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?
- A \(r\)\({\omega _0}\)
- B \(\frac{{{\rm{r}}{\omega _0}}}{4}\)
- C \({\rm{\;}}\frac{{{\rm{r}}{\omega _0}}}{3}\)
- D \({\rm{\;}}\frac{{{\rm{r}}{\omega _0}}}{2}\)
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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