Discussion Forum

A hoop of radius r and mass m rotating with an angular velocity ω₀  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:

Explanation:

ω = v/r

From conservation of angular momentum about bottommost  point

  $mr^{2}\omega_{0}=mvr+mr^{2}\times\frac{v}{r}$

$\Rightarrow$    $v=\frac{\omega_{0}r}{2}$