Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)
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The constrained pendulum multiplier follows from tangency

Example

For a planar pendulum of mass m>0 and rod length ℓ>0, let M=mI2, V(q)=mgq2, and c(q)=(∥q∥22−ℓ2)/2. If c(q)=0 and qTp=0, the multiplier in the constrained equations is λ(q,p)=∥p∥22/m−mgq2ℓ2. At the bottom rest state q=(0,−ℓ), p=0, this gives λ=mg/ℓ and zero acceleration.

Facts & Assumptions

Given: M=mI2, the constraint and potential in the Example, and consistent data c(q)=0, qTp=0.

[F1]

Regular holonomic phase space and constrained dynamics: A(q)=J(q)M−1J(q)T and λ=A−1(D2c[v,v]−JM−1∇V), where v=M−1p.

[F2]

The constrained Hamiltonian vector field has a unique local flow: these equations define the unique local constrained flow for consistent data in the regular chart.

Verification

1.1givenF1algebra

Here J(q)=qT, A(q)=qTq/m=ℓ2/m, D2c[v,v]=∥p∥22/m2, and JM−1∇V=gq2. Substitution into [F1] gives λ=(∥p∥22/m−mgq2)/ℓ2.

2.1F1F2step 1.1algebra∎

At q=(0,−ℓ) and p=0, the multiplier is mg/ℓ. The acceleration is p˙=−mg(0,1)−λ(0,−ℓ)=(0,0); moreover c(q)=0 and qTp=0, so this is a consistent equilibrium of the local flow [F2].

Depends on

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