A body rotating at 20 rad /s is acted upon by a constant torque provid

1.	A body rotating at 20 rad /s is acted upon by a constant torque providing it a deceleration of 2 rad/s², At what time will the body have kinetic energy same as the initial value if the torque continues to act . ?
Answer» *Given in thew question :-* Initial velocity = 20 rad / second. $\alpha$ therefore = 20 rad/second² *Due to the TORQUE the effect of DECLARATION on the body , the body will start accelerating in OPPOSITE direction.* Therefore the initial Kinetic ENERGY = K.e at any instant. Hence Init. velocity = velocity at any instant $\omega'$ Now Fin. velocity =(-ω)= -20 rad/ second. $\omega = 20 rad/second$ Init. velocity *Therefore we know the formula ,* $\omega' = \omega - \alpha t$ $\omega' = -\omega$ since Hence, $-\omega = \omega - \alpha t$ 2.ω = α t t = (2 × ω) / α t = (2 × 20 )/2 *t = 20 second.* *Hope it Helps :-)*