Solve the following :


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


$\omega_{0}=20 \frac{\mathrm{rad}}{\sec } ; \alpha=-\frac{2 \mathrm{rad}}{\mathrm{s}^{2}} ; \omega=0$

$\omega=\omega_{0}+\alpha t_{1}$

Now, wheel accelerates and when angular velocity becomes $20 \mathrm{rad} / \mathrm{sec}$ it will have same KE.

$\omega=\omega_{0}+\alpha t_{2}$

$20=0+2 t_{2}$

$t_{2}=10 \mathrm{sec}$


$T=20$ sec

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