Discussion
A uniform disc of radius \( a \) and surface density \( \sigma \) has, about its own axis, a second moment of
Answer: C
Weight each element by the square of its distance from the axis:
\[ I \;=\; \sigma\int_{0}^{2\pi}\!\!\int_{0}^{a} r^{2}\cdot r\,dr\,d\theta \;=\; 2\pi\sigma\left[\frac{r^{4}}{4}\right]_{0}^{a} \;=\; \frac{\pi\sigma a^{4}}{2}. \]
Since the mass is \( m = \pi\sigma a^{2} \), this is \( \tfrac{1}{2}ma^{2} \), the familiar disc result
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