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

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(A) The number of revolutions per minute is given as $N = \frac{3000}{\pi} \ rpm$.The frequency $n$ in Hertz is $n = \frac{N}{60} = \frac{3000}{60 	i

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$2 	imes 10^6 \ J$

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(A) The number of revolutions per minute is given as $N = \frac{3000}{\pi} \ rpm$.The frequency $n$ in Hertz is $n = \frac{N}{60} = \frac{3000}{60 	imes \pi} = \frac{50}{\pi} \ Hz$.The angular velocity $\omega$ is given by $\omega = 2 \pi n = 2 \pi 	imes \frac{50}{\pi} = 100 \ rad/s$.The moment of inertia $I$ is $400 \ kg \ m^2$.The rotational kinetic energy $K$ is calculated using the formula $K = \frac{1}{2} I \omega^2$.Substituting the values: $K = \frac{1}{2} 	imes 400 	imes (100)^2 = 200 	imes 10000 = 2 	imes 10^6 \ J$.

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

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