Question

A flywheel is rotating at a rate of 60rev/min.A n amount of 98.7J of work is needed to increase the rate of rotation to 180rev/min.Find the moment of inertia of flywheel about its axis of rotation

Answer 1

Method: Work done is equal to change in rotational kinetic energy.

Formula:
Rotational kinetic energy, KE = $\frac{1}{2}I\omega^2$.

Angular speed, $\omega\:=\:2\pi\frac{{RPM}}{60}$.

Initial angular speed,$\omega_o\:=\:2\pi\frac{60}{60}\:=\:2\pi\:rads^{-1}$

Final angular speed,$\omega_o\:=\:2\pi\frac{180}{60}\:=\:6\pi\:rads^{-1}$.

work done = change in kinetic energy.

W = $\frac{1}{2}I\omega^2\:-\:\frac{1}{2}I\omega_o^2$.

W = $\frac{1}{2}I(\omega^2\:-\:\omega_o^2)$.
Then, moment of inertia, I = $\frac{2W}{(\omega^2\:-\:\omega_o^2)}$.
Substituting values, 

I = $\frac{2\times 98.7}{(6\pi)^2\:-\:(2\pi)^2}\:=\:\frac{2\times 98.7}{36\pi^2\:-\:4\pi^2}$.

or, I = $\frac{197.4}{32\pi^2}= 0.6250\:kgm^2$