Question

If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.69 rev/s , friction in the bearings causes the wheel to stop in just 13 s . If the moment of inertia of the wheel about its axle is 0.33 kg⋅m2 , what is the magnitude of the frictional torque?

Answer 1

Answer:

magnitude of the frictional torque is 0.11 Nm

Explanation:

Moment of inertia I = 0.33 kg⋅m2

Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s

Final angular velocity w = 0 (since it stops)

Time t = 13 secs

Using w = w° + §t

Where § is angular acceleration

O = 4.34 + 13§

§ = -4.34/13 = -0.33 rad/s2

The negative sign implies it's a negative acceleration.

Frictional torque that brought it to rest must be equal to the original torque.

Torqu = I x §

T = 0.33 x 0.33 = 0.11 Nm