A meter stick is pivoted at a point a distance a from its center and swings as a physical pendulum.Of the following values for a, which results in the shortest period of oscillation?
A)a = 0.1 m
B)a = 0.2 m
C)a = 0.3 m
D)a = 0.4 m
E)a = 0.5 m

Question

Quizplus · Accepted Answer

The period of oscillation of a physical pendulum is given by T=2π√(I/mgd), where I is the moment of inertia, m is the mass, g is the acceleration due to gravity, and d is the distance between the pivot point and the center of mass. Since the meter stick is uniform, its center of mass is at its center, which is 0.5 m from the pivot point. Therefore, d=0.5-a. Plugging in the given values and simplifying, we get T=2π√((1/12)(0.5)^2/9.81(0.5-a))=2π√(5/294.3(0.5-a)). To minimize T, we need to maximize the square root expression, which means we need to minimize the denominator (0.5-a). The largest value of a that does not exceed 0.5 is 0.3, so the best choice is C.

A Meter Stick Is Pivoted at a Point a Distance