From a circular disc of radius $R$ and mass $9 \mathrm{M}$, a small disc of mass $M$ and radius…
From a circular disc of radius $R$ and mass $9 \mathrm{M}$, a small disc of mass $M$ and radius $\frac{R}{3}$ is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is
$\frac{40}{9} \mathrm{MR}^2$
$\mathrm{MR}^2$
$4 \mathrm{MR}^2$
$\frac{4}{9} \mathrm{MR}^2$
Solution
The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre
$\begin{aligned}
\mathrm{I} & =\mathrm{I}_1-\mathrm{I}_2 \\
& =\frac{9 \mathrm{MR}^2}{2}-\frac{\mathrm{MR}^2}{18} \\
& =\frac{81 \mathrm{MR}^2-\mathrm{MR}^2}{18} \\
& =\frac{40 \mathrm{MR}^2}{9}
\end{aligned}$