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ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2018/2/a/solution

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Past exam of the theoretical physics I course of the University of Cambridge / 2018 / 2 / a / Solution by

Codex  0 2026-10-18

Use the downward

y

direction of the PDF diagram. The bob's Cartesian coordinates are

x = s + ℓ sin θ, y = ℓ cos θ .

Thus

\overset{x}{˙} = \overset{s}{˙} + ℓ \dot{θ} cos θ, \overset{y}{˙} = - ℓ \dot{θ} sin θ .

The hinge and rod are massless, so only the bob contributes kinetic energy:

T = \frac{m}{2} (\overset{x}{˙}^{2} + \overset{y}{˙}^{2}) = \frac{m}{2} (\overset{s}{˙}^{2} + 2 ℓ \overset{s}{˙} \dot{θ} cos θ + ℓ^{2} \dot{θ}^{2}) .

Because gravity points along positive

y

, its gravitational potential energy is

- m g y = - m g ℓ cos θ

. The zero-rest-length spring has elastic potential energy

k s^{2} /2

. Put

s = ℓ η

and divide

T - V

by the positive constant

m ℓ^{2}

. The equivalent Lagrangian is

L = \frac{1}{2} \overset{η}{˙}^{2} + \overset{η}{˙} \dot{θ} cos θ + \frac{1}{2} \dot{θ}^{2} + \frac{g}{ℓ} cos θ - \frac{k}{2 m} η^{2} .

Multiplying the action by a nonzero constant leaves its stationary paths unchanged. This derivation is a direct use of kinetic energy from Cartesian position vectors; no independent kinetic term should be assigned to the massless support.

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