 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ii/paper-4/12c/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

2 E = \sum I_{i} ω_{i}^{2}

and

L^{2} = \sum I_{i}^{2} ω_{i}^{2}

\dot{E} = ω_{1} ω_{2} ω_{3} [(I_{2} - I_{3}) + (I_{3} - I_{1}) + (I_{1} - I_{2})] = 0.

For the squared angular momentum,

\frac{1}{2} \frac{d}{d t} L^{2} = ω_{1} ω_{2} ω_{3} [I_{1} (I_{2} - I_{3}) + I_{2} (I_{3} - I_{1}) + I_{3} (I_{1} - I_{2})] = 0.

Thus both $E$ and $L^{2}$ are conserved, even though the body-frame components of

L

generally change.

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