A loop of radius $\mathrm{r}$ and mass $\mathrm{m}$ rotating with an angular velocity $\omega_{0}$ is placed…

A loop of radius $\mathrm{r}$ and mass $\mathrm{m}$ 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. $\frac{r \omega_{0}}{4}$
  2. $\frac{\mathrm{r} \omega_{0}}{3}$
  3. $\frac{\mathrm{r} \omega_{0}}{2}$
  4. $\mathrm{r} \omega_{0}$

Solution



From conservation of angular momentum about any fix point on the surface, $m r^{2} \omega_{0}=2 m r^{2} \omega$
$\Rightarrow \omega=\omega_{0} / 2 \Rightarrow v=\frac{\omega_{0} r}{2}[\because v=r \omega]$

Asked in: JEE Mains - Rotational Motion - Test 3

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