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JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion

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?

  1. A \(r\)\({\omega _0}\)
  2. B \(\frac{{{\rm{r}}{\omega _0}}}{4}\)
  3. C \({\rm{\;}}\frac{{{\rm{r}}{\omega _0}}}{3}\)
  4. D \({\rm{\;}}\frac{{{\rm{r}}{\omega _0}}}{2}\)
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Answer & Solution

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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