Single Choice

A hoop of radius $$r$$ and mass $$m$$ rotating with an angular velocity $$\omega_{o}$$ 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$$\displaystyle \frac{r\omega _{0}}{3}$$
B$$\displaystyle \frac{r\omega _{0}}{2}$$
Correct Answer
C$$\displaystyle r\omega _{0}$$
D$$\displaystyle \frac{r\omega _{0}}{4}$$

Solution

From conservation of angular momentum
$$mr^2\omega_0 = mvr+mr^2\dfrac{v}{r}$$.
or
$$mr^2\omega_0 = 2mvr$$
or
$$ \therefore v= \dfrac{r\omega_0}{2}$$

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