Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 78 / 4 / i / Solution

Set $V=\Delta \dot{\alpha }$, positive downwards. Then $h_t=-V\cos \theta$. The leading axisymmetric lubrication flux is $q_{\theta }=-h^3{p}_{\theta }/(12\mu a)$. Wall Couette flux from the translating sphere is smaller by $\Delta /a$ than the squeeze flux and does not affect this leading pressure. The spherical Reynolds lubrication equation isRegularity at the pole gives $q_{\theta }=(aV/2)\sin \theta$, soThe apparent singularity at $\alpha =0$ is a constant-pressure gauge term: subtract $3\mu a^{2}V/(\alpha {\Delta }^3)$ before taking the limit. The remaining pressure tends to $6\mu a^{2}V\cos \theta /{\Delta }^3$.

To leading lubrication order the force is the pressure integral. Its downward component on the inner sphere isThe original PDF's first integration hint is incorrect: the rational term needs $1-b^2$, not $(1-b{)}^2$. From the antiderivative $1/(1-t)+\log (1-t)$,Thus the squeeze resistance of eccentric nested spheres givesIt opposes the velocity for either sign of $\alpha$. The continuous concentric limit is $F_z=-8\pi \mu a^{4}V/{\Delta }^3$, since the bracket is $4{\alpha }^3/3+O({\alpha }^5)$. As $\alpha \to 1$, its leading magnitude is $6\pi \mu a^4|V|/[{\Delta }^3(1-\alpha )]$.