 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2017/8/a/solution

For a uniform thin disc, the surface mass density is

σ = m / (π a^{2})

. An annulus of radius

r

and thickness

d r

has mass

d m = 2 πσ r d r

r

from the normal central axis. Its contribution to the moment of inertia is

r^{2} d m

, so

I = \int_{0}^{a} r^{2} (2 πσ r d r) = \frac{2 m}{a ^{2}} [\frac{r ^{4}}{4}]_{0}^{a}, I = \frac{1}{2} m a^{2} .

This establishes the moment of inertia of a uniform disc; the normal axis and uniform surface mass density are part of the result.

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