The angular momentum of a flywheel havinga rotational inertia of 0.641 kgm2 abou

Question

The angular momentum of a flywheel havinga rotational inertia of 0.641 kgm2 about its centralases from 4.00 to 0.800 kg m2/s in 4.20 s. (a) What is the magnitude of the average torque acting on the lywheel 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 (b) Number (c) Number (d) Number Units Units Units Units

Explanation / Answer

(A) I = 0.641 kg m^2

average torque = rate of change of angular momentum

= (4 - 0.800) / 4.20

= 0.762 N m  

(B) torque = I alpha

0.762 = (0.641) alpha

alpha = 1.19 rad/s^2

wi = 4 / 0.641 = 6.24 rad/s

wf = 0.80 / 0.641 = 1.248 rad/s

wf^2 - wi^2 = 2 alpha theta

1.248^2 - 6.24^2 = 2(-1.19)(theta)

theta = 15.7 rad  

(C) Work done = change in KE

= I (wf^2 - wi^2) /2

= (0.641) (1.248^2 - 6.24^2) /2

= - 11.98 J OR - 12 J  

(D) Power = work done / time

P = 12 / 4.20

P = 2.85 W

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