 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-2/14b/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Solving the two momentum equations gives

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

Substitution into the conserved energy reduces the heavy symmetric top to one-dimensional effective potential motion:

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

where

V_{eff} (θ) = \frac{( L _{z} - L _{3} cos θ ) ^{2}}{2 I _{1} sin ^{2} θ} + M g l cos θ, E^{'} = E - \frac{L _{3}^{2}}{2 I _{3}} .

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