Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 4 / 11B (Dynamics)

Let ${\mathbf{e}}_1^{\prime }(t),{\mathbf{e}}_2^{\prime }(t),{\mathbf{e}}_3^{\prime }(t)$ be an orthonormal basis fixed in the rotating reference frame $S^{\prime }$. A small rotation through $\mathbf{\omega }dt$ changes each basis vector by $\mathbf{\omega }\times {\mathbf{e}}_j^{\prime }dt$, so its derivative in the inertial frame isWrite the arbitrary vector as $\mathbf{a}={\sum }_j{a}_j(t){\mathbf{e}}_j^{\prime }(t)$. Differentiating the components and the moving basis separately givesThis proves the rotating-frame derivative formula, including when the angular velocity varies with time.

Use $D=(d/dt{)}_S$, $D^{\prime }=(d/dt{)}_{S^{\prime }}$, and ${\mathbf{v}}^{\prime }=D^{\prime }\mathbf{r}$. By the rotating-frame derivative formula, $D\mathbf{r}={\mathbf{v}}^{\prime }+\mathbf{\omega }\times \mathbf{r}$. Apply the product rule for the cross product and the same rotating-frame derivative formula again:We have $D{\mathbf{v}}^{\prime }=D^{\prime }{\mathbf{v}}^{\prime }+\mathbf{\omega }\times {\mathbf{v}}^{\prime }$ and $D\mathbf{\omega }=D^{\prime }\mathbf{\omega }$, since $\mathbf{\omega }\times \mathbf{\omega }=0$. ThereforeRearranging this identity into an equation of motion in a rotating frame produces the apparent Coriolis acceleration, Euler acceleration and centrifugal acceleration, with signs opposite to their corresponding terms on the right of this inertial acceleration identity. No translational term is needed because the two origins coincide.

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.