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 .
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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 .
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(i) Moment of Inertia of a Thin Uniform Spherical Shell
Consider a thin uniform spherical shell of radius [tex]\sf{R}[/tex] and mass [tex]\sf{M}[/tex] (Figure 1). Consider an elemental ring of radius [tex]\sf{r}[/tex] and mass [tex]\sf{dM}[/tex] whose center is at a distance [tex]\sf{x}[/tex] from the center of the shell.
Here [tex]\sf{dl}[/tex] is edge width of ring which subtends an angle [tex]\sf{d\theta}[/tex] at center of the shell. So,
[tex]\sf{\longrightarrow dl=R\ d\theta}[/tex]
From the triangle,
[tex]\sf{\longrightarrow r=R\sin\theta}[/tex]
The surface area of the ring,
[tex]\sf{\longrightarrow dA=2\pi r\ dl}[/tex]
[tex]\sf{\longrightarrow dA=2\pi R^2\sin\theta\ d\theta}[/tex]
Since the shell is uniform,
[tex]\sf{\longrightarrow\dfrac{M}{A}=\dfrac{dM}{dA}}[/tex]
[tex]\sf{\longrightarrow dM=\dfrac{M}{4\pi R^2}\cdot2\pi R^2\sin\theta\ d\theta}[/tex]
[tex]\sf{\longrightarrow dM=\dfrac{1}{2}\,M\sin\theta\ d\theta}[/tex]
The moment of inertia of the ring about an axis passing through its center and perpendicular to the plane is,
[tex]\sf{\longrightarrow dI=dM\cdot r^2}[/tex]
[tex]\sf{\longrightarrow dI=\dfrac{1}{2}\,M\sin\theta\ d\theta\,(R\sin\theta)^2}[/tex]
[tex]\sf{\longrightarrow dI=\dfrac{1}{2}\,MR^2\sin^3\theta\ d\theta}[/tex]
Then the moment of inertia of the whole shell is,
[tex]\displaystyle\sf{\longrightarrow I=\dfrac{1}{2}\,MR^2\int\limits_0^{\pi}\sin^3\theta\ d\theta}[/tex]
Solving the integral,
[tex]\displaystyle\sf{\longrightarrow\int\limits_0^{\pi}\sin^3\theta\ d\theta=\int\limits_0^{\pi}\sin^2\theta\cdot\sin\theta\ d\theta}[/tex]
[tex]\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}[/tex]
Take [tex]\sf{u=\cos\theta.}[/tex] Then,
[tex]\displaystyle\sf{\longrightarrow\int\limits_0^{\pi}\sin^3\theta\ d\theta=-\int\limits_1^{-1}\left(1-u^2\right)\ du}[/tex]
[tex]\displaystyle\sf{\longrightarrow\int\limits_0^{\pi}\sin^3\theta\ d\theta=\dfrac{4}{3}}[/tex]
Hence,
[tex]\displaystyle\sf{\longrightarrow I=\dfrac{1}{2}\,MR^2\cdot\dfrac{4}{3}}[/tex]
[tex]\displaystyle\sf{\longrightarrow\underline{\underline{I=\dfrac{2}{3}\,MR^2}}}[/tex]
(ii) Moment of Inertia of a Solid Uniform Sphere
Consider a solid uniform sphere of radius [tex]\sf{R}[/tex] and mass [tex]\sf{M}[/tex] (Figure 2). Consider an elemental disc of radius [tex]\sf{r}[/tex] and mass [tex]\sf{dM}[/tex] whose center is at a distance [tex]\sf{x}[/tex] from the center of the shell.
From the triangle,
[tex]\sf{\longrightarrow x=R\cos\theta}[/tex]
[tex]\sf{\longrightarrow dx=-R\sin\theta\ d\theta}[/tex]
and,
[tex]\sf{\longrightarrow r=R\sin\theta}[/tex]
Volume of disc,
[tex]\sf{\longrightarrow dV=\pi r^2\ dx}[/tex]
[tex]\sf{\longrightarrow dV=-\pi R^3\sin^3\theta\ d\theta}[/tex]
Since the sphere is uniform,
[tex]\sf{\longrightarrow\dfrac{M}{V}=\dfrac{dM}{dV}}[/tex]
[tex]\sf{\longrightarrow dM=\dfrac{M}{\dfrac{4}{3}\pi R^3}\cdot -\pi R^3\sin^3\theta\ d\theta}[/tex]
[tex]\sf{\longrightarrow dM=-\dfrac{3}{4}\,M\sin^3\theta\ d\theta}[/tex]
Moment of inertia of the disc about an axis passing through its center and perpendicular to the plane is,
[tex]\displaystyle\sf{\longrightarrow dI=\dfrac{1}{2}\,dM\cdot r^2}[/tex]
[tex]\sf{\longrightarrow dI=-\dfrac{3}{8}\,M\sin^3\theta\ d\theta\cdot(R\sin\theta)^2}[/tex]
[tex]\sf{\longrightarrow dI=-\dfrac{3}{8}\,MR^2\sin^5\theta\ d\theta}[/tex]
Then moment of inertia of the whole sphere will be,
[tex]\displaystyle\sf{\longrightarrow I=\int\limits_{\pi}^0-\dfrac{3}{8}\,MR^2\sin^5\theta\ d\theta }[/tex]
[tex]\displaystyle\sf{\longrightarrow I=\dfrac{3}{8}\,MR^2\int\limits_0^{\pi}\sin^5\theta\ d\theta }[/tex]
Solving the integral,
[tex]\displaystyle\sf{\longrightarrow \int\limits_0^{\pi}\sin^5\theta\ d\theta=\int\limits_0^{\pi}\sin^4\theta\cdot\sin\theta\ d\theta}[/tex]
[tex]\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}[/tex]
Take [tex]\sf{u=\cos\theta.}[/tex] Then,
[tex]\displaystyle\sf{\longrightarrow \int\limits_0^{\pi}\sin^5\theta\ d\theta=-\int\limits_1^{-1}\left(1-u^2\right)^2\ du}[/tex]
[tex]\displaystyle\sf{\longrightarrow \int\limits_0^{\pi}\sin^5\theta\ d\theta=\dfrac{16}{15}}[/tex]
Hence,
[tex]\displaystyle\sf{\longrightarrow I=\dfrac{3}{8}\,MR^2\cdot\dfrac{16}{15}}[/tex]
[tex]\displaystyle\sf{\longrightarrow\underline{\underline{I=\dfrac{2}{5}\,MR^2}}}[/tex]
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