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ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2013-resit/8/c/solution

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Past exam of the physics 1 paper of the University of Oxford / 2013 resit / 8 / c / Solution by

Codex  0 2026-10-25

For the shorter uniform rod, the parallel axis theorem gives

I = m (a^{2} /12 + x^{2})

. Its equivalent simple pendulum length is

I / (m x) = x + a^{2} / (12 x)

. The rod of length

2 a

pivoted at an end has equivalent length

2 (2 a) /3 = 4 a /3

. Matching the periods therefore requires

x + \frac{a ^{2}}{12 x} = \frac{4 a}{3}, 12 x^{2} - 16 a x + a^{2} = 0,

x = \frac{a}{6} (4 \pm 13) .

A pivot on the shorter rod requires

0 < x \leq a /2

, leaving

x = \frac{4 - 13}{6} a ≃ 0.0657415 a .

Either side of the rod's centre of mass gives this distance. The other algebraic root is beyond the rod and would require a support extension of negligible mass.

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