In the diagram shown, a rod of mass M has been fixed on a ring of the same mass. The whole system has  been placed on a perfectly rough surface. The system is gently displaced so that the ring starts rolling. 
 The velocity of the centre of the ring when the rod becomes horizontal is $\sqrt{\frac{6gR}{x}}.$
 The value of x is _______ ( the length of the rod is equal to the radius of the ring )

The moment of inertia of the system about instantaneous axis of rotation is given by
$I=I_{ring}+I_{rod}$

$\Rightarrow I=(M{R}^2+M{R}^2)+(\frac{1}{12}M{R}^2+M{R}_1^2)$   ( from parallel axis theorem ) 
 $\Rightarrow I=2M{R}^2+\frac{1}{12}M{R}^2+M{\left(\frac{\sqrt{5}R}{2}\right)}^2$          $(\because R_1=\sqrt{R^2+{\left(\frac{R}{2}\right)}^2}=\frac{\sqrt{5}R}{2})$  
$\Rightarrow I=2M{R}^2+\frac{1}{12}M{R}^2+\frac{5}{4}M{R}^2=2M{R}^2+\frac{4}{3}M{R}^2=\frac{10}{3}M{R}^2$
Hence,  $I_{system}=\frac{10}{3}M{R}^2$
Now from work – energy theorem, we get,
$Mg\frac{R}{2}=$ Rotational kinetic energy of the system about instantaneous axis of rotation 
( here $\frac{R}{2}$  is the distance through which the center of mass of the rod descends)
$$\begin{gathered} \Rightarrow Mg\frac{R}{2}=\frac{1}{2}{I}_{system}{\omega }^2 \\ \Rightarrow Mg\frac{R}{2}=\frac{1}{2}\times \frac{10}{3}M{R}^2\times {\omega }^2 \\ \Rightarrow \omega =\sqrt{\frac{3g}{10R}} \end{gathered}$$
Now velocity of center of mass is given as, $v=R\omega$ .
$\Rightarrow v=R\times \sqrt{\frac{3g}{10R}}$
 $=\sqrt{\frac{6gR}{20}}$