Past exam of the physics 1 paper of the University of Oxford / 2015 / 9 / b / ii

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 $\sqrt{3}f/2$ 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 givesThus $fR=\beta m{R}^2|\alpha |$ gives $f=2\beta mw$. The lower cylinder's outward translation equation and the top cylinder's downward equation areUsing $u=\sqrt{3}w$ gives $N=2\sqrt{3}m(1+\beta )w$ and henceFor 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 ${\mu }_c\ge f/N=\beta /[\sqrt{3}(1+\beta )]$ initially.

These are the two cases of the initial acceleration of three stacked cylinders. As a check, as $\beta \to 0$ the rotational contribution disappears in both formulas and they approach the smooth-contact release result $w=g/7$.