Q.7 A ceiling fan has moment of inertia of 2 kg m\(^2\). It attains maximum frequency of 60 r.p.m. in \(2\pi\) seconds. Calculate its power rating.
Best answer
Answer: \(P = 4\pi \approx 12.57\) W
Given: \(I = 2\) kg m\(^2\), \(n = 60\) r.p.m. \(= 1\) r.p.s., \(t = 2\pi\) s, starting from rest.
Maximum angular speed: \(\omega = 2\pi n = 2\pi\) rad/s.
Angular acceleration: \(\alpha = \dfrac{\omega - 0}{t} = \dfrac{2\pi}{2\pi} = 1\) rad/s\(^2\).
Torque: \(\tau = I\alpha = 2 \times 1 = 2\) N m.
Power rating (power delivered at the working speed):
\[P = \tau\,\omega = 2 \times 2\pi = 4\pi = 4(3.142) \approx 12.57\ \text{W}\](Note: the average power while the fan speeds up, \(\tfrac{1}{2}I\omega^2/t = \dfrac{\tfrac12\times2\times(2\pi)^2}{2\pi} = 2\pi \approx 6.28\) W, is half of this. The question asks for the power rating, which the textbook takes as \(P = \tau\omega\) at full speed.)
Log in or create a free account to answer this question.
