A uniform meter stick is pivoted at the 75-cm mark and hangs vertically. If it is displaced from equilibrium by a small amount and then released, what is the period of the resulting oscillations?
A)2.01 s
B)1.26 s
C)1.53 s
D)1.00 s
E)1.74 s

Question

Quizplus · Accepted Answer

The period of oscillation for a physical pendulum is given by $ T = 2\pi \sqrt{\frac{I}{mgd}} $, where $ I $ is the moment of inertia about the pivot, $ m $ is the mass, $ g $ is the acceleration due to gravity, and $ d $ is the distance from the pivot to the center of mass. For a uniform meter stick pivoted at the 75-cm mark, $ I = \frac{1}{12}mL^2 + md^2 $ and $ d = 0.25 $ m. Solving gives $ T \approx 1.53 $ s.

A Uniform Meter Stick Is Pivoted at the 75-Cm Mark