Question

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

• A. $\frac{37}{8}$
• B. $\frac{37}{13}$
• C. $\frac{8}{13}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B $\frac{37}{13}$

Assume the mass of bigger disc is $M$
Hence, mass of removed disc is $m$
Since disc has uniform mass density,
$\frac{M}{\pi (2R{)}^2}=\frac{m}{\pi R^2}$
$M=4m$ ... (1)
MOI about point $\text{O}$ for bigger disc $=\frac{M(2R{)}^2}{2}=2M{R}^2$
MOI about point $\text{O}$ for removed disc $=\frac{m(R{)}^2}{2}+m{R}^2$
$=\frac{3m{R}^2}{2}=\frac{3M{R}^2}{8}$ $(\because M=4m)$
Hence $I_0=2M{R}^2-\left[\frac{3M{R}^2}{8}\right]$
(-ve because we removed that portion)
$I_0=\frac{13M{R}^2}{8}$ ... (2)

MOI about point $\text{P}$ for bigger disc $=\frac{M(2R{)}^2}{2}+M(2R{)}^2$
$=6M{R}^2$
Distance between point $\text{P}$ and center of removed disc is
$=\sqrt{(2R{)}^2+(R{)}^2}=\sqrt{5}R$
MOI about point $\text{P}$ for removed disc
$=\frac{m{R}^2}{2}+m(\sqrt{5}R{)}^2$
$=\frac{11m{R}^2}{2}$ $(\because m=\frac{M}{4})$
$=\frac{11M{R}^2}{8}$
Hence $I_P=6M{R}^2-\left[\frac{11M{R}^2}{8}\right]$ {-ve because we removed that portion}
$I_P=\frac{37M{R}^2}{8}$
$\therefore \frac{I_P}{I_O}=\frac{\frac{37M{R}^2}{8}}{\frac{13M{R}^2}{8}}=\frac{37}{13}$