The angular momentum of a flywheel having a rotational inertia of 0.132 kgm^2 ab

Question

The angular momentum of a flywheel having a rotational inertia of 0.132 kgm^2 about its central axis decreases from 6.40 to 0.560 kgm^2/s in 1.70 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? (a) Number Units N-m (b) Number Units (c) Number Units j (d) Number Units w

Explanation / Answer

(a)
torque = rate of change of angular momentum
= (.56 -6.4)/1.7
= -3.43 Nm

(b)
angular velocity = angular momentum / rotational inertia
initial angular velocity = 6.40/0.132 = 48.48 rad/s
final angular velocity = 0.56/0.132 = 4.24 rad/s
angular displacement = (average angular velocity)(time)
= (1/2)(48.48 + 4.24)(1.70)
= 44.812 rad

(c)
work = (torque)(angular displacement)
= (3.43)(44.812)
= 153.70 J

(d)
power = work/time
= 153.70/1.7
= 90.41 W

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