 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-3/8b/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

p_{ϕ} = I_{1} \dot{ϕ} sin^{2} θ + I_{3} (\dot{ψ} + \dot{ϕ} cos θ) cos θ

and

p_{ψ} = I_{3} (\dot{ψ} + \dot{ϕ} cos θ) .

Since the Lagrangian contains neither

ϕ

nor

ψ

, both momenta are constant in time.

Eliminating the two angular velocities gives

\dot{ϕ} = \frac{p _{ϕ} - p _{ψ} c o s θ}{I _{1} s i n ^{2} θ}, \dot{ψ} + \dot{ϕ} cos θ = \frac{p _{ψ}}{I _{3}} .

\frac{1}{2} I_{1} \dot{θ}^{2} + V_{eff} (θ) = E,

where

V_{eff} (θ) = \frac{( p _{ϕ} - p _{ψ} cos θ ) ^{2}}{2 I _{1} sin ^{2} θ} + \frac{p _{ψ}^{2}}{2 I _{3}} + M g l cos θ .

θ \to 0

, the first term diverges unless

p_{ϕ} = p_{ψ} .

Writing their common value as

p

, the small-angle expansion is

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

By the effective potential stability criterion, the vertical state is stable exactly when

p^{2} > 4 I_{1} M g l .

At the vertical state

p = I_{3} (\dot{ψ} + \dot{ϕ})

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