8.6

Rolling Motion

Rolling motion is a familiar situation that involves rotation, as Figure 8-15 illustrates for the case of an automobile tire. The essence of rolling motion is that there is no slipping at the point of contact where the tire touches the ground. To a good approximation, the tires on a normally moving automobile roll and do not slip. In contrast, the squealing tires that accompany the start of a drag race are rotating, but they are not rolling while they rapidly spin and slip against the ground.

Figure 8-15 (a) An automobile moves with a linear speed v. (b) If the tires roll and do not slip, the distance d through which an axle moves equals the circular arc length s along the outer edge of a tire.

When the tires in Figure 8-15 roll, there is a relationship between the angular speed at which the tires rotate and the linear speed (assumed constant) at which the car moves forward. In part b of the drawing, consider the points labeled A and B on the left tire. Between these points we apply a coat of red paint to the tread of the tire; the length of this circular arc of paint is s. The tire then rolls to the right until point B comes in contact with the ground. As the tire rolls, all the paint comes off the tire and sticks to the ground, leaving behind the horizontal red line shown in the drawing. The axle of the wheel moves through a linear distance d, which is equal to the length of the horizontal strip of paint. Since the tire does not slip, the distance d must be equal to the circular arc length s, measured along the outer edge of the tire: d = s. Dividing both sides of this equation by the elapsed time t shows that d/t = s/t. The term d/t is the speed at which the axle moves parallel to the ground—namely, the linear speed v of the car. The term s/t is the tangential speed vT at which a point on the outer edge of the tire moves relative to the axle. In addition, vT is related to the angular speed about the axle according to (Equation 8.9). Therefore, it follows that

| (8.12)

If the car in Figure 8-15 has a linear acceleration parallel to the ground, a point on the tire’s outer edge experiences a tangential acceleration relative to the axle. The same kind of reasoning as that used in the last paragraph reveals that the magnitudes of these accelerations are the same and that they are related to the angular acceleration of the wheel relative to the axle:

| (8.13)

Equations 8.12 and 8.13 may be applied to any rolling motion, because the object does not slip against the surface on which it is rolling. Example 8 illustrates the basic features of rolling motion.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 8 | | | An Accelerating Car

An automobile, starting from rest, has a linear acceleration to the right whose magnitude is 0.800 m/s2 (see Figure 8-16a). During the next 20.0 s, the tires roll without slipping. The radius of each wheel is 0.330 m. At the end of this time, what is the angle through which each wheel has rotated?

Figure 8-16 As the car accelerates to the right, each wheel rotates through an angle .

Reasoning Since the magnitude a of the car’s linear acceleration is constant and the tires roll without slipping, the relation (Equation 8.13) reveals that the magnitude of the angular acceleration is also constant. Therefore, the equations of kinematics for rotational motion apply, and we will use one of them to find the angle through which each tire rotates. Two kinematic variables are known (the initial angular velocity and the time). However, to use such an equation we also need to know the angular acceleration of each tire. To find the angular acceleration, we will use Equation 8.13.
Knowns and Unknowns The data for this problem are:

Description | Symbol | Value | Comment
Explicit Data
Time | t | 20.0 s |
Magnitude of linear acceleration | a | 0.800 m/s2 |
Tire radius | r | 0.330 m |
Implicit Data
Initial angular velocity |  | 0 rad/s | Car starts from rest.
Unknown Variable
Angle through which each wheel rotates |  | ? |

Modeling the Problem

| Step 1 Angular Displacement
| Since the initial angular velocity and the time t are known, we will begin by selecting Equation 8.7 from the equations of kinematics for rotational motion. This equation relates the angular displacement to these variables as shown at the right. Equation 8.6 could also be used, but it provides a less convenient starting point. At this point we do not know the angular acceleration , so it will be evaluated in the next step.

| (8.7)

| Step 2 Angular Acceleration
| Angular acceleration, being a vector, has a magnitude and a direction. Because each wheel rolls without slipping, the magnitude of its angular acceleration is equal to a/r (Equation 8.13), where a is the magnitude of the car’s linear acceleration and r is the radius of the wheel.
As the car accelerates to the right, the tires rotate faster and faster in the clockwise, or negative, direction (see Figure 8-16), so the angular acceleration is also negative. Inserting a minus sign into the expression a/r to denote the direction of the angular acceleration gives

All the variables on the right side of this equation are known, and we substitute it into Equation 8.7 (see the right column).

Solution Algebraically combining the results of the steps above, we have:
The angle through which each wheel rotates is

The minus sign in this result means that the wheel has rotated in the clockwise direction.
Related Homework: Problem 50

Test your understanding of the material in Sections 8.4–8.6:

· | Angular Variables and Tangential Variables
· | Centripetal Acceleration and Tangential Acceleration
· | Rolling Motion

Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.