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

Answer:

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

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

        $T = 2 \pi * \sqrt{ \frac{I}{ m * g * d } }$

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

=>    $I = [ \frac{T}{2 \pi } ]^2 * m* g * d$

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

        $T = \frac{1}{f}$

substituting values

      $T = \frac{1}{0.460}$

      $T = 2.174 \ s$

So

      $I = [ \frac{2.174}{2 * 3.142 } ]^2 * 2.40* 9.8 * 0.380$

      $I =1.0697 \ kg m^2$