Question:
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?
Solution:
$\omega_{0}=20 \frac{\mathrm{rad}}{\sec } ; \alpha=-\frac{2 \mathrm{rad}}{\mathrm{s}^{2}} ; \omega=0$
$\omega=\omega_{0}+\alpha t_{1}$
Now, wheel accelerates and when angular velocity becomes $20 \mathrm{rad} / \mathrm{sec}$ it will have same KE.
$\omega=\omega_{0}+\alpha t_{2}$
$20=0+2 t_{2}$
$t_{2}=10 \mathrm{sec}$
$\mathrm{T}=t_{1}+t_{2}$
$T=20$ sec
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.