Question

A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.680 Hz. The pendulum has a mass of 2.00 kg, and the pivot is located 0.340 m from the center of mass. Determine the moment of inertia of the pendulum about the pivot point.

Answer 1

Answer:

Therefore, the moment of inertia is:

$I=0.37 \: kgm^{2}$

Explanation:

The period of an oscillation equation of a solid pendulum is given by:

$T=2\pi \sqrt{\frac{I}{Mgd}}$ (1)

Where:

• I is the moment of inertia
• M is the mass of the pendulum
• d is the distance from the center of mass to the pivot
• g is the gravity

Let's solve the equation (1) for I

$T=2\pi \sqrt{\frac{I}{Mgd}}$

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

Before find I, we need to remember that

$T = \frac{1}{f}=\frac{1}{0.680}=1.47\: s$

Now, the moment of inertia will be:

$I=2*9.81*0.340(\frac{1.47}{2\pi})^{2}$  

Therefore, the moment of inertia is:

$I=0.37 \: kgm^{2}$

I hope it helps you!