 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-4/15e/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

The initial relation implies

L^{2} = 2 I_{2} T

, and conservation preserves it. With

I_{3} = I_{1} + I_{2}

, this identity reduces to

ω_{3}^{2} = \frac{I _{2} - I _{1}}{I _{1} + I _{2}} ω_{1}^{2} .

The energy then gives

ω_{2}^{2} = \frac{2 T}{I _{2}} - ω_{1}^{2} .

The first Euler equation is

\overset{ω}{˙}_{1} = - ω_{2} ω_{3}

. Squaring and substituting the preceding identities yields

\overset{ω}{˙}_{1}^{2} = (\frac{2 T}{I _{2}} - ω_{1}^{2}) \frac{I _{2} - I _{1}}{I _{2} + I _{1}} ω_{1}^{2} .

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