 Solution

Rough ground, smooth cylinder contacts. Let $N$ be either upper-to-lower normal force, and $f_g$ the magnitude of the inward ground friction force on each lower cylinder. Pure rolling on the ground requires $|\alpha |=u/R$. The torque equation about a lower cylinder's centre isThe outward translation equation is $mu=N/2-f_g$, giving $N=2m(1+\beta )u$. The top cylinder's vertical equation is $mw=mg-\sqrt{3}N$. Using $u=\sqrt{3}w$ yieldsFor uniform solid cylinders, $\beta =1/2$ and $a_{\mathrm{top},\mathrm{down}}=g/10$. The left lower cylinder rotates counterclockwise while rolling left; the right rotates clockwise while rolling right.

The assumed no-slip ground contact is possible if its coefficient of static friction satisfies ${\mu }_g\ge \beta \sqrt{3}/(10+9\beta )$ at release. This follows from $f_g=\beta m\sqrt{3}w$ and ground normal force $N_g=mg+\sqrt{3}N/2=mw(10+9\beta )$.