Passage VII

A physical pendulum is an object that is suspended vertically from a rigid support (see example in Figure 1).

L is the distance between the rod (the rigid support) and the pendulum's center of mass (CM). When the pendulum is displaced from its equilibrium position and then released, it oscillates (swings back and forth) about the rod, which acts as the axis of rotation.

Students tested several types of physical pendulums. In each trial, first, they computed the moment of inertia, I, in kilogram meters2 (kg m2), of a pendulum. (I is a measure of the mass of a pendulum and how the mass is distributed. ) Next, they displaced the pendulum slightly from its equilibrium position, released it, and measured the time, in seconds, it took the pendulum to oscillate through 20 cycles (1 cycle is 1 round-trip back and forth). Finally, they computed the pendulum's average period, P (the average time it took to complete 1 cycle) .

Study 1

In Trials 1-5, each pendulum was a 1.0 kg solid sphere with a diameter, D, of 0.20 m(see Figure 2).

Each pendulum, however, had a different L. The results are shown in Table 1.

Study 2

In Trials 6-10, each pendulum was a 1.0 kg thin, flat circular plate with D = 0.20 m (see Figure 3).

Each pendulum had a different L (see Table 2) .

Study 3

In Trials 11-15, each pendulum was a 1. 0 kg thin, spherical shell with D = 0.20m (see Figure 4).

Each pendulum had a different L (see Table 3).

37. Object A and Object B each consists of a pair of spheres connected by a rigid wire of insignificant mass. The 4 spheres are identical to each other. Each object is suspended vertically from a support at a distance L above the object's CM, as shown in the figure below.

L is the same for the 2 objects. Based on Studies 1 and 3, which object has the greater moment of inertia?

Answer and Explanation

Your Answer is

Correct Answer is A

Explanation

It can be seen from tables 1-3 that when L (that is, the distance from the center of gravity to the support) increases, I increases. Corresponding to the picture in this question, A is farther away from support, so sphere A should have a larger I

(Note that the L in the 37 question picture and the L in the previous tables are actually not the same concept. The previous In the experiment, the rod/support passes through the entire sphere, so the calculated distance between the center of gravity of the sphere and the support; but the support in this question does not run through the ball, so when doing the question, you need to consider Sphere A and The distance from sphere B to support)