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

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

Codex  0 2026-10-25

For the rolling disk with a rim point mass, impose rolling without slipping and take the clockwise angular displacement

θ

from the configuration with the extra mass at the bottom. The disk centre has

x_{C} = Rθ

and

y_{C} = R

. The extra mass has

x_{M} = R (θ - sin θ), y_{M} = R (1 - cos θ), v_{M}^{2} = 2 R^{2} (1 - cos θ) \dot{θ}^{2} .

The disk's translational and rotational kinetic energies sum to

3 M R^{2} \dot{θ}^{2} /2

. Including the rim mass gives

T = \frac{1}{2} M R^{2} (5 - 2 cos θ) \dot{θ}^{2}, U = M g R (1 - cos θ) .

The exact Euler-Lagrange equation is

M R^{2} (5 - 2 cos θ) \ddot{θ} + M R^{2} sin θ \dot{θ}^{2} + M g R sin θ = 0.

To first order,

3 M R^{2} \ddot{θ} + M g Rθ = 0

. Therefore

ω_{roll} = \frac{g}{3 R}, \frac{T _{roll}}{T _{fixed}} = \frac{3}{2} .

At the bottom the extra mass is instantaneously at the contact point and has no first-order velocity. The increased effective inertia comes from the moving disk. Static friction enforces the rolling constraint without doing work at the stationary contact.

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