Physics - Rotation Question with Solution | TestHub

Given, $\mathrm{m}=1\mathrm{kg}$
$\begin{array}{rl}& {\omega }_{\mathrm{i}}=1800\mathrm{rpm}=1800\times \frac{2\pi }{60}=60\pi \frac{\mathrm{rad}}{\sec }\\ & {\omega }_{\mathrm{f}}=2100\mathrm{rpm}=2100\times \frac{2\pi }{60}=70\pi \frac{\mathrm{rad}}{\sec }\\ & {\tau }_{ext}=25\pi \mathrm{Nm}\\ & \mathrm{t}=40\sec \end{array}$
Using equation of motion
$\begin{array}{rl}& {\omega }_{\mathrm{f}}={\omega }_{\mathrm{i}}+\alpha \mathrm{t}\\ & 70\pi =60\pi +\alpha (40)\\ & \alpha =\frac{\pi }{4}\mathrm{rad}/{\sec }^2\end{array}$
Also, $\tau =\mathrm{I}\alpha$
$\tau =\frac{{\mathrm{mR}}^2}{4}\alpha$
$\begin{array}{rl}& 25\pi =\frac{1\times {\mathrm{R}}^2}{4}\times \frac{\pi }{4}\\ & \mathrm{R}=20\mathrm{m}\end{array}$
Hence, diameter of disk $=2\mathrm{R}=2\times 20=40\mathrm{m}$