= Solution
\b[Smooth ground, rough cylinder contacts.] Write $f$ for the tangential contact <force> magnitude at each upper-to-lower contact. On the top cylinder, its tangential components point upward and outward, while their horizontal components and <torques> cancel by symmetry. For a lower cylinder, the tangential <force> has an inward horizontal component $\sqrt3f/2$ and a downward vertical component $f/2$.
The rolling condition is now at the contact between cylinders, not at the ground. At the initial zero <velocity>, equality of the tangential contact-point <accelerations> gives
$$
R|\alpha|=\frac{\sqrt3u+w}{2}=2w.
$$
Thus $fR=\beta mR^2|\alpha|$ gives $f=2\beta mw$. The lower cylinder's outward translation equation and the top cylinder's downward equation are
$$
mu=\frac N2-\frac{\sqrt3}2f,\qquad mw=mg-\sqrt3N-f.
$$
Using $u=\sqrt3w$ gives $N=2\sqrt3m(1+\beta)w$ and hence
$$
(7+8\beta)w=g,
\qquad
\boxed{a_{\mathrm{top,down}}=\frac{g}{7+8\beta}.}
$$
For solid cylinders this is $g/11$. Here the left lower cylinder rotates clockwise while moving left, and the right rotates counterclockwise while moving right. They may slip freely on the smooth ground, so imposing the ground-rolling relation $R|\alpha|=u$ would be incorrect. No slip at the cylinder contacts requires $\mu_c\geq f/N=\beta/[\sqrt3(1+\beta)]$ initially.
These are the two cases of the <initial acceleration of three stacked cylinders>. As a check, as $\beta\to0$ the rotational contribution disappears in both formulas and they approach the smooth-contact release result $w=g/7$.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2015-three-cylinder-contacts-and-release.png]
{title=Three-cylinder contact forces and the two distinct rolling constraints on release}
{description=Original geometry and rotation diagrams. Contact normals in part (a) must remain nonnegative. In part (b), the lower cylinders start moving apart; rough-ground rolling and rough-cylinder rolling produce opposite lower-cylinder rotation senses.}
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