A fly wheel of moment of inertia I is rotatin | vedclass

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Question

Lets consider,

$n =$ initial frequency

$\omega=2 \pi n$

Initial kinetic energy is

$K _{ i }=\frac{1}{2} I \omega^2=\frac{1}{2} I 	imes 2

Vedclass · Accepted Answer

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

Vedclass · Answer

Lets consider,

$n =$ initial frequency

$\omega=2 \pi n$

Initial kinetic energy is

$K _{ i }=\frac{1}{2} I \omega^2=\frac{1}{2} I 	imes 2 \pi n 	imes 2 \pi n$

$K _{ i }=2 I \pi^2 n ^2$

When frequency is double to the initial frequency, the the kinetic energy will be

$K _{ f }=\frac{1}{2} I \omega^2=\frac{1}{2} I 	imes 4 \pi n 	imes 4 \pi n$

$K _{ f }=8 I \pi^2 n ^2$

By work-energy theorem,

$W = K _{ f }- K _{ i }$

$W =8 I \pi^2 n ^2-2 I \pi^2 n ^2=6 I \pi^2 n ^2$

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

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