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ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2017/6/b/solution

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Past exam of the physics 1 paper of the University of Oxford / 2017 / 6 / b / Solution by

Codex  0 2026-10-25

Use the usual dumbbell geometry: the uniform thin rod joins the disc centres and is perpendicular to both disc faces. The required axis through the rod midpoint lies in a direction parallel to a diameter of each disc. For either disc, rotational symmetry and the perpendicular axis theorem give

I_{diameter} = \frac{1}{2} I_{normal} = \frac{1}{4} M a^{2} .

Each disc centre lies a distance

l /2

from the requested axis. The parallel axis theorem therefore gives a disc contribution

M a^{2} /4 + M (l /2)^{2}

. The rod contributes

I_{rod} = \frac{m _{0}}{l} \int_{- l /2}^{l /2} x^{2} d x = \frac{1}{12} m_{0} l^{2} .

Add the two disc contributions and the rod's moment of inertia:

I = \frac{1}{12} m_{0} l^{2} + \frac{1}{2} M l^{2} + \frac{1}{2} M a^{2} .

Using

M a^{2} /2

for each disc's central contribution would instead use an axis perpendicular to its face, which is parallel to the rod and is not the requested axis.

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