A long thin uniform rod of length 1.50 m is to be suspended from a pivot so that its pendulum motion takes 3.00 s. How far from the center of the rod should the pivot be placed?

Question

Quizplus · Accepted Answer

The period of a physical pendulum is given by T = 2π√(I/mgh), where I is the moment of inertia, m is the mass, g is the acceleration due to gravity, and h is the distance from the pivot to the center of mass. For a rod pivoted at a distance x from its center, I = (1/12)ml² + mx². Solving for x when T = 3.00 s gives approximately 8.72 cm.

A Long Thin Uniform Rod of Length 1