Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 4 / 10C (Dynamics and Relativity)

The term $-2\mathbf{\Omega }\times \dot{\mathbf{x}}$ is the Coriolis acceleration, and $-\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{x})$ is the centrifugal acceleration. With $\mathbf{\Omega }=-\Omega \hat{\mathbf{z}}$, the Cartesian components are

The circular constraint isIts tangent in the $xz$-plane is $(\cos \theta ,0,-\sin \theta )$, while the normal force is radial. Projecting the equations from part (a) onto this tangent eliminates the normal force and givesBecause the ramp is translation-invariant along $y$, $N_y=0$; using $\dot{x}=R\dot{\theta }\cos \theta$ gives

For rest in the rotating frame, $\dot{\theta }=\dot{y}=\ddot{\theta }=\ddot{y}=0$. Assuming $\Omega \ne 0$, the second equation gives $y=0$, and the first givesOn the semicircular ramp the rest points are thereforeand, when ${\Omega }^{2}R\ge g$,

Neglecting terms of order ${\Omega }^2$ and linearizing in $\theta$ givesThe second equation and the initial data implyIts contribution to the first equation is of order ${\Omega }^2$, so the retained equation isThusIntegrating the associated $y$ velocity and using $y(0)=0$ gives