 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-4/12a/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Apply part (a) first to the position

r

(\dot{r})_{S} = (\dot{r})_{S^{'}} + ω \times r .

Applying it again to both terms on the right gives

(\ddot{r})_{S} = (\ddot{r})_{S^{'}} + ω \times (\dot{r})_{S^{'}} + (\dot{ω})_{S^{'}} \times r + ω \times (\dot{r})_{S},

where

(\dot{ω})_{S} = (\dot{ω})_{S^{'}}

because

ω \times ω = 0

. Substituting the first formula into the last term produces

(\ddot{r})_{S} = (\ddot{r})_{S^{'}} + 2 ω \times (\dot{r})_{S^{'}} + (\dot{ω})_{S^{'}} \times r + ω \times (ω \times r) .

In the Earth-fixed rotating reference frame, the sideways apparent force is the Coriolis acceleration multiplied by the ski-doo's mass:

F_{side} = - 2 m ω \times v .

At the South Pole,

ω

is vertical and antiparallel to the local outward normal, while

v

is horizontal, so

∣ F_{side} ∣ = 2 mω v .

The force points to the left of the direction of travel, which is westward when the ski-doo is moving north after crossing the pole.

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