= Solution
The cup's rotation relative to the radial basis has <angular speed> $\phi$, so its absolute orientation has <speed> $\alpha=\Omega+\phi$. With $\mathcal R(\theta)$ the <planar rotation> matrix,
$$
\boxed{\binom{x}{y}=R\binom{\cos\Omega t}{\sin\Omega t}+\mathcal R(\alpha t)\binom{x_1}{y_1}.}
$$
At $\phi=-\Omega$, $\alpha=0$: 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.
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