Q.A ceiling fan having moment of inertia 2 kg-m² attains its maximum frequency of 60 rpm in '2π' seconds. Calculate its power rating.

Uniform spin-up from rest to 60 rpm ($n=1$ rps, $\omega=2\pi$ rad/s) in $2\pi$ s gives $\alpha=1$ rad/s², and power $P=\tau\omega=I\alpha\cdot\omega$. …

The fan starts from rest ($\omega_0=0$) and attains its maximum frequency of 60 rpm, i.e. $n=1$ rev/s, so $\omega=2\pi n=2\pi$ rad/s, in $t=2\pi$ s. The uniform angular acceleration is $\alpha=\frac{\omega-\omega_0}{t}=\frac{2\pi-0}{2\pi}=1\ \text{rad/s}^2$ so the driving torque is $\tau=I\alpha=2\times1=2$ N m, and the power delivered at the full (maximum) angular speed — the rated power — is $P=\tau\cdot\omega=I\alpha\cdot\omega=2\times1\times2\pi=4\pi\ \text{W}\approx12.6\ \text{W}$ …

Energy route: the fan's final kinetic energy is $\frac{1}{2}I\omega^2 = \frac{1}{2}(2)(2\pi)^2 = 4\pi^2$ J, delivered over $t=2\pi$ s → average power $= 4\pi^2/2\pi = 2\pi$ W; the rated power at full speed, $\tau\o …

Reading 60 rpm as 2 rev/s (it is 1 rev/s) — the book's own printed Solution makes exactly this slip, which is why its printed 50 W corresponds to a 120 rpm fan; als …