= Solution
Treat the marble as a frictionless point particle; no size, rolling constraint or <friction> law is supplied. Neglect its vertical <kinetic energy> as requested, but retain the gravitational <potential energy> $mgz=m\omega^2(x_1^2+y_1^2)/2$.
For completeness keep the general constant cup <rotation> rate $\nu=\Omega+\phi$. In cup components the center's <velocity> is $\mathbf w=R\Omega(\sin\phi t,\cos\phi t)$. With $\mathsf J(x,y)=(-y,x)$, the horizontal <velocity> norm is $|\dot{\mathbf q}+\nu\mathsf J\mathbf q+\mathbf w|$. Thus
$$
\boxed{L=\frac m2\left[(\dot x_1-\nu y_1+R\Omega\sin\phi t)^2+
(\dot y_1+\nu x_1+R\Omega\cos\phi t)^2\right]
-\frac{m\omega^2}{2}(x_1^2+y_1^2).}
$$
In the nonrotating translating case $\phi=-\Omega$, this reduces to
$$
\boxed{L=\frac m2\left[(\dot x_1-R\Omega\sin\Omega t)^2+
(\dot y_1+R\Omega\cos\Omega t)^2\right]-\frac{m\omega^2}{2}(x_1^2+y_1^2).}
$$
The time-only constant $mR^2\Omega^2/2$ may be dropped. Integrating the cross-velocity term by parts gives an equivalent <Lagrangian>
$$
L'=\frac m2(\dot x_1^2+\dot y_1^2)-\frac{m\omega^2}{2}(x_1^2+y_1^2)
+mR\Omega^2(x_1\cos\Omega t+y_1\sin\Omega t).
$$
It differs by a <total derivative> and a time-only term, so gives the same <trajectories> but different expressions for the <canonical momenta>. The small-vertical-velocity approximation is typically consistent at small surface slope $\omega^2|\mathbf q|/g$; it need not remain valid during unlimited resonant growth.
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