Question

A student holds a bike wheel and starts it spinning with an initial angular speed of 7.0 rotations per second. The wheel is subject to some friction, so it gradually slows down. In the 10-s period following the inital spin, the bike wheel undergoes 60.0 complete rotations.

Assuming the frictional torque remains constant, how much more time ?ts will it take the bike wheel to come to a complete stop?

The bike wheel has a mass of 0.725 kg and a radius of 0.315 m. If all the mass of the wheel is assumed to be located on the rim, find the magnitude of the frictional torque ?f that was acting on the spinning wheel.

Answer 1

If torque is constant, the angular deceleration is constant.

Average angular speed over 10s = 60 rev / 10 s = 6 rev/s.

It starts at 7 rev/s, so after 10s the speed is 5 rev/s (making the average 6 rev/s).

The wheel loses 2 rev/s every 10s, or 1 rev/s every 5 s.

So after slowing to 5 rev/s the wheel takes a further 5 x 5 = 25 s to come to a complete stop.

The angular deceleration is 1 rev/s every 5 s, ie 0.2 rev/s per second.

In radians this is 2pi x 0.2 = 1.2566 rad/s2.

The moment of inertia of the wheel is 0.725kg x (0.315m)^2 = 0.071938 kg m2.

So the frictional torque is

T = MI x angular deceleration

= 0.071938 kg m2 x 1.2566 rad/s2

= 0.090397 N-m