Past exam of the classical mechanics course of Part B of the physics course of the University of Oxford / 2016 / 1 / d

For the original Lagrangian of solution, differentiation with respect to the generalized velocities givesIn contrast, using $L^{\prime }$ in the nonrotating case gives $p_{x_1}^{\prime }=m{\dot{x}}_1$, $p_{y_1}^{\prime }=m{\dot{y}}_1$; this is the expected momentum shift under adding a total derivative.

The Euler-Lagrange equations in the general cup frame areThese include the Coriolis and centrifugal terms when the cup axes really rotate. Set $z_q=x_1+i{y}_1$ to obtainFor ${\omega }^2\ne {\Omega }^2$, its complete solution iswhere the two complex constants provide the four real initial data. The driving denominator is independent of cup-axis rotation: rotation changes the coordinate description, not the physical trap frequency.

In the requested special case $\nu =0$, an equivalent real form isFor example $a_x=x_1(0)-D$, $b_x={\dot{x}}_1(0)/\omega$, $a_y=y_1(0)$ and $b_y=[{\dot{y}}_1(0)-D\Omega ]/\omega$ determine the constants.

At resonance ${\omega }^2={\Omega }^2$ with nonzero $\Omega$, replace the particular term byIn the nonrotating cup frame this has components $(R\Omega t\sin \Omega t/2,-R\Omega t\cos \Omega t/2)$, plus the homogeneous solution. Secular growth eventually invalidates a finite cup or the neglected vertical kinetic energy. If the ride does not rotate, its forcing vanishes and ordinary free oscillations remain.