The angular momentum of a flywheel having a rotational inert

The angular momentum of a flywheel having a rotational inertia of 0.902 kg·m2 about its central axis decreases from 4.90 to 0.710 kg·m2/s in 3.70 s. (a) What is the magnitude of the average torque acting on the flywheel about its central axis during this period? (b) Assuming a constant angular acceleration, through what angle does the flywheel turn? (c) How much work is done on the wheel? (d) What is the magnitude of the average power done on the flywheel?

Solution

Here,

moment of inertia , I1 = 0.902 Kg.m^2

angular momentum = 0.710 Kg.m^2

time , t = 3.70 s

a)

magnitude of average torque acting = (4.90 - 0.710)/3.70 N.s

magnitude of average torque acting = 1.13 N.m

b)

intial angular speed , wi = 4.90/0.902 rad/s

wi = 5.43 rad/s

final angular speed, wf = 0.710/.902 = 0.787 rad/s

angular acceleration ,a = 1.13/.902 = 1.25 rad/s^2

let the angle rotated is theta

Using third equation of motion

wf^2 - wi^2 = 2 * a * theta

1.25^2 - 5.43^2 = 2 * 1.25 * theta

theta = 11.17 rad

the angle rotated by the flywheel is 11.17 rad

c)

work done on the wheel = theta * torque

work done on the wheel = 11.17 * 1.13 * (-1)

work done on the wheel = -12.62 J

d)

magnitude of average power = -12.62/3.70 W

magnitude of average power = -3.41 W