Popular in Course
Popular in Physics 2
This 15 page Class Notes was uploaded by Dr. Carissa Rowe on Tuesday October 20, 2015. The Class Notes belongs to PHYS608 at San Diego State University taught by M.Bromley in Fall. Since its upload, it has received 28 views. For similar materials see /class/225318/phys608-san-diego-state-university in Physics 2 at San Diego State University.
Reviews for PHYS608
Report this Material
What is Karma?
Karma is the currency of StudySoup.
Date Created: 10/20/15
Lecture 7 Outline Multi Pliers Lagrange Multipliers Section 24 Constants of Motion Conservation of Momentum Sections 26 Conservation of Energy Sections 27 Hamilton s Variational Principle 0 For monogenic systems in which all forces are derivable from scalar potentials except constraints 8 a a0d04 with independent Virtual displacements 6 The motion of the system from time 251 to 252 is such that t2 t2 1 ltqrqmtgtdtj T vmt t1 151 has a stationary value for the actual path of motion t2 6 d ac so far coord constraints built into generalised coords System with constraints How does Hamilton s principle treat constraint forces Consider number of holonomic constraints faQ177Qn t0 Apply the variational principle to obtain eqn s of motion Via method of Lagrange undetermined multipliers First introduce functions Aaq1 gm 2 such that m 752 m ZAafa0 gt 5 lt Mfg dt0 oz1 t1 a1 and combining with the action 6 fit L dt 0 5 lt 3Aafagt dtOz tt2 dt oz1 Lagrange undetermined multipliers o from the effective Lagrangian L L 221 Aafa a consider variations in 6 along with choice of Aa 0 since we may be coupled eg fq1 q2 Sql 6q2 0 t2 d an an m af 5 can gt A 0 dz 8 8 Z A Qt a but it means that we now have to solve n m eqns 0 solving for each of the n coordinates ql gm 0 including m eqn s faq1 gm 1 0 a get the forces of constraint Qk through the solution Lagrange Multiplier example a Single particle with constraint i g ky 0 semiholonomic multipliers effective Lagrangian Lquot L 2271 Aafa also works when semiholonomic faq1qnq 1q nt in particular fa 2k aaqu a0 0 then with m multipliers 050 ia a dt q k 8 a aq k oz1 again that we now have to solve n m eqns solving for each of the n coordinates ql qn m 8 gives Forces of Constraint Via Qk Z Mala J Qk oz1 Please Explain I 0 Consider a 2 D system with one holonon1io constraint fay AxBy C 0 then fi3j AdsBy 0 6f a f6x a f6y A6aB6y 81 8y a variation of effective Lagrangian 0 5 12 3 A f dt Q d 8 83 9 f 0 5 a 5 1 11 2k d15an 8 Qk 2k 5 Qk 0 ie the variation of the sum of the integrals is zero e d8 an af M ll dt i 8x xai amp dt t2 dac ac of A 5 dt 11 dtay 8y 81 y Please Explain II while variation of sum is zero individual terms are not snuxaconstraHNJsaysiA556y220 but we can choose the unknown Mt such that 52 132 48 1 0 8y then immediately we must have the other term ddL 93 Ag0 89 if we have more constraints Aa still ok we are still doing no work against the constraints here though example I Truer Semiholonomic a Single particle with constraint i g ky 0 example II Holonomic a Motion of hoop 0n inclined plane spare spare example III Holonomic timedep o bead moving on rotating Wire spare Constants of motion 0 are constant combinations of qz and or qi 0 eg in homework 2 Spherical Pendulum Goldstein 119 2 2 o Lagrangian L mr2 sin2 6 gb mgr cos6 a has two Lagrange Equations the rst is d d d mr2 sin2 Q gb 0 thus mr2 sin2 Q gb constant dt dt dt 0 ie a constant of motion is Z mum mr2 sin2 6