M and R be the mass and radius of a disc. A small disc of radius $R / 3$ is removed from the bigger disc as shown in figure. The moment of inertia of remaining part of bigger disc about an axis AB passing through the centre O and perpendicular to the plane of disc is $\frac{4}{x} M R^2$. The value of $x$ is ________.

Question

Aicharya · Accepted Answer

Without cavity $\mathrm{I}_1=\frac{\mathrm{MR}^2}{2}$
$\begin{aligned}
\text { Mass of removed disc }= & \frac{\mathrm{M}}{\pi \mathrm{R}^2} \times\left(\frac{\mathrm{R}}{3}\right)^2 \pi \\ & =\left(\frac{\mathrm{M}}{9}\right)
\end{aligned}$
M.I. of removed disc $I_2=\frac{\frac{M}{9}\left(\frac{R}{3}\right)^2}{2}+\frac{M}{9} \times\left(\frac{2 R}{3}\right)^2$
$\begin{aligned}
& =\frac{\mathrm{MR}^2}{18} \\ & \mathrm{I}=\mathrm{I}_1-\mathrm{I}_2=\frac{\mathrm{MR}^2}{2}-\frac{\mathrm{MR}^2}{18}=\frac{4 \mathrm{MR}^2}{9} \\ & (\mathrm{n}=9)
\end{aligned}$

M and R be the mass and radius of a disc. A small disc of radius $R / 3$ is removed from the bigger disc as…

Solution