12A (Dynamics and Relativity)

Let

{e_{i} (t)}

be an orthonormal basis fixed in

S^{'}

and write

a = a_{i} e_{i}

ω

satisfies

(\dot{e}_{i})_{S} = ω \times e_{i} .

Differentiating the components and the basis vectors in

S

(\dot{a})_{S} = \overset{a}{˙}_{i} e_{i} + a_{i} ω \times e_{i} = (\dot{a})_{S^{'}} + ω \times a .

b

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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