Question

A holiday ornament in the shape of a hollow sphere with mass 0.010 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 with negligible friction. Calculate its period. (Use the parallel-axis theorem to find the moment of inertia of the sphere about the pivot at the tree limb.)

Answer 1

Answer:

0.608 s

Explanation:

Given that

m = 0.01 kg

r = 0.055 m

the period of a pendulum is primarily stated as

T = 2π √(I/mgd), where

I = moment of inertia

m = mass of pendulum

g = acceleration due to gravity

d = radius

moment of inertia, I is given as

I = ⅔MR² + MR²

I = 5/2 MR²

also, going forward, we assume d = R

next, we substitute each into the equation for period, i.e 2π√(I/mgd)

T = 2π √[(5/3MR²) / MgR]

T = 2π √[(5/3R) / g]

T = 2π √(5R/3g)

Next, we plug in the values of each, and we have

T = 2π √[(5 * 0.055) / (3 * 9.8)]

T = 2π √(0.275/29.4)

T = 2π √0.00935

T = 2π * 0.0967

T = 2 * 3.142 * 0.0967

T = 0.608 s

Therefore, the period of the hollow sphere is 0.608 s