Physics - Rotation Question with Solution | TestHub

M.I. of complete disc about its centre $O$. $$I_{\mathrm{Total}}=\frac{1}{2}M{R}^2$$
Mass of circular hole (removed) $$=\frac{M}{4}\left(\text{ As }M=\pi R^{2}t\therefore M\propto R^2\right)$$ M.I. of removed hole about its own axis $$=\frac{1}{2}\left(\frac{M}{4}\right){\left(\frac{R}{2}\right)}^2=\frac{1}{32}M{R}^2$$ M.I. of removed hole about $O^{\prime }$ $$\begin{array}{rl}{I}_{\text{removed hole }}& =I_{\mathrm{cm}}+m{x}^2\\ & =\frac{M{R}^2}{32}+\frac{M}{4}{\left(\frac{R}{2}\right)}^2\\ & =\frac{M{R}^2}{32}+\frac{M{R}^2}{16}=\frac{3M{R}^2}{32}\end{array}$$ M.I. of complete disc can also be written as $I_{\text{Total }}=I_{\text{removed hole }}+I_{\text{remaining disc }}$ $I_{\text{Total }}=\frac{3M{R}^2}{32}+I_{\text{remaining disc }}$ From eq. (i) and (ii), $$\begin{array}{rl}& \frac{1}{2}M{R}^2=\frac{3M{R}^2}{32}+I_{\text{remaining disc }}\\ & \Rightarrow I_{\text{remaining disc }}\\ & =\frac{M{R}^2}{2}-\frac{3M{R}^2}{32}=\left(\frac{13}{32}\right)M{R}^2\end{array}$$