Answer to Question #164634 in Physics for Treasure Enyinnaya

Question

A flywheel of mass 400 kg and radius of gyration 1 m losses its speed from 300 r.p.m. to 240 r.p.m. m 120 seconds. Determine the retardmg torque acting on it.

(Ans, 20.8 N-m).

Answer 1

Let's first find the angular acceleration of the flywheel:

$\alpha=\dfrac{\omega_f-\omega_i}{t},$ $\alpha=\dfrac{240\ \dfrac{rev}{min}\cdot\dfrac{1\ min}{60\ s}\cdot2\pi\ rad-300\ \dfrac{rev}{min}\cdot\dfrac{1\ min}{60\ s}\cdot2\pi\ rad}{120\ s},$ $\alpha=-0.052\ \dfrac{rad}{s^2}.$

The sign minus means that the flywheel is decelerates.

Next, let's calculate the moment of inertia of the flywheel:

$I=mk^2=400\ kg\cdot(1\ m)^2=400\ kg\cdot m^2.$

Finally, we can find the retarding torque acting on the flywheel from the formula:

$\tau=I\alpha=400\ kg\cdot m^2\cdot(-0.052\ \dfrac{rad}{s^2})=-20.8\ N\cdot m.$

The magnitude of the retarding torque is $20.8\ N\cdot m$. The sign minus means that the torque acts in the opposite direction to the motion of the flywheel.