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ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2012/paper-1/1/c/solution

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Past exam of the theoretical physics I course of the University of Cambridge / 2012 / Paper 1 / 1 / c / Solution by

Codex  0 2026-10-20

Write

M = m_{1} + m_{2}

. Integrating the zero-momentum condition gives

M x + m_{2} l sin ϕ = M X_{c}

, where

X_{c}

is the constant horizontal centre-of-mass position. Therefore the bob coordinates are

X = x + l sin ϕ = X_{c} + \frac{m _{1} l}{M} sin ϕ, Y = - l cos ϕ .

Eliminating the angle gives the elliptic bob path at zero horizontal momentum,

\frac{( X - X _{c} ) ^{2}}{( m _{1} l / M ) ^{2}} + \frac{Y ^{2}}{l ^{2}} = 1.

The semiaxes are

m_{1} l / M

horizontally and

l

vertically. An oscillation traverses only the allowed arc, reversing at its turning points. The support counter-moves to keep the horizontal centre of mass fixed. The heavy-support limit tends to the familiar fixed-pivot circle.

Figure 1.
Pendulum ellipse and support motion
. Details.

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