Question

A hoop of radius and mass m rotating with an angular velocity omega 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?

Answer 1

$r\frac{\omega }{2}$
Radius of hoop = r, Initial angular velocity $={\omega }_0$, Initial velocity (u) = 0, velocity of centre of hoop = ?
By momentum conservation, initial momentum of hoop will be same as final momentum
i.e., $L_i=L_f$
$m{r}^2{\omega }_0=mvr+\frac{m{r}^{2}v}{r}$
$r{\omega }_0=v+v$
$v=\frac{r{\omega }_0}{2}$