Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 4 / 11B / Solution

Let $\Omega$ be the Earth's angular velocity magnitude. Use the local east, north and upward unit vectors ${\mathbf{e}}_E,{\mathbf{e}}_N,{\mathbf{e}}_U$, with ${\mathbf{e}}_E\times {\mathbf{e}}_N={\mathbf{e}}_U$. At latitude $\lambda$, the Earth's angular velocity and the train's ground-relative velocity areThus the east component of the Coriolis force isFor a north–south track, the train has no eastward acceleration relative to Earth. Its path curvature lies in the local north–vertical plane. The centrifugal acceleration and Newtonian gravity also lie in that plane, and the Earth's constant angular velocity gives no Euler acceleration. Hence the sideways force from the track must cancel the eastward Coriolis force:Its magnitude is $2m\Omega V\sin \lambda$, and its direction is west in the Northern hemisphere. The eastward tendency is the apparent Coriolis force; the requested track force has the opposite direction.