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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-2/8a/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2 / 8A / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

With

x = ℓ sin θ

and

y = - a t^{2} /2 - ℓ cos θ

, discard a total derivative from the kinetic energy to obtain

L = \frac{1}{2} m ℓ^{2} \dot{θ}^{2} + m ℓ (g - a) cos θ .

(1)

Thus $\ddot{θ} + (g - a) sin θ / ℓ = 0$ . If $a = g$ , then $\ddot{θ} = 0$ : the pendulum is weightless in the lift and rotates uniformly (or remains fixed).

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