Question #144459
The angular momentum of a flywheel having a rotational inertia of 0.125 kgm2decreasesfrom 3.0 to 2.0 kgm2/s in a period of 1.5 s. (a) What is the average torque acting on the flywheelduring this period? (b) Assuming a uniform angular acceleration, through how many revolutions will the flywheel have turned? (c) How much work was done on the wheel? (d) What was theaverage power supplied by the flywheel?
1
Expert's answer
2020-11-17T07:10:04-0500

Tavg=LfLiδtT_{avg}=\frac{L_f-L_i}{\delta t}


a) 3.02.01.5=1.15N.m\frac {3.0-2.0}{1.5}= 1.15N.m


b) θ=(3)(1.5s)+(1.467)(1.5)220.125\theta= \frac{(3)(1.5s)+(-1.467) \frac{(1.5)^{2}}{2}}{0.125}


θ=22.8rad\theta=22.8 rad


c)W=τθ=(1.47)(22.8)W=\tau\theta= (-1.47)(22.8)


W=33.51JW=-33.51J


Pavg=33.511.50=22.3WP_{avg}=\frac {-33.51}{1.50}= 22.3W


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