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?

Expert's answer

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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