Question

A lamina is made by removing a small disc of diameter $2R$ from a bigger disc of uniform mass density and radius $2R$, as shown in the figure. The moment of inertia of this lamina about axes passing through $O$ and $P$ is $I_0$ and $I_P$ respectively. Both these axes are perpendicular to the plane of the lamina. The ratio $\frac{I_P}{I_0}$

• A. $13/37$
• B. $37/13$
• C. $73/31$
• D. $8/13$

Answer 1

Answer: (B)

Let $M$ be mass of the whole disc.
Then, the mass of the removed disc $=\frac{M}{\pi {\left(2R\right)}^2}\pi R^2=\frac{M}{4}$
So, moment of inertia of the remaining disc about an axis passing through $O$ ,br> $I_0=\frac{1}{2}M{\left(2R\right)}^2-\left[\frac{1}{2}\left(\frac{M}{4}\right)R^2+\frac{M}{4}{R}^2\right]$
$=2M{R}^2-\left[\frac{M{R}^2+2M{R}^2}{8}\right]$
$=M{R}^2\left[2-\frac{3}{8}\right]=\frac{13}{8}M{R}^2$
Moment of the inertia of the remaining disc about an axis passing through $P$ is
$I_p=\left[\frac{1}{2}M\left(2R^2\right)+M{\left(2R\right)}^2\right]-\left[\frac{1}{2}\left(\frac{M}{4}\right)R^2+\frac{M}{4}{\left(\sqrt{5}R\right)}^2\right]$
$=\left[2M{R}^2+4M{R}^2\right]-\left[\frac{M{R}^2}{8}+\frac{5M{R}^2}{4}\right]$
$=6M{R}^2-\frac{11}{8}M{R}^2$
$=\frac{37}{8}M{R}^2$
$\therefore \frac{I_P}{I_O}=\frac{37}{8}\times \frac{8}{13}=\frac{37}{13}$