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Take positive $\Omega$ counterclockwise in the $(x,y)$ plane and put $W=\Omega a$, $d=ϵa$. The bottom cylinder surface moves locally at $W$; the wall moves at $-U$. In the lubrication approximation,The leading equations are $p_y=0$, $\mu u_{yy}=p_x$ and incompressibility. No slip gives the Couette-Poiseuille flow in a thin gapThe steady gap flux is independent of $x$. Equal leading pressures at the two matched outer ends require $\int p_{x}dx=0$. Let $J_n=\int h^{-n}dx=\ell d^{-n}{I}_n$. Then $J_2/J_3=4d/3$, and $\boxed{q=\frac{2}{3}d(W-U)}$. The velocity field and flux determine the normal velocity by incompressibility and no penetration.

At the wall and cylinder, respectively,Here $\tau ={\sigma }_{xy}$; the fluid force on the cylinder has the opposite traction sign from that on the wall. Since $J_2=J_1/(2d)$,The shear and pressure force partition in cylinder lubrication explains why these two shear forces need not balance: pressure on the sloping cylinder surface supplies an equally important tangential component. Its contribution isBoundary terms vanish under outer matching. Thus the total cylinder force is $-2\mu U{J}_1$, and it balances the wall force once pressure is included. Pressure acts normally through the circular cylinder's center and gives no torque; leading shear torque is $a{F}_{\mathrm{cylinder}}^{\mathrm{shear}}=-2\mu \Omega a^2{J}_1$.

The effective weight per unit length is $+\pi a^2\Delta \rho g$ in the falling direction, with zero applied couple. Force and couple balance give $\boxed{U=\Delta \rho g{a}^2\sqrt{ϵ}/(2\sqrt{2}\mu ),\Omega =0\text{ at leading order}}$. Finite outer-region torque can generate smaller rotation; the leading singular gap resistance is what fixes this result.

For $\Omega =0$, $q=-2dU/3$ andThe pressure is odd, negative for $x<0$ and positive for $x>0$ with the wall moving toward decreasing $x$, and tends to the common reference pressure at either end. The cylinder shear changes sign at $\boxed{x=\pm \ell =\pm a\sqrt{2ϵ}}$. For $\zeta =y/h$, the velocity isFlow is entirely wall-directed near the neck, while a reversed layer next to the cylinder exists for $|\xi |>1$. These changes are the separation and reattachment points shown in the stream-function sketch.

For the final Couette flow, the printed center $(\lambda ϵa,0)$ is an offset parallel to the walls. Translational invariance removes it; the centered two-gap symmetry gives the literal answer $\boxed{V=0}$ for every $\lambda$. A dependence on $\lambda$ requires a transverse position, evidently the intended $(0,\lambda ϵa)$. With that repaired interpretation, the lower and upper gaps are $d_b=ϵa(1+\lambda )$ and $d_t=ϵa(1-\lambda )$.

In the translating cylinder frame the wall velocities are $-U-V$ and $U-V$. Each leading tangential drag is $2\mu J_1$ times its wall velocity, independent of the rotation. Hence zero net force requires $(U-V)J_t-(U+V)J_b=0$, where $J_{t,b}=\pi \sqrt{2a/d_{t,b}}$. For the intended transverse offset,This formula assumes both gaps remain lubrication gaps, with fixed $|\lambda |<1$. It gives $V=0$ at $\lambda =0$ and motion toward the faster nearby wall otherwise.

At $\lambda =0$, the lower-gap flow is the previous $-U$ flow and the upper gap its reversed mirror, with fluxes $\mp 2dU/3$. Their leading translational shear torques vanish. Any finite outer-fluid Couette torque is $O(\mu Ua)$, while the sum of the gap rotation torques is $O(\mu \Omega a^2/\sqrt{ϵ})$. Balancing them explains $\boxed{\Omega =O(\sqrt{ϵ}U/a)}$. Leading gap analysis alone does not determine its prefactor.