Work and Rotational Kinetic Energy — Question 7

Question 7

Constant power makes squared angular speed grow linearly

A rotor of inertia $0.50\,\mathrm{kg\,m^2}$ begins at positive $\omega_0=4\,\mathrm{rad/s}$. It receives constant net mechanical power $8\,\mathrm W$ for $3\,\mathrm s$.

1. Find its final angular speed and the net torque at the beginning and end.

2. Explain why an ideal constant-power law cannot start it from exact rest with bounded torque.

Question 7 - Solution

Constant power makes squared angular speed grow linearly

(a) Derivation and calculation

Integrate constant net power as an energy rate: $$\frac{dK}{dt}=P\Rightarrow\frac12I(\omega^2-\omega_0^2)=Pt.$$ $$\omega^2=16+\frac{2(8)}{0.50}t=16+32t,\qquad \boxed{\omega_f=\sqrt{112}=10.583\ \mathrm{rad/s}}.$$ $$\boxed{\tau_i=P/\omega_0=2.0\ \mathrm{N\,m}},\qquad \boxed{\tau_f=8/\sqrt{112}=0.756\ \mathrm{N\,m}}.$$

(b) Result and interpretation

A finite positive constant power requires $$\tau=\frac P\omega\longrightarrow\infty\quad\text{as}\quad\omega\to0^+.$$ Equivalently, bounded torque gives $P=\tau\omega\to0$ at rest. A bounded-torque start therefore requires a different control regime before constant-power operation.

Physical check

The 24 J energy input raises $K$ from 4 to 28 J, consistent with the final squared speed.