Question

A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency f. The pendulum has a mass m, and the pivot is located a distance d from the center of mass. Determine the moment of inertia of the pendulum about the pivot point

Answer 1

Answer:

$I = \frac{mgd}{4\pi^{2}f^{2}}$

Explanation:

The moment of inertia (I) of the pendulum can be found using the following equation:

$T = 2\pi\sqrt{\frac{I}{mgd}}$    

Where:

T: is the period of the pendulum

m: is the pendulum's mass

g: is the gravity

d: is the distance of the pivot from the center of mass

Solving the above equation for I:

$I = \frac{T^{2}*mgd}{(2\pi)^{2}}$     (1)

We have that the pendulum moves in simple harmonic motion with a frequency f, and this f is equal to:

$f = \frac{1}{T} \rightarrow T = \frac{1}{f}$

By entering T into equation (1) we have:

$I = \frac{T^{2}*mgd}{(2\pi)^{2}}$

$I = \frac{(1/f)^{2}*mgd}{(2\pi)^{2}}$  

$I = \frac{mgd}{4\pi^{2}f^{2}}$

Therefore, the moment of inertia of the pendulum about the pivot point is $I = \frac{mgd}{4\pi^{2}f^{2}}$.

I hope it helps you!