 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ii/paper-2/7d/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

Eliminate the cyclic velocities using

\dot{ϕ} = (L_{z} - L_{3} cos θ) / (I_{1} sin^{2} θ)

. The conserved energy becomes

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

Differentiating the effective potential gives

- \frac{d V _{eff}}{d θ} = I_{1} \dot{ϕ}^{2} sin θ cos θ - L_{3} \dot{ϕ} sin θ + M g l sin θ,

which is also the

θ

Euler-Lagrange equation. Hence $I_{1} \ddot{θ} = - V_{eff}^{'} (θ)$ . Deriving it directly from the Euler-Lagrange equation ensures it also holds at turning points, where division by

\dot{θ}

in an energy argument would be invalid.

 New to topics? Read the docs here!