Question

Q. Find the moment of inertia of a uniform annular disc of inner radius 2r and outer radius 3r with mass M about an axis lying in the plane of the disc and passing through its center.

Answer 1

Dear Student,

$$\begin{gathered} firstfindtheMI\hspace{0.17em}abouttheaxisperpendiculartotheplaneofthedisc \\ massdensity\sigma =\frac{M}{{\mathrm{\pi r}}^2\left(9-4\right)}=\frac{M}{5{\mathrm{\pi r}}^2} \\ MIabouttheaxis=\frac{\sigma \mathrm{\pi }{\left(3\mathrm{r}\right)}^2{\left(3\mathrm{r}\right)}^2}{2}-\frac{\sigma \mathrm{\pi }{\left(2\mathrm{r}\right)}^2{\left(2\mathrm{r}\right)}^2}{2}=\frac{M}{5{\mathrm{\pi r}}^2}.\frac{\mathrm{\pi }{\left(\mathrm{r}\right)}^4}{2}\left(81-16\right) \\ =\frac{M}{5}.\frac{{\left(\mathrm{r}\right)}^2}{2}\left(65\right)=\frac{13M{r}^2}{2} \\ MIbouttheaxisintheplane=\frac{13M{r}^2}{4} \\ Regards \end{gathered}$$