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Question

A bike tire with a radius of 0.30 m and a mass of 1.1 kg is rotating at 93.6 rad/s.

What torque is necessary to stop the tire in 1.7 s (in units of N . m)?

Answer 1

Assuming the mass of the tire to be concentrated at the given radius and taking inertia of the tire as m*r*r = 0.099 kg m2.
Angulare momentum of the tire is (I * w) which comes out to be 9.2664. Torque for stopping the tire = 9.2664 / 1.7 = 5.45 Nm.

Answer 2

You probably can assume that the bike tire's mass is distributed so that the rotational inertia of a hoop,
I = M*R^2, is close enough.

The kinetic energy is given by (1/2)*I*w^2. The torque will have to do that amount of work to stop it and
rotational work = torque*angular displacement in radians.

And you can get the angular displacement from realizing that since the deceleration will be linear you can use (1/2)*93.6 rad/s as the average angular velocity during the 1.7 s.

Answer 3

The angular momentum will be the rotation speed (given) times the moment of inertia (calculable). Assume all the mass is in the tire to calculate it. Divide the angular momentum by the stopping time to get the required torque.

Answer 4

v = angular velocity
a = angular acceleration
t = time
T = tourque
I = moment of inertia
w = intial angular velocity
M = mass of tire
r = radius of tire

v = w + at so a=w/t
T = Ia
I = 1/2Mr^2 assumes tire is uniform disk. You may need to use a different moment.

so:

T = 1/2Mr^2(w/t) plug and chug!