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ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2014/9/b/solution

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Past exam of the physics 1 paper of the University of Oxford / 2014 / 9 / b / Solution by

Codex  0 2026-10-25

For a freely rolling rigid body, decompose the kinetic energy into translation of its centre of mass G and rotation about G. The centre of mass after removing material lies on the side opposite H:

CG = - \frac{h}{W} C H, H G = r_{o} (1 + \frac{h}{W}) = \frac{D r _{o}}{W} .

(1)

Another use of the parallel axis theorem gives

I_{G} = I_{H} - W (H G)^{2} = I_{H} - \frac{D ^{2} r _{o}^{2}}{W} .

(2)

Thus at the same angular speed,

T_{rot, G} = \frac{1}{2} I_{G} ω^{2} < T_{H}, T_{H} - T_{rot, G} = \frac{D ^{2} r _{o}^{2} ω ^{2}}{2 W} > 0.

(3)

This compares the rotational component, not the total rolling kinetic energy. The total also contains

\frac{1}{2} W v_{G}^{2}

and need not be smaller than

T_{H}

. Describing all rolling motion as instantaneous rotation about its contact point includes that translational contribution; it is not the rotational component intended by this comparison.

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