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(B) 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 $\omeg

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(B) 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: $K = 500 	ext{ J}$ and frequency $f = 5 	ext{ Hz}$.The angular velocity $\omega$ is related to frequency $f$ by the formula: $\omega = 2 \pi f$.Substituting the value of $f$: $\omega = 2 	imes \pi 	imes 5 = 10 \pi 	ext{ rad/s}$.Now, substitute the values into the kinetic energy formula: $500 = \frac{1}{2} 	imes I 	imes (10 \pi)^2$.$500 = \frac{1}{2} 	imes I 	imes 100 \pi^2$.$500 = 50 	imes I 	imes \pi^2$.Given $\pi^2 = 10$, substitute this value: $500 = 50 	imes I 	imes 10$.$500 = 500 	imes I$.Therefore, $I = 1 	ext{ kg m}^2$.

$A$ flywheel rotating about a fixed axis has a kinetic energy of $500 \text{ J}$. If its angular frequency is $5 \text{ Hz}$, calculate the moment of inertia of the wheel about the axis of rotation. (Take $\pi^2 = 10$) (in $\text{ kg m}^2$)

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