A flywheel is making frac{3000}{pi} revo | vedclass

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Question

Revolutions per minute $(\mathrm{N})=\frac{3000}{\pi} \mathrm{rpm}$

so, frequency, $n=\frac{N}{60}=\frac{3000}{60 	imes \pi}=\frac{50}{\pi} \mathr

Vedclass · Accepted Answer

$2 	imes 10^6\, J$

Vedclass · Answer

Revolutions per minute $(\mathrm{N})=\frac{3000}{\pi} \mathrm{rpm}$

so, frequency, $n=\frac{N}{60}=\frac{3000}{60 	imes \pi}=\frac{50}{\pi} \mathrm{Hz}$

and $\quad \omega=2 \pi n=2 \pi 	imes \frac{50}{\pi}=100 \mathrm{rad} / \mathrm{sec}$

$\mathrm{I}=400 \mathrm{kg} \mathrm{m}^{2}$

so rotational kinetic energy $(K)$ is

$\mathrm{K}=\frac{1}{2} \mathrm{I} \omega^{2}=\frac{1}{2} 	imes 400 	imes(100)^{2}=2 	imes 10^{6} \mathrm{Joule}$

A flywheel is making $\frac{3000}{\pi}$ revolutions per minute about its axis. If the moment of inertia of the flywheel about that axis is $400\, kgm^2$, its rotational kinetic energy is

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