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.]
Hint
Reveal
Solution
Reveal
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is the moment of inertia of the
flywheel about the axis of rotation, and α is its angular
acceleration.

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 
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.)