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
Answers
Answer:
(i) Moment of Inertia of a Thin Uniform Spherical Shell
Consider a thin uniform spherical shell of radius \sf{R}R and mass \sf{M}M (Figure 1). Consider an elemental ring of radius \sf{r}r and mass \sf{dM}dM whose center is at a distance \sf{x}x from the center of the shell.
Here \sf{dl}dl is edge width of ring which subtends an angle \sf{d\theta}dθ at center of the shell. So,
\sf{\longrightarrow dl=R\ d\theta}⟶dl=R dθ
From the triangle,
\sf{\longrightarrow r=R\sin\theta}⟶r=Rsinθ
The surface area of the ring,
\sf{\longrightarrow dA=2\pi r\ dl}⟶dA=2πr dl
\sf{\longrightarrow dA=2\pi R^2\sin\theta\ d\theta}⟶dA=2πR
2
sinθ dθ
Since the shell is uniform,
\sf{\longrightarrow\dfrac{M}{A}=\dfrac{dM}{dA}}⟶
A
M
=
dA
dM
\sf{\longrightarrow dM=\dfrac{M}{4\pi R^2}\cdot2\pi R^2\sin\theta\ d\theta}⟶dM=
4πR
2
M
⋅2πR
2
sinθ dθ
\sf{\longrightarrow dM=\dfrac{1}{2}\,M\sin\theta\ d\theta}⟶dM=
2
1
Msinθ dθ
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}⟶dI=dM⋅r
2
\sf{\longrightarrow dI=\dfrac{1}{2}\,M\sin\theta\ d\theta\,(R\sin\theta)^2}⟶dI=
2
1
Msinθ dθ(Rsinθ)
2
\sf{\longrightarrow dI=\dfrac{1}{2}\,MR^2\sin^3\theta\ d\theta}⟶dI=
2
1
MR
2
sin
3
θ dθ
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}⟶I=
2
1
MR
2
0
∫
π
sin
3
θ dθ
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}⟶
0
∫
π
sin
3
θ dθ=
0
∫
π
sin
2
θ⋅sinθ dθ
\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}⟶
0
∫
π
sin
3
θ dθ=−
0
∫
π
(1−cos
2
θ)⋅−sinθ dθ
Take \sf{u=\cos\theta.}u=cosθ. Then,
\displaystyle\sf{\longrightarrow\int\limits_0^{\pi}\sin^3\theta\ d\theta=-\int\limits_1^{-1}\left(1-u^2\right)\ du}⟶
0
∫
π
sin
3
θ dθ=−
1
∫
−1
(1−u
2
) du
\displaystyle\sf{\longrightarrow\int\limits_0^{\pi}\sin^3\theta\ d\theta=\dfrac{4}{3}}⟶
0
∫
π
sin
3
θ dθ=
3
4
Hence,
\displaystyle\sf{\longrightarrow I=\dfrac{1}{2}\,MR^2\cdot\dfrac{4}{3}}⟶I=
2
1
MR
2
⋅
3
4
\displaystyle\sf{\longrightarrow\underline{\underline{I=\dfrac{2}{3}\,MR^2}}}⟶
I=
3
2
MR
2
(ii) Moment of Inertia of a Solid Uniform Sphere
Consider a solid uniform sphere of radius \sf{R}R and mass \sf{M}M (Figure 2). Consider an elemental disc of radius \sf{r}r and mass \sf{dM}dM whose center is at a distance \sf{x}x from the center of the shell.
From the triangle,
\sf{\longrightarrow x=R\cos\theta}⟶x=Rcosθ
\sf{\longrightarrow dx=-R\sin\theta\ d\theta}⟶dx=−Rsinθ dθ
and,
\sf{\longrightarrow r=R\sin\theta}⟶r=Rsinθ
Volume of disc,
\sf{\longrightarrow dV=\pi r^2\ dx}⟶dV=πr
2
dx
\sf{\longrightarrow dV=-\pi R^3\sin^3\theta\ d\theta}⟶dV=−πR
3
sin
3
θ dθ
Since the sphere is uniform,
\sf{\longrightarrow\dfrac{M}{V}=\dfrac{dM}{dV}}⟶
V
M
=
dV
dM
\sf{\longrightarrow dM=\dfrac{M}{\dfrac{4}{3}\pi R^3}\cdot -\pi R^3\sin^3\theta\ d\theta}⟶dM=
3
4
πR
3
M
⋅−πR
3
sin
3
θ dθ
\sf{\longrightarrow dM=-\dfrac{3}{4}\,M\sin^3\theta\ d\theta}⟶dM=−
4
3
Msin
3
θ dθ
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}⟶dI=
2
1
dM⋅r
2
\sf{\longrightarrow dI=-\dfrac{3}{8}\,M\sin^3\theta\ d\theta\cdot(R\sin\theta)^2}⟶dI=−
8
3
Msin
3
θ dθ⋅(Rsinθ)
2
\sf{\longrightarrow dI=-\dfrac{3}{8}\,MR^2\sin^5\theta\ d\theta}⟶dI=−
8
3
MR
2
sin
5
θ dθ
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 }⟶I=
π
∫
0
−
8
3
MR
2
sin
5
θ dθ
\displaystyle\sf{\longrightarrow I=\dfrac{3}{8}\,MR^2\int\limits_0^{\pi}\sin^5\theta\ d\theta }⟶I=
8
3
MR
2
0
∫
π
sin
5
θ dθ
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}⟶
0
∫
π
sin
5
θ dθ=
0
∫
π
sin
4
θ⋅sinθ dθ
\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}⟶
0
∫
π
sin
5
θ dθ=−
0
∫
π
(1−cos
2
θ)
2
⋅−sinθ dθ
Take \sf{u=\cos\theta.}u=cosθ. Then,
\displaystyle\sf{\longrightarrow \int\limits_0^{\pi}\sin^5\theta\ d\theta=-\int\limits_1^{-1}\left(1-u^2\right)^2\ du}⟶
0
∫
π
sin
5
θ dθ=−
1
∫
−1
(1−u
2
)
2
du
\displaystyle\sf{\longrightarrow \int\limits_0^{\pi}\sin^5\theta\ d\theta=\dfrac{16}{15}}⟶
0
∫
π
sin
5
θ dθ=
15
16
Hence,
\displaystyle\sf{\longrightarrow I=\dfrac{3}{8}\,MR^2\cdot\dfrac{16}{15}}⟶I=
8
3
MR
2
⋅
15
16
\displaystyle\sf{\longrightarrow\underline{\underline{I=\dfrac{2}{5}\,MR^2}}}⟶
I=
5
2
MR
2
Answer:
above is the correct answer..