A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so that its pendulum motion takes 3.00 s.How far from the center of the rod should the pivot be located?
A) 7.98 cm
B) 7.52 cm
C) 8.73 cm
D) 8.40 cm
E) 23.4 cm

Question

Quizplus · Accepted Answer

The period of a physical pendulum is given by $ T = 2\pi \sqrt{\frac{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 from the pivot to the center of mass. For a rod pivoted at a distance $ x $ from its center, $ I = \frac{1}{12}mL^2 + mx^2 $. Solving for $ x $ with $ T = 3.00 $ s and $ L = 1.50 $ m gives $ x \approx 8.73 $ cm.

A Long Thin Uniform Rod of Length 1