Question

A uniform circular disc placed on a rough horizontal surface has initially 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. $1/2$
• B. $1$
• C. $3/2$
• D. $2$

Answer 1

The correct option is A $1/2$

$v_p=v_0+5w_0$

If disk stops,

$v$ and $w=0$

$\therefore L_i=m{v}_{0}r-I_0{w}_0=L_{final}=0$

$\therefore m{v}_{0}r=\frac{m{r}^2}{2}{w}_0$

$\therefore 2v_0=r{w}_0$

$\Rightarrow \frac{v_0}{r{w}_0}=\frac{1}{2}$

Hence, the answer is $\frac{1}{2}.$