Lab 4 - Uniform Circular Motion

Introduction

If you have ever been on an amusement park ride that travels in a curved or circular path, then you have experienced a force, called a centripetal force, pushing you into the ride. Whether it's the back wall of the "Roundup" or "Rotor", the ride where the floor drops from beneath your feet, or the seat belt of the "roller coaster" that supplies the force, you are constantly being accelerated toward the ride's center of curvature. If—what you fear most on such a ride—this force were suddenly removed, you would move off in a direction tangent to the circular path. That's what happens when you go over a hill on the roller coaster just before the seat belt takes effect. Figure 1 compares the motion in the presence of a centripetal force to the resulting motion of a body if the centripetal force were to cease suddenly. As another example consider a ball, attached to a string and whirled in a circle as shown in Fig. 2a. The tension in the string applies the centripetal force to the ball, causing it to move in a circular path. The string pulls the ball toward the center of the circle while the ball pulls outward on the string and hence on your hand in accordance with Newton's third law of action and reaction. So this outward force does not act on the ball, though it is commonly and incorrectly referred to as the centrifugal force acting on the ball. When the centripetal force is discontinued, such as in Fig. 2b when the string breaks, the object moves in the direction of the velocity at that instant. This direction is tangential to the circle at that point. The centripetal force that holds you in the ride can be determined with a few measurements and calculations. In this experiment you will determine what variables must be known to determine the centripetal force required to keep a mass moving in a circular path with a constant speed.

Discussion of Principles

Newton's second law states that the net force

F_net 

exerted on a moving body is equal to the product of a body's mass

m 

and its acceleration

a 

When applying Newton's second law to circular motion, it is convenient to use coordinates that are parallel and perpendicular to the object's motion. For a body moving in a straight line, the acceleration is due to a change in the magnitude of the velocity. For a body moving in a circular path with constant speed the magnitude

| v | 

of the velocity does not change, but the direction of the velocity vector

v 

constantly changes. In this case the object has no acceleration in the direction parallel to its motion and

F_|| 

in Eq. (2)

F_|| = ma_|| 

 is zero. This motion is called uniform circular motion - motion in a circular path at constant speed. Since the velocity vector is changing in time, the object in uniform circular motion is accelerating. Conceptually, using parallel and perpendicular coordinates is convenient because the parallel force is responsible for changes in speed and the perpendicular force (or centripetal force) is responsible for changes in direction.

Centripetal Acceleration

What acceleration is needed to keep the body moving in a circle with constant speed? Refer to Fig. 3a below. At a certain instant the particle is located at the tip of the radius vector

r 

. At another instant

Δt 

later the particle is located at the tip of the radius vector

r' 

. In the small time interval

Δt 

the body moves along an arc of length

vΔt. 

If

Δt 

is small enough, the arc length is approximately equal to the distance between the position of the body at the beginning of the interval

Δt 

and the position at the end of the interval. This straight-line segment, together with the two radii, forms the isosceles triangle shown in Fig. 3a. Figure 3b shows the velocity vectors at the two positions. Notice that the magnitude of the velocity vector is not changing. In Fig. 3c the two velocity vectors are redrawn, without changing their lengths or orientations. The arrow joining the tips of the velocity vectors represents the change

Δv 

in the velocity of the particle due to the change in direction during the time interval

Δt 

. Note that as

Δt 

gets smaller, the angle between the direction of

Δv 

and both

v 

and

v' 

comes close to a right angle 90°. Since the tangent to a circle is perpendicular to its radius, it follows that the angle

θ 

between

v 

and

v' 

as shown in Fig. 3c is the same as the angle between the two radii in Fig. 3a. From trigonometry we know that these two triangles are congruent, and therefore the ratios of their sides must be equal. So we can write the following expression Dividing both sides by

Δt / v 

gives Of course

Δv / Δt 

is the acceleration of the body, and it is in the direction of

Δv. 

From Fig. 3c we can see that the acceleration is directed toward the center of the circle. The magnitude of the centripetal acceleration

a_c 

is given by and the centripetal force is Since it is difficult to measure the velocity of the body directly, you will instead compute the velocity from quantities that are easier to measure. The magnitude of the velocity vector can be determined by measuring the distance that the object travels per unit time. If

T 

is the period (length of time needed for the object to make one complete revolution), then the speed is equal to the distance traveled in that one revolution divided by the period. Substituting this into Eq. (6)

a_c =

v2

r

 

 for the centripetal acceleration gives where

f = 1/T 

is the number of revolutions per second measured in Hertz. Now Eq. (7)

F_c = ma_c = m

v2

r

. 

 can be written in terms of the measurable quantities

m, 

T, 

and

r 

as

Objective

In this experiment you will measure the period of rotation for an object and calculate the centripetal force acting on it. You will compare this centripetal force with an equivalent force needed to maintain the object at the same radius.

Equipment

Procedure

While it might be more fun to carry out this study on a midway ride, a simple laboratory apparatus shown in Fig. 4 below will be used to examine the nature of centripetal force. By using the centripetal force apparatus you can measure the frequency of rotation of an object moving in a circular path of a known radius. Eq. (10)

F_c = 4π²mf²r 

 can then be used to compute the centripetal force acting on the object.

Description of apparatus

Figure 4 shows the apparatus and its various components. The bottom of the base is provided with adjustable screws that can be used to level the entire apparatus. The crossbar with the counter-weight at its end can be moved to change the position of the rotating mass. The pointer can be moved along the slot and positioned just under the tip of the rotating mass. This helps to measure the radius of the circular path of the rotating mass. In this apparatus the centripetal force is provided by the spring. When the spring is attached to the rotating mass

M_R, 

it gets pulled in as shown in Fig. 5a. To bring the rotating mass back above the pointer or radius indicator, the string is passed over the pulley and enough mass added at the end of the string to bring the rotating mass back above the radius indicator. In Fig. 5b the weight

m_hg 

of the hanging mass

m_h 

just balances the elastic force of the spring when the rotating object

M_R 

rests above the pointer or radius indicator post. In Fig. 7 the spring is stretched by the same amount as the mass

M_R 

rotates in a path with the same radius as that set in Fig. 5b. The magnitude of the two forces

F_c 

and

m_hg 

should agree within experimental error since both forces produce the same elongation of the spring.

Procedure A: Measuring the period of rotation

Procedure B: Plot of

m_hg 

versus

a_c 

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