Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 329 / 1

Put $\delta =H/L\ll 1$. In a thin gap, the streamwise Stokes flow balance isso $P\sim \mu UL/H^2$. The characteristic shear stress is $\tau \sim \mu U/H$, and thereforeThis is the pressure-dominance scaling of lubrication theory.

Let the centre displacement be $\lambda \Delta$ in the downward direction. To first order in $\Delta /a$, projecting this displacement onto the radial direction at polar angle $\theta$ changes the concentric gap $\Delta$ by $-\lambda \Delta \cos \theta$. Hence

Axisymmetric mass conservation in the spherical gap givesSince $h_t=-V\cos \theta$ and regularity requires $q\sin \theta =0$ at $\theta =0$, integration gives

The leading pressure-driven lubrication flux isSubstitution of $h$ and $q$ givesand therefore

Take downward as the positive vertical direction. The constant pressure contributes no resultant, while the pressure force on the inner sphere is opposite its outward normal. ThusWith $t=\lambda \cos \theta$ and the supplied integral,The sign is upward for $V>0$, so this is a drag force. As $\lambda \to 0$, the bracket is $4{\lambda }^3/3+O({\lambda }^5)$ and $F_z\to -8\pi \mu a^{4}V/{\Delta }^3$.

Rotation about the vertical axis gives the inner surface the azimuthal speed $\Omega a\sin \theta$. The leading Couette flow shear traction is opposing and has magnitudeIts moment arm about the vertical axis is $a\sin \theta$, and $dS=2\pi a^2\sin \theta d\theta$. ConsequentlyPutting $t=\lambda \cos \theta$ and using the supplied integral givesThis viscous shear torque opposes the rotation. Its concentric limit is $-8\pi \mu \Omega a^4/(3\Delta )$, agreeing with the thin-gap limit of Torque in rotational Stokes flow between concentric spheres.

A rotation by $\pi$ about the vertical symmetry axis maps a horizontal angular velocity ${\mathbf{\Omega }}^{\prime }$ to $-{\mathbf{\Omega }}^{\prime }$ but leaves a possible vertical force unchanged. Linearity and uniqueness of Stokes flow therefore require that vertical force to equal its own negative, so it vanishes. A horizontal force and a horizontal couple are allowed by the same symmetry.

Write $\lambda =1-ϵ$ and examine the lower pole, where $\theta \ll 1$. Since $\cos \theta =1-{\theta }^2/2+O({\theta }^4)$,Thus $h=O(ϵ\Delta )$ for $\theta =O({ϵ}^{1/2})$, a circular patch of radius

In the broad region, the surface speed is $O({\Omega }^{\prime }a)$, the shear is $O(\mu {\Omega }^{\prime }a/\Delta )$, the area is $O(a^2)$, and the moment arm is $O(a)$. HenceIn the narrow patch, the shear rises to $O(\mu {\Omega }^{\prime }a/(ϵ\Delta ))$, while its area falls to $O(ϵa^2)$; the moment arm remains $O(a)$. Therefore

Dropping the moment arm gives the shear-force scale. Pressure produces the same horizontal order after multiplication by the small surface slope. Both regions consequently contribute