Question

do it with proper explain​

Answer 1

So we've a uniform circular ring of radius $\sf{R}$ and mass $\sf{M}$ and we need to find its moment of inertia about an axis passing through one of its diameters

⭐ Required Diagram ⭐

$\setlength{\unitlength}{2mm}\begin{picture}(5,5)\thicklines\qbezier(-10,0)(-9.5,9.5)(0,10)\qbezier(0,10)(9.3,9.3)(10,0)\qbezier(10,0)(9.5,-9.5)(0,-10)\qbezier(0,-10)(-9.5,-9.5)(-10,0)\put(0,0){\vector(0,-1){20}}\put(0,0){\vector(0,1){20}}\put(0,0){\circle*{0.75}}\multiput(0,0)(1.33,0){8}{\line(1,0){0.67}}\put(0,0){\vector(4,3){8}}\put(2,4){\sf{R}}\qbezier(2.9,0)(2.5,1.2)(2,2)\put(0,0){\vector(1,1){7.07}}\put(9.3,4.5){\vector(-3,4){1.2}}\put(6.46,7.88){\vector(1,-1){0.707}}\put(3,0.5){ \theta }\put(8,7){\sf{dL}}\put(1,-2){ \sf{R\cos\theta} }\put(8.5,2){ \sf{R\sin\theta} }\multiput(7.5,0)(0,1){7}{\line(0,1){0.5}}\put(7.5,0){\vector(1,0){0}}\put(7.5,7){\vector(0,1){0}}\tiny\put(4,3.6){\sf{d \theta }}\end{picture}$

Consider an elemental length $\sf{dL}$ of mass $\sf{dM}$ which subtends an angle $\sf{d\theta}$ at the center and is inclined at an angle $\theta$ with the horizontal.

So the length $\sf{dL}$ will be given by,

$\sf{\longrightarrow dL=R\,d\theta}$

Since the ring is of uniform, the linear density of the whole ring and that of elemental length should be same.

$\sf{\longrightarrow \dfrac{dM}{dL}=\dfrac{M}{2\pi R}}$

$\sf{\longrightarrow \dfrac{dM}{R\,d\theta}=\dfrac{M}{2\pi R}}$

$\sf{\longrightarrow \dfrac{dM}{d\theta}=\dfrac{M}{2\pi}}$

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

As we see in the figure, this elemental length is at a perpendicular distance of $\sf{R\cos\theta}$ from the axis.

So its moment of inertia about that axis is,

$\sf{\longrightarrow dI=dM(R\cos\theta)^2}$

$\sf{\longrightarrow dI=\dfrac{M}{2\pi}\,R^2\cos^2\theta\,d\theta}$

And so the moment of inertia of the whole ring will be,

$\displaystyle\sf{\longrightarrow I=\int\limits_0^{2\pi}\dfrac{M}{2\pi}\,R^2\cos^2\theta\,d\theta}$

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

$\displaystyle\sf{\longrightarrow I=\dfrac{MR^2}{4\pi}\int\limits_0^{2\pi}(1+\cos(2\theta))\,d\theta\quad\left[\because\,1+\cos(2x)=2\cos^2x\right]}$

$\displaystyle\sf{\longrightarrow I=\dfrac{MR^2}{4\pi}\left[\theta+\dfrac{1}{2}\sin(2\theta)\right]_0^{2\pi}}$

$\displaystyle\sf{\longrightarrow I=\dfrac{MR^2}{4\pi}\left(2\pi-0+\dfrac{1}{2}\big(\sin(4\pi)-\sin0\big)\right)}$

$\displaystyle\sf{\longrightarrow I=\dfrac{MR^2}{4\pi}\cdot2\pi}$

$\displaystyle\sf{\longrightarrow\underline{\underline{I=\dfrac{1}{2}\,MR^2}}}$

Hence the moment of inertia of the ring about an axis passing through one of its diameters is $\sf{\dfrac{1}{2}\,MR^2.}$