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

Solution

Subtracting

I_{2}

times the energy identity from the angular-momentum identity, under

L^{2} = 2 E I_{2}

, gives

I_{1} (I_{2} - I_{1}) ω_{1}^{2} = I_{3} (I_{3} - I_{2}) ω_{3}^{2} .

Combining this with

I_{1} ω_{1}^{2} + I_{3} ω_{3}^{2} = 2 E - I_{2} ω_{2}^{2} = \frac{L ^{2}}{I _{2}} - I_{2} ω_{2}^{2}

yields

ω_{1}^{2} = \frac{I _{3} - I _{2}}{I _{1} ( I _{3} - I _{1} )} (2 E - I_{2} ω_{2}^{2})

and

ω_{3}^{2} = \frac{I _{2} - I _{1}}{I _{3} ( I _{3} - I _{1} )} (2 E - I_{2} ω_{2}^{2}) .

The second Euler equation now gives, after squaring or choosing one orientation of the orbit,

\overset{ω}{˙}_{2} = \pm \frac{( I _{3} - I _{2} ) ( I _{2} - I _{1} )}{I _{1} I _{3}} (μ^{2} - ω_{2}^{2}), μ = \frac{2 E}{I _{2}} .

Indeed, the factor

I_{3} - I_{1}

I_{2} \overset{ω}{˙}_{2} = (I_{3} - I_{1}) ω_{3} ω_{1}

cancels the denominator from the product of the two preceding expressions.

For the plus sign, separation gives

artanh \frac{ω _{2}}{μ} = μ \frac{( I _{3} - I _{2} ) ( I _{2} - I _{1} )}{I _{1} I _{3}} (t - t_{0}) .

After choosing the time origin

t_{0} = 0

ω_{2} (t) = μ tanh (λ t), λ = μ \frac{( I _{3} - I _{2} ) ( I _{2} - I _{1} )}{I _{1} I _{3}} .

For example, a compatible choice of the other components is

ω_{1} (t) = \frac{2 E ( I _{3} - I _{2} )}{I _{1} ( I _{3} - I _{1} )} sech (λ t),

ω_{3} (t) = \frac{2 E ( I _{2} - I _{1} )}{I _{3} ( I _{3} - I _{1} )} sech (λ t) .

t

runs from

- \infty

+ \infty

, the solution approaches the rotations

ω_{2} = - μ

and

ω_{2} = + μ

. It is the intermediate-axis separatrix, a heteroclinic trajectory that exhibits the instability found in part (b).

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