Physics - Circular Motion Question with Solution | TestHub

For uniform angular acceleration we can write,

$$\begin{gathered} \omega ={\omega }_0+\alpha t \\ \Rightarrow \alpha =\frac{\omega -{\omega }_0}{t} \end{gathered}$$

Given, $\omega =3120 \mathrm{rpm}=\frac{3120\times 2\mathrm{\pi }}{60}=104\mathrm{\pi } \mathrm{rad} {\mathrm{s}}^{-1}$

and ${\omega }_0=1200 \mathrm{rpm}=\frac{1200\times 2\mathrm{\pi }}{60}=40\mathrm{\pi } \mathrm{rad} {\mathrm{s}}^{-1}$

Therefore, $\alpha =\frac{104\mathrm{\pi } \mathrm{rad} {\mathrm{s}}^{-1}-40\mathrm{\pi } \mathrm{rad} {\mathrm{s}}^{-1}}{16}=4\pi \mathrm{rad} {\mathrm{s}}^{-2}$