Question

A flat, rigid object oscillates as a physical pendulum in simple harmonic motion with a frequency f. The mass of the pendulum is m, and the pivot point is a distance d from the center of mass. What is the moment of inertia of the pendulum about its pivot point

Answer 1

Answer:

I = mgd/4π²f²

Explanation:

The period of the physical pendulum T = 1/f where f = frequency,

T = 1/f = 2π√(I/mgd) where I = moment of inertia of the physical pendulum, m =mass of pendulum, g = acceleration due to gravity and d = distance of center of mass of pendulum from pivot point.

1/f = 2π√(I/mgd)

dividing both sides by 2π, we have

1/2πf = √(I/mgd)

squaring both sides, we have

(1/2πf) = [√(I/mgd)]²

1/4π²f² = I/mgd

multiplying both sides by mgd, we have

I = mgd/4π²f²