= Solution
Take downward <velocity> as positive. Lubrication theory reduces the axial <momentum> equation to a constant <pressure> <gradient>, and the no-slip conditions are $u(0)=0$, $u(h)=U$. Twice integrating gives
$$
\boxed{u(y)=-\frac{\Delta p}{2\mu L}y(h-y)+U\frac yh.}
$$
The flux per unit circumference and the total gap flux are
$$
q=\int_0^hu(y)dy
=-\frac{\Delta p\,h^3}{12\mu L}+\frac{Uh}{2},
\qquad
Q\simeq2\pi a q.
$$
The shear at the cylinder is
$$
\boxed{
\tau_h=\mu u_y(h)=\frac{\Delta p\,h}{2L}+\frac{\mu U}{h},}
$$
with the fluid traction on the falling cylinder opposing its motion.
The descending solid displaces volume at rate $\pi a^2U$, so closed-container mass conservation gives
$$
2\pi a\left(-\frac{\Delta p\,h^3}{12\mu L}+\frac{Uh}{2}\right)
=-\pi a^2U.
$$
Since $h\ll a$, the Couette-flux term is smaller, and
$$
\boxed{\Delta p\simeq\frac{6\mu aLU}{h^3}.}
$$
The <pressure> or form drag is
$$
F_p\simeq\pi a^2\Delta p
=\frac{6\pi\mu a^3LU}{h^3}.
$$
Using the pressure-dominated part of $\tau_h$, the side shear force is of order
$$
F_\mu\simeq2\pi aL\frac{\Delta p h}{2L}
=\frac{6\pi\mu a^2LU}{h^2},
$$
so $F_p/F_\mu\simeq a/h\gg1$. Balancing form drag against the excess weight $\Delta\rho g\pi a^2L$ gives
$$
\boxed{U\simeq\frac{\Delta\rho\,g h^3}{6\mu a}.}
$$
This is <annular lubrication drag on a settling cylinder>.
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