Question

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

$\omega$ = 100 rev/min

= $\frac{5}{3}$ rev/s = $\frac{10\pi }{3}$ rad/s

$\theta$ = 10 rev. = 20$pi$ rad r = 0.2 m

After 10 revolutions, the wheel will come to rest by a tangential force.

Therefore the angular deceleration produced by the force

$=K=\frac{K2}{2K}$

Therefore the torque by which the wheel will come to an rest = I cm $\times$ K

$\Rightarrow F\times r=I_{cm}\times \alpha$

$\Rightarrow F\times 0.2=\frac{1}{2}m{r}^2\times$

${\left(\frac{10\pi }{3}\right)}^2(2\times 20\pi )$

$\Rightarrow F=\frac{1}{2}\times 10\times 0.2\times \frac{100{\pi }^2}{(9\times 2\times 20\pi )}$

$=\frac{5\pi }{18}=\frac{15.7}{18}=0.87$