Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2 / 39C (Fluid Dynamics II)

Stokes flow obeys $-\nabla p+\mu {\nabla }^{2}u=0$, $\nabla \cdot u=0$. Reversing all applied forces reverses $u$ while leaving streamlines unchanged. Reflection in the vertical plane containing the cylinder axis rules out rotation normal to that plane; reflection combined with reversibility rules out rotation within it and spin about the axis. Thus its orientation remains fixed.

Lubrication gives $u_r=(p_r/2\mu )z(z-h)$ and continuity gives ${\int }_0^h{u}_{r}dz=-Vr/2$. Hence $p_r=6\mu Vr/h^3$ and $p-p_0=3\mu V(r^2-a^2)/h^3$. Balancing the upward pressure force with $W$ yieldsTherefore $h^{-2}=h_0^{-2}+4Wt/(3\pi \mu a^4)$ and $h\sim (3\pi \mu a^4/4Wt{)}^{1/2}$.