Past exam of the short options paper shared by Parts A and B of the physics course of the University of Oxford / 2019 / S7 / 2

Write the geometric vector as $\mathbf{V}={\sum }_i{V}_i^{\prime }{\mathbf{e}}_i^{\prime }(t)$. A rotating orthonormal basis obeys ${\dot{\mathbf{e}}}_i^{\prime }=\mathbf{\Omega }\times {\mathbf{e}}_i^{\prime }$. The product rule gives the transport identityBoth terms must be expressed in the same basis before adding them. If $R(t)$ maps rotating components into inertial components, the precise component equation is $d(R{\mathbf{V}}^{\prime })/dt=R({\dot{\mathbf{V}}}^{\prime }+{\mathbf{\Omega }}^{\prime }\times {\mathbf{V}}^{\prime })$. The printed expression suppresses this basis conversion.

The inertial Lagrangian is $L=m|\dot{\mathbf{r}}{|}^2/2-U(\mathbf{r},t)$. Put $\mathbf{r}=R(t){\mathbf{r}}^{\prime }$ and define $U^{\prime }({\mathbf{r}}^{\prime },t)=U(R(t){\mathbf{r}}^{\prime },t)$. The transport identity and invariance of the Euclidean norm under a rotation giveIts expansion contains the ordinary kinetic energy, $m{\dot{\mathbf{r}}}^{\prime }\cdot ({\mathbf{\Omega }}^{\prime }\times {\mathbf{r}}^{\prime })$, and $m|{\mathbf{\Omega }}^{\prime }\times {\mathbf{r}}^{\prime }{|}^2/2$. These velocity and position terms generate the apparent forces in the rotating frame.

Suppress primes on the angular velocity while retaining rotating components. The canonical momentum is $\partial L^{\prime }/\partial {\dot{\mathbf{r}}}^{\prime }=m({\dot{\mathbf{r}}}^{\prime }+\mathbf{\Omega }\times {\mathbf{r}}^{\prime })$. Varying $L^{\prime }$ with respect to position and applying the Euler-Lagrange equations givesThe terms are the physical force, the velocity-dependent Coriolis force, the outward centrifugal force, and the Euler force caused by changing angular velocity. The common origin is inertial; an accelerating origin would introduce another translational apparent force.

At latitude $\ell$, the Earth's angular velocity in east–north–up coordinates is $\mathbf{\Omega }=(0,\Omega \cos \ell ,\Omega \sin \ell )$. Assume a small horizontal patch, constant $\Omega$, negligible vertical velocity, and a vertical constraint balancing effective gravity. Absorb the steady centrifugal force into that effective gravity. With $f=2\Omega \sin \ell$ and physical horizontal force per unit mass $F_x^{\prime },F_y^{\prime }$, the Coriolis force givesLarge-scale atmospheric and oceanic flows deflect to their right in the northern hemisphere and left in the southern hemisphere; their balance with a horizontal pressure gradient produces geostrophic flow. A Foucault pendulum has an oscillation plane that precesses relative to the ground at angular rate $-\Omega \sin \ell$. Both effects vanish at the equator within this horizontal approximation.

Let $m$ be the ice mass for which the stated drag coefficients are defined, $K=k_1+k_2$ and $f=2\Omega \sin \ell$. With stationary water, the steady horizontal ice equations areThe linear system givesFor fixed wind speed $V=10\mathrm{m}{\mathrm{s}}^{-1}$, the Cauchy-Schwarz inequality minimizes this component whenThus the wind must blow towards a direction east of south by ${\tan }^{-1}(mf/K)$ in the northern hemisphere, west of south in the southern hemisphere, and due south at the equator. This compensates the Coriolis force; at the optimum $v_x=0$ and the ice moves due south. If the drag coefficients are per unit mass, replace $K/m$ by the corresponding total drag rate. “Blows towards” avoids the opposite convention used when naming winds by their origin.