= Solution
\b[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 is
$$
f_gR=(\beta mR^2)\frac uR,\qquad f_g=\beta mu.
$$
The outward translation equation is $mu=N/2-f_g$, giving $N=2m(1+\beta)u$. The top cylinder's vertical equation is $mw=mg-\sqrt3N$. Using $u=\sqrt3w$ yields
$$
(7+6\beta)w=g,
\qquad
\boxed{a_{\mathrm{top,down}}=\frac{g}{7+6\beta}.}
$$
For uniform solid cylinders, $\beta=1/2$ and $a_{\mathrm{top,down}}=g/10$. 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 $\mu_g\geq\beta\sqrt3/(10+9\beta)$ at release. This follows from $f_g=\beta m\sqrt3w$ and ground normal <force> $N_g=mg+\sqrt3N/2=mw(10+9\beta)$.
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