Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 39D / c

At vertical displacement $x$ from the axis, the gap isFor $|x|\ll a$,soThe narrow-gap length scale is therefore $(a{h}_0{)}^{1/2}$.

Take $x$ upward, let $y$ measure distance from the wall, and work in the cylinder frame. The wall at $y=0$ moves upward with speed $V$. Choose the sign of $\Omega$ so that the near-cylinder surface at $y=h$ moves upward with speed $a\Omega$. Lubrication theory givesThusThe vertical volume flux per unit cylinder length is independent of $x$ and equalsThereforeBecause $p(+\infty )=p(-\infty )$, its integral over $x$ vanishes. Withthe quoted integrals giveIt follows that

Pressure acts normally to the circular cylinder and so has no moment about its axis. To leading lubrication order, the tangential shear at the cylinder isSubstitution of the pressure gradient givesA torque-free cylinder requires ${\int }_{-\infty }^{\infty }{\tau }_{c}dx=0$. Also,Using $Q=2h_0(V+a\Omega )/3$, the torque condition reduces toHence