Physics - Rotation Question with Solution | TestHub

Mass per unit area$=\frac{\mathit{\text{M}}}{\pi \left({\mathit{\text{R}}}^2-{\mathit{\text{r}}}^2\right)}$

$\therefore$mass of the whole disc$=\frac{\mathit{\text{M}}·\pi {\mathit{\text{R}}}^2}{\pi \left({\mathit{\text{R}}}^2-{\mathit{\text{r}}}^2\right)}=\frac{{\mathit{\text{MR}}}^2}{\left({\mathit{\text{R}}}^2-r^2\right)}$

Moment of inertia of the disc of radius$r$

$=\frac{1}{2}\frac{M·\pi R^2}{2\pi \left(R^2-r^2\right)}·r^2=\frac{1}{2}\frac{{\mathit{\text{Mr}}}^4}{\left({\mathit{\text{R}}}^2-r^2\right)}$

Moment of inertia of the whole disc$=\frac{1}{2}\frac{{\mathit{\text{Mr}}}^2·{\mathit{\text{r}}}^2}{\left({\mathit{\text{R}}}^2-{\mathit{\text{r}}}^2\right)}=\frac{{\mathit{\text{Mr}}}^4}{2\left(R^2-r^2\right)}$

$\therefore$Moment of inertia of the annular disc$=\frac{\mathit{\text{M}}}{2}\left[\frac{{\mathit{\text{R}}}^4-r^4}{{\mathit{\text{R}}}^2-r^2}\right]=\frac{M\left(R^2+r^2\right)}{2}$