Question

A holiday ornament in the shape of a hollow sphere with mass 1.5×10−2 kg and radius 4.5×10−2 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.

Answer 1

Answer:

0.55 s

Explanation:

We are given that

Mass of holiday ornament=$1.5\times 10^{-2}$kg

Radius of hollow sphere=$4.5\times 10^{-2}$ m

We have to find the period of ornament.

Moment of inertia of the sphere about the pivot at the tree limb

$I=\frac{5mR^2}{3}$

Time period,T=$2\pi\sqrt{\frac{I}{mgR}}$

T=$2\pi\sqrt{\frac{5mR^2}{3mgR}}$

$T=2\pi\sqrt{\frac{5R}{3g}}$

g=$9.8m/s^2$

Substitute the values then, we get

$T=2\times \frac{22}{7}\times \sqrt{\frac{5\times 4.5\times 10^{-2}}{3\times 9.8}}$

$T=0.55 s$

Hence, the time period of ornament=0.55 s