Question:
To maintain a rotor at a uniform angular speed of $200 \mathrm{rad} \mathrm{s}^{-1}$, an engine needs to transmit a torque of $180 \mathrm{Nm}$. What is the power required by the engine?
(Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100 % efficient.
Solution:
Angular speed of the rotor, $\omega=200 \mathrm{rad} / \mathrm{s}$
Torque required, $\mathrm{T}=180 \mathrm{Nm}$
The power of the rotor $(P)$ is related to torque and angular speed by the relation:
$P=\mathrm{T} \omega$
$=180 \times 200=36 \times 10^{3}$
= 36 kW
Hence, the power required by the engine is 36 kW.