= Solution
Let $x^\mu(\tau)$ denote the spacetime coordinates, with dots meaning differentiation with respect to an <affine parameter>; for a massive particle take <proper time> $\tau$. The <metric tensor> determines the quadratic <geodesic Lagrangian> $L=g_{\mu\nu}\dot x^\mu\dot x^\nu$. The <Euler-Lagrange equations> are
$$
\frac d{d\tau}\frac{\partial L}{\partial\dot x^\mu}-\frac{\partial L}{\partial x^\mu}=0.
$$
Symmetry of the <metric tensor> gives $\partial L/\partial\dot x^\mu=2g_{\mu\nu}\dot x^\nu$. Hence the covariant <four-momentum> is $p_\mu=mg_{\mu\nu}\dot x^\nu=(m/2)\partial L/\partial\dot x^\mu$. If $x^\mu$ is a <cyclic coordinate>, its <Euler-Lagrange equation> immediately gives $\dot p_\mu=0$.
For the rotating metric, $t,\phi,z$ are all cyclic. Thus
$$
\boxed{\begin{aligned}
p_t&=m[(c^2-\omega^2r^2)\dot t-\omega r^2\dot\phi],\\
p_\phi&=-mr^2(\dot\phi+\omega\dot t),\qquad p_z=-m\dot z.
\end{aligned}}
$$
Combining the first two equations cancels their $r$ dependence:
$$
\boxed{\dot t=\frac{p_t-\omega p_\phi}{mc^2}=\mathrm{constant}.}
$$
The radial <Euler-Lagrange equation> is $-2\ddot r=-2r(\dot\phi+\omega\dot t)^2$, so
$$
\boxed{\ddot r=r(\dot\phi+\omega\dot t)^2=\frac{p_\phi^2}{m^2r^3}.}
$$
This is the <geodesic equation in a uniformly rotating Minkowski chart>. With inertial angle $\Phi=\phi+\omega t$, $-p_\phi$ is the inertial axial <angular momentum>. The outward radial-coordinate acceleration expresses the familiar centrifugal term associated with angular motion: the underlying free trajectory is a straight line in <Minkowski spacetime>. It does not imply an external radial force. Holding a particle at rest in the <rotating reference frame> instead requires a real inward force.
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