Question

A uniform circular disc placed on a rough horizontal surface has initially a velocity $v_0$ and an angular velocity ${\omega }_0$ as shown in the figure. The disc comes to rest after moving some distance in the direction of motion. Then $\frac{v_0}{r{\omega }_0}$ is :

• A. $\frac{1}{2}$
• B. $1$
• C. $\frac{3}{2}$
• D. $2$

Answer 1

The correct option is A $\frac{1}{2}$

Slipping will take place and to oppose the relative motion between the disc and the ground, friction will act in the shown direction.
So we can write
$f=ma$ So $a=\frac{f}{m}$
$f.r=I\alpha$
For the disc $I=\frac{1}{2}m{r}^2$
$\alpha =\frac{2f}{mr}$
If the disc stops moving after travelling for a while, both $v_0$ and ${\omega }_0$ become zero at the same time.
If time $t$ required to stop, we can write
$v_0-at=0$ and ${\omega }_0-\alpha t=0$

$\therefore \frac{v_0}{a}=\frac{{\omega }_0}{\alpha }$

$\therefore \frac{v_{0}m}{f}=\frac{{\omega }_{0}rm}{2f}$

$\therefore \frac{v_0}{{\omega }_{0}r}=\frac{1}{2}$