Foucault's pendulum without Coriolis force

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In spherical coordinates, Foucault's pendulum without Coriolis force can be described by the Lagrangian action
$$S_L[\theta, \phi] = \int_{t_0}^{t_1} \mathbb{d}t\left\{\frac{1}{2}m l^2 \left(\dot\theta^2 + \dot\phi^2\sin^2\theta\right) + m l^2 \omega_0^2 \cos\theta\right\}\,.$$

$\phi$ is a cyclic coordinate,
$$\phi = \omega_\phi\cdot (t-t_0)\,.$$
We arrive at
$$\frac{1}{2} \left(\left(\frac{1}{t'(\theta)}\right)^2 + \omega_\phi^2\sin^2\theta\right) - \omega_0^2 \cos\theta = \omega_0^2 \cos\theta_0\,.$$
Mathematica suggests this is integrable.

The spherical coordinates, in my humble opinion, is more natural than Cartesian coordinates in this use case, which was recorded in https://link.springer.com/book/10.1007/978-3-642-03434-3. My argument is that the system is constrained in the sphere with radius $l$.

However, if we add Coriolis force, things might become different. We may be forced back to Cartesian coordinates.

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The Lagrangian of a particle in a rotating reference system is given in sec. 39, https://doi.org/10.1016/B978-0-08-050347-9.50011-3. We can use it to include the Coriolis force.