Question

A wheel of mass $10kg$ and radius $20cm$ is rotating at an angular speed of $100rev/min$ 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.

Answer 1

Rotation of the wheel $=100rev/min$

$w=\frac{5}{3}rev/s$

Now, $w^1=0$

Total $rev$, $\theta =10$ revolutions

$w^1=w^2-2\propto \theta$

$0={\left(\frac{10}{6}\right)}^2-2a\times 10$

$0=100/36-20a$

$\therefore 0=\frac{5}{36}rev{s}^{-2}$

or $a=2\pi \left(\frac{5}{36}\right)=\frac{10\pi }{36}rad{s}^{-2}$

Now, $I=\frac{{mr}^2}{2}$

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

$=0.2kg{m}^{-2}$

$\tau =F.r$

$\tau =0.2F$

Now Torque, $\tau =I\propto$

$=\frac{0.20}{10\pi /36}=\frac{2\pi }{36}$

$\therefore 0.2F=\frac{2\pi }{36}$

$F=\frac{2\pi }{7.2}=0.87N$