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ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2014/paper-1/1/a/solution

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Past exam of the theoretical physics I course of the University of Cambridge / 2014 / Paper 1 / 1 / a / Solution by

Codex  0 2026-10-20

Measure both angles from the downward vertical, as in the source diagram. With upward

y

, the cylinder centres are

R = (2 a sin θ, - 2 a cos θ), r = R + (a sin ϕ, - a cos ϕ) .

The large hoop has central inertia

4 M a^{2}

and angular velocity

\dot{θ}

, while the small hoop has no spin in this part. Translation and rotation therefore give

T = \frac{1}{2} M a^{2} [12 \dot{θ}^{2} + \dot{ϕ}^{2} + 4 cos (θ - ϕ) \dot{θ} \dot{ϕ}], V = - M g a (4 cos θ + cos ϕ) .

Thus

L = T - V

. Counting each angle and velocity as first order, replacing the cosine in the velocity product by one changes only fourth-order terms. Expanding the potential and dropping its constant gives

L_{2} = \frac{1}{2} M a^{2} (12 \dot{θ}^{2} + \dot{ϕ}^{2} + 4 \dot{θ} \dot{ϕ}) - \frac{1}{2} M g a (4 θ^{2} + ϕ^{2}) .

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