Question

A holiday ornament in the shape of a hollow sphere with mass M = 1.50×10−2 kg and radius R = 5.50×10−2 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. (You can ignore friction at the pivot. The moment of inertia of the sphere about the pivot at the tree limb is 5MR^{2}/3.)
Take the free fall acceleration to be 9.80 m/s^2. Express your answer using two significant figures.

Answer 1

Answer:

T = 0.607 seconds

Explanation:

Given:

Mass, M = 1.50 × 10⁻² kg

Radius, R = 5.50 × 10⁻² m

Now,

the time period in terms of moment of inertia is given as:

$T = 2\pi\sqrt\frac{I}{mgR}$    .....................1

where, T is the time period

g is the acceleration due to gravity

I is the moment of inertia

Now,

Moment of inertia, I is given as:

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

on substituting the moment of inertia in the equation 1, we get

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

or

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

on substituting the valeus, we get

$T = 2\pi\sqrt\frac{{5\times5.50\times10^{-2}}}{3\times9.8}$

or

T = 0.607 seconds

Hence, the time period is 0.607 seconds