DYNAMICS
Kinematics of Particles
Given: Find: t when v = 0 v, a and d when x = 0
Assumption: (a)
Velocity is zero at time 6 s. (b)
1
Chapter 11
DYNAMICS
Kinematics of Particles
Chapter 11
At
Distance traveled from
Distance traveled from
Distance traveled from |
|
|
|
â „
When x = 0,
2
â „
DYNAMICS
Kinematics of Particles
11.11
At t = 0 s, v=16 in/s ∫
∫
At
⁄
3
Chapter 11
DYNAMICS
Kinematics of Particles
At t = 1 s, x = 20 in ∫
∫
(
)
(
)
At t = 7 s
Distance traveled from ⁄
(
) Distance traveled from
Distance traveled from |
|
|
| 4
Chapter 11
DYNAMICS
Kinematics of Particles
Chapter 11
11.19
The packing material can be treated as a spring, and the equipment as a particle in simple harmonic motion. Its velocity will be maximum when the particle passes through the equilibrium position (x = 0), and its acceleration will be maximum at the turning points (first turning point at x = 20 mm). While the box is falling, the packing material is not being compressed, so x = 0. Compression starts when the box hits the ground at 4 m/s, so we will considered this velocity to be v0 , and x0 = 0 m. Final velocity is vf = 0 m/s, and xf = 0.02 m. Maximum acceleration of the equipment is reached at the time of maximum compression. Acceleration is given as a function of x. Positive x direction is up. Given:
⁄
⁄
Find: Maximum acceleration of the equipment. 1. Find value of k.
∫
∫
∫
∫
(
2. Use known values of k and xmax to find amax.
⁄
(upwards)
5
)
(
)
DYNAMICS
Kinematics of Particles
11.24
Given: ⁄
Find:
∫
∫ |
|
| (
|
|
| )
(
|
| )
|
√
6
Chapter 11
DYNAMICS
Kinematics of Particles
11.27
(a)
∫
∫ [ [
] ]
⁄ ⁄
(b)
[
]
[
]
(c) ⁄
⁄
7
Chapter 11
DYNAMICS
Kinematics of Particles
11.28
Given:
Find: a at x = 2 m when v0 = 3.6 m/s t from x = 1 to x = 3 m when v0 = 3.6 m/s
(a) (
)
(
)
(
)(
For v0 = 3.6 m/s and x = 2:
(b) ( ∫
) ∫
[
]
[ ]
8
)
Chapter 11
DYNAMICS
Kinematics of Particles
Chapter 11
11.40
Given: ⁄ ⁄
Find:
and time when runner B
should begin to run. (a) Runner A:
⁄ Velocity of runner A at t = 1.82 s = velocity of runner B
Runner B:
⁄ (b)
Find how long it takes runner B to go from 0 m/s to 9.08 m/s:
Therefore, runner B should start running 2.59 seconds before runner A reaches the exchange zone. 9
DYNAMICS
Kinematics of Particles
11.45
Given: ⁄
From
From ⁄
At
10
Chapter 11
DYNAMICS
Kinematics of Particles
At
(
)
(
)
(
)
(
)
( (
)
)
answer to (b) (
)
â „
(c)
(
)
11
Chapter 11
DYNAMICS
Kinematics of Particles
11.61
Velocity: ⁄ ⁄ ⁄ ⁄ ⁄ ⁄ ⁄ Position:
12
Chapter 11
DYNAMICS
Kinematics of Particles
Chapter 11
(a)
v(t) vs t x(t) vs t
(b) â „
11.101
13
DYNAMICS
Given:
Kinematics of Particles
â „
Find:
Find time when
Find
(a) Find time when 14
Chapter 11
DYNAMICS
Kinematics of Particles
Find x when t = 1.2712 s:
(b)
Therefore, the ball will land 7.01 m away from the net.
15
Chapter 11
DYNAMICS
Kinematics of Particles
Chapter 11
11.112
â „
Given: Find:
Largest value (less than 45o) of the angle
for which
.
Time required for the puck to reach
Time when
â „
Angle when
(
)
:
(
)
. At this angle, the puck will hit the crossbar. (a)
The largest value (less than 45o) of the angle
(b)
16
for which the puck will enter the net is
14.66o.
DYNAMICS
Kinematics of Particles
11.114
Given: Find: Minimum value for
(
.
)
(
)
√ According to the plot to the right of , 0 has a minimum value between 0 and 90 at 2.53 m/s when Therefore:
17
Chapter 11
DYNAMICS
Kinematics of Particles
11.119
Given: ̂
Velocity of boat:
̂
Velocity of boat relative to river: ̂
̂
̂
̂
̂
̂ ̂
̂
18
Chapter 11
DYNAMICS
Kinematics of Particles
11.125
Given: Find:
⃗⃗⃗⃗
⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗⃗⃗⃗
When
19
Chapter 11
DYNAMICS
Kinetics of Particles: Newton’s Second law
Chapter 12
12.4
Given: Load on spring scale: Find:
Weight of packages. Load indicated by the spring scale and the mass needed to balance the lever scale when the elevator moves upward with an acceleration of 4 ft/s2.
(a) Resolve forces on y-axis: ∑
(
) The load on the spring (force đ??šđ?‘ ) is equal to the mass of the package times the
(
)
acceleration of the elevator and the acceleration due to gravity combined:
đ?‘š đ??šđ?‘
đ?‘šđ?‘Ž
đ?‘Š đ?‘” 05
6
0 5 đ?‘ đ?‘™đ?‘˘đ?‘” đ?&#x;?đ?&#x;’ đ?&#x;? đ?’?đ?’ƒ
when elevator is going down
1
DYNAMICS
Kinetics of Particles: Newton’s Second law
(b)
Chapter 12
The load on the spring (force đ??šđ?‘ ) is equal to the mass of the package times
∑
the acceleration of the elevator and the acceleration due to gravity combined:
đ?‘š
( 6 (
)
đ??šđ?‘
đ?‘šđ?‘Ž
đ?‘Š đ?‘” 05
6
0 5 đ?‘ đ?‘™đ?‘˘đ?‘” đ?&#x;?đ?&#x;– đ?&#x;? đ?’?đ?’ƒ
when elevator is going up
)
(Load indicated by the spring scale when the elevator accelerates upwards.)
In order for the lever scale to be in equilibrium, the mass of the weights has to equal the mass of the package. If the lever scale is in equilibrium, any effects caused by acceleration will be equal on both sides of the scale, so they can be neglected.
Therefore, Mass of the weights = Mass of the package
6
2
DYNAMICS
Kinetics of Particles: Newton’s Second law
Chapter 12
12.7
Given:
60 50
Find: x at F.B.D. of bus on level road. Forces acting on the bus: Weight W Normal N Traction force P
Level: ∑
3
Kinetics of Particles: Newton’s Second law
DYNAMICS
Chapter 12
Uphill: ∑
(
) F.B.D. of bus on the incline (
)
0
0 0 66
4
Kinetics of Particles: Newton’s Second law
DYNAMICS 12.10
Given: ⁄ Find:
F.B.D. of package at point A, with x-axis along the slope:
Find an expression for the force of friction: 0 0
0
0 0
0 0
0
5
Chapter 12
Kinetics of Particles: Newton’s Second law
DYNAMICS 0
0 0 0
0
F.B.D. of package at point B:
0 5
0
5 5
0
5
5 5 5
0
5
0
5 5
0
5
6
Chapter 12
Kinetics of Particles: Newton’s Second law
DYNAMICS
Chapter 12
12.13
0 0
Given:
0 0
5
66
0
Find: Shortest distance in which the rig can be brought to a stop if the load is not to shift. F.B.D. of load while braking (load wants to move fwd). Sliding is impending. 0 0
0
0 0
0
66 66
7
DYNAMICS
Kinetics of Particles: Newton’s Second law
Chapter 12
12.14
Given: 60 0 5 000 00 600
Find:
00
and
(a) the distance traveled by the tractor-trailer before it comes to a stop. (b) the horizontal component of the force in the hitch between the tractor and the trailer while they are slowing down. F.B.D. of tractor and trailer combined:
(a)
600
00
5 000
00
00 00
0 8
Kinetics of Particles: Newton’s Second law
DYNAMICS
(b)
Chapter 12
Replace cab with a coupling. Assume a tensile force on the coupling. Force is acting on the coupling.
00 00
00
00
9
DYNAMICS
Kinetics of Particles: Newton’s Second law
Chapter 12
12.22 To transport a series of bundles of shingles A to a roof, a contractor uses a motordriven lift consisting of a horizontal platform BC which rides on rails attached to the sides of a ladder. The lift starts from rest and initially moves with a constant acceleration a1 as shown. The lift then decelerates at a constant rate a2 and comes to rest at D, near the top of the ladder. Knowing that the coefficient of static friction between a bundle of shingles and the horizontal platform is 0.30, determine the largest allowable acceleration a1 and the largest allowable deceleration a2 if the bundle is not to slide on the platform. Given:
0 0
Find: The largest allowable acceleration a1 and the largest allowable deceleration a2 if the bundle is not to slide on the platform. (a)When the bundle accelerates, the force of friction on the platform is directed to the right: F.B.D.
65 65 65 65
[1]
65 65 65 65
10
Kinetics of Particles: Newton’s Second law
DYNAMICS
Chapter 12
65 65 65
65
65
65
65 65
65 0 0 65 0 0
65
65
(b) When the bundle decelerates, the force of friction left: Repeat part (a) changing the sign for a:
65 65
[2]
65 65
65
0 0 65 0 0
65
11
on the platform is directed to the
DYNAMICS
Kinetics of Particles: Newton’s Second law
Chapter 12
12.24 The propellers of a ship of weight W can produce a propulsive force F0; they produce a force of the same magnitude but of opposite direction when the engines are reversed. Knowing that the ship was proceeding forward at its maximum speed v0 when the engines were put into reverse, determine the distance the ship travels before coming to a stop. Assume that the frictional resistance of the water varies directly with the square of the velocity. Given: 0 Find: The distance the ship travels before coming to a stop. Since the frictional force is not dependent on the ship’s weight, forces acting on the vertical plane are omitted, and all calculations occur in the horizontal plane, with the positive direction being the direction in which the ship is moving. full steam ahead
full steam reverse
When the ship is moving full steam ahead at maximum speed (a constant speed), its acceleration is zero: 0 0
When the ship’s engines are put on reverse, the ship starts to decelerate until it comes to a stop:
12
Kinetics of Particles: Newton’s Second law
DYNAMICS
(
)
(
∫
)
∫
∫
∫
(
)[
(
)
(
)
(
)
(
)
]
(
)
(
13
)
Chapter 12
DYNAMICS
Kinetics of Particles: Newton’s Second law
12.27 Determine the maximum theoretical speed that a 2700lb automobile starting from rest can reach after traveling a quarter of a mile if air resistance is considered. Assume that the coefficient of static friction between the tires and the pavement is 0.70, that the automobile has front-wheel drive, that the front wheels support 62 percent of the automobile’s weight, and that the aerodynamic drag D has a magnitude 00 expressed in pounds and ft/s, respectively. 00
Given:
0
06
Chapter 12
, where D and v are 0 0
0
00
Find: The maximum theoretical speed. F.B.D.
As long as the traction tires are
=
not skidding, the force of friction between them and the ground is
0 0 06
(
00
)
equal to the car’s force forward.
00
6
00 0
0 00
6
The car’s acceleration is a function of velocity, therefore not constant. Since the problem involves �, and not �, integrate using �
0
0 âˆŤ
âˆŤ
0 14
đ?‘Ł đ?‘‘đ?‘Ł đ?‘‘đ?‘Ľ .
Kinetics of Particles: Newton’s Second law
DYNAMICS ∫
0 000
∫
∫
0 000 ∫
∫
0 000 ∫
06 650 5 650 5
0 000 5 000 5000
650 5 650 5 650 5
√
650 5
Insert value for : √
650 5
15
Chapter 12
Kinetics of Particles: Newton’s Second law
DYNAMICS
12.35 A 500-lb crate B is suspended from a cable attached to a 40-lb trolley A which rides on an inclined I-beam as shown. Knowing that at the instant shown the trolley has an acceleration of 1.2 ft/s2 up and to the right, determine (a) the acceleration of B relative to A, (b) the tension in cable CD. 500
Given:
0
Find: The acceleration of B relative to A, and the tension in cable CD. F.B.D. of B
= (a) For crate B:
0
5 5 5
5
(b) Find tension in cable AB:
5 5 (
5)
500 (
5)
16
50
Chapter 12
Kinetics of Particles: Newton’s Second law
DYNAMICS (b) F.B.D. of A
For trolley A:
50
5
5
5
5 5
0
5
0
17
Chapter 12
Kinetics of Particles: Newton’s Second law
DYNAMICS
Chapter 12
12.37
Given: mass of ball
50
0 5
constant speed Find: F.B.D. 0 0
(
) 0 5
Solving for
6
0 06
0 5
:
0 5 5
18
Kinetics of Particles: Newton’s Second law
DYNAMICS
Chapter 12
12.44 A child having a mass of 22 kg sits on a swing and is held in the position shown by a second child. Neglecting the mass of the swing, determine the tension in rope AB (a) while the second child holds the swing with his arms outstretched horizontally, (b) immediately after the swing is released.
5
Given:
Find: T when swing is not moving T immediately after the swing is released
F.B.D. when swing is held:
0 5
0 5
5
6
Swing is held by two ropes, so tension of each rope is: 6 F.B.D. when swing is released: Immediately after the swing is released, the swing’s acceleration is still zero: 0 5 5 5 6 Swing is held by two ropes, so tension of each rope is: 6
19
Kinetics of Particles: Newton’s Second law
DYNAMICS
Chapter 12
12.46 During a high-speed chase, a 2400-lb sports car traveling at a speed of 100 mi/h just loses contact with the road as it reaches the crest A of a hill. (a) Determine the radius of curvature of the vertical profile of the road at A. (b) Using the value of found in part a, determine the force exerted on a 160-lb driver by the seat of his 3100-lb car as the car, traveling at a constant speed of 50 mi/h, passes through A. Given:
00
00
66
60
50 Find: The force exerted on a 160-lb driver by the seat of his 3100-lb car as the car, traveling at a constant speed of 50 mi/h, passes through A. When the car loses contact with the road at point A, Normal and frictional forces are zero: F.B.D of Maserati Quattroporte as it loses contact with the road:
20
Kinetics of Particles: Newton’s Second law
DYNAMICS
(b) Driver is not moving horizontally with respect to her seat, so
F.B.D. of driver at point A
(
60 (
)
66
)
21
0.
Chapter 12
Kinetics of Particles: Newton’s Second law
DYNAMICS
Chapter 12
12.51 A curve in a speed track has a radius of 1000-ft and a rated speed of 120 mi/h. (See Sample Prob. 12.6 for the definition of rated speed.) Knowing that a racing car starts skidding on the curve when traveling at a speed of 180 mi/h, determine (a) the banking angle θ, (b) the coefficient of static friction between the tires and the track under the prevailing conditions, (c) the minimum speed at which the same car could negotiate the curve.
Given: radius of track rated speed
000 0
skidding speed
6 0
6
Find: (a) the banking angle θ (b) the coefficient of static friction between the tires and the track under the prevailing conditions (c) the minimum speed at which the same car could negotiate the curve.
= The car travels in a horizontal circular path of radius đ?œŒ. The normal component đ?‘Žđ?‘› of the acceleration is directed toward the center of the path; its magnitude is đ?‘Žđ?‘›
đ?‘Ł
đ?œŒ, where đ?‘Ł is the speed of the car
in ft/s. Between 120 mph and 180 mph, đ??šđ?‘“ is what keeps the car from skidding.
(
)
22
Kinetics of Particles: Newton’s Second law
DYNAMICS (a) 0
At rated speed, (
0
)
(
) 6 000
(
0 6
)
Substituting [1] and [2]: (
)
(
)
6 6
000 000 23
Chapter 12
DYNAMICS
Kinetics of Particles: Newton’s Second law
Chapter 12
(c) As speed decreases below 120 mph, Ff is what keeps the car from sliding down, so Ff in this case is pointing in the opposite direction as the force of friction that kept the car from sliding up the bank. Therefore,
Substituting [1] and [2]: (
)
(
)
000 000
0
5 05 6
0 0
0
050
0
0
5
0 √
0
5
6
24
Kinetics of Particles: Newton’s Second law
DYNAMICS
Chapter 12
12.59 Three seconds after a polisher is started from rest, small tufts of fleece from along the circumference of the 225-mmdiameter polishing pad are observed to fly free of the pad. If the polisher is started so that the fleece along the circumference undergoes a constant tangential acceleration of 4 m/s2, determine (a) the speed v of a tuft as it leaves the pad, (b) the magnitude of the force required to free a tuft if the average mass of a tuft is 1.6 mg. 5
Given:
0
5
0
5
6
6
0
Find: (a) the speed v of a tuft as it leaves the pad (b) the magnitude of the force required to free a tuft if the avg mass of a tuft is 1.6 mg. F.B.D.
(a)
0
(b)
6 6
0
6
0
0
05 √ 25
DYNAMICS Kinetics of Particles: Energy and Momentum Methods 13.5 Determine the maximum theoretical speed that may be achieved over a distance of 360 ft by a car starting from rest assuming there is no slipping. The coefficient of static friction between the tires and pavement is 0.75, and 60 percent of the weight of the car is distributed over its front wheels and 40 percent over its rear wheels. Assume (a) front-wheel drive, (b) rear-wheel drive. Given: Find: Maximum theoretical speed.
Principle of work and energy:
Kinetic energy: Position 1: Position 2: Work from 1 to 2:
(a) Front-wheel drive: đ??š
đ??šđ?‘“
đ?œ‡đ?‘ đ?‘“
(
)
đ?‘ đ?‘“ đ?‘ đ?‘“
đ??ž ( đ?‘Ł
đ?‘Š)( (
(b) Rear-wheel drive:
)(
1 2 1 đ?‘“đ?‘Ą đ?‘
đ??š đ?‘Š
đ?‘ˆ đ?‘“đ?‘Ą)
đ??šđ?‘“
đ?œ‡đ?‘ đ?‘&#x;
(
)
đ?‘ đ?‘&#x;
đ??ž
đ??ž đ?‘Š đ?‘Ł 2đ?‘”
(
8
đ?&#x;”đ?&#x;— đ?&#x;” đ?’Žđ?’‘đ?’‰
1
đ?‘Š)( (
đ?‘Ł
)2( 2 2)
đ?‘ đ?‘&#x;
)( đ?‘“đ?‘Ą đ?‘
đ?‘Š
đ?‘ˆ đ?‘“đ?‘Ą)
đ??ž đ?‘Š đ?‘Ł 2đ?‘”
)2( 2 2) đ?&#x;“đ?&#x;” đ?&#x;— đ?’Žđ?’‘đ?’‰
Chapter 13
DYNAMICS Kinetics of Particles: Energy and Momentum Methods
Chapter 13
13.6 Skid marks on a drag race track indicate that the rear (drive) wheels of a car slip for the first 60 ft of the 1320-ft track. (a) Knowing that the coefficient of kinetic friction is 0.60, determine the speed of the car at the end of the first 60-ft portion of the track if it starts from rest and the front wheels are just off the ground. (b) What is the maximum theoretical speed for the car at the finish line if, after skidding for 60 ft, it is driven without the wheels slipping for the remainder of the race? Assume that while the car is rolling without slipping, 60 percent of the weight of the car is on the rear wheels and the coefficient of static friction is 0.85. Ignore air resistance and rolling resistance. Given: Point 1 at Point 2 at 1 2
Point 3 at
8
Find:
and
rear-wheel drive
under given conditions.
(b) đ??š
(a) All weight on rear traction wheels. Spinning means đ??š đ??šđ?‘“ đ?œ‡đ?‘˜ đ?‘ đ?‘&#x; đ?œ‡đ?‘˜ đ?‘Š
đ??šđ?‘Ľ (
đ?‘Šđ?‘Ľ )đ?‘Š(
đ??ž
1đ?‘Š 81 2đ?‘”
đ?‘ˆ
đ??šđ?‘Ľ
đ?œ‡đ?‘
đ?‘Šđ?‘Ľ )đ?‘Š(12
)
đ?‘źđ?&#x;?
đ?&#x;?
1đ?‘Š đ?‘Ł 2đ?‘”
đ??ž
đ?‘˛đ?&#x;?
đ?‘Š đ?‘Ł 2đ?‘” 2đ?‘”
đ?‘Š 12 đ?‘“đ?‘Ą
2 đ?‘Š đ?‘˛đ?&#x;?
đ?‘Š
đ?œ‡đ?‘
đ?‘Š
( 8 )(
đ?‘Š
1đ?‘Š đ?‘Ł 2đ?‘”
đ??ž
đ?‘Ł
)
đ?œ‡đ?‘ đ?‘ đ?‘&#x; 1 2 −
đ?‘Ľ
đ??ž đ?‘ˆ
đ??šđ?‘“
đ?‘˛đ?&#x;? 8 1 đ?‘“đ?‘Ą đ?‘
đ?‘Š
đ?&#x;‘đ?&#x;? đ?&#x;– đ?’Žđ?’‘đ?’‰ đ?‘Ł 2
đ?‘źđ?&#x;?
2 đ?‘Š 1
đ?‘”
đ?&#x;?
đ?‘˛đ?&#x;?
đ?‘Š đ?‘Ł 2đ?‘”
2 9 đ?‘“đ?‘Ą đ?‘
đ?&#x;?đ?&#x;’đ?&#x;? đ?&#x;“ đ?’Žđ?’‘đ?’‰
DYNAMICS Kinetics of Particles: Energy and Momentum Methods
Chapter 13
13.15 The subway train shown is traveling at a speed of 30 mi/h when the brakes are fully applied on the wheels of cars B and C, causing them to slide on the track, but are not applied on the wheels of car A. Knowing that the coefficient of kinetic friction is 0.35 between the wheels and the track, determine (a) the distance required to bring the train to a stop, (b) the force in each coupling. Given: 8 1 8
Find: x and the force in each coupling.
(a)
1 ( ) 2 1 2 ( ) 2 −
81 1 9 (
)
−
81 1 9 − 81 1 9
3
DYNAMICS Kinetics of Particles: Energy and Momentum Methods (b) Force in coupling AB: Replace wagon A with a coupling. Assume a tensile force on the coupling. Force đ??šđ??´đ??ľ is acting on the coupling.
1 ( ) 2
18 2
11 18 − 12
11 18
(
)
− 81
12
2
12
−
)12
8
− 81
8
19 82
12
(18
(
)
Force in coupling BC: Replace wagons A and B with a coupling. Assume a tensile force on the coupling. Force đ??šđ??ľđ??ś is acting on the coupling.
1 ( ) 2
8
2
2
9 9
− 12
(8
− 12
2
9 9 1
12 8 11
12
−
− 8 1
)12 8
8 ( 4
)
Chapter 13
DYNAMICS Kinetics of Particles: Energy and Momentum Methods
Chapter 13
13.17 A trailer truck enters a 2 percent downhill grade traveling at 108 km/h and must slow down to 72 km/h in 300 m. The cab has a mass of 1800 kg and the trailer 5400 kg. Determine (a) the average braking force that must be applied, (b) the average force exerted on the coupling between cab and trailer if 70 percent of the braking force is supplied by the trailer and 30 percent by the cab. 11
Given:
8
18
1 8 Find:
2
2
and force on coupling
(a) (18
)9 81
2
1 12 1 2
1 2 2
(
1 2
1 2 2
(2 )
1
2
−
−
2
1 12
1
2 2 22
)
−
2
1
1
1
5
DYNAMICS Kinetics of Particles: Energy and Momentum Methods (b)
Replace cab with a coupling. Assume a tensile force on the coupling. Force đ??šđ??´đ??ľ is acting on the coupling.
(
)9 81
29
1
9 8
1
1 2
1 2
(
1 2
1 2
(2 )
)
−
2
1
2 1 8 1 8
−12 8
−1
1 8
1
−111 2 −111 2
−
9 8
1
6
2
−12 8
Chapter 13
DYNAMICS Kinetics of Particles: Energy and Momentum Methods
Chapter 13
13.27 A 10-lb block is attached to an unstretched spring of constant 12 . The coefficients of static and kinetic friction between the block and the plane are 0.60 and 0.40, respectively. If a force F is applied to the block until the tension in the spring reaches 20 lb and then suddenly removed, determine (a) how far the block will move to the left before coming to a stop, (b) whether the block will then move back to the right. The force on the spring is 20 lb, therefore the reaction force by the spring is -20 lb.
Given:
1
−
−2
12
Find: (a) how far the block will move to the left before coming to a stop, (b) whether the block will then move back to the right. 1. Find how far to the right the spring is stretched by force F: −
−2
−2 −12
1
2. Draw F.B.D. with forces and points:
Point 1: Block is released from rest after spring has been stretched by force F. Point 2: Block stops after spring has been compressed to its maximum. Point 3: Block possibly moves back to the right.
The work done by a spring is negative when the spring is being stretched or compressed, and positive when it is released.
7
DYNAMICS Kinetics of Particles: Energy and Momentum Methods
Chapter 13
From point 1 to point 2:
− 1 ( 2
)−
(
1 12(1 2
−
)−(
−
−1
−
−
)
)(1 )(1
−
)
(a)
−
−1
−
1
Therefore, the block moves from +1.667 to –1, a distance of 2.667 in or 0.222 ft. (b) When the block stops at point 2, it will move again to the right if the force exerted by the spring is greater than the static frictional force exerted by the surface on which the block stands. −
−(12)(−1) (
12
)(1 ) , therefore the block will move back to the right.
8
DYNAMICS Kinetics of Particles: Energy and Momentum Methods
Chapter 13
13.28 A 3-kg block rests on top of a 2-kg block supported by but not attached to a spring of constant 40 N/m. The upper block is suddenly removed. Determine (a) the maximum speed reached by the 2-kg block, (b) the maximum height reached by the 2-kg block. 2
Given:
Find: (a) the maximum speed reached by the 2-kg block, (b) the maximum height reached by the 2-kg block. When the 3-kg block is removed, the spring will stretch a certain distance while pushing the 2-kg block. At x = 0, the spring and the block will reach their maximum speed. At that point the spring will slow down until its speed is zero at its maximum stretch, and the 2-kg block will keep going upwards until the force of gravity makes it reach a speed of zero before it makes it fall down.
1. Find the force on the spring by the weight of the masses: (
2)(−9 81)
− 9
2. Find the distance that a weight of 49.05 N will compress the given spring: −
9
−
−1 22
3. Define points: Point 1: 3-kg block has been removed. Speed of 2-kg block is zero and its position is −1 22 . Point 2: Spring and block reach their maximum speed at
.
Point 3: Spring has been left behind and block reaches its maximum height.
9
DYNAMICS Kinetics of Particles: Energy and Momentum Methods
Chapter 13
4. Apply the method of work and energy to find maximum velocity of the spring-mass system at
1 2
1 2 2 − 1 ( 2 1 2
)−
− (−1 22
(
)
− ) − (2)(9 81)(1 22
)
2−2 1 2
1 2
5. Apply the method of work and energy to find height of block at point 3.
− 1 ( 2 1 2
)−
− (−1 22
(
)
− ) − (2)(9 81)(
)
2 − 19 2
The spring stops doing work on the block at y = 0.
2 − 19 2 2 19 2
1
29
10
DYNAMICS Kinetics of Particles: Energy and Momentum Methods 13.44 A small block slides at a speed 8 on a horizontal surface at a height above the ground. Determine (a) the angle θ at which it will leave the cylindrical surface BCD, (b) the distance x at which it will hit the ground. Neglect friction and air resistance. 8
Given: Find:
and
At point B: [speed is constant] −
At point C, using n and t coordinates:
(
1 2
1 2
1 2
1 2 (
2 2
(
−
8
)
2
)
(
−
)
−
)
1 2
(
−
)
1 2
11
Chapter 13
DYNAMICS Kinetics of Particles: Energy and Momentum Methods 2 2
9
1 22 2
)
2 2( − − 22
8
1 1
128 2
2
2
88
88 Find velocity at point C:
2
(
−
)
−
1 2
2
(
)
1 2
2
2 2( − 2
)
2
2 2( − 2
)
1 2
1 2
92 8 21
2
− 2 1 1
2
2
− 2
−2
(8 21 )(
−
1 22 2 2
)
2 12
Chapter 13
DYNAMICS Kinetics of Particles: Energy and Momentum Methods
Chapter 13
13.46 (a) A 120-lb woman rides a 15-lb bicycle up a 3-percent slope at a constant speed of 5 ft/s. How much power must be developed by the woman? (b) A 180-lb man on an 18-lb bicycle starts down the same slope and maintains a constant speed of 20 ft/s by braking. How much power is dissipated by the brakes? Ignore air resistance and rolling resistance.
Given:
12
1
1
18
1 18
(a)
Forces in the direction of v: −1 (−1
(a)
)( )
Forces in the direction of v: 198 (198
)(2 )
13
18
198
DYNAMICS Kinetics of Particles: Energy and Momentum Methods
Chapter 13
13.49 In an automobile drag race, the rear (drive) wheels of a l000-kg car skid for the first 20 m and roll with sliding impending during the remaining 380 m. The front wheels of the car are just off the ground for the first 20 m, and for the remainder of the race 80 percent of the weight is on the rear wheels. Knowing that the coefficients of friction are 9 and 8, determine the power developed by the car at the drive wheels (a) at the end of the 20-m portion of the race, (b) at the end of the race. Give your answer in kW and in hp. Ignore the effect of air resistance and rolling friction.
1
Given:
9
8
Find: (a) The power developed by the car at the drive wheels at the end of the 20-m portion of the race, (b) at the end of the race
(a) All weight on rear traction wheels. (
Spinning means
( 1 2
1 1 √
8)(2 )
1
8)(1
)(9 81)
1
1 1 2
1 1
(
1 8 )(1
1 kW = 1.34102209 hp )
1 89
14
8
DYNAMICS Kinetics of Particles: Energy and Momentum Methods (b)
Roll with sliding impending means: ( 9 )( 8)(1
8
)(9 81)
2
8 1 2
(
)(1
(
)
2)( 8 )
1
1
2 8
1
1 2
1
1 √
2 8
1
9 (
2)(
)
2
11
15
Chapter 13
DYNAMICS Kinetics of Particles: Energy and Momentum Methods
Chapter 13
13.69 A spring is used to stop a 50-kg package which is moving down a 20° incline. The spring has a constant and is held by cables so that it is initially compressed 50 mm. Knowing that the velocity of the package is 2 m/s when it is 8 m from the spring and the kinetic coefficient of friction between the package and the incline is 0.2., determine the maximum additional deformation of the spring in bringing the package to rest. 2
Given:
Find: The maximum additional deformation of the spring in bringing the package to rest. 1 2
(
)(
)(2)
1 2
(
)(
)
− (
1
2 −
1)
9 9)
2
2 (
2 1 (
)
1
(
)
)
)
(
1 (
(
(
1
(
(
2281
)
1
) −1 )
2
2 2 (
1 ( 2
2 )
2 )
)(9 81)( 1
2 2
2
( 2 −
(8)(
1
)− )
2 12 2281
16
)
DYNAMICS Kinetics of Particles: Energy and Momentum Methods 13.144 The design for a new cementless hip implant is to be studied using an instrumented implant and a fixed simulated femur. Assuming the punch applies an average force of 2 kN over a time of 2 ms to the 200 g implant, determine (a) the velocity of the implant immediately after impact, (b) the average resistance of the implant to penetration if the implant moves 1 mm before coming to rest. (
2
Given:
−
)
2
2
Find: (a) the velocity of the implant immediately after impact, (b) the average resistance of the implant to penetration if the implant moves 1 mm before coming to rest.
(
−
)
(2
)(
2)
2
2
1 2
(
−
2−
)( 2)(2 )
−
2
1
1
17
Chapter 13
DYNAMICS Kinetics of Particles: Energy and Momentum Methods 13.174 A 1-kg block B is moving with a velocity v0 of magnitude 2 as it hits the 0.5-kg sphere A, which is at rest and hanging from a cord attached at O. Knowing that between the block and the horizontal surface and 8 between the block and the sphere, determine after impact (a) the maximum height h reached by the sphere, (b) the distance x traveled by the block. 1
Given: 2
8
Find: (a) the maximum height h reached by the sphere, (b) the distance x traveled by the block.
(1)(2)
1
− − −
− −2 −1
1
(a)
2
−1
8 1 −1 ] 1 2
1 [
1
Sphere: 1 2
(
(
1
}[
2
)(
)(9 81)
)(2 )
1
9
9
18
2
] 8
2 8
Chapter 13
DYNAMICS Kinetics of Particles: Energy and Momentum Methods (b)
Block: 1 2
(
−
−
)(1)( 8)
2
(
)(1)(9 81)
2 − 88
19
− 88
Chapter 13
DYNAMICS Kinetics of Particles: Energy and Momentum Methods
Chapter 13
13.180 Two cars of the same mass run head-on into each other at C. After the collision, the cars skid with their brakes locked and come to a stop in the positions shown in the lower part of the figure. Knowing that the speed of car A just before impact was 5 mi/h and that the coefficient of kinetic friction between the pavement and the tires of both cars is 0.30, determine (a) the speed of car B just before impact, (b) the effective coefficient of restitution between the two cars. Given:
−12 − Find: (a) the speed of car B just before impact, (b) the effective coefficient of restitution between the two cars. 1.
Conservation of linear momentum: [general form] CAR A 1 đ?‘šđ?‘Ł 2
đ??ž
đ??ž
đ?‘ˆ
−đ??šđ?‘“ đ?‘Ľ
đ?‘ˆ
−(
đ?‘˛đ?&#x;?
đ?‘źđ?&#x;?
CAR B
đ?&#x;?
đ??ž
−đ?œ‡đ?‘˜ đ?‘ đ?‘Ľ )đ?‘š( 2 2)(12)
11 92 đ?‘š
đ?‘˛đ?&#x;?
(2)(11 92)
đ??ž
đ?‘ˆ
−đ??šđ?‘“ đ?‘Ľ
đ?‘ˆ
−(
đ?‘˛đ?&#x;?
1 đ?‘šđ?‘Ł − 11 92 đ?‘š 2 đ?‘Łđ??´
1 đ?‘šđ?‘Ł 2
đ?‘źđ?&#x;?
−đ?œ‡đ?‘˜ đ?‘ đ?‘Ľ )đ?‘š( 2 2)( )
đ?&#x;?
28 98 đ?‘š
đ?‘˛đ?&#x;?
1 đ?‘šđ?‘Ł − 28 98 đ?‘š 2 1 22
đ?‘“đ?‘Ą đ?‘
đ?‘Łđ??ľ
20
(2)(28 98)
1 1 đ?‘“đ?‘Ą đ?‘
DYNAMICS Kinetics of Particles: Energy and Momentum Methods −
−
−
[masses are equal,
(a) 1 22
1 1
2
(b) − −
−
1 1 − (−1 22 ) − (− 1 2 )
1 2 1
21
]
Chapter 13
DYNAMICS Kinetics of Particles: Energy and Momentum Methods
Chapter 13
Class Example Vehicles A and B collide as shown. 1
Given:
Find:
final position (brakes applied, both vehicles skid).
1. TANGENTIAL AXIS Tangential velocities don’t change after impact: (
)
(
)
(
)
(
)
2
2
2
2. NORMAL AXIS (
)
(
(
( (
)
)
(
2 ) (
2 )
(
2 )
)
(
)
(
)
(
2 )
(
)
2
2 − (
)
(
) (
( )
1 1
[B has no normal velocity]
( (
) )
[g cancelled]
)
[divide both sides by 1000]
)
[1 equation, 2 unknowns]
3. USE e TO FIND ANOTHER EQUATION WITH SAME VARIABLES ( (
− − ( 2
) −(
)
2 − (
) −( ) −(
− )
2 (
)
) ) (
( )
(
) (
2
22
) −( ) − 2 2 )
21
2
DYNAMICS Kinetics of Particles: Energy and Momentum Methods (
)
(
)
2
2
2
(
9
2 − ( 2
2 )
2
)
2
2
2
(
2
Chapter 13
)
4. FIND RESULTING VELOCITIES: 9
2
2
− −
5. WITH INITIAL VELOCITIES AFTER COLLISION, FINAL VELOCITIES (ZERO), AND WORK DONE (FRICTION), THE WORK AND ENERGY METHOD CAN BE USED TO FIND FINAL POSITIONS.
23
DYNAMICS
Systems of Particles
Chapter 14
14.3 A 180-lb man and a 120-lb woman stand side by side at the same end of a 300-lb boat, ready to dive, each with a 16-ft/s velocity relative to the boat. Determine the velocity of the boat after they have both dived, if (a) the woman dives first, (b) the man dives first.
Given: Find: The velocity of the boat after they have both dived, if (a) the woman dives first, (b) the man dives first.
Principle of Conservation of Linear Momentum L: (a) đ?‘Łđ?‘‘
L before anyone dives = L after woman dives:
đ?‘Łđ?‘‘
(
(
)
L after woman dives = L after man dives
(
)(
)
(
)
1
)
đ?‘Łđ?‘‘
đ?‘Łđ?‘?
đ?‘Łđ?‘‘
đ?‘Łđ?‘? đ?‘Łđ?‘?
đ?‘?
DYNAMICS
Systems of Particles
Chapter 14
(b) L before anyone dives = L after man dives:
(
(
)
)
L after man dives = L after woman dives
(
)(
)
(
)
2
DYNAMICS
Systems of Particles
14.5 A bullet is fired with a horizontal velocity of 1500 ft/s through a 6-lb block A and becomes embedded in a 4.95-lb block B. Knowing that blocks A and B start moving with velocities of 5 ft/s and 9 ft/s, respectively, determine (a) the weight of the bullet, (b) its velocity as it travels from block A to block B. Given: Find:
and
between A and B
(a) L of bullet = L of block A + L of block B and bullet (
)
1491
(b) Initial L of bullet = L of bullet after going through block A (
)
(
) (
( (
)
)
)
3
Chapter 14
DYNAMICS
Systems of Particles
Chapter 14
14.8 At an amusement park there are 200-kg bumper cars A, B, and C that have riders with masses of 40 kg, 60 kg, and 35 kg respectively. Car A is moving to the right with a velocity vA = 2 m/s when it hits stationary car B. The coefficient of restitution between each car is 0.8. Determine the velocity of car C so that after car B collides with car C the velocity of car B is zero. Given:
Find:
(
)( )
so that after car B collides with car C, the velocity of car B is zero.
(
[
(
[
)( )
]
)(
[
]
)
(
]
[
)( )
] 4
DYNAMICS
Systems of Particles
Chapter 14
14.17 A small airplane of mass 1500 kg and a helicopter of mass 3000 kg flying at an altitude of 1200 m are observed to collide directly above a tower located at O in a wooded area. Four minutes earlier the helicopter had been sighted 8.4 km due west of the tower and the airplane 16 km west and 12 km north of the tower. As a result of the collision the helicopter was split into two pieces, H1 and H2, of mass m1 = 1000 kg and m2 = 2000 kg, respectively; the airplane remained in one piece as it fell to the ground. Knowing that the two fragments of the helicopter were located at points H1 (500 m, –100 m) and H2 (600 m, –500 m), respectively, and assuming that all pieces hit the ground at the same time, determine the coordinates of the point A where the wreckage of the airplane will be found. Given:
(
(
) ̂
)
̂ ( ̂
Find: (
( ̂
) ̂
) ̂
̂
)
1. Find velocity of airplane and helicopter at time of collision: ( ̂
̂)
(
̂ (
( ̂
̂)
̂) ̂
)
( ( )
̂)
̂
̂
2. Find velocity of mass center G of the fragments after the collision: The mass center G
( (
)(
̂) ̂
̂
̂ ̂
̂ ̂
of a system of
) (
particles moves as
)(
̂)
(
̂
)
if the entire mass of the system and all the external forces were concentrated
̂
at that point.
̂
̂
5
DYNAMICS
Systems of Particles
Chapter 14
3. Find the time it takes the fragments to fall freely from
to
√
√
4. Find position of G at time of impact with the ground: (
̂) ̂
̂
̂
5. Use Equation 14.12 to find position of plane where it hits the ground: ̅
∑ (
(
)
)(
(
(
)
(
̂) ̂
̂
(
(
̂) ̂
)(
(
)
( (
)
̂
)
̂
) )
(
)
( )(
( ̂
)(
̂) ̂
̂)
̂
̂
̂ (
)
( ̂
6
̂) ̂
̂)
(
̂
(
)(
( )(
̂ ̂)
̂
̂)
̂)
DYNAMICS
Systems of Particles
Chapter 14
14.19 Car A was traveling east at high speed when it collided at point O with car B, which was traveling north at 72 km/h. Car C, which was traveling west at 90 km/h, was 10 m east and 3 m north of point O at the time of the collision. Because the pavement was wet, the driver of car C could not prevent his car from sliding into the other two cars, and the three cars, stuck together, kept sliding until they hit the utility pole P. Knowing that the masses of cars A, B, and C are, respectively, 1500 kg, 1300 kg, and 1200 kg, and neglecting the forces exerted on the cars by the wet pavement solve the following problem: Knowing that the coordinates of the utility pole are xp = 18 m and yp = 13.9 m, determine (a) the time elapsed from the first collision to the stop at P, (b) the speed of car A. Given:
̂ ̂ (
̂) ̂
Find: and (
Time from O to
)
Use Equation 14.12 to work backwards from final position: ̅
∑
(
)
(
)(
( )( ̂
̂
̂)
(
̂
̂ ̂
̂
̂)
( ̂
(
)
̂)
(
̂
̂
̂
̂ ̂
(
)̂
7
̂
̂ ̂ )
)
DYNAMICS
Systems of Particles
Equating coefficients of ̂:
Equating coefficients of ̂: (
)
(
)
8
Chapter 14
DYNAMICS
Systems of Particles
Chapter 14
14.73 A floor fan designed to deliver air at a maximum velocity of 6 m/s in a 400-mm-diameter slipstream is supported by a 200-mm-diameter circular base plate. Knowing that the total weight of the assembly is 60 N and that its center of gravity is located directly above the center of the base plate, determine the maximum height h at which the fan may be operated if it is not to tip over. Assume Ď = 1.21 kg/m3 for air and neglect the approach velocity of the air. Given: Find: Maximum height h at which the fan may be operated if it is not to tip over. 1. Find mass flow rate dm/dt
Mass flow rate
(
)(
(
)
(
)
đ?‘‘đ?‘š đ?‘‘đ?‘Ą
đ?œŒđ?‘„ đ?œŒđ??´đ?‘Ł
2. Find force of the fluid Ff (
)
)(
� � �
)
đ?›ž đ??´đ?‘Ł đ?‘”
3. Find force of thrust Fth
4. Sum forces on an axis or moments. In this case, we will sum moments about the center of gravity in line with the base of the fan. When the fan is in “impending tip�, the normal force N is applied at the “tipping hinge�.
( ) (
)(
)
9
DYNAMICS
Systems of Particles
14.74 The helicopter shown can produce a maximum downward air speed of 80 ft/s in a 30-ft-diameter slipstream. Knowing that the weight of the helicopter and its crew is 3500 lb and assuming lb/ft3 for air, determine the maximum load that the helicopter can lift while hovering in midair.
Given: (specific weight of air) Find: The maximum load L that the helicopter can lift while hovering in midair. 1. Find mass flow rate dm/dt (
)
)(
)
2. Find force of the fluid Ff (
)
(
3. Find force of thrust Fth
4. Sum forces on an axis or moments. [acceleration is zero when hovering]
10
Chapter 14
DYNAMICS
Systems of Particles
Chapter 14
14.75 A jet airliner is cruising at a speed of 600 mi/h with each of its three engines discharging air with a velocity of 2000 ft/s relative to the plane. Determine the speed of the airliner after it has lost the use of (a) one of its engines, (b) two of its engines. Assume that the drag due to air friction is proportional to the square of the speed and that the remaining engines keep operating at the same rate. Given:
Find: The speed of the airliner after it has lost the use of (a) one of its engines, (b) two of its engines.
(
) [acceleration is zero when at cruising speed]
Moving reference frame:
(
đ?‘Łđ?‘–đ?‘›
)
With three engines running: (
) (
(
)
)
With two engines running: (
)
(
) (
(
) )
(
) 11
đ?‘Łđ?‘Žđ?‘–đ?‘&#x;đ?‘?đ?‘™đ?‘Žđ?‘›đ?‘’
DYNAMICS
Systems of Particles
With one engine running: (
)
(
) (
(
) )
(
)
12
Chapter 14
Dynamics
Kinematics of Rigid Bodies
Chapter 15
15.48 In the planetary gear system shown, the radius of gears A, B, C, and D is 3 in. and the radius of the outer gear E is 9 in. Knowing that gear E has an angular velocity of 120 rpm clockwise and that the central gear has an angular velocity of 150 rpm clockwise, determine (a) the angular velocity of each planetary gear, (b) the angular velocity of the spider connecting the planetary gears. Given:
Find:
Gear A:
[1]
Spider:
[2]
Gear B:
[3] [4]
Gear E:
[5]
[ ] [ ]
( )(
[ ]
)
( )(
( )(
)
)
( )
[ ]
( )
1
Dynamics
Kinematics of Rigid Bodies
15.57 In the engine system shown, l = 160 mm and b = 60 mm. Knowing that the crank AB rotates with a constant angular velocity of 1000 rpm clockwise, determine the velocity of the piston P and the angular velocity of the connecting rod when (a) , (b) . Given:
Find:
(a) (
)(
)
( )
⁄ ⁄
⁄
( )
(b) (
(
)
( ⁄
)
)(
(
)
)
(
)
(
( ) ( )
2
)
Chapter 15
Dynamics
Kinematics of Rigid Bodies
Chapter 15
15.111 An automobile travels to the left at a constant speed of 48 mi/h. Knowing that the diameter of the wheel is 22 in., determine the acceleration (a) of point B, (b) of point C, (c) of point D. Given:
Find: The wheel is rolling and not sliding, therefore point C is the instantaneous center.
(
)(
)
( ) ( ) ( )
3
Dynamics
Kinematics of Rigid Bodies
15.125 Knowing that crank AB rotates about point A with a constant angular velocity of 900 rpm clockwise, determine the acceleration of the piston P when Given:
Find:
Rod AB: ( (
)(
)
)
(
)(
)
Rod BD: (
)
(
)
(
) (
(
)
(
)
(
)
(
)(
(
)
)
)
(
(
)(
)
)
Horizontal components (
)
Vertical components (
)(
)
4
Chapter 15
DYNAMICS
Plane Motion of Rigid Bodies: Forces and Acceleration
16.7 A 20-kg cabinet is mounted on casters that allow it to move freely ( ) on the floor. If a 100-N force is applied as shown, determine (a) the acceleration of the cabinet, (b) the range of values of h for which the cabinet will not tip.
Given: Find: (a) the acceleration of the cabinet, (b) the range of values of h for which the cabinet will not tip.
When tipping is impending
When tipping is impending
Therefore, the range of values of h for which the cabinet will not tip is
1
Chapter 16
DYNAMICS
Plane Motion of Rigid Bodies: Forces and Acceleration
16.33 In order to determine the mass moment of inertia of a flywheel of radius 600 mm, a 12-kg block is attached to a wire that is wrapped around the flywheel. The block is released and is observed to fall 3 m in 4.6 s. To eliminate bearing friction from the computation, a second block of mass 24 kg is used and is observed to fall 3 m in 3.1 s. Assuming that the moment of the couple due to friction remains constant, determine the mass moment of inertia of the flywheel. Given: When When
Find: the mass moment of inertia of the flywheel.
̅ ̅ Case 1: ̅ ̅ Case 2: ̅ ̅ ̅ ̅
2
Chapter 16
DYNAMICS
Plane Motion of Rigid Bodies: Forces and Acceleration
16.62 The 3-oz yo-yo shown has a centroidal radius of gyration of 1.25 in. The radius of the inner drum on which a string is wound is 0.25 in. Knowing that at the instant shown the acceleration of the center of the yo-yo is 3 ft/s2 upward, determine (a) the required tension T in the string, (b) the corresponding angular acceleration of the yo-yo. ̅
⁄
Given:
⁄ ⁄
⁄
Find: (a) the required tension T in the string, (b) the corresponding angular acceleration of the yo-yo.
̅ ̅ (
)
3
Chapter 16
DYNAMICS
Plane Motion of Rigid Bodies: Forces and Acceleration
16.69 A bowler projects an 8-in.-diameter ball weighing 12 lb along an alley with a forward velocity v0 of 15 ft/s and a backspin of 9 rad/s. Knowing that the coefficient of kinetic friction between the ball and the alley is 0.10, determine (a) the time t1 at which the ball will start rolling without sliding, (b) the speed of the ball at time t1, (c) the distance the ball will have traveled at time t1. Given: Ě… At the instant , the ball starts to roll, point C becomes the instantaneous center of rotation, and . At
Ě…
Kinematics at
4
Chapter 16
DYNAMICS
Plane Motion of Rigid Bodies: Forces and Acceleration
(
)r (
)
5
Chapter 16
DYNAMICS
Plane Motion of Rigid Bodies: Forces and Acceleration
Chapter 16
16.82 A turbine disk of mass 26 kg rotates at a constant rate of 9600 rpm. Knowing that the mass center of the disk coincides with the center of rotation O, determine the reaction at O immediately after a single blade at A, of mass 45 g, becomes loose and is thrown off. Given:
Find: the reaction at O immediately after a single blade at A, of mass 45 g, becomes loose and is thrown off.
6
DYNAMICS
Plane Motion of Rigid Bodies: Forces and Acceleration
16.97 A homogeneous sphere S, a uniform cylinder C, and a thin pipe P are in contact when they are released from rest on the incline shown. Knowing that all three objects roll without slipping, determine, after 4 s of motion, the clear distance between (a) the pipe and the cylinder, (b) the cylinder and the sphere. (SP 16.8) ̅
Given:
̅
Find: The clear distance, after 4 s of motion, between (a) the pipe and the cylinder, (b) the cylinder and the sphere. General case: ̅ ̅ ̅ ̅
̅
For pipe: ̅ For cylinder: ̅ For sphere: ̅
(a) (b)
7
Chapter 16
DYNAMICS
Plane Motion of Rigid Bodies: Forces and Acceleration
Chapter 16
16.104 A drum of 60-mm radius is attached to a disk of 120-mm radius. The disk and drum have a total mass of 6 kg and a combined radius of gyration of 90 mm. A cord is attached as shown and pulled with a force P of magnitude 20 N. Knowing that the disk rolls without sliding, determine (a) the angular acceleration of the disk and the acceleration of G, (b) the minimum value of the coefficient of static friction compatible with this motion. (See Sample Problem 16.9) Given:
̅
Find: (a) the angular acceleration of the disk and the acceleration of G, (b) the minimum value of the coefficient of static friction compatible with this motion.
̅
̅
̅
̅ ̅
̅
̅
8
DYNAMICS
Plane Motion of Rigid Bodies: Energy and Momentum Methods
Chapter 17
17.9 Each of the gears A and B has a mass of 2.4 kg and a radius of gyration of 60 mm, while gear C has a mass of 12 kg and a radius of gyration of 150 mm. A couple M of constant magnitude 10 N·m is applied to gear C. Determine (a) the number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm, (b) the corresponding tangential force acting on gear A. Given:
̅
̅
̅
Find: (a) the number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm, (b) the corresponding tangential force acting on gear A. All three gears are in mesh, so their contact velocity is the same:
Moment of inertia of the gears: ̅
̅ ̅
̅
̅
̅
̅ 1
DYNAMICS
Plane Motion of Rigid Bodies: Energy and Momentum Methods
Find kinetic energy of system at position 1:
( ̅
)
( ̅
( ̅
)
[
)
]
Find kinetic energy of system at position 2:
( ̅
( ̅
)
( ̅
)
[
)
]
Find work from position 1 to position 2 (work of the couple):
From the Principle of Conservation of Energy:
(a) Number of revolutions in 39.898 radians: (b)
( ̅ [
Gear A:
( ̅
)
)
]
[
2
]
Chapter 17
DYNAMICS
Plane Motion of Rigid Bodies: Energy and Momentum Methods
Chapter 17
17.24 A 20-kg uniform cylindrical roller, initially at rest, is acted upon by a 90-N force as shown. Knowing that the body rolls without slipping, determine (a) the velocity of its center G after it has moved 1.5 m, (b) the friction force required to prevent slipping. Given: Find: (a) the velocity of its center G after it has moved 1.5 m, (b) the friction force required to prevent slipping.
[
]
Ě…
(
)(
)
√
3
DYNAMICS
Plane Motion of Rigid Bodies: Energy and Momentum Methods
Chapter 17
17.69 A wheel of radius r and centroidal radius of gyration ̅ is released from rest on the incline shown at time t = 0. Assuming that the wheel rolls without sliding, determine (a) the velocity of its center at time t, (b) the coefficient of static friction required to prevent slipping.
The external forces W, F, and N form a system equivalent to the system of effective forces represented by the vector maG and the couple IGα. No sliding means aG = r α
̅ ̅ ̅ ̅ ̅
̅
(a)
̅
4
DYNAMICS
Plane Motion of Rigid Bodies: Energy and Momentum Methods
(b)
̅ ̅ (
̅ ̅
(
̅ ̅
(
̅
(
(
̅ ̅
̅
)
)
)
̅ )
)
5
Chapter 17
DYNAMICS
Plane Motion of Rigid Bodies: Energy and Momentum Methods
Chapter 17
17.99 A 45-g bullet is fired with a velocity of 400 m/s at θ = 5° into a 9-kg square panel of side b = 200 mm. Knowing that the panel is initially at rest, determine (a) the required distance h if the horizontal component of the impulsive reaction at A is to be zero, (b) the corresponding velocity of the center of the panel immediately after the bullet becomes embedded.
̅
Given:
Find: (a) the required distance h if the horizontal component of the impulsive reaction at A is to be zero, (b) the corresponding velocity of the center of the panel immediately after the bullet becomes embedded. F.B.D
Apply Principle of Impulse and Momentum for the plane motion of a rigid body: Syst Momenta1 + Syst Ext Imp1→2 = Syst Momenta2 Moments about A
: ̅ ̅
̅
6
DYNAMICS
Plane Motion of Rigid Bodies: Energy and Momentum Methods
[ ]
If the horizontal component of the impulsive reaction at A is to be zero:
Value of
into equation [1]:
(a)
(b)
7
Chapter 17