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A sphere of radius $a$ at gap $aϵ$ translates at $U{\mathbf{e}}_x$ and rotates at $\omega {\mathbf{e}}_y$ above a stationary wall in a right-handed coordinate system. At leading order its gap is $h=aϵ+r^2/(2a)$. In its translating frame the lower and upper tangential velocities are $-U{\mathbf{e}}_x$ and $-a\omega {\mathbf{e}}_x$. The Reynolds lubrication equation givesIntegrating the upper-boundary traction, including pressure on the sloping surface, and its moment about the sphere centre gives the sphere-on-fluid force and torqueThe off-diagonal symmetry is required by the Lorentz reciprocal theorem for Stokes flow. A torque-free sphere therefore has $a\omega =U/4$ and leading drag $3\pi \mu aU\log (1/ϵ)$. Compare the rigid-wall terms of Bertin et al., equations (4.4)–(4.5); those forces act on the sphere and have the opposite sign.

This is a squeezing motion, not the rotational motion of the previous part. The gap changes at rate $h_t=V\sin \theta$. The Reynolds lubrication equation therefore givesThe outer wall also has tangential speed $V\cos \theta$, but its local Couette flow flux is $O(V\Delta )$ compared with the squeezing flux $O(Va)$ and does not affect the leading force. Thus $p_{\theta }=-12\mu aq/h^3$ at leading order. Periodic pressure gives $C=0$, because the integral of $\cos \theta /h^3$ vanishes over a full circle. Balancing pressure traction on the fixed inner cylinder and using integration by parts,The supplied integral gives the squeeze resistance of an eccentric journal bearing:The fluid tends to pull the inner cylinder upwards with the moving outer wall, so a downward holding force is required when $V>0$. At concentricity the translation resistance is $R_0=12\pi \mu a^3/{\Delta }^3$. The omitted wall-tangential flux and shear traction give smaller thin-gap contributions.

Reflection in $y=0$ preserves the prescribed translational velocity and axial angular velocity. A force transforms as a polar vector, whereas a torque transforms as an axial vector; consequently $F_y=G_x=G_z=0$. Reflection in $x=0$ reverses both imposed motions but preserves the normal force component. The linearity of Stokes flow then gives $F_z=0$. Thus only $F_x$ and $G_y$ can be nonzero.

There is a sign inconsistency in the printed question. In right-handed coordinates, $\mathbf{\Omega }=\Omega {\mathbf{e}}_y$ gives a rotational velocity $-\Omega a{\mathbf{e}}_x$ at the sphere's lowest point. The printed shear and force formulas instead use the opposite rotational sense. Below, let $\omega$ be the right-handed component along $+{\mathbf{e}}_y$; the paper's displayed formulas are recovered by setting $\omega =-\Omega$.

In the translating frame, the parabolic gap and leading boundary velocities areThe no-slip boundary condition and lubrication theory giveThe gap is stationary in this frame, so ${\nabla }_{\parallel }\cdot \mathbf{q}=0$. The Reynolds lubrication equation is thereforeSince $h_x=x/a$ and ${\nabla }_{\parallel }^{2}h=2/a$,Differentiating the velocity profile gives the leading shear stressesThe first expression becomes the printed formula after $\omega =-\Omega$.

For the logarithmic lubrication resistance of a sphere near a wall, write $\ell =\log (1/ϵ)$. The logarithmic region is $a\sqrt{ϵ}\ll r\ll a$, where $h\sim r^2/(2a)$. ConsequentlyThese follow from $\int r/hdr=a\ell +O(a)$ and $\int r^3/h^{2}dr=2a^2\ell +O(a^2)$, with a fixed small outer cutoff. The general logarithmic radial numerator is $r^{2n-1}$ over $h^n$; the numerator in the printed integration hint appears to have a typographical error.

The force exerted on the fluid through the upper gap boundary includes pressure acting on its slope:The pressure cancels the $x^2/h^2$ term, givingPressure exerts no torque about the centre of a sphere because its traction is radial. To logarithmic order, the shear has lever arm $-a{\mathbf{e}}_z$, soThe other components vanish by parity: $G_x$ involves an integral of the $xy$ term, and $G_z$ has an integrand odd in $y$. The combined hydrodynamic resistance matrix isIts equal off-diagonal coefficients are required by the Lorentz reciprocal theorem for Stokes flow. It is a positive-definite matrix, as required by viscous dissipation.

In the rotational convention of the printed force formula, set $\Omega =-\omega$ and measure the scalar couple along $-{\mathbf{e}}_y$, so $G=-G_y$. ThenFor sedimentation under the buoyancy-adjusted weight $F$, a uniform sphere is torque-free. Thus $\omega a=U/4$, and the force balance givesThe rotational scalar in the paper's displayed-force convention is $\Omega =-U/(4a)$. The sphere rolls in the right-handed $+{\mathbf{e}}_y$ sense but still slips: its lowest point has laboratory velocity $U-a\omega =3U/4$. The signs and coefficients of the right-handed resistance agree with the rigid-wall terms of Bertin et al., equations (4.4)–(4.5), after reversing the forces and torques there from fluid-on-sphere to sphere-on-fluid.

Near the closest point, the cylinder–plane gap isFor translation parallel to the cylinder axis, the leading flow is Couette flow. Its shear traction is $\mu U/h$, soHence $A_{yy}=2\pi {\epsilon }^{-1/2}\mu L$.

For transverse translation, the local Couette-Poiseuille flow in a thin gap and mass conservation give the Reynolds lubrication equation. Writing $x=a\sqrt{\epsilon }X$ and using pressure recovery at both ends determines its integration constant. The pressure and viscous contributions to the horizontal traction reduce towhere the final term is the direct Couette shear contribution. Therefore

Integrating the incompressible continuity equation $u_x+v_y=0$ across the moving gap, with $v(x,0)=0$ and $v(x,h)=-V$, gives the Reynolds lubrication equationHencefor a constant $C$, and the flux formula from part (b) givesDefine the spatial meanBecause both ends meet fluid at pressure $p_0$, the pressure recovery condition in lubrication flow isIt determinesTherefore the requested pressure gradient in a translating and squeezing finite gap is

Let the inner-centre displacement from a fixed outer centre have magnitude $r=\Delta e$, with $0\le e<1$. The leading full-film journal bearing resistance to translation in the displacement direction and perpendicular to it isFor $h=\Delta (1-e\cos \theta )$, the Reynolds lubrication equation gives radial squeezing flux $a{v}_r\sin \theta$ and tangential squeezing flux $a{v}_t[-\cos \theta +3e/(2+e^2)]$. The additive constant enforces periodic pressure. Integrating the pressure traction gives these coefficients; at $e=0$ they coincide, as required by rotational symmetry.