Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2 / 39C / b

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 $h\sim (3\pi \mu a^4/4Wt{)}^{1/2}$.