Work and Rotational Kinetic Energy — Question 6

Question 6

Shaft power and electrical input are different quantities

A motor delivers $120\,\mathrm W$ of mechanical shaft power at steady angular speed $20\,\mathrm{rad/s}$. Its electrical-to-mechanical efficiency is $0.80$. The only mechanical load is a resisting torque.

1. Find shaft torque, electrical input power and motor loss rate.

2. If the same mechanical power is delivered at $10\,\mathrm{rad/s}$, what torque is required? Does the original steady load automatically permit that operating point?

Question 6 - Solution

Shaft power and electrical input are different quantities

(a) Derivation and calculation

At steady speed, shaft power is torque times angular velocity: $$\boxed{\tau=\frac{P_{\mathrm{shaft}}}\omega=\frac{120}{20}=6.0\ \mathrm{N\,m}}.$$ $$\boxed{P_{\mathrm{elec}}=\frac{P_{\mathrm{shaft}}}\eta=\frac{120}{0.80}=150\ \mathrm W},\qquad \boxed{P_{\mathrm{loss}}=150-120=30\ \mathrm W}.$$

(b) Result and interpretation

At the lower rate with the same mechanical power, $$\boxed{\tau_{10}=\frac{120}{10}=12\ \mathrm{N\,m}}.$$ Steady operation requires $\tau_{\mathrm{load}}=-12$ N m. Retaining only a $-6$ N m load leaves $\tau_{\mathrm{net}}=+6$ N m and causes acceleration.

Physical check

Power and speed determine the delivered torque, but a matching load is also needed for equilibrium.