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A hoop of radius r and mass m rotating with an angular velocity $ω_{0}$ is placed on rough horizontal surface. The initial velocity of centre of mass of the hoop is zero. What will be the velocity of the center of hoop when it ceases to slip?

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$ω_{o} r$

$2 ω_{o} r$

$\frac{ω_{o} r}{2}$

$\frac{1}{3} ω_{o} r$

answer is B.

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Detailed Solution

$\begin{array}{l} m r^{2} ω_{o} = m v r + m r^{2} \frac{v}{r} \\ ⇒ v = \frac{ω_{o} r}{2} \end{array}$

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