 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2019/3/a/solution

x

along the ideal thin rod from its centre of mass. Its perpendicular distance from the inclined axis is

∣ x ∣ sin α

M / L

. The moment of inertia is therefore

I = \frac{M}{L} \int_{- L /2}^{L /2} x^{2} sin^{2} α d x = \frac{M L ^{2}}{12} sin^{2} α .

This tends to zero for an axis along an infinitely thin rod and to

M L^{2} /12

for a perpendicular central axis. A finite rod thickness would add a contribution absent from the stated thin-rod model.

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