Question

A coat hanger of mass m = 0.248 kg oscillates on a peg as a physical pendulum as shown in the figure below. The distance from the pivot to the center of mass of the coat hanger is d = 18.0 cm and the period of the motion is T = 1.16 s. Find the moment of inertia of the coat hanger about the pivot.

Pivot CM

_______________________=kg · m²

Answer 1

Given,

m = 0.248 kg ; d = 18 cm = 0.18 m ; T = 1.16 s

We need to determine the moment of inertia of coat hanger about the pivot.

As oscillates on a peg as a physical pendulum. We know that the time period and Moment of inertia of a physical pendulum are related as:

T = 2 pi sqrt ( I / m g l ) ; In our case, l = d = 0.18 m

Squaring and solving the above eqn for I we get:

I = T² x m g d / 4 pi²

Putting in the values we have:

I = 1.16 x 1.16 x 0.248 x 9.8 x 0.18 / 4 x 3.14 x 3.14 = 0.015 kg-m²

Hence, I = 0.015 kg-m².