Question

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

• A.
• B. MR²
• C.
• D.

Answer 1

The correct option is C

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 $RSin\theta$.

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

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

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

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

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

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

= $\frac{M}{2}{R}^{2}Si{n}^3\theta d\theta$

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

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