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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2/7/a/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 7 / a / Solution by

Codex  0 2026-10-18

Take downward vertical distances from the support. The climber positions are

q_{1} = L_{1}

,

q_{2} = L_{1} + L_{2}

and

q_{3} = L_{1} + L_{3}

. Their kinetic energy is therefore

T = \frac{1}{2} m_{1} \dot{L}_{1}^{2} + \frac{1}{2} m_{2} (\dot{L}_{1} + \dot{L}_{2})^{2} + \frac{1}{2} m_{3} (\dot{L}_{1} + \dot{L}_{3})^{2} .

Integrating the tension-extension relation gives the elastic potential energy

k (L_{i} - L_{0})^{2} /2

for each taut rope. Because downward displacement lowers gravitational potential,

U = \frac{k}{2} i = 1 \sum 3 (L_{i} - L_{0})^{2} - g [(m_{1} + m_{2} + m_{3}) L_{1} + m_{2} L_{2} + m_{3} L_{3}], L = T - U .

An arbitrary additive constant in

U

is immaterial. This is the vertical-motion model implicit in using only rope lengths as generalized coordinates; swinging angles are not additional degrees of freedom in this model.

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