Question

A wheel of mass $10kg$ and radius $20cm$ is rotating at an angular speed pf $100$ revolution$/$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$?$

• A. 0.02
• B. 0.23
• C. 0.05
• D. 0.87

Answer 1

The correct option is D 0.87

Rotation of the wheel $=100$ $revolution/min$ or $=5/3$

$revolution/second$ $.$

Now$,w^{\prime }=0$

Total rev. $10$revolution

Now we have to find $\alpha =?$

We know the formula$,$

$w^{\prime }=w^2-2\alpha \theta$

$0=(10/6{)}^2-2\times \alpha \times 10$

$0=100/36-20\alpha$

$\alpha =5/36rev/secon{d}^2$

or$,\alpha =2\pi \times \left(5/36\right)$

$\alpha =10\pi /36radian/secon{d}^2.$

Now$,$ we have to moment of inertia of the wheel$.$

$I=\frac{m{r}^2}{2}$

$I=1/2({0.2}^2\times 10)$

$I=0.2kg.m^2$

Since the force is F, Now torque $,$

$F=F.r$

$0.2F$

Now torque $=I\alpha$

$0.20 / (10 π /36)44

$=2\pi /36$

$0.2F=2\pi /36$

$F=2\pi /7.2$

$F=0.87N$

Hence,

option $\left(D\right)$ is correct answer.