JEE Main · 2013 · 07 Apr 2013 · Offline · Rotational Mechanics

Put the reference point on the ground where the hoop touches; friction acts there and so exerts no torque about that point, which makes the angular momentum about it conserved. Initially the hoop only spins, $L_i=m{r}^2{\omega }_0$. When it finally rolls without slipping the centre moves at $v=r\omega$, and its angular momentum about the contact point is the spin part plus the translational part, $L_f=m{r}^2\omega +mvr=2m{r}^2\omega$. Setting $L_i=L_f$ gives $\omega ={\omega }_0/2$, so the centre moves at $v=r\omega =\frac{r{\omega }_0}{2}$.