91Ó°ÊÓ

The moment of inertia of a thin ring of radius $\mathbf{f}\left(x\right)$about the axis perpendicular to it and passing through its centre is expressed as follows,

$\mathbf{dl}\mathbf{=}{\left(f\left(x\right)\right)}^{\mathbf{2}}\mathbf{}\mathbf{dM}$

The value of $\mathbf{dM}$is as follows,

$$\begin{gathered} \mathbf{dM}\mathbf{=}\mathbf{2}\mathbf{Ï€´Ú}\left(x\right)\mathbf{δ»å²õ}\left(x\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{2}\mathbf{πδ}\sqrt{\mathbf{x}}\sqrt{\mathbf{1}\mathbf{+}{\left(f\left(x\right)\right)}^{\mathbf{2}}} \end{gathered}$$

Determine the value of mass element.

$$\begin{gathered} dM\mathit{=}2\mathrm{π}f\left(x\right)\mathrm{δ}ds\left(x\right) \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}2\mathrm{π}\mathrm{δ}\sqrt{\mathit{x}}\sqrt{\mathit{1}\mathit{+}{\left(f^{\text{'}}\left(x\right)\right)}^{\mathit{2}}} \\ =2\mathrm{π}\mathrm{δ}\sqrt{\mathit{x}\mathit{+}\frac{\mathit{1}}{\mathit{4}}} \end{gathered}$$

Here, $\mathrm{δ}$is the surface mass density

Perform integration in the limits 0 to 2.

$$\begin{gathered} l={∫}_0^{2}dl\left(x\right) \\ ={∫}_0^2{\left(f\left(x\right)\right)}^{2}dM \\ =2\mathrm{πδ}{∫}_0^2\mathrm{x}\sqrt{\mathrm{x}+\frac{1}{4}}\mathrm{dx} \\ =2\mathrm{πδ}\left[{∫}_0^2\left(\mathrm{x}+\frac{1}{4}\right)\sqrt{\mathrm{x}+\frac{1}{4}}\mathrm{d}\left(\mathrm{x}+\frac{1}{4}\right)-\frac{1}{4}{∫}_0^2\sqrt{\mathrm{x}+\frac{1}{4}}\mathrm{d}\left(\mathrm{x}+\frac{1}{4}\right)\right] \end{gathered}$$

Further, simplify the expression.

$$\begin{gathered} l=2\mathrm{πδ}\left[{\overline{)\frac{2}{5}{\left(\mathrm{x}+\frac{1}{4}\right)}^{5/2}}}_0^2-\frac{1}{4}.\frac{2}{3}{\left(\mathrm{x}+\frac{1}{4}\right)}^{3/2}\right] \\ =2\mathrm{πδ}\left[\frac{2}{5}\left({\left(\frac{9}{4}\right)}^{5/2}-{\left(\frac{1}{4}\right)}^{5/2}\right)-\frac{1}{6}\left({\left(\frac{9}{4}\right)}^{3/2}-{\left(\frac{1}{4}\right)}^{3/2}\right)\right] \\ =2\mathrm{πδ}\left(\frac{149}{60}\right) \\ =\mathrm{πδ}\left(\frac{149}{30}\right) \end{gathered}$$

From the earlier calculation, the value of $S$is$\frac{13\mathrm{π}}{3}$, and$M=\mathrm{δ}S$.

Divide role="math" localid="1658895170104" $l$by $M$to find the value of inertia in terms of $M$.

$$\begin{gathered} \frac{l}{M}=\frac{\mathrm{πδ}{\displaystyle \frac{149}{30}}}{{\displaystyle \frac{13\mathrm{πδ}}{3}}} \\ \frac{l}{M}=\frac{149}{30} \\ l=\frac{149}{30}M \end{gathered}$$

Thus, the moment of inertia of thin shell is $l=\frac{149}{30}M$.