Question

A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.690 Hz. The pendulum has a mass of 2.40 kg, and the pivot is located 0.260 m from the center of mass. Determine the moment of inertia of the pendulum about the pivot point.

Answer 1

Moment of inertia of the pendulum about the pivot point is 0.326kgm²

Step-by-step explanation:

The formula for calculating the period T as a function of the moment of inertia I is expressed as shown:

T = 2π√(I/mgd) ... (1)

Also T = 1/f ... (2)

Substituting equation 2 into 1, we have;

1/f = 2π√I/mgd...(3)

Making the moment of inertia I the subject of the formula from equation 3 since that's what we are asked to find

First we will have to square both sides;

(1/f)² = {2π√I/mgd}²

1/f² = 4π²(I/mgd)

1/f² = 4Iπ²/mgd

Cross multiplying:

mgd = 4Iπ²f²

Dividing both sides by 4π²f²

mgd/4π²f² = I

Given the following data

Frequency f = 0.690Hz

Pivot distance d = 0.260m

Mass of pendulum m = 2.40kg

acceleration due to gravity g = 9.81m/s²

I = 2.4×9.81×0.26/4π²×0.69²

I = 6.12/18.79

I = 0.326kgm²