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 perpendicular to the plane of the lamina and passing through the points $O$ and $P$ is $I_O$ and $I_P$ ​ respectively. If the ratio $\frac{I_P}{I_O}=\frac{m}{n}$ , where $m$ and $n$ are the smallest integers, then what is the value of $m+n$ ?

Answer 1

Answer: 50

${\left(\text{I}\right)}_{\text{P}}=\left[\frac{4\text{m}{\left(\text{R}\right)}^2}{2}+\text{m}\left(4{\left(\text{R}\right)}^2\right)\right]-\left[\frac{\text{m}}{4}\frac{{\left(\text{R}\right)}^2}{2}+\frac{\text{m}}{4}5{\left(\text{R}\right)}^2\right]$
${\left(\text{I}\right)}_{\text{P}}={\left(\text{mR}\right)}^2\left[\left(2+4\right)-\left(\frac{1}{8}+\frac{5}{4}\right)\right]$
${\left(\text{I}\right)}_{\text{P}}={\left(\text{mR}\right)}^2\left(6-\frac{11}{8}\right)=\frac{37}{8}{\left(\text{mR}\right)}^2\dots \left(1\right)$
${\left(\text{I}\right)}_O=\left(\frac{4{\left(\text{mR}\right)}^2}{2}\right)-\left(\frac{\text{m}}{4}\frac{{\left(\text{R}\right)}^2}{2}+\frac{\text{m}}{4}{\left(\text{R}\right)}^2\right)$
${\left(\text{I}\right)}_O={\left(\text{mR}\right)}^2\left[2-\left(\frac{1}{8}+\frac{1}{4}\right)\right]={\left(\text{mR}\right)}^2\left[2-\frac{3}{8}\right]={\left(\text{mR}\right)}^2\left(\frac{13}{8}\right)\dots \left(2\right)$
$So\frac{{\text{I}}_{\text{P}}}{{\text{I}}_{\text{O}}}=\frac{37/8}{13/8}=\frac{37}{13}$
$\Rightarrow m+n=37+13=50$