Moment of Inertia of a Uniform Hollow Sphere about a Diameter

The moment of inertia of a sphere can be calculated in two situations. The first is when the sphere is hollow and when it is a solid sphere.

The moment of inertia of a uniform hollow sphere of radius R and mass M is given by

$\displaystyle I=\frac{2}{3}M{{R}^{2}}$

Derivation

Let $M$ be the mass and $R$ be the radius of a hollow sphere with center $O$. We have to calculate the moment of inertia of this sphere about the diameter $XOX’$ of the sphere. As the inside of the sphere is hollow, the mass of the sphere is spread uniformly over its surface.

Surface area of the sphere = $4 \pi R^2$

Mass per unit area of the sphere,

\[
\rho = \frac{M}{4 \pi R^2}
\]

Consider a small element of the hollow sphere in the form of a ring of radius $x$ with center $C$ and width $dx$ as shown in Fig. 5(b).21.

Let $\angle OAC = \theta$, $\angle AOB = d\theta$

In $\Delta AOC$:

$\displaystyle \sin \theta =\frac{x}{R}$ so $ \displaystyle x=R\sin \theta $

This is the radius of the elementary ring. Circumference of the elementary ring is given by:

\[
2 \pi x = 2 \pi R \sin \theta
\]

Width of the ring: $AB = dx = R d\theta$

Therefore, area of the elementary ring:

\[
= (2 \pi R \sin \theta) R d\theta
\]

Calculating mass of Elementary Ring:

Mass of the elementary ring $\displaystyle =\rho \cdot (2\pi R\sin \theta )Rd\theta $

$ \displaystyle m$ $\displaystyle =\frac{M}{{4\pi {{R}^{2}}}}\cdot (2\pi R\sin \theta )(Rd\theta )$

\[
m = \frac{M}{2} \sin \theta d\theta
\]

Moment of Inertia of the Elementary Ring

Moment of inertia of the elementary ring about $XOX’$ is given by:

$ \displaystyle dI=m{{x}^{2}}$ $ \displaystyle =\left( {\dfrac{M}{2}\sin \theta d\theta } \right){{(R\sin \theta )}^{2}}$

\[
dI = \frac{MR^2}{2} \sin^3 \theta d\theta
\]

As $\theta$ increases from $0$ to $\pi$, the elemental circular rings cover the whole spherical surface. Therefore, the moment of inertia of the hollow sphere about the given axis is:

$ \displaystyle I=\int_{0}^{\pi }{d}I$

$\displaystyle I=\frac{{M{{R}^{2}}}}{2}\int_{0}^{\pi }{{{{{\sin }}^{3}}}}\theta d\theta $

Simplifying the Integral

We know that:

\[
\int_0^\pi \sin^3 \theta d\theta = \frac{4}{3}
\]

Thus:

\[
I = \frac{MR^2}{2} \times \frac{4}{3}
\]

\[
I = \frac{2}{3} MR^2
\]

$\displaystyle I=\frac{2}{3}M{{R}^{2}}$