The angular speed of a motor wheel is increased from \(1200 \mathrm{rpm}\) to \(3120 \mathrm{rpm}\) in \(16 \mathrm{~s}\). The angular acceleration of the motor wheel is

Given, initial angular frequency of wheel,
\(f_0=1200 \mathrm{rpm}=\frac{1200}{60} \mathrm{rps}=20 \mathrm{rps}\)
\(\therefore\) Initial angular velocity,
\(\omega_0=2 \pi f_0=2 \pi \times 20=40 \pi \mathrm{rad} / \mathrm{s}\)
Similarly, final angular speed,
\(\omega=2 \pi \times\left(\frac{3120}{60}\right) \mathrm{rad} / \mathrm{s}=104 \pi \mathrm{rad} / \mathrm{s}\)
Time, \(\quad t=16 \mathrm{~s}\)
If \(\alpha\) be the angular acceleration of the wheel, then
\(\omega =\omega_0+\alpha t \)
\(\Rightarrow \alpha =\frac{\omega-\omega_0}{t}=\frac{104 \pi-40 \pi}{16} \)
\(=\frac{64 \pi}{16}=4 \pi \mathrm{rad} / \mathrm{s}^2\)