Physics - Rotation Question with Solution | TestHub

From parallel axis theorem, ${\mathrm{I}}_{\mathrm{o}}={\mathrm{I}}_{\mathrm{c}}+{\mathrm{Mh}}^2$ Let the moment of inertia of sphere 1 be ${\mathrm{I}}_1=\frac{2}{5}\mathrm{M}{\left(\frac{\mathrm{R}}{2}\right)}^2+\mathrm{M}(2\mathrm{R}{)}^2$ and, Let the moment of inertia of sphere 2 be ${\mathrm{I}}_2=\frac{2}{5}\mathrm{M}{\left(\frac{\mathrm{R}}{2}\right)}^2$ Moment of inertia of the rod ${\mathrm{I}}_3=0$ $\therefore$ Moment of inertia of the system, $\begin{array}{rl}\mathrm{I}& ={\mathrm{I}}_1+{\mathrm{I}}_2+{\mathrm{I}}_3\\ \mathrm{I}& =\frac{2}{5}\mathrm{M}{\left(\frac{\mathrm{R}}{2}\right)}^2+\mathrm{M}(2\mathrm{R}{)}^2+\frac{2}{5}\mathrm{M}{\left(\frac{\mathrm{R}}{2}\right)}^2\\ & =\frac{4}{5}\mathrm{M}{\left(\frac{\mathrm{R}}{2}\right)}^2+4{\mathrm{MR}}^2\\ & =\frac{1}{5}{\mathrm{MR}}^2+4{\mathrm{MR}}^2\\ & =\frac{21}{5}{\mathrm{MR}}^2\end{array}$