= Solution
Let $\mathsf R(\alpha)$ denote the planar <rotation matrix>, and choose the initial cup center on the positive $x$ axis, as in the requested equations. Its <trajectory> is $\mathbf C(t)=R(\cos\Omega t,\sin\Omega t)$. The handle rotates relative to the radial line at signed rate $\phi$, so its absolute <angular velocity> is $\nu=\Omega+\phi$. A positive recurrence time is $2\pi/|\phi|$; the sign records <rotation> direction.
Writing $\mathbf q=(x_1,y_1)$, the relation is
$$
\boxed{\mathbf r=\mathbf C+\mathsf R(\nu t)\mathbf q,}
$$
or explicitly
$$
x=R\cos\Omega t+x_1\cos\nu t-y_1\sin\nu t,
\qquad
y=R\sin\Omega t+x_1\sin\nu t+y_1\cos\nu t.
$$
One obtains the same result by rotating cup coordinates through $\phi t$ into the radially aligned axes, then rotating through $\Omega t$ into the inertial axes.
At $\phi=-\Omega$, $\nu=0$, and
$$
\boxed{x=R\cos\Omega t+x_1,\qquad y=R\sin\Omega t+y_1.}
$$
The axes maintain a fixed orientation in inertial space while their origin accelerates around a circle. This is translation without <rotation> of the cup frame; it is not an inertial frame because the origin accelerates. In particular its inertial correction is a spatially uniform fictitious force, rather than a <Coriolis force> from <rotation> of these axes.
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