= Solution
<Pressure> acts normally to the circular cylinder and so has no moment about its axis. To leading lubrication order, the tangential shear at the cylinder is
$$
\tau_c=\mu\left.\frac{\partial u}{\partial y}\right|_{y=h}
=\frac{\mu(a\Omega-V)}h+\frac h2\frac{dp}{dx}.
$$
Substitution of the <pressure> <gradient> gives
$$
\tau_c=\mu\left[\frac{4a\Omega+2V}{h}-\frac{6Q}{h^2}\right].
$$
A torque-free cylinder requires $\int_{-\infty}^{\infty}\tau_c\,dx=0$. Also,
$$
\int h^{-1}dx=\pi\sqrt{2ah_0}\,h_0^{-1},
\qquad
\int h^{-2}dx=\frac{\pi}{2}\sqrt{2ah_0}\,h_0^{-2}.
$$
Using $Q=2h_0(V+a\Omega)/3$, the <torque> condition reduces to
$$
(4a\Omega+2V)-2(V+a\Omega)=2a\Omega=0.
$$
Hence
$$
\boxed{\Omega=0.}
$$
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