Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 329 / 3

Near the closest point, the cylinder–plane gap isFor translation parallel to the cylinder axis, the leading flow is Couette flow. Its shear traction is $\mu U/h$, soHence $A_{yy}=2\pi {\epsilon }^{-1/2}\mu L$.

For transverse translation, the local Couette-Poiseuille flow in a thin gap and mass conservation give the Reynolds lubrication equation. Writing $x=a\sqrt{\epsilon }X$ and using pressure recovery at both ends determines its integration constant. The pressure and viscous contributions to the horizontal traction reduce towhere the final term is the direct Couette shear contribution. Therefore

Treat each short torus segment as a straight cylinder. For translation with velocity $U{\mathbf{e}}_x$, its components normal and tangent to the circular centreline are $U\cos \theta$ and $-U\sin \theta$. Integrating the local resistance per unit length around $ds=Rd\theta$ gives

Rotation at angular velocity $\Omega {\mathbf{e}}_z$ gives the everywhere tangential speed $\Omega R$. Only the axial-cylinder resistance $2\pi \mu {\epsilon }^{-1/2}$ contributes. Multiplying the force by its moment arm $R$ and integrating around the centreline gives

In the torus frame the two planes translate with velocity $U{\mathbf{e}}_x$. Away from the $O(a)$ neighborhood of the torus, the depth-averaged Hele–Shaw flow between planes separated by $2a$ isNegligible leakage imposes ${\overline{u}}_r=0$ at $r=R$. The harmonic pressure that decays at infinity in the exterior and the regular harmonic pressure in the interior are thereforeup to a common constant. The interior velocity is zero, while the exterior flow is the uniform stream diverted around a circular obstacle. In plan view the inside has high pressure on the $+x$ side and low pressure on the $-x$ side; the immediately adjacent exterior has the opposite signs, producing the pressure jump across the torus.

The jump at $r=R$ isIntegrating it over the projected vertical area $2aRd\theta$ gives the global pressure resistanceThere are two narrow gaps, so their local resistance is twice the one-plane result from part b:Consequently the local gap resistance dominates when $a\ll R\ll a{\epsilon }^{-1/2}$, whereas the global Hele–Shaw pressure resistance dominates when

To interpret the upper bound, the pressure jump has scale $\Delta p\sim \mu UR/a^2$. Each narrow gap has thickness $a\epsilon$ and streamwise lubrication length $a\sqrt{\epsilon }$. Its pressure-driven leakage flux per unit centreline length therefore scales asThe blocked Hele–Shaw flux has scale $Ua$. Leakage is negligible precisely when $UR{\epsilon }^{5/2}\ll Ua$, or