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 to slip?

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Correct Answer

(D) \({\rm{\;}}\frac{{{\rm{r}}{\omega _0}}}{2}\)

Step-by-step Solution

Detailed explanation

From conservation of angular momentum about any fix point on the surface,

\(\begin{array}{l}
m{r^2}{\omega _0} = 2m{r^2}\omega \\
\Rightarrow \,\omega = {\omega _0}\backslash 2 \Rightarrow v = \frac{{{\omega _0}r}}{2}
\end{array}\)

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