APSI 2018 AP Physics 1: Chapter 3.1 by Edvantage Science

3.1 Motion in a Circle Warm Up Imagine you are swinging a tin can on a string in a circle above your head as shown in the diagram. Suddenly, the string breaks! Draw on the diagram the direction in which the can will move.

In a previous course, you may have solved problems involving the force of gravity and 

Gravity and Motion the acceleration due to gravity ( g ). You assumed g remains constant as a body falls from a height, and you considered many situations, most of which happened on or near the surface of Earth. Here are some questions you should be able to answer after you study this chapter:  • Is g the same for a satellite orbiting Earth several hundred kilometres above the surface as it is at Earth’s surface? • Does Earth exert a force of gravity on the Moon? • Why does the Moon not “fall down” to Earth?

The force that keeps you firmly attached to this planet is the type of force that keeps Earth in orbit around the Sun. The force of gravity exists between any two masses in the universe. All the planets orbit the Sun in elliptical orbits. Satellites (both artificial and our Moon) orbit Earth in elliptical paths. The ellipses are usually very close to being circular, however, so we begin our study of gravity by learning about objects moving in circular paths.

Uniform Circular Motion

Imagine you are driving a car around a circular track, maintaining the same speed all the way around the track. Any object that moves in a circle at steady speed is said to be in uniform circular motion. Is such an object accelerating? It may seem to have zero acceleration, but in fact an object moving in a circle has a constant acceleration  not because of a change in speed, but because of its constantly changing direction. Remember: acceleration is defined as change in velocity divided by change in time, and velocity is a vector quantity. ! ! Δv a= Δt

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Identifying Centripetal Acceleration

If a body is moving in a circular path, in what direction does it accelerate? Figure 3.1.1 shows two consecutive positions of a body moving in a circle. Velocity vectors are   labelled v 0 and v1 . To find out the direction of the acceleration, we need the direction !   ! of Δv . Since Δv is the vector difference between v1 and v 0 , we use the rules for vector subtraction. v0 Figure 3.1.2 shows how to find the vector   ! difference, Δv , between v1 and v 0 . ! ! ! Δv = v1 – v 0 v1 ! ! ! Δv = v1 +(−v 0 ) Figure 3.1.1 Two consecutive positions of !   Vector Δv is the resultant of v1 and (– v 0 ). a body moving in a circle v0

In Figure 3.1.2, the velocity vectors chosen represent the velocity of the body at two different times. The time interval between the occurrences of the velocities is relatively long.

Δv

–v0

If this time interval (∆t) is shortened, the ! direction of Δv becomes closer and closer to being toward the centre of the circle as in Figure 3.1.3. ! In fact, as ∆t → 0, the direction of Δv and  therefore the direction of the acceleration a for all practical purposes, is toward the centre of the circle.

Figure 3.1.2 The vector difference, Δv ,



between v1 and v 0 Δv

v0 –v0

Figure 3.1.3 As ∆t is shortened, the

direction of Δv becomes closer and closer to being toward the centre of the circle.

Direction of Centripetal Acceleration

Since the direction of the acceleration of a body moving at uniform speed in a circle is toward the centre of the circle, the acceleration is called centripetal acceleration. Centripetal means directed toward a centre. A device called an accelerometer can be used to show that a body moving in a circle accelerates toward the centre of the circle (Figure 3.1.4). If the accelerometer is attached to a lab cart accelerating in a straight line, the colored water inside the cart forms a wedge pointing in the direction of the acceleration. If the same accelerometer is attached to a toy train travelling at constant speed on a circular track, the accelerometer shows no acceleration in the direction of travel, but a definite acceleration perpendicular to the direction of travel (Figure 3.1.5). In accelerometer other words, the train on the circular track accelerates toward the centre of the track. F

Figure 3.1.4 An accelerometer on a lab cart

Figure 3.1.5 An accelerometer on a train going

around a curve. Note the direction of the force is pointed inward to the centre of the circle.

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 In Figure 3.1.6, a body is moving in a circle of radius R0 . The radius is a displacement  vector. The velocity of the body is v 0 . Following a very short time interval ∆t, the body has  moved through a small angle θ, and the body has a new velocity v1 , the same magnitude  as v 0 , but in a new direction. v0 R0

v1 θ R 1

Figure 3.1.6 A body moving at

  v 0 in a circle of radius R0

  ! In Figure 3.1.7(a), ΔR is the vector difference between R1 and R0 . ! ! ! Defining Centripetal R1 +(−R0 ) = ΔR Acceleration or ! ! ! R 0 + ΔR = R1 ΔR R0

(a)

v0 θ v1

Δv

(b) Figure 3.1.7 Radius vectors

(a) and velocity vectors (b) in circular motion

 Figure 3.1.7(b) shows vector subtraction of the velocity vectors. Vector v is the   vector difference between v1 and v 0 . ! ! ! v1 +(−v 0 ) = Δv or ! ! ! v 0 + Δv = v1 Now, consider the triangles formed by the radius vectors in Figure 3.1.7(a) and the velocity vectors in Figure 3.1.7(b). Both are isosceles triangles with a common angle θ, so they are similar triangles. Corresponding sides of similar triangles are proportional; therefore, Δv v = ΔR R

(The subscripts have been dropped because the magnitudes of the speeds and the radii do not change.) v Δv = ΔR R Δv During a time interval ∆t, the average acceleration is , so Δt ! Δv v ΔR a= = i Δt R Δt Consider what happens during the motion of one complete circle of the body in Figures 3.1.6 and 3.1.7.

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If!the time interval ∆t is chosen to be very small (∆t → 0), Figure 3.1.8 shows that each ΔR becomes closer to being equal to the section of corresponding arc ∆s on the ! circumference of the circle. If ∆s is the arc in question, then ΔR → ∆s as ∆t → 0. Δs The speed v of the body as it moves around the circle is v = , and if ∆t → 0, Δt ΔR v= Δt R0 ΔR Δs

θ R1

Figure 3.1.8 If ∆t → 0, ΔR → ∆s.

It can therefore be said that the average acceleration during time ∆t is ! Δv v ΔR v v2 a= = i = iv =

Δt

R Δt

If ∆t → 0, the magnitude of the average acceleration approaches the instantaneous  centripetal acceleration, ac. The magnitude of the centripetal acceleration is, therefore, ac =

v2 R

To summarize: when a body moves in a circle with uniform speed, it accelerates v2 toward the centre of the circle, and the acceleration has a magnitude of . R Newton’s second law suggests that since the acceleration is toward the centre of the circle, the net force causing it should also be a centripetal force. If the centripetal force Another Way to Calculate Centripetal causing the centripetal acceleration is “turned off,” the body will travel off in a direction Acceleration that is along a tangent to the circle. Another useful equation for calculating centripetal acceleration can be derived from v2 ac = . If one full revolution of the body is considered, its speed will equal the R 2π R circumference of the circle divided by the period of the revolution; that is, v = , T where R is the radius of the circle, and T is the period of one revolution. 2πR 2 1 v2 Since ac = , then ac = R , therefore, T R

( )

ac =

4π 2R T2

Both equations for centripetal acceleration are useful in many situations. You will use them when studying the motion of planets around the Sun, satellites around Earth, electrons in a magnetic field, and any kind of circular motion.

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Quick Check 1. What is the centripetal acceleration of the Moon toward Earth? Given: R = 3.84 × 108 m and T = 2.36 × 106 s

2. A skater travels at 2.0 m/s in a circle of radius 4.0 m. What is her centripetal acceleration?

3. A 20.0 g rubber stopper is attached to a 0.855 m string. The stopper is spun in a horizontal circle making one revolution in 1.36 s. What is the acceleration of the stopper?

Centripetal Force

The net force that causes centripetal acceleration is called centripetal force. Acceleration and net force are related by Newton’s second law, which says that F = ma. So the magnitude of the centripetal force can be calculated by: centripetal force Therefore,

Fc = m

Fc = mac

v2 4π 2R or Fc = m 2 R T

Sample Problem 3.1.1 — Centripetal Force In a local playground a merry-go-round is turning at 4.50 m/s. If a 50.0 kg person is standing on the platforms edge, which is 5.80 m from the centre, what force of friction is necessary to keep her from falling off the platform?

What to Think About

How to Do It

1. This is a circular motion question and the force of friction between the person and the platform is the centripetal force.

ac =

2. Find the centripetal acceleration.

ac =

v2 R ( 4.50 m s)2

5.80 m ac = 3.49 m s 2

3. Find the centripetal force, which is the frictional force keeping her on the platform as it spins.

Ffr = Fc = mac Ffr = ( 50.0 kg)( 3.49 m s 2 ) Ffr = 175 N

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Practice Problems 3.1.1 — Centripetal Force 1. A 61 kg skater cuts a circle of radius 4.0 m on the ice. If her speed is 4.00 m/s, what is the centripetal force? What exerts this force?

2. What centripetal force is needed to keep a 12 kg object revolving with a frequency of 5.0 Hz in an orbit of radius 6.0 m?

3. A 1.2 x 103 kg car rounds a curve of radius 50.0 m at a speed of 80.0 km/h (22 m/s). (a) What is the centripetal acceleration of the car?

(b) How much centripetal force is needed to cause this acceleration?

(c) If the coefficient of kinetic friction µ is 0.25 on a slippery road, will the force of friction between the road and the wheels of the car be enough to keep the car from skidding?

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