An experimental bicycle wheel is placed on a test stand so that it is free to turn on its axle. If a constant net torque of 6.00 N * m is applied to the tire for 2.00 s, the angular speed of the tire increases from zero to 120 rev/min. The external torque is then removed, and the wheel is brought to rest in 145 s by friction in its bearings.

Compute the moment of inertia of the wheel about the axis of rotation.
I = ? Units= kg * m^2

Compute the friction torque.
Units = N * m

Compute the total number of revolutions made by the wheel in the 145 s time interval.
N= ?

Explanation:

Given:

Torque applied on the wheel,

time for which the torque is applied,

initial angular speed,

final angular speed,

time after which the wheel comes to rest due to friction,

As most mass of the tyre is on its periphery so we consider it as a ring:

Moment of inertia of the ring is given as,

....................(1)

where:

angular acceleration

given as:

Now

The torque when the wheel comes to rest form the final velocity to zero:

The angular acceleration during the deceleration will be:

(negative sign denotes deceleration)

From eq. (1)

Now the no. of revolutions made in 145 seconds:

Using equation of motion:

(keep values in revolutions)

           (∵ )

i'm thinking more like energy

d is the correct answer