Mechanical stability
Objective. Analyse equilibrium through energy curvature.
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- 2. Worked example
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Learn: the key idea
Mechanical stability determines whether an equilibrium point returns a system to rest after small disturbance, using the curvature of potential energy rather than forces directly. An equilibrium at q₀ satisfies V'(q₀)=0, and it is stable if V''(q₀)>0 (a local minimum, restoring force pushes back), unstable if V''(q₀)<0 (local maximum). Example: for V(θ)=mgl(1−cosθ), equilibrium at θ=0 has V''(0)=mgl>0, so hanging straight down is stable, while θ=π has V''(π)=−mgl<0, the inverted pendulum, unstable. This curvature test generalises the single-variable second-derivative test for minima and maxima to mechanics.
Worked example
A system has potential energy V(x) = x⁴ − 8x² near equilibrium points. Determine whether the equilibrium at x = 2 is stable.
- 1Find equilibria: V'(x) = 4x³ − 16x = 0, so x(x²−4)=0, giving x=0,±2
- 2Compute second derivative: V''(x) = 12x² − 16
- 3Evaluate at x = 2: V''(2) = 48 − 16 = 32
- 4Since V''(2) = 32 > 0, the equilibrium at x = 2 is stable
Answer: Stable, since V''(2) = 32 > 0
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Start practisingCommon mistake. Students conclude stability just from V'(q₀)=0 without checking the sign of V''(q₀); a stationary point of potential energy can be a maximum, giving an unstable equilibrium despite zero net force there.
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