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

Let the tangential velocity scale be $U$, gap scale $H$ and along-gap length $L\gg H$. The tangential Stokes equation balances $P/L$ with $\mu U/H^2$, while wall shear is $\tau \sim \mu U/H$. ThereforeThis is lubrication pressure dominates shear stress; pressure force projections may nevertheless be small in a particular direction or may give no torque on a spherical surface.

Write the signed centre displacement along the downward axis as $d=\alpha \Delta$. A point $an$ on the inner sphere has distance $|an+d{e}_z|=a+d\cos \theta +O({\Delta }^2/a)$ from the outer centre. The leading normal gap to radius $a+\Delta$ is consequentlyThe local-curvature corrections are smaller by $\Delta /a$. This approximation also follows by solving the outer-sphere intersection along the ray $n$ from the inner centre and expanding its positive root.

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 $6\pi \mu a^4|V|/[{\Delta }^3(1-\alpha )]$.

Axisymmetric rotation about the offset axis gives wall velocity $v_{\phi }=a\Omega \sin \theta$. Its azimuthal Couette flow has no pressure-driving divergence, so pressure remains constant. The fluid-on-inner-sphere tangential traction is $-\mu a\Omega \sin \theta /h$. Multiply it by the moment arm $a\sin \theta$ and integrate:Putting $t=\alpha z$ and using the second supplied identity with its generic parameter equal to $\alpha$ gives the axial rotational resistance of eccentric nested spheres:The continuous limit at $\alpha =0$ is $-8\pi \mu \Omega a^4/(3\Delta )$, consistent with a concentric narrow spherical shell. At $\alpha \to 1$, the torque tends to $-4\pi \mu \Omega a^4/\Delta$: it remains finite because the rotational wall speed vanishes at the closest pole.

Let $ϵ=1-\alpha \ll 1$, and use local distance $\rho =a\theta$ from the closest pole. The gap isHorizontal rotation has nonzero wall speed $V_s\sim |{\Omega }^{\prime }|a$ at this pole. In the broad region with $h=O(\Delta )$, shear is $O(\mu |{\Omega }^{\prime }|a/\Delta )$, the area is $O(a^2)$ and the lever arm is $O(a)$. Its torque is therefore $O(\mu |{\Omega }^{\prime }|a^4/\Delta )$. The smallest gap patch has radius $O(a\sqrt{ϵ})$ and area $O(a^2ϵ)$; its shear is $O(\mu |{\Omega }^{\prime }|a/(ϵ\Delta ))$. It gives the same bounded torque scale.

These are the two estimates requested in the paper. They omit an intermediate annulus, so the printed uniform $O(\mu |{\Omega }^{\prime }|a^4/\Delta )$ assertion is not correct as $ϵ\to 0$. For $a\sqrt{ϵ}\ll \rho \ll a$, $h\sim \Delta {\rho }^2/a^2$. An annular shear-torque contribution scales asIntegrating between the inner and outer scales produces $\log (1/ϵ)$.

A viscous dissipation lower bound makes this conclusion independent of possible signs of local tractions. A fluid velocity changing by $V_s$ between two no-slip walls dissipates at least $\mu V_s^2/h$ per unit area, by the Cauchy-Schwarz inequality in the cross-gap coordinate. The lubrication velocity profile gives more explicitlywith no cross term. The second term is nonnegative. Thus the annulus entails a logarithmic divergence in total power and hence in $|G^{\prime }|$, since the centre is held fixed and all boundary work is $G^{\prime }{\Omega }^{\prime }$.

The horizontal rotation in a nearly touching spherical shell calculation also gives the corrected force scale. Choose local $x$ along $v_s$ and put $V_s=a{\Omega }^{\prime }$ with its signed value. At leading order the stationary-gap Reynolds lubrication equation isFor $h=h_0+(x^2+y^2)/(2R)$ it has the particular pressurewhich satisfies the equation by direct differentiation. The pressure itself gives no torque about the centre of an exactly spherical inner body, because its normal is radial. It does contribute to tangential shear through the pressure gradient.

For any fixed local outer cutoff, the logarithmically divergent integrals areThe resisting tangential traction is $\mu V_s/h+(h/2)p_x$. Integration by parts gives torque magnitude $a\mu |V_s|(I_1+3I_2/(5R))$ to logarithmic order. Consequently, at leading order in $\Delta /a$,The strict corrected order is $\Theta ((\mu |{\Omega }^{\prime }|a^4/\Delta )\log (1/ϵ))$; the paper's scale is the bounded contribution of each of its two nominated regions, not the full asymptotic answer.

For the horizontal force, pressure dominates shear by a factor $a/\Delta$. Its horizontal projection on the inner sphere has normal component $n_x\simeq x/a$, giving $|F_p|\sim (6\mu |V_s|/(5a))I_2$. Hence the required holding-force magnitude isThe broad region and smallest patch each supply $O(\mu |{\Omega }^{\prime }|a^4/{\Delta }^2)$; the intermediate annulus again supplies the extra logarithm. The shear force is smaller by $\Delta /a$. These expressions assume the completely filled lubrication film of the problem.

The vertical holding force is zero. The geometry is invariant under a half-turn about the vertical axis. That operation reverses the horizontal angular velocity but leaves a vertical force component unchanged. Linearity would reverse that force component too, forcing it to vanish. Equivalently, the induced pressure has the first azimuthal harmonic, whose vertical-force integral over azimuth is zero. The horizontal holding force is the negative of the fluid's horizontal force on the body.