Moment of inertia of a uniform annular disc o | vedclass

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Question

Suppose $M$ be the mass of the annular disc of outer radius $\mathrm{R}$ and inner radius $\mathrm{r}$.

$	ext { Then } \quad \frac{	ext { mass }}{\

Vedclass · Accepted Answer

$\frac{1}{2}M\left( {{R^2} + {r^2}} \right)$

Vedclass · Answer

Suppose $M$ be the mass of the annular disc of outer radius $\mathrm{R}$ and inner radius $\mathrm{r}$.

$	ext { Then } \quad \frac{	ext { mass }}{	ext { area }}=\frac{M}{\pi\left(R^{2}-r^{2}ight)}$

Mass of elementary ring of radius $\mathrm{x}$ and thickness

$\mathrm{dx}$

$=\frac{\mathrm{M}}{\pi\left(\mathrm{R}^{2}-\mathrm{r}^{2}ight)} 	imes 2 \pi \mathrm{x} \mathrm{dx}=\frac{2 \mathrm{Mx} \mathrm{dx}}{\left(\mathrm{R}^{2}-\mathrm{r}^{2}ight)}$

$M.I.$ of ring about an axis passing through centre of mass and perpendicular to its plane is

$\mathrm{dI}=\frac{2 \mathrm{Mx} \mathrm{dx}}{\left(\mathrm{R}^{2}-\mathrm{r}^{2}ight)} \mathrm{x}^{2}=\frac{2 \mathrm{Mx}^{3} \mathrm{dx}}{\left(\mathrm{R}^{2}-\mathrm{r}^{2}ight)}$

$	herefore \quad \mathrm{I}=\int_{\mathrm{r}}^{\mathrm{R}} \frac{2 \mathrm{Mx}^{3} \mathrm{dx}}{\left(\mathrm{R}^{2}-\mathrm{r}^{2}ight)}$

$=\frac{2 \mathrm{M}}{\left(\mathrm{R}^{2}-\mathrm{r}^{2}ight)}\left[\frac{\mathrm{R}^{4}-\mathrm{r}^{4}}{4}ight]=\frac{1}{2} \mathrm{M}\left(\mathrm{R}^{2}+\mathrm{r}^{2}ight)$

Moment of inertia of a uniform annular disc of internal radius $r$ and external radius $R$ and mass $M$ about an axis through its centre and perpendicular to its plane is

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