Question

A student holds a bike wheel and starts it spinning with an initial angular speed of 9.0 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 65.0 complete rotations. Assuming the frictional torque remains constant, how much more time delta 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 τf that was acting on the spinning wheel.

Answer 1

Answer:

T = 0.00889 N*m

Explanation:

Given the initial speed

Vo = 9.0rev/s.

V = 65 rev/10s

V = 6.5 rev/s.

V = Vo + a*t

Solve to acceleration knowing the initial velocity and the velocity at 10 s

6.5 rev/s - 9 rev/s = a*10s

a = -0.25 rev/s^2.

Now the solve the time at stop time so V=0

V = Vo + a*t

0 = 9.0 - 0.25 rev/s *t,

t = 36 s The Stopping time.

36s -  10s = 26s

The torque can be find using the acceleration using the equation

T = I*a

I = 1/2*m*r^2

I = 1/2*0.725kg*(0.315m)^2= 0.0359kg*m^2

T = 0.0359kg*m^2*-0.25rev/s^2

T = 0.00889 N*m