Q: A hoop of radius r and mass m rotating with an angular velocity ‘ω_{0}’ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop when it ceases to slip.

Sol: $\large m r^2 \omega_0 = m v r + m r^2 \times \frac{v}{r} $

$\large m r^2 \omega_0 = 2 m v r $

$\large r^2 \omega_0 = 2 v r $

$\large v = \frac{\omega_0 r}{2} $