A particle of mass moves in a plane under the potential with

PHYSICS · Multi-part · Fermi University Physics · Mechanics & Dynamical Systems · Lagrangian mechanics · Lagrangian mechanics

Action and Symmetry, Ch The action, the first variation and the boundary term

Problem

A particle of mass m=0.50 kgm = 0.50\,\mathrm{kg} moves in a plane under the potential V(r)=12kr2V(r) = \tfrac12 kr^2 with k=8.0 N m−1k = 8.0\,\mathrm{N\,m^{-1}}, so that L=12m(r˙2+r2ϕ˙2)−12kr2.L = \tfrac12 m(\dot r^2 + r^2\dot\phi^2) - \tfrac12 kr^2. At some instant r=0.20 mr = 0.20\,\mathrm m, r˙=0\dot r = 0 and ϕ˙=5.0 rad s−1\dot\phi = 5.0\,\mathrm{rad\,s^{-1}}. ((a)) Explain why LL has no explicit dependence on the independent variable tt, and use this to write down the conserved quantity h=r˙ ∂L∂r˙+ϕ˙ ∂L∂ϕ˙−Lh = \dot r\,\dfrac{\partial L}{\partial\dot r} + \dot\phi\,\dfrac{\partial L}{\partial\dot\phi} - L, simplifying it in terms of the kinetic energy TT and potential energy VV. ((b)) Evaluate hh at the given instant, in joules.

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