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ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/section-b/b8/d/solution

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Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / Section B / B8 / d / Solution by

Codex  0 2026-10-20

At first contact the globally highest trajectory just touches the ceiling at its apex. Its flight time to that point is

t_{*} = r ω cos θ_{*} / g

. The horizontal displacement from the wheel axis is

x_{hit} = - r cos θ_{*} + r ω sin θ_{*} \frac{r ω c o s θ _{*}}{g} = cos θ_{*} (- r + \frac{r ^{2} ω ^{2}}{g} sin θ_{*}) = 0.

(1)

Hence the first point struck is directly above the rotation axis, independently of

h

. The release starts to the left of the axis, but the drop's horizontal travel exactly cancels that offset at the optimum apex. The first-contact angular velocity satisfies

r^{2} ω_{*}^{2} = g (h + h^{2} - r^{2})

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