= Solution
Work per unit axial length, with $y$ vertically upward and $x$ across the thin gap. In the frame of the downward-translating cylinder, the wall moves upward at speed $U$ and the cylinder surface is stationary. Let $\eta$ be <viscosity> and use the leading <lubrication theory>
$$
h(y)=h_0+\frac{y^2}{2a},\qquad h_0\ll a.
$$
The transverse <pressure> <gradient> is negligible; solving $\eta u_{xx}=p_y$ with $u(0)=U$, $u(h)=0$ gives
$$
u(x,y)=U\left(1-\frac xh\right)+\frac{p_y}{2\eta}x(x-h),\qquad
q=\int_0^h u\,dx=\frac{Uh}2-\frac{h^3p_y}{12\eta}.
$$
The flux $q$ is constant by <incompressibility>. Equal ambient <pressure> at both ends implies $\int p_y\,dy=0$. With $I_j=\int_{-\infty}^{\infty}h^{-j}\,dy$ and $I_2/I_3=4h_0/3$, this yields
$$
q=\frac{2Uh_0}3,\qquad p_y=\frac{6\eta U}{h^2}-\frac{8\eta Uh_0}{h^3},\qquad
\boxed{p(y)=-\frac{2\eta Uy}{h(y)^2}.}
$$
The tangential traction on the cylinder is
$$
\tau=-\eta u_x(h)=-\frac{2\eta U}h+\frac{4\eta Uh_0}{h^2}.
$$
Since $I_2=I_1/(2h_0)$, its integrated force is $-2\eta UI_1+4\eta Uh_0I_2=0$. <Pressure> acts radially on a circular cylinder and contributes no torque; the tangential traction has moment arm $a$ to leading order. Hence
$$
\boxed{F_{\rm tangential}=0,\qquad\text{torque}=0\quad\text{to leading lubrication order}.}
$$
For clarity, the total vertical hydrodynamic force is not zero. Its leading <pressure> contribution is
$$
F_{\rm drag}=-\int p h'\,dy=\boxed{2\pi\eta U\sqrt{\frac{2a}{h_0}}}
$$
upward, while the leading shear contribution just calculated vanishes. Thus \b[this model predicts no induced rolling direction]: a freely rotating cylinder can remain nonrotating, rather than rolling like a wheel in contact with the wall. This is the <torque-free cylinder in a lubrication gap> cancellation; finite ends or different boundary conditions would require a different model.
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