The angular speed of a motor wheel is increased from $1200 \; rpm$ to $3120 \; rpm$ in $16 \; s$.
$(i)$ What is its angular acceleration,assuming the acceleration to be uniform?
$(ii)$ How many revolutions does the engine make during this time?

(N/A) $(i)$ We use the equation $\omega = \omega_{0} + \alpha t$.
Initial angular speed $\omega_{0} = \frac{2 \pi \times 1200}{60} \; rad/s = 40 \pi \; rad/s$.
Final angular speed $\omega = \frac{2 \pi \times 3120}{60} \; rad/s = 104 \pi \; rad/s$.
Angular acceleration $\alpha = \frac{\omega - \omega_{0}}{t} = \frac{104 \pi - 40 \pi}{16} = \frac{64 \pi}{16} = 4 \pi \; rad/s^{2}$.
$(ii)$ The angular displacement $\theta$ is given by $\theta = \omega_{0} t + \frac{1}{2} \alpha t^{2}$.
$\theta = (40 \pi \times 16) + \frac{1}{2} \times (4 \pi) \times (16)^{2} = 640 \pi + 512 \pi = 1152 \pi \; rad$.
Number of revolutions $n = \frac{\theta}{2 \pi} = \frac{1152 \pi}{2 \pi} = 576$.