= Solution
The floor contributes <gravitational potential> $mgz=m\omega^2(x_1^2+y_1^2)/2$. Neglecting <vertical velocity> as specified, for the nonrotating cup orientation $\phi=-\Omega$ the <Lagrangian> is
$$
\boxed{L=\frac m2[(\dot x_1-R\Omega\sin\Omega t)^2+(\dot y_1+R\Omega\cos\Omega t)^2]-\frac{m\omega^2}{2}(x_1^2+y_1^2).}
$$
For general $\alpha=\Omega+\phi$, replace the <velocity> components by $\dot x_1-\alpha y_1+R\Omega\sin\phi t$ and $\dot y_1+\alpha x_1+R\Omega\cos\phi t$. The potential remains the same because the paraboloid is axisymmetric.
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