Past exam of the physics 1 paper of the University of Oxford / 2016 resit / 4 / Solution

Let the motor's constant power be $P>0$ and put $\omega =\dot{\theta }$. The braking coefficient in the proposed formula must describe a retarding torque $k\omega$. This constant-power rotational acceleration obeys the energy balance:Consequently the steady angular velocity is ${\omega }_0=\sqrt{P/k}$, and the ordinary rotational frequency is ${\omega }_0/(2\pi )$. Solving the linear equation for ${\omega }^2$ from rest givesThe printed time formula is missing a minus sign. Its logarithm is negative for $0<\omega <{\omega }_0$, whereas elapsed time must be positive. The terminal speed is approached only as $t\to \infty$.

A constant-power motor starting exactly from rest is an idealization: its driving torque $P/\omega$ would diverge. The energy equation has a regular solution for ${\omega }^2$, although the initial angular acceleration is singular. If $k\omega$ were literally a tangential force, a lever arm would also be needed; the dimensions of the requested expression identify $k$ as a rotational damping coefficient.