A ring of mass M hangs from a thread and two beads of mass m slides on it without friction. The beads are released simultaneously from the top of the ring and slides down in opposite sides, Show that the ring will start to rise, if $\mathrm{m}>\frac{3\mathrm{M}}{2}$. 

Let R be the radius of the ring

$\mathrm{h}=\mathrm{R}(1-\mathrm{cos\theta })$

${\mathrm{v}}^2=2\mathrm{gh}=2\mathrm{gR}(1-\mathrm{cos\theta })$

$\frac{{\mathrm{mv}}^2}{\mathrm{R}}=\mathrm{N}+\mathrm{mgcos\theta }$

or $\mathrm{N}=2\mathrm{mg}(1-\mathrm{cos\theta })-\mathrm{mgcos\theta }$

$\mathrm{N}=2\mathrm{mg}-3\mathrm{mgcos\theta }$

In the critical condition, tension in the string is zero and net upward force on the ring

$\mathrm{F}=2\mathrm{Ncos\theta }=2\mathrm{mg}\left(2\mathrm{cos\theta }-3{\cos }^2\mathrm{\theta }\right)$

F is maximum when $\frac{\mathrm{dF}}{\mathrm{d\theta }}=0$

or $-2\mathrm{sin\theta }+6\mathrm{sin\theta cos\theta }=0$

or

$\mathrm{cos\theta }=\frac{1}{3}$

Substituting in Eq. (i), we get

${\mathrm{F}}_{\text{max }}=2\mathrm{mg}\left(2\times \frac{1}{3}-3\times \frac{1}{9}\right)=\frac{2}{3}\mathrm{mg}$

${\mathrm{F}}_{\max }>\mathrm{Mg}$ or $\frac{2}{3}\mathrm{mg}>\mathrm{Mg}$

or

$\mathrm{m}>\frac{3}{2}\mathrm{M}$

Hence proved.