 Solution

The cup's rotation relative to the radial basis has angular speed $\varphi$, so its absolute orientation has speed $\alpha =\Omega +\varphi$. With $\mathcal{R}(\theta )$ the planar rotation matrix,At $\varphi =-\Omega$, $\alpha =0$: the cup translates on a circle while its axes remain parallel to fixed inertial axes. Thus $x=R\cos \Omega t+x_1$, $y=R\sin \Omega t+y_1$. Cancellation of orientation rotation does not remove acceleration of the cup's center.