 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-4/11c/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

With total mass

M = \sum_{i} m_{i}

R = \frac{1}{M} \sum_{i} m_{i} x_{i}, P = \sum_{i} m_{i} \overset{x}{˙}_{i}, L = \sum_{i} x_{i} \times m_{i} \overset{x}{˙}_{i} .

The angular momentum about the centre of mass is

L_{CoM} = \sum_{i} (x_{i} - R) \times m_{i} (\overset{x}{˙}_{i} - \dot{R}) .

Expanding, using

\sum_{i} m_{i} (x_{i} - R) = 0

and

P = M \dot{R}

, gives

L_{CoM} = L - R \times P .

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