Question 233 of 280

A hoop of radius $\text{r}$ and mass $\text{m}$ rotating with an angular velocity ${\omega }_{0}$ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?

A$\dfrac{\text{r}{\omega }_{0}}{2}$B$\text{r}{\omega }_{0}$C$\dfrac{\text{r}{\omega }_{0}}{4}$D$\dfrac{\text{r}{\omega }_{0}}{3}$

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Answer: (A) $\dfrac{\text{r}{\omega }_{0}}{2}$

By conservation of angular momentum at the point of contact

$m{r}^{2}{\omega }_{0}=mvr+m{r}^{2}\omega =mvr+m{r}^{2}.v/r$
$\Rightarrow {r}^{2}{\omega }_{0}=2vr\Rightarrow v=r{\omega }_{0}/2$