To maintain a rotor at a uniform angular speed of 200 rad s–1,

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.

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