 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 $\nu =\Omega +\varphi$. In cup components the center's velocity is $\mathbf{w}=R\Omega (\sin \varphi t,\cos \varphi t)$. With $\mathsf{J}(x,y)=(-y,x)$, the horizontal velocity norm is $|\dot{\mathbf{q}}+\nu \mathsf{J}\mathbf{q}+\mathbf{w}|$. ThusIn the nonrotating translating case $\varphi =-\Omega$, this reduces toThe time-only constant $m{R}^2{\Omega }^2/2$ may be dropped. Integrating the cross-velocity term by parts gives an equivalent LagrangianIt 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.