Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2016 / S7 / 1

In a frame rotating with angular velocity $\mathbf{\Omega }$, the Coriolis force is ${\mathbf{F}}_C=-2m\mathbf{\Omega }\times {\mathbf{v}}_{\mathrm{rot}}$. In the northern hemisphere horizontal flow is deflected to the right: balanced circulation is counterclockwise about a low-pressure center and clockwise about a high-pressure center. The deflection reverses in the southern hemisphere. Jupiter's counterclockwise southern Great Red Spot is consequently anticyclonic, a high-pressure system. These statements assume a large-scale rotating atmosphere; they do not explain tiny vortices dominated by local boundary conditions.

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.

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.

For $\varphi =-\Omega$, $p_1=m({\dot{x}}_1-R\Omega \sin \Omega t)$ and $p_2=m({\dot{y}}_1+R\Omega \cos \Omega t)$. The equations are ${\ddot{x}}_1+{\omega }^2{x}_1=R{\Omega }^2\cos \Omega t$, ${\ddot{y}}_1+{\omega }^2{y}_1=R{\Omega }^2\sin \Omega t$. For $\Omega \ne \omega$,At resonance the forced terms are $R\omega t\sin \omega t/2$ and $-R\omega t\cos \omega t/2$. Constants depend on the marble's initial position and velocity. A general cup orientation is obtained by multiplying the vector solution by $\mathcal{R}(-\alpha t)$; the physical inertial solution is unchanged by relabeling the cup axes. Resonant growth eventually invalidates the small-horizontal-motion/neglected-vertical-speed approximation.