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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/2021/section-a/2/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 / 2021 / Section A / 2 / Solution by

Codex  0 2026-10-18

The original diagram shows two equal balls, each of mass

m

, attached to the fixed upper pivot and light sliding collar by four equal rods. Each ball describes a horizontal circle of radius

r = L sin θ

. The pivot-to-collar distance is

2 L cos θ

, so the upward collar displacement compresses the spring by

δ = 2 L (1 - cos θ) .

Thus Hooke's law gives a downward spring force

F_{s} = k δ

on the collar. Let

T_{u}, T_{l}

be the tensions in the upper and lower rods attached to one ball. Vertical force balance on the light collar and then on either ball gives

2 T_{l} cos θ = F_{s}, T_{u} cos θ - T_{l} cos θ = m g .

The radial force on one ball supplies its centripetal acceleration:

(T_{u} + T_{l}) sin θ = m ω^{2} L sin θ .

For

θ \neq = 0

, adding the vertical-balance results and eliminating the rod tensions gives

ω^{2} = \frac{m g + F _{s}}{m L cos θ} = \frac{g}{L cos θ} + \frac{2 k}{m} \frac{1 - cos θ}{cos θ} .

For the given centrifugal governor,

cos 3 0^{\circ} = 3 /2

and

δ = 0.0535898 m, F_{s} = 26.7949 N, ω^{2} = 133.988 s^{- 2} .

Hence

ω = 11.58 rad s^{- 1} ≃ 110.5 revolutions per minute .

The factor of two in the collar balance matters: both lower rods support the same light collar. The spring force is not an extra downward force of this full magnitude on each ball; its effect on each ball is transmitted through its own lower rod. The formula is the steady spring-loaded centrifugal governor balance.

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