A heavy symmetric top has transverse moment of inertia about the pivot

PHYSICS · Multi-part · Fermi University Physics · Mechanics & Dynamical Systems · Rotation, rigid bodies and accelerated frames · Hamiltonian mechanics and canonical structure

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

Problem

A heavy symmetric top has transverse moment of inertia about the pivot I1=5.0×10−3 kg m2I_1=5.0\times10^{-3}\,\mathrm{kg\,m^2}, symmetry-axis moment I3=1.0×10−3 kg m2I_3=1.0\times10^{-3}\,\mathrm{kg\,m^2}, mass M=0.50 kgM=0.50\,\mathrm{kg}, and pivot-to-centre-of-mass distance d=0.10 md=0.10\,\mathrm{m}; take g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}}. The cyclic angles ϕ\phi and ψ\psi of Chapter 7's Lagrangian give the conserved momenta pϕ=0.15 kg m2 s−1p_\phi=0.15\,\mathrm{kg\,m^2\,s^{-1}} and pψ=0.20 kg m2 s−1p_\psi=0.20\,\mathrm{kg\,m^2\,s^{-1}}. At the instant θ=40.0∘\theta=40.0^\circ the nutation rate is θ˙=0.30 rad s−1\dot\theta=0.30\,\mathrm{rad\,s^{-1}}. ((a)) Using the Routhian R=−12I1θ˙2+Veff(θ)+pψ2/2I3R=-\tfrac12I_1\dot\theta^2+V_{\rm eff}(\theta)+p_\psi^2/2I_3 with Chapter 7's Veff(θ)=(pϕ−pψcos⁡θ)2/(2I1sin⁡2θ)+Mgdcos⁡θV_{\rm eff}(\theta)=(p_\phi-p_\psi\cos\theta)^2/(2I_1\sin^2\theta)+Mgd\cos\theta, evaluate VeffV_{\rm eff} at θ=40.0∘\theta=40.0^\circ. ((b)) Evaluate the top's Hamiltonian H=12I1θ˙2+Veff(θ)+pψ2/2I3H=\tfrac12I_1\dot\theta^2+V_{\rm eff}(\theta)+p_\psi^2/2I_3 at this instant, and state whether HH is conserved and whether it equals the mechanical energy T+VT+V, giving the reason.

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