Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 14B / iv

Table of contents
- Solution iv

Solution

Put

L_{z} = L_{3} (1 - ϵ^{2} /2)

and seek

θ = O (ϵ)

. Using

cos θ = 1 - θ^{2} /2 + O (θ^{4})

and

sin^{2} θ = θ^{2} + O (θ^{4})

, the

θ

-dependent part of the effective potential is

V_{eff} (θ) = constant + \frac{L _{3}^{2} - 4 M g l I _{1}}{8 I _{1}} θ^{2} + \frac{L _{3}^{2} ϵ ^{4}}{8 I _{1} θ ^{2}} + O (ϵ^{4}) .

It has a local minimum precisely when

L_{3}^{2} > 4 M g l I_{1},

which is the gyroscopic stabilization of an inverted symmetric top condition. Differentiating the displayed approximation gives

θ_{eq}^{4} = \frac{L _{3}^{2} ϵ ^{4}}{L _{3}^{2} - 4 M g l I _{1}},

and hence

θ_{eq} = ϵ (\frac{L _{3}^{2}}{L _{3}^{2} - 4 M g l I _{1}})^{1/4} .

Taking

L_{3} > 0

without loss of orientation, the steady precession rate is

\dot{ϕ} = \frac{L _{z} - L _{3} c o s θ _{eq}}{I _{1} s i n ^{2} θ _{eq}} = \frac{L _{3}}{2 I _{1}} (1 - \frac{ϵ ^{2}}{θ _{eq}^{2}}) + O (ϵ^{2}),

so

\dot{ϕ} = \frac{L _{3} - L _{3}^{2} - 4 M g l I _{1}}{2 I _{1}} + O (ϵ^{2}) .

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