Question

From a uniform disc of radius $R$ and mass $9M$, four small discs of radius $\frac{R}{3}$ are removed as shown in figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane and passing through the centre of the disc is $nM{R}^2$. Then, $n$ is

Answer 1

Given, mass of complete disc $=9M$
and radius $=R$
Let $\sigma$ be the mass per unit area.
$\therefore 9M=\sigma \times \pi R^2=\sigma \pi R^2$

The mass of each small disc
$=\sigma \times \pi {\left(\frac{R}{3}\right)}^2=\sigma \frac{\pi R^2}{9}=M$

Now, let us consider the above system as a complete disc of mass $9M$ and 4 discs of negative mass $M$ super imposed on it.
$I_1=$ Moment of inertia of bigger disc along axis perpendicular to the plane passing through centre.
$I_2=$ MOI of each smaller disc along axis perpendicular to the plane passing through centre.
Then, $I_1=\frac{1}{2}\left(9M\right)R^2=\frac{9M{R}^2}{2}$
$I_2=\frac{1}{2}\left(M\right){\left(\frac{R}{3}\right)}^2+M{\left(\frac{2R}{3}\right)}^2$
$=\frac{M{R}^2}{18}+\frac{4M{R}^2}{9}=\frac{9M{R}^2}{18}=\frac{M{R}^2}{2}$

$\therefore$ MOI of remaining disc about the axis $=I_1-4I_2$
$=\frac{9}{2}M{R}^2-2M{R}^2=2.5M{R}^2$