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

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$20$

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(A) The rotational kinetic energy of a body is given by $K = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.For the kinetic energy to be the same as the initial value,the magnitude of the angular velocity must be the same as the initial angular velocity,i.e.,$|\omega| = |\omega_0| = 20 \ rad/s$.Since the torque provides a constant deceleration,the angular velocity at time $t$ is given by $\omega = \omega_0 - \alpha t$.We want the final angular velocity to be $\omega = -20 \ rad/s$ (since it is decelerating and we need the same magnitude of kinetic energy).Substituting the values: $-20 = 20 - 2t$.Rearranging the equation: $2t = 40$.Therefore,$t = 20 \ s$.

$A$ body rotating at $20 \ rad/s$ is acted upon by a constant torque providing it a deceleration of $2 \ rad/s^2$. At what time will the body have kinetic energy same as the initial value if the torque continues to act?

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