A lamina is made by removing a small disc of diameter $2 R$ from a bigger disc of uniform mass density and…
A lamina is made by removing a small disc of diameter $2 R$ from a bigger disc of uniform mass density and radius $2 R$, as shown in the figure. The moment of inertia of this lamina about axes passing though $O$ and $P$ is $I_{O}$ 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
Solution
Let $\sigma$ be the surface mass density.
Moment of inertia of the lamina about axes passing through 'O'
$\begin{array}{l}
I_{O}=\frac{1}{2} \sigma\left[\pi(2 R)^{2}\right] \times(2 R)^{2}- \\
=\frac{13}{2} \pi \sigma R^{4} \quad\left[\frac{1}{2}\left(\sigma \pi R^{2}\right)^{2}+\sigma\left(\pi R^{2}\right) \times R^{2}\right]
\end{array}$
Moment of inertia of the lamina about axes passing through 'P'
$\begin{array}{l}
I_{P}=8 \pi \sigma \mathrm{R}^{4}+\sigma \pi(2 \mathrm{R})^{2} \times(2 \mathrm{R})^{2} \\
{\left[\frac{1}{2} \sigma\left(\pi R^{2}\right) R^{2}+\sigma\left(\pi R^{2}\right)\left(\sqrt{(2 R)^{2}+R^{2}}\right)^{2}\right]}
\end{array}$
$\begin{array}{l}
=24 \pi \sigma \mathrm{R}^{4}-5.5 \sigma \pi \mathrm{R}^{4}=18.5 \pi \sigma \mathrm{R}^{4} \\
\therefore \quad \frac{I_{P}}{I_{O}}=\frac{18.5 \pi \sigma R^{4}}{\frac{13}{2} \pi \sigma R^{4}}=\frac{37}{13} \approx 3
\end{array}$
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