advanced31 min studyLesson 130 in Physics

Lagrangian stability and effective motion

Analyze local equilibria using effective potentials while distinguishing conservative stability, attraction, and a motor-driven constraint.

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What you’ll understand

  • Use potential curvature to classify nondegenerate one-coordinate equilibria.
  • Distinguish Lyapunov stability from asymptotic attraction.
  • State whether effective motion fixes momentum or an externally imposed angular speed.

An effective potential can turn a complicated constrained motion into a one-coordinate stability question. The simplification is powerful only if its assumptions remain visible. A positive local curvature gives small oscillations in a regular conservative model; it does not make the equilibrium attract nearby trajectories. Moreover, a motor holding an angular speed fixed creates a different reduced problem from conserving angular momentum. We will test those distinctions with a bead on a rotating hoop.

Curvature predicts local motion under stated assumptions

Take a one-coordinate autonomous Lagrangian with positive effective mass and a smooth potential. At a rest equilibrium the potential derivative vanishes. Linearize about that point: the effective mass at equilibrium multiplies the displacement acceleration, and the potential curvature supplies the linear restoring coefficient. Positive curvature gives a positive squared oscillation frequency; negative curvature produces an exponentially growing direction. The derivative of a position-dependent mass does not contribute at linear order about rest because it multiplies products of small quantities.

In this one-coordinate conservative setting, a strict local potential minimum with a regular positive kinetic term gives Lyapunov stability through the conserved energy: sufficiently low nearby energy confines motion to a neighborhood. It is not asymptotic stability. A nearby nonzero-energy orbit generally keeps oscillating rather than approaching rest. Damping can create attraction, but that changes the equations and the energy balance. If curvature vanishes, the quadratic approximation is inconclusive and the first relevant higher-order term must be examined.

V′(q0)=0,M(q0)η¨+V′′(q0)η=0,ω2=V′′(q0)M(q0)V'(q_0)=0,\qquad M(q_0)\ddot\eta+V''(q_0)\eta=0,\qquad \omega^2=\frac{V''(q_0)}{M(q_0)}
This test applies to a regular one-coordinate rest equilibrium. At zero curvature it does not decide nonlinear stability.↔ Scroll wide equations horizontally.

A prescribed-speed hoop is an externally driven constraint

Let a vertical circular hoop of radius a rotate about its vertical diameter with angular speed Ω maintained by a motor. Measure the bead angle θ from the bottom. Its kinetic energy includes both motion along the hoop and azimuthal motion imposed by rotation. Eliminating the prescribed azimuthal coordinate gives an autonomous θ Lagrangian when Ω is constant. Its effective potential contains a negative centrifugal term. This is not fixed-angular-momentum Routh reduction: the motor supplies whatever torque is needed to maintain its commanded speed.

The effective θ energy is conserved for constant Ω in this ideal model, but the bead's laboratory mechanical energy need not be. The motor can do work through the moving hoop. At the bottom, the curvature is proportional to g−aΩ². Below the threshold the bottom is stable; above it the bottom is unstable and new off-bottom equilibria satisfy cos θ=g/(aΩ²). Their existence and stability should be checked in the effective problem rather than inferred by borrowing the positive centrifugal barrier from a different fixed-momentum model.

Veff(θ)=mga(1−cos⁡θ)−12ma2Ω2sin⁡2θ,ωbottom2=ga−Ω2V_{\mathrm{eff}}(\theta)=mga(1-\cos\theta)-\frac12ma^2\Omega^2\sin^2\theta,\qquad \omega_{\mathrm{bottom}}^2=\frac ga-\Omega^2
The hoop speed Ω is prescribed externally. The effective potential differs from a fixed-angular-momentum potential because the apparatus exchanges angular momentum and energy with the bead.↔ Scroll wide equations horizontally.

Test the threshold and the meaning of the lab

At the threshold Ω²=g/a, the quadratic term at the bottom vanishes. Expanding the stated effective potential gives a positive quartic leading term, so the ideal bottom is still a strict local minimum there, although the small-oscillation frequency is zero and the harmonic approximation fails. The period then depends on amplitude. This example shows why a zero eigenvalue should trigger a higher-order calculation rather than an automatic stable or unstable label. Real friction, finite bead size, and imperfect speed control require additional model terms.

The oscillator lab illustrates the distinction between conservative oscillations and damped attraction after a stable linearization. At damping ratio zero, predict persistent oscillation; at damping ratio one, predict a nonoscillatory return for its fixed initial displacement and zero velocity. Natural frequency is directly controlled, so the lab does not compute the hoop threshold, quartic potential, or unstable inverted dynamics. Its normalized amplitude must be separately scaled to remain a small physical perturbation. Use it to inspect a linear model only where that model is justified.

The same bottom changes character with motor speed

A 2 kg bead moves on a hoop with a=1 m and g=10 m/s². Compare the bottom equilibrium at prescribed Ω=2 rad/s and Ω=4 rad/s. Find the linear oscillation or growth rate.

  1. The generalized mass for θ is ma²=2 kg m². The bottom curvature is ma(g−aΩ²), so it is 2(10−4)=12 at Ω=2 and 2(10−16)=−12 at Ω=4.

  2. Divide each curvature by the mass coefficient. At Ω=2, ω²=6 s⁻² and the perturbation obeys ηddot+6η=0, giving frequency √6 rad/s.

  3. At Ω=4, the equation is ηddot−6η=0. One linear solution grows as exp(√6 t), so the bottom is unstable with growth rate √6 s⁻¹; there is also a decaying solution.

The result

A sign change in curvature changes oscillation into a growing and decaying pair. Neither the negative curvature nor its growth rate should be represented by the lab's stable oscillator controls.

Explore the model

Interactive lab

Interactive lab · oscillator

Compare stability with attraction

Set Natural frequency to one. Predict persistent oscillation at Damping ratio zero and nonoscillatory relaxation at one, then play each. This stable normalized oscillator does not simulate the hoop or a negative squared frequency.

Compare stability with attractionUnit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Current readings: Displacement x(t): 1; Velocity v(t): 0; Regime: underdamped. All parameters are adjustable with the labeled controls below.02.557.510−1.15−0.5700.571.15Time · sDisplacementunderdamped
Displacement x(t)Current time
Displacement x(t)
1
Velocity v(t)
0
Regime
underdamped
t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forConservative stability confines nearby motion; attraction requires a dissipative mechanism in this model.

Check your understanding

A regular autonomous one-coordinate conservative system has a strict local potential minimum. What follows near rest equilibrium?

Try it yourself

At Ω²=g/a, expand the hoop effective potential through θ⁴ using cos θ=1−θ²/2+θ⁴/24+… and sin²θ=θ²−θ⁴/3+…. What does the leading term imply, and what does the harmonic approximation miss?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Substitute the threshold relation before combining the quadratic and quartic coefficients.

Compare with the solution

The quadratic terms cancel. The quartic coefficient is mga(−1/24+1/6)=mga/8, so Veff=mga θ⁴/8+O(θ⁶). The bottom remains a strict local minimum in the ideal effective model, but there is no nonzero linear oscillation frequency. Finite-amplitude motion requires the quartic restoring term and has an amplitude-dependent time scale.

Take it with you

  • State which quantities are conserved and which are externally controlled.
  • Positive curvature predicts oscillation, not automatic attraction.
  • Zero curvature demands a higher-order stability analysis.

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