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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 15B / a / Solution

The rotational kinetic energy and squared angular momentum are the torque-free rigid-body invariants

E = \frac{1}{2} (I_{1} ω_{1}^{2} + I_{2} ω_{2}^{2} + I_{3} ω_{3}^{2}),

and

L^{2} = I_{1}^{2} ω_{1}^{2} + I_{2}^{2} ω_{2}^{2} + I_{3}^{2} ω_{3}^{2} .

Differentiating the energy and inserting the Euler equations gives

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

Similarly,

\frac{1}{2} \frac{d L ^{2}}{d t} = I_{1}^{2} ω_{1} \overset{ω}{˙}_{1} + I_{2}^{2} ω_{2} \overset{ω}{˙}_{2} + I_{3}^{2} ω_{3} \overset{ω}{˙}_{3} = 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.

Solved by gpt-5.6-sol high.

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