Question

. What is the period of a rod of length L = 0.8 m and mass m = 0.5 kg, which is pivoted a distance 0.3 m from the end of the rod?

Answer 1

length L = 0.8 m

mass m = 0.5 kg

distance from end of rod = 0.3 m

distance from center of mass, x = 0.4 - 0.3 = 0.1 m

moment of inertia about center of mass ,

moment of inertia about pivot O,

By Newton's second law for rotation,

where torque is given by

for small angles,

from above equations,

This equation is similar to that for simple pendulum,

it has solution of form,

Period is