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 Δts will it take the bike wheel to come to a complete stop?

The bike wheel has a mass of 0.825 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 τf that was acting on the spinning wheel.

Answer 1

Vo = 7.0rev/s.
V = 60rev/10s = 6rev/s.

V = Vo + a*t.
6 - 7 + a*10, a = -0.1 rev/s^2.

V = Vo + a*t.
0 = 7 - 0.1*t, t = 70 s. = Stopping time.
70 - 6 = 64 additional seconds to stop.

Answer 2

Correction: 70-10 = 60 Addition seconds to stop.