rad/s. (a) What is the angular acceleration of the cylinder relative to the Earth-fixed reference frame? (b) What is the acceleration relative to the Earth-fixed reference frame of the point of the cylinder with coordinates x = 0, y = R, z = L /2? x z

ω 0 cos β

k L ω cos β 2R 0 1 L 2

1 1 = − L ω 0 cos β i − L ω 0 cos β j + Rω 0 cos β k 2 2 = −0.919 i − 0.919 j + 0.613k (m/s). (a) 3.06 i + 4.60 k (rad/s). (b) − 0.919 i − 0.919 j + 0.613k (m/s).

but we don’t know the value of ω rel . To determine it, we will use the fact that the velocity of the point of contact P of the cylinder with the surface is zero: v P = v G + ω × r P /G :

0 = 0+

i j k ω 0 cos β 0 ω 0 sin β + ω rel 1 −R 0 − L 2

(

)

1 =  (ω 0 sin β + ω rel )(−R) − (ω 0 cos β ) − L  j.  2  Solving this equation for ω rel and substituting the result into Eq. (1), we obtain L ω = ω 0 cos β i + ω cos β k. 2R 0 The components of ω are constant. From Eq. (20.4), the angular acceleration of the disk relative to the Earth-fixed frame is

α = Ω×ω =

i j ω 0 cos β 0 ω 0 cos β 0

(

= ω 0 2 sin β cos β −

k ω 0 sin β L ω cos β 2R 0

)

L cos 2 β j 2R

= −6.21 j (rad/s 2 ).

Solution: (a) The figure shows a side view of the cylinder. Because the y axis rotates in the horizontal plane and the center of mass is stationary, the angular velocity vector of the secondary reference frame is

Æ x

z b

Ω = ω 0 cos β i + ω 0 sin β k. G

Relative to the secondary reference frame, the cylinder rotates about the z axis, so

ω rel = ω rel k. The angular velocity of the cylinder relative to the Earth-fixed reference frame is ω = Ω + ω rel = ω 0 cos β i + (ω 0 sin β + ω rel )k, (1)

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20.24* (Continued) (b) Let A denote the point of the cylinder with coordinates x = 0, y = R, z = L /2. The cross product i

The acceleration of A relative to the primary frame is a A = a G + α × r A / G + ω × (ω × r A / G )

k L ω × r A /G = ω 0 cos β 0 ω cos β 2R 0 1 R L 0 2 1 1 = − L ω 0 cos β i − L ω 0 cos β j + Rω 0 cos β k. 2 2

i = 0+

0 ω0

(

L sin β cos β − cos 2 β 2R

)

0 1 L 2

R i

ω 0 cos β

k L ω cos β 2R 0

1 1 − Lω 0 cos β − Lω 0 cos β 2 2

(

Rω 0 cos β

)

1 L L2 cos 2 β + ω 2 cos 2 β  i =  Lω 0 2 sin β cos β − 2  2R 4R 0 L2 1   2 2 2 2 2 2 − ω cos β + Rω 0 cos β  j − Lω 0 cos β k  4R 0  2 = 2.36 i − 6.10 j − 2.82 k (m/s 2 ). (a) − 6.21 j (rad/s 2 ). (b) 2.36 i − 6.10 j − 2.82k (m/s 2 ).

Problem 20.25* The landing gear of the P-40 airplane used in World War II retracts by rotating the strut 90° from the deployed position to the retracted position with constant angular velocity. As the wheel retracts, the strut rotates 90° about its axis with constant angular velocity so that the wheel is horizontal when in the retracted position. (Viewed from the horizontal axis toward the wheel, the strut rotates in the clockwise direction.) The x-axis of the secondary coordinate system remains parallel to the horizontal axis as the landing gear retracts. The y-axis remains parallel to the strut. The horizontal axis is 3 ft from the center of the wheel and the radius of the wheel is 1 ft. At time t = 0, the airplane takes off at 90 knots and the wheel requires 6 s to retract. Assume that the wheel’s angular velocity ω W about the wheel axis remains constant. Determine the components of the wheel’s angular velocity relative to the airplane as functions of time as it retracts. Retracted position y

Solution: 90 knots

The airplane’s velocity at takeoff in ft/s is

ft/s ( 1.689 ) = 152 ft/s. 1 knot

The angular velocity of its wheels as it takes off is ωW =

152 ft/s = 152 rad/s. 1 ft

During the retraction, the secondary reference frame rotates 90 ° about the x axis in six seconds. The angular velocity of the secondary frame about the x axis is 15 deg/s

( π180rad° ) = 0.262 rad/s.

Let ω R = 0.262 rad/s. The angular velocity of the secondary coordinate system is Ω = ω R i = 0.262 i (rad/s). Relative to the secondary coordinate system, as the gear retracts the wheel rotates 90 ° about the y axis. This rotation is in the clockwise direction when viewing the wheel from the horizontal axis, so expressed as a vector it is ω R j = −0.262 j (rad/s). Æ

x vRt

z vW

Horizontal axis Strut

Deployed position

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20.25* (Continued) At takeoff, the angular velocity vector ω W representing the wheel’s rotation about its axis points in the x-axis direction, but as the gear retracts that vector rotates in the x-z plane and in the retracted position it points in the z -axis direction. Relative to the secondary frame the vector rotates 90 ° in six seconds, so the angle shown in the figure (in radians) equals ω R t. The vector

Finally, the angular velocity of the wheel relative to a frame fixed with respect to the airplane is

ω W = ω W cos ω R ti + ω W sin ω R tk,

(0.262 + 152 cos 0.262t ) i + 0.262 j + 152sin 0.262tk (rad/s).

ω = Ω + ω rel = (ω R + ω W cos ω R t ) i + ω R j + ω W sin ω R tk = (0.262 + 152 cos 0.262t ) i + 0.262 j + 152sin 0.262tk (rad/s).

and the angular velocity of the wheel relative to the secondary frame is ω rel = ω R j + ω W = ω W cos ω R ti + ω R j + ω W sin ω R tk.

Problem 20.26 The moments and products of inertia of the 4-kg rectangular plate are given in Example 20.6. At the instant shown, the plate is horizontal, its angular velocity is zero, and its angular acceleration vector is α = 200 i − 320 j + 100 k (rad/s 2 ). Determine the couple C exerted on the plate at O by the robotic manipulator at that instant. Solution:

dω y dω z dω x − I xy − I xz dt dt dt = (0.48 kg-m 2 )(200 rad/s 2 ) − (0.18 kg-m 2 )

C x − 11.8 N-m = I xx

(−320 rad/s 2 ) − (0)(100 rad/s 2 ) = 154 N-m, dω y dω z dω x + I yy − I yz dt dt dt = −(0.18 kg-m 2 )(200 rad/s 2 ) + (0.12 kg-m 2 )

C y + 5.89 N-m = −I yx

The moment about O due to the weight of the plate is

(−320 rad/s 2 ) − (0)(100 rad/s 2 ) = −74.4 N-m,

[(0.15 m) i + (0.3 m) j] × [−(4 kg)(9.81 m/s 2 )k] =

dω y dω z dω x − I zy + I zz dt dt dt = −(0)(200 rad/s 2 ) − (0)(−320 rad/s 2 ) + (0.6 kg-m 2 )

i j k 0.15 m 0.3 m 0 0 0 −(4 kg)(9.81 m/s 2 )

C z = −I zx

(100 rad/s 2 )

= −11.8i + 5.89 j (N-m).

= 60 N-m,

We use Eqs. (20.12) to determine the couple C exerted by the manipulator. The components of the angular velocity are zero, so from Eq. (20.4) the time derivatives of the components of the angular velocity equal the components of the angular acceleration. We obtain the equations

which yield C x = 165 N-m, C y = −80 N-m, C z = 60 N-m. C = 165i − 80.3 j + 60 k N-m.

150 mm 150 mm 300 mm O 300 mm x

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Problem 20.27 The moments and products of inertia of the 4-kg rectangular plate are given in Example 20.6. At the instant shown, the plate is horizontal, its angular velocity is zero, and the couple exerted on the plate at O by the robotic manipulator is C = 150 i + 80 j (N-m). What is the angular acceleration of the plate at that instant? Solution:

The moment about O due to the weight of the plate is

[(0.15 m) i + (0.3 m) j] i j k × [−(4 kg)(9.81 m/s 2 ) k] = 0.15 m 0.3 m 0 0 0 −(4 kg)(9.81 m/s 2 )

the time derivatives of the components of the angular velocity equal the components of the angular acceleration. We obtain the equations dω y dω dω − (0) z , C x − 11.8 N-m = (0.48 kg-m 2 ) x − (0.18 kg-m 2 ) dt dt dt dω y dω dω x 2 2 C y + 5.89 N-m = −(0.18 kg-m ) + (0.12 kg-m ) − (0) z , dt dt dt dω y dω z dω x 2 − (0) + (0.6 kg-m ) , C z = −(0) dt dt dt yielding dω y dω z dω x = α x = 1270 rad/s 2 , = α y = 2620 rad/s 2 , = α z = 0. dt dt dt 1270 i + 2620 j (rad/s 2 ).

= −11.8i + 5.89 j (N-m). We use Eqs. (20.12) to determine components of the angular acceleration. The components of the angular velocity are zero, so from Eq. (20.4) y

150 mm 150 mm 300 mm O 300 mm x

Problem 20.28 A robotic manipulator moves a casting. The inertia matrix of the casting in terms of a body-fixed coordinate system with its origin at the center of mass is shown. At the present instant, the angular velocity and angular acceleration of the casting are ω = 1.2 i + 0.8 j − 0.4 k (rad/s) and α = 0.26 i − 0.07 j + 0.13k (rad/s 2 ). What moment is exerted about the center of mass of the casting by the manipulator?

Solution:  M   0.05 −0.03 x  0   0.26      0   −0.07  N-m  M y  =  −0.03 0.08      0 0 0.04   0.13    M z     0 0  0.4 0.8   0.05 −0.03   0  +  −0.4 0 −1.2   −0.03 0.08    0 0 0.04  0    −0.8 1.2  1.2  ×  0.8  N-m    −0.4  M = ( M x i + M y j + M z k) = (0.0135i − 0.0086 j + 0.01k) N-m. Ixx 2Iyx 2Izx

546

2Ixy Iyy 2Izy

2Ixz 0.05 20.03 0 kg-m2. 2Iyz 5 20.03 0.08 0 0 Izz 0 0.04

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Problem 20.29 A robotic manipulator holds a casting. The inertia matrix of the casting in terms of a body-fixed coordinate system with its origin at the center of mass is shown. At the present instant, the casting is stationary. If the manipulator exerts a moment ∑ M = 0.042 i + 0.036 j + 0.066k (N-m) about the center of mass, what is the angular acceleration of the casting at that instant?

Solution:  α  0.042   0.05 −0.03  x 0       0  kg-m 2  α y  0.036  N-m =  −0.03 0.08      0 0 0.04    0.066   α z 

      

Solving we find α = (α x i + α y j + α z k) = (1.43i + 0.987 j + 1.65k) rad/s 2 .

Ixx 2Iyx 2Izx

2Ixy Iyy 2Izy

2Ixz 0.05 20.03 0 kg-m2. 2Iyz 5 20.03 0.08 0 0 Izz 0 0.04

Problem 20.30 The rigid body rotates about the fixed point O. Its inertia matrix in terms of the body-fixed coordinate system is shown. At the present instant, the rigid body’s angular velocity is ω = 6 i + 6 j − 4 k (rad/s) and its angular acceleration is zero. What total moment about O is being exerted on the rigid body? Solution:  M x   My   M z 

  0 4 6   4 −2 0   6        −4 0 −6   −2 3 1   6  ft-lb =        1 5   −4   −6 6 0   0  

M = ( M x i + M y j + M z k) = (−76 i + 36 j − 60 k) ft-lb.

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O z x Ixx 2Iyx 2Izx

2Ixy Iyy 2Izy

2Ixz 4 2Iyz 5 22 0 Izz

22 3 1

0 2 1 slug-ft . 5

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Problem 20.31 The rigid body rotates about the fixed point O. Its inertia matrix in terms of the body-fixed coordinate system is shown. At the present instant, the rigid body’s angular velocity is ω = 6 i + 6 j − 4 k (rad/s). The total moment about O due to the forces and couples acting on the rigid body is zero. What is its angular acceleration?

Solution:

O z

 α   4 −2 0   0   x   0  =  −2 3 1  slug-ft 2  α   y          1 5   0  α z   0     0 4 6   4 −2 0   6      +  −4 0 −6   −2 3 1   6  1b-ft      1 5   −4   −6 6 0   0

x Ixx 2Iyx 2Izx

2Ixy Iyy 2Izy

2Ixz 4 2Iyz 5 22 0 Izz

22 3 1

0 2 1 slug-ft . 5

Solving we find α = (16.2 i − 5.56 j + 13.1k) rad/s 2 .

Problem 20.32 The object consists of three rigidly connected slender bars, each with length L = 1 m and mass m = 5 kg. The object is stationary relative to an Earth-fixed reference frame when the force F = 60 i + 40 j + 20 k (N) is applied at A. No other forces act on the object. At that instant, what is the acceleration of point B relative to the Earth-fixed frame? y

The moment due to the applied force about the center of mass is

M O = r A /O × F =

i j k 1 L 0 0 2 60 N 40 N 20 N

1 1 = − L (20 N) j + L (40 N)k 2 2 = −10 j + 20 k (N-m). From Eqs. (20.18), the components of the angular acceleration of the object are

αx =

M Oy M Ox −10 N-m = 0, α y = = = −12 rad/s 2 , 0.833 kg-m 2 I xx I yy

αz =

M Oz 20 N-m = = 24 rad/s 2 . I zz 0.833 kg-m 2

The acceleration of point B is

x L

A a B = a O + α × rB /O = a Ox i + a Oy j + a Oz k +

B z

i αx

j αy

k αz

1 L 2

= −2 i + 2.67 j + 1.33k (m/s 2 ). a B = −2 i + 2.67 j + 1.33k (m/s 2 ).

Solution:

From Appendix C, the moment of inertia about the x axis due to the two bars that are perpendicular to it is I xx =

1 1 mL2 + mL2 = 0.833 kg-m 2 . 12 12

Similarly, I yy = I zz = (1/6)mL2 = 0.833 kg-m 2 . The products of inertia are zero. We can use Newton’s second law to determine the acceleration of the center of mass O of the object: Fy Fx 60 N 40 N = = 4 m/s 2 , a Oy = = = 2.67 m/s 2 , 3m 15 kg 3m 15 kg F 20 N a Oz = z = = 1.33 m/s 2 . 3m 15 kg a Ox =

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Problem 20.33 In terms of the coordinate system shown, the inertia matrix of the 6-kg slender bar is

Applying Newton’s second law, the acceleration of the center of mass due to the force F is

 I   0.500 0.667 0   xx −I xy −I xz       −I yx I yy −I yz  =  0.667 2.667 0  kg-m 2 .     0 3.167   0  −I zx −I zy I zz   

aG =

The bar is stationary relative to an inertial reference frame when the force F = 20 i + 8 j + 12k (N) is applied at the right end of the bar. No other forces or couples act on it. Determine (a) the bar’s angular acceleration relative to the inertial reference frame and (b) the acceleration of the top of the 1-m segment of the bar at the instant the force is applied. y

1m x 2m

Solution:

1 1 F = [20 i + 8 j + 12k (N)] m 6 kg = 3.33i + 1.33 j + 2k (m/s 2 ).

The moment of the force F about the center of mass is

M = r A /G × F =

i j k 2 m − x −y 0 20 N 8 N 12 N

= −2 i − 16 j + 14 k (N-m). We use Eqs. (20.18) to determine components of the angular acceleration. The components of the angular velocity are zero, so from Eq. (20.4) the time derivatives of the components of the angular velocity equal the components of the angular acceleration. We obtain the equations dω y dω dω −2 N-m = (0.5 kg-m 2 ) x + (0.667 kg-m 2 ) − (0) z , dt dt dt dω y dω dω x 2 2 −16 N-m = (0.667 kg-m ) + (2.667 kg-m ) − (0) z , dt dt dt dω y dω z dω x 2 14 N-m = −(0) − (0) + (3.167 kg-m ) , dt dt dt yielding dω y dω x = α x = 6.01 rad/s 2 , = α y = −7.50 rad/s 2 , dt dt dω z = α y = 4.42 rad/s 2 . dt (b) The acceleration of the top of the 1-m segment is

a T = a G + α × rT /G = 3.33i + 1.33 j + 2k (m/s 2 ) i +

6.01 rad/s 2 −7.50 rad/s 2 −x 1m−y

k 4.42 rad/s 2 0

= −0.351i − 1.61 j + 2.00 k (m/s 2 ). x 2m

(a) 6.01i − 7.50 j + 4.42k (rad/s 2 ). (b) − 0.351i − 1.61 j + 2.00 k (m/s 2 ).

(a) We first place the coordinate system as shown. In terms of it, the position of the center of mass is (0)(2 kg) + (1 m)(4 kg) = 0.667 m, 6 kg (0.5 m)(2 kg) + (0)(4 kg) y = = 0.167 m. 6 kg x =

Now we convert to the coordinate system with its origin at the center of mass. y T

F G

x A

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Problem 20.34 In terms of the coordinate system shown, the inertia matrix of the 12-kg slender bar is  I xx    −I yx   −I zx 

−I xy I yy −I zy

Solution: y

−I xz   2 −3 0     −I yz  =  −3 8 0  kg-m 2 .    0 10   0 I zz  

The bar is stationary relative to an inertial reference frame when the force F = 40 i + 16 j + 24 k (N) is applied at the point with coordinates x = −1 m, y = −1 m. No other forces or couples act on it. Determine (a) the bar’s angular acceleration relative to the inertial reference frame and (b) the acceleration of the point with coordinates x = 1 m, y = 1 m at the instant the force is applied.

F A

(a) Applying Newton’s second law, the acceleration of the center of mass due to the force F is 1 1 F = [40 i + 16 j + 24 k (N)] m 12 kg = 3.33i + 1.33 j + 2k (m/s 2 ).

aG = y 1m

The moment of the force F about the center of mass is x M = r A /G × F =

i j k −1 m −1 m 0 40 N 16 N 24 N

= −24 i + 24 j + 24 k (N-m). We use Eqs. (20.18) to determine components of the angular acceleration. The components of the angular velocity are zero, so from Eq. (20.4) the time derivatives of the components of the angular velocity equal the components of the angular acceleration. We obtain the equations

1m z 1m

dω y dω dω x − (3 kg-m 2 ) − (0) z , dt dt dt dω y dω dω x 2 2 24 N-m = −(3 kg-m ) + (8 kg-m ) − (0) z , dt dt dt dω y ω d dω x z 24 N-m = −(0) − (0) + (10 kg-m 2 ) , dt dt dt

−24 N-m = (2 kg-m 2 )

yielding dω y dω x = α x = −17.1 rad/s 2 , = α y = −3.43 rad/s 2 , dt dt dω z = α y = 2.40 rad/s 2 . dt (b) The acceleration of point B is a B = a G + α × rB /G = 3.33i + 1.33 j + 2k (m/s 2 ) i +

−17.1 rad/s 2 −3.43 rad/s 2 1m 1m

k 2.40 rad/s 2 0

= 0.933i + 3.73 j − 11.7k (m/s 2 ). (a) −17.1 i − 3.43 j + 2.40 k (rad/s 2 ). (b) 0.933i + 3.73 j − 11.7k (m/s 2 ).

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Problem 20.35 The inertia matrix of the Hubble telescope in terms of the body-fixed coordinate system shown is  I xx    −I yx   −I zx 

−I xy I yy −I zy

−I xz   36046 −706 1491     −I yz  =  −706 86868 449     449 93848   1491 I zz   kg-m 2 .

The telescope’s attitude control system uses a combination of internal reaction wheels and magnetic torquer rods that interact with Earth’s magnetic field. If the system is causing the telescope to rotate with a constant angular velocity ω = 4 i (degrees/minute), what moment must the system exert about the center of mass during this motion? Solution: We use Eqs. (20.18) to determine the moment. The constant components of the angular velocity are ω x = 4 deg/min ω y = 0,

( π180rad° )( 160mins ) = 0.00116 rad/s, ω z = 0.

Source: Courtesy of NASA. ∑ M x = 0, ∑ M y = Ω x I zx ω x = (0.00116 rad/s)(−1491 kg-m 2 )(0.00116 rad/s) = −0.00202 N-m, ∑ M z = −Ω x I yx ω x = −(0.00116 rad/s)(706 kg-m 2 )(0.00116 rad/s) = −0.000956 N-m.

The time derivatives of these components are zero, and because the coordinate system is body-fixed, Ω = ω. From Eqs. (20.18) we obtain

Problem 20.36 The Hubble telescope is not rotating when its attitude control system applies a moment 0.010 i − 0.020 j + 0.015k (N-m) about the center of mass. At that instant, what are the components of its angular acceleration? (See the inertia matrix in Problem 20.35.)

ΣM x = 0, ΣM y = −0.00202 N-m, ΣM z = −0.000956 N-m.

Solution:

At the instant the moment is applied, the components of the angular velocity are zero, and because the coordinate system is bodyfixed, Ω = ω = 0. Knowing the components of the moment, Eqs. (20.18) provide three equations for the three components of the angular acceleration: ∑ M x = 0.010 N-m = (36046 kg-m 2 )

− (−1491 kg-m 2 )

dω z , dt

dω y dω x − (706 kg-m 2 ) dt dt

∑ M y = −0.020 N-m = −(706 kg-m 2 ) − (−449 kg-m 2 )

dω z , dt

dω y dω x + (86868 kg-m 2 ) dt dt

∑ M z = 0.015 N-m = −(−1491 kg-m 2 ) + (93848 kg-m 2 )

dω z , dt

dω y dω x − (−449 kg-m 2 ) dt dt

yielding

dω y dω x = 2.66E-7 rad/s 2 , = −2.29E-7 rad/s 2 , dt dt dω z = 1.57E-7 rad/s 2 . dt (2.66 i − 2.29 j + 1.57k)E-7 rad/s 2 .

Source: Courtesy of NASA.

BandF_6e_ISM_C20_Pt1.indd 551

 I −I xy −I xz xx    −I yx I yy −I yz   −I zx −I zy I zz 

  0.06 kg-m 2    0  =     0  

0.78 kg-m 2

0.84 kg-m 2

  .   

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Problem 20.37 A 4-kg slender bar is rigidly attached to a 6-kg thin circular disk. In terms of the body-fixed coordinate system shown, the inertia matrix of the composite object is  I −I xy −I xz xx    −I yx I yy −I yz   −I zx −I zy I zz  2   0.06 kg-m  0   0 

    =    0

0.78 kg-m 2

0.84 kg-m 2

Solution:  ΣM x    ΣM y   ΣM z 

  .   

At the present instant, the angular velocity of the object relative to an Earth-fixed reference frame is ω = 12 i + 6 j − 4 k (rad/s) and its angular acceleration is zero. What are the components of the total moment exerted on the object about its center of mass?

Applying Eq. (20.19),

  0 −ω z ω y   I xx −I xy −I xz      −ω x   −I yx I yy −I yz 0  =  ωz      −ω y ω x 0   −I zx −I zy I zz      0 4 rad/s 6 rad/s   =  −4 rad/s −12 rad/s  0   0   −6 rad/s 12 rad/s

   ωx    ωy    ωz  

 0.06 kg-m 2 0   0 0.78 kg-m 2   0 0   −1.44 N-m    =  37.4 N-m  .    51.8 N-m 

  12 rad/s     6 rad/s     −4 rad/s    

0 0 0.84 kg-m 2

      

−1.44 i + 37.4 j + 51.8 k (N-m).

200 mm x 600 mm

Problem 20.38 A 4-kg slender bar is rigidly attached to a 6-kg thin circular disk. In terms of the body-fixed coordinate system shown, the inertia matrix of the composite object is given in Problem 20.37. At t = 0, the composite object is stationary and is subjected to the couple M = 2 i + 28 j − 40 k (N-m) about its center of mass. No other forces or couples act on the object. Determine the acceleration of the left end of the bar at t = 0. y

Applying Eqs. (20.18), dω y dω dω x − (0) − (0) z , dt dt dt dω y dω dω x 2 28 N-m = −(0) + (0.78 kg-m ) − (0) z , dt dt dt dω y ω d dω x −40 N-m = −(0) − (0) + (0.84 kg-m 2 ) z , dt dt dt 2 N-m = (0.06 kg-m 2 )

the angular acceleration of the object is α = 33.3i + 35.9 j − 47.6k (rad/s 2 ). There is no force, so the acceleration of the center of mass G is zero. The acceleration of the left end A of the bar is a A = a G + α × r A /G + ω × (ω × r A /G )

200 mm

i = 0+

33.3 rad/s 2 −0.6 m

35.9 rad/s 2 −47.6 rad/s 2 0 0

= 28.6 j + 21.5k (m/s 2 ).

600 mm

28.6 j + 21.5k (m/s 2 ).

Solution:

Let L = 0.6 m, R = 0.2 m, m B = 4 kg, m D = 6 kg. In terms of the x ′-y ′ coordinate system, the distance to the common center of mass is dB =

(0)m B +

( 12 L + R )m = 0.3 m, D

x, x

mB + mD

1 L + R − d B = 0.2 m. The center of mass is 2 located where the bar and disk are joined. so the distance d D =

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Problem 20.39 The vertical shaft supporting the dish antenna is rotating with a constant angular velocity of 1 rad/s. The angle θ = 30 °, d θ /dt = 20 °/s, and d 2 θ /dt 2 = −40 °/s 2 . The mass of the antenna is 280 kg, and its moments and products of inertia, in kg-m 2 , are I xx = 140, I yy = I zz = 220, and I xy = I yz = I zx = 0. Determine the couple exerted on the antenna by its support at A at the instant shown.

Solution:

The reactions at the support arise from (a) the Euler moments about the point A, and (b) the weight unbalance due to the offset center of mass. The Euler Equations: Express the reactions in the x, y, z system. The angular velocity in the x, y, z system ω = i sin θ + j cos θ + (d θ /dt )k = 0.5i + 0.866 j + 0.3491k (rad/s).

The angular acceleration is dω dθ d 2θ = ( i cos θ − j sin θ) + 2k dt dt dt = 0.3023i − 0.1745 j − 0.6981k (rad/s 2 ).

α =

Since the coordinates are body fixed Ω = ω, and Eq. (20.13) is  ΣM ox    ΣM oy   ΣM oz 

  140 0  0   0.302      =  0 220 0   −0.1745      0 220   −0.6981   0    −0.3491 0.866  0   +  0.3491 −0.5  0   0.5 0   −0.866

y 0.8 m

u A

 140 0 0   0.5    ×  0 220 0   0.866  .    0 220   0.3491   0  ΣM ox    ΣM oy   ΣM oz 

1 rad/s

   42.32  −76.79 190.52  0       −110  0  =  −38.39  +  48.87     110 0   −121.2  −153.6     0.5    ×  0.866     0.3491 

x C Base

 42.32     42.32  0       =  −38.39  +  −13.96  =  −52.36  ,        −118.9   34.64   −153.6 

u A

M 0 = 42.32 i − 52.36 j − 118.9k N -m mg

The unbalance exerted by the offset center of mass: The weight of the antenna acting through the center of mass in the x, y, z system is W = mg(− i sin θ − j cos θ) = −1373.4 i − 2378.8 j (N). The vector distance to the center of mass is in the x, y, z system is rG /O = 0.8i (m). The moment exerted by the weight is  i j k    M W = rG /O × W =  0.8 0 0   −1373.4 −2378.8 0    = 1903.0 k.

The couple exerted by the base:  C  x   Cy   Cz 

 ΣM    42.32      x    0 0            =  −52.36  +  0 0  =  ΣM y  −            −118.9   +1903.0   −1903.0   ΣM z       42.32    =  −52.36  (N-m),    1784.1 

C Base = 42.35i − 52.36 j + 1784.1k (N-m).

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Problem 20.40 The 5-kg triangular plate is connected to a ball-and-socket support at O. If the plate is released from rest in the horizontal position, what are the components of its angular acceleration at that instant?

0.6 m

0.9 m

Solution:

From the appendix to Chapter 20 on moments of inertia, the inertia matrix in terms of the reference frame shown is  m 1 m 1 2 2  bh 3 b h 0 −  A 12 A 8  m 1 m 1 b 2h 2 hb 3 0 [ I ] =  − A 8 A 4   m 1 1  0 0 bh 3 + hb 3  A 12 4  0.300 −0.675  0    kg-m 2 . =  −0.675 2.025 0   0 0 2.325  

(

) )

( (

)

(

)

          h z F

The moment exerted by the weight about the fixed pt. 0 is

(

)

2 1 Σm 0 = b i + h j × (−mg k) 3 3 = −9.81i + 29.43 j (N-m).

From Eq. (20.13) with ω = Ω = 0,    0.300 −0.675   d ω x /dt   −9.81  0        d ω y /dt  .  29.43  =  −0.675 2.025 0     0 0 0 2.325   d ω /dt     z   Solving, we obtain α = 14.5 j (rad/s 2 ).

Problem 20.41 If the 5-kg plate is released from rest in the horizontal position, what force is exerted on it by the ball-and-socket support at that instant? Solution:

See the solution of Problem 20.40. Let G denote the center of mass and Let F be the force exerted by the support.

The acceleration of the center of mass is

0.6 m

0.9 m

a G = a O + α × rG /O + ω × (ω × rG /O )

= 0+

i j k 0 14.5 0 2 1 b h 0 3 3

= −8.72k (m/s 2 ) From Newton’s second law, ΣF = ma G : F − (5)(9.81)k = (5)(−8.72k), we obtain F = 5.45k (N).

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Problem 20.42 The 5-kg triangular plate is connected to a ball-and-socket support at O. If the plate is released in the horizontal position with angular velocity ω = 4 i (rad/s), what are the components of its angular acceleration vector at that instant? Solution:

See the solution of Problem 20.40. From Eq. (20.13) with ω = Ω = 4 i (rad/s);

0.6 m

0.9 m

  d ω x /d t   0.300 −0.675  −9.81  0        d ω y /d t   29.43  =  −0.675 2.025 0     0 0 0 2.325   d ω /d t     z    4   0 0 0   0.300 −0.675 0       0 . +  0 0 −4   −0.675 2.025 0     0 0 2.325   0   0 4 0   Solving, we obtain α = 14.53 j + 4.65k (rad/s 2 ).

Problem 20.43 A subassembly of a space station can be modeled as two rigidly connected slender bars, each with a mass of 5000 kg. The subassembly is not rotating at t = 0, when a reaction control motor exerts a force F = 400 k (N) at B. What is the acceleration of point A relative to the center of mass of the subassembly at t = 0?

where m = 5000 kg, and L = 20 m. The moment of inertia matrix is   0.6667 0 0    Mg-m 2 . [I A ] =  0 0.6667 0   0 0 1.333   From the parallel axis theorem, Eq. (20.42), the moments and products of inertia about the center of mass are: I xx = I xxA − (d z2 + d y2 )(2 m) = 0.4167 Mg-m 2 , I yy = I yyA − (d x2 + d z2 )(2 m) = 0.4167 Mg-m 2 , I zz = I zzA − (d x2 + d y2 )(2 m) = 0.8333 Mg-m 2 .

I xy = I xyA − d x d y (2 m) = −0.2500 Mg-m 2 ,

20 m

from which the inertia matrix is   0.4167 0.2500 0    Mg-m 2 . [ I ] =  0.2500 0.4167 0   0 0 0.8333  

B 20 m

The vector distance from the center of mass to the point B is rB /G = (20 − 5) i + (0 − 5) j = 15i − 5 j (m). The moment about the center of mass is

Solution:

Choose a x ′, y ′, z ′ coordinate system with the origin at A and the x ′ axis parallel to the horizontal bar, and a parallel x, y, z system with origin at the center of mass. The Euler Equations: The center of mass in the x ′, y ′, z ′ system has the coordinates 10(5000) + 0(5000) xG = = 5 m, 10000 10(5000) + 0(5000) yG = = 5 m, 10000 z G = 0, from which (d x , d y , d z ) = (5,5,0) m. From Appendix C, the moments and products of inertia of each bar about A are I xxA = I yyA =

mL2 , 3

2mL2 3 I xyA = I xzA = I yzA = 0, I zzA = I xxA + I yyA =

BandF_6e_ISM_C20_Pt1.indd 555

 i j k    M G = rB /G × F =  15 −5 0   0 0 400    = −2000 i − 6000 j (N-m). The coordinates are body-fixed, and the object is initially stationary, from which Ω = ω = 0, and Eq. (20.19) reduces to  −2000     −6000  = [ I ]   0    4.167 × 10 5  =  2.5 × 10 5   0 

2.5 × 10 5 0 4.167 × 10 5 0 0 8.333 × 10 5

  α x    αy   α  z

   .   

Carry out the matrix multiplication to obtain: 4.167 × 10 5 α x + 2.5 × 10 5 α y = −2000, 2.5 × 10 5 α x + 4.167 × 10 5 α y = −6000, and α z = 0 . Solve: α = 0.006 i − 0.018 j (rad/s 2 ).

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20.43 (Continued) Newton’s second law: The acceleration of the center of mass of the object from Newton’s second law is aG =

( 21m )F = 0.04k (m/s ). 2

20 m

The acceleration of point A: The vector distance from the center of mass to the point A is r A /G = −5i − 5 j (m). The acceleration of point A is

a A = a G + α × r A /G + ω × (ω × r A /G ). Since the object is initially stationary, ω = 0. y

 i j k   a A = a G + α × r A /G = 0.04 k +  0.006 −0.018 0   −5 −5 0    = −0.08k (m/s 2 ), a A = −0.08k (m/s 2 ).

Problem 20.44 A subassembly of a space station can be modeled as two rigidly connected slender bars, each with a mass of 5000 kg. If the subassembly is rotating about the x-axis at a constant rate of 1 revolution every 10 minutes, what is the magnitude of the couple its reaction control system is exerting on it?

x z

20 m

(See Figure in solution to Problem 20.100.) The angular acceleration of the disk is given by j ωO 0

k  0   0  

Solution:

 i  dω d = (ω i + ω O j) + ω O × ω d = 0 +  0 dt dt d  ω  d = −ω O ω d k.

B 20 m

The velocity of point A relative to O is a A /O = α × r A /O + ω × (ω × r A /O ) = (−ω O ω d )(k × r A /O ) + ω × (ω × r A /O ). Term by term:   i j k    −ω O ω d (k × r A /O ) = −ω O ω d  0 0 1   b R sin θ −R cos θ    = ω O ω d R sin θ i − ω O ω d b j,   i j k    ω × (ω × r A /O ) = ω ×  ω d ωO 0   b R sin θ −R cos θ      i j k    =  ωd ωO 0     −Rω O cos θ Rω d cos θ Rω d sin θ − bω O  = ( Rω d sin θ − bω O )(ω o i − ω d j) + ( R cos θ )(ω d2 + ω O2 )k. Collecting terms: a A /O = (2 Rω O ω d sin θ − bω O2 ) i − ( Rω d2 sin θ) j + ( Rω d2 cos θ + Rω O2 cos θ)k.

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Problem 20.45 The thin circular disk of radius R = 0.2 m and mass m = 4 kg is rigidly attached to the vertical shaft. The plane of the disk is slanted at an angle β = 30 ° relative to the horizontal. The shaft rotates with constant angular velocity ω 0 = 25 rad/s. Determine the magnitude of the couple exerted on the disk by the shaft.

Solution: y

v0 z

b x

R b

In terms of the body-fixed reference frame shown, the disk’s inertia matrix is  1  mR 2  4  [ I ] =  0    0  

1 mR 2 4

1 mR 2 2

    0.04 0 0     =  0 0.04 0  kg-m 2 .     0 0 0.08      

The disk’s angular velocity is ω = Ω = ω 0 sin β j + ω 0 cos β k = 12.50 j + 21.65k (rad/s). From Eq. (20.19) with d ω x /dt = d ω y /dt = d ω z /dt = 0,  ΣM x    ΣM y   ΣM z 

Problem 20.46 The slender bar of mass m = 10 kg and length L = 1 m is welded to a horizontal shaft that rotates with constant angular velocity ω 0 = 30 rad/s. The constant angle β = 25 °. Determine the magnitude of the couple C exerted on the bar by the shaft. Solution:

In terms of the body-fixed coordinate system, the moments of inertia of the slender bar are I xx = 0,

I yy = I zz =

  −21.65 12.5   0.04 0  0 0      0.04 0  0 0  0  =  21.65    0 0.08  0 0   0  −12.5      10.8  0     ×  12.5  =  0  N-m.      21.65   0 

x y v0

L 2

1 mL2 = 0.833 kg-m 2 , 12

and the products of inertia are zero. The angular velocity of the bar is ω = ω 0 cos θ i − ω 0 sin θ j = 27.2 i − 12.7 j (rad/s). The bar’s angular acceleration is zero. From Eqs. (20.12), the components of the couple exerted on the bar by the shaft are C x = 0, C y = 0, C z = I yy ω x ω y = −287 N-m. C = 287 N-m.

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Problem 20.47* The slender bar of mass m = 10 kg and length L = 1 m is pinned to the rotating shaft so that it can rotate freely about the z-axis of the body-fixed coordinate system. (That is, no couple is exerted on the bar about the z-axis.) The horizontal shaft rotates with constant angular velocity ω 0 = 30 rad/s. At the instant shown, the angle β = 25 ° and d β /dt = 0. What is the bar’s angular acceleration at that instant?

The angular velocity of the bar is ω = ω x i + ω y j + ω zk = ω 0 cos θ i − ω 0 sin θ j + ω z k. Its angular acceleration is dω y dω z dω x dω i+ j+ k α = = dt dt dt dt dω z dθ dθ = −ω 0 sin θ i − ω 0 cos θ j+ k. dt dt dt Because d θ /dt = 0 at the present instant, the angular acceleration is

x y

α =

dω z k. dt

No couple is exerted on the bar about the z axis, so from Eq. (20.12) 3 ,

0 = I zz

dω z + I yy ω x ω y , dt

and we obtain dω z = −ω x ω y dt = ω 0 2 sin θ cos θ

L 2

= 345 rad/s 2 .

Solution:

In terms of the body-fixed coordinate system, the moments of inertia of the slender bar are 1 I xx = 0, I yy = I zz = mL2 = 0.833 kg-m 2 , 12

α = 345 k (rad/s 2 ).

and the products of inertia are zero.

Problem 20.48 The slender bar of length l and mass m is pinned to the vertical shaft at O. The vertical shaft rotates with a constant angular velocity ω 0 . Show that the value of ω 0 necessary for the bar to remain at a constant angle β relative to the vertical is ω0 =

3g /(2l cos β ).

Solution:

This is motion about a fixed point so Eq. (20.13) is applicable. Choose a body-fixed x, y, z coordinate system with the origin at O, the positive x axis parallel to the slender bar, and z axis out of the page. The angular velocity of the vertical shaft is Ω = ω 0 (−i cos β + j sin β ).

The vector from O to the center of mass of the bar is rG /O = ( L /2) i. The weight is W = mg( i cos β − j sin β ). The moment about the point O is  i j k     mgL L M G = rG /O × W =  0 0  = − sin β k   2 2    mg cos β −mg sin β 0   

(

)

The moments and products of inertia about O in the x, y, z system are I xx = 0,

I yy = I zz = mL2 3, l

I xy = I xz = I yz = 0. The body-fixed coordinate system rotates with angular velocity

Ω = ω = ω 0 (−i cos β + j sin β ). Eq. (20.13) reduces to  M Ox    M Oy   M Oz 

558

  0 0 ω 0 sin β      0 0 ω 0 cos β   =     0    −ω 0 sin β −ω 0 cos β       0 0 0   −ω cos β  0       mL2 × 0 0   ω 0 sin β  . 3     0    mL2    0 0   3  

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20.48 (Continued) Carry out the matrix multiplication,  M Ox    M Oy   M Oz 

  0 0       0  =  0     ω mL2 cos β   0 − 0  3

ω 0 mL2 sin β 3 ω 0 mL2 cos β 3

   0  =  0   − ω 02 mL2 cos β sin β  3 

    .    

      −ω 0 cos β      ω 0 sin β       0     

Equate the z components: MOz = −

ω 02 mL2 cos β sin β . 3

The pin-supported joint at O cannot support a couple, C = 0, from which ω 2 mL2 cos β sin β mgL MO = MG ⋅ − sin β = − 0 2 3

b mg

Assume that β ≠ 0, from which sin β ≠ 0, and the equation can be solved for ω0 =

3g . 2l cos β

Problem 20.49 The vertical shaft rotates with constant angular velocity ω 0 . The 35° angle between the edge of the 10-lb thin rectangular plate pinned to the shaft and the shaft remains constant. Determine ω 0 .

358

10 lb

2 ft x 358 The moment about the center of mass is 1 ft

Solution:

This is motion about a fixed point, and Eq. (20.13) is applicable. Choose an x, y, z coordinate system with the origin at the pinned joint O and the x axis parallel to the lower edge of the plate, and the y axis parallel to the upper narrow edge of the plate. Denote β = 35 °. The plate rotates with angular velocity ω = ω 0 (−i cos β + j sin β ) = ω 0 (−0.8192 i + 0.5736 j) (rad/s).

The vector from the pin joint to the center of mass of the plate is rG /O = i + (0.5) j (ft). The weight of the plate is W = 10( i cos β − j sin β )

 i j k   M G = rG /O × W =  1 0.5 0   8.192 −5.736 0    = −9.832k (ft-lb). From Appendix C, the moments and products of inertia of a thin plate about O are mh 2 = 0.1036 slug-ft 2 , 3 mb 2 I yy = = 0.4145 slug-ft 2 , 3 m I zz = (h 2 + b 2 ) = 0.5181 slug-ft 2 , 3 mbh I xy = = 0.1554 slug-ft 2 , 4 I xz = I yz = 0. I xx =

= 8.192 i − 5.736 j (lb).

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20.49 (Continued) At a constant rate of rotation, the angle β = 35 ° = const, α = 0. The body-fixed coordinate system rotates with angular velocity Ω = ω = ω 0 (−i cos β + j sin β ) = −0.8191ω 0 i + 0.5736ω 0 j (rad/s),

Carry out the matrix operations to obtain:  M Ox    M Oy   M Oz 

   0      0  = ω 02      −0.1992   

and Eq. (20.13) reduces to:

The pin support cannot support a couple, from which

 M Ox    M Oy   M Oz 

M Oz = M Gz , − 9.832 = −ω 02 (0.1992),

  0 0 0.5736      2 0 0 0.8191  ω =  0    0  −0.5736 −0.8191     0.1036 −0.1554   −0.8191  0      0.5736  . 0 ×  −0.1554 0.4145    0 0 0 0.5181    

Problem 20.50 The radius of the 20-lb thin circular disk is R = 1 ft. The disk is mounted on the horizontal shaft and rotates with constant angular velocity ω d = 10 rad/s relative to the shaft. The horizontal shaft is 2 ft in length. The vertical shaft rotates with constant angular velocity ω 0 = 4 rad/s. Determine the force and couple exerted at the center of the disk by the horizontal shaft.

from which ω 0 = 7.025 rad/s.

Solution:

Using Newton’s Second Law

F − mg j = ma = −mr ω 02 k F = (20 lb) j −

lb ( 32.220 ft/s )(2ft)(4rad/s) k 2

F = (20 j − 19.9k) lb. In preparation to use Euler’s Equations we have ω = ω 0 j + ω d k = (4 j + 10 k) rad/s α = ω 0 j × ω d k = (40rad/s 2 ) i

 1  mR 2  4  [ I ] =  0    0  

vd z

x R

1 mR 2 4

1 mR 2 2

    0.155  0 0    =   slug-ft 2 0 0.155 0     0 0 0.311       

Euler’s Equations are now v0

M = [ I ]α + ω × [ I ]ω   40   0.155 0 0       0  M =  0 0.155 0     0 0 0.311   0     0   0 −10 4   0.155 0 0        4  +  10 0 0  0 0.155 0      0 0   0 0 0.311   10   −4 M = 12.4 i lb-ft.

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Problem 20.51 The object shown in Fig. a consists of two 1-kg vertical slender bars welded to the 4-kg horizontal slender bar. In Fig. b, the object is supported by bearings at A and B. The horizontal circular disk is supported by a vertical shaft that rotates with constant angular velocity ω 0 = 6 rad/s. The horizontal bar rotates with constant angular velocity ω = 10 rad/s. At the instant shown, determine the y and z components of the forces exerted on the object at A and B. Solution:

The nonzero inertias are

1 I xx = 2 (1 kg)(0.1 m) 2 = 0.00667 kg-m 2 , 3 1 I yy = (4 kg)(0.4 m) 2 + 2(1 kg)(0.1 m) 2 = 0.0733 kg-m 2 , 12 I zz = I xx + I yy = 0.08 kg-m 2 ,

The angular velocity and angular acceleration are ω = (10 i + 6 j) rad/s α = 6 j × 10 i = −60 k rad/s 2 . The moment about the center of mass is M x = 0, M y = B z (0.2 m) − Az (0.2 m), M z = B y (0.2 m) − A y (0.2 m). Euler’s equations are now   0   B z − Az  B − A y  y

 0.00667 −0.01 0   0    =  −0.01 0.0733 0   0     0 0 0.08   −60    0 6 0   0.00667 −0.01 0   10    +  −6 0   6  0 10   −0.01 0.0733     0 0 0.08   0   0 −10 0  

I xy = 2(1 kg)(0.1 m)(0.05 m) = 0.01 kg-m 2 . Newton’s Second Law gives (the acceleration of the center of mass is zero). ΣFy : A y + B y − (6 kg)(9.81 m/s 2 ) = 0,

    (0.2)  

Solving Euler’s equations along with Newton’s Second Law we find

ΣFz : Az + B z = 0.

A y = 33.0 N, Az = 0, B y = 25.8 N, B z = 0.

0.1 m

0.2 m

0.1 m

(a)

(b)

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Problem 20.52 The distance h from the 0.135-kg symmetric top’s point to its center of mass is 50 mm. The top’s moments of inertia are I xx = I yy = 2.5E−4 m 4 , I zz = 4.30E−5 m 4 , and its products of inertia are zero. The point of the spinning top remains at a fixed point on the floor, which is the origin O of the secondary reference frame shown. The angle θ = 30 ° between the z-axis and the vertical is constant. The x-axis remains parallel to the floor and rotates about O with constant angular velocity ω 0 = 6.3 rad/s. The constant angular velocity of the top relative to the secondary frame, its “spin rate,” is ω rel = ω rel k. Apply Euler’s equations to the top’s motion to determine the spin rate.

Solution:

The secondary reference frame rotates about the vertical axis with angular velocity ω 0 = 6.3 rad/s. Applying the right-hand rule, its angular velocity vector is Ω = ω 0 sin θ j + ω 0 cos θ k. The angular velocity of the top relative to the secondary frame is ω rel = 250 k (rad/s) = ω rel k, so the angular velocity of the top relative to the Earth-fixed frame is ω = Ω + ω rel = ω 0 sin θ j + (ω 0 cos θ + ω rel )k. The moment about the x axis due to the top’s weight is ΣM Ox = mgh sin θ. Equation (20.12) 1 is ΣM Ox = −Ω z I yy ω y + Ω y I zz ω z : mgh sin θ = −ω 0 cos θ I yy ω 0 sin θ + ω 0 sin θ I zz (ω 0 cos θ + ω rel ),

mgh = −ω 0 cos θ I yy ω 0 + ω 0 I zz (ω 0 cos θ + ω rel ) = ω 0 2 cos θ( I zz − I yy ) + ω 0 ω rel I zz .

Observe that Eqs. (20.12) 2 and (20.12) 3 are identically satisfied. Solving this equation with ω 0 = 6.3 rad/s, we obtain ω rel = 271 rad/s.

ω rel = 271 rad/s.

O v0

Problem 20.53 The thin rectangular plate (mass m = 10 kg, dimensions b = 0.4 m, h = 0.3 m ) is welded to the rectangular frame. The constant angle θ = 25 °. The frame rotates with constant angular velocity ω 0 = 30 rad/s. Determine the magnitude of the couple C exerted on the plate by the frame.

Solution: In terms of the body-fixed coordinate system shown, the moments of inertia of the thin plate are I xx =

1 1 mh 2 = 0.075 kg-m 2 , I yy = mb 2 = 0.133 kg-m 2 , 12 12 1 I zz = m(b 2 + h 2 ) = 0.208 kg-m 2 , 12

and the products of inertia are zero. z

The angular velocity of the plate is

ω = ω xi + ω y j + ω zk = −ω 0 sin θ i + ω 0 cos θk

= −12.7 i + 27.2 k (rad/s). The bar’s angular acceleration is zero. From Eqs. (20.12), the components of the couple exerted on the plate by the frame are C x = 0, C y = I xx ω x ω z = −25.9 N-m, C z = 0.

C = 25.9 N-m.

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Problem 20.54* The thin rectangular plate (mass m = 10 kg, dimensions b = 0.4 m, h = 0.3 m ) is attached to the rectangular frame by pins. (No couple is exerted on the plate about the y-axis.) The frame rotates with constant angular velocity ω 0 = 30 rad/s. At the instant shown, the angle θ = 25 ° and d θ /dt = 0. What is the plate’s angular acceleration at that instant?

Solution:

In terms of the body-fixed coordinate system shown, the moments of inertia of the thin plate are I xx =

1 mh 2 , 12

I yy =

1 1 mb 2 , I zz = m(b 2 + h 2 ) = I xx + I yy , 12 12

and the products of inertia are zero. The angular velocity of the plate is ω = ω xi + ω y j + ω z k

= −ω 0 sin θ i + ω y j + ω 0 cos θ k. Its angular acceleration is dω y dω z dω x dω i+ j+ k = dt dt dt dt dω y dθ dθ j − ω 0 sin θ k. = −ω 0 cos θ i + dt dt dt

α =

Because d θ /dt = 0 at the present instant, the angular acceleration is α =

dω y j. dt

No couple is exerted on the plate about the y axis, so from Eq. (20.12) 2 , 0 = I yy

dω y + I xx ω x ω z − I zz ω x ω z , dt

and we obtain  I − I xx  dω y  ω x ω z =  zz  dt I yy  = −ω 0 2 sin θ cos θ = −345 rad/s 2 . α = −345 k (rad/s 2 ). α = 345 k (rad/s 2 ).

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Problem 20.55* The axis of the right circular cone of mass m, height h, and radius R spins about the vertical axis with constant angular velocity ω 0 . The center of mass of the cone is stationary, and its base rolls on the floor. Show that the angular velocity necessary for this motion is ω 0 = 10 g /3 R . (See Example 20.7.)

x v0 z

W 3h 4 h

v0 The moments and products of inertia of a cone about its center of mass in the x, y, z system are, from Appendix C,

z R

I xx = I yy = m h

I zz =

Solution:

This a problem of general motion, and Eq. (20.19) applies. The vector distance from the center of mass to the base of the cone is h r B /G = k 4

(see Appendix C). The angular velocity of rotation of the body fixed coordinate system is Ω = ω 0 i. The velocity of the center of the base is    i  v B = Ω × r B /G =  ω 0    0 

ω 0h j, 4

ω h from which φ = − 0 . 4R

( ω4h − Rφ ) j = 0,

dω 0 = 0. dt The body-fixed coordinate system rotates with angular velocity Ω = ω 0 i, and

 M Gx    M Gy   M Gz 

hω 0 k. 4R

   0 0 0   I xx       = ω 02  0 0 −1   0     0 1 0   0   

The center of mass of the cone is at a zero distance from the axis of rotation, from which the acceleration of the center of mass is zero. The angular velocity about the z -axis, ω = ω 0i −

Since the rotation rate is constant, and the z axis remains horizontal, the angular acceleration is zero,

from which 0 = v + ω × (−Ri) = −

Eq. (9.26) becomes:

Let the spin rate about the z axis be φ , so that the angular velocity, from which ω = Ω + φ k. The point of contact with the surface is stationary, and the velocity of the center of the base of the cone is v = −

3mR 2 , I xy = I xz = I yz = 0. 10

ω = ω 0i −

  j k   ω h 0 0  = − 0 j.  4  h  0 4 

( 803 h + 203 R ),

ω 0h k (rad/s). 4R

 0   hI zz  4R   0

ω 02 

0 I yy

0 0

I zz

   1   0   − h  4R 

        

   .   

For equilibrium, M Oy = M Gy , from which mgh 3mhR 2 = ω0 . 4 40 Solve ω 0 =

10 g . 3R

The weight of the cone is W = −mg i. The reaction of the floor on the cone is N = −W. The moment about the center of mass exerted by the weight is  i j k    mgh h  = + M G = r B /G × N =  0 0 j.   4 4    mg 0 0   

(

564

)

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Problem 20.56 The tilted homogeneous cone undergoes a steady motion in which its flat end rolls on the floor while the center of mass remains stationary. The angle β between the axis and the horizontal remains constant, and the axis rotates about the vertical axis with constant angular velocity ω 0 . The cone has mass m, radius R, and height h. Show that the angular velocity ω 0 necessary for this motion satisfies ω0 2 =

g ( R sin β − 14 h cos β ) . 3 1 2 3 2 2 20 ( R + 4 h ) sin β cos β − 40 hR cos β

(See Example 20.7.) Solution:

Following Example 20.6, we use a coordinate system with the z axis pointing along the cone axis, the y axis remains horizontal (out of the paper) and the x axis completes the set Ω = ω 0 cos β i + ω 0 sin β k ω = Ω + ω rel k = ω 0 cos β i + (ω 0 sin β + ω rel )k

The point P of the cone that is in contact with the ground does not move, therefore v P = v C + ω × r P /C 1 0 = 0 + [ω 0 cos β i + (ω 0 sin β + ω rel )k] ×  −Ri − h k   4  1   =  hω 0 cos β − R(ω 0 sin β + ω rel )  j. 4  Solving yields h h ω rel =  ω cos β k. cos β − sin β  ω 0 , ω = ω 0 cos β i +  4R  4R 0

Since the center of mass is stationary, the floor exerts no horizontal force, and the vertical force is equal to the weight ( N = mg). The moment about the center of mass due to the normal force is

(

M = mg R sin β −

)

1 h cos β j 4

The moments and products of inertia for the cone are  3 3  mR 2 + mh 2  20 80  [ I ] =  0    0  

3 3 mR 2 + mh 2 20 80

3 mR 2 10

          

Substituting these expressions into Eq. (20.19), and evaluating the matrix products, we obtain

(

mg R sin β −

)

(

)

1 3 2 3 2 h cos β =  h + R cos β sin β  80 4 20 3 − hR cos 2 β  mω 02  40

Solving, we find that

ω 02 =

(

)

1 h cos β 4 . 3 3 1 R 2 + h 2 sin β cos β − hR cos 2 β 40 20 4 g R sin β −

(

)

h R

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Problem 20.57* The two thin disks are rigidly connected by a slender bar. The radius of the large disk is 200 mm and its mass is 4 kg. The radius of the small disk is 100 mm and its mass is 1 kg. The bar is 400 mm in length and its mass is negligible. The composite object undergoes a steady motion in which it spins about the vertical y-axis through its center of mass with angular velocity ω 0 . The bar is horizontal during this motion and the large disk rolls on the floor. What is ω 0? y v0

Solution: The z axis remains aligned with the bar and the y axis remains vertical. The center of mass (measured from the large disk) is located a distance d =

(1 kg)(0.4 m) = 0.08 m (5 kg)

The inertias are 1 1 (4 kg)(0.2 m) 2 + (1 kg)(0.1 m) 2 = 0.085 kg-m 2 2 2 1 I xx = I yy = (4 kg)(0.2 m) 2 + (4 kg)(0.08 m 2 ) 4 1 + (1 kg)(0.1 m) 2 + (1 kg)(0.32 m) 2 = 0.1705 kg-m 2 4 I xy = I xz = I yz = 0 I zz =

The angular velocity of the coordinate system is Ω = ω 0 j Define ω z to be the rate of rotation of the object about the z axis. Thus ω = ω 0 j + ω zk To find ω z , require that the velocity of the point in contact with the floor be zero

v P = ω × r = (ω 0 j + ω z k) × (−0.08k − 0.2 j) m = [(0.2 m)ω z − (0.08 m)ω 0 ]i = 0 ⇒ ω z = 0.4 ω 0

Since the center of mass does not move, the normal force on the contact point is equal to the weight. Therefore the moment about the center o mass is given by M = [(−0.2 m) j − (0.08 m)k] × [(5 kg)(9.81 m/s 2 ) j] = (3.924 N) i Equation 20.19 now gives (d ω x /dt = d ω y /dt = d ω z /dt = 0)  0  3.924 N     0   =  0    −ω 0 0   

0 ω0 0 0 0 0

  I xx   0     0

0 I yy

0 0

I zz

 0     ω0     0.4 ω 0 

    

Putting in the values and solving we find ω 0 = 10.7 rad/s.

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Problem 20.58 The view of an airplane’s landing gear as seen looking from behind the airplane is shown in Fig. a. The radius of the wheel is 300 mm, and its moment of inertia is 2 kg-m 2 . The airplane takes off at 30 m/s. After takeoff, the landing gear retracts by rotating toward the right side of the airplane, as shown in Fig. b. Determine the magnitude of the couple exerted by the wheel on its support. (Neglect the airplane’s angular motion.)

Solution:

Choose a coordinate system with the origin at the center of mass of the wheel and the z axis aligned with the carriage, as shown. Assume that the angular velocities are constant, so that the angular accelerations are zero. The moments and products of inertia of the wheel are I xx = mR 2 /2 = 2 kg-m 2 , from which m = 44.44 kg. I yy = I zz = mR 2 /4 = 1 kg-m 2 .

The angular velocities are Ω = −(45(π /180)) j = −0.7853 j rad/s. v 30 ω = − i+ Ω = − i − 0.7853 j = −100 i − 0.7853 j. R 0.3

( )

Eq. (20.19) becomes

45 deg/s

 M x    My   Mz 

0 Ω y   I xx   0 0   0  0 0   0 

  0     0 =     −Ω y   

0 I yy

0 0

I zz

   ω x    ωy    0 

  0    =  0     −Ω y ω x I xx  

  ,   

from which M 0 = −Ω y ω x I xx k. Substitute: M = (0.7854)(100)2 = 157 N-m. z 300 mm (a)

Problem 20.59 If the rider turns to his left, will the couple exerted on the motorcycle by its wheels tend to cause the motorcycle to lean toward the rider’s left side or his right side?

45 deg/s

(b)

Solution:

Choose a coordinate system as shown in the front view, with y positive into the paper. The Eqs. (20.19) in condensed notation are ∑M =

For

dH + Ω × H. dt

dH = 0, dt ∑ M = Ω × H.

If the rider turns to his left, the angular velocity is Ω = +Ωk rad /s. The angular momentum is H = H x i + H z k, where H x > 0. The cross product  i  Ω × H =  0  Hx 

j k  0 +Ω  = +ΩH x j. 0 H z  

For a left turn the moment about y is positive, causing the cycle to lean to the left.

Z æ

X v0

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Problem 20.60* By substituting the components of H O from Eqs. (20.9) into the equation ΣM O =

dH Oy dH Oz dH Ox i+ j+ k + Ω × HO , dt dt dt

Substituting Eqs. (20.9) and assuming that the moments and products of inertia are constants, we obtain Eqs. (20.12): dω y dω z dω x − I xy − I xz dt dt dt + Ω y (−I zx ω x − I zy ω y + I zz ω z )

∑ M Ox = I xx

− Ω z (−I yx ω x + I yy ω y − I yz ω z ),

derive Eqs. (20.12).

dω y dω z dω x + I yy − I yz dt dt dt − Ω x (−I zx ω x − I zy ω y + I zz ω z )

∑ M Oy = −I yx

Solution: dH Oy dH Oz dH Ox i+ j+ k dt dt dt

∑M0 =

i Ωx

j Ωy

k Ωz

H Ox

H Oy

H Oz

+ Ω z ( I xx ω x − I xy ω y − I xz ω z ), dω y dω z dω x − I zy + I zz dt dt dt + Ω x (−I yx ω x + I yy ω y − I yz ω z )

∑ M Oz = −I zx

− Ω y ( I xx ω x − I xy ω y − I xz ω z ).

The components of this equation are dH Ox + Ω y H Oz − Ω z H Oy , dt dH Oy ∑ M Oy = − Ω x H Oz + Ω z H Ox , dt dH Oz ∑ M Oz = + Ω x H Oy − Ω y H Ox . dt ∑ M Ox =

Problem 20.61 A ship has a turbine engine. The spin axis of the axisymmetric turbine is horizontal and aligned with the ship’s longitudinal axis. The turbine rotates at 10,000 rpm. Its moment of inertia about its spin axis is 1000 kg-m 2 . If the ship turns at a constant rate of 20 degrees per minute, what is the magnitude of the moment exerted on the ship by the turbine? Strategy: Treat the turbine’s motion as steady precession with nutation angle θ = 90 °.

Solution: Choose a coordinate system with the z axis parallel to the axis of the turbine, and y positive upward. From Eq. (20.29), sin θ, ∑ M x = ( I zz − I xx )ψ 2 sin θ cos θ + I zzφψ where I xx =

1 I = 500 kg-m 2 2 zz

ψ = 20

π ( 180 )( 601 ) = 0.005818 (rad/s),

φ = 10000(2π /60) = 1047.2 rad/s, θ = 90 °. M x = 6092 N-m.

x c = 208/min (top view of turbine)

f = 10,000 rpm

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Problem 20.62 The center of the car’s wheel A travels in a circular path about O at 15 mi/h. The wheel’s radius is 1 ft, and the moment of inertia of the wheel about its axis of rotation is 0.8 slug-ft 2 . What is the magnitude of the total external moment about the wheel’s center of mass? Strategy: Treat the wheel’s motion as steady precession with nutation angle θ = 90 °. Solution:

18 ft

From Eq. (20.29)

sin θ, ∑ M x = ( I zz − I xx )ψ 2 sin θ cos θ + I zzφψ where the spin is v 15 5280 22 φ = = = = 1.222 rad/s, R 18 3600 18

(

)

the precession rate is v 22 ψ = = = 22 rad/s, Rw 1

f c

1 and the nutation angle is θ = 90 °. Using I xx = I zz = 0.4 slug-ft 2 , 2 from which M x = 21.51 ft-1b.

Problem 20.63 The radius of the 5-kg disk is R = 0.2 m. The disk is pinned to the horizontal shaft and rotates with constant angular velocity ω d = 6 rad/s relative to the shaft. The vertical shaft rotates with constant angular velocity ω 0 = 2 rad/s. By treating the motion of the disk as steady precession, determine the magnitude of the couple exerted on the disk by the horizontal shaft. (See Practice Example 20.8.)

R v0

Solution:

In Problem 20.50 a thin circular disk of mass m is mounted on a horizontal shaft and rotates relative to the shaft with constant angular velocity ω d . The horizontal shaft is rigidly attached to the vertical shaft rotating with constant angular velocity ω 0 . The magnitude of the couple exerted on the disk by the horizontal shaft is to be determined. The nutation angle is θ = 90 °. The precession rate is ψ = ω 0 , and the spin rate is φ = ω d . The moments and products of inertia of the disk: I xx = I zz =

mR 2 , 2

mR 2 , 4 I xy = I xz = I yz = 0. I yy =

Eq. (20.29) is sin θ, M y = ( I xx − I yy )ψ 2 sin θ cos θ + I xx ψφ from which My =

mR 2 ω 0 ω d = 1.2 N-m. 2

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Problem 20.64 The helicopter is stationary. The z-axis of the body-fixed coordinate system points downward and is coincident with the axis of the helicopter’s rotor. The moment of inertia of the rotor about the z-axis is 8600 kg-m 2 . Its angular velocity is −258k (rpm). If the helicopter begins a pitch maneuver during which its angular velocity is 0.02 j (rad/s), what is the magnitude of the gyroscopic moment exerted on the helicopter by the rotor? Does the moment tend to cause the helicopter to roll about the x-axis in the clockwise direction (as seen in the photograph) or the counterclockwise direction? Solution:

x y

The spin rate is φ = −258 rpm = −27.0 rad/s.

The pitch rate is ψ = 0.02 rad/s.

Source: Courtesy of Stocktrek Images/Alamy Stock Photo.

In eq. 20.29, the moment exerted on the rotor is = (8600 kg-m 2 )(−27.0 rad/s)(0.02 rad/s) = −4650 N-m M = I zzφψ The motor exerts a moment on the helicopter in the opposite direction which tends to roll the helicopter in the counterclockwise direction. Answer: M = 4650 N-m counterclockwise.

Problem 20.65 The bent bar is rigidly attached to the vertical shaft, which rotates with constant angular velocity ω 0 = 10 rad/s. The thin circular disk of mass m = 10 kg and radius R = 0.2 m is pinned to the bent bar and rotates with constant angular velocity ω d = 30 rad/s relative to the bar. The dimensions b = h = 0.5 m and the angle β = 40 °. (a) What is the magnitude of the force exerted on the disk by the bar? (b) Use Eq. (20.29) to determine the magnitude of the couple exerted on the disk by the bar.

vd b v0

Solution: h

(a) The center of mass of the disk moves in a circular path with radius r = h + b cos β = 0.5 m + (0.5 m) cos 40 ° = 0.883 m. The acceleration of the center of mass toward the center of the circular path is

a n = ω 0 2r = (10 rad/s) 2 (0.883 m) = 88.3 m/s 2 . The horizontal force exerted on the disk by the bar is

u z

Fn = ma n , so the magnitude of the total force exerted on the disk by the bar is x

(ma n ) 2 + (mg) 2 = 888 N. (b) Following Fig. 20.13, we orient the coordinate system as shown, with the z axis perpendicular to the disk and the x axis horizontal. With this orientation, 1 1 I xx = mR 2 , I zz = mR 2 . 4 2 The spin rate φ = ω , the precession rate ψ = ω , and d

θ = 90 ° − β , and Eq. (20.29) gives

ΣM x = ( I zz − I xx )ω 0 2 sin(90 ° − β ) cos(90 ° − β ) + I zz ω d ω 0 sin(90 ° − β ) = 50.9 N-m. (a) 888 N. (b) 50.9 N-m.

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Problem 20.66 The bent bar is rigidly attached to the vertical shaft, which rotates with constant angular velocity ω 0 = 10 rad/s. The thin circular disk of mass m = 10 kg and radius R = 0.2 m is pinned to the bent bar and rotates with constant angular velocity ω d relative to the bar. The dimensions b = h = 0.5 m and the angle β = 40 °. Use Eq. (20.29) to determine the value of ω d for which the bar exerts no couple on the disk.

vd b v0

Solution:

Following Fig. 20.13, we orient the coordinate system as shown, with the z axis perpendicular to the disk and the x axis horizontal. With this orientation, I xx =

1 mR 2 , 4

I zz =

b h

1 mR 2 . 2

The spin rate φ = ω d , the precession rate ψ = ω 0 , and θ = 90 ° − β , and Eq. (20.29) gives ΣM x = ( I zz − I xx )ω 0

u z

2 sin(90 ° − β ) cos(90 ° − β )

+ I zz ω d ω 0 sin(90 ° − β ).

Setting ΣM x = 0 and solving, we obtainω d = −3.21 rad/s.

ω d = −3.21 rad/s.

Problem 20.67 A thin circular disk undergoes moment-free steady precession. The z-axis is perpendicular to the disk. Show that the disk’s precession rate is ψ = −2φ / cos θ. (Notice that when the nutation angle is small, the precession rate is approximately two times the spin rate.)

? f Y

Solution:

Moment free steady precession is described by Eq. (20.33), ( I zz − I xx )ψ cos θ + I zzφ = 0, where ψ is the precession rate, φ is the spin rate, and θ is the nutation angle. For a thin circular disk, the moments and products of inertia are I xx = I yy =

mR 2 , 4

mR 2 , 2 I xy = I xz = I yz = 0. I zz =

Substitute: mR 2

( 12 − 14 )ψ cos θ + ( mR2 )φ = 0. 2

Reduce, to obtain 2φ ψ = − . cos θ When the nutation angle is small, θ → 0, cos θ → 1, and ψ ≅ −2φ .

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Problem 20.68 The rocket is in moment-free steady precession with nutation angle θ = 40 ° and spin rate φ = 4 revolutions per second. Its moments of inertia are I xx = 10,000 kg-m 2 and I zz = 2000 kg-m 2 . What is the rocket’s precession rate ψ in revolutions per second?

Moment-free steady precession is described by Eq. (20.33), ( I zz − I xx )ψ cos θ + I zzφ = 0, where ψ is the precession rate, φ is the spin rate, and θ is the nutation angle. Solve for the precession rate: I zzφ ψ = = 1.31 rev/s. ( I xx − I zz ) cos θ

Solution:

The angle θ = 40 °. The angle β defined by

 I   β = tan −1   zz  tan θ  = 9.53 °   I xx   satisfies the condition β < θ. The body cone with an axis along the z axis, rolls on a space cone with axis on the Z axis. The result is shown.

? f

Solution:

Problem 20.69 Sketch the body and space cones for the motion of the rocket in Problem 20.68.

u Z

? f

Y x

Space cone

f Body cone

Y x

c X

Problem 20.70 The top is in steady precession with nutation angle θ = 15 ° and precession rate ψ = 1 revolution per second. The mass of the top is 8 × 10 −4 slug, its center of mass is 1 in from the point, and its moments of inertia are I xx = 6 × 10 −6 slug-ft 2 and I zz = 2 × 10 −6 slug-ft 2 . What is the spin rate φ of the top in revolutions per second?

Z z

? f

Solution:

The steady precession is not moment-free, since the weight of the top exerts a moment mg Mx = sin θ. 12 The motion of a spinning top is described by Eq. (20.32), , mgh = ( I − I ) ψ 2 cos θ + I ψφ

( ) zz

c x

where ψ is the rate of precession, φ is the spin rate, and θ is the nutation angle and h =

( 121 )ft

is the distance from the point to the center of mass. Solve: mg − ( I zz − I xx ) ψ 2 cos θ φ = 12 . I zz ψ Substitute numerical values (using ψ = 2π rad/s for dimensional con-

z u

( )

sistency) to obtain φ = 182.8 rad/s, from which φ = 29.1 rev/s.

572

1 ft 12 1 sin u ft 12

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Problem 20.71 Suppose that top described in Problem 20.70 has a spin rate φ = 15 revolutions per second. Draw a graph of the precession rate (in revolutions per second) as a function of the nutation angle θ for values of θ from zero to 45 °.

10 9 8 7 6 c, 5 rev/s 4 3 2 1 0

Z z

? f

Y x

15 20 25 30 Nutation u, deg

− mgh = 0. center of mass. Rearrange: ( I zz − I xx )ψ 2 cos θ + I zz ψφ The velocity of the center of the base is v = −

c X

Precession rate vs Nutation angle

ω 0h , 4

from which the spin axis is the z axis and the spin rate is

, mgh = ( I xx − I yy )ψ 2 cos θ + I xx ψφ

ω h v = − 0 . φ = R 4R The solution, ψ = −b ±

where ψ is the rate of precession, φ is the spin rate, and θ is the nutation angle and h = (1/12) ft is the distance from the point to the

The two solutions, which are real over the interval, are graphed as a function of θ over the range 0 ≤ θ ≤ 45 °. The graph is shown.

Solution:

The behavior of the top is described in Eq. (20.32),

Problem 20.72 The rotor of a tumbling gyroscope can be modeled as being in moment-free steady precession. The moments of inertia of the gyroscope are I xx = I yy = 0.04 kg-m 2 and I zz = 0.18 kg-m 2 . The gyroscope’s spin rate is φ = 1500 rpm and its nutation angle is θ = 20 °. (a) What is the precession rate of the gyroscope in rpm? (b) Sketch the body and space cones. Solution: (a) The motion in moment-free, steady precession is described by Eq. (20.33), ( I zz − I xx )ψ cos θ + I zzφ = 0, where ψ is the precession rate, φ = 1500 rpm is the spin rate, and θ = 20 ° is the nutation angle. Solve: ψ = −

1,2

b 2 − c.

Space cone

Body cone

I zzφ = −2050 rpm. ( I zz − I xx ) cos θ

(b) The apex angle for the body cone is given by I  tan β =  zz  tan θ,  I xx  from which β = 58.6 °. Since β > θ, the space cone lies inside the body cone as the figure.

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Problem 20.73 A satellite can be modeled as an 800-kg cylinder 4 m in length and 2 m in diameter. If the nutation angle is θ = 20 ° and the spin rate φ is one revolution per second, what is the satellite’s precession rate ψ in revolutions per second?

Z z

? f

Solution:

From Appendix C, the moments and products of inertia of a homogenous cylinder are

(

)

L2 R2 I xx = I yy = m + = 1267 kg-m 2 , 12 4 I zz = mR 2 2 = 400 kg-m 2 . I xy = I xz = I yz = 0.

c X

The angular motion of an axisymmetric moment-free object in steady precession is described by Eq. (20.33), ( I zz − I xx )ψ cos θ + I zzφ = 0, where ψ is the precession rate, θ = 20 ° is the nutation angle, and φ = 1 rps is the spin rate. Solve: ψ = −

I zzφ = 0.49 rps. ( I zz − I xx ) cos θ

Problem 20.74* The top consists of a thin disk bonded to a slender bar. The radius of the disk is 30 mm and its mass is 0.008 kg. The length of the bar is 80 mm and its mass is negligible compared to the disk. When the top is in steady precession with a nutation angle of 10 °, the precession rate is observed to be 2 revolutions per second in the same direction the top is spinning. What is the top’s spin rate? 108

Solution:

The inertias about the base are

1 (0.008 kg)(0.03 m) 2 = 3.6 × 10 −6 kg-m 2 2 1 I xx = I yy = (0.008 kg)(0.03 m) 2 2 + (0.008 kg)(0.08 m) 2 = 53 × 10 −6 kg-m 2 I zz =

The precession rate is ψ = 2(2π ) = 12.6 rad/s The moment about the base is M = M x i = (0.008 kg)(9.81 m/s 2 )(0.08 m) i = (6.28 × 10 −3 N-m) i Eq. 20.32 is M x = ( I zz − I xx )ψ 2 cos10 ° + I zz ψφ Solving we find φ = 309 rad/s (49.1 rev/s).

z 108

30 mm

h = 80 mm y

c x

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Problem 20.75 Solve Problem 20.58 by treating the motion as steady precession.

Solution:

The view of an airplane’s landing gear looking from behind the airplane is shown in Fig. (a). The radius of the wheel is 300 mm and its moment of inertia is 2 kg-m 2 . The airplane takes off at 30 m/s. After takeoff, the landing gear retracts by rotating toward the right side of the airplane as shown in Fig. (b). The magnitude of the couple exerted by the wheel on its support is to be determined.

45 deg/s z

Choose X , Y , Z with the Z axis parallel to the runway, X perpendicular to the runway, and Y parallel to the runway. Choose the x, y, z coordinate system with the origin at the center of mass of the wheel and the z axis aligned with the direction of the axis of rotation of the wheel and the y axis positive upward. The Eq. (20.29) is

angle between the X and x, which is increasing in value, from which ψ = 45 °/s = 0.7853 rad/s. The spin vector is aligned with the z axis, from which φ =

sin θ. ∑ M x = (I zz − I xx )ψ 2 sin θ cos θ + I zzφψ

30 ( vR ) = ( 0.3 ) = 100 rad/s

The moments and products of inertia of the wheel are I zz = mR 2 /2 = sin 90 ° = 2(0.7854) (100) = 2 kg-m 2 . The moment is M x = I zz ψφ

The nutation angle and rates of precession: The nutation angle is the angle between Z and z, θ = 90 °. The precession angle is the

157 N-m.

Problem 20.76* The two thin disks are rigidly connected by a slender bar. The radius of the large disk is 200 mm and its mass is 4 kg. The radius of the small disk is 100 mm and its mass is 1 kg. The bar is 400 mm in length and its mass is negligible. The composite object undergoes a steady motion in which it spins about the vertical y-axis through its center of mass with angular velocity ω 0 . The bar is horizontal during this motion and the large disk rolls on the floor. Determine ω 0 by treating the motion as steady precession. Solution:

y v0

Use the data from 20.57

Set the nutation angle to 90 ° and the precession rate to ω 0 3.924 N-m = (0.085 kg-m 2 )(0.4 ω 0 )ω 0 Solving we obtain ω 0 = 10.7 rad/s.

Problem 20.77* Suppose that you are testing a car and use accelerometers and gyroscopes to measure its Euler angles and their derivatives relative to a reference coordinate system. At a particular instant, ψ = 15 °, θ = 4 °, φ = 15 °, the rates of change of the Euler angles are zero, and their second deriva = 0, θ = 1 rad/s 2 , tives with respect to time are ψ 2 and φ = −0.5 rad/s . The car’s principal moments of inertia, in kg-m 2 , are I xx = 2200, I yy = 480, and I zz = 2600. What are the components of the total moment about the car’s center of mass?

Solution:

The description of the motion of an arbitrarily shaped object is given by the Eqs. (20.36): cos θ sin φ M x = I xx (ψ sin θ sin φ + θ cos φ + ψθ + ψφ sin θ cos φ − θφ sin φ) − ( I yy − I zz )(ψ sin θ cos φ − θ sin φ)(ψ cos θ + φ ), cos θ cos φ M y = I yy (ψ sin θ cos φ − θ sin φ + ψθ sin θ sin φ − θφ cos φ) − ψφ −( I − I )(ψ sin θ sin φ + θ cos φ)(ψ cos θ + φ ), zz

sin θ) − ( I − I )(ψ sin θ sin φ M z = I zz (ψ cos θ + φ − ψθ xx yy + θ cos φ)(ψ sin θ cos φ − θ sin φ).

z Z

= 0, ψ = θ = φ = 0, to obtain Substitute ψ M y = −I yy θ sin φ, M z = I zzφ . Substitute values:

y x

M x = I xx θ sin φ,

M x = 2125 N-m, M y = −124.2, M z = −1300 N-m.

Y X

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Problem 20.78* If the Euler angles and their second derivatives for the car described in Problem 20.77 have the given values, but their rates of change are ψ = 0.2 rad/s, θ = −2 rad/s, and φ = 0, what are the components of the total moment about the car’s center of mass? Solution:

z Z

y x

Use Eqns. (20.36)

cos θ sin φ M x = I xx (ψ sin θ sin φ + θ cos φ + ψθ + ψφ sin θ cos φ − θφ sin φ) − ( I yy − I zz )(ψ sin θ cos φ − θ sin φ)(ψ cos θ + φ ), cos θ cos φ M y = I yy (ψ sin θ cos φ − θ sin φ + ψθ sin θ sin φ − θφ cos φ) − ψφ − ( I − I )(ψ sin θ sin φ + θ cos φ)(ψ cos θ + φ ), zz

Substitute values: M x = 2123 N-m, M y = −155.4 N-m,

M z = 534 N-m.

sin θ) − ( I − I )(ψ sin θ sin φ M z = I zz (ψ cos θ + φ − ψθ xx yy + θ cos φ)(ψ sin θ cos φ − θ sin φ)

Problem 20.79* Suppose that the Euler angles of the car described in Problem 20.77 are ψ = 40 °, θ = 20 °, and φ = 5 °, their rates of change are zero, and the components of the total moment about the car’s center of mass are ΣM x = −400 N-m, ΣM y = 200 N-m, and ΣM z = 0. What are the x, y, and z components of the car’s angular acceleration? Solution:

Eq. (20.36) simplifies when ψ = φ = θ = 0 to

z Z

y x

Y X

sin θ sin φ + θ cos φ), M x = I xx (ψ sin θ cos φ − θ sin φ), M = I (ψ y

cos θ + φ ). M z = I zz (ψ These three simultaneous equations have the solutions, My Mx cos φ − sin φ = −0.2174 rad/s 2 , θ = I xx I yy =  M xx  sin φ +  M y  cos φ = 1.167 rad/s 2 , ψ   I yy  sin θ I xx  sin θ cos θ = −1.097 rad/s 2 . φ = ( M z /I zz ) − ψ From Eq. (20.35), when ψ = φ = θ = 0: dω x sin θ sin φ + θ cos φ = −0.1818 rad/s 2 , = ψ dt dω y sin θ cos φ − θ sin φ = 0.417 rad/s, = ψ dt dω z cos φ + φ = 0. = ψ dt

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Problem 20.80 The mass of the bar is 6 kg. Determine the moments and products of inertia of the bar in terms of the coordinate system shown. y

x9 y9 x

O 1m

mv L2v + mv (0.5 2 ) = 0.6667 kg-m 2 . 12 (1) = I (1) + (d 2 + d 2 )m = 0. I yy x z v y ′y ′

(1) = I (1) + (d 2 + d 2 )m = I xx y z v x ′x ′

x 2m

The products of inertia are

Solution:

One strategy (not the simplest, see Check note following the solution) is to determine the moment of inertia matrix for each element of the bar, and then to use the parallel axis theorem to transfer each to the coordinate system shown. (a) The vertical element Oy of the bar. The mass density per unit vol6 ume is ρ = kg/m 3 , where A is the (unknown) cross section 3A of the bar, from which ρ A = 2 kg/m. The element of mass is dm = ρ Ad L, where dL is an element of length. The mass of the vertical element is m v = ρ AL v = 2 kg, where L v = 1 m. From Appendix C the moment of inertia about an x ′ axis passing through the center of mass is m L2 I x(1)′x ′ = v v = 0.1667 kg-m 2 . 12 Since the bar is slender, I y(1)′y ′

= 0.

(1) = I (1) + d d ρ A(1) = 0, I xz x z x ′z ′ (1) = I (1) + d d ρ A(1) = 0. I yz y z y ′z ′

The inertia matrix for the vertical element:  ρA  0  3  (1) [I ] =  0 0   0 0  

    0 .  ρA  3  0

For the horizontal element, the coordinates of the center of mass relative to O are (d x , d y , d z ) = (1, 0, 0) m. From the parallel axis theorem, (2) = I (2) + (d 2 + d 2 )m = 0. I xx y z h x ′x ′

2 2 2 I zz(2) = I z(2) ′z ′ + (d x + d y )m h = 5.333 kg-m .

Since the bar is slender, the products of inertia vanish: I x(1)′y ′ = ∫ x ′y ′ dm = 0, I x(1)′z ′ = ∫ x ′z ′ dm = 0, m

I y(1)′z ′ = ∫ y ′z ′ dm = 0, m

from which the inertia matrix for the element Oy about the x ′ axis is   0.1667 0 0    kg-m 2 . 0 0 0 [ I (1) ] =    0 0 0.1667   (b) The horizontal element Ox of the bar. The mass of the horizontal element is m h = ρ AL h = 4 kg, where L h = 2 m. From Appendix C the moments and products of inertia about the y ′ axis passing through the center of mass of the horizontal element are: I x(2) ′x ′ = 0, m h L2h = 1.333 kg-m 2 , 12 m h L2h = 1.333 kg-m 2 . I z(2) ′z ′ = 12 Since the bar is slender, the cross products of inertia about the y ′ axis through the center of mass of the horizontal element of the bar vanish: I y(2) ′y ′ =

(2) (2) I x(2) ′y ′ = 0, I x ′z ′ = 0, I y ′z ′ = 0.

The inertia matrix is   0 0 0    kg-m 2 . 0 [ I (2) ] =  0 1.333   0 1.333   0 Use the parallel axis theorem to transfer the moment of inertia matrix to the origin O: For the vertical element the coordinates of the center of mass O are (d x , d y , d z ) = (0, 0.5, 0) m. Use the parallel axis theorem (see Eq. (20.42)).

BandF_6e_ISM_C20_Pt2.indd 577

(1) = I (1) + d d ρ A(1) = 0, I xy x y x ′y ′

(2) = I (2) + (d 2 + d 2 )m = 5.333 kg-m 2 . I yy x z h y ′y ′

m L2 I z(1)′z ′ = v v = 0.1667 kg-m 2 . 12

I zz(1) = I z(1)′z ′ + (d x2 + d y2 )mv = 0.6667 kg-m 2 .

By inspection, the products of inertia vanish. The inertia matrix is   0 0 0   . [ I (2) ] =  0 5.333 0   0 5.333   0 Sum the two inertia matrices:  0.6667 0 0    [ I ]O = [ I (1) ] + [ I (2) ] =  0 5.333 0  kg-m 2 .   0 0 6   [Check: The moment of inertia in the coordinate system shown can be derived by inspection by taking the moment of inertia of each element about the origin: From Appendix C the moments of inertia about the origin of the slender bars are m L3 m L2 I xx = v v , I yy = h h , 3 3 and I zz = I xx + I yy , where the subscripts v and h denote the vertical and horizontal bars respectively. Noting that the masses are mL h mL v mv = , mh = , Lv + L h Lv + L h the moment of inertia matrix becomes:   6(1) 3   0   3(3) 0    3  6(2)  [ I ] =  0 0 3(3)      0 0 6.0      0.6667 0 0    =  0 5.333 0  kg-m 2 . check.   0 0 6  

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Problem 20.81 The object consists of two 1-kg vertical slender bars welded to a 4-kg horizontal slender bar. Determine its moments and products of inertia in terms of the coordinate system shown.

0.1 m

Solution:

1 I xx = 2 (1 kg)(0.1 m) 2 = 0.00667 kg-m 2 , 3 1 I yy = (4 kg)(0.4 m) 2 + 2(1 kg)(0.1 m) 2 = 0.0733 kg-m 2 , 12 I zz = I yy + I zz = 0.08 kg-m 2 ,

0.1 m

I xy = 2(1 kg)(0.1 m)(0.05 m) = 0.01 kg-m 2 ,

0.1 m

0.2 m

I xz = 0, I yz = 0. I xx = 0.00667 kg-m 2 , I yy = 0.0733 kg-m 2 , I zz = 0.08 kg-m 2 , I xy = 0.01 kg-m 2 ,

I xz = 0,

I yz = 0.

Problem 20.82 The 4-kg thin rectangular plate lies in the x – y plane. Determine the moments and products of inertia of the plate in terms of the coordinate system shown. (See Example 20.10.) z

Solution:

From Appendix B, the moments of inertia of the plate’s

area are 1 (0.3)(0.6) 3 = 0.00540 m 4 , 12 1 Iy = (0.6)(0.3) 3 = 0.00135 m 4 , 12 I xyA = 0. Ix =

Therefore the moments of inertia of the plate are y

300 mm 600 mm

m 4 I = (0.00540) A x (0.3)(0.6) = 0.12 kg-m 2 , 4 m (0.00135) I yy = I y = (0.3)(0.6) A 2 = 0.03 kg-m , I xx =

I zz = I x + I y = 0.15 kg-m 2 ,

I xy = I yz = I zx = 0.

Problem 20.83 If the 4-kg plate is rotating with angular velocity ω = 6 i + 4 j − 2k (rad/s), what is its angular momentum about its center of mass? (See Example 20.10.) Solution:

Angular momentum is

 0.12 0 0  6     0.03 0  4  [H ] =  0    0 0.15   −2   0  0.72    =  0.12  kg-m 2 /s,    −0.3 

300 mm 600 mm x

from which H = 0.72 i + 0.12 j − 0.3k (kg-m 2 /s).

578

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Problem 20.84 The 30-lb triangular plate lies in the x – y plane. Determine the moments and products of inertia of the plate in terms of the coordinate system shown. z 4 ft

x 6 ft

Solution:

From Appendix B, the moments of inertia of the plate’s

area are 1 (6)(4) 3 = 32 ft 4 , 12 1 I y = (4)(6) 3 = 216 ft 4 , 4 1 2 2 A I xy = (6) (4) = 72 ft 4 . 8 Ix =

The plate’s area and mass are A =

1 (6)(4) = 12 ft 2 2

and m = 30/32.2 = 0.932 slug. The plate’s moments and products of inertia are m I = 2.48 slug-ft 2 , A x m I yy = I y = 16.77 slug-ft 2 , A m I xy = I xyA = 5.59 slug-ft 2 , A I yz = I zx = O, I xx =

I zz = I xx + I yy = 19.25 slug-ft 2 .

Problem 20.85 The 30-lb triangular plate lies in the x – y plane. (a) Determine its moments and products of inertia in terms of a parallel coordinate system x ′ y ′ z ′ with its origin at the plate’s center of mass. (b) If the plate is rotating with angular velocity ω = 20 i − 12 j + 16k (rad/s), what is its angular momentum about its center of mass? Solution:

z 4 ft

x 6 ft

See the solution of Problem 20.84

(a) The coordinates of the plate’s center of mass are 2 (6) = 4 ft, 3 1 d y = (4) = 1.33 ft, 3 d z = 0. dx =

y y9

From the parallel-axis theorems (Eq. 20.42), we obtain I x′x′ = 2.48 − (1.33) 2 (0.932) = 0.828 slug-ft 2 , I y′y′ = 16.77 − (4) 2 (0.932) = 1.863 slug-ft 2 ,

x9 x

I z ′z ′ = 19.25 − [(1.33) 2 + (4) 2 ](0.932) = 2.692 slug-ft 2 , I x′y′ = 5.59 − (1.33)(4)(0.932) = 0.621 slug-ft 2 , I y′z ′ = I z ′x′ = 0.   20   0.828 −0.621 0      −12  0 (b) [I ]ω =  −0.621 1.863    0 0 2.692   16    24.0    =  −34.8     43.1  H = 24.0 i − 34.8 j + 43.1k (slug-ft 2 /s).

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Problem 20.86 Determine the inertia matrix of the 2.4-kg steel plate in terms of the coordinate system shown. Solution:

220 mm

Equation (20.39) gives the plate’s moments and products of inertia in terms of the moments and product of inertia of its area. Treating the area as a quarter-circle using Appendix B, the moments and products of inertia of the area are 150 mm 1 1 Ix = π (0.22) 4 − (0.05)(0.15) 3 = 0.000404 m 4 16 3 1 1 Iy = π (0.22) 4 − (0.05)(0.05) 3 = 0.000454 m 4 16 3 1 1 I xyA = (0.22) 4 − (0.05) 2 (0.15) 2 = 0.000279 m 4 . 8 4

The moments of inertia of the plate are

The area is

I xx =

A =

1 π (0.22) 2 − (0.05)(0.15) = 0.0305 m 2 . 4

x 50 mm

m I = 0.0318 kg-m 2 , A x m I yy = I y = 0.0357 kg-m 2 , A I zz = I xx + I yy = 0.0674 kg-m 2 , I xy =

m A I = 0.0219 kg-m 2 , and I yz = I zx = 0. A xy

Problem 20.87 The mass of the steel plate is 2.4 kg. (a) Determine its moments and products of inertia in terms of a parallel coordinate system x ′ y ′ z ′ with its origin at the plate’s center of mass. (b) If the plate is rotating with angular velocity ω = 20 i + 10 j − 10 k (rad/s), what is its angular momentum about its center of mass?

220 mm

150 mm

Solution: (a) The x and y coordinates of the center of mass coincide with the centroid of the area: 1 π (0.22) 2 = 0.0380 m 2 , 4 A2 = (0.05)(0.15) = 0.0075 m 2 4(0.22) A1 − (0.025) A2 3π x = = 0.1102 m, A1 − A2 4(0.22) A1 − (0.075) A2 3π y = = 0.0979 m. A1 − A2 A1 =

x 50 mm

Using the results of the solution of Problem 20.86 and the parallel axis theorems, I x ′x ′ = I xx − my 2 = 0.00876 kg-m 2 : I y ′y ′ = I yy − mx 2 = 0.00655 kg-m 2 : I z ′z ′ = I x ′x ′ + I y ′y ′ = 0.01531 kg-m 2 I x ′y ′ = I xy − mxy = −0.00396 kg-m 2 , and I y ′z ′ = I z ′x ′ = 0. The angular momentum is  H x′    H y′   H z′ 

 I  −I x′y′ x ′x ′     = −  I x′y′ I y′y′      0 0  

0 0 I z ′z ′

   20      10    −10     

 0.215    =  0.145  (kg-m 2 /s).    −0.153 

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Problem 20.88 The slender bar of mass m rotates about the fixed point O with angular velocity ω = ω y j + ω z k. Determine its angular momentum (a) about its center of mass and (b) about O.

Alternatively, in terms of the i, j, k , HG =

mL2 (ω y j + ω z k). 12

(b) From Appendix C and by inspection, the moments and products of inertia about O are

I xx = 0, mL2 , 3 I xy = I xz = I yz = 0 I yy = I zz =

O z

The angular momentum about O is

l x

Solution: (a) From Appendix C and by inspection, the moments and products of inertia about the center of mass of the bar are: I xx = 0, mL2 , 12 I xy = I xz = I yz = 0. I yy = I zz =

 I −I xy −I xz   0  xx     [ H ]O =  −I yx I yy −I yz   ω y      −I zx −I zy I zz   ω z          0 0 0   0         0    mL2 ω    mL2  y   =  0 0   ωy  =  , 3     3     ωz  2   2 mL     0  mL ω z  0    3  3    or, alternatively, in terms of the unit vectors i, j, k,

The angular momentum about its center of mass is  I −I xy −I xz   0 xx    I yy −I yz   ω y [ H ] G =  −I yx    −I zx −I zy I zz   ω z        0  0 0    0  2 mL    =  0 0   ωy  12       mL2   ω z   0 0   12  

HO =

      

Problem 20.89 The slender bar of mass m is parallel to the x-axis. If the coordinate system is body fixed and its angular velocity about the fixed point O is ω = ω y j, what is the bar’s angular momentum about O? y

mL2 (ω y j + ω z k). 3

Solution:

From Appendix C and by inspection, the moments and products of inertia about the center of mass of the bar are: I x ′x ′ = 0, mL2 , 12 I x ′y ′ = I x ′z ′ = I y ′z ′ = 0. I y ′y ′ = I z ′z ′ =

L h

O x

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20.89 (Continued) The coordinates of the center of mass are (d x , d y , d z ) =

(

The angular momentum about O is

)

L , h, 0 . 2

From Eq. (20.42), I xx = I x ′x ′ + (d y2 + d z2 )m = mh 2 mL2 mL2 mL2 + = 12 4 3 L2 I zz = I z ′z ′ + (d x2 + d y2 )m = m h 2 + , 3 mLh I xy = I x ′y ′ + d x d y m = 0 + , 2 I xz = I x ′z ′ + d x d z m = 0 I yy = I y ′y ′ + (d x2 + d z2 )m =

(

)

I yz = I y ′z ′ + d y d z m = 0

 I −I xy −I xz   0  xx     [ H ]O =  −I yx I yy −I yz   ω y     −I zx −I zy I zz   0     mLh 2  mh − 0  2  2  mLh mL =  − 0 2 3   L2  m h2 + 0 0  3    mLh ω y   −   2    mL2 ω  y . =   3       0    

(

)

    0    ωy    0   

     

Alternatively, HO = −

Problem 20.90 In Example 20.9, the moments and products of inertia of the object consisting of the booms AB and BC were determined in terms of the xyz coordinate system shown in the example. Determine the moments and products of inertia of the object in terms of a parallel coordinate system x ′ y ′ z ′ with its origin at the center of mass of the object. Solution:

From Example 20.8, the inertia matrix for the two booms in the x, y, z system is

 I −I xy −I xz  xx    I yy −I yz  [ I ]O =  −I yx    −I zx −I zy I zz      19, 200 86, 400 0    kg-m 2 . =  86, 400 1, 036,800 0    0 0 1, 056, 000  

mLh mL2 ω yi + ω y j. 2 3

I z(′Gz ′) = I zz( o ) − (d x2 + d y2 )m = 242,400 kg-m 2 , (0) − d d m = − 32,400 kg-m 2 , I x(G′y′) = I xy x y

I x(G′z )′ = 0 , I y(G′z )′ = 0. The inertia matrix for the x ′, y ′, z ′ system is   15,600 32,400 0    kg-m 2 . 0 [ I ] G =  32,400 226,800   0 0 242,400  

The mass of the boom AB is m AB = 4800 kg. The mass of the boom BC is m BC = 1600 kg. The coordinates of the center of mass of the two booms are

6m B

z ′ = 0, from which (d x , d y , d z ) = (11.25, − 0.75, 0) m. Algebrai cally rearrange Eq. (20.42) to obtain the moments and products of inertia about the parallel axis passing through the center of mass of the two booms when the moments and products of inertia in the x, y, z system are known:

(O ) − (d 2 + d 2 )m = 226,800 kg-m 2 . I y(G′y)′ = I yy x z

582

18 m

508

L + m BC L 2 = 11.25 m. x′ = m AB + m BC m (0) + m BC (−3) = −0.75 m. y ′ = AB m AB + m BC m AB

(0) − (d 2 + d 2 )m = 15,600 kg-m 2 . I x(G′x)′ = I xx y x

B 18 m

6m C

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Problem 20.91 A 4-kg slender bar is rigidly attached to a 6-kg thin circular disk. Determine the moments of products of inertia of the combined object in terms of the coordinate system shown.

600 mm

Solution: y

R x, x9

 I −I xy −I xz  xx  −I yz  −I yx I yy   −I zx −I zy I zz 

  0.06 kg-m 2    0  =     0  

0.78 kg-m 2

0.84 kg-m 2

  .   

I xy = I xz = I yz = 0.

Problem 20.92 A 4-kg slender bar is rigidly attached to a 6-kg thin circular disk. In terms of the body-fixed coordinate system shown, the angular velocity of the composite object relative to an Earth-fixed reference frame is ω = 12 i + 6 j − 4 k (rad/s). What is the object’s angular momentum about its center of mass? y

200 mm x 600 mm

1 L + R − d B = 0.2 m. 2 For the bar, applying the parallel-axis theorems, 1 I xx = 0, I yy = I zz = m L2 + d B 2 m B = 0.48 kg-m 2 , 12 B and the products of inertia are zero. For the disk, so the distance d D =

1 m R 2 = 0.06 kg-m 2 , 4 D 1 I yy = m D R 2 + d D 2 m D = 0.30 kg-m 2 , 4 1 I zz = m D R 2 + d D 2 m D = 0.36 kg-m 2 , 2 I xx =

and the products of inertia are zero. Summing the results for the bar and disk, the inertia matrix for the object is  I −I xy −I xz  xx  I I yy − −I yz  yx   −I zx −I zy I zz 

Solution: y

  0.06 kg-m 2     0 =      0  

0.78 kg-m 2

0.84 kg-m 2

  .   

The angular momentum is

R x, x9

L Let L = 0.6 m, R = 0.2 m, m B = 4 kg, m D = 6 kg. In terms of the x '-y ' coordinate system, the distance to the common center of mass is

( 12 L + R )m = 0.3 m,

 H x    Hy   Hz 

 I  −I xy −I xz   ω x  xx        =  −I yx I yy −I yz   ω y       −I zx −I zy I zz   ω z         0.06 kg-m 2 0 0  =  0 0.78 kg-m 2 0  2 0 0 0.84 kg-m   0.72 kg-m 2 /s    =  4.68 kg-m 2 /s  .  −3.36 kg-m 2 /s   

  12 rad/s     6 rad/s     −4 rad/s    

mB + mD

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mB + mD

I xx = 0.06 kg-m 2 , I yy = 0.78 kg-m 2 , I zz = 0.84 kg-m 2 ,

(0)m B +

( 12 L + R )m = 0.3 m,

and the products of inertia are zero. For the disk, 1 I xx = m D R 2 = 0.06 kg-m 2 , 4 1 I yy = m D R 2 + d D 2 m D = 0.30 kg-m 2 , 4 1 I zz = m D R 2 + d D 2 m D = 0.36 kg-m 2 , 2 and the products of inertia are zero. Summing the results for the bar and disk, the inertia matrix for the object is

dB =

(0)m B +

so the distance d D =

200 mm

dB =

1 L + R − d B = 0.2 m. 2 For the bar, applying the parallel-axis theorems, 1 I xx = 0, I yy = I zz = m L2 + d B 2 m B = 0.48 kg-m 2 , 12 B

Let L = 0.6 m, R = 0.2 m, m B = 4 kg, m D = 6 kg. In terms of the x '-y ' coordinate system, the distance to the common center of mass is

H = 0.72 i + 4.68 j − 3.36k (kg-m 2 /s).

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Problem 20.93* The mass of the homogeneous slender bar is m. If the bar rotates with angular velocity ω = ω 0 (24 i + 12 j − 6k), what is its angular momentum about its center of mass?

y b x

Solution:

The strategy is to transfer the moments of inertia of the ends about their attached ends to the center of mass, and sum the resulting moments and products of inertia. The mass of the central element m m is mC = , and the mass of each end element is m E = . For the 2 4 central element about its center of mass:

b b

I xx = 0, mC (2b) 2 mb 2 = , 12 6 I xy = I xz = I yz = 0.

I yy = I zz =

For each end element about its center of mass: mEb 2 mb 2 = , 12 48 I yy = 0, I xx =

The sum of the matrices:

mEb 2 mb 2 = , 12 48 I xy = I xz = I yz = 0. I zz =

[ I ] G = [ I ] G C + 2[ I ] end

The coordinates of the center of mass (at the origin) relative to the cenb ter of mass of the left end is (d x , d y , d z ) = −b, − , 0 , and from 2 b the right end (d x , d y , d z ) = b, , 0 . From Eq. (20.42), the moments 2 of inertia of each the end pieces about the center of mass (at the origin) are

(

( L ) + (d 2 + d 2 )m I x(G′x)′ = I xx y z E =

)

(

)

m b2 mb 2 mb 2 + E = . 48 4 12

mb 2 . 4 b2 mb 2 mb 2 + mE b 2 + = I z(G′z ′) = I zz( L ) + (d x2 + d y2 )m E = 48 4 3 (G ) (G ) (L) + d d m 2 I x(G′y)′ = I xy x y E = mb /8, I x ′z ′ = 0, I y ′z ′ = 0. ( L ) + (d 2 + d 2 )m I y(G′y)′ I yy x z E =

(

)

  0.1667 −0.25 0   . = mb 2  −0.25 0.6667 0   0 0 0.8333   The angular momentum about the center of mass is   24   0.1667 −0.25 0      12  [ H ] G = ω 0 mb 2  −0.25 0.6667 0    0 0 0.8333   −6    1    = ω 0 mb 2  2  .    −5  H = ω 0 mb 2 ( i + 2 j − 5k).

Problem 20.94* The 8-kg homogeneous slender bar has ball-and-socket supports at A and B. (a) What is the bar’s moment of inertia about the axis AB? (b) If the bar rotates about the axis AB at 4 rad/s, what is the magnitude of its angular momentum about its axis of rotation?

Solution:

Divide the bar into three elements: the central element, and the two end element. The strategy is to find the moments and products of inertia in the x, y, z system shown, and then to use Eq. (20.43) to find the moment of inertia about the axis AB . Denote the total mass of the bar by m = 8 kg, the mass of each end elem ment by m E = = 2 kg, and the mass of the central element by 4 m mC = = 4 kg. 2 The left end element: The moments and products of inertia about point A are: m E (1 2 ) m = = 0.6667 kg-m 2 , 3 12 m E (1 2 ) m ( LA ) = = = 0.6667 kg-m 2 , I yy 3 12 I zz( LA) = 0, ( LA ) = I xx

( LA ) = I ( LA ) = I ( LA ) = 0. I xy xz yz

584

A 2m 1m z The right end element: The moments and products of inertia about its center of mass are m E (1 2 ) m ( RG ) = I xx = = 0.1667 kg-m 2 , 12 48 ( RG ) = 0, I yy m E (12 ) m = = 0.1667 kg-m 2 , 12 48 ( RG ) = I ( RG ) = I ( RG ) = 0. I xx xz yz I zz( RG ) =

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20.94 (Continued) The coordinates of the center of mass of the right end element are (d x , d y , d z ) = (2, 0.5, 1). From Eq. (20.42), the moments and products of inertia in the x, y, z system are

The coordinates of the center of mass of the central element are (d x , d y , d z ) = (1, 0, 1). From Eq. (20.42) the moments and products of inertia in the x, y, z system are:

m = 2.667 kg-m 2 , 4 ( RA ) = I ( RG ) + (d 2 + d 2 ) m = 10.0 kg-m 2 I yy yy x z 4 m ( ) ( ) 2 2 RA RG I zz = I zz + (d x + d y ) = 8.667 kg-m 2 , 4 m ( ) ( ) RA RG I xy = I xy + d x d y = 2.0 kg-m 2 4 ( RA ) = I ( RG ) + d d m = 4.0 kg-m 2 , I xz xz x z 4 ( RA ) = I ( RG ) + d d m = 1.0 kg-m 2 I yz yz y z 4

m = 4 kg-m 2 , 2 (CA ) = I (CG ) + (d 2 + d 2 ) m = 9.333 kg-m 2 I yy yy x z 2 m ( ) ( ) 2 2 CA CG I zz = I zz + (d x + d y ) = 5.333 kg-m 2 , 2 m ( ) ( ) CA CG I xy = I xy + d x d y = 0, 2 m (CA ) = I (CG ) + d d I xz = 4 kg-m 2 , xz x z 2 (CA ) = I (CG ) + d d m = 0 I yz yz y z 2 (CA ) = I (CG ) + (d 2 + d 2 ) I xx xx y z

( RA ) = I ( RG ) + (d 2 + d 2 ) I xx xx y z

Sum the inertia matrices:

Sum the two inertia matrices:  3.333 −2 −4    [ I ] RLA = [ I ] RA + [ I ] LA =  −2 10.67 −1  kg-m 2 ,   −1 8.667   −4 where the negative signs are a consequence of the definition of the inertia matrix. The central element: The moments and products of inertia of the central element about its center of mass are:

From Eq. (20.43), the moment of inertia about the AB axis is

(2 2 )

mC m = = 1.333 kg-m 2 , 12 6 m (2 2 ) m = = 1.333 kg-m 2 . I zz(CG ) = C 12 6 (CG ) = I (CG ) = I (CG ) = 0. I xy xz yz (CG ) = 0, I (CG ) = I xx yy

 7.333 −2 −8    [ I ] A = [ I ] RLA + [ I ]CA =  −2 20.00 −1  kg-m 2 .   −1 14.00   −8 (a) The moment of inertia about the axis AB: The distance AB is parallel to the vector r AB = 2 i + 1 j + 1k(m). The unit vector parallel to r AB is r e AB = AB = 0.8165i + 0.4082 j + 4082k. r AB ( A) e 2 + I ( A) e 2 + I ( A) e 2 − 2 I ( A) e e − 2 I ( A) e e I AB = I xx x yy y zz z xy x y xz x z ( A) e e , I 2 − 2 I yz y z AB = 3.56 kg − m .

(b) The angular momentum about the AB axis is H AB = I AB ω = 3.56(4) = 14.22 kg-m 2 /s.

Problem 20.95* The 8-kg homogeneous slender bar is released from rest in the position shown. (The x – z plane is horizontal.) What is the magnitude of the bar’s angular acceleration about the axis AB at the instant of release? B

Solution: 1 m 1 yG = m 1 zG = m xG =

The center of mass of the bar has the coordinates:

(( m4 )0 + ( m2 )(1) + ( m4 )(2) ) = 1 m. (( m4 )(0) + ( m2 )(0) + ( m4 )(0.5) ) = 0.125 m. (( m4 )(0.5) + ( m2 )(1) + ( m4 )(1) ) = 0.875 m.

The line from A to the center of mass is parallel to the vector r AG = i + 0.125 j + 0.875k(m). From the solution to Problem 20.94 the unit vector parallel to the line AB is e AB = 0.8165i + 0.4082 j + 0.4082k. The magnitude of the moment about line AB due to the weight is 0.8165 0.4082 0.4082 1.000 0.125 0.875 0 −78.48 0

e[r AB × (−mg j)] =

= 24.03 N-m.

From the solution to Problem 20.94, I AB = 3.556 kg-m 2 . From the equation of angular motion about axis AB, M AB = I ABα, from which

A 2m 1m

α =

24.03 = 6.75 rad/s 2 . 3.556 B

y x A

G rAG –mgj z

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Problem 20.96 In terms of a coordinate system x ′ y ′ z ′ with its origin at the center of mass, the inertia matrix of a rigid body is  20 10 −10    [ I ′ ]=  10 60 0  kg-m 2 .   80   −10 0

The function f ( I ) = AI 3 + BI 2 + CI + D is graphed to find the zero crossings, and these values are refined by iteration. The graph is shown. The principal moments of inertia are: I 1 = 16.15 kg-m 2 , I 2 = 62.10 kg-m 2 ,

Determine the principal moments of inertia and unit vectors parallel to the corresponding principal axes. (See Example 20.11.)

40000

Solution:

20000

I 3 = 81.75 kg-m 2 .

Roots of Cubic 30000

Principal Moments of Inertia: The moments and

products:

10000

I x ′x ′ = 20 kg-m 2 ,

f (I)

I y ′y ′ = 60 kg-m 2 ,

0 –10000

I z ′z ′ = 80 kg-m 2 , I x ′y ′ = −10 kg-m 2 ,

–20000

I x ′z ′ = 10 kg-m 2 , I y ′z ′ = 0,

–30000 –40000

From Eq. (20.45), the principal values are the roots of the cubic equation. AI 3 + BI 2 + CI + D = 0, where

100

A = +1, B = −( I x ′x ′ + I y ′y ′ + I z ′z ′ ) = −160, C = ( I x ′x ′ I y ′y ′ + I y ′y ′ I z ′z ′ + I z ′z ′ I x ′x ′ − I x2′y ′ − I x2′z ′ − I y2′z ′ ) = 7400, D = − ( I x ′x ′ I y ′y ′ I z ′z ′ − I x ′x ′ I y2′z ′ − I y ′y ′ I x2′z ′ − I z ′z ′ I x2′y ′ − 2 I x ′y ′ I y ′z ′ I x ′z ′ ) = −82000.

Problem 20.97 For the object in Problem 20.81, determine the principal moments of inertia and unit vectors parallel to the corresponding principal axes. Draw a sketch of the object showing the principal axes. Solution:

In Problem 20.81, we found the inertia matrix to be

 0.00667 −0.01 0    0  kg-m 2 [ I ] =  −0.01 0.0733   0 0 0.08   To find the principal inertia we expand the determinant to produce the characteristic equation det

0.00667 − I −0.01 0

−0.01 0.0733 − I 0

0 0 0.08 − I

= (0.08 − I )[(0.00667 − I )(0.0733 − I ) − (0.01) 2 ] = 0 Solving this equation, we have I 1 = 0.08 kg-m 2 , I 2 = 0.0748 kg-m 2 , I 3 = 0.0052 kg-m 2 . Substituting these principal inertias into Eqs. (20.46) and dividing the resulting vector V by its magnitude, we find a unit vector parallel to the corresponding principal value. e 1 = 0.989 i + 0.145 j, e 2 = −0.145i + 0.989 j, e 3 = k. The principal axes are shown on the figure at the top of the page with e 1 pointing out of the paper. The axes are rotated counterclockwise through the angle θ = cos −1 (0.989) = 8.51 °.

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Problem 20.98 The 1-kg, 1-m long slender bar lies in the x ′ – y ′ plane. Its inertia matrix in kg-m 2 is 1 2  − 121 sin β cos β 12 sin β   1 1 2 [ I ′ ] =  − 12 sin β cos β 12 cos β   0 0 

0   0 .  1  12 

b x9

Use Eqs. (20.45) and (20.46) to determine the principal moments of inertia and unit vectors parallel to the corresponding principal axes. Solution:

[Preliminary Discussion: The moment of inertia about an axis coinciding with the slender rod is zero; it follows that one principal value will be zero, and the associated principal axis will coincide with the slender bar. Since the moments of inertia about the axes normal to the slender bar will be equal, there will be two equal principal values, and Eq. (20.46) will fail to yield unique solutions for the associated characteristic vectors. However the problem can be solved by inspection: the unit vector parallel to the axis of the slender rod will be e 1 = i cos β + j sin β . A unit vector orthogonal to e 1 is e 2 = −i sin β + j cos β . A third unit vector orthogonal to these two is e 3 = k. The solution based on Eq. (20.46) must agree with these preliminary results.] Principal Moments of Inertia: The moments and products: sin 2 β cos 2 β 1 I x ′x ′ = , I y ′y ′ = , I z ′z ′ = . 12 12 12 sin β cos β I x ′y ′ = + , I x ′z ′ = 0, I y ′z ′ = 0. 12

A = +1,

( sin12 β + cos12 β + 121 ) = − 16 , 2

I 1 = 0, V (1) =

cos β sin β cos β cos 2 β i+ j = (cos β i + sin β j). 144 144 144

The magnitude: V (1) =

cos β 144

cos 2 β + sin 2 β =

cos β , 144

and the unit vector is e 1 = sgn( cos β )(cos β i + sin β j), where sgn( cos β ) is equal to the sign of cos β . Without loss of generality, β can be restricted to lie in the first or fourth quadrants, from which e 1 = (cos β i + sin β j).

From Eq. (20.45), the principal values are the roots of the cubic, AI 3 + BI 2 + CI + D = 0. The coefficients are:

B = − ( I x ′x ′ + I y ′y ′ + I z ′z ′ ) = −

For the first root,

For I 2 = I 3 = ( 121 ), V (2) = 0, and V (3) = 0, from which the equation Eq. (20.46) fails for the repeated principal values, and the characteristic vectors are to determined from the condition of orthogonality with e 1 . From the preliminary discussion, e 2 = −i sin β + j cos β and e 3 = 1k.

C = ( I x ′x ′ I yy ′ + I y ′y ′ I z ′z ′ + I z ′z ′ I x ′x ′ − I x2′y ′ − I x2′z ′ − I y2′z ′ ), sin 2 β cos 2 β cos 2 β sin 2 β sin 2 β cos 2 β 1 + + − = . 144 144 144 144 144 2 2 2 D = − ( I x ′x ′ I y ′y ′ I z ′z ′ − I x ′x ′ I y ′z ′ − I y ′y ′ I x ′z ′ − I z ′z ′ I x ′y − 2 I x ′y ′ I y ′z ′ I x ′z ′ ), C =

β cos β sin β cos β − ( sin 12 ) = 0. 12

D = −

The cubic equation reduces to 1 2 1 1 1 I3 − I + I = I2 − I + I = 0. 6 144 6 144 By inspection, the least root is I 1 = 0. The other two roots

( )

(

)

(

( )

(

))

are the solution of the quadratic I 2 + 2bI + c = 0 where 1 1 1 b = − ,c = , from which I 1,2 = −b ± b 2 − c = , 12 144 12 1 1 from which I 2 = , I3 = . 12 12 Principal axes: The characteristic vectors parallel to the principal axes are obtained from Eq. (20.46), V x(′j ) = ( I y ′y ′ − I j )( I z ′z ′ − I j ) − I y2′z ′ V y( j ) = I x ′y ′ ( I z ′z ′ − I j ) + I x ′z ′ I y ′z ′ Vz(′ j ) = I x ′z ′ ( I y ′y ′ − I j ) + I x ′y ′ I y ′z ′ .

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Problem 20.99* The mass of the homogeneous thin plate is 3 slugs. For the coordinate system shown, determine the plate’s principal moments of inertia and unit vectors parallel to the corresponding principal axes.

y9 3 ft

2 ft

Solution: Divide the plate into A and B sheets, as shown. Denote m = 3 slugs, a = 2 ft, b = 4 ft, and c = 3 ft. The mass of 6m m plate A is m A = = = 1 slug. The mass of plate B is 18 3 12m mB = = 2 slugs. The coordinates of the center of mass of A are 18 c a (d x( A) , d y( A) , d z( A) ) = , , 0 = (1.5, 1, 0) ft. 2 2

(

4 ft

)

The coordinates of the center of mass of B are c b , 0, = (1.5, 0, 2) ft. (d x( B ) , d y( B ) , d z( B ) ) = 2 2

(

)

Principal Values: From Appendix C, the moments and products of inertia for plate A are m a2 = A + (d y2 + d z2 )m A = 1.333 slug-ft 2 , 12 m c2 I y( A′y)′ = A + (d x2 + d z2 )m A = 3 slug-ft 2 , 12 m (c 2 + a 2 ) + (d x2 + d y2 ) m A = 4.333 slug-ft 2 , I z( ′Az )′ = A 12 I x( A′y)′ = d x d y m A = 1.5slug-ft 2 , I x( A′x)′

z9 The inertia matrix is the sum of the two matrices:  I −I x ′y ′ −I x ′z ′   12 x ′x ′  −15 −6      . [ I ′] =  −I y ′x ′ I y ′y ′ −I y ′z ′  =  −1.5 19.67 0     − 6 0 10.33    −I z ′x ′ −I z ′y ′ I z ′z ′     From Eq. (20.45), the principal values are the roots of the cubic equation AI 3 + BI 2 + CI + D = 0, where

I x( A′z)′ = 0, I y( A′z)′ = 0.

A = +1, B = −( I x′x′ + I y′y′ + I z ′z ′ ) = −42,

The moments and products of inertia for plate B are

C = ( I x′x′ I y′y′ + I y′y′ I z ′z ′ + I z ′z ′ I x′x′ − I x2′y′ − I x2′z ′ − I y2′z ′ )

m Bb 2 + (d y2 + d z2 )m B = 10.67 slug-ft 2 , 12 m (c 2 + b 2 ) + (d x2 + d z2 )m B = 16.67 slug-ft 2 , I y( B′y)′ = B 12 m c2 I z( B′z )′ = B + (d x2 + d y2 )m B = 6 slug-ft 2 , I (x B′y)′ = 0, 12 I x( B′z)′ = d x d z m B = 6 slug-ft 2 , I y( B′z)′ = 0. I x( B′x)′ =

= 524.97, D = −( I x′x′ I y′y′ I z ′z ′ − I x′x′ I y2′z ′ − I y′y′ I x2′z ′ − I z ′z ′ I x2′y′ − 2 I x′y′ I y′z ′ I x′z ′ ) = −1707.4 The function f ( I ) = AI 3 + BI 2 + CI + D is graphed to determine the zero crossings, and the values refined by iteration. The graph is shown. The principal values are I 1 = 5.042 slug-ft 2 ,

I 3 = 20.17 slug-ft 2 ,

c A

Principal axes: The characteristic vectors parallel to the principal axes are obtained from Eq. (9.20),

x9 B

I 2 = 16.79 slug-ft 2 ,

V x(′j ) = ( I y ′y ′ − I j )( I z ′z ′ − I j ) − I y2′z ′

V y( j ) = I x ′y ′ ( I z ′z ′ − I j ) + I x ′z ′ I y ′z ′

Vz(′ j ) = I x ′z ′ ( I y ′y ′ − I j ) + I x ′y ′ I y ′z ′ . 400 350 300 250 200 f (I) 150 100 50 0 –50 –100

Zero Crossings

For I 1 = 5.042, V (1) = 77.38i + 7.937 j + 87.75k, and e 1 = 0.6599 i + 0.06768 j + 0.7483k. For I 2 = 16.79, V (2) = −18.57 i − 9.687 j − 17.25k, and e 2 = −0.6843i − 0.3570 j + 0.6358k. For I 3 = 20.17, V (3) = 4.911i − 14.75 j − 2.997k,

and e 3 = +0.3102 i − 0.9316 j − 0.1893k.

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Problem 20.100 The disk is pinned to the horizontal shaft and rotates relative to it with angular velocity ω d . Relative to an Earth-fixed reference frame, the vertical shaft rotates with angular velocity ω 0 . (a) Determine the disk’s angular velocity vector ω relative to the Earthfixed reference frame. (b) What is the velocity of point A of the disk relative to the Earth-fixed reference frame? Solution:

y b A

From Problem 20.43, the inertia matrix is

 4.167 × 10 5  [ I ] =  2.5 × 10 5   0 

2.5 × 10 5 0 4.167 × 10 5 0 0 8.333 × 10 5

   kg-m 2 .   

For rotation at a constant rate, the angular acceleration is zero, α = 0. The body-fixed coordinate system rotates with angular velocity Eq. (20.19) reduces to:  ∑M Ox    ∑ M Oy   ∑ M Oz 

   0  0 0    −0.01047  0  =  0    0 0.01047 0      4.167 × 10 5 2.5 × 10 5 0  ×  2.5 × 10 5 4.167 × 10 5 0   × 10 5 0 0 8.333 

     

 0.01047     × 0   0     0.01047   0 0 0     =  −8726.7   0 0 0    0 0    2618.0 4363.3    0    N-m, =  0    27.42 

M 0 = 27.4 (N-m).

Problem 20.101 The disk is pinned to the horizontal shaft and rotates relative to it with constant angular velocity ω d . Relative to an Earth-fixed reference frame, the vertical shaft rotates with constant angular velocity ω 0 . What is the acceleration of point A of the disk relative to the Earth-fixed reference frame? y b

u v0

vd (See figure in solution to Problem 20.100.) The angular acceleration of the disk is given by j ωO 0

k  0   0  

BandF_6e_ISM_C20_Pt2.indd 589

= (−ω O ω d )(k × r A / O ) + ω × (ω × r A / O ). Term by term:   i j k    −ω o ω d (k × r A / O ) = −ω O ω d  0 0 1   b R sin θ −R cos θ    = ω o ω d R sin θ i − ω o ω d b j,

     

+ ( R cos θ)( ω d2 + ω O2 )k.

Solution:

 i  dω d (ω i + ω O j) + ω O × ω d = 0 +  0 = dt dt d   ω d = −ω O ω d k.

a A / O = α × r A / O + ω × (ω × r A / O )

  i j k    ω × (ω × r A / O ) = ω ×  ω d ωO 0   b R sin θ −R cos θ     i j k  =  ωd ωO 0   −Rω O cos θ Rω d cos θ Rω d sin θ − bω O = ( Rω d sin θ − bω O )(ω O i − ω d j)

A z

The velocity of point A relative to O is

Collecting terms: a A / O = (2 Rω O ω d sin θ − bω O2 ) i − ( Rω d2 sin θ) j + ( Rω d2 cos θ + Rω O2 cos θ)k.

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Problem 20.102 The cone is connected by a ball-andsocket joint at its vertex to a 100-mm post. The radius of its base is 100 mm, and the base rolls on the floor. The velocity of the center of the base is v C = 2k (m/s). (a) What is the cone’s angular velocity vector ω ? (b) What is the velocity of point A?

y 400 mm z

A 100 mm 608 C x

Solution:

Denote θ = 60 °, h = 0.1 m, L = 0.4 m, R = 0.1 m.

(a) The strategy is to express the velocity of the center of the base of the cone, point C, in terms of the known (zero) velocities of O and P to formulate simultaneous equations for the angular velocity vector components. The line OC is parallel to the vector rC /O = iL (m). The line PC is parallel to the vector rC /P = Rj (m). The velocity of the center of the cone is given by the two expressions: v C = v O + ω × rC /O , and v C = v P + ω × rC /P , where v C = 2k (m/s), and v O = v P = 0.

y 0.4 m z

A 608

O 0.1 m

Expanding:  i  v C = ω × rC / O =  ω x   L

j ωy

 i  v C = ω × rC / P =  ω x   0

j ωy

   = ω Lj − ω Lk. z y   0  k  ω z  = −Rω z i + Rω x k.  0  k ωz

Solve by inspection: vC v = 20 rad/s, ω y = − C = −s 5 m/s, R L ω z = 0. ω = 20 i − 5 j (rad/s).

ωx =

(b) The line OA is parallel to the vector r A /O = iL + jR sin θ − kR cos θ. The velocity of point A is given by: v A = v O + ω × r A /O , where v O = 0.  i  j k   , v A =  20 −5 0   L R sin θ −R cos θ    v A = 0.25i + 1 j + 3.73k (m/s).

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Problem 20.103 The mechanism shown is a type of universal joint called a yoke and spider. The axis L lies in the x – z plane. Determine the angular velocity ω L and the angular velocity vector ω S of the cross-shaped “spider” in terms of the angular velocity ω R at the instant shown. Solution:

Denote the center of mass of the spider by point O, and denote the line coinciding with the vertical arms of the spider (the y axis) by P ′P, and the line coinciding with the horizontal arms by Q ′Q. The line P ′P is parallel to the vector rP /O = b j. The angular velocity of the right hand yoke is positive along the x axis, ω R = ω R i, from which the angular velocity ω L is positive toward the right, so that ω L = ω L ( i cos φ + k sin φ). The velocity of the point P on the extremities of the line P ′P is v P = v O + ω R × rP / O , where v O = 0, from which  i  v P =  ω R  0 

j k  0 0  = bω R k. b 0 

f 2b z

from which ω Sz = 0, ω Sx = ω R . The line Q ′Q is parallel to the vector rQ /O = ib sin φ − kb cos φ. The velocity of the point Q is   i j k   ,  v Q = v O + ω S × rQ / O = 0 +  ω Sx ω Sy 0     b sin φ 0 −b cos φ    v Q = −i(ω Sy b cos φ) + j(ω Sx b cos φ) − k(ω Sy b sin φ).

The velocity v P is also given by  i  v P = v O + ω S × rP / O =  ω Sx   0 = −ibω Sz + kbω Sx ,

j ω Sy

k ω Sz

     

y P

The velocity v Q is also given by v Q = v O + ω L × rQ / O , from which   i j k    v Q =  ω L cos φ 0 ω L sin φ  = jbω L (cos 2 φ + sin 2 φ)    b sin φ 0 −b cos φ  = j bω L , from which ω Sy = 0, from which

vR Q9

ωS = ω R, ω S = ω Ri ,

ω L = ω Sx cos φ = ω R cos φ .

Problem 20.104 The inertia matrix of a rigid body in terms of a body-fixed coordinate system with its origin at the center of mass is  4 1 −1    1 2 0  kg-m 2 .   6   −1 0

[I] = 

If the rigid body’s angular velocity is ω = 10 i − 5 j + 10 k (rad/s), what is its angular momentum about its center of mass? Solution:  H Ox    H Oy   H Oz 

The angular momentum is

  25   4 1 −1   10         0   −5  =  0  kg-m 2 /s  =  1 2       6   10   50   −1 0  

In terms of the unit vectors i, j, k, H = 25i + 50 k kg-m 2 /s.

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Problem 20.105 What is the moment of inertia of the rigid body described in Problem 20.104 about the axis that passes through the origin and the point (4, −4, 7) m ? Strategy: Determine the components of a unit vector parallel to the axis and use Eq. (20.43).

The inertia matrix in Problem 20.104 is  I −I xy −I xz  4 1 −1  xx     0  =  −I xy I yy −I yz [I ] =  1 2    6   −I xz −I yz I zz  −1 0 

   ,   

The unit vector parallel to the line passing through (0, 0, 0) and (4, − 4, 7) is

where advantage is taken of the symmetric property of the inertia matrix. From Eq. (20.43), the new moment of inertia about the line through (0, 0, 0) and (4, − 4, 7) is

4 i − 4 j + 7k = 0.4444 i − 0.4444 j + 0.7778k. 42 + 42 + 72

I O = 4 e x2 + 2e y2 + 6e z2 + 2(1)(e x e y ) + 2(−1)(e x e z ) + 2(0)(e y e z ),

Solution: e =

I O = 3.728 kg-m 2 .

Problem 20.106 Determine the inertia matrix of the 0.6-slug thin plate in terms of the coordinate system shown.

Solution:

The strategy is to determine the moments and products for a solid thin plate about the origin, and then subtract the moments and products of the cut-out. The mass density is ρ =

(

0.6 π (0.5 2 ) − π

( ))

3 2 T 24

0.8149 slug/ft 3 , T

from which ρT = 0.8149 slug/ft 2 , where T is the (unknown) thickness of the plate. From Appendix C and by inspection, the moments and products of inertia for a thin plate of radius R are: 6 in

I xx = I yy =

1.5 in x

3 in

mR 2 mR 2 , I zz = , I xy = I xz = I yz = 0. 4 2

For a 6 in. radius solid thin plate, m = ρT π (0.5 2 ) = 0.64 slug. I xx = I yy = 0.04 slug-ft 2 , I zz = 0.08 slug-ft 2 , I xy = I xz = I yz = 0. The coordinates of the 1.5 in. radius cut-out are (d x , d y , d x ) = (3, 0, 0). The mass removed by the cut-out is mC = ρT π RC2 = 0.04 slug. The moments and products of inertia of the cut-out are mC RC2 C = I xx = 1.563 × 10 −4 slug-ft 2 , 4 mC RC2 3 2 C = I yy mC = 2.656 × 10 −3 slug-ft 2 , + 4 12 m R2 I zzC = C C + d x2 mC = 2.813 × 10 −3 slug-ft 2 , 2 C = 0, I C = 0, I C = 0. I xy xz yz

( )

The inertia matrix of the plate with the cut-out is   0.04 0 0    0.04 0 [I ] 0 =  0   0 0.08   0  1.563 × 10 −4  − 0   0 

0 2.656 × 10 −3 0

0 0 2.813 × 10 −3

  ,   

  0.0398 0 0    slug-ft 2 . [I ] 0 =  0 0.0373 0   0 0 0.0772  

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Problem 20.107 At t = 0, the 0.6-slug thin plate has angular velocity ω = 10 i + 10 j (rad/m) and is subjected to the force F = −10 k (lb) acting at the point (0, 6, 0) in. No other forces or couples act on the plate. What are the components of its angular acceleration at that instant?

6 in

1.5 in

Solution:

The coordinates of the center of mass are (−0.01667, 0, 0) ft. The vector from the center of mass to the point of application of the force is rF /G = 0.01667 i + 0.5 j (ft). The moment about the center of mass of the plate is

3 in

 i j k   M G = rF /G × F =  0.01667 0.5 0   0 0 −10    = −5 i + 0.1667 j (lb-ft). Eq. (20.19) reduces to   −5    αx  0.03984 0 −0       0.1667  =   0 0.03718 0 w  αy     0 0 0 0.07702   α     z   0   . 0 +     −0.267 

      

Carry out the matrix multiplication to obtain the three equations: 0.03984α x = −5, 0.03718α y = 0.1667, 0.07702α z − 0.267 = 0. α = −125.5i + 4.484 j + 3.467k rad/s 2 .

Solve:

dω = 0. The coordinate system is rotating with angular dt velocity ω, from which ω = ω. Eq. (20.19) reduces to

Problem 20.108 The inertia matrix of a rigid body in terms of a body-fixed coordinate system with its origin at the center of mass is

with α =

  ∑ M φx   ∑ M φy    ∑ M φz 

 4 1 −1    [I] =  1 2 0  kg-m 2 .    −1 0 6  If

the

rigid

body’s

angular

velocity

ω = 10 i − 5 j + 10 k (rad/s) and its angular acceleration

is zero, what are the components of the total moment about its center of mass? Solution:  ∑M φx    ∑ M φy   ∑M φz 

   0 −10 −5   4 1 −1   10        =  10 0 −10   1 2 0   −5        10 0   −1 0 6   10   5    −5 −20 −30   10     =  50 10 −70   −5  N-m,    25 −5   10   30

M = −250 i − 250 j + 125k (N-m).

Use general motion, Eq. (20.19),

  I −I xy −I xz  xx    I yy −I yz  =  −I yx     −I zx −I zy I zz    0 −Ω z   +  Ωz 0   −Ω y Ω x 

    αx    α  y      αz     Ω y   I xx −I xy −I xz  −Ω x   −I yx I yy −I yz  0   −I zx −I zy I zz  

   ωx    ωy    ωz  

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      

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Problem 20.109 If the total moment about the center of mass of the rigid body described in Problem 20.108 is zero, what are the components of its angular acceleration? Solution:

Use general motion, Eq. (20.19), with the moment components equated to zero,  I −I xy −I xz   α x   0  xx        0  =  −I yx I yy −I yz   α y         −I zx −I zy I zz   α z   0      I  0 − Ω Ω −I xy −I xz   ω x z y xx      +  Ωz −Ω x   −I yx 0 I yy −I yz   ω y     −Ω y Ω x 0   −I zx −I zy I zz   ω z      α     4 1 −1   x  −10 −5   0    =  1 2 0   α y  +  10 0 −10       6   α  10 0   5  −1 0  z   4 1 −1   10     × 1 2 0   −5  ,    6   10   −1 0   4 1 −1   α x  0       0  =  1 2 0   αy     6   α  −1 0  0   z

Carry out the matrix multiplication to obtain the three simultaneous equations in the unknowns: 4α x + α y − α z = 250, α x + 2α y + 0 = 250, − α x + 0 + 6α z = −125. Solve: α = 31.25i + 109.4 j − 15.63k (rad/s 2 ).

      

  −250       +  −250  .    125  

Problem 20.110 The slender bar of length l and mass m is pinned to the L-shaped bar at O. The L-shaped bar rotates about the vertical axis with a constant angular velocity ω 0 . Determine the value of ω 0 necessary for the bar to remain at a constant angle β relative to the vertical.

Solution: Since the point O is not fixed, this is general motion, in which Eq. (20.19) applies. Choose a coordinate system with the origin at O and the x axis parallel to the slender bar.

The moment exerted by the bar. Since axis of rotation is fixed, the acceleration must be taken into account in determining the moment. The vector distance from the axis of rotation in the coordinates system shown is

r0 = b( i cos(90 ° − β ) + j sin (90 ° − β )) = b( i sin β + j cos β ). The vector distance to the center of mass of the slender bar is L rG /O = i . The angular velocity is a constant and the coordinate sys2 tem is rotating with an angular velocity ω = ω 0 (−i cos β + j sin β ). The acceleration of the center of mass relative to O is

( )

a G = ω × (ω × (r0 + rG / O ))   i   = ω ×  −ω 0 cos β    b sin β + L  2

  j k   ω 0 sin β 0  ,   b cos β 0  

   i j k  a G =  −ω 0 cos β ω 0 sin β 0   ω L sin β  0 0 −ω 0 b − 0  2

b W

        

L sin β ) i ( 2 L −ω ( b cos β + sin β cos β ) j. 2

a G = −ω 02 +b sin β +

2 0

= a Gx i + a Gy j

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20.110 (Continued) From Newton’s second law, ma G = A + W, from which A = ma G − W. The weight is W = mg ( i cos β − j sin β ). The moment about the center of mass is M G = rO / G × A = rO / G × (ma G − W)  i j k     L =  − 0 0  . 2    ma Gx − mg cos β ma Gy + mg sin β 0    mω o2 bL mω 02 L2 mgL MG = + cos β − sin β + sin β cos β k 2 2 4 = M zk

(

  0     0      0 

mL2 12

       0     0 .  =     2 mL2 ω  − 0  cos β sin β    12  

mω 02 bL mω 02 L2 mgL cos β − sin β + sin β cos β 2 2 4 mL2 = −ω 02 sin β cos β . 12

)

  0 0 + sin β    2 0 0 cos β = ω  o    − sin β − cos β 0   

 M x    My   Mz 

Substitute:

The Euler Equations: The moments of inertia of the bar about the cenmL2 ter of mass are I xx = 0, I yy = I zz = , I xy = I xz = I yz = 0. 12 Eq. (20.19) becomes:  M x    My   Mz 

Carry out the matrix multiplication:

g sin β . 2 L sin β cos β + b cos β 3

ω0 =

Solve:

( )

          

 − cos β     ×  sin β  .   0  

Problem 20.111 A slender bar of length l and mass m is rigidly attached to the center of a thin circular disk of radius R and mass m. The composite object undergoes a motion in which the bar rotates in the horizontal plane with constant angular velocity ω 0 about the center of mass of the composite object and the disk rolls on the floor. Show that ω 0 = 2 g /R .

rate is the rotation about the Z axis, ψ = ω 0 rad/s. The velocity of the center of mass of the disk is v G = ( L /4)ψ , from which the spin rate is v L φ = = ψ. From Eq. (20.29), the moment about the x-axis is R 4R 2 sin θ = I φψ = ω0 L I . M x = ( I zz − I xx )ψ 2 sin θ cos θ + I zz ψφ zz 4 R zz

( )

The moment of inertia is I zz = Mx =

v0 R

( mRL )ω . 8 2 0

The normal force acting at the point of contact is N = 2mg. The moment exerted about the center of mass MG = N

mR 2 , from which 2

( )

L L = mg . 4 2

Equate the moments: mg ω0 = 2

( L2 ) = m( RL8 )ω , from which 2 0

g . R

Solution:

y, Z

Measuring from the left end of the slender bar, the distance to the center of mass is dG =

( )

L m + mL 3L 2 = . 2m 4

Choose an X , Y , Z coordinate system with the origin at the center of mass, the Z axis parallel to the vertical axis of rotation and the X axis parallel to the slender bar. Choose an x, y, z coordinate system with the origin at the center of mass, the z axis parallel to the slender bar, and the y axis parallel to the Z axis. By definition, the nutation angle is the angle between Z and z, θ = 90 °. The precession

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3L 4

X, R W

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Problem 20.112* The thin plate of mass m spins about a vertical axis with the plane of the plate perpendicular to the floor. The corner of the plate at O rests in an indentation, so that it remains at the same point on the floor. The plate rotates with constant angular velocity ω 0 and the angle β is constant. (a) Show that the angular velocity ω 0 is related to the angle β by

y mg

Expand, M z = hmg

( sin2 β − cos β )

( sin β3cos β + cos2 β − sin2 β + 4 sin β3cos β ) 2

= mh 2 ω a2 − Solve:

hω 0 2 2 cos β − sin β . = g sin 2 β − 2 sin β cos β − cos 2 β (b) The equation you obtained in (a) indicates that ω 0 = 0 when 2 cos β − sin β = 0. What is the interpretation of this result?

ω 02 h 2 cos β − sin β = g sin 2 β − 2sin β cos β − cos 2 β

(b) The perpendicular distance from the axis of rotation to the center of mass of the plate is h

ω d = rO /G × ω

(

= h cos β − b

 i j k     h =  0 − −h    2    0 cos β sin β   

)

sin β . 2

If this distance is zero, β = tan −1 (2) = 63.43 °, the accelerations of the center of mass and the external moments are zero (see equasin β tions above, where for convenience the term cos β − has been 2 kept as a factor) and the plate is balanced.

Solution:

Choose a body-fixed coordinate system with its origin at the fixed point O and the axes aligned with the plate’s edges. Using the moments of inertia for a rectangular area,

The angular velocity of rotation is zero (the plate is stationary) if β = tan −1 (2) = 63.435 °, since the numerator of the right hand term in the boxed expression vanishes (the balance at this point would be very unstable, since an infinitesimally small change in β would induce a destabilizing moment.).

mh 2 4 mh 2 5mh 2 , I zz = I xx + I yy = , , I yy = 3 3 3 2 mh I xy = , I xz = I yz = 0. 2 I xx =

The plate’s angular velocity is ω = ω 0 sin β i + ω 0 cos β j , and the moment about O due to the plate’s weight is ∑ M x = 0, h ∑ M y = 0, ∑ M z = mg sin β − hmg cos β . 2 Choose a coordinate system with the x axis parallel to the right lower edge of the plate and the y axis parallel to the left lower edge of the plate, as shown. The body fixed coordinate system rotates with angular velocityω = ω 0 ( j sin β + k cos β ) . From Eq. (20.13),  0   0   Mz 

596

  0 0 ω 0 cos β     =  0 0 −ω 0 sin β      cos sin 0 β ω0 β   −ω 0  2   mh 2 mh  0  −   ω cos β  3 2  0  2 2   4 mh mh 0   ω 0 cos β × − 2 3   0  5m 2    0 0   3  

  .   

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Problem 20.113* In Problem 20.112, determine the range of values of the angle β for which the plate will remain in the steady motion described.

Solution: ω 02 h g

From the solution to Problem 20.112

2 cos β − sin β . sin 2 β − 2sin β cos β − cos 2 β

The angular velocity is a real number, from which ω 02 ≥ 0, from which 2 cos β − sin β f (β ) = ≥ 0. sin 2 β − 2 cos β sin β − cos 2 β

A graph of f (β ) for values of 0 ≤ β ≤ 90 ° is shown. The function is positive over the half-open interval 63.4348 ° ≤ β < 67.50 °. The angular velocity is zero at the lower end of the interval, and “blows up” (becomes infinite) when the denominator vanishes, which occurs at 3π exactly β = = 67.50 °. 8

h 2h

f (b )

Problem 20.114 Arm BC has a mass of 12 kg, and its moments and products of inertia, in terms of the coordinate system shown, are I xx = 0.03 kg-m 2 , I yy = I zz = 4 kg-m 2 , and I xy = I yz = I xz = 0. At the instant shown, arm AB is rotating in the horizontal plane with a constant angular velocity of 1 rad/s in the counterclockwise direction viewed from above. Relative to arm AB, arm BC is rotating about the z-axis with a constant angular velocity of 2 rad/s. Determine the force and couple exerted on arm BC at B. x 2 rad/s y

1 rad/s

0 30

f( b ) vs b

40 50 60 b, deg

0.3 m G

y G

mg B

In terms of the angle θ, the angular velocity of the coordinate system is ω = (1)sin θ i + (1) cos θ j + (2)k (rad/s), so its angular acceleration is Eq. (20.4) α =

dθ dθ dω = cos θ i − sin θ j. dt dt dt

Setting θ = 40 ° and d θ /dt = 2 rad/s, we obtain

ω = 0.643i + 0.766 j + 2k (rad/s), (1) 408 B

5 4 3 2 1 0 –1 –2 –3 –4 –5

α = 1.532 i − 1.286 j (rad/s 2 ). (2) Equation (20.19) is

In terms of the body-fixed coordinate system shown, the moments and products of inertia are

    M Bx   αx   0.03 0   0        αy  2.92 0  M By + 0.3FBz  =  0      0 0 2.92   α    M Bz − 0.3FBy   z      0 −ω ω y   0.03 0 z 0   ωx     2.92 0   ωy 0 −ω x   0 + ωz    0 2.92   ω  −ω y ω x 0   0  z  

I xx = 0.03 kg-m 2 ,

Using Eqs. (1) and (2), this gives the equations

700 mm

Solution:

I yy = I zz =

4 − (0.3) 2 (12) =

2.92 kg-m 2 ,

I xy = I yz = I zx = 0.

   .   

M Bx = 0.046, M By + 0.3FBz = −7.469, M Bz − 0.3FBy = 1.423,

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20.114 (Continued) From Newton’s second law (see free-body diagram),

From which

FBx − mg sin 40 ° = m(−1.912),

M B = 0.046 i − 10.25 j + 30.63k (N-m).

FBy − mg cos 40 ° = m(0.598),

The acceleration of B toward A is (1 rad/s) 2 (0.7 m) = 0.7 m/s 2 , so a B = −0.7 cos 40 °i + 0.7sin 40 ° j

FBz = m(0.771). Solving, FB = 52.72 i + 97.35 j + 9.26k(N). The moment about the

= −0.536 i + 0.450 j (m/s 2 ).

center of mass is

The acceleration of the center of mass is

∑ M = M B + (−0.3i) × FB

a G = a B + α × rG /B + ω × (ω × rG /B )

= M B + 0.3FBz j − 0.3FBy k .

= −1.912 i + 0.598 j + 0.771k (m/s 2 ).

Problem 20.115 Suppose that you throw a football in a wobbly spiral with a nutation angle of 25 °. The football’s moments of inertia are I xx = I yy = 0.003 slug-ft 2 and I zz = 0.001 slug-ft 2 . If the spin rate is φ = 4 revolutions per second, what is the magnitude of the precession rate (the rate at which the football wobbles)?

z 258

Solution: This is modeled as moment-free, steady precession of an axisymmetric object. From Eq. (20.33), ( I zz − I xx )ψ cos θ + I zzφ = 0, from which ψ = −

I zzφ . ( I zz − I xx ) cos θ

Substitute ψ = −

(0.001)(4) = 2.21 rev/s. (0.001 − 0.003) cos 25 °

ψ = 2.21 rev/s.

Problem 20.116 Sketch the body and space cones for the motion of the football in Problem 20.115.

I  The angle β is given by tan β =  zz  tan θ, from which  I xx  β = 8.8 °. β < θ, and the body cone revolves outside the space cone. The sketch is shown. The angle θ = 25 °.

Solution:

z 258

Space cone

u z

b f

Body cone

598

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Problem 20.117 The mass of the homogeneous thin plate is 1 kg. For the coordinate system shown, determine the plate’s principal moments of inertia and the directions of unit vectors parallel to the corresponding principal axes.

The principal moments of inertia. The principal values are given by the roots of the cubic equation AI 3 + BI 2 + CI + D = 0, where A = 1, B = −( I xx + I yy + I zz ) = −0.1274, 2 − I 2 − I 2 = 4.71 × 10 −3 , C = I xx I yy + I yy I zz + I xx I zz − I xy xz yz 2 − I I 2 − I I 2 − 2I I I ) D = − ( I xx I yy I zz − I xx I yz yy xz zz xy xy yz xz

= −4.158 × 10 −5.

The function f ( I ) = AI 3 + BI 2 + CI + D is graphed to get an estimate of the roots, and these estimates are refined by iteration. The graph is shown. The refined values of the roots are I 1 = 0.01283 kg-m 2 , I 2 = 0.05086 kg-m 2 , I 3 = 0.06370 kg-m 2 .

160 mm 160 mm

The principal axes. The principal axes are obtained from a solution of the equations 2 V x = ( I yy − I )( I zz − I ) − I yz

160 mm x9

O 400 mm z9

Solution:

The moment of inertia is determined by the strategy of determining the moments and products of a larger plate, and then subtracting the moments and products of inertia of a cutout, as shown in the sketch. Denote h = 0.32 m, b = 0.4 m, c = 0.16 m, d = b − c = 0.24 m. The area of the plate is A = hb − cd = 0.0896 m 2 . Denote the mass density by ρ kg/m 3 . The mass is ρ AT = 1 kg, from which ρT = 1A = 11.16 kg/m 2 , where T is the (unknown) thickness. The moments and products of inertia of the large plate: The moments and products of inertia about O are (see Appendix C): 3 3 ( p ) = ρTbh , I ( p ) = ρThb , I xx yy 3 3 2 2 ( p ) + I ( p ) = ρTbh(h + b ) , I zz( p) = I yy xx 3 2 2 ( p ) = ρTh b , I ( p ) = I ( p ) = 0. I xy xz yz 4

The moments and products of inertia for the cutout: The moments and products of inertia about the center of mass of the cutout are: ρTdc 3 ( c ) ρTcd 3 ( c ) ρTcd (c 2 + d 2 ) , I yy = , I zz = , 12 12 12 ( ) ( ) ( ) c c c I xy = I xz = I yz = 0. (c) = I xx

The distance from O to the center of mass of the cutout is (d x , d y , 0) = (0.28, 0.24, 0) m. The moments and products of inertia about the point O are (0) = I ( c ) + ρTcd (d 2 ), I (0) = I ( c ) + ρTcd (d 2 ), I xx xx y yy yy x

I zz(0) = I zz( c ) + ρTcd (d x2 + d y2 ),

V y = I xy ( I zz − I ) + I xz I yz Vz = I xz ( I yy − I ) + I xz I yz . Since I xz = I yz = 0, Vz = 0, and the solution fails for this axis, and the vector is to be determined from the orthogonality condition. Solving for V x , V y , the unit vectors are: for I = I 1 , V x(1) = 0.00141, V y(1) = 8.6 × 10 −4 , from which the unit vectors are e 1 = 0.8535i + 0.5212 j . For

I = I 2 , V x(2) = −1.325 × 10 −4 , V y(2) = 2.170 × 10 −4 ,

from

which e 2 = −0.5212 i + 0.8535 j. The third unit vector is determined from orthogonality conditions: e 3 = k .

0.24 m 0.16 m

0.32 m

0.4 m

f (I) vs I .00002 .000015 .00001 .000005 f (I) 0 –.000005 –.00001 –.000015

.02

.04

.06

.08

I, kg-m2

(0) = I ( c ) + ρTcd (d d ), I (0) = I (0) = 0. I xy xy x y xz yz

The moments and products of inertia of the object: The moments and products of inertia about O are ( p ) − I (0) = 0.02316 kg-m 2 , I xx = I xx xx ( p ) − I (0) = 0.04053 kg-m 2 , I yy = I yy yy

I zz = I zz( p) − I zz(0) = 0.06370 kg-m 2 , ( p ) − I (0) = 0.01691 kg-m 2 , I xy = I xy xy

I xz = I yz = 0.

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Problem 20.118 The airplane’s principal moments of inertia, in slug-ft 2 , are I xx = 8000, I yy = 48, 000, and I zz = 50, 000. (a) The airplane begins in the reference position shown and maneuvers into the orientation ψ = θ = φ = 45 °. Draw a sketch showing the plane’s orientation relative to the XYZ system. (b) If the airplane is in the orientation described in (a), the rates of change of the Euler angles are ψ = 0, θ = 0.2 rad/s, and φ = 0.2 rad/s, and the second derivatives of the angles with respect to time are zero, what are the components of the total moment about the airplane’s center of mass?

Z, z X, x

Solution: (a) (b) The Eqs. (20.36) apply. cos θ sin φ M x = I xx (ψ sin θ sin φ + θ cos φ + ψθ + ψφ sin θ cos φ − θφ sin φ) − ( I − I ) yy

u = 458 z

× (ψ sin θ cos φ − θ sin φ)(ψ cos θ + φ), cos θ cos φ = I (ψ sin θ cos φ − θ sin φ + ψθ

x Y

sin θ sin φ − θφ cos φ) − (I − I ) − ψφ zz xx × (ψ sin θ sin φ + θ cos φ)(ψ cos θ + φ ), sin θ ) − ( I − I )(ψ sin θ sin φ M z = I zz (ψ cos θ + φ − ψθ xx yy

f = 458 X c = 458

+ θ cos φ)(ψ sin θ cos φ − θ sin φ). Substitute I xx = 8000, I yy = 48,000, and I zz = 50,000, in slug-ft 2 , and ψ = 0, θ = 0.2 rad/s, and φ = 0.2 rad/s, and = θ = φ = 0. The moments: ψ M x = −282.84 ft-lb ,

M y = −2545.6 ft-lb ,

M z = −800 ft-lb.

Problem 20.119 What are the x, y, and z components of the angular acceleration of the airplane described in Problem 20.118? Solution:

The angular accelerations are given by Eq. (20.35):

Z, z X, x

dω x cos θ sin φ + ψφ sin θ cos φ = ψ sin θ sin φ + ψθ dt sin φ. + θ cos φ − θφ dω y cos θ cos φ − ψφ sin θ sin φ = ψ sin θ cos φ + ψθ dt cos φ. − θ sin φ − θφ dω z sin θ + φ . = ψ cos θ − ψψθ dt Substitute I xx = 8000, I yy = 48,000, and I zz = 50,000, in slug-ft 2 , and ψ = 0, θ = 0.2 rad/s, and φ = 0.2 rad/s, and = θ = φ = 0, to obtain ψ α x = −0.0283 rad/s 2 ,

600

α y = −0.0283 rad/s ,

α z = 0.

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Problem 20.120 If the orientation of the airplane in Problem 20.118 is ψ = 45 °, θ = 60 °, and φ = 45 °, the rates of change of the Euler angles are ψ = 0, θ = 0.2 rad/s, and φ = 0.1 rad/s, and the components of the total moment about the center of mass of the plane are ΣM x = 400 ft-lb, ΣM y = 1200 ft-lb, and ΣM z = 0, what are the x, y, and z components of the airplane’s angular acceleration?

Z, z X, x

, The strategy is to solve Eqs. (20.36) for θ , φ , and ψ and then to use Eqs. (20.35) to determine the angular accelerations.

Solution:

cos θ sin φ M x = I xx (ψ sin θ sin φ + θ cos φ + ψθ sin θ cos φ − θφ sin φ) − ( I − I ) + ψφ yy

× (ψ sin θ cos φ − θ sin φ)(ψ cos θ + φ ), cos θ cos φ = I (ψ sin θ cos φ − θ sin φ + ψθ yy

sin θ sin φ − θφ cos φ) − (I − I ) − ψφ zz xx × (ψ sin θ sin φ + θ cos φ)(ψ cos θ + φ ), sin θ ) − ( I − I )(ψ sin θ sin φ M z = I zz (ψ cos θ + φ − ψθ xx yy + θ cos φ)(ψ sin θ cos φ − θ sin φ). Substitute numerical values and solve to obtain φ = −0.03266 rad/s 2 , = 0.09732 rad/s 2 . These are to be used θ = 0.01143 rad/s 2 , ψ (with the other data) in Eqs. (20.35), dω x cos θ sin φ + ψφ sin θ cos φ = ψ sin θ sin φ + ψθ dt sin φ. + θ cos φ − θφ dω y cos θ cos φ − ψψ sin θ sin φ = ψ sin θ cos φ + ψθ dt cos φ, − θ sin φ − θφ dω z sin θ + φ . = ψ cos θ − ψθ dt Substitute, to obtain: α x = 0.05354 rad/s 2 ,

α y = 0.03737 rad/s 2 ,

α z = 0.016 rad/s 2 .

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Chapter 21 Problem 21.1 The mass m = 5 kg and the spring constant k = 2000 N/m. (a) How many cycles does the vibrating mass go through in 10 s? (b) How long does the mass take to go through a single cycle? That is, what is the period of the vibrations?

x k

Solution: (a) The circular frequency is ω =

k = m

2000 N/m = 20 rad/s, 5 kg

(b) The period is the inverse of the frequency. That is, the inverse of cycles per second is seconds per cycle: 1 τ = = 0.314 s. f

so the frequency is ω f = = 3.18 Hz. 2π

(a) 31.8 cycles. (b) τ = 0.314 s.

The frequency is the number of cycles per second, so in 10 seconds the mass goes through 31.8 cycles.

Problem 21.2 The mass weighs 10 lb and the spring constant is k = 24 lb/ft. (a) What is the frequency of vibration of the mass? (b) If the mass is displaced a distance x = 1 ft and released from rest, what is the displacement at t = 2 s?

x k

Solution: (a) The mass is m =

10 lb = 0.311 slug. 32.2 ft/s 2

The circular frequency is ω =

k = m

Substituting the initial conditions x = 1 ft and dx /dt = 0 at t = 0, we obtain the equations

24 lb/ft = 8.79 rad/s, 0.311 slug

1 ft = 0 + B, 0 = Aω,

so the frequency is ω f = = 1.40 Hz. 2π

so the displacement as a function of time is x = (1 ft) cos8.79t.

(b) Eq. (21.8) is

At t = 2 s this yields x = 0.298 ft.

x = A sin ωt + B cos ωt,

x = 0.298 ft.

and the velocity is dx = Aω cos ωt − Bω sin ωt. dt

Problem 21.3 The mass m = 5 kg and the spring constant k = 2000 N/m. Describe the position of the mass by Eq. (21.9): x = E sin(ωt − φ).

Equation (21.9) is x = E sin(ωt − φ) and the velocity is dx = E ω cos(ωt − φ). dt Substituting the initial conditions x 0 = 0.15 m and dx /dt = v 0 = 5 m/s at t = 0, we obtain the equations

At t = 0, the position is x = 0.15 m and the velocity is dx /dt = 5 m/s. (a) Determine the amplitude E and phase angle φ. (b) What is the position of the mass at t = 2 s?

(1) (2)

Dividing Eq. (1) by Eq. (2), we obtain the equation ωx 0 , tan(−φ) = v0

x k

x 0 = E sin(−φ) v 0 = E ω cos(−φ).

which yields φ = −0.540 rad. Then from Eq. (1) we obtain E = 0.292 m. (b) The position at t = 2 s is

Solution:

x = E sin(ωt − φ)

(a) The circular frequency is ω =

602

k = m

2000 N/m = 20 rad/s. 5 kg

= (0.292 m)sin[(20 rad/s)(2 s) + 0.540 rad] = 0.0862 m. (a) E = 0.292 m, φ = −0.540 rad. (b) x = 0.0862 m.

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Problem 21.4 An engineer is studying a conservative one-degree-of-freedom system with the coordinate denoted by x. Subjecting it to an initial velocity dx /dt = 0.25 m/s with the displacement x = 0, the data shown is obtained. Choose values of the mass m and spring constant k of the spring-mass oscillator so that it would simulate the motion of the system.

Solution:

From the given data, we observe that the system goes through 4 cycles in approximately 10 seconds. Therefore the frequency is f =

4 cycles ω = = 0.4 Hz, 2π 10 s

which yields ω = 2.51 rad/s. Alternatively, let the initial velocity of the system be denoted by v 0 = 0.25 m/s. Equation (21.8) is x = A sin ωt + B cos ωt, and the velocity is dx = Aω cos ωt − Bω sin ωt. dt

x k

Substituting the initial conditions x = 0 and dx /dt = v 0 at t = 0, we obtain the equations

0 = 0 + B, v 0 = Aω, so A = v 0 /ω, B = 0 and the displacement as a function of time is v x = 0 sin ωt. ω

0.15

We see that the maximum displacement is v 0 /ω. From the data, the maximum displacement is approximately 0.1 m, so v0 0.25 m/s = = 0.1 m, ω ω which yields ω = 2.5 rad/s.

Displacement, meters

0.1 0.05 0

Either way, we obtain (approximately) k = ω 2 = (2.5 rad/s) 2 = 6.25 rad 2 /s 2 . m

20.05

For example, m = 1 kg, k = 6.25 N/m could be used. 20.1 Values of k and m such that k /m = 6.25 rad 2 /s 2 . 20.15 0

6 t, seconds

Problem 21.5 The mass m = 4 kg, and the spring constant is k = 64 N/m. For vibration of the spring– mass oscillator relative to its equilibrium position, determine (a) the frequency in Hz and (b) the period. Solution:

k x

Since the vibration is around the equilibrium position,

we have k = m

(

ω =

(b)

2π 2π τ = = = 1.57 s. ω 4 rad/s

M21_BEDF2222_06_SE_C21_Part1.indd 603

)

1 cycle 64 N/m = 4 rad/s = 0.637 Hz. 4 kg 2π rad

(a)

208

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Problem 21.6* The mass m = 4 kg, and the spring constant is k = 64 N/m. The spring is unstretched when x = 0. At t = 0, x = 0, and the mass has a velocity of 2 m/s down the inclined surface. What is the value of x at t = 0.8 s?

Solution: m

The equation of motion is

( )

d 2x k d 2x + kx = mg sin 20 ° ⇒ + x = g sin 20 ° dt 2 dt 2 m

Putting in the numbers we have d 2x + (16 s −2 ) x = 3.36 m/s 2 dt 2 The solution to this non-homogeneous equation is x = A sin([4 s −1 ]t ) + B cos([4 s −1 ]t ) + 0.210 m Using the initial conditions we have

k x m 208

0 = B + 0.210 m   ⇒ A = 0.5 m, B = −0.210 m 2 m/s = A(4 s −1 )  The motion is x = (0.5 m)sin([4 s −1 ]t ) − (0.210 m)cos([4 s −1 ]t ) + 0.210 m At t = 0.8 s we have x = 0.390 m.

Problem 21.7 The 10-kg sphere has radius R = 0.1 m. It is rigidly attached to the 4-kg slender bar with length L = 0.25 m. What is the frequency of small oscillations of the pendulum? u

R Setting

Solution:

Let m B = 4 kg and m S = 10 kg. The moment of inertia of the bar about the pin support is 1 I B = m B L2 = 0.0833 kg-m 2 . 3 Applying the parallel-axis theorem, the moment of inertia of the sphere about the pin support is 2 I S = m S R 2 + ( L + R) 2 m S = 1.265 kg-m 2 . 5 The moment of inertia of the pendulum is I = I B + I S = 1.35 kg-m 2 . In terms of the angular velocity ω = d θ /dt, the kinetic energy of the pendulum is 1 T = Iω 2. 2 Placing the datum at the level of the pin support, the potential energy of the pendulum is 1 V = −m B g L cos θ − m S g( L + R) cos θ. 2

604

{

}

d d 1 2 1 (T + V ) = I ω −  m B g L + m S g( L + R)  cos θ   2 dt dt 2 1 dω   = Iω +  m B g L + m S g( L + R)  sin θω   dt 2 = 0 and using the approximation sin θ = θ, we obtain the equation of motion d 2θ + ω 2θ = 0, dt 2 where 1 m B g L + m S g ( L + R) 2 = 5.39 rad/s. I The frequency is ω f = = 0.859 Hz. 2π ω =

f = 0.859 Hz.

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Problem 21.8* The 10-kg sphere has radius R = 0.1 m. It is rigidly attached to the slender bar with length L. The mass per unit length of the slender bar is 16 kg/m. Determine the value of L if the frequency of small oscillations of the pendulum is 0.7 Hz. Strategy: Plot a graph of the frequency as a function of L and use it to estimate the value of the length for which f = 0.7 Hz.

In terms of the angular velocity ω = d θ /dt, the kinetic energy of the pendulum is 1 T = Iω 2. 2 Placing the datum at the level of the pin support, the potential energy of the pendulum is 1 V = −m B g L cos θ − m S g( L + R) cos θ. 2 Setting

{

d d 1 2 1 (T + V ) = I ω −  m B g L + m S g( L + R)  cos θ   dt dt 2 2 dω 1 = Iω +  m B g L + m S g( L + R)  sin θω   dt 2 = 0

}

and using the approximation sin θ = θ, we obtain the equation of motion R

d 2θ + ω 2θ = 0, dt 2 where

Solution: Let m S = 10 kg. In terms of its length in meters, the mass of the bar is m B = (16 kg/m) L. (1) The moment of inertia of the bar about the pin support is IB =

1 m L2 . 3 B

Applying the parallel-axis theorem, the moment of inertia of the sphere about the pin support is IS =

1 m B g L + m S g ( L + R) 2 . I

ω =

The frequency is f =

ω 1 = 2π 2π

For a given value of the length L, the mass of the bar is given by Eq. (1). The moment of inertia of the pendulum can be obtained from Eq. (2), then the frequency determined from Eq. (3). The graph of f as a function of L is shown:

2 m R 2 + ( L + R) 2 m S . 5 S

0.95

0.9

1 2 I = I B + I S = m B L2 + m S R 2 + ( L + R) 2 m S . (2) 3 5

0.85 Frequency, Hertz

The moment of inertia of the pendulum is

1 m B g L + m S g ( L + R) 2 . (3) I

0.8

0.75

0.7

0.65

0.6 0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.55

0.6

Length L, meters

From the graph, we estimate that the frequency is 0.7 Hz when L = 0.46 m. Using software designed to solve nonlinear algebraic equations, we obtain L = 0.4606 m. L = 0.461 m.

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Problem 21.9 The spring constant is k = 785 N/m. The spring is unstretched when x = 0. Neglect the mass of the pulley, that is, assume that the tension in the rope is the same on both sides of the pulley. The system is released from rest with x = 0. Determine x as a function of time.

Solution:

We have the equations

T − (4 kg)(9.81 m/s 2 ) − (785 N/m) x = (4 kg) x T − (20 kg)(9.81 m/s 2 ) = −(20 kg) x If we eliminate the tension T from these equations, we find (24 kg) x + (785 N/m) x = (16 kg)(9.81 m/s 2 ) x + (5.72 rad/s) 2 x = 6.54 m/s 2 . The solution of this equation is x = A sin ωt + B cos ωt + (0.200 m), v = Aω cos ωt − Bω sin ωt. Using the initial conditions, we have x (t = 0) = B + (0.200 m) ⇒ B = −0.200 m, v (t = 0) = Aω = 0 ⇒ A = 0.

20 kg

4 kg

Thus the equation is x = (0.200 m)(1 − cos[5.72 rad/s t ]).

Problem 21.10 The spring constant is k = 785 N/m. The spring is unstretched when x = 0. The radius of the pulley is 125 mm, and moment of inertia about its axis is I = 0.05 kg-m 2 . The system is released from rest with x = 0. Determine x as a function of time. Solution:

Let T1 be the tension in the rope on the left side, and T2 be the tension in the rope on the right side. We have the equations T1 − (4 kg)(9.81 m/s 2 ) − (785 N/m) x = (4 kg) x

4 kg

T2 − (20 kg)(9.81 m/s 2 ) = −(20 kg) x (T2 − T1 )(0.125 m) =

(0.05 kg − m 2 )

x (0.125 m)

20 kg x

If we eliminate the tensions T1 and T2 from these equations, we find x + (785 N/m) x = (16 kg)(9.81 m/s 2 ) (27.2 kg) x + (5.37 rad/s) 2 x = 5.77 m/s 2 . The solution of this equation is x = A sin ωt + B cos ωt + (0.200 m), v = Aω cos ωt − Bω sin ωt. Using the initial conditions, we have x (t = 0) = B + (0.200 m) ⇒ B = −0.200 m, v (t = 0) = Aω = 0 ⇒ A = 0. Thus the equation is x = (0.200 m)(1 − cos [5.37 rad/st ]).

606

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Problem 21.11 A “bungee jumper” who weighs 160 lb leaps from a bridge above a river. The bungee cord has an unstretched length of 60 ft, and it stretches an additional 40 ft before the jumper rebounds. Model the cord as a linear spring. When his motion has nearly stopped, what are the period and frequency of his vertical oscillations? (You can’t model the cord as a linear spring during the early part of his motion. Why not?) Solution:

Use energy to find the spring constant

T1 = 0, V1 = 0, T2 = 0, V2 = −(160 lb)(100 ft) +

1 k (40 ft) 2 2

T1 + V1 = T2 + V2 ⇒ k = 20 lb/ft

Source: Courtesy of Judith Lienert/Shutterstock.

1 k 1 20 lb/ft = = 0.319 Hz, 2π m 2π (160 lb)/(32.2 ft/s 2 ) 1 τ = = 3.13 s. f f =

The cord cannot be modeled as a linear spring during the early motion because it is slack and does not support a compressive load.

Problem 21.12 The spring constant is k = 800 N/m, and the spring is unstretched when x = 0. The mass of each object is 30 kg. The inclined surface is smooth. Neglect the mass of the pulley. The system is released from rest with x = 0. (a) Determine the frequency and period of the resulting vibration. (b) What is the value of x at t = 4 s?

k 208

Solution:

The equations of motion are

T − mg sin θ − kx = mx , T − mg = −mx . If we eliminate T, we find 2mx + kx = mg(1 − sin θ), x +

1 k x = g(1 − sin θ), 2m 2

1  800 N/m  1 x +   x = (9.81 m/s 2 )(1 − sin 20 °), 2  30 kg  2 x + (3.65 rad/s) 2 x = 3.23 m/s 2 . (a) The natural frequency, frequency, and period are ω 1 ω = 3.65 rad/s, f = = 0.581 Hz, τ = = 1.72 s. 2π f f = 0.581 s, τ = 1.72 s. (b) The solution to the differential equation is x = A sin ωt + B cos ωt + 0.242 m, v = Aω cos ωt − Bω sin ωt. Putting in the initial conditions, we have x (t = 0) = B + 0.242 m = 0 ⇒ B = −0.242 m, v (t = 0) = Aω = 0 ⇒ A = 0. Thus the equation is x = (0.242 m)(1 − cos[3.65 rad/st]) At t = 4 s we have x = 0.351 m.

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Problem 21.13 The spring constant is k = 800 N/m, and the spring is unstretched when x = 0. The mass of each object is 30 kg. The inclined surface is smooth. The radius of the pulley is 120 mm and its moment of inertia is I = 0.03 kg-m 2 . At t = 0, x = 0 and dx /dt = 1 m/s. (a) Determine the frequency and period of the resulting vibration. (b) What is the value of x at t = 4 s?

Solution: Let T1 be the tension in the rope on the left of the pulley, and T2 be the tension in the rope on the right of the pulley. The equations of motion are x T1 − mg sin θ − kx = mx , T2 − mg = −mx , (T2 − T1 )r = I . r If we eliminate T1 and T2 , we find

( 2m + rI ) x + kx = mg(1 − sin θ), 2

x +

mgr 2 kr 2 x = x + (3.59 rad/s) 2 (1 − sin θ), 2mr 2 + I 2mr 2 + I x = 3.12 m/s 2 .

(a) The natural frequency, frequency, and period are k

ω = 3.59 rad/s, f =

ω 1 = 0.571 Hz, τ = = 1.75 s. 2π f

f = 0.571 s,

208

τ = 1.75 s.

(b) The solution to the differential equation is x = A sin ωt + B cos ωt + 0.242 m, v = Aω cos ωt − Bω sin ωt. Putting in the initial conditions, we have x (t = 0) = B + 0.242 m = 0 ⇒ B = −0.242 m, 1 m/s v (t = 0) = Aω = (1 m/s) ⇒ A = = 0.279 m. 3.59 rad/s Thus the equation is x = (0.242 m)(1 − cos[359 rad/st ]) + (0.279 m)sin[359 rad/s t] At t = 4 s we have x = 0.567 m.

Problem 21.14 The 20-lb disk rolls on the horizontal surface. Its radius is R = 6 in. Determine the spring constant k so that the frequency of vibration of the system relative to its equilibrium position is f = 1 Hz.

Solution:

Given m =

The equations of motion  d 2x  dt 2  ⇒ d 2 x + 2k x = 0 2x  dt 2 3m 1 1 d  FR = − mR 2 2 R dt 2 

−kx + F = m

(

( )

)

We require that f =

Problem 21.15 The 20-lb disk rolls on the horizontal surface. Its radius is R = 6 in. The spring constant is k = 15 lb/ft. At t = 0, the spring is unstretched and the disk has a clockwise angular velocity of 2 rad/s. What is the amplitude of the resulting vibrations of the center of the disk?

20 lb , R = 6 in 32.2 ft/s 2

Solution:

1 2k = 1 Hz ⇒ k = 36.8 lb/ft. 2π 3m

See the solution to 21.14

( )

d 2x 2k d 2x + x = 0 ⇒ + (4.01 rad/s) 2 x = 0 dt 2 3m dt 2 The solution is x = A cos([4.01 rad/s]t ) + B sin([4.01 rad/s]t ) Using the initial conditions 0 = A

(2 rad/s)(6 in) = B(4.01 rad/s) Thus

x = (2.99 in)sin([4.01 rad/s]t)

The amplitude is

608

} ⇒ A = 0, B = 2.99 in

B = 2.99 in = 0.249 ft.

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Problem 21.16 The 2-lb bar is pinned to the 5-lb disk. The disk rolls on the circular surface. What is the frequency of small vibrations of the system relative to its vertical equilibrium position? Solution: T =

Use energy methods.

(

1 1  2 lb   15  ft 2 3  32.2 ft/s 2   12 

)ω

(

1 1  5 lb   4  2 ft + 2 2  32.2 ft/s 2   12  V = −(2 lb)

2 bar

15 in +

(

1 5 lb 2 32.2 ft/s 2

15 ft ) ω )( 12

2 bar

4 in

)( )

15 2 2 2 ω bar = (0.198 lb-s 2 -ft)ω bar 4

15 ft ) cos θ − (5 lb)( ft ) cos θ = −(7.5 ft-lb) cos θ ( 7.5 12 12

Differentiating and linearizing we find (0.396 lb-s 2 -ft) f =

d 2θ d 2θ + (7.5 ft-lb)θ = 0 ⇒ + (4.35 rad/s) 2 θ = 0 dt 2 dt 2

4.35 rad/s = 0.692 Hz. 2π rad

Problem 21.17 The mass of the suspended object A is 4 kg. The mass of the pulley is 2 kg and its moment of inertia is 0.018 N-m 2 . The spring constant is k = 150 N/m. For vibration of the system relative to its equilibrium position, determine (a) the frequency in Hz and (b) the period.

Solution:

Use energy methods

2 1 1 v (6 kg)v 2 + (0.018 kg-m 2 ) = (3.625 kg)v 2 2 2 0.12 m 1 V = −(6 kg)(9.81 m/s 2 ) x + (150 N/m)(2 x ) 2 2 = (300 N/m) x 2 − (58.9 N) x

(

T =

)

Differentiating we have (7.25 kg) k

120 mm

⇒

d 2x + (600 N/m) x = 58.9 N dt 2

d 2x + (9.10 rad/s) 2 x = 8.12 m/s 2 dt 2

(a)

f =

9.10 rad/s = 1.45 Hz. 2π rad

(b)

τ =

1 = 0.691 s. f

A x

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Problem 21.18 The mass of the suspended object A is 4 kg. The mass of the pulley is 2 kg and its moment of inertia is 0.018 N-m 2 . The spring constant is k = 150 N/m. The spring is unstretched when x = 0. At t = 0, the system is released from rest with x = 0. What is the velocity of the object A at t = 1 s?

Solution:

See the solution to 21.17

The motion is given by x = A cos([9.10 rad/s]t ) + B sin ([9.10 rad/s]t ) + 0.0981 m v =

dx = (9.10 rad/s){− A sin ([9.10 rad/s]t ) + B cos([9.10 rad/s]t)} dt

Use the initial conditions A + 0.0981 m = 0 B(0.910 rad/s) = 0

} ⇒ A = −0.0981 m, B = 0

Thus we have k

x = (0.0981 m)(1 − cos([9.10 rad/s]t )) v =

dx = −(0.892 m/s)sin ([9.10 rad/s]t ) dt v = 0.287 m/s.

At t = 1 s, 120 mm

A x

Problem 21.19 The suspended mass m = 12 kg and the mass of the disk is m d = 6 kg. The radius R = 0.2 m. The coordinate x measures the displacement of the mass relative to its position when the spring is unstretched. Determine the value of the spring constant k so that the frequency of vertical oscillations of the suspended mass is 2 Hz.

We know for this conservative system that

(

)

d d 1 1 1 (T + V ) = m + m d v 2 − mgx + kx 2   2 2 dt dt  2 1 dv = m + md v − mgv + kxv 2 dt = 0.

(

)

Dividing this equation by v, we obtain the equation of motion

( m + 12 m ) ddt x + kx = mg. 2

k R

When the system is stationary and in equilibrium, the acceleration is zero and the equation of motion gives the equilibrium position x = mg /k. Let us define a new variable that is the position relative to the equilibrium position:

mg x = x − . k In terms of this variable, the equation of motion is

m x

d 2 x + ω 2 x = 0, dt 2 where ω2 =

Solution:

Let v = dx /dt. The clockwise angular velocity of the disk is ω = v /R. The moment of inertia of the disk is I = (1/2)m d R 2 . The kinetic energy of the system is 1 2 1 mv + I ω 2 2 2 1 1 = m + md v 2 . 2 2

k m+

1 m 2 d

The frequency is f =

ω 1 = 2π 2π

T =

(

)

Placing the datum of the suspended mass at x = 0, the potential energy of the system is 1 V = −mgx + kx 2 . 2

610

k m+

1 m 2 d

Setting f = 2 Hz and solving, we obtain k = 2370 N/m. k = 2.37 kN/m.

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Problem 21.20 The suspended mass m = 12 kg and the mass of the disk is m d = 6 kg. The spring constant k = 1200 N/m. The radius R = 0.2 m. The coordinate x measures the displacement of the mass relative to its position when the spring is unstretched. The system is released from rest with x = 0. What is the velocity of the suspended mass at t = 1 s? k

Dividing this equation by v, we obtain the equation of motion

( m + 12 m ) ddt x + kx = mg. 2

d 2 x + ω 2 x = 0, dt 2

where ω =

m x

Solution:

Let v = dx /dt. The clockwise angular velocity of the disk is ω = v /R.

The moment of inertia of the disk is I = (1/2)m d R 2 . The kinetic energy of the system is 1 2 1 mv + I ω 2 2 2 1 1 = m + md v 2 . 2 2

T =

(

)

1 m 2 d

= 8.94 rad/s.

The solution for x is x = A sin ωt + B cos ωt and the velocity is dx = Aω cos ωt − Bω sin ωt. dt At t = 0, x = −mg /k and dx /dt = 0. Using these conditions to determine A and B, we obtain A = 0 and B = −mg /k, so the velocity as a function of time is mgω dx = sin ωt. dt k At t = 0 we obtain

Placing the datum of the suspended mass at x = 0, the potential energy of the system is V = −mgx +

k m+

1 2 kx . 2

mgω (12 kg)(9.81 m/s 2 )(8.94 rad/s) dx = sin ωt = sin(8.94 rad/s)(1 s) dt k 1200 N/m = 0.406 N/s. 0.406 N/s downward.

We know for this conservative system that

(

)

d d 1 1 1 (T + V ) = m + m d v 2 − mgx + kx 2   dt  2 2 2 dt 1 dv = m + md v − mgv + kxv 2 dt 0. =

(

)

Problem 21.21 A slender bar of mass m = 4 kg and length L = 2 m is pinned to a fixed support. A torsional spring of constant k = 60 N-m/rad attached to the bar at the support is unstretched when θ = 0. Determine the frequency of small angular vibrations of the bar.

Solution:

The kinetic energy of the bar is

1 T = I 0ω 2 , 2 Where I 0 = (1/3)mL2 = 5.33 kg-m 2 is the moment of inertia about the support. Placing the datum of the bar’s center of mass at the level of the support, the potential energy of the system is 1 1 V = mg L cos θ + kθ 2 . 2 2

u k L/2

mg k

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21.21 (Continued) We know for this conservative system that

where

d d 1 1 1 (T + V ) = I ω 2 + mg L cos θ + kθ 2   dt dt  2 0 2 2 dω 1 = I 0ω − mg L sin θω + kθω dt 2 = 0.

The frequency is

Dividing this equation by the angular velocity ω and using the smallangle approximation sin θ = θ, we obtain the equation of motion

f =

d 2θ + ω 2θ = 0, dt 2

0.314 Hz.

ω =

k −

1 mgL 2 = 1.97 rad/s. I0

ω 1 = 2π 2π

k −

Problem 21.22 A slender bar of mass m = 4 kg and length L = 2 m is pinned to a fixed support. A torsional spring of constant k = 60 N-m/rad attached to the bar at the support is unstretched when θ = 0. At t = 0, the bar is displaced through an angle θ = 0.1 rad and released from rest. Determine the value of θ at t = 6 s.

1 mgL 2 = 0.314 Hz. I0

L/2

mg k

u The equation for θ as a function of time is θ = A sin ωt + B cos ωt, k

Solution: T =

The kinetic energy of the bar is

1 I ω 2, 2 0

and the angular acceleration is dθ = Aω cos ωt − Bω sin ωt. dt Substituting the initial conditions θ = 0.1 rad and d θ /dt = 0 at t = 0, we obtain the equations 0.1 rad = 0 + B, 0 = Aω,

Where I 0 = (1/3)mL2 = 5.33 kg-m 2 is the moment of inertia about the support.

so the angle as a function of time is

Placing the datum of the bar’s center of mass at the level of the support, the potential energy of the system is

At t = 6 s this yields θ = 0.0746 rad.

1 1 V = mg L cos θ + kθ 2 . 2 2

θ = 0.0746 rad.

θ = (0.1 rad) cos 1.97t.

We know for this conservative system that 1 1 d d 1 (T + V ) = I ω 2 + mg L cos θ + kθ 2   2 2 dt dt  2 0 1 dω = I 0ω − mg L sin θω + kθω 2 dt = 0. Dividing this equation by the angular velocity ω and using the smallangle approximation sin θ = θ, we obtain the equation of motion d 2θ + ω 2θ = 0, dt 2 where ω =

612

k −

1 mgL 2 = 1.97 rad/s. I0

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Problem 21.23 Engineers use the device shown to measure an astronaut’s moment of inertia. The horizontal board is pinned at O and supported by the linear spring with constant k = 12 kN/m. When the astronaut is not present, the frequency of small vibrations of the board about O is measured and determined to be 6.0 Hz. When the astronaut is lying on the board as shown, the frequency of small vibrations of the board about O is 2.8 Hz. What is the astronaut’s moment of inertia about the z-axis? y

Mb Fs

O 1.90 m

1.9 u Fs – k (1.9)

O 1.90 m

x9 x

O 1.90 m

Solution:

When the astronaut is not present: Let Fs be the spring force and M b be the moment about 0 due to the board’s weight when the system is in equilibrium. The moment about 0 equals zero, ΣM (pt 0) = (1.9) Fs − M b = 0(1). When the system is in motion and displaced by a small counterclockwise angle θ, the spring force decreases to Fs − k(1.9θ): The equation of angular motion about 0 is d 2θ ΣM (pt0) = (1.9)[ Fs − k (1.9θ)] − M b = I b 2 , dt

where I b is the moment of inertial of the board about the z axis. Using Equation (1), we can write the equation of angular motion as d 2θ + ω 12θ = 0, dt 2

where ω 12 =

(12,000)(1.9) 2 k (1.9) 2 = . Ib Ib

We know that f1 = ω 1 /2 π = 6 Hz, so ω 1 = 12π = 37.7 rad/s and we can solve for I b : I b = 30.48 kg-m 2 . When the astronaut is present: Let Fs be the spring force and M ba be the moment about 0 due to the weight of the board and astronaut when the system is in equilibrium. The moment about 0 equals zero, ΣM ( pt 0) = (1.9) Fs − M ba = 0(2). When the system is in motion and displaced by a small counterclockwise angle θ, the spring force decreases to Fs − k(1.9 θ): The equation of angular motion about 0 is ΣM (pt0) = (1.9)[ Fs − k (1.9 θ)] − M ba = ( I b + I a )

where I a is the moment of inertia of the astronaut about the z axis. Using Equation (2), we can write the equation of angular motion as d 2θ + ω 22θ = 0, dt 2

From the solution of Problem 21.23, his moment of inertial about the z axis is I z = 109.48 kg-m 2 . From the parallel-axis theorem, I z′ =

I z − (d x2

+ d y2 )m

= 109.48 − [(1.01) 2

where ω 22 =

k (1.9) 2 (12,000)(1.9) 2 = . Ib + Ia Ib + Ia

In this case f 2 = ω 2 /2π = 2.8 Hz, so ω 2 = 2.8(2π ) = 17.59 rad/s. Since we know I b , we can determine I a , obtaining I a = 109.48 kg-m 2 .

Problem 21.24 In Problem 21.23, the astronaut’s center of mass is at x = 1.01 m, y = 0.16 m, and his mass is 81.6 kg. What is his moment of inertia about the z ′-axis through his center of mass? Solution:

d 2θ 2 , dt

x9 x

O 1.90 m

+ (0.16) 2 ](81.6)

= 24.2 kg-m 2 .

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Problem 21.25 A floating sonobuoy (sound-measuring device) is in equilibrium in the vertical position shown. (Its center of mass is low enough that it is stable in this position.) The device is an 18-kg cylinder 1.2 m in length and 150 mm in diameter. The water density is 1024 kg/m 3 . The buoyancy force supporting the buoy equals the weight of the volume of water that would occupy the volume of the part of the cylinder below the surface. If you push the sonobuoy slightly deeper and release it, what is the frequency of the resulting vertical vibrations? Solution:

Denote the mass of the cylinder by m = 18 kg, and let x denote the length of the cylinder that is under water:

Newton’s second law for the cylinder is therefore m

d 2x = mg − ρ gAx. dt 2

From this equation we can see that the equilibrium position is x = m /(ρ A). Let us define a new variable that is the position relative to the equilibrium position: m x = x − . ρA

In terms of this variable, the equation of motion is d 2 x + ω 2 x = 0, dt 2 (Notice that the coordinate x is also a measure of the position of the cylinder.) The diameter of the cylinder D = 0.15 m. Its cross-sectional area is A = π D 2 /4 = 0.0177 m 2 . The volume of the cylinder that is under water is V = Ax. In terms of the density ρ = 1024 kg/m 3 of the water, the (upward) buoyancy force on the cylinder is

where ω =

ρ gA = 3.14 rad/s. m

The frequency is f =

ρVg = ρ gAx.

ω = 0.500 Hz. 2π

0.500 Hz.

Problem 21.26 The disk rotates in the horizontal plane with constant angular velocity Ω = 12 rad/s. The mass m = 2 kg slides in a smooth slot in the disk and is attached to a spring with constant k = 860 N/m. The radial position of the mass when the spring is unstretched is r = 0.2 m. (a) Determine the “equilibrium” position of the mass, the value of r at which it will remain stationary relative to the center of the disk. (b) What is the frequency of vibration of the mass relative to its equilibrium position? Strategy: Apply Newton’s second law to the mass in terms of polar coordinates.

Solution:

Using polar coordinates, Newton’s second law in the r

direction is

(

)

k k ΣFr : −k (r − r0 ) = m(r − rΩ 2 ) ⇒ r + − Ω 2 r = r0 m m (a) The “equilibrium” position occurs when r = 0 req =

k r m 0

k − Ω2 m = 0.301 m.

kr0 (860 N/m)(0.2 m) = k − mΩ 2 (860 N/m) − (2 kg)(12 rad/s) 2

req = 0.301 m. (b) The frequency of vibration is found k 860 N/m − Ω2 = − (12 rad/s) 2 = 16.9 rad/s, m 2 kg ω f = = 2.69 Hz. 2π

Ω

ω = m

f = 2.69 Hz.

k r

614

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Problem 21.27 The disk rotates in the horizontal plane with constant angular velocity Ω = 12 rad/s. The mass m = 2 kg slides in a smooth slot in the disk and is attached to a spring with constant k = 860 N/m. The radial position of the mass when the spring is unstretched is r = 0.2 m. At t = 0, the mass is in the position r = 0.4 m and dr /dt = 0. Determine the position r as a function of time. Ω

Solution:

Using polar coordinates, Newton’s second law in the r

direction is ΣFr : −k (r − r0 ) = m(r − rΩ 2 ), r +

( mk − Ω )r = mk r 2

r + (16.9 rad/s) 2 r

= 86 m/s 2 .

The solution is r = A sin ωt + B cos ωt + 0.301 m,

dr = Aω cos ωt − Bω sin ωt. dt

Putting in the initial conditions, we have r (t = 0) = B + 0.301 m = 0.4 m ⇒ B = 0.0993 m,

dr (t = 0) = Aω = 0 ⇒ A = 0. dt

Thus the equation is

r = (0.0993 m) cos[16.9 rad/s t] + (0.301 m).

Problem 21.28 A homogeneous 100-lb disk with radius R = 1 ft is attached to two identical cylindrical steel bars of length L = 1 ft. The relation between the moment M exerted on the disk by one of the bars and the angle of rotation, θ, of the disk is M =

GJ θ, L

Solution:

The moment exerted by two bars on the disk is M = kθ, where the spring constant is k =

∂M 2GJ = . L ∂θ

The polar moment of the cross section of a bar is J =

πr 4 , 2

from which k =

where J is the polar moment of inertia of the cross section of the bar and G = 1.7 × 10 9 lb/ft 2 is the shear modulus of the steel. Determine the required radius of the bars if the frequency of rotational vibrations of the disk is to be 10 Hz.

From the equation of angular motion, I

d 2θ = −kθ. dt 2

The moment of inertia of the disk is I =

πr 4 G . L

W 2 R , 2g d

from which d 2θ + ω 2θ = 0, dt 2

Solve:

r =

where ω =

 r 2  2πGg 2πGgr 4 =  .   R d  WL WLR d2

WL ω = 2πGg

2πWL f. Gg

Substitute numerical values: R d = 1 ft, L = 1 ft, W = 100 lb, G = 1.7 × 10 9 lb/ft 2 , g = 32.17 ft/s 2 , from which r = 0.03274 ft = 0.393 in. L

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Problem 21.29 The moments of inertia of gears A and B are I A = 0.1 kg-m 2 and I B = 0.4 kg-m 2 . Gear A is connected to a torsional spring with constant k. If the frequency of angular oscillation of the gears is f = 2 Hz, what is the value of k ?

Its potential energy is V =

1 kθ 2 . 2 A

We evaluate the derivative

(

)

d d 1 1 1 (T + V ) = I ω 2 + I B ω B 2 + kθ A 2 dt dt 2 A A 2 2 dω A dω B dθ = I Aω A + I Bω B + kθ A A dt dt dt = 0 and divide the resulting equation by ω A , obtaining the equation

200 mm A

d 2θ A + ω 2θ A = 0, dt 2 where ω =

140 mm

IA +

k . rA 2 I rB 2 B

The frequency is

Solution: Let r A = 0.14 m and r A = 0.20 m. Let θ A be the counterclockwise angular displacement and θ B the clockwise angular displacement of the gears relative to their positions when the spring is unstretched. Because of the contact of the gears, the angular displacements, angular velocities and angular accelerations are related by r A θ A = rB θ B , r A ω A = r B ω B , r A α A = r B α B .

f =

ω 1 = 2π 2π

IA +

k . rA 2 I rB 2 B

Setting f = 2 Hz and solving, we obtain k = 46.7 N/m. k = 46.7 N/m.

The kinetic energy of the system is T =

1 1 I ω 2 + I Bω B 2 . 2 A A 2

Problem 21.30 The moments of inertia of gears A and B are I A = 0.1 kg-m 2 and I B = 0.4 kg-m 2 . Gear A is connected to a torsional spring with constant k = 20 N-m/rad. If the spring is unstretched at t = 0 and gear A is given a counterclockwise angular velocity of 30 rad/s, what is the angle of rotation at t = 4 s of gear B relative to its position at t = 0?

Its potential energy is 1 V = kθ A 2 . 2 We evaluate the derivative 1 1 d d 1 (T + V ) = I ω 2 + I B ω B 2 + kθ A 2 2 2 dt 2 A A dt dω A dω B dθ = I Aω A + I B ωB + kθ A A dt dt dt = 0

(

)

and divide the resulting equation by ω A , obtaining the equation d 2θ A + ω 2θ A = 0, dt 2 where ω = 200 mm A

140 mm

Solution: Let r A = 0.14 m and r A = 0.20 m. Let θ A be the counterclockwise angular displacement and θ B the clockwise angular displacement of the gears relative to their positions when the spring is unstretched. Because of the contact of the gears, the angular displacements, angular velocities and angular accelerations are related by r A θ A = r B θ B , r A ω A = r B ω B , r A α A = rB α B . The kinetic energy of the system is 1 1 T = I Aω A 2 + I B ω B 2 . 2 2

616

IA +

k = 8.22 rad/s. rA 2 IB 2 rB

The angle of gear A as a function of time is θ A = A sin ωt + B cos ωt, and the angular acceleration is dθ A = Aω cos ωt − Bω sin ωt. dt Substituting the initial conditions θ A = 0 and d θ A /dt = 30 rad/s at t = 0, we obtain the equations 0 = 0 + B, 30 rad/s = Aω, so the angle as a function of time is 30 rad/s θA = sin8.22t. 8.22 rad/s At t = 4 s we obtain θ A = 3.63 rad, and the rotation of gear B is θ B = (r A /rB )θ A = 2.54 rad. 2.54 rad clockwise.

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Problem 21.31 Each 2-kg slender bar is 1 m in length. What are the period and frequency of small vibrations of the system? u

so the (linearized) equation of motion is

Solution:

6g d 2θ + θ = 0. dt 2 5L

The total energy is

1 1 2 dθ 2 1 1 2 dθ 2 mL ml + 2 3 2 3 dt dt 2 1 dθ L L + m L − mg cos θ − mg cos θ − mgL cos θ 2 2 2 dt 5 2 dθ 2 = mL − 2mgL cos θ. dt 6 2 d dθ d θ dθ 5 (T + V ) = mL2 + 2mgL sin θ = 0, dt dt dt 2 dt 3 T +V =

(

)( ) ( ) ( )

)( )

(

Problem 21.32* The masses of the slender bar and the homogeneous disk are m and m d , respectively. The spring is unstretched when θ = 0. Assume that the disk rolls on the horizontal surface. (a) Show that the motion of the system is governed by the equation

( 13 + 32mm cos θ ) ddt θ − 32mm sin θ cos θ( ddtθ ) d

−

g k sin θ + (1 − cos θ) sin θ = 0. 2l m

(b) If the system is in equilibrium at the angle θ = θ e and θ = θ − θ e, show that the equation governing small vibrations relative to the equilibrium position is 3m d d 2 θ 1 cos 2 θ e + dt 2 3 2m g k +  (cos θ e − cos 2 θ e + sin 2 θ e ) − cos θ e  θ = 0. m  2l

(

)

(See Example 21.3.)

Therefore ω =

6g = 5L

so τ =

2π = 1.83 s, f = 0.546 Hz. ω

Solution:

6(9.81) = 3.43 rad/s, 5(1)

(See Example 21.2.) The system is conservative.

(a) The kinetic energy is T =

1 1 1 1 I θ + mv 2 + I d θ d2 + m dv d2 , 2 2 2 2 2

mL2 is the moment of inertia of the bar about its center 12 mR 2 of mass, v is the velocity of the center of mass of the bar, I d = 2 is the polar moment of inertia of the disk, and v d is the velocity of the center of mass of the disk. The height of the center of mass of the bar L cos θ is h = , and the stretch of the spring is S = L(1 − cos θ), 2 from which the potential energy is where I =

V =

mgL 1 cos θ + kL2 (1 − cos θ) 2 . 2 2

The system is conservative, v L cos θ T + V = const. θ l d = = θ. R R Choose a coordinate system with the origin at the pivot point and the x axis parallel to the lower surface. The instantaneous center of rotation of the bar is located at ( L sin θ, L cos θ). The center of mass of the L L bar is located at sin θ, cos θ . The distance from the center of 2 2 mass to the center of rotation is

(

r =

)

( L − L2 ) sin θ + ( L − L2 ) cos θ = L2 , 2

L θ. The velocity of the center of mass of 2 the disk is v d = ( L cos θ)θ . The angular velocity of the disk is v L cos θ θ l d = = θ. R R

from which v = l u

Substitute and reduce:

(

)

3m d g 1 1 k + cos 2 θ 2 + cos θ + (1 − cos θ) 2 = const. 2 3 2m 2L 2m R

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21.32* (Continued)

Substitute and reduce: (1) k

( 13 + 32mm cos θ ) ddt θ → ( 13 + 3mm θ sin θ cos θ ) 3m d θ 1 d θ × → ( + cos θ ) . dt 3 2m dt d

(2) = − L

( ) → − 32mm θ sin θ cos θ

3m d dθ cos θ sin θ 2m dt

2  d θ  ×   → 0.  dt 

u R

(3) − (4)

g g g sin θ → − θ cos θ e − sin θ e . 2L 2L 2L

k k (1 − cos θ)sin θ → (1 − cos θ e )sin θ e m m k θ (cos θ e − cos 2 θ e + sin 2 θ e ), m 2  d θ  d 2θ where the terms θ 2 → 0, θ   → 0, and terms in θ 2 have   dt dt been dropped.

L sin u

L cos u

Collect terms in (1) to (4) and substitute into the equations of motion:

( 13 + 32mm cos θ ) d θ +  k (cos θ − cos θ + sin θ )

sin θ e

Take the time derivative:

(

)

(

g k sin θ + (1 − cos θ)sin θ = 0 2L m g (b) The non-homogenous term is , as can be seen by dividing the 2L equation of motion by sin θ ≠ 0, −

( 13 + 32mm cos θ ) d θ − 3m cos θ dθ ( dt ) dt sin θ 2m d

 m sin θ e g cos θ e  g k  − (1 − cos θ e )  . −  =   2L  m 2 L sin θ e  dt 2

g k − (1 − cos θ e ) = 0, as shown by 2L m mg substituting the value cos θ e = 1 − , from which 2 Lk 3m d 1 d 2θ + cos 2 θ e 3 2m dt 2 g k +  (cos θ e − cos 2 θ e + sin 2 θ e ) − cos θ e  θ = 0 m  2L

( )

from which d

The term on the right

3m d 3m d  1 d 2θ dθ 2 θ  + − cos 2 θ sin θ cos θ 2  3 2m 2m dt dt g k  − sin θ + (1 − cos θ)sin θ  = 0,  2L m

( 13 + 32mm cos θ ) ddt θ − 32mm sin θ cos θ ( ddtθ )

)

is the equation of motion for small amplitude oscillations about the equilibrium point.

g k (1 − cos θ) = . m 2L

Since the non-homogenous term is independent of time and angle, the equilibrium point can be found by setting the acceleration mg and the velocity terms to zero, cos θ e = 1 − . [Check: This 2 Lk is identical to Eq. (21.15) in Example 21.2, as expected. check.] Denote θ = θ − θ e . For small angles: cos θ = cos θ cos θ e − sin θ sin θ e → cos θ e − θ sin θ e . cos 2 θ → cos 2 θ − 2θ sin θ cos θ . e

sin θ = sin θ cos θ e + cos θ sin θ e → θ cos θ e + sin 2 θ e . (1 − cos θ)sin θ → (1 − cos θ )sin θ + θ (cos θ − cos θ e

+ sin 2 θ e ). sin θ cos θ → θ (cos 2 θ e − sin 2 θ e ).

618

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Problem 21.33* The masses of the bar and disk in Problem 21.32 are m = 2 kg and m d = 4 kg, respectively. The dimensions l = 1 m and R = 0.28 m, and the spring constant is k = 70 N/m. (a) Determine the angle θ e at which the system is in equilibrium. (b) The system is at rest in the equilibrium position, and the disk is given a clockwise angular velocity of 0.1 rad/s. Determine θ as a function of time. (See Example 21.3.)

Solution: (a) From the solution to Problem 21.32, the static equilibrium angle is

(

θ e = cos −1 1 −

)

mg = 30.7 ° = 0.5358 rad. 2kL

(b) The canonical form (see Eq. (21.4)) of the equation of motion is d 2θ + ω 2θ = 0, where dt 2 g k (cos θ e − cos 2 θ e + sin 2 θ e ) − cos θ e m 2L 3m d 1 cos 2 θ e + 3 2m = 1.891 rad/s.

ω =

(

)

Assume a solution of the form θ = θ − θ e = A sin ωt + B cos ωt, from which θ = θ e + A sin ωt + B cos ωt. l

From the solution to Problem 21.32, the angular velocity of the disk is

v ( L cos θ e ) θ d = d = θ. R R The initial conditions are t = 0, θ = θ e , θ = R

Rθ d (0.1)(0.28) = = 0.03256 rad/s, L cos θ e 0.86

from which B = 0, A = 0.03256/ω = 0.01722, from which the solution is θ = 0.5358 + 0.01722sin(1.891t ).

Problem 21.34 The mass of each slender bar is 1 kg. If the frequency of small vibrations of the system is 0.935 Hz, what is the mass of the object A?

V =

mgL mgL (1 − cos θ) + (1 − cos θ) + mgL (1 − cos θ) 2 2 + MgL A (1 − cos θ),

V = ( MgL A + 2mgL )(1 − cos θ). 1 (2 I θ 2 + mv 2 + Mv A2 ) + ( MgL A + 2mgL ) 2 × (1 − cos θ).

T + V = const. =

280 mm

350 mm A

350 mm

Solution:

The system is conservative. Denote L = 0.350 m, L A = 0.280 m, m = 1 kg, and M the mass of A. The moments of inertia about the fixed point is the same for the two vertical bars. The kinetic energy is T =

1 2 1 1 1 I θ + I θ 2 + mv 2 + Mv 2A , 2 2 2 2

where v is the velocity of the center of mass of the lower bar and v A is the velocity of the center of mass of A, from which 1 T = (2 I θ 2 + mv 2 + Mv 2A ). The potential energy is 2

From kinematics, v = L cos θ(θ ), and v A = L A cos θ(θ ). Substitute: 1 (2 I + (mL2 + ML2A ) cos 2 θ)θ 2 + ( MgL A + 2mgL )(1 − cos θ) 2 = const.

(

M21_BEDF2222_06_SE_C21_Part1.indd 619

)

MgL A 1 (2 I + mL2 + ML2A )θ 2 + + mgL θ 2 = const. 2 2 Take the time derivative: d 2θ θ  (2 I + mL2 + ML2A ) 2 + ( MgL A + 2mgL )θ  = 0.   dt Ignore the possible solution θ = 0, from which d 2θ + ω 2θ = 0, dt 2

where ω =

2mgL + MgL A . 2 I + mL2 + ML2A

The moment of inertia of a slender bar about one end (the fixed point) is mL2 I = , from which 3 ω =

θ2 , from which 2

For small angles: cos 2 θ → 1, (1 − cos θ) →

g(2 + η ) , 5 L + ηL A 3

where η =

ML A . mL

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21.34 (Continued)

u A

The frequency is f =

3g(2 + η ) = 0.935 Hz. 5 L + 3η L A

ω 1 = 2π 2π

Solve: η =

5Lω 2 − 6g = 3.502, 3( g − L A ω 2 ) M =

from which

mL (3.502) = 4.38 kg. LA

Problem 21.35* The 4-kg slender bar is 2 m in length. It is held in equilibrium in the position θ 0 = 35° by a torsional spring with constant k. The spring is unstretched when the bar is vertical. Determine the period and frequency of small vibrations of the bar relative to the equilibrium position shown. Solution:

The total energy is

1 1 2 dθ 2 1 L mL + kθ 2 + mg cos θ. 2 3 dt 2 2 d 1 2 d θ d 2θ dθ L dθ (T + v ) = mL + kθ − mg sin θ = 0, dt dt dt 2 dt dt 3 2 T +v =

)( )

(

so the equation of motion is 3g d 2θ 3k + θ− sin θ = 0. dt 2 mL2 2L

(1)

Let θ 0 = 35 ° be the equilibrium position: 3g 3k θ − sin θ 0 = 0. mL2 0 2L

u (2)

Solving for k, k=

mgL sin θ 0 (4)(9.81)(2) sin 35 ° = = 36.8 N-m. 2 2 (35π /180) θ0

Let θ = θ − θ 0 . Then sin θ = sin(θ 0 + θ ) = sin θ 0 + (cos θ 0 )θ + Using this expression and Eq. (2), Eq. (1) (linearized) is 3g d 2θ 3k + − cos θ 0 θ = 0. dt 2 mL2 2L

(

)

Therefore ω =

3g 3k − cos θ 0 mL2 2L

(3)(36.8) (3)(9.81) − cos35 ° (4)(2) 2 (2)(2)

= 0.939 rad/s, 2π = 6.69 s, ω 1 f = = 0.149 Hz. τ

τ = so

620

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Problem 21.36 The mass m = 2 slug, the spring constant is k = 72 lb/ft, and the damping constant is c = 8 lb-s/ft. The spring is unstretched when x = 0. The mass is displaced to the position x = 1 ft and released from rest. (a) If the damping is subcritical, what is the frequency of the resulting damped vibrations? (b) What is the value of x at t = 1 s? (See Practice Example 21.4.)

Solution:

The equation of motion is

c k 8 72 mx + cx + kx = 0, x + x + x = 0, x + x + x = 0, m m 2 2 2 x + 2(2) x + (6) x = 0. We recognize ω = 6, d = 2, d < ω ⇒ yes, the motion is subcritical. ω , (a) ω d = ω 2 − d 2 = 6 2 − 2 2 = 5.66 rad/s, f = 2π f = 0.900 Hz. (b) The solution to the differential equation is x = e −dt ( A sin ω d t + B cos ω d t )

x c

Putting in the initial conditions we have

dx = e −dt ([ Aω d − Bd ]cos ω d t − [ Bω d + Ad ]sin ω d t ) dt

x (t = 0) = B = 1 ft, dx d 2 (t = 0) = ( Aω d − Bd ) = 0 ⇒ A = B = (1 ft) dt ωd 5.66 = 0.354 ft. The equation of motion is now x = e −2t (0.354 sin[5.66t ] + cos[5.66t ]). At t = 1 s, we have x = 0.0816 ft.

Problem 21.37 The mass m = 2 slug, the spring constant is k = 72 lb/ft, and the damping constant is c = 32 lb-s/ft. The spring is unstretched when x = 0. The mass is displaced to the position x = 1 ft and released from rest. (a) If the damping is subcritical, what is the frequency of the resulting damped vibrations? (b) What is the value of x at t = 1 s? (See Practice Example 21.4.)

Putting in the initial conditions, we have x (t = 0) = C + D = 1 ft,   dx (t = 0) = −(d − h)C − (d + h) D = 0   dt d + h h−d ⇒ C = ,D = 2h 2h C = 1.26 ft, D = −0.256 ft. The general solution is then

x = (1.26)e −2.71t − (0.256)e −13.3t .

At t = 1 s we have x = 0.0837 ft. x c m

Solution:

The equation of motion is

c k 32 72 mx + cx + kx = 0, x + x + x = 0, x + x + x = 0, m m 2 2 x + 2(8) x + (6) 2 x = 0. (a) We recognize (b) We have

ω = 6, d = 8, d > ω ⇒ Damping is supercritical.

h =

d2 − ω2 =

8 2 − 6 2 = 5.29 rad/s.

The solution to the differential equation is x = Ce −( d − h )t + De −( d + h )t ,

dx = −(d − h)Ce −( d − h )t dt

− (d + h) De −( d + h )t .

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Problem 21.38 The mass m = 4 slug and the spring constant is k = 72 lb/ft. The spring is unstretched when x = 0. (a) What value of the damping constant c causes the system to be critically damped? (b) Suppose that c has the value determined in part (a). At t = 0, x = 1 ft and dx /dt = 4 ft/s. What is the value of x at t = 1 s? (See Practice Example 21.4.)

Solution:

The equation of motion is

c k c 72 mx + cx + kx = 0, x + x + x = 0, x + x + x = 0, m m 4 4 c x + 2 x + (4.24) 2 x = 0. 8

( )

(a) The system is critically damped when c d = = ω = 4.24 ⇒ c = 33.9 lb-s/ft. 8 (b) The solution to the differential equation is dx x = Ce −dt + Dte −dt , = ( D − Cd − Ddt )e −dt . dt Putting in the initial conditions, we have

x c

x (t = 0) = C = 1 ft, dx (t = 0) = D − Cd = 4 ft/s ⇒ D = (1)(4.24) + 4 = 8.24 ft. dt

The general solution is then x = (1 ft)e −4.24 t + (8.24 ft)te −4.24 t . At t = 1 s we have x = 0.133 ft.

Problem 21.39 The mass m = 2 kg, the spring constant is k = 8 N/m, and the damping coefficient is c = 12 N-s/m. The spring is unstretched when x = 0. At t = 0, the mass is released from rest with x = 0. Determine the value of x at t = 2 s.

Solution: (2 kg)

d 2x dt 2

The equation of motion is

+ (12 N-s/m)

dx + (8 N/m) x = (2 kg)(9.81 m/s 2 )sin 20 ° dt

d 2x dx + 2(3 rad/s) + (2 rad/s) 2 x = 3.36 m/s 2 dt 2 dt We identify ω = 2 rad/s, d = 3 rad/s, h =

d 2 − ω 2 = 2.24 rad/s.

Since d > ω, we have a supercritical case. The solution is x = Ce −( d − h )t + De −( d + h )t + 0.839 m v =

c x k

dx = −C (d − h)e −( d − h )t − D(d + h)e −( d + h )t dt

Using the initial conditions we have 0 = C + D + 0.839 m   ⇒ C = −0.982 m, D = 0.143 m 0 = −C (d − h) − D(d + h) 

The motion is

208

x = −(0.982 m)e −(0.764 rad/s)t + (0.143 m)e −(5.24 rad/s)t + 0.839 m At t = 2 s

Problem 21.40 The mass m = 0.15 slug, the spring constant is k = 0.5 lb/ft, and the damping coefficient is c = 0.8 lb-s/ft. The spring is unstretched when x = 0. At t = 0, the mass is released from rest with x = 0. Determine the value of x at t = 2 s.

Solution:

x = 0.626 m.

The equation of motion is

dx + (0.8 lb-s/ft) + (0.5 lb/ft) x dt 2 dt 2 = (0.15 slugs)(32.2 ft/s )sin 20 °

(0.15 slugs)

d 2x

d 2x dx + 2(2.67 rad/s) + (1.826 rad/s) 2 x = 11.01 ft/s 2 dt 2 dt We identify ω = 1.83 rad/s, d = 2.67 rad/s, h =

d 2 − ω 2 = 1.94 rad/s

Since d > ω, we have the supercritical case. The solution is x = Ce −( d − h )t + De −( d + h )t + 3.30 ft

c x k m 208

dx = −C (d − h)e −( d − h )t − D(d + h)e −( d + h )t dt Using the initial conditions we find v =

0 = C + D + 3.30 ft

  ⇒ C = −3.92 ft, D = 0.615 ft 0 = −C (d − h) − D(d + h)  Thus the solution is x = −(3.92 ft)e −(0.723 rad/s)t + (0.615 ft)e −(4.61 rad/s)t + 3.30 ft At t = 2 s,

622

x = 2.38 ft.

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Problem 21.41 A 1500-kg test car moving with velocity v 0 = 2 m/s collides with a rigid barrier at t = 0. Engineers model the response of its energyabsorbing bumper by the spring–mass oscillator shown with k = 30 kN/m and c = 20 kN-s/m. Assume that at t = 0 the model mass is moving with velocity v 0 = 2 m/s and the spring is unstretched. (a) Is the damping of the car’s motion subcritical or supercritical? (b) Determine the car’s position at t = 1 s. v0

Solution: (a) Let m = 1500 kg. In terms of the coordinate x shown, the equation of motion for the model mass is m

d 2x dx = −kx − c , dt 2 dt

or, in the notation of Eq. (21.16), m

d 2x dx + 2d + ω 2x = 0 dt 2 dt

where ω = k /m = 4.47 rad/s and d = c /2m = 6.67 s −1 . The car’s motion is supercritically damped. (b) For supercritical damping, the car’s position as a function of time is described by Eq. (21.41), x = Ce −( d − h )t + De −( d + h )t , where h =

d 2 − ω 2 = 4.94 s −1 . The car’s velocity is

dx = −(d − h)Ce −( d − h )t − (d + h) De −( d + h )t . dt

Substituting the initial conditions x = 0 and dx /dt = 2 m/s at t = 0, we obtain the equations

x Car colliding with a rigid barrier

0 = C + D, 2 m/s = −(d − h)C − (d + h) D,

which yield C = 0.2023, D = −0.2023. The displacement as a function of time is

x = 0.2023e −(6.67 − 4.94)t − 0.2023e −(6.67 + 4.94)t .

At t = 1 s we obtain x = 0.0361 m (1.42 in). (a) Supercritical. (b) x = 0.0361 m.

x Simulation model

The car’s displacement is shown:

0.14

Displacement, meters

0.12 0.1 0.08 0.06 0.04 0.02 0

0.5

1.5

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2 t, seconds

2.5

3.5

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Problem 21.42 A 1500-kg test car moving with velocity v 0 = 2 m/s collides with a rigid barrier at t = 0. Engineers model the response of its energyabsorbing bumper by the spring–mass oscillator shown with k = 30 kN/m and c = 20 kN-s/m. Assume that at t = 0 the model mass is moving with velocity v 0 = 2 m/s and the spring is unstretched. Determine the magnitude of the car’s deceleration (a) immediately after it contacts the barrier; (b) at t = 0.1 s.

Solution: (a) Let m = 1500 kg. In terms of the coordinate x shown, the equation of motion for the model mass is m

dx d 2x = −kx − c , dt 2 dt

or, in the notation of Eq. (21.16), m

dx d 2x + 2d + ω 2x = 0 dt 2 dt

x = Ce −( d − h )t + De −( d + h )t , where h =

d 2 − ω 2 = 4.94 s −1 . The car’s velocity is

dx = −(d − h)Ce −( d − h )t − (d + h) De −( d + h )t . dt

Substituting the initial conditions x = 0 and dx /dt = 2 m/s at t = 0, we obtain the equations

x Car colliding with a rigid barrier

0 = C + D, 2 m/s = −(d − h)C − (d + h) D, which yield C = 0.2023, D = −0.2023.

The car’s acceleration is d 2x = (d − h) 2 Ce −( d − h )t + (d + h) 2 De −( d + h )t . dt 2

From this expression we obtain the magnitudes x

(a) 26.7 m/s 2 (2.72 g’s). (b) 8.03 m/s 2 .

Simulation model

Problem 21.43 The motion of the car’s suspension can be modeled by the damped spring–mass oscillator with m = 36 kg, k = 22 kN/m, and c = 2.2 kN-s/m. Assume that no external forces act on the tire and wheel. At t = 0, the spring is unstretched and the tire and wheel are given a velocity dx /dt = 10 m/s. Determine the position x as a function of time.

Since d > ω, the motion is supercritically damped and Equation (21.24) is the solution, where h =

Equation (21.24) is x = Ce −( d − h )t + De −( d + h )t , or

Coil spring

d 2 − ω 2 = 17.96 rad/s.

x = Ce −12.6t + De −48.5t .

The time derivative is

x m

dx = −12.6Ce −12.6t − 48.5 De −48.5t . dt At t = 0, x = 0 and dx /dt = 10 m/s: Hence, 0 = C + D and 10 = −12.6C − 48.5 D. Solving for C and D, we obtain C = 0.278 m, D = −0.278 m. The solution is

x = 0.278(e −12.6t − e −48.5t ) m.

Shock absorber

Solution:

624

Calculating ω and d, we obtain

ω =

k = m

and

d =

22,000 = 24.72 rad/s 36

c 2200 = = 30.56 rad/s. 2m 2(36)

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Problem 21.44 The 4-kg slender bar is 2 m in length. Aerodynamic drag on the bar and friction at the support exert a resisting moment about the pin support of magnitude 1.4(d θ /dt ) N-m, where d θ /dt is the angular velocity in rad/s. (a) What are the period and frequency of small vibrations of the bar? (b) How long does it take for the amplitude of vibration to decrease to one-half of its initial value?

1.4 du N-m dt

This is of the form of Eq. (21.16) with u

d =

4.2 4.2 = = 0.131 rad/s, 2mL2 2(4)(2) 2

ω =

Solution: (a)

3(9.81) = 2.71 rad/s. 2(2)

From Eq. (21.18), ω d = (21.20),

∑ M 0 = I 0α: −1.4

3g = 2L

dθ L 1 − mg sin θ = mL2α. dt 2 3

The (linearized) equation of motion is 3g d 2θ 4.2 d θ + + θ = 0. dt 2 mL2 dt 2L

Problem 21.45 The bar described in Problem 21.44 is given a displacement θ = 2° and released from rest at t = 0. What is the value of θ (in degrees) at t = 2 s?

ω 2 − d 2 = 2.71 rad/s, and from Eqs.

2π = 2.32 s, ωd 1 fd = = 0.431 Hz. τd

τd =

(b) Setting e −dt = e −0.131t = 0.5, we obtain t = 5.28 s.

Solution:

From the solution of Problem 21.44, the damping is subcritical with d = 0.131 rad/s, ω = 2.71 rad/s. From Eq. (21.19), θ = e −0.131t ( A sin 2.71t + B cos 2.71t ),

so d θ = −0.131e −0.131t ( A sin 2.71t + B cos 2.71t ) dt + e −0.131t (2.71 A cos 2.71t − 2.71B sin 2.71t ). At t = 0, θ = 2 ° and d θ /dt = 0. Substituting these conditions, 2 ° = B, 0 = −0.131B + 2.71 A, we see that B = 2 °, A = 0.0969 °, so θ = e −0.131t (0.0969 ° sin 2.71t + 2 ° cos 2.71t ). u

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At t = 2 s, we obtain θ = 0.942 °.

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Problem 21.46 The radius of the pulley is R = 100 mm and its moment of inertia is I = 0.1 kg-m 2 . The mass m = 5 kg, and the spring constant is k = 135 N/m. The cable does not slip relative to the pulley. The coordinate x measures the displacement of the mass relative to the position in which the spring is unstretched. Determine x as a function of time if c = 60 N-s/m and the system is released from rest with x = 0.

m k

Solution: Denote the angular rotation of the pulley by θ. The moment on the pulley is ΣM = R(kx ) − RF , where F is the force acting on the right side of the pulley. From the equation of angular motion for the pulley, I

d 2θ dt 2

F m

= Rkx − RF,

from which F = −

I d 2θ + kx. R dt 2

The force on the mass is −F + f + mg, where the friction force dx f = −c acts in opposition to the velocity of the mass. From dt Newton’s second law for the mass,

dx Apply the initial conditions: at t = 0, x = 0, = 0, from which dt mg 0 = B+ , and 0 = −d [ x c ] t = 0 + ω d A = −dB + ω d A, from k dB which B = −0.3633, A = = −0.3250, and ωd

d 2x dx I d 2θ dx + mg. = −F − c + mg = − kx − e R dt 2 dt dt 2 dt x From kinematics, θ = − , from which the equation of motion for R the mass is

  dmg mg mg x (t ) = e −dt  − sin ω d − cos ω d t  +   kω d k k

( )

x (t ) = e −2t (−0.325sin(2.236 t ) − 0.363cos(2.236 t )) + 0.3633 (m).

( RI + m ) dt + c dxdt + kx = mg. d 2x

The canonical form (see Eq. (21.16)) of the equation of motion is R 2 mg d 2x dx + 2d + ω 2x = , dt 2 dt I + R 2m where d =

cR 2 kR 2 = 2 rad/s, ω 2 = = 9(rad/s) 2 . 2( I + R 2 m) ( I + R 2 m)

The damping is sub-critical, since d 2 < ω 2 . The solution is the sum of the solution to the homogenous equation of motion, of the form x c = e −dt ( A sin ω d t + B cos ω d t ), where ω d = ω 2 − d 2 = 2.236 rad/s, and the solution to the nonhomogenous equation, of the form xp =

mgR 2 mg = = 0.3633. ( I + R 2 m) ω 2 k

(The particular solution x p is obtained by setting the acceleration and velocity to zero and solving, since the non-homogenous term mg is not a function of time or position.) The solution is x = x c + x p = e −dt ( A sin ω d t + B cos ω d t ) +

626

mg . k

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Problem 21.47 For the system described in Problem 21.46, determine x as a function of time if c = 120 N-s/m and the system is released from rest with x = 0.

Solution:

From the solution to Problem 21.46 the canonical form of the equation of motion is

R 2 mg d 2x dx + 2d + ω 2x = , dt 2 dt I + R 2m where d =

cR 2 = 4 rad/s, 2( I + R 2 m)

and ω 2 =

kR 2 = 9 (rad/s) 2 . ( I + R 2 m)

The system is supercritically damped, since d 2 > ω 2 . The homogenous solution is of the form (see Eq. (21.24)) x c = Ce −( d − h )t + De −( d + h )t , where h =

d 2 − ω 2 = 2.646 rad/s, (d − h) = 1.354, (d + h) = 6.646.

mg = 0.3633. The solution is k mg x (t ) = x c + x pc = Ce −( d − h )t + De −( d + h )t + . k

The particular solution is x p =

Problem 21.48 For the system described in Problem 21.46, choose the value of c so that the system is critically damped, and determine x as a function of time if the system is released from rest with x = 0.

dx Apply the initial conditions, at t = 0, x = 0, = 0, from which dt mg 0 = C + D+ , and 0 = −(d − h)C − (d + h) D. Solve: k (d + h) mg C = − , 2h k (d − h) mg and D = , 2h k

( ) ( )

from which

(

)

mg (d + h) −( d − h )t (d − h) −( d + h )t e + 1− e k 2h 2h = 0.3633(1 − 1.256e −1.354 t + 0.2559e −6.646t ) (m).

x (t ) =

Solution:

From the solution to Problem 21.46, the canonical form of the equation of motion is d 2x dt

+ 2d 2

R 2 mg

dx + ω 2x = , dt I + R 2m

where d =

m k

cR 2 kR 2 , and ω 2 = = 9 (rad/s) 2 . 2( I + R 2 m) ( I + R 2 m)

For critical damping, d 2 = ω 2 , from which d = 3 rad/s. The homogenous solution is (see Eq. (21.25)) x c = Ce −dt + Dte −dt and mg the particular solution is x p = = 0.3633 m. The solution: k mg x (t ) = x c + x p = Ce −dt + Dte −dt + . k dx conditions at t = 0, x = 0, = 0, dt mg = 0, and −dC + D = 0. Solve: from which C + k mg C = − = −0.3633, D = dC = −1.09. The solution is k mg x (t ) = (1 − e −dt (1 + dt )), k Apply

the

initial

x (t ) = 0.363(1 − e −3t (1 + 3t )) m.

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Problem 21.49 The spring constant is k = 800 N/m, and the spring is unstretched when x = 0. The mass of each object is 30 kg. The inclined surface is smooth. The radius of the pulley is 120 mm and its moment of inertia is I = 0.03 kg-m 2 . Determine the frequency and period of vibration of the system relative to its equilibrium position if (a) c = 0 and (b) c = 250 N-s/m.

c k

208

Solution: Let T1 be the tension in the rope on the left of the pulley, and T2 be the tension in the rope on the right of the pulley. The equations of motion are x T1 − mg sin θ − cx − kx = mx , T2 − mg = −mx , (T2 − T1 )r = I . r If we eliminate T1 and T2 , we find

( 2m + rI ) x + cx + kx = mg(1 − sin θ), 2

x +

mgr 2 cr 2 kr 2 x + x = (1 − sin θ), 2 2 2mr + I 2mr + I 2mr 2 + I

x + (0.016c) x + (3.59 rad/s) 2 x = 3.12 m/s 2 . (a) If we set c = 0, the natural frequency, frequency, and period are ω = 3.59 rad/s, f =

ω 1 = 0.571 Hz, τ = = 1.75 s. f 2π

f = 0.571 s, τ = 1.75 s.

(b) If we set c = 250 N/m, then x + 2(2.01) x + (3.59 rad/s) 2 x = 3.12 m/s 2 . We recognize ω = 3.59, d = 2.01, ω d = f =

ωd , 2π

τ =

1 , f

ω 2 − d 2 = 2.97 rad/s.

f = 0.473 Hz, τ = 2.11 s.

Problem 21.50 The spring constant is k = 800 N/m, and the spring is unstretched when x = 0. The damping constant is c = 250 N-s/m. The mass of each object is 30 kg. The inclined surface is smooth. The radius of the pulley is 120 mm and its moment of inertia is I = 0.03 kg-m 2 . At t = 0, x = 0, and dx /dt = 1 m/s. What is the value of x at t = 2 s?

Solution:

From Problem 21.49 we know that the damping is subcritical and the key parameters are ω = 3.59 rad/s, d = 2.01 rad/s, ω d = 2.97 rad/s.

The Equation of motion is x + 2(2.01) x + (3.59 rad/s) 2 x = 3.12 m/s 2 . The solution is x = e −dt ( A sin ω d t + B cos ω d t ) + 0.242 dx = e −dt ([ Aω d − Bd ]cos ω d t − [ Ad + Bω d ]sin ω d t ) dt Putting in the initial conditions, we have

x (t = 0) = B + 0.242 = 0 ⇒ B = −0.242, k

208

dx (t = 0) = Aω d − Bd = 1 ⇒ A = 0.172 dt Thus the motion is governed by x = e −2.01t (0.172sin[2.97t ] − 0.242 cos[2.97t ]) + 0.242 At time t = 2 s we have

628

x = 0.237 m.

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Problem 21.51 The homogeneous disk weighs 100 lb and its radius is R = 1 ft. It rolls on the plane surface. The spring constant is k = 100 lb/ft and the damping constant is c = 3 lb-s/ft. Determine the frequency of small vibrations of the disk relative to its equilibrium position.

R k

Solution:

Choose a coordinate system with the origin at the center of the disk and the positive x axis parallel to the floor. Denote the angle of rotation by θ. The horizontal forces acting on the disk are ΣF = −kx − c

dx + f. dt

W f

From Newton’s second law, m

dx d 2x = ΣF = −kx − c + f. dt 2 dt

The moment about the center of mass of the disk is ΣM = Rf . From d 2θ I d 2θ the equation of angular motion, I 2 = Rf , from which f = , dt R dt 2 mR 2 where the moment of inertia is I = = 1.554 slug-ft 2 . 2 Substitute: dx d 2x I d 2θ . = −kx − c + dt 2 dt R dt 2 x From kinematics, θ = − , from which the equation of motion is R I d 2x dx m+ 2 + c + kx = 0. R dt 2 dt m

(

)

where

d =

cR 2 c = = 0.3217 rad/s 2(mR 2 + I ) 3m

and

ω2 =

kR 2 2k = = 21.45 (rad/s) 2 . ( I + R 2 m) 3m

The damping is sub-critical, since d 2 < ω 2 . The frequency is fd =

1 ω 2 − d 2 = 0.7353 Hz. 2π

The canonical form (see Eq. (21.16)) is d 2x dx + 2d + ω 2 x = 0, dt 2 dt

Problem 21.52 In Problem 21.51, the spring is unstretched at t = 0 and the disk has a clockwise angular velocity of 2 rad/s. What is the angular velocity of the disk when t = 3 s?

c R k

Solution:

From the solution to Problem 21.51, the canonical form of the equation of motion is

d 2x dx + 2d + ω 2 x = 0, dt 2 dt c 2k where d = = 0.322 rad/s and ω 2 = = 21.47 (rad/s) 2 . 3m 3m The system is sub-critically damped, so that the solution is of the form x = e −dt ( A sin ω d t + B cos ω d t ), where ω d = ω 2 − d 2 = 4.622 rad/s. Apply the initial conditions: x 0 = 0, and from kinematics, θ 0 = x 0 /R = −2 rad/s, from which x 0 = 2 ft/s, from which B = 0 and A = x 0 /ω d = 0.4327. The  x  solution is x (t ) = e −dt  0  sin ω d t, and x (t ) = − dx + x 0 e −dt cos ω d t.  ωd  At t = 3 s, x = 0.1586 ft and x = 0.1528 ft/s. From kinematics, x (t ) θ(t ) = − . At t = 3 s, θ = −0.153 rad/s clockwise. R

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Problem 21.53 The radius of the 30-kg homogeneous disk is R = 0.2 m. The spring constant is k = 800 N/m. The disk rolls on the inclined surface. Determine the frequency and period of vibration of the disk relative to its equilibrium position if the damping constant is (a) c = 0 and (b) c = 400 N-s/m. Solution: I =

k c

The moment of inertia of the disk is

1 mR 2 = 0.6 kg-m 2 . 2

Let x denote the displacement of the center of the disk down the inclined plane and let θ denote its clockwise rotation relative to the disk’s position when the spring is unstretched. Consider the free-body diagram: We have Newton’s second law, mg sin 30 ° − c

308

dx d 2x − 2kx − f = m 2 , dt dt

2kx

and the equation of angular motion, −2kxR + fR = I

c dx dt

d 2θ . dt 2

Eliminating f between these two equations and using the rolling relation d 2θ 1 d 2x = , dt 2 R dt 2 f

we can write the disk’s equation of motion as mg sin 30° d 2x dx , + 2d + ω 2x = I dt 2 dt m+ 2 R where

(

ω =

I m+ 2 R

)

= 8.43 rad/s,

308

d =

(

2 m+

I R2

)

The frequency and period of vibration are fd =

ωd = 2π

ω2 − d2 , 2π

τd =

1 . fd

When c = 0, we obtain f = 1.34 Hz, τ = 0.745 s, and when c = 400 N-s/m, we obtain f d = 1.14 Hz, τ d = 0.877 s. (a) f = 1.34 Hz, τ = 0.745 s. (b) f d = 1.14 Hz, τ d = 0.877 s.

Problem 21.54== The radius of the 30-kg homogeneous disk is R = 0.2 m. The spring constant is k = 800 N/m and the damping constant is c = 50 N-s/m. The disk rolls on the inclined surface. At t = 0, the system is in its equilibrium position and the disk is given a clockwise angular velocity of 10 rad/s. What is the magnitude of the angular velocity of the disk at t = 3 s?

k c

Solution: Let x denote the displacement of the center of the disk down the inclined plane and let θ denote its clockwise rotation relative 308

630

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21.54*

(Continued) We can see from the equation of motion that the equilibrium position of the system is

2kx

x =

dx dt

mg sin 30 ° I R2

(

ω2 m +

)

mg sin 30 ° . 4k

Expressed in terms of a new variable that is the position relative to the equilibrium position, mg sin 30 ° x = x− , 4k f

the equation of motion becomes

308

d 2 x dx + 2d + ω 2 x = 0. dt 2 dt

to the disk’s position when the spring is unstretched. Because the disk is rolling, they are related by x = Rθ. Consider the free-body diagram: We have Newton’s second law, d 2x dx mg sin 30 ° − c − 2kx − f = m 2 , dt dt and the equation of angular motion, d 2θ −2kxR + fR = I 2 . dt

The system is subcritically damped. From Eq. (21.19). the position and velocity are x = e −dt ( A sin ω d t + B cos ω d t ), dx = −de −dt ( A sin ω d t + B cos ω d t ) + e −dt ( Aω d cos ω d t − Bω d sin ω d t ), dt where ω d =

ω 2 − d 2 = 8.41 rad/s.

Eliminating f between these two equations and using the relation

The initial conditions at t = 0 are x = 0, dx /dt = R(10 rad/s) = 2 m/s. Substituting these into the equations for the position and velocity,

d 2θ 1 d 2x = , dt 2 R dt 2

2 m/s = −dB + ω d A,

0 = B,

we can write the disk’s equation of motion as we obtain A = 0.238 m, B = 0. Then from the equation for the velocity as a function of time, at t = 3 s we obtain dx /dt = 0.373 m/s, and the angular velocity is d θ /dt = dx /dt /R = 1.86 rad/s.

mg sin 30° d 2x dx , + 2d + ω 2x = I dt 2 dt m+ 2 R where

(

ω =

I m+ 2 R

)

= 8.43 rad/s,

d =

(

I 2 m+ R

)

= 0.556 rad/s.

1.86 rad/s.

The clockwise angular velocity as a function of time is shown: 10 8

Clockwise angular velocity, rad/s

6 4 2 0 –2 –4 –6 –8 –10

0.5

1.5

2.5

3.5

t, seconds

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Problem 21.55 The moments of inertia of gears A and B are I A = 0.025 kg-m 2 and I B = 0.100 kg-m 2 . Gear A is connected to a torsional spring with constant k = 10 N-m/rad. The bearing supporting gear B incorporates a damping element that exerts a resisting moment on gear B of magnitude 2(d θ B /dt ) N-m, where d θ B /dt is the angular velocity of gear B in rad/s. What is the frequency of small angular vibrations of the gears?

200 mm A

Solution:

The sum of the moments on gear A is ΣM = −kθ A + R A F, where the moment exerted by the spring opposes the angular displacement θ A . From the equation of angular motion, d 2θ A IA = ΣM = −kθ A + R A F , dt 2

140 mm

 I  d 2θ  k  from which F =  A  2A +  θ .  R A  dt  R A  A

The sum of the moments acting on gear B is ΣM = − 2

dθB + RB F, dt

where the moment exerted by the damping element opposes the angular velocity of B. From the equation of angular motion applied to B, IB

dθ d 2θ B = ΣM = − 2 B + R B F . dt 2 dt

The canonical form of the equation of motion is

Substitute the expression for F, IB

d 2θ

dt 2

+ 2

(

)

R  dθB A −  B  I A + kθ A = 0.  RA  dt dt 2 d 2θ

dθ d 2θ B + 2d B + ω 2θ B = 0, dt 2 dt

From kinematics,

where

R  θ A = − B  θ B ,  RA 

and

from which the equation of motion for gear B is

d =

1 = 6.622 rad/s M

 R 2 k ω 2 =  B  = 135.1 (rad/s) 2 .  RA  M

  R 2  R  2  d 2θ B dθB +  B  kθ B = 0.  I B +  B  I A  2 + 2  RA    R dt dt   A

The system is sub critically damped, since d 2 < ω 2 , from which ω d = ω 2 − d 2 = 9.555 rad/s, from which the frequency of small vibrations is

 R 2 Define M = I B +  B  I A = 0.1510 kg-m 2 .  RA 

fd =

Problem 21.56 At t = 0, the torsional spring in Problem 21.55 is unstretched and gear B has a counterclockwise angular velocity of 2 rad/s. Determine the counterclockwise angular position of gear B relative to its equilibrium position as a function of time.

Solution:

From the solution to Problem 21.55, the canonical form of the equation of motion for gear B is d 2θ B dθ + 2d B + ω 2θ B = 0, dt 2 dt

where d = and

200 mm A 140 mm

ωd = 1.521 Hz. 2π

 R 2 M = I B +  B  I A = 0.1510 kg-m 2 ,  RA  1 = 6.622 rad/s, M  R 2 k ω 2 =  B  = 135.1 (rad/s) 2 .  RA  M

The system is sub critically damped, since d 2 < ω 2 , from which ω d = ω 2 − d 2 = 9.555 rad/s. The solution is of the form θ B (t ) = e −dt ( A sin ω d t + B cos ω d t ). Apply the initial conditions, [θ B ] t = 0 = 0, [θ B ] t = 0 = 2 rad/s, from which B = 0, and 2 A = = 0.2093, from which the solution is ωd θ B (t ) = e −6.62t (0.209sin(9.55t )).

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Problem 21.57 For the case of critically damped motion, confirm that the expression x

= Ce −dt

+ Dte −dt

Solution:

Eq. (21.16) is

dx d 2x + 2d + ω 2 x = 0. dt 2 dt We show that the expression is a solution by substitution. The individual terms are:

is a solution of Eq. (21.16).

(1) x = e −dt (C + Dt ), dx = −dx + De −dt , dt dx d 2x (3) = −d − dDe −dt = d 2 x − 2dDe −dt . dt 2 dt (2)

Substitute: d 2x dx + 2d + ω 2 x = (d 2 x − 2 De −dt ) + 2d (−dx + De −dt ) dt 2 dt + ω 2 x = 0. Reduce: d 2x dx + 2d + ω 2 x = (−d 2 + ω 2 ) x = 0. dt 2 dt This is true if d 2 = ω 2 , which is the definition of a critically damped system. Note: Substitution leading to an identity shows that x = Ce −dt + Dte −dt is a solution. It does not prove that it is the only solution.

Problem 21.58 The mass m = 2 slug and the spring constant is k = 72 lb/ft. The spring is unstretched when x = 0. The mass is initially stationary with the spring unstretched, and at t = 0 the force F (t ) = 10 sin 4t lb is applied to the mass. What is the position of the mass at t = 2 s?

x k

F(t)

Solution:

The equation of motion is F (t ) k 72 10 mx + kx = F (t ) ⇒ x + x = ⇒ sin 4t x + x = m 2 2 m 2 x + (6) x = 5sin 4t

We recognize the following: ω = 6, ω 0 = 4, d = 0, a 0 = 5, b 0 = 0. Ap =

(6 2 − 4 2 )5 = 0.25, B p = 0 (6 2 − 4 2 ) 2

Therefore, the complete solution (homogeneous plus particular) is x = A sin 6t + B cos6t + 0.25sin 4t,

dx = 6 A cos6t − 6 B sin 6t + cos 4t. dt

Putting in the initial conditions, we have x (t = 0) = B = 0,

dx (t = 0) = 6 A + 1 ⇒ A = −0.167. dt

The complete solution is now x = −0.167sin 6t + 0.25sin 4t At t = 6 s we have x = 0.337 ft.

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Problem 21.59 The mass m = 2 slug and the spring constant is k = 72 lb/ft. The spring is unstretched when x = 0. At t = 0, x = 1 ft, dx /dt = 1 ft/s, and the force F (t ) = 10 sin 4t + 10 cos 4t lb is applied to the mass. What is the position of the mass at t = 2 s? Solution:

F(t)

The equation of motion is

F (t ) k 72 10 10 mx + kx = F (t ) ⇒ x + x = ⇒ sin 4t + cos 4t x + x = m 2 2 2 m 2 x + (6) x = 5sin 4t + 5cos 4t. We recognize the following: ω = 6, ω 0 = 4, d = 0, a 0 = 5, b 0 = 5. Ap =

(6 2 − 4 2 )5 (6 2 − 4 2 )5 = 0.25, B p = 2 = 0.25. (6 2 − 4 2 ) 2 (6 − 4 2 ) 2

Therefore, the complete solution (homogeneous plus particular) is x = A sin 6t + B cos6t + 0.25sin 4t + 0.25cos 4t, dx = 6 A cos6t − 6 B sin 6t + cos 4t − sin 4t. dt Putting in the initial conditions, we have x (t = 0) = B + 0.25 = 1 ⇒ B = 0.75 dx (t = 0) = 6 A + 1 = 1 ⇒ A = 0. dt The complete solution is now x = 0.75cos6t + 0.25sin 4t + 0.25cos 4t. At t = 6 s we have x = 0.844 ft.

Problem 21.60 The damped spring–mass oscillator is initially stationary with the spring unstretched. At t = 0, a force F (t ) = 6 N is applied to the mass. (a) What is the steady-state (particular) solution? (b) Determine the position of the mass as a function of time. Solution:

x 12 N/m 6 N-s/m

3 kg

F(t)

Writing Newton’s second law for the mass, the equation

of motion is dx d 2x − kx = m 2 or dt dt dx d 2x F (t ) − 6 − 12 x = 3 2 , dt dt which we can write as F (t ) − c

The time derivative is dx = −e −1 ( A sin1.73t + B cos1.73t ) dt + e −t (1.73 A cos1.73t − 1.73t − 1.73B sin1.73t ).

F (t ) d 2x dx + 2 + 4x = . (1) dt 2 dt 3

At t = 0, x = 0, and dx /dt = 0 0 = B + 0.5, and 0 = −B + 1.73 A. We see that B = −0.5 and A = −0.289 and the solution is

(a) If F (t ) = 6 N, we seek a particular solution of the form x p = A0 , a F (t ) constant. Substituting it into Equation (1), we get 4 x p = = 2 3 and obtain the particular solution: x p = 0.5 m.

x = e −t (−0.289sin1.73t − 0.5cos1.73t ) + 0.5 m.

(b) Comparing equation (1) with Equation (21.26), we see that d = 1 rad/s and ω = 2 rad/s. The system is subcritically damped and the homogeneous solution is given by Equation (21.19) with ω d = ω 2 − d 2 = 1.73 rad/s. The general solution is x = x h + x p = e −1 ( A sin 1.73t + B cos 1.73t ) + 0.5 m.

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Problem 21.61 The damped spring–mass oscillator is initially stationary with the spring unstretched. At t = 0, a force F (t ) = 6 cos1.6t N is applied to the mass. (a) What is the steady-state (particular) solution? (b) Determine the position of the mass as a function of time. (See Practice Example 21.6.) Solution:

Writing Newton’s second law for the mass, the equation of motion can be written as F (t ) d 2x dx + 2 + 4x = = 2 cos1.6t. dt 2 dt 3

(a) Comparing this equation with Equation (21.26), we see that d = 1 rad/s, ω = 2 rad/s, and the forcing function is a(t ) = 2 cos1.6t. This forcing function is of the form of E quation (21.27) with a 0 = 0, b 0 = 2 and ω 0 = 1.6. Substituting these values into Equation (21.30), the particular solution is x p = 0.520 sin1.6t + 0.234 cos1.6t (m). (b) The system is subcritically damped so the homogeneous solution is given by Equation (21.19) with ω d = ω 2 − d 2 = 1.73 rad/s. The general solution is

x 12 N/m 3 kg

6 N-s/m

F(t)

The time derivative is dx = −e −t ( A sin1.73t + B cos1.73t ) dt + e −t (1.73 A cos1.73t − 1.73B sin1.73t ) + (1.6)(0.520) cos1.6t − (1.6)(0.234)sin1.6t At t = 0, x = 0 and dx /dt = 0:0 = B + 0.2340 = −B + 1.73 A + (1.6)(0.520). Solving, we obtain A = − 0.615 and B = −0.234, so the solution is x = e −t (−0.615sin1.73t − 0.234 cos1.73t ) + 0.520 sin1.6t + 0.234 cos1.6t m.

x = x k + x p = e −t ( A sin1.73t + B cos1.73t ) + 0.520 sin1.6t + 0.234 cos1.6t.

Problem 21.62 The disk with moment of inertia I = 3 kg-m 2 rotates about a fixed shaft and is attached to a torsional spring with constant k = 20 N-m/rad. At t = 0, the angle θ = 0, the angular velocity is d θ /dt = 4 rad/s, and the disk is subjected to a couple M (t ) = 10 sin 2t N-m. Determine θ as a function of time. Solution:

M(t)

The equation of angular motion for the disk is

M (t ) − kθ = 1

d 2θ d 2θ : or 10 sin 2t − (20)θ = 3 2 , dt 2 dt

or, rewriting in standard form, we have d 2θ 20 10 + θ = sin 2t dt 2 3 3 (a) Comparing this equation with equation (21.26), we see that d = 0, ω = 20/3 = 2.58 rad/s and the forcing function is 10 a(t ) = sin 2t. This forcing function is of the form of Equation 3 (21.27), with a 0 = 10/3, b 0 = 0 and ω 0 = 2. Substituting these values into Equation (21.30), the particular solution is θ p = 1.25sin 2t. (b) The general solution is θ = θ h + θ p = A sin 2.58t + B cos 2.58t + 1.25sin 2t The time derivative is dθ = 2.58 A cos 2.58t − 2.58B sin 2.58t + 2.50 cos 2t. dt At t = 0, θ = 0 and d θ /dt = 4 rad/s, 0 = B, and 4 = 2.58 A + 2.50. Solving, we obtain A = 0.581 and B = 0. The solution is θ = 0.581sin 2.58t + 1.25sin 2t rad.

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Problem 21.63 The stepped disk weighs 20 lb and its moment of inertia is I = 0.6 slug-ft 2 . It rolls on the horizontal surface. The disk is initially stationary with the spring unstretched, and at t = 0 a constant force F = 10 lb is applied as shown. Determine the position of the center of the disk as a function of time. (See Example 21.7.)

16 lb/ft

16 in

8 lb-s/ft

8 in

Solution:

The strategy is to apply the free body diagram to obtain equations for both θ and x, and then to eliminate one of these. An essential element in the strategy is the determination of the stretch of the spring. Denote R = 8 in. = 0.6667 ft, and the stretch of the spring by S. Choose a coordinate system with the positive x axis to the right. The sum of the moments about the center of the disk is ΣM C = RkS + 2 Rf − 2 RF . From the equation of angular motion, I

d 2θ = ΣM C = RkS + 2 Rf − 2 RF . dt 2

Solve for the reaction at the floor: f =

I d 2θ k − S + F. 2 R dt 2 2

The sum of the horizontal forces: ΣFx = −kS − c

dx + F + f. dt

From Newton’s second law: m

dx d 2x = ΣFx = −kS − c + F + f. dt 2 dt

Substitute for f and rearrange: m

kS cx·

f P

The particular solution is found by setting the acceleration and veloca 8F ity to zero and solving: x p = 2 = = 0.5556 ft. Since ω 9k 2 2 d < ω , the system is sub-critically damped, so the homogenous solution is x h = e −dt ( A sin ω d t + B cos ω d t ), where ω d = ω 2 − d 2 = 4.488 rad/s. The complete solution is 8F x = e −dt ( A sin ω d t + B cos ω d t ) + . Apply the initial conditions: 9k 8F = −0.5556, and at t = 0, x 0 = 0, x 0 = 0, from which B = − 9k dB 2 A = = −0.5162. Adopting g = 32.17 ft/s the solution is ωd x =

  d 8F  sin ω d t + cos ω d t    1 − e dt     ωd 9k 

= 0.5556 − e −4.170 t (0.5162sin 4.488t + 0.5556 cos 4.488t ) ft ,

dx d 2x I d 2θ 3 + kS = 2 F . + + c dt 2 2 R dt 2 dt 2

From kinematics, the displacement of the center of the disk is x = −2 Rθ. The stretch of the spring is the amount wrapped around the disk plus the translation of the disk, S = −Rθ − 2 Rθ = −3 Rθ = 32 x. Substitute:  I  d 2x dx 3 2  m + (2 R) 2  dt 2 + c dt + 2 kx = 2 F .

( )

Define M = m +

I = 0.9592 slug. (2 R) 2

The canonical form of the equation of motion is d 2x dx + 2d + ω 2 x = a, dt 2 dt c where d = = 4.170 rad/s, 2M 3 2 k ω2 = = 37.53 (rad/s) 2 , 2 M 2F and a = = 20.85 ft/s 2 . M

( )

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Problem 21.64* An electric motor is bolted to a metal table. When the motor is on, it causes the tabletop to vibrate horizontally. Assume that the legs of the table behave like linear springs, and neglect damping. The total weight of the motor and the tabletop is 150 lb. When the motor is not turned on, the frequency of horizontal vibration of the tabletop and motor is 5 Hz. When the motor is running at 600 rpm, the amplitude of the horizontal vibration is 0.01 in. What is the magnitude of the oscillatory force exerted on the table by the motor at its 600-rpm running speed?

Solution:

For d = 0, the canonical form (see Eq. (21.26)) of the equation of motion is d 2x + ω 2 x = a(t ), dt 2 where ω 2 = (2π f ) 2 = (10 π ) 2 = 986.96 (rad/s) 2 and a(t ) =

gF (t ) F (t ) = . m W

The forcing frequency is f0 =

( 600 ) = 10 Hz, 60

from which ω 0 = (2π )10 = 62.83 rad/s. Assume that F (t ) can be written in the form F (t ) = F0 sin ω 0 t. From Eq. (21.31), the amplitude of the oscillation is x0 =

a0 gF0 = . W (ω 2 − ω 02 ) ω 2 − ω 02

Solve for the magnitude: F0 =

W (ω 2 − ω 02 ) x 0 g

Substitute numerical values: W = 150 lb, g = 32.17 ft/s 2 , (ω 2 − ω 02 ) = 2960.9 (rad/s) 2 , x 0 = 0.01 in. = 8.33 × 10 −4 ft, from which F0 = 11.5 lb.

Problem 21.65 The moments of inertia of gears A and B are I A = 0.014 slug-ft 2 and I B = 0.100 slug-ft 2 . Gear A is connected to a torsional spring with constant k = 2 ft-lb/rad. The system is in equilibrium at t = 0 when it is subjected to an oscillatory force F (t ) = 4 sin 3t lb. What is the downward displacement of the 5-lb weight as a function of time?

Solution: Choose a coordinate system with the x axis positive upward. The sum of the moments on gear A is ∑ M = −kθ A + R A F. From Newton’s second law, IA

d 2θ A = ∑ M = −kθ A + R A F , dt 2

 I  d 2θ  k  from which F =  A  2A +  θ .  R A  dt  R A  A The sum of the moments acting on gear B is ∑ M = R B F − RW Fd . From the equation of angular motion applied to gear B, IB

10 in 3 in

d 2θ B = ∑ M B = R B F − RW Fd . dt 2

Substitute for F to obtain the equation of motion for gear B: IB

 R  d 2θ A R  d 2θ B −  B  I A −  A  kθ A − RW Fd = 0.  RA   RB  dt 2 dt 2

Solve: A

 R B   R B  d 2θ A  I  d 2θ −  Fd =  B  2B −   kθ .   R A RW  A  R A RW  dt 2  RW  dt

6 in

The sum of the forces on the weight are ∑ F = +Fd − W − F (t ). From Newton’s second law applied to the weight, 5 lb F(t)

 W  d 2 x  g  dt 2 = Fd − W − F (t ). Substitute for Fd , and rearrange to obtain the equation of motion for the weight: I d 2θ B R I d 2θ A RB W d 2x − B + B A + kθ 2 2 g dt RW dt RW R A dt 2 RW R A A = −W − F (t ).

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21.65

(Continued) Fd

The particular solution is x p = −

4 sin 3t = −0.05927sin 3t ft. M (ω 2 − 3 2 )

The solution to the homogenous equation is MA

F(t)

R  From kinematics, θ A = − B  θ B , and x = −RW θ B , from which  RA   R  d 2θ d 2θ B RB d 2 x 1 d 2 x d 2θ A = − = − B  2B = , ,  R A  dt dt 2 RW dt 2 dt 2 RW R A dt 2 and

θA =

RB x. RW R A

x h = Ah sin ωt + B h cos ωt, and the complete solution is x (t ) = Ah sin ωt + B h cos ωt −

Apply the initial conditions: at t = 0, x 0 = 0, x 0 = 0, from which 0 = B, 12 0 = ω Ah − , M (ω 2 − 3 2 ) from which Ah =

Define

RB η = = 6.667 ft −1 , RW R A

and M =

 R B  2 IB W + +   I = 2.378 slug,  RW R A  A g ( RW ) 2

4 sin 3t. M (ω 2 − 3 2 )

12 = 0.02908. ω M (ω 2 − 3 2 )

The complete solution for vibration about the equilibrium point is:

from which the equation of motion for the weight about the unstretched spring position is: d 2x M 2 + (k η 2 ) x = −W − F (t ). dt

x (t ) =

(

4 3 sin ωt − sin 3t M (ω 2 − ω 02 ) ω

)

= 0.02908 sin 6.114t − 0.05927sin 3t ft. The downward travel is the negative of this: x down = −0.02908 sin6.114t + 0.05927sin3t ft.

For d = 0, the canonical form (see Eq. (21.16)) of the equation of motion is d 2x + ω 2 x = a(t ), dt 2

where ω 2 =

η 2k = 37.39 (rad/s) 2 . M

[Check: This agrees with the result in the solution to Problem 21.53, as it should, since nothing has changed except for the absence of a damping F (t ) W element. check.] The non-homogenous terms are a(t ) = + . M M W Since is not a function of t, M x pw = −

W W = − 2 = −0.05625 ft, ω 2M η k

which is the equilibrium point. Make the transformation x p = x p − x pω . The equation of motion about the equilibrium point is d 2x 4 sin 3t + ω 2 x = − . dt 2 M Assume a solution of the form x p = A p sin 3t + B p cos3t. Substitute: (ω 2 − 3 2 )( A p sin 3t + B p cos3t ) = −

3sin 3t , M

from which B p = 0, and Ap = −

638

4 = −0.05927 ft. M (ω 2 − 3 2 )

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Problem 21.66* A 1.5-kg cylinder is mounted on a sting in a wind tunnel with the cylinder axis transverse to the direction of flow. When there is no flow, a 10-N vertical force applied to the cylinder causes it to deflect 0.15 mm. When air flows in the wind tunnel, vortices subject the cylinder to alternating lateral forces. The velocity of the air is 5 m/s, the distance between vortices is 80 mm, and the magnitude of the lateral forces is 1 N. If you model the lateral forces by the oscillatory function F (t ) = (1.0) sin ω 0 t N, what is the amplitude of the steady-state lateral motion of the sphere? Solution:

The time interval between the appearance of an upper 0.08 vortex and a lower vortex is δ t = = 0.016 s, from which the 5 period of a sinusoidal-like disturbance is τ = 2(δt ) = 0.032 s, 1 from which f 0 = = 31.25 Hz. [Check: Use the physical relaτ tionship between frequency, wavelength and velocity of propagation of a small amplitude sinusoidal wave, λ f = v . The wavelength of a traveling sinusoidal disturbance is the distance between two peaks or two troughs, or twice the distance between adjacent peaks and troughs, λ = 2(0.08) = 0.16 m, from which the frev quency is f 0 = = 31.25 Hz. check.] The circular frequency is λ ω 0 = 2π f 0 = 196.35 rad/s. The spring constant of the sting is k =

F 10 = = 66667 N/m. δ 0.00015

80 mm

from which ω = 2π f = 210.82 rad/s. For d = 0, the canonical form (see Eq. 21.16)) of the equation of motion is a(t ) =

d 2x + ω 2 x = a(t ), where dt 2

F 1 = sin ω 0 t = 0.6667sin196.3t. m 15

From Eq. (21.31) the amplitude is E =

a0 0.6667 = = 1.132 × 10 −4 m. ω 2 − ω 02 5891

[Note: This is a small deflection (113 microns) but the associated aerodynamic forces may be significant to the tests (e.g. Famp = 7.5 N), since the sting is stiff. Vortices may cause undesirable noise in sensitive static aerodynamic loads test measurements.]

The natural frequency of the sting-cylinder system is f =

1 2π

k 1 66667 = = 33.55 Hz, m 2π 1.5

Problem 21.67 Show that the amplitude of the particular solution given by Eq. (21.31) is a maximum when the frequency of the oscillatory forcing function is ω 0 = ω 2 − 2d 2 . Solution: Ep =

Eq. (21.31) is a 02 + b 02

(ω 2 − ω 02 ) 2

+ 4 d 2 ω 02

Ep a 02 + b 02

[(4 ω 0 )(ω 02 − ω 2 + 2d 2 )] ω2 0 = ω 0  d 2η  0 = 3 = ω   5  d ω 02  ω 0 4 [(ω 2 − ω 02 ) 2 + 2d 2 ω 02 ] ω2 0 = ω 0 −

2 2 2 1 4[3ω 0 − ω + 2d ] ω 0 = ω 0 , 3 2 [(ω 2 − ω 02 ) 2 + 2d 2 ω 02 ] ω2 0 = ω 0

d 2η from which  2  = −  dt  ω 0 = ω 0

Since the numerator is a constant, rearrange: η =

0 = ω 2 − 2d 2 be the maximizing value. To show that η is Let ω indeed a maximum, take the second derivative:

−

= [(ω 2 − ω 02 ) 2 + 4 d 2 ω 02 ] 2 .

4 ω 02

< 0,

[2ω 2 d 2 ] 2

which demonstrates that it is a maximum.

The maximum (or minimum) is found from dη = 0. dω 0 dη 1 4[−(ω 2 − ω 02 )(ω 0 ) + 2d 2 ω 0 ] = − = 0, 3 dω 0 2 [(ω 2 − ω 02 ) 2 + 4 d 2 ω 02 ] 2 from which −(ω 2 − ω 02 )(ω 0 ) + 2d 2 ω 0 = 0. Rearrange, (ω 02 − ω 2 + 2d 2 )ω 0 = 0. Ignore the possible solution ω 0 = 0, from which ω0 =

ω 2 − 2d 2 .

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Problem 21.68* A floating sonobuoy (sound- measuring device) consists of an 18-kg cylinder 1.2 m in length and 0.15 m in diameter. The water density is 1024 kg/m 3 . The buoyancy force supporting the buoy equals the weight of the volume of water that would occupy the volume of the part of the cylinder below the surface. The buoy is initially in equilibrium in a wave tank (Fig. a). Waves are then created in the tank (Fig. b), and, at the location of the buoy, the downward displacement of the surface of the water relative to its position when the water was stationary is s = (0.1 m)sin 6t. The coordinate x measures the position of the bottom of the buoy relative to the position of the water surface when it was stationary. What is the amplitude of the particular solution describing vertical vibrations of the buoy? Solution:

s x

(a)

From this equation we can see that in the absence of waves, the equilibrium position is x = m /(ρ A). Let us define a new variable that is the position relative to the equilibrium position: m . x = x − ρA In terms of this variable, we can write the equation of motion as d 2 x + ω 2 x = a 0 sin ω 0 t, dt 2 where ω =

V = A( x − s).

(b)

ρ gA = 3.14 rad/s, m

a0 =

ρ gAs 0 = 0.986 m/s 2 . m

In terms of the density ρ = 1024 kg/m 3 of the water, the (upward) buoyancy force on the cylinder is

In Eq. (21.30) for the particular solution, notice that the damping coefficient d = 0 and the coefficient b 0 = 0. The particular solution reduces to

ρVg = ρ gA( x − s).

x p =

Newton’s second law for the cylinder is therefore m

a0 sin ω 0 t, ω 2 − ω0 2

and its amplitude is

d 2x = mg − ρ gA( x − s) dt 2 = mg − ρ gAx + ρ gAs 0 sin ω 0 t.

a0 = 0.0377 m. ω 2 − ω0 2 37.7 mm.

Problem 21.69* A floating sonobuoy (sound-measuring device) consists of an 18-kg cylinder 1.2 m in length and 0.15 m in diameter. The water density is 1024 kg/m 3 . The buoyancy force supporting the buoy equals the weight of the volume of water that would occupy the volume of the part of the cylinder below the surface. The buoy is initially stationary in equilibrium in a wave tank (Fig. a). At t = 0, waves are created (Fig. b) and, at the location of the buoy, the downward displacement of the surface of the water relative to its position when the water was stationary is s = (0.1 m)sin 6t. The coordinate x measures the position of the bottom of the buoy relative to the position of the water surface when it was stationary. What is the magnitude of the buoy’s vertical velocity at t = 8 s?

Solution:

Let the water surface displacement be written s = s 0 sin ω 0 t, where s 0 = 0.1 m and ω 0 = 6 rad/s. Denote the mass of the buoy by m = 18 kg and its diameter by D = 0.15 m. Its cross-sectional area is A = π D 2 /4 = 0.0177 m 2 . When the waves begin, the volume of the cylinder that is under water is V = A( x − s). In terms of the density ρ = 1024 kg/m 3 of the water, the (upward) buoyancy force on the cylinder is ρVg = ρ gA( x − s). Newton’s second law for the cylinder is therefore d 2x m 2 = mg − ρ gA( x − s) dt = mg − ρ gAx + ρ gAs 0 sin ω 0 t. From this equation we can see that in the absence of waves, the equilibrium position is x = m /(ρ A). Let us define a new variable that is the position relative to the equilibrium position: m x = x − . ρA

In terms of this variable, we can write the equation of motion as x

d 2 x + ω 2 x = a 0 sin ω 0 t, dt 2 where

(a)

640

(b)

ω =

ρ gA = 3.14 rad/s, m

a0 =

ρ gAs 0 = 0.986 m/s 2 . m

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21.69

(Continued)

In Eq. (21.30) for the particular solution, notice that the damping coefficient d = 0 and the coefficient b 0 = 0. The particular solution reduces to x p =

a0 sin ω 0 t ω 2 − ω0 2

0 = D, 0 = Cω + C 0ω 0 . We see that the constants C = −(ω 0 /ω ) C 0 , D = 0, and the equation for the velocity becomes

= C 0 sin ω 0 t,

dx = C 0 ω 0 (− cos ωt + cos ω 0 t ) dt = (−0.0377 m)(6 rad/s)[− cos(3.14 rad/s)t + cos(6 rad/s)t ].

where C0 =

From the initial conditions we obtain the equations

a0 = −0.0377 m. ω 2 − ω0 2

We see that the general solution of the equation of motion is

At t = 8 s, we obtain dx /dt = 0.371 m/s. The (interesting) velocity history is shown:

x = C sin ωt + D cos ωt + C 0 sin ω 0 t.

0.371 m/s.

The velocity is dx = C ω cos ωt − Dω sin ωt + C 0 ω 0 cos ω 0 t. dt 0.5 0.4 0.3

Velocity, m/s

0.2 0.1 0 –0.1 –0.2 –0.3 –0.4 –0.5

Problem 21.70 The mass weighs 50 lb. The spring constant is k = 200 lb/ft, and c = 10 lb-s/ft. If the base is subjected to an oscillatory displacement x i of amplitude 10 in and frequency ω i = 15 rad/s, what is the resulting steady-state amplitude of the displacement of the mass relative to the base? (See Example 21.8.) xi

5 t, seconds

Solution:

The canonical form of the equation of motion is d 2x

dt + ω 2 x = a(t ), dt gc c d = = = 3.217 rad/s 2m 2W kg = 128.7 (rad/s) 2 , ω2 = W dt 2

Where

+ 2d

and (see Eq. (21.38),

x c

a(t ) = − m

Ep = =

d 2xi = x i ω i2 sin(ω i t − φ). dt 2

The displacement of the mass relative to its base is

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Problem 21.71 The mass is 100 kg. The spring constant is k = 4 N/m, and c = 24 N-s/m. The base is subjected to an oscillatory displacement of frequency ω i = 0.2 rad/s. The steady-state amplitude of the displacement of the mass relative to the base is measured and determined to be 200 mm. What is the amplitude of the displacement of the base? (See Example 21.8.) Solution:

From Example 21.7 and the solution to Problem 21.70, the canonical form of the equation of motion is

d =

The displacement of the mass relative to its base is

c k = 0.12 rad/s, ω 2 = = 0.04 (rad/s) 2 , 2m m

and (see Eq. (21.38)) a(t ) = −

m k

ω i2 x i (ω 2 − ω i2 ) 2 + 4 d 2 ω i2 (0.2 2 ) x i = 0.8333 x i m, (0.2 2 − 0.2 2 ) 2 + 4(0.12 2 )(0.2 2 )

E p = 0.2 =

d 2x dx + 2d + ω 2 x = a(t ) dt 2 dt where

x c

from which x i =

d 2xi = x i ω i2 sin(ω i t − φ). dt 2

Problem 21.72 A team of engineering students builds the simple seismograph shown. The coordinate x i measures the local horizontal ground motion. The coordinate x measures the position of the mass relative to the frame of the seismograph. The spring is unstretched when x = 0. The mass m = 1 kg, the spring constant k = 10 N/m, and c = 2 N-s/m. Suppose that the seismograph is initially stationary and that at t = 0 it is subjected to an oscillatory ground motion x i = 10 sin 2t mm. What is the amplitude of the steady-state response of the mass? (See Example 21.8.)

0.2 = 0.24 m. 0.8333

Solution:

From Example 21.7 and the solution to Problem 21.70, the canonical form of the equation of motion is d 2x dx + 2d + ω 2 x = a(t ) dt 2 dt

where

d =

c k = 1 rad/s, ω 2 = = 10 (rad/s) 2 , 2m m

and (see Eq. (21.38)) a(t ) = −

d 2xi = x i ω i2 sin ω i t dt 2

where x i = 10 mm, ω i = 2 rad/s.

The amplitude of the steady state response of the mass relative to its base is ω i2 x i = (ω 2 − ω i2 ) 2 + 4 d 2 ω i2 = 5.55 mm.

Ep =

(2 2 )(10) (3.162 2 − 2 2 ) 2 + 4(1 2 )(2 2 )

TOP VIEW

k m c

642

SIDE VIEW

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Problem 21.73 In Problem 21.72, determine the position x of the mass relative to the base as a function of time. (See Example 21.8.)

TOP VIEW

k m c

Solution:

SIDE VIEW

From Example 21.7 and the solution to Problem 21.70, the canonical form of the equation of motion is

and

d 2x dx + 2d + ω 2 x = a(t ) dt 2 dt c k where d = = 1 rad/s, ω 2 = = 10 (rad/s) 2 , 2m m

Ap =

(ω 2 − ω i2 )ω i2 x i , (ω 2 − ω i2 ) + 4 d 2 ω i2

Bp =

−2d ω i3 x i , check.] (ω 2 − ω i2 ) 2 + 4 d 2 ω i2

and

a(t ) = −

d 2xi = x i ω i2 sin ω i t dt 2

where x i = 10 mm, ω i = 2 rad/s. From a comparison with Eq. (21.27) and Eq. (21.30), the particular solution is x p = A p sin ω i t + B p cos ω i t, where Ap =

(ω 2 − ω i2 )ω i2 x i = 4.615. (ω 2 − ω i2 ) 2 + 4 d 2 ω i2

Bp = −

2d ω i3 x i = −3.077. (ω 2 − ω i2 ) 2 + 4 d 2 ω i2

[Check: Assume a solution of the form x p = A p sin ω i t + B p cos ω i t. Substitute into the equation of motion: [(ω 2 − ω i2 ) A p − 2d ω i B p ]sin ω i t + [(ω 2 − ω i2 ) B p + 2d ω i Ap]cos ω i t = x i ω i2 sin ω i t.

2d ω i A p + (ω 2 − ω i2 ) B p = 0.

Solve:

Since d 2 < ω 2 , the system is sub-critically damped, and the homogenous solution is x h = e −dt ( A sin ω d t + B cos ω d t ), where ω d = ω 2 − d 2 = 3 rad/s. The complete solution is x = x h + x p . Apply the initial conditions: at t = 0, x 0 = 0, from which 0 = B + B p , and 0 = −dB + ω d A + ω i A p . Solve: B = −B p = 3.077, A =

ωi Ap dB − = −2.051 ωd ωd

The solution is x = e −dt ( A sin ωt + B cos ωt ) + A p sin ω i t + B p cos ω i t: x = −e −t (2.051sin3t − 3.077cos3t ) + 4.615sin 2t − 3.077cos2t mm.

Equate like coefficients: (ω 2 − ω i2 ) A p − 2d ω i B p = x i ω i2 ,

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Problem 21.74 The coordinate x measures the displacement of the mass relative to the position in which the spring is unstretched. The mass is given the initial conditions  x = 0.1 m,  t = 0  dx  = 0.  dt (a) Determine the position of the mass as a function of time. (b) Draw graphs of the position and velocity of the mass as functions of time for the first 5 s of motion. Solution: d 2x dt 2

The canonical equation of motion is

x 90 N/m 10 kg

(b) The graphs are shown.

+ ω 2 x = 0,

where

ω =

k = m

.4 .3 .2 .1 0 –.1 –.2 –.3 –.4

90 = 3 rad/s. 10

(a) The position is x (t ) = A sin ωt + B cos ωt, and the velocity is dx (t ) = ω A cos ωt − ω B sin ωt. dt At t = 0, x (0) = 0.1 = B, and

Position and velocity vs time velocity, m/s

position, m

1.5

2.5 t, s

3.5

4.5

dx (0) = 0 = ω A, dt from which A = 0, B = 0.1 m, and x (t ) = 0.1cos3t m , dx (t ) = −0.3sin 3t m/s. dt x

Problem 21.75 When t = 0, the mass is in the position in which the spring is unstretched and has a velocity of 0.3 m/s to the right. Determine the position of the mass as a function of time and the amplitude of the vibration (a) by expressing the solution in the form given by Eq. (21.8) and (b) by expressing the solution in the form given by Eq. (21.9) Solution:

From Eq. (21.5), the canonical form of the equation of

motion is d 2x + ω 2 x = 0, dt 2 k where ω = = m

90 N/m 10 kg

(b) From Eq. (21.7) the position is x (t ) = E sin(ωt − φ), and the velocity is

90 = 3 rad/s. 10

(a) From Eq. (21.6) the position is x (t ) = A sin ωt + B cos ωt, and the velocity is

dx (t ) = ω E cos(ωt − φ). dt At t = 0, x (0) = −E sin φ, and the velocity is dx (0) = 0.3 = ω E cos φ. dt

dx (t ) = ω A cos ωt − ω B sin ωt. dt dx (0) At t = 0, x (0) = 0 = B, and = 0.3 = ω A m/s, from dt 0.3 which B = 0 and A = = 0.1 m. The position is ω x (t ) = 0.1sin 3t m. The amplitude is

Solve: φ = 0, E = from which

0.3 = 0.1, ω

x (t ) = 0.1sin 3t .

The amplitude is x (t ) = E = 0.1 m.

x (t ) max = 0.1 m.

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Problem 21.76 A homogeneous disk of mass m and radius R rotates about a fixed shaft and is attached to a torsional spring with constant k. (The torsional spring exerts a restoring moment of magnitude kθ, where θ is the angle of rotation of the disk relative to its position in which the spring is unstretched.) Show that the period of rotational vibrations of the disk is τ = π R 2m /k .

Solution:

From the equation of angular motion, the equation of motion is I α = M, where M = −kθ, from which I

d 2θ + kθ = 0, dt 2

and the canonical form (see Eq. (21.4)) is d 2θ + ω 2θ = 0, dt 2

where

ω =

k . I

For a homogenous disk the moment of inertia about the axis of rotation is I =

mR 2 , 2

from which

The period is τ =

ω =

2k 1 2k = . mR 2 R m

2π m 2m = 2π R = πR . ω 2k k

Problem 21.77 Assigned to determine the moments of inertia of astronaut candidates, an engineer attaches a horizontal platform to a vertical steel bar. The moment of inertia of the platform about L is 7.5 kg-m 2 , and the frequency of torsional oscillations of the unloaded platform is 1 Hz. With an astronaut candidate in the position shown, the frequency of torsional oscillations is 0.520 Hz. What is the candidate’s moment of inertia about L?

Solution:

f = from which

1 ω = 2π 2π

k = 1 Hz, I

k = (2π f ) 2 I = (2π ) 2 7.5 = 296.1 N-m/rad.

The natural frequency of the loaded platform is f1 = from which

The natural frequency of the unloaded platform is

ω1 1 = 2π 2π

k = 0.520 Hz, I1

 1  2 I 1 =   k = 27.74 kg-m 2 ,  2π f1 

from which

I A = I 1 − I = 20.24 kg-m 2 .

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Problem 21.78 The 22-kg platen P rests on four roller bearings that can be modeled as 1-kg homogeneous cylinders with 30-mm radii. The spring constant is k = 900 N/m. What is the frequency of horizontal vibrations of the platen relative to its equilibrium position? Solution:

The kinetic energy is the sum of the kinetic energies of translation of the platen P and the roller bearings, and the kinetic energy of rotation of the roller bearings. Denote references to the platen by the subscript P and references to the ball bearings by the subscript B:

T =

( ) + 12 (4m )( dxdt ) + 12 (4I )( ddtθ ) . 2

dx P 1 m 2 P dt

The potential energy is the energy stored in the spring: V =

1 2 kx . 2 P

xB =

and

( dxdt ) = 0 I d x or ( m + m + + kx = 0. R )( dt ) P

xP . 2

B 2

P 2

The first can be ignored, from which the canonical form of the equation of motion is d 2xP + ω 2 x P = 0, dt 2 k where ω = R . R 2 (m P + m B ) + I B

IB =

( 12 )( m + m + RI )( dxdt ) + ( 12 )kx = const. B 2

For a homogenous cylinder,

Since the system is conservative, T + V = const. Substitute the kinematic relations and reduce: B

This has two possible solutions,

−Rθ = x B

From kinematics,

2 P

from which ω =

mBR 2 , 2 k mP +

The frequency is f =

3 m 2 B

= 6.189 rad/s.

ω = 0.985 Hz . 2π

Take the time derivative:

( dxdt )  ( m + m + RI )( ddtx ) + kx  = 0. P

B 2

P 2

Problem 21.79 At t = 0, the platen described in Problem 21.78 is 0.1 m to the left of its equilibrium position and is moving to the right at 2 m/s. What are the platen’s position and velocity at t = 4 s? Solution:

The position is

x (t ) = A sin ωt + B cos ωt, and the velocity is dx = ω A cos ωt − ω B sin ωt, dt where, from the solution to Problem 21.78, ω = 6.189 rad/s. At t = 0, x(0) = −0.1 m, and  dx  = 2 m/s,  dt  t=0 from which B = −0.1 m, A =

2 = 0.3232 m. ω

The position and velocity are x (t ) = 0.3232sin(6.189 t ) − 0.1cos(6.189 t )(m), dx = 2 cos(6.189 t ) + 0.6189sin(6.189 t )(m/s). dt At t = 4 s,

x = −0.2124 m , dx = 1.630 m/s. dt

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Problem 21.80 The moments of inertia of gears A and B are I A = 0.014 slug-ft 2 and I B = 0.100 slug-ft 2 . Gear A is connected to a torsional spring with constant k = 2 ft-lb/rad. What is the frequency of angular vibrations of the gears relative to their equilibrium position?

10 in 3 in

6 in

5 lb

Solution:

The system is conservative. The strategy is to determine the equilibrium position from the equation of motion about the unstretched spring position. Choose a coordinate system with the y axis positive upward. Denote R A = 6 in. = 0.5 ft, R B = 10 in. = 0.8333 ft, and R M = 3 in. = 0.25 ft, and W = 5 lb. The kinetic energy of the system is T =

1 2 1 1W 2 I θ + I Bθ B2 + v , 2 A A 2 2 g

1 2 kθ + W y . 2 A

v = R M θ B ,

F = ω =

WR M  R A  , M  R B  k = 6.114 rad/s, M

is the equation of motion about the unstretched spring position. Note that R  F W = R  A  = 0.375 rad ω2 k M  R B 

F θ = θ A − 2 , ω

R  θ B = − A  θ A ,  RB 

from which the canonical form (see Eq. (21.4)) of the equation of motion about the equilibrium point is

R  y = R M θ B = −R M  A  θ A .  RB 

d 2θ + ω 2θ = 0, dt 2

T + V = const. =

and the natural frequency is



( 12 )  I +  RR  I A

f =

R   W  +   ( R M2 ) A   θ A2 g  R B   +

Define

where

is the equilibrium position of θ A , obtained by setting the acceleration to zero (since the non-homogenous term F is a constant). Make the change of variable:

From kinematics,

Substitute,

d 2θ A + ω 2θ A = F, dt 2

and

where θ A , θ B are the angular velocities of gears A and B respectively, and v is the velocity of the 5 lb weight. The potential energy is the sum of the energy stored in the spring plus the energy gain due to the increase in the height of the 5 lb weight: V =

Ignore the possible solution θ A = 0, to obtain

ω = 0.9732 Hz. 2π

( 12 )kθ − (R ) RR θ W . 2 A

2  R 2 W 2  R A  M = I A +  A  I B + (R )   RB  g M  R B 

= 0.05350 slug-ft 2 , and take the time derivative:

(

)

  R  d 2θ A θ A  M + kθ A − W ( R M ) A   = 0.  R B    dt 2

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Problem 21.81 The 5-lb weight in Problem 21.80 is raised 0.5 in from its equilibrium position and released from rest at t = 0. Determine the counterclockwise angular position of gear B relative to its equilibrium position as a function of time.

10 in 3 in

A 6 in

5 lb

Solution:

From the solution to Problem 21.80, the equation of motion for gear A is d 2θ A + ω 2θ A = F, dt 2 2

R  W 2  R A  M = I A +  A  I B + (R )   RB  g M  R B 

where

Assume a solution of the form θ A = A sin ωt + B cos ωt. The displacement from the equilibrium position is, from kinematics,

R  θ A (t = 0) =  B  θ A  RA  1  R B   y (t = 0) R M  R A  = −0.2778 rad,

= 0.05350 slug-ft 2 , ω =

k = 6.114 rad/s, M

R  WR M  A   RB  F = = 14.02 rad/s 2 . M

and

As in the solution to Problem 21.80, the equilibrium angular position θ A associated with the equilibrium position of the weight is [θ A ] eq =

F = 0.375 rad. ω2

= −

from which the initial conditions are θ A (t = 0) = −0.2778 rad

and

 d θ A    = 0,  dt  t = 0

from which B = −0.2778, A = 0. The angular position of gear A is θ a = −0.2778cos(6.114t ) rad, from which the angular position of gear B is

Make the change of variable:

R  θ B = − A θ A = 0.1667cos(6.114t )rad.  RB 

θ A = θ A − [θ A ] eq ,

about the equilibrium position.

from which the canonical form of the equation of motion about the equilibrium point is d 2θ A + ω 2θ A = 0. dt 2

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Problem 21.82 The mass of the slender bar is m. The spring is unstretched when the bar is vertical. The light collar C slides on the smooth vertical bar so that the spring remains horizontal. Determine the frequency of small vibrations of the bar.

k C

Solution:

The system is conservative. Denote the angle between the bar and the vertical by θ. The base of the bar is a fixed point. The kinetic energy of the bar is T =

1 dθ 2 I . 2 dt

( )

Denote the datum by θ = 0. The potential energy is the result of the change in the height of the center of mass of the bar from the datum and the stretch of the spring, V = −

L sin u

mgL 1 (1 − cos θ) + k ( L sin θ) 2 . 2 2

The system is conservative, T + V = const =

mgL 1 dθ 2 kL2 I + sin 2 θ − (1 − cos θ). 2 dt 2 2

( )

Take the time derivative: sin θ  = 0. ( ddtθ )  I ddt θ + kL sin θ cos θ − mgL  2 2

From which I

d 2θ dt 2

+ kL2 sin θ cos θ −

mgL sin θ = 0. 2

3g 3k − . m 2L

where ω =

For small angles, sin θ → θ, cos θ → 1. The moment of inertia about the fixed point is I =

d 2θ + ω 2θ = 0, dt 2

from which

mL2 , 3

The frequency is f =

3g ω 1 3k = − . 2π 2π m 2L

Problem 21.83 A homogeneous hemisphere of radius R and mass m rests on a level surface. If you rotate the hemisphere slightly from its equilibrium position and release it, what is the frequency of its vibrations?

Solution:

The system is conservative. The distance from the center of mass to point O is h = 3 R /8. Denote the angle of rotation about P by θ. Rotation about P causes the center of mass to rotate relative to the radius center OP, suggesting the analogy with a pendulum suspended from O. The kinetic energy is T = (1/2) I pθ 2 . The potential energy is V = mgh(1 − cos θ), where h(1 − cos θ) is the increase in height of the center of mass. T + V = const = I pθ 2 + mgh(1 − cos θ). Take the time derivative: d 2θ θ  I p 2 + mgh sin θ  = 0,  dt  from which

where

For small angles sin θ → θ, and the moment of inertia about P is 83 5 2 13 mR 2 + mR 2 = mR 2 , 320 8 20

( )

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3R 8

from which

mgh d 2θ + sin θ = 0. dt 2 Ip

I p = I CM + m( R − h) 2 =

d 2θ + ω 2θ = 0, dt 2

ω =

20(3) g = 13(8) R

The frequency is f =

15g . 26 R

1 15g . 2π 26 R

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Problem 21.84 The frequency of the spring–mass oscillator is measured and determined to be 4.00 Hz. The oscillator is then placed in a barrel of oil, and its frequency is determined to be 3.80 Hz. What is the logarithmic decrement of vibrations of the mass when the oscillator is immersed in oil? Solution:

The undamped and damped frequencies are f = 4 Hz and f d = 3.8 Hz, so τd =

1 = 0.263 s, fd

ω = 2π f = 25.13 rad/s,

k 10 kg

From the relation ωd =

ω2 − d2,

we obtain d = 7.85 rad/s, so the logarithmic decrement is δ = d τ d = (7.85)(0.263) = 2.07. δ = 2.07.

ω d = 2π f d = 23.88 rad/sa

Problem 21.85 Consider the oscillator immersed in oil described in Problem 21.84. If the mass is displaced 0.1 m to the right of its equilibrium position and released from rest, what is its position relative to the equilibrium position as a function of time?

k 10 kg

Solution:

The mass and spring constant are unknown. The canonical form of the equation of motion (see Eq. (21.16)) is d 2x dx + 2d + ω 2 x = 0, dt 2 dt

where, from the solution to Problem 21.84, d = 7.848 rad/s, and ω = 2π (4) = 25.13 rad/s. The solution is of the form (see Eq. (21.19)) x = e −dt ( A sin ω d t + B cos ω d t ), where ω d = 2π (3.8) = 23.88 rad/s. Apply the initial conditions: at t = 0, x 0 = 0.1 m, and x 0 = 0, from which B = x 0 , and

from which A =

dx 0 , ωd

The solution is  d  x = x 0 e −dt  sin ω d + cos ω d   ωd  = 0.1e −7.848t (0.3287sin 23.88t + cos 23.88t ).

0 = −dB + ω d A,

Problem 21.86 The stepped disk weighs 20 lb, and its moment of inertia is I = 0.6 slug-ft 2 . It rolls on the horizontal surface. If c = 8 lb-s/ft, what is the frequency of vibration of the disk?

16 lb/ft

16 in

c 8 in

Solution:

kS cx f

∑ M C = RkS + 2 Rf .

From the equation of angular motion, I

d 2θ = ∑ M C = RkS + 2 Rf . dt 2

Solve for the reaction at the floor: f =

650

I d 2θ k − S. 2 R dt 2 2

The sum of the horizontal forces: ∑ Fx = −kS − c

dx + f. dt

From Newton’s second law: m

d 2x dx = ∑ Fx = −kS − c + f. dt 2 dt

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21.86

(Continued) The canonical form of the equation of motion is

Substitute for f and rearrange: m

d 2x dx + 2d + ω 2 x = 0, dt 2 dt

dx d 2x I d 2θ 3 + kS = 0 + + c dt 2 2 R dt 2 dt 2

From kinematics, the displacement of the center of the center of the disk is x = −2 Rθ. The stretch of the spring is the amount wrapped around the disk plus the translation of the disk, S = −Rθ − 2 Rθ = −3 Rθ =

3 x. 2

( )

M = m+

ω2 =

c = 4.170 rad/s, 2M

fd =

( 32 ) Mk = 37.53 (rad/s) .

Therefore ω d =

 I  d 2 x dx 3 2 + kx = 0. Substitute:  m +  2 + c 2  (2 R) dt dt 2 Define

where d =

ω 2 − d 2 = 4.488 rad/s, and the frequency is

ωd = 0.714 Hz. 2π

I = 0.9592 slug. (2 R) 2

Problem 21.87 The stepped disk described in Problem 21.86 is initially in equilibrium, and at t = 0 it is given a clockwise angular velocity of 1 rad/s. Determine the position of the center of the disk relative to its equilibrium position as a function of time.

16 lb/ft

16 in

c 8 in

Solution:

From the solution to Problem 21.86, the equation of

motion is d 2x dx + 2d + ω 2 x = 0, dt 2 dt where d = 4.170 rad/s, ω 2 = 37.53(rad/s) 2 . Therefore ωd =

ω 2 − d 2 = 4.488 rad/s.

Since d 2 < ω 2 , the system is sub-critically damped. The solution is of the form x = e −dt ( A sin ω d t + B cos ω d t ). Apply the initial conditions: at t = 0, θ 0 = 0, andθ 0 = −1 rad/s.. From kinematics, x 0 = −2 Rθ 0 = 2 R ft. Substitute, to obtain B = 0 and A =

x 0 = 0.2971 ωd

and the position of the center of the disk is x = 0.2971e −4.170 t sin 4.488t.

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Problem 21.88 The stepped disk described in Problem 21.86 is initially in equilibrium, and at t = 0 it is given a clockwise angular velocity of 1 rad/s. Determine the position of the center of the disk relative to its equilibrium position as a function of time if c = 16 lb-s/ft.

Solution:

From the solution to Problem 21.86, the equation of

motion is d 2x dx + 2d + ω 2 x = 0, dt 2 dt c where d = = 8.340 rad/s, 2M 2 3 k ω2 = = 37.53 (rad/s) 2 . 2 M

( )

16 lb/ft

16 in

c 8 in

Since d 2 > ω 2 , the system is supercritically damped. The solution is of the form (see Eq. (21.24)) x = e −dt (Ce ht + De −ht ), where h = d 2 − ω 2 = 5.659 rad/s. Apply the initial conditions: at t = 0, θ 0 = 0, and θ 0 = −1 rad/s. From kinematics, x 0 = −2 Rθ 0 = 2 R ft. Substitute, to obtain 0 = C + D and x 0 = −(d − h)C − (d + h) D. Solve: x0 = 0.1178, 2h x D = − 0 = −0.1178, 2h C =

from which the position of the center of the disk is x = 0.1179e −8.340 t (e 5.659t − e −5.659t ) = 0.118(e −2.680 t − e −14.00 t ) ft.

Problem 21.89 The 22-kg platen P rests on four roller bearings that can be modeled as 1-kg homogeneous cylinders with 30-mm radii. The spring constant is k = 900 N/m. The platen is subjected to a force F (t ) = 100 sin 3t N. What is the magnitude of the platen’s steady-state horizontal vibration? Solution:

Choose a coordinate system with the origin at the wall and the x axis parallel to the plane surface. Denote the roller bearings by the subscript B and the platen by the subscript P.

F(t) FB

The roller bearings: The sum of the moments about the mass center of a roller bearing is ∑ M B − cm = + RFB + Rf B . From Newton’s second law: IB

d 2θ = RFB + Rf B . dt 2

Solve for the reaction at the floor: I d 2θ f B = B 2 − FB . R dt

From Newton’s second law mB

652

d 2xB = −FB + f B , dt 2

where x B is the translation of the center of mass of the roller bearing. Substitute f p, mB

d 2xB I d 2θ = B 2 − 2 FB . dt 2 R dt

From kinematics, θ B = − from which

The sum of the horizontal forces on each roller bearing: ∑ Fx = −FB + f P .

F(t)

xB , R

( m + RI ) ddtx = −2F . B

B 2

The platen: The sum of the forces on the platen are ∑ FP = −kx + 4 FB + F (t ). From Newton’s second law, mP

d 2xP = −kx p + 4 FB + F (t ). dt 2

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21.89

(Continued) For d = 0, the canonical form of the equation of motion (see Eq. (21.26)) is

Substitute for FB and rearrange: mp

d 2x

dt 2

(

+ kx P + 2 m B +

From kinematics, x B =

(

)

I B d 2xB = F (t ). R 2 dt 2

xP , 2

from which m P + m B +

)

I B d 2xP + kx P = F (t ). R 2 dt 2

where

ω 2 = 38.30 (rad/s) 2 ,

and

a(t ) =

F (t ) = 4.255sin 3t (m/s 2 ). M

The amplitude of the steady state motion is given by Eq. (21.31):

For a homogenous cylinder, mBR 2 , 2

IB =

d 2x p + ω 2 x p = a(t ), dt 2

Ep =

4.255 = 0.1452 m. (ω 2 − 3 2 )

from which we define M = mp +

3 m = 23.5 kg. 2 B

Problem 21.90 At t = 0, the platen described in Problem 21.89 is 0.1 m to the right of its equilibrium position and is moving to the right at 2 m/s. Determine the platen’s position relative to its equilibrium position as a function of time. Solution:

F(t)

From the solution to Problem 21.89, the equation of

motion is d 2x p + ω 2 x p = a(t ), dt 2 where ω 2 = 38.30 (rad/s) 2 , and a(t ) =

F (t ) = 4.255sin 3t m/s 2 . M

The solution is in the form x = x h + x p , where the homogenous solution is of the form x h = A sin ωt + B cos ωt and the particular solution x p is given by Eq. (21.30), with d = 0 and b 0 = 0. The result: x = A sin ωt + B cos ωt +

a0 sin ω 0 t, (ω 2 − ω 02 )

where a 0 = 4.255 m, ω = 6.189 rad/s, and ω 0 = 3 rad/s. Apply the initial conditions: at t = 0, x 0 = 0.1 m, and x 0 = 2 m/s, from which B = 0.1, and A =

( )

a0 ω0 2 − = 0.2528, ω ω (ω 2 − ω 02 )

from which x = 0.253sin 6.19t + 0.1cos6.19t + 0.145sin3t m.

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Problem 21.91 The moments of inertia of gears A and B are I A = 0.014 slug-ft 2 and I B = 0.100 slug-ft 2 . Gear A is connected to a torsional spring with constant k = 2 ft-lb/rad. The bearing supporting gear B incorporates a damping element that exerts a resisting moment on gear B of magnitude 1.5(d θ B /dt ) ft-lb, where d θ B /dt is the angular velocity of gear B in rad/s. What is the frequency of angular vibration of the gears?

10 in 3 in

A 6 in

5 lb

Solution: Choose a coordinate system with the x axis positive downward. The sum of the moments on gear A is ∑ M = −kθ A + R A F, where the moment exerted by the spring opposes the angular displacement θ A . From the equation of angular motion, IA

d 2θ A = ∑ M = −kθ A + R A F , dt 2

from which

 I  d 2θ  k  F =  A  2A +  θ .  R A  dt  R A  A

The sum of the moments acting on gear B is

F FW

The canonical form of the equation of motion is

dθ ∑ M = −1.5 B + R B F − RW FW , dt

d 2θ B dθ + 2 d B + ω 2 θ B = P, dt 2 dt

where W = 5 lb, and the moment exerted by the damping element opposes the angular velocity of B. From the equation of angular motion applied to B, IB

dθ d 2θ B = ∑ M = −1.5 B + R B F − RW FW . dt 2 dt

The sum of the forces on the weight are ∑ F = +FW − W. From Newton’s second law applied to the weight,  W  d 2 x  g  dt 2 = FW − W,

where d =

1.5 = 5.047 rad/s, 2M

 R B  2  R  k A ω2 = = 37.39 (rad/s) 2 , M and P =

RW W = 8.412 (rad/s) 2 . M

The system is sub critically damped, since d 2 < ω 2 , from which ω d = ω 2 − d 2 = 3.452 rad/s, and the frequency is

 W  d 2x from which FW =   2 + W .  g  dt

fd =

ωd = 0.5493 Hz. 2π

Substitute for F and FW to obtain the equation of motion for gear B: IB

(

R  d 2θ B dθ d 2θ A + 1.5 B −  B  I A + kθ A  RA  dt 2 dt dt 2

)

 W  d 2 x  − RW    2 + W  = 0.   g  dt  R ( RA )θ , and x = −R θ , from which

From kinematics, θ A = −

W B

 R 2 d 2θ B dθ + 1.5 B +  B  kθ B = RW W , 2  RA  dt dt

 R 2 W  where M = I A +  B  I A + RW2   = 0.1486 slug-ft 2 g  RA 

654

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Problem 21.92 The 5-lb weight in Problem 21.91 is raised 0.5 in from its equilibrium position and released from rest at t = 0. Determine the counterclockwise angular position of gear B relative to its equilibrium position as a function of time.

10 in 3 in

A 6 in

5 lb

Solution:

From the solution to Problem 81.91,

dθ d 2θ B + 2 d B + ω 2 θ B = P, dt 2 dt where d = 5.047 rad/s, ω 2 = 37.39 (rad/s) 2 , and P = 8.412 (rad/s) 2 . Since the non homogenous term P is independent of time and angle, the equilibrium position is found by setting the acceleration and velocity to zero in the equation of motion and solving: P θ eq = 2 . Make the transformation θ B = θ B − θ eq , from which, by ω substitution, d θ d 2θ B + 2d B + ω 2θ = 0 dt 2 dt

is the equation of motion about the equilibrium point. Since d 2 < ω 2 , the system is sub critically damped, from which the solution is θ B = e −dt ( A sin ω d t + B cos ω d t ). Apply the initial conditions: x 0 = −0.5 in. = −0.04167 ft, from which [θ B ] t = 0 = −

x0 d θ B = 0.1667 rad, = 0, Rω dt

from which B = 0.1667, A =

d (0.1667) = 0.2437. ωd

The solution is θ (t ) = e −5.047t (0.244sin(3.45t ) + 0.167cos(3.45t )).

Problem 21.93 The base and mass m are initially stationary. The base is then subjected to a vertical displacement h sin ω i t relative to its original position. What is the magnitude of the resulting steady-state vibration of the mass m relative to the base?

Base

d 2x + ω 2 x = a(t ), dt 2

Solution:

From Eq. (21.26), for d = 0,

where ω 2 =

k , and a(t ) = ω i2 h sin ω i t. From Eq. (21.31), the steady m

state amplitude is Ep =

ω i2 h = (ω 2 − ω i2 )

(

ω i2 h . k − ω i2 m

)

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Problem 21.94* The mass of the trailer, not including its wheels and axle, is m, and the spring constant of its suspension is k. To analyze the suspension’s behavior, an engineer assumes that the height of the road surface relative to its mean height is h sin(2π x /λ). Assume that the trailer’s wheels remain on the road and its horizontal component of velocity is v . Neglect the damping due to the suspension’s shock absorbers. (a) Determine the magnitude of the trailer’s vertical steady-state vibration relative to the road surface. (b) At what velocity v does resonance occur?

x l

Solution:

Since the wheels and axle act as a base that moves with the disturbance, this is analogous to the transducer problem (Example t

21.7). For a constant velocity the distance x = ∫ v dt = v t, from 0

which the movement of the axle-wheel assembly as a function of time 2πv is h f (t ) = h sin(ω 0 t ), where ω 0 = rad/s. [Check: The velocity λ of the disturbing “waves” in the road relative to the trailer is v. Use the physical relationship between frequency, wavelength and velocity of propagation λ f = v . The wavelength of the road disturbance is λ, v from which the forcing function frequency is f 0 = , and the circuλ 2πv lar frequency is ω 0 = 2π f 0 = . check.] The forcing function on λ the spring-mass oscillator (that is, the trailer body and spring) is (see Example 21.7) F (t ) = −m

d 2 h f (t ) 2πv 2 2πv = m h sin t . 2 dt λ λ

(

)

(

)

For d = 0, the canonical form of the equation of motion is d 2y k + ω 2 y = a(t ), where ω = = 8.787 rad/s, and m dt 2 a(t ) =

F (t ) 2πv 2 2πv = h sin t λ λ m

(

= a0

)

(

)

( 2λπv ) sin( 2λπv t ), 2

where a 0 = h. (a) The magnitude of the steady state amplitude of the motion relative to the wheel-axle assembly is given by Eq. (21.31) and the equations in Example 21.7 for d = 0 and b 0 = 0,

( 2λπv ) h . E = ( mk − ( 2λπv ) ) 2

(b) Resonance, by definition, occurs when the denominator vanishes, from which v =

656

λ k . 2π m

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Problem 21.95* The trailer in Problem 21.94, not including its wheels and axle, weighs 1000 lb. The spring constant of its suspension is k = 2400 lb/ft, and the damping coefficient due to its shock absorbers is c = 200 lb-s/ft. The road surface parameters are h = 2 in and λ = 8 ft. The trailer’s horizontal velocity is v = 6 mi/h. Determine the magnitude of the trailer’s vertical steady-state vibration relative to the road surface, (a) neglecting the damping due to the shock absorbers and (b) not neglecting the damping.

x l

Solution: (a) From the solution to Problem 21.94, for zero damping, the steady state amplitude relative to the road surface is

( 2λπv ) h . E = ( mk − ( 2λπv ) ) 2

Substitute numerical values: v = 6 mi/h = 8.8 ft/s, λ = 8 ft, k = 2400 lb/ft, W m = = 31.08 slug, g to obtain

E p = 0.2704 ft = 3.25 in.

(b) From Example 21.7, and the solution to Problem 21.94, the canonical form of the equation of motion is d 2y dy + 2d + ω 2 y = a(t ), dt 2 dt where d =

and

c 200 = = 3.217 rad/s, ω = 8.787 rad/s, 2m 2(31.08)

a(t ) = a 0

( 2λπv ) sin( 2λπv t ), 2

where a 0 = h. The system is sub-critically damped, from which ω d = ω 2 − d 2 = 8.177 rad/s. From Example 21.7, the magnitude of the steady state response is

( 2λπv ) h E = ( ω − ( 2λπv ) ) + 4d ( 2λπv ) 2

2 2

= 0.149 ft = 1.791 in.

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Problem 21.96* A disk with moment of inertia I rotates about a fixed shaft and is attached to a torsional spring with constant k. The angle θ measures the angular position of the disk relative to its position when the spring is unstretched. The disk is initially stationary with the spring unstretched. At t = 0, a time-dependent moment M (t ) = M 0 (1 − e −t ) is applied to the disk, where M 0 is a constant. Show that the angular position of the disk as a function of time is θ =

M0 I

 1 1 sin ωt − 2 cos ωt −  ω (1 + ω 2 ) ω (1 + ω 2 )  1 1 e −t  . + 2 −  (1 + ω 2 ) ω

θ p = Ap + B p e −t ,

Solution:

The sum of the moments on the disk are ∑ M = −kθ + M (t ). From the equation of angular motion, d = 0

For

the

canonical

form

M0 k d 2θ + ω 2θ = a(t ), where ω = (1 − e −t ). , and a(t ) = dt 2 I I (a) For the particular solution, assume a solution of the form θ p = A p + B p e −t . Substitute into the equation of motion, d 2θ p M0 (1 − e −t ). + ω 2θ p = B p e −t + ω 2 ( A p + B p e −t ) = dt 2 I Rearrange:

θ = θ h + θ p = A sin ωt + B cos ωt +

M0  1  e −t  − . I  ω 2 (1 + ω 2 ) 

M  1  1 θ 0 = 0: θ 0 = 0 = B + 0  2 − , I ω (1 + ω 2 )  M   1 θ 0 = 0 = ω A + 0  , I  (1 + ω 2 )  from which M0   1  , I  ω (1 + ω 2 )  M  1 M0    1 1  B = − 0  2 − , = − (1 + ω 2 )  I ω I  ω 2 (1 + ω 2 )  A = −

and the complete solution is θ =

ω 2 A p + B p (1 + ω 2 )e −t =

M(t)

Apply the initial conditions: at t = 0, θ 0 = 0 and

where Ap and B p are constants that you must determine.

d 2θ = −kθ + M (t ). dt 2

The system is undamped. The solution to the homogenous equation has the form θ h = A sin ωt + B cos ωt. The trial solution is:

Strategy: To determine the particular solution, seek a solution of the form

M0 M − 0 e −t . I I

M0  1 1 sin ωt − 2 cos ωt − ω (1 + ω 2 ) I  ω (1 + ω 2 )  1 1 + 2 − e −t  .  (1 + ω 2 ) ω

Equate like coefficients: ω 2 Ap =

M0 M , B p (1 + ω 2 ) = − 0 , I I

from which A p =

M0 M0 . , and B p = − ω 2I (1 + ω 2 ) I

The particular solution is θp =

658

M0  1  e −t −  . I  ω 2 (1 + ω 2 ) 

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