= Solution
If $\mathbf r$ is measured in the rotating frame, its inertial velocity is $\dot{\mathbf r}+\boldsymbol\omega\times\mathbf r$. The masses producing the gravitational field are stationary in this frame, so their <potential energy> is a time-independent function $V(\mathbf r)$. Thus kinetic minus potential energy gives the stated <Lagrangian>
$$
L=\frac m2(\dot{\mathbf r}+\boldsymbol\omega\times\mathbf r)^2-V(\mathbf r).
$$
Put $\mathbf v=\dot{\mathbf r}+\boldsymbol\omega\times\mathbf r$. Then
$$
\frac{\partial L}{\partial\dot{\mathbf r}}=m\mathbf v,
\qquad
\frac{\partial L}{\partial\mathbf r}=-m\boldsymbol\omega\times\mathbf v-\nabla V.
$$
Substitution in the <Euler-Lagrange equation> yields
$$
\boxed{m\left(\ddot{\mathbf r}+2\boldsymbol\omega\times\dot{\mathbf r}
+\dot{\boldsymbol\omega}\times\mathbf r
+\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r)\right)
=-\nabla V}.
$$
The extra terms are the <Coriolis acceleration>, <Euler acceleration>, and <centrifugal acceleration> of a <rotating reference frame>.
The <canonical momentum> and <Hamiltonian> are
$$
\boxed{\mathbf p=m(\dot{\mathbf r}+\boldsymbol\omega\times\mathbf r)},
$$
$$
\boxed{H(\mathbf r,\mathbf p,t)
=\frac{\mathbf p^2}{2m}
-\boldsymbol\omega(t)\mathbin\cdot(\mathbf r\times\mathbf p)
+V(\mathbf r)}.
$$
Accordingly, <Hamilton's equations> are
$$
\boxed{\dot{\mathbf r}=\frac{\mathbf p}{m}-\boldsymbol\omega\times\mathbf r,
\qquad
\dot{\mathbf p}=-\boldsymbol\omega\times\mathbf p-\nabla V}.
$$
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