Question

A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.290 Hz. The pendulum has a mass of 2.40 kg, and the pivot is located 0.300 m from the center of mass. Determine the moment of inertia of the pendulum about the pivot point.

Answer 1

Answer:

1.4584 kg$m^2$

Explanation:

Time period of a physical pendulum is given by $T=2Π\sqrt{\frac{I}{mgd}}$

Here f=0.290 so $T=\frac{1}{F}=\frac{1}{0.29}=3.44827$

Mass =2.40 kg

d=0.300 m

g =9.8 m$sec^2$

So $3.448=2\times \pi \sqrt{\frac{I}{2.4\times 9.81\times 0.300}}=1.4584$  kg-$m^2$

So the moment of inertia of the pendulum about the pivot point will be 1.4584 kg-$m^2$