Question

A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.460 Hz. The pendulum has a mass of 2.40 kg, and the pivot is located 0.380 m from the center of mass. Determine the moment of inertia of the pendulum about the pivot point.

Answer 1

The moment of inertia is
`I =1.0697 \ kg m^2`

Step-by-step explanation:

From the question we are told that

The frequency is
`f = 0.460 \ Hz`

The mass of the pendulum is
`m = 2.40 \ kg`

The location of the pivot from the center is
`d = 0.380 \ m`

Generally the period of the simple harmonic motion is mathematically represented as

`T = 2 \pi * \sqrt{ (I)/( m * g * d ) }`

Where I is the moment of inertia about the pivot point , so making I the subject of the formula it

=>
`I = [ (T)/(2 \pi ) ]^2 * m* g * d`

But the period of this simple harmonic motion can also be represented mathematically as

`T = (1)/(f)`

substituting values

`T = (1)/(0.460)`

`T = 2.174 \ s`

So

`I = [ (2.174)/(2 * 3.142 ) ]^2 * 2.40* 9.8 * 0.380`

`I =1.0697 \ kg m^2`