Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

Solve the following :

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

Leave a comment

None
Free Study Material