 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-iv/6/a/solution

All time derivatives in this part are taken relative to the rotating reference frame, so its basis is held fixed during differentiation. Put

f (ϕ) = p (ϕ) sin ϕ

and

q (ϕ) = p (ϕ) cos ϕ

\ddot{r} = \ddot{ϕ} (f^{'} e_{1} - q^{'} e_{3}) + \dot{ϕ}^{2} (f^{''} e_{1} - q^{''} e_{3}) .

Explicitly,

f^{'} q^{'} = p^{'} sin ϕ + p cos ϕ, = p^{'} cos ϕ - p sin ϕ, f^{''} q^{''} = p^{''} sin ϕ + 2 p^{'} cos ϕ - p sin ϕ, = p^{''} cos ϕ - 2 p^{'} sin ϕ - p cos ϕ .

The separate Coriolis force and centrifugal force terms in the stated equation account for the motion of the basis relative to an inertial observer.

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