If a body completes one revolution in pi s,t | vedclass

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Question

(C) The rotational kinetic energy $(K)$ of a body is given by the formula: $K = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega

Vedclass · Accepted Answer

Half of rotational kinetic energy

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(C) The rotational kinetic energy $(K)$ of a body is given by the formula: $K = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.Given that the body completes one revolution in $T = \pi \ s$,the angular velocity is $\omega = \frac{2\pi}{T} = \frac{2\pi}{\pi} = 2 \ rad/s$.Substituting the value of $\omega$ into the kinetic energy formula:$K = \frac{1}{2} I (2)^2 = \frac{1}{2} I (4) = 2I$.Therefore,$I = \frac{K}{2}$,which means the moment of inertia is half of the rotational kinetic energy.

If a body completes one revolution in $\pi \ s$,then the moment of inertia is related to the rotational kinetic energy $(K)$ by which of the following expressions?

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