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 Incoming links: Heavy symmetric top

Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 2 / 7D / a / Solution Created 2026-09-24 Updated 2026-10-06

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The two cyclic Euler angles give conserved canonical momenta

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

(1)

Here

L_{3}

is the body-axis angular momentum and

L_{z}

the vertical angular momentum. Thus $ω_{3} = L_{3} / I_{3}$ is constant. Since the Lagrangian has no explicit time dependence, the energy is also conserved:

E = \frac{I _{1}}{2} \dot{θ}^{2} + \frac{I _{1}}{2} \dot{ϕ}^{2} sin^{2} θ + \frac{L _{3}^{2}}{2 I _{3}} + M g l cos θ .

(2)

These are the three independent conserved quantities furnished by the continuous symmetries of the heavy symmetric top.

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Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 14B / iii / Solution Created 2026-09-24 Updated 2026-09-29

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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}} .

(1)

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^{'},

(2)

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}} .

(3)

This is the heavy symmetric top reduction.

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Uniform precession of a heavy symmetric top 2026-10-06

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A heavy symmetric top precesses uniformly when its inclination is constant and its azimuthal speed is constant. For a nonvertical inclination, the condition is

I_{1} cos θ Ω^{2} - I_{3} ω_{3} Ω + M g l = 0

. The real-root condition depends on whether the body axis points above or below the horizontal.

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