Two identical discs having same radii R, rotating in opposite sense are joined with light rod having frictionless grooves A and B and placed on rough horizontal surface as shown in figure. Initially, translational velocity of each disc is zero. Then:-

Angular momentum conservation gives

$$\begin{gathered} \left(2{\omega }_0\right)\frac{m{R}^2}{2}-\frac{m{R}^2}{2}\left({\omega }_0\right)=2m{v}_{C}R+\frac{2m{R}^2}{2}\omega \text{'} \\ \frac{m{R}^2}{2}{\omega }_0=2m{v}_{C}R+mR{v}_C(\text{Since\hspace{0.17em}}{v}_C=\omega \text{'}R) \\ {v}_C=\frac{R{\omega }_0}{6} \end{gathered}$$

$\to$ Second figure shows that rod is under stress & hence its length will increase
$\to$ Finally when the discs roll purely, there is no friction & hence no tensile stress
$\to$ Since friction is kinetic therefore mechanical energy will decrease.