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

1.4584 kg
`m^2`

Step-by-step explanation:

Time period of a physical pendulum is given by
`T=2Π\sqrt{(I)/(mgd)}`

Here f=0.290 so
`T=(1)/(F)=(1)/(0.29)=3.44827`

Mass =2.40 kg

d=0.300 m

g =9.8 m
`sec^2`

So
`3.448=2* \pi \sqrt{(I)/(2.4* 9.81* 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`