A flywheel of moment of inertia I is rotating | vedclass

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(C) Let the initial frequency be $n$.The initial angular velocity is $\omega_i = 2\pi n$.The initial rotational kinetic energy is $K_i = \frac{1}{2} I

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$6{\pi ^2}I{n^2}$

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(C) Let the initial frequency be $n$.The initial angular velocity is $\omega_i = 2\pi n$.The initial rotational kinetic energy is $K_i = \frac{1}{2} I \omega_i^2 = \frac{1}{2} I (2\pi n)^2 = 2 I \pi^2 n^2$.When the frequency is doubled,the new frequency is $n' = 2n$.The new angular velocity is $\omega_f = 2\pi (2n) = 4\pi n$.The final rotational kinetic energy is $K_f = \frac{1}{2} I \omega_f^2 = \frac{1}{2} I (4\pi n)^2 = 8 I \pi^2 n^2$.According to the work-energy theorem,the work done $W$ is equal to the change in kinetic energy:$W = K_f - K_i = 8 I \pi^2 n^2 - 2 I \pi^2 n^2 = 6 I \pi^2 n^2$.

$A$ flywheel of moment of inertia $I$ is rotating at $n$ revolutions per second. The work needed to double the frequency would be

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