Bead on a rotating hoop

The setup

Imagine a circular wire hoop of radius R oriented vertically, like a Ferris wheel. A bead is threaded onto the hoop and can slide along it without friction. Now the entire hoop rotates about its vertical diameter at angular velocity ω. The bead feels gravity pulling it down and the centrifugal effect pushing it outward from the rotation axis. What equilibrium position does the bead settle into?

The bifurcation

At low ω, the bottom of the hoop (θ = 0) is the only stable equilibrium. But there is a critical angular velocity ω_c = √(g/R) at which the nature of this equilibrium changes. Above ω_c, the bottom becomes unstable and two new stable equilibria appear at θ_eq = ±arccos(g/ω²R). This is a supercritical pitchfork bifurcation — the single equilibrium splits into three, with the original one becoming unstable.

Spontaneous symmetry breaking

The system has left–right symmetry: nothing about the physics favors the bead going left versus right. Yet above the critical speed, the bead must choose one side or the other. This is spontaneous symmetry breaking — the same mechanism that appears in phase transitions, the Higgs mechanism, and many other areas of physics. The symmetry of the equations is preserved, but the solution breaks it.

The equation of motion

The Lagrangian analysis gives θ̈ = ω² sinθ cosθ − (g/R) sinθ. Adding damping γ, we get θ̈ = ω² sinθ cosθ − (g/R) sinθ − γθ̇. Setting θ̈ = θ̇ = 0 gives the equilibrium condition cosθ = g/(ω²R), which has real solutions only when ω > ω_c.

The bifurcation diagram

The diagram on the right plots the equilibrium angle θ_eq as a function of ω. Below ω_c, only θ = 0 exists (stable). Above ω_c, the zero solution persists but is unstable (shown dashed), while two new stable branches curve away symmetrically. The shape of this diagram — like a pitchfork — gives the bifurcation its name.