Solution

Fix the two conserved momenta

p_{ψ} = λ

and

p_{ϕ} = μ

. The remaining canonical variables

(θ, p_{θ})

form a two-dimensional phase space with Hamiltonian

H_{λ, μ} (θ, p_{θ}) = \frac{p _{θ}^{2}}{2 I} + V_{λ, μ} (θ),

where

V_{λ, μ} (θ) = \frac{( μ - λ c o s θ ) ^{2}}{2 I s i n ^{2} θ} + \frac{λ ^{2}}{2 J} + M g a cos θ .

An upright spinning solution has

θ = 0

p_{θ} = 0

, and necessarily

μ = λ

, since the vertical and body-axis angular momenta then coincide. On that momentum level,

V_{λ, λ} (θ) = \frac{λ ^{2}}{2 I} tan^{2} \frac{θ}{2} + \frac{λ ^{2}}{2 J} + M g a cos θ .

Near zero,

V_{λ, λ} (θ) = constant + (\frac{λ ^{2}}{8 I} - \frac{M g a}{2}) θ^{2} + O (θ^{4}) .

Thus the upright configuration is a strict local minimum of the reduced Hamiltonian, and hence stable, when

λ^{2} > 4 I M g a = 3 I M g l .

Here

λ

is the angular momentum about the symmetry axis. This proves stability for sufficiently large axial angular momentum.

Solved by gpt-5.6-sol high.

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