 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/8b/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Take

θ

to be the angle between

e_{3}

and the upward vertical. From the supplied angular-velocity components,

T = \frac{I}{2} (\dot{θ}^{2} + sin^{2} θ \dot{ϕ}^{2}) + \frac{J}{2} (\dot{ψ} + cos θ \dot{ϕ})^{2}, V = M g a cos θ .

Therefore

L = \frac{I}{2} (\dot{θ}^{2} + sin^{2} θ \dot{ϕ}^{2}) + \frac{J}{2} (\dot{ψ} + cos θ \dot{ϕ})^{2} - M g a cos θ .

Both

ψ

and

ϕ

are cyclic. Their conserved conjugate momenta are

p_{ψ} = J (\dot{ψ} + cos θ \dot{ϕ}), p_{ϕ} = I sin^{2} θ \dot{ϕ} + p_{ψ} cos θ,

and

p_{θ} = I \dot{θ}

. Solving for the velocities and taking the Legendre transform gives

H = \frac{p _{θ}^{2}}{2 I} + \frac{( p _{ϕ} - p _{ψ} cos θ ) ^{2}}{2 I sin ^{2} θ} + \frac{p _{ψ}^{2}}{2 J} + M g a cos θ .

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