Q6.9 Slowing flywheel (S)

A flywheel of mass 10 kg and radius 0.3 m is set spinning at a rate of 10 revolutions per second. After 2 minutes it has slowed down to 5 revolutions per second. Assuming that the slowing down is caused by a constant frictional torque at the bearing, calculate the magnitude of this torque.

[The moment of inertia of a disc about its axis is I=Mr2/2 where M is the mass of the disc and r its radius.]

Hide

Hint

The constant torque provides a constant angular acceleration (Key Point 5.11).
 

Solution

Reveal
Hide

Solution

If the frictional torque at the bearing is τ, then the angular acceleration is parallel to the torque (Key Point 5.19) and

τ=Iα
where $I=0.45\,{\rm kg\,m^{2}}$ is the moment of inertia of the flywheel about the axis of rotation, and α is its angular acceleration.

Since the torque is constant, the angular acceleration is also constant, and given by the first constant acceleration equation

ω=ω0+αt
giving α=-0.26 rad s-2.

Thus

τ=Iα=-0.12 Nm.
(Note : the minus signs are a reminder that the torque and the angular acceleration it causes are in the opposite sense to the angular motion.)