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 $l_o$ and $l_p$,respectively. Both these axes are perpendicular to the plane of the lamina. The ratio $\frac{l_p}{l_o}$ to the nearest integer is :

Answer 1

$m_{60}=m$

$m_{SD}=\frac{m}{4}$

$I_o=\frac{m{\left(2R\right)}^2}{2}-\frac{m}{4}[\frac{R^2}{2}+R^2]-m{R}^2(2-\frac{3}{8})=\frac{13m{R}^2}{8}$

$I_p=m[\frac{{\left(2R\right)}^2}{2}+{\left(2R\right)}^2]-\frac{m}{4}[\frac{R^2}{2}+5R^2]$

$=m{R}^2(6-\frac{11}{8})$

$=\frac{37}{8}m{R}^2$

$\therefore \frac{I_p}{I_o}=\frac{\frac{37}{8}}{\frac{13}{8}}=2.8\simeq 3$