Torque-free cylinder in a lubrication gap

For a circular cylinder translating parallel to a nearby wall at zero Reynolds number, leading lubrication shear and pressure recovery make the zero-torque condition force zero angular velocity.

For a cylinder falling parallel to a wall through a minimum gap $ϵa/2$, the parabolic lubrication gap has length scale $a\sqrt{ϵ}$. The leading tangential drag scales as $\mu U/\sqrt{ϵ}$, while the leading rotational resistance scales as $\mu \Omega a/\sqrt{ϵ}$.

In the translating-cylinder frame, a wall moves at $-U$ and the nearby cylinder surface at $W$. With gap $d+x^2/(2a)$ and equal outer pressures, lubrication theory gives the displayed forces with $J_1=\pi \sqrt{2a/d}$. Wall and cylinder shear forces alone do not balance because the sloping cylinder's pressure force is of the same order. Their complete sum is zero. Pressure exerts no moment about the circular cylinder center, while shear supplies the leading singular rotational resistance.

A nearly confined cylinder in Couette flow is driven by the unequal lubrication resistances of its two gaps. If their minimum widths are proportional to $1+\lambda$ and $1-\lambda$, force balance weights the relative wall velocities by the inverse square roots of those widths.