 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 $\varphi =-\Omega$ the Lagrangian isFor general $\alpha =\Omega +\varphi$, replace the velocity components by ${\dot{x}}_1-\alpha y_1+R\Omega \sin \varphi t$ and ${\dot{y}}_1+\alpha x_1+R\Omega \cos \varphi t$. The potential remains the same because the paraboloid is axisymmetric.