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

`I = (mgd)/(4\pi^(2)f^(2))`

Step-by-step explanation:

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

`T = 2\pi\sqrt{(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 = (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 = (1)/(T) \rightarrow T = (1)/(f)`

By entering T into equation (1) we have:

`I = (T^(2)*mgd)/((2\pi)^(2))`

`I = ((1/f)^(2)*mgd)/((2\pi)^(2))`

`I = (mgd)/(4\pi^(2)f^(2))`

Therefore, the moment of inertia of the pendulum about the pivot point is
`I = (mgd)/(4\pi^(2)f^(2))`.

I hope it helps you!