A wheel of mass $10 \mathrm{~kg}$ and radius $20 \mathrm{~cm}$ is rotating at an angular speed of $100 \mathrm{rev} / \mathrm{min}$ when the motor is turned off. Neglecting the friction at the axle, calculate the force that must be applied tangentially to the wheel to bring it to rest in 10 revolutions.
$\omega_{0}=100 \frac{r e v}{\min }=100 \times \frac{2 \pi}{60}$
$\omega_{0}=\frac{10 \pi}{3} \frac{\mathrm{rad}}{\sec }$
$\theta=10 \mathrm{rev}=20 \pi \mathrm{rad}$
$\omega=0$
Using, $\omega^{2}=\omega_{0}^{2}+2 \alpha \theta$
$\tau=\operatorname{Fr}=\mathrm{I} \alpha$
$F=\left(\frac{m r^{2}}{2}\right)\left(\frac{\sigma}{2}\right)$
$=\frac{10 \times 0.2}{2} \times\left(\frac{10 \pi}{3}\right)^{2} \times \frac{1}{2 \times 20 \pi}$
$F=0.87 N$
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.