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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

Keeping one resonant harmonic and treating orbital eccentricity as fixed reduces mean-motion resonance dynamics to a physical pendulum equation. For the inner first-order term $\mathcal{R}=(G{M}_2/a_2)f(\alpha )e\cos \varphi$, leading Lagrange planetary equations give $\ddot{\varphi }\simeq 3j^2{\mu }_2{n}^2\alpha fe\sin \varphi$. Around its stable phase, the small-amplitude libration frequency is $jn\sqrt{3{\mu }_2\alpha |f|e}$. Evolution of eccentricity, other harmonics, and approach to a separatrix change this approximation.