A double pendulum has two bobs

PHYSICS · Multi-part · Fermi University Physics · Mechanics & Dynamical Systems · Hamiltonian mechanics and canonical structure · Hamiltonian mechanics and canonical structure

Action and Symmetry, Ch The Hamiltonian, phase space and the variational principle in phase space

Problem

A double pendulum has two bobs, each of mass m=0.50 kgm=0.50\,\mathrm{kg}, on light rods each of length l=0.40 ml=0.40\,\mathrm{m}, swinging in one vertical plane. With θ1\theta_1 and θ2\theta_2 the angles of the upper and lower rods from the downward vertical, and g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}}, its Lagrangian is L=12ml2[2θ˙12+θ˙22+2θ˙1θ˙2cos⁡(θ1−θ2)]+mgl (2cos⁡θ1+cos⁡θ2).L=\tfrac12ml^2\left[2\dot\theta_1^2+\dot\theta_2^2+2\dot\theta_1\dot\theta_2\cos(\theta_1-\theta_2)\right]+mgl\,(2\cos\theta_1+\cos\theta_2). ((a)) Write the velocity Hessian ∂2L/∂θ˙i∂θ˙j\partial^2L/\partial\dot\theta_i\partial\dot\theta_j and its determinant, and say for which configurations the Legendre transform can be inverted. ((b)) Invert the momentum relations and write H(θ1,θ2,p1,p2)H(\theta_1,\theta_2,p_1,p_2). ((c)) At an instant with θ1=60.0∘\theta_1=60.0^\circ, θ2=0\theta_2=0, p1=0.20 kg m2 s−1p_1=0.20\,\mathrm{kg\,m^2\,s^{-1}} and p2=0.10 kg m2 s−1p_2=0.10\,\mathrm{kg\,m^2\,s^{-1}}, find θ˙1\dot\theta_1, θ˙2\dot\theta_2 and the value of HH.

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