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 7.00 N · m is applied to the tire for 2.00 s, the angular speed of the tire increases from 0 to 100 rev/min. The external torque is then removed, and the wheel is brought to rest by friction in its bearings in 125 s. Compute (a) the moment of inertia of the wheel about the rotation axis; (b) the friction torque; (c) the total number of revolutions made by the wheel in the 125-s time interval.

Solution 58 P Step 1: Data given Torque = 7.00 N Time t 1 2.00 s Angular velocity = 100 rev /min = 100 rev/60sec × ( 2 rad/1 rev) = 10.47 rad /sec This can be approximated to = 10.5 rad /sec Time to stop t = 125 s 2 We need to find the moment of inertia of the wheel about the rotation axis We shall the acceleration of the wheel It is given by = / t Substitute the values as = 10.5 rad /sec , t = 2.0 s We get = 10.5 rad /sec/ 2.0 s 2 = 5.25 rad/sec Step 2 : The moment of inertia is given by = I Rearrange to find I we have I = / Substitute the values as = 7.00 N = 5.25 rad/sec 2 We get I = 7.00 N/5.25 rad/sec 2 2 I = 1.33 Kg m 2 Hence we have the moment of inertia as 1.33 Kg m