 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/ia/paper-4/11b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-07

Write the position in the rotating basis as

r = \sum_{i} x_{i}^{'} e_{i}^{'}

. Differentiating with respect to time in the inertial frame,

\frac{d r}{d t} = \sum_{i} \overset{x}{˙}_{i}^{'} e_{i}^{'} + \sum_{i} x_{i}^{'} \frac{d e _{i}^{'}}{d t} = v^{'} + ω \times \sum_{i} x_{i}^{'} e_{i}^{'} .

Here

v^{'} = \sum_{i} \overset{x}{˙}_{i}^{'} e_{i}^{'}

is the relative velocity measured in the rotating reference frame. Consequently

v = v^{'} + ω \times r .

This is the vector differentiation rule for a rotating reference frame with a common origin: the extra term accounts for the changing basis, not a translational motion of that origin.

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