= Solution
The <moment of inertia> of the uniform disc about its central perpendicular axis is
$$
I_O=\int_0^a r^2\frac{m}{\pi a^2}2\pi r\,dr=\frac12ma^2.
$$
The <parallel axis theorem> gives $I_A=I_O+ma^2=3ma^2/2$. For angle $\beta$ from the downward resting position, the gravitational <torque> is $-mga\sin\beta$, so $I_A\ddot\beta=-mga\sin\beta$. Linearizing at $\beta=0$ gives $\ddot\beta+(2g/(3a))\beta=0$. Thus the <uniform disc pivoted at its rim> is a <physical pendulum> with period
$$
\boxed{T=2\pi\sqrt{\frac{3a}{2g}}.}
$$
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