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 though $O$ and $P$ is $I_0$ and $I_P$ respectively. Both these axes are perpendicular to the plane of the lamina. The ratio $I_P/I_O$ to the nearest integer is

Answer 1

Answer: 3

$I_p=\left[\frac{4m{R}^2}{2}+m\left(4R^2\right)\right]-\left[\frac{m}{4}\frac{R^2}{2}+\frac{m}{4}5R^2\right]$
$I_p=m{R}^2\left[(2+4)-\left(\frac{1}{8}+\frac{5}{4}\right)\right]$
$I_p=m{R}^2\left(6-\frac{11}{8}\right)=\frac{37}{8}m{R}^2\dots$(1)
$I_0=\left(\frac{4m{R}^2}{2}\right)-\left(\frac{m}{4}\frac{R^2}{2}+\frac{m}{4}{R}^2\right)$
$I_0=m{R}^2\left[2-\left(\frac{1}{8}+\frac{1}{4}\right)\right]=m{R}^2\left[2-\frac{3}{8}\right]=m{R}^2\left(\frac{13}{8}\right)\dots$ (2)
So $\frac{I_p}{I_0}=\frac{37/8}{13/8}3$ (Nearest Integer)