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Let $\theta$ measure the spoke's displacement from downward equilibrium. The total moment of inertia about the axle is $I=M{R}_2^2+m{R}_1^2$: the rim is a thin thin hoop, and the attached body is a point mass. The rim's weight has no axle torque, while the attached mass gives restoring torque $-mg{R}_1\sin \theta$. HenceEquivalently conservation of energy gives $\frac{1}{2}I{\dot{\theta }}^2=mg{R}_1(\cos \theta -\cos {\theta }_0)$. For small angles, $\sin \theta \simeq \theta$, so the motion is approximately $\theta ={\theta }_0\cos (\omega t)$ with ${\omega }^2=mg{R}_1/I$. The small-amplitude period isAt finite amplitude the nonlinear pendulum equation remains valid but this period approximation does not. The $M\to 0$ limit recovers a simple pendulum of length $R_1$.

Consider a uniform thin hoop of radius $a$ and mass $m$, suspended from a point on its rim, with a bead of equal mass sliding frictionlessly around it. Let $\theta$ be the angle of the line from the pivot to the hoop's centre, and let $\varphi$ be the angle from the hoop's centre to the bead, both measured from the downward vertical. The stable equilibrium is $\theta =\varphi =0$.

The hoop's pivot moment of inertia is $2m{a}^2$. Adding its kinetic energy to the bead's gives the quadratic small oscillation LagrangianThus the mass matrix and stiffness matrix are $m{a}^2\left(\begin{array}{cc}3& 1\\ 1& 1\end{array}\right)$ and $mga\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)$. Their generalized eigenvalue problem gives $2c^2-5c+2=0$ for $c=a{\omega }^2/g$. The angular frequencies are therefore $\sqrt{g/(2a)}$ and $\sqrt{2g/a}$. The slow normal mode has $\varphi =\theta$; the fast normal mode has $\varphi =-2\theta$. Specifying the two absolute angles explains the off-diagonal coupling and restores the physical scale hidden by dimensionless matrices.