Question

Derive the moment of inertia of

(i)Thin,uniform spherical shell
(ii)Solid,uniform sphere

Explain each step and with proper explanation with correct integration .​

Answer 1

(i) Moment of Inertia of a Thin Uniform Spherical Shell

Consider a thin uniform spherical shell of radius $\sf{R}$ and mass $\sf{M}$ (Figure 1). Consider an elemental ring of radius $\sf{r}$ and mass $\sf{dM}$ whose center is at a distance $\sf{x}$ from the center of the shell.

Here $\sf{dl}$ is edge width of ring which subtends an angle $\sf{d\theta}$ at center of the shell. So,

$\sf{\longrightarrow dl=R\ d\theta}$

From the triangle,

$\sf{\longrightarrow r=R\sin\theta}$

The surface area of the ring,

$\sf{\longrightarrow dA=2\pi r\ dl}$

$\sf{\longrightarrow dA=2\pi R^2\sin\theta\ d\theta}$

Since the shell is uniform,

$\sf{\longrightarrow\dfrac{M}{A}=\dfrac{dM}{dA}}$

$\sf{\longrightarrow dM=\dfrac{M}{4\pi R^2}\cdot2\pi R^2\sin\theta\ d\theta}$

$\sf{\longrightarrow dM=\dfrac{1}{2}\,M\sin\theta\ d\theta}$

The moment of inertia of the ring about an axis passing through its center and perpendicular to the plane is,

$\sf{\longrightarrow dI=dM\cdot r^2}$

$\sf{\longrightarrow dI=\dfrac{1}{2}\,M\sin\theta\ d\theta\,(R\sin\theta)^2}$

$\sf{\longrightarrow dI=\dfrac{1}{2}\,MR^2\sin^3\theta\ d\theta}$

Then the moment of inertia of the whole shell is,

$\displaystyle\sf{\longrightarrow I=\dfrac{1}{2}\,MR^2\int\limits_0^{\pi}\sin^3\theta\ d\theta}$

Solving the integral,

$\displaystyle\sf{\longrightarrow\int\limits_0^{\pi}\sin^3\theta\ d\theta=\int\limits_0^{\pi}\sin^2\theta\cdot\sin\theta\ d\theta}$

$\displaystyle\sf{\longrightarrow\int\limits_0^{\pi}\sin^3\theta\ d\theta=-\int\limits_0^{\pi}\left(1-\cos^2\theta\right)\cdot-\sin\theta\ d\theta}$

Take $\sf{u=\cos\theta.}$ Then,

$\displaystyle\sf{\longrightarrow\int\limits_0^{\pi}\sin^3\theta\ d\theta=-\int\limits_1^{-1}\left(1-u^2\right)\ du}$

$\displaystyle\sf{\longrightarrow\int\limits_0^{\pi}\sin^3\theta\ d\theta=\dfrac{4}{3}}$

Hence,

$\displaystyle\sf{\longrightarrow I=\dfrac{1}{2}\,MR^2\cdot\dfrac{4}{3}}$

$\displaystyle\sf{\longrightarrow\underline{\underline{I=\dfrac{2}{3}\,MR^2}}}$

(ii) Moment of Inertia of a Solid Uniform Sphere

Consider a solid uniform sphere of radius $\sf{R}$ and mass $\sf{M}$ (Figure 2). Consider an elemental disc of radius $\sf{r}$ and mass $\sf{dM}$ whose center is at a distance $\sf{x}$ from the center of the shell.

From the triangle,

$\sf{\longrightarrow x=R\cos\theta}$

$\sf{\longrightarrow dx=-R\sin\theta\ d\theta}$

and,

$\sf{\longrightarrow r=R\sin\theta}$

Volume of disc,

$\sf{\longrightarrow dV=\pi r^2\ dx}$

$\sf{\longrightarrow dV=-\pi R^3\sin^3\theta\ d\theta}$

Since the sphere is uniform,

$\sf{\longrightarrow\dfrac{M}{V}=\dfrac{dM}{dV}}$

$\sf{\longrightarrow dM=\dfrac{M}{\dfrac{4}{3}\pi R^3}\cdot -\pi R^3\sin^3\theta\ d\theta}$

$\sf{\longrightarrow dM=-\dfrac{3}{4}\,M\sin^3\theta\ d\theta}$

Moment of inertia of the disc about an axis passing through its center and perpendicular to the plane is,

$\displaystyle\sf{\longrightarrow dI=\dfrac{1}{2}\,dM\cdot r^2}$

$\sf{\longrightarrow dI=-\dfrac{3}{8}\,M\sin^3\theta\ d\theta\cdot(R\sin\theta)^2}$

$\sf{\longrightarrow dI=-\dfrac{3}{8}\,MR^2\sin^5\theta\ d\theta}$

Then moment of inertia of the whole sphere will be,

$\displaystyle\sf{\longrightarrow I=\int\limits_{\pi}^0-\dfrac{3}{8}\,MR^2\sin^5\theta\ d\theta }$

$\displaystyle\sf{\longrightarrow I=\dfrac{3}{8}\,MR^2\int\limits_0^{\pi}\sin^5\theta\ d\theta }$

Solving the integral,

$\displaystyle\sf{\longrightarrow \int\limits_0^{\pi}\sin^5\theta\ d\theta=\int\limits_0^{\pi}\sin^4\theta\cdot\sin\theta\ d\theta}$

$\displaystyle\sf{\longrightarrow \int\limits_0^{\pi}\sin^5\theta\ d\theta=-\int\limits_0^{\pi}(1-\cos^2\theta)^2\cdot-\sin\theta\ d\theta}$

Take $\sf{u=\cos\theta.}$ Then,

$\displaystyle\sf{\longrightarrow \int\limits_0^{\pi}\sin^5\theta\ d\theta=-\int\limits_1^{-1}\left(1-u^2\right)^2\ du}$

$\displaystyle\sf{\longrightarrow \int\limits_0^{\pi}\sin^5\theta\ d\theta=\dfrac{16}{15}}$

Hence,

$\displaystyle\sf{\longrightarrow I=\dfrac{3}{8}\,MR^2\cdot\dfrac{16}{15}}$

$\displaystyle\sf{\longrightarrow\underline{\underline{I=\dfrac{2}{5}\,MR^2}}}$

Answer 2

Answer:

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