 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/15b/b/solution

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

Setting

ω_{1} = ω_{3} = 0

makes all three Euler equations stationary. The invariants then give

E = \frac{1}{2} I_{2} ω_{2}^{2}, L^{2} = I_{2}^{2} ω_{2}^{2} = 2 E I_{2},

so the two rotations are

ω_{2} = \pm \frac{2 E}{I _{2}} = \pm \frac{L}{I _{2}},

where

L = L^{2}

Linearize about

(0, Ω, 0)

, where

Ω

is either of these values. Writing the transverse perturbations as

(ξ, ζ) = (δ ω_{1}, δ ω_{3})

gives

I_{1} \dot{ξ} = (I_{2} - I_{3}) Ω ζ, I_{3} \dot{ζ} = (I_{1} - I_{2}) Ω ξ .

Consequently

\ddot{ξ} = Ω^{2} \frac{( I _{3} - I _{2} ) ( I _{2} - I _{1} )}{I _{1} I _{3}} ξ .

The transverse eigenvalues are therefore

s = \pm ∣Ω∣ \frac{( I _{3} - I _{2} ) ( I _{2} - I _{1} )}{I _{1} I _{3}},

one of which is positive. Both rotations are linearly unstable, which is the intermediate axis theorem.

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