Question

A holiday ornament in the shape of a hollow sphere with mass 0.015 kg and radius 0.055 m is hung from a tree limb by a small loop of wire attached to the surface of the sphere. If the ornament is displaced a small distance and released, it swings back and forth as a physical pendulum.
Calculate its period. (You can ignore friction at the pivot. The moment of inertia of the sphere about the pivot at the tree limb is 5MR²/3.)
Take the free-fall acceleration to be 9.80 m/s². Express your answer using two significant figures.

Answer 1

Answer: 0.61 s

Explanation:

Given

Mass of object, m = 0.015 kg

Radius of object, r = 0.055 m

Acceleration of object, g = 9.8 m/s²

In a pendulum,

T = 2π * √[I /(mgd)]

The moment of Inertia, I of a hollow sphere is given by

I(sphere) = 2/3MR² + MR²

I(sphere) = 5/3MR²

Also, d = R

Substituting these into the first equation, we have

T = 2π * √[(5/3MR²) / (mgr)]

T = 2π * √[(5/3r) / (g)]

T = 2 * 3.142 * √(5/3 * 0.055) / (9.8)]

T = 6.284 * √(0.092/9.8)

T = 6.284 * √0.00939

T = 6.284 * 0.097

T = 0.6095 s

To 2 significant figures,

The period is 0.61 s