9.7

Concepts & Calculations

In this chapter we have seen that a rotational or angular acceleration results when a net external torque acts on an object. In contrast, when a net external force acts on an object, it leads to a translational or linear acceleration, as Chapter 4 discusses. Torque and force, then, are fundamentally different concepts, and Example 16 focuses on this fact.

| Example 16 | | | Torque and Force

Figure 9-30a shows a crate resting on a horizontal surface. It has a square cross section and a weight of W = 580 N, which is uniformly distributed. At the bottom right edge is a small obstruction that prevents the crate from sliding when a horizontal pushing force is applied to the left side. However, if this force is great enough, the crate will begin to tip or rotate over the obstruction. Determine the minimum pushing force that leads to tipping.

Figure 9-30 (a) A horizontal pushing force is applied to a uniform crate, which has a square cross section and a weight . The crate rests on the ground, up against a small obstruction. (b) Some of the forces acting on the crate and their lever arms.

Concept Questions and Answers What causes tipping—the force or the torque that it creates?

Answer Tipping is a rotational or angular motion. Since the crate starts from rest, an angular acceleration is needed, which can only be created by a net external torque. Thus, the torque created by the force causes the tipping.

A given force can create a variety of torques, depending on the lever arm of the force with respect to the rotational axis. In this case, the rotational axis is located at the small obstruction in Figure 9-30a and is perpendicular to the page. For this axis, the lever arm of the force is P, as the drawing shows. Where should be applied so that a minimum force will give the necessary torque? In other words, should the lever arm be a minimum or a maximum?

Answer The magnitude of the torque is the product of the magnitude of the force and the lever arm. Thus, for a minimum force, the lever arm should be a maximum, and the force should be applied at the upper left corner of the crate, as in Figure 9-30b.

Consider the crate just at the instant before it begins to rotate. At this instant, the crate is in equilibrium. What must be true about the sum of the external torques acting on the crate?

Answer Because the crate is in equilibrium, the sum of the external torques must be zero.

Solution Only the pushing force and the weight of the crate produce external torques with respect to a rotational axis through the lower right corner of the crate. The obstruction also applies a force to the crate, but it creates no torque, since its line of action passes through the axis. Since the sum of the external torques is zero, we refer to Figure 9-30b for the lever arms and write

The lever arm for the force is , where L is the length of the side of the crate. The lever arm for the weight is . This is because the crate is uniform, and the center of gravity is at the center of the crate. Substituting these lever arms in the torque equation, we obtain

For rotational motion, the moment of inertia plays a key role. We have seen in this chapter that the moment of inertia and the net external torque determine the angular acceleration of a rotating object, according to Newton’s second law. The angular acceleration, in turn, can be used in the equations of rotational kinematics, provided that it remains constant, as Chapter 8 discusses. When applied together in this way, Newton’s second law and the equations of rotational kinematics are particularly useful in accounting for a wide variety of rotational motion. The following example reviews this approach in a situation where a rotating object is slowing down.

| Example 17 | | | Which Sphere Takes Longer to Stop?

Two spheres are each rotating at an angular speed of 24 rad/s about axes that pass through their centers. Each has a radius of 0.20 m and a mass of 1.5 kg. However, as Figure 9-31 shows, one is solid and the other is a thin-walled spherical shell. Suddenly, a net external torque due to friction (magnitude = 0.12 N · m) begins to act on each sphere and slows the motion down. How long does it take each sphere to come to a halt?

Figure 9-31 The two spheres have identical radii and masses; initially, they also have the same angular velocities. Which one comes to rest first as they slow down due to the same frictional torque?

Concept Questions and Answers Which sphere has the greater moment of inertia and why?

Answer Referring to Table 9-1, we see that the solid sphere has a moment of inertia of while the shell has a moment of inertia of . Since the masses and radii of the spheres are the same, it follows that the shell has the greater moment of inertia. The reason is that more of the mass of the shell is located farther from the rotational axis than is the case for the solid sphere. In the solid sphere, some of the mass is located close to the axis and, therefore, does not contribute as much to the moment of inertia.

Which sphere has the angular acceleration (a deceleration) with the smaller magnitude?

Answer Newton’s second law for rotation (Equation 9.7) specifies that the angular acceleration is where is the net external torque and I is the moment of inertia. Since the moment of inertia is in the denominator, the angular acceleration is smaller when I is greater. Because it has the greater moment of inertia, the shell has the angular acceleration with the smaller magnitude.

Which sphere takes the longer time to come to a halt?

Answer Since the angular acceleration of the shell has the smaller magnitude, the shell requires a longer time for the deceleration to reduce its angular velocity to zero.

Solution According to Equation 8.4 from the equations of rotational kinematics, the time is given by where and are, respectively, the final and initial angular velocities. From Newton’s second law as given in Equation 9.7 we know that the angular acceleration is . Substituting this into the expression for the time gives

In applying this result, we arbitrarily choose the direction of the initial rotation to be positive. With this choice, the torque must be negative, since it causes a deceleration. Using the proper moments of inertia, we find the following times for the spheres to come to a halt:

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