Physics - Rotation Question with Solution | TestHub

Moment of inertia of disc is given by ${\mathrm{I}}_{\text{disc }}={\mathrm{I}}_{\mathrm{f}}+{\mathrm{I}}_{\text{hole }}$ .... $\{{\mathrm{I}}_{\mathrm{f}}=$ M.I. of remaining part $\}$ $\begin{array}{rl}& \therefore {\mathrm{I}}_{\mathrm{r}}={\mathrm{I}}_{\text{dise }}-{\mathrm{I}}_{\text{hole }}\\ & {\mathrm{I}}_{\text{disc }}=\frac{{\mathrm{MR}}^2}{2}\end{array}$ By parallel axes theorem we get, ${\mathrm{I}}_{\text{hole }}=\left[\frac{\frac{\mathrm{M}}{4}{\left(\frac{\mathrm{R}}{2}\right)}^2}{2}+\frac{\mathrm{M}}{4}{\left(\frac{\mathrm{R}}{2}\right)}^2\right]$ $\dots \left\{\begin{array}{l}\because {\mathrm{M}}_{\text{bole }}=\frac{{\mathrm{M}}_{\text{dive }}}{4}\\ \because \text{ the surface density is same }\end{array}\right\}$ $\therefore {\mathrm{I}}_{\text{hole }}=\left[\frac{{\mathrm{MR}}^2}{32}+\frac{{\mathrm{MR}}^2}{16}\right]$ Substituting eq (iii) and eq (ii) in eq (i) we get, $\begin{array}{rl}{\mathrm{I}}_{\mathrm{r}}& =\frac{{\mathrm{MR}}^2}{2}-\frac{{\mathrm{MR}}^2}{32}-\frac{{\mathrm{MR}}^2}{16}\\ & ={\mathrm{MR}}^2\left[\frac{1}{2}-\frac{1}{32}-\frac{1}{16}\right]\\ & =\frac{13}{32}{\mathrm{MR}}^2\end{array}$