The moment of inertia of a uniform disc of radius R from which a circular portion of radius r is removed from the periphery, about a tangent in the plane at the point of removal, if the remaining mass is m, is

Area of the remaining disc is $\mathrm{\pi }\left({\mathrm{R}}^2-{\mathrm{r}}^2\right)$

$\mathrm{mass} \mathrm{per} \mathrm{unit} \mathrm{area} = \frac{\mathrm{m}}{\mathrm{\pi }\left({\mathrm{R}}^2-{\mathrm{r}}^2\right)}=\mathrm{\sigma }$

$$\begin{gathered} {\text{moment of inertia of the hole disc I}}_0\text{ about a tangent in the plane} \\ {I}_0=\frac{5}{4}\sigma \pi R^4 \\ \mathrm{and} \mathrm{moment} \mathrm{of} \mathrm{inertia} \mathrm{of} \mathrm{the} \mathrm{removed} \mathrm{disc} \mathrm{about} \mathrm{the} \mathrm{same} \mathrm{tangent} \\ {\mathrm{I}}_1=\frac{5}{4}{\mathrm{\sigma \pi r}}^4 \\ \mathrm{moment} \mathrm{of} \mathrm{inertia} \mathrm{of} \mathrm{the} \mathrm{remaing} \mathrm{part} \\ {\mathrm{I}}_0-{\mathrm{I}}_1=\frac{5}{4}{\mathrm{\sigma \pi R}}^4-\frac{5}{4}{\mathrm{\sigma \pi r}}^4 \\ {\mathrm{I}}_0-{\mathrm{I}}_1=\frac{5}{4}\frac{\mathrm{m}}{\mathrm{\pi }\left({\mathrm{R}}^2-{\mathrm{r}}^2\right)}\pi \left(R^2+r^2\right)\left(R^2-r^2\right) \\ {\mathrm{I}}_{\mathrm{remaining}}=\frac{5}{4}\mathrm{m}\left({\mathrm{R}}^2+{\mathrm{r}}^2\right) \end{gathered}$$