A hoop of radius r and mass 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 ot slip?

Options

A. `(romega_(0))/(4)`
B. `(romega_(0))/(3)`
C. `(romega_(0))/(2)`
D. `romega_(0)`

Answer

C

Explanation

From conservation of angular momentum,
`mr^(2)omega_(0)=mvr+mr^(2)xx(v)/(r) rArr v=(omega_(0)r)/(2)`