It's a bit late and I'm a little tired, but I tried to calculate the moment of the Coriolis force experienced by a rotating ring of mass. Let ##\{ \mathbf{i}, \mathbf{j}, \mathbf{k} \}## be a vector basis with ##\mathbf{k}## in the direction of the local vertical at a point on the Earth's surface at polar angle ##\theta##, and ##\{ \mathbf{i}, \mathbf{j} \}## tangent to the surface. The wheel has an angular velocity of approximately ##\boldsymbol{\omega} = \omega \mathbf{k}## so long as ##\omega \gg \Omega##.
Let ##\hat{\mathbf{z}}## be a unit vector aligned with the rotation axis of the Earth, then ##\boldsymbol{\Omega} = \Omega \hat{\mathbf{z}}##. Let ##\mathbf{r}## be the radius vector from the centre of the ring to an arbitrary point on the edge. The moment of the Coriolis force on the wheel about its centre is\begin{align*}
\mathbf{M} &= -2\int_C \mathbf{r} \times (\boldsymbol{\Omega} \times (\boldsymbol{\omega} \times \mathbf{r})) dm \\ \\
&= -2 \int_{0}^{2\pi} \mathbf{r} \times ((\boldsymbol{\Omega} \cdot \mathbf{r}) \boldsymbol{\omega} - (\boldsymbol{\Omega} \cdot \boldsymbol{\omega})\mathbf{r}) \rho r d\varphi \\ \\
&= -2 \int_{0}^{2\pi} (\boldsymbol{\Omega} \cdot \mathbf{r}) (\mathbf{r} \times \boldsymbol{\omega}) \rho r d\varphi \\ \\
&= 2 \rho \int_0^{2\pi} (\Omega r \cos{\alpha} ) (r\omega \hat{\boldsymbol{\varphi}}) r d\varphi \\ \\
&= 2 \rho \Omega \omega r^3 \int_0^{2\pi} \cos{\alpha} \hat{\boldsymbol{\varphi}} d\varphi
\end{align*}where ##\cos{\alpha} = \sin{\theta} \sin{\varphi}##. Therefore\begin{align*}
\mathbf{M} &= 2 \rho \Omega \omega r^3 \sin{\theta} \int_{0}^{2\pi} \sin{\varphi} \hat{\boldsymbol{\varphi}} d\varphi \\ \\
&= 2 \rho \Omega \omega r^3 \sin{\theta} \left[ -\mathbf{i} \int_0^{2\pi} \sin^2{\varphi} d\varphi + \mathbf{j}\int_0^{2\pi} \sin{\varphi} \cos{\varphi} d\varphi\right] \\ \\
&= -2\pi \rho \Omega \omega r^3 \sin{\theta} \mathbf{i}
\end{align*}If I didn't screw up the calculation, then ##\mathbf{M}## points in the ##- \mathbf{i}## direction [tangent to the surface, anti-parallel to the direction of motion of the local surface of Earth w.r.t a space-fixed frame]. Thus it doesn't seem as if the Coriolis torque points in the right direction to change ##\omega##...
... but I reckon I should actually have written ##\mathbf{v}' = \boldsymbol{\omega} \times \mathbf{r} + V \hat{\boldsymbol{\theta}}##, to account for the small velocity parallel to the lines of longitude as the ring is moved slowly Southward, which might give rise to a term in ##\mathbf{M}## parallel to ##\mathbf{k}##, or maybe not? I'll check tomorrow! Perhaps we actually require an external couple ##\mathbf{C}## applied to the axis.