Question

A bicycle wheel, of radius 0.3100 m and mass 2.000 kg (concentrated on the rim), is rotating at 4.110 rev/s. After 41.00 s the wheel comes to a stop because of friction. What is the magnitude of the average torque due to frictional forces? N · m

Answer 1

Solution-

Here T is torque
T = I*a

I = moment of inertia of the object

a= angular acceleration

for the wheel I = MR^2 so for the given wheel,

I = 2.0 kg * 0.31m^2 =0.62kgm^2

Now
angular acceleration = change in angular velocity/time

change in angular velocity = final ang vel - initial ang vel
= 0 revs - 4.11rev/s = -4.11revs/s

In radians,

since there are 2 pi rad/rev, we have that change in angular vel = - 25.8 rad/s

therefore, ang accel = -25.8rad/s / 41 s = -0.63 rad/s/s

=>.torque = 0.62 kgm^2 x (-0.63rad/s/s)
torque = --0.39 Nm (here the minus sign shows that the torque causes the wheel to slow)