Physics - Rotation Question with Solution | TestHub

Given, $\begin{array}{rl}& {\omega }_1=60\mathrm{rpm}=60\times \frac{2\pi }{60}=2\pi \mathrm{rad}/\mathrm{s}\\ & {\omega }_2=360\mathrm{rpm}=360\times \frac{2\pi }{60}=12\pi \mathrm{rad}/\mathrm{s}\end{array}$ Change in kinetic energy, $\begin{array}{rl}\Delta \mathrm{K}.\mathrm{E}& =\frac{1}{2}\mathrm{I}\left({\omega }_2^2-{\omega }_1^2\right)\\ 484& =\frac{1}{2}\times \mathrm{I}\left({\omega }_2+{\omega }_1\right)\left({\omega }_2-{\omega }_1\right)\\ \Rightarrow \mathrm{I}& =\frac{484\times 2}{14\pi \times 10\pi }=0.7\mathrm{kg}\cdot {\mathrm{m}}^2\end{array}$ (where, the symbols have their usual meanings)