Question:9: The angular speed of a fly wheel moving with uniform angular acceleration changes from 1200 rpm to 3120 rpm in 16 seconds. The angular acceleration in \mathrm{rad} / \mathrm{s}^{2} is
(1) 12 \pi
(2) \quad 104 \pi
(3) 2 \pi
(4) \quad 4 \pi
Answer: Option (4)
Explanation:
The initial angular speed is given as 1200 rpm. Converting it into \mathrm{rad\,s^{-1}},
\omega_1 = 1200 \times \frac{2\pi}{60} = 40\pi \ \mathrm{rad\,s^{-1}}.
The final angular speed is 3120 rpm. Converting it into \mathrm{rad\,s^{-1}},
\omega_2 = 3120 \times \frac{2\pi}{60} = 104\pi \ \mathrm{rad\,s^{-1}}.
The time interval is t = 16 \ \mathrm{s}.
For uniform angular acceleration, the relation between angular speeds is
\alpha = \frac{\omega_2 - \omega_1}{t}.
Substituting the values,
\alpha = \frac{104\pi - 40\pi}{16} = \frac{64\pi}{16} = 4\pi \ \mathrm{rad\,s^{-2}}.
Hence, the angular acceleration of the fly wheel is 4\pi \ \mathrm{rad\,s^{-2}}.