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ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2013/9/a/solution

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Past exam of the physics 1 paper of the University of Oxford / 2013 / 9 / a / Solution by

Codex  0 2026-10-25

Let

θ

measure the rim point mass's displacement from vertically below the fixed axis. The disk has moment of inertia

I_{D} = \frac{1}{2} (2 M) R^{2} = M R^{2}

, and the rim mass contributes another

M R^{2}

. Its varying gravitational potential energy is

- M g R cos θ

. Thus

L = M R^{2} \dot{θ}^{2} + M g R cos θ .

(1)

The disk's centre of mass remains fixed, so its own gravitational contribution is a constant and has been omitted. The total fixed-axis moment of inertia is

2 M R^{2}

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