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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2 / 14A / a / Solution

Let

r

be the cylindrical radius of the point of contact and

ϕ

its azimuth. Rolling without slip or twist gives

∣ ω ∣ = v / a

and therefore

L = \frac{7 m}{10} (\overset{r}{˙}^{2} csc^{2} α + r^{2} \dot{ϕ}^{2}) - m g r cot α,

up to an irrelevant additive constant. The factor

7/5

combines translational and rotational inertia.

Solved by gpt-5.6-sol high.

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