Question:
A wheel rotating at a speed of 600 rpm (revolutions per minute) about its axis is brought to rest by applying a constant torque for 10 seconds. Find the angular deceleration and the angular velocity 5 seconds after the application of the torque.
Solution:
$\omega_{0}=600 \mathrm{rpm}=\frac{600}{60} \frac{\mathrm{rev}}{\mathrm{sec}}$
$\omega_{0}=10 \frac{\mathrm{rev}}{\mathrm{sec}} ; \omega=0 ; t=10 \mathrm{sec}$
$\omega=\omega_{0}+\alpha t$
$\alpha=-1 \mathrm{rev} / \mathrm{sec}^{2}$
For $\mathrm{t}=5 \mathrm{sec}$,
$\omega=\omega_{0}+\alpha t$
$=10+(-1)(5)$
$\omega=5 \mathrm{rev} / \mathrm{sec}$
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.