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

$\left(\text{I}\right)_{\text{P}} = \left[\frac{4 \text{m} \left(\text{R}\right)^{2}}{2} + \text{m} \left(4 \left(\text{R}\right)^{2}\right)\right] - \left[\frac{\text{m}}{4} \frac{\left(\text{R}\right)^{2}}{2} + \frac{\text{m}}{4} 5 \left(\text{R}\right)^{2}\right]$
$\left(\text{I}\right)_{\text{P}} = \left(\text{mR}\right)^{2} \left[\left(2 + 4\right) - \left(\frac{1}{8} + \frac{5}{4}\right)\right]$
$\left(\text{I}\right)_{\text{P}} = \left(\text{mR}\right)^{2} \left(6 - \frac{1 1}{8}\right) = \frac{3 7}{8} \left(\text{mR}\right)^{2} \ldots \left(1\right)$
$\left(\text{I}\right)_{O}=\left(\frac{4 \left(\text{mR}\right)^{2}}{2}\right)-\left(\frac{\text{m}}{4} \frac{\left(\text{R}\right)^{2}}{2} + \frac{\text{m}}{4} \left(\text{R}\right)^{2}\right)$
$\left(\text{I}\right)_{O}=\left(\text{mR}\right)^{2}\left[2 - \left(\frac{1}{8} + \frac{1}{4}\right)\right]=\left(\text{mR}\right)^{2}\left[2 - \frac{3}{8}\right]=\left(\text{mR}\right)^{2}\left(\frac{13}{8}\right)\ldots \left(2\right)$
$So\frac{\text{I}_{\text{P}}}{\text{I}_{\text{O}}}=\frac{3 7 / 8}{1 3 / 8}=\frac{37}{13}$
$\Rightarrow m+n=37+13=50$