 Solution

The moment of inertia of the uniform disc about its central perpendicular axis isThe parallel axis theorem gives $I_A=I_O+m{a}^2=3m{a}^2/2$. 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