Question

What is the moment of Inertia of a uniform hollow sphere of mass $M$ and radius $R$, about its diameter

• A. $\frac{1}{2}M{R}^2$
• B. $M{R}^2$
• C. $\frac{2}{3}M{R}^2$
• D. $\frac{2}{5}M{R}^2$

Answer 1

The correct option is C

$\frac{2}{3}M{R}^2$

Let us consider a small area element of $Rd\theta$, at an angle $\theta$ with $x$ axis as shown in figure.

It forms a ring of radius $RCos\theta$.

The width of this ring is $Rd\theta$ and its periphery is $2\pi Rcos\theta$. Hence, the area of the ring = $\left(2\pi Rcos\theta \right)\left(Rd\theta \right)$

Mass per unit Area of the given sphere = $\frac{M}{4\pi R^2}$

$\therefore$ Mass of ring element = $\left(\frac{M}{4\pi R^2}\right)\left(2\pi RCos\theta \right)\left(Rd\theta \right)$

= $\frac{M}{2}Cos\theta d\theta$

Now, the moment of Inertia of this elemental Ring about $OX$ is

$dI$ = $\left(\frac{M}{2}Cos\theta d\theta \right)(RCos\theta {)}^2$

= $\frac{M}{2}{R}^{2}Co{s}^3\theta d\theta$

As $\theta$ increases from $0$ to $\pi$, the elemental ring cover the whole spherical Hollow surface.

Therefore, $I$ = $\underset{0}{\overset{\pi }{\int }}\frac{M}{R^2}Co{s}^3\theta d\theta$ = $\frac{2}{3}M{R}^2$