Moment of Inertia of a Sphere – Formula and Derivation
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.
Moment of Inertia of a Uniform Hollow Sphere about a Diameter
The moment of inertia of a uniform hollow sphere of radius R and mass M is given by
$\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\displaystyle I=\frac{2}{3}M{{R}^{2}}}}$
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:
\[
I = \frac{MR^2}{2} \times \frac{4}{3}
\]
\[
I = \frac{2}{3} MR^2
\]
$\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\displaystyle I=\frac{2}{3}M{{R}^{2}}}}$
Moment of Inertia of a Uniform Solid Sphere about a Diameter
The moment of inertia of a uniform solid sphere of radius R and mass M is given by
Let $M$ be the mass and $R$ be the radius of a uniform solid sphere with center $O$. We have to calculate the moment of inertia of this sphere about its diameter $XOX’$.
Volume of the solid sphere = $\dfrac{4}{3} \pi R^3$
Mass per unit volume of the solid sphere:
\[
\rho = \frac{M}{\dfrac{4}{3} \pi R^3}
\]
\[
\rho = \frac{3M}{4 \pi R^3} \tag{i}
\]
With $O$ as the center and radii $x$ and $(x + dx)$, draw two spheres within the given solid sphere. The thin spherical shell trapped between these spheres can be treated as a hollow sphere of radius $x$ and thickness $dx$.
Surface area of the elementary hollow sphere:
\[
4 \pi x^2
\]
Volume of the elementary hollow sphere:
\[
4 \pi x^2 dx
\]
Mass of the elementary hollow sphere:
\[
m = \rho (4 \pi x^2 dx)
\]
\[
m = \frac{3M}{4 \pi R^3} (4 \pi x^2 dx)
\]
\[
m = \frac{3M}{R^3} x^2 dx
\]
Moment of inertia of the elementary hollow sphere about its diameter is:
$\displaystyle dI=m{{x}^{2}}$ $\displaystyle =\left( {\frac{{3M}}{{{{R}^{3}}}}{{x}^{2}}dx} \right){{x}^{2}}$
\[
dI = \frac{3M}{R^3} x^4 dx
\]
When $x = 0$, the spherical shell is formed at the center of the solid sphere. As $x$ increases from 0 to $R$, the shells cover the entire solid sphere.
Moment of inertia of the solid sphere about its diameter is:
\[
I = \int_0^R dI = \frac{3M}{R^3} \int_0^R x^4 dx
\]
\[
I = \frac{3M}{R^3} \cdot \frac{x^5}{5} \bigg|_0^R
\]
\[
I = \frac{3M}{R^3} \cdot \frac{R^5}{5}
\]
\[
I = \frac{3M}{5} R^2
\]
$\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\displaystyle I=\frac{2}{5}M{{R}^{2}}}}$
Using the theorem of parallel axes, moment of inertia of a solid sphere about a tangent to the sphere is:
$ \displaystyle \begin{array}{l}{I}’=I+M{{R}^{2}}\\{I}’=\dfrac{2}{5}M{{R}^{2}}+M{{R}^{2}}\\{I}’=\dfrac{7}{5}M{{R}^{2}}\end{array}$
