Question

A hoop of radius $r$ and mass $m$ rotating with an angular velocity ${\omega }_0$ is placed on a rough horizontal surface. The initial velocity of the center of the hoop is zero. What will be the velocity of the center of the hoop when it ceases to slip?

• A. $\frac{r{\omega }_0}{2}$
• B. $r{\omega }_0$
• C. $\frac{r{\omega }_0}{4}$
• D. $\frac{r{\omega }_0}{3}$

Answer 1

The correct option is A $\frac{r{\omega }_0}{2}$

Let $v$ be the velocity of the center of the hoop, when it ceases to slip.
By applying the principle of conservation of angular momentum, we get
$\left(m{r}^2\right){\omega }_0=mv\times r+m{r}^2\omega$$=mv\times r+m{r}^2\left(\frac{v}{r}\right)=2mvr$
$\Rightarrow v=\frac{\left(m{r}^2\right){\omega }_0}{2mr}=\frac{{\omega }_{0}r}{2}$