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by: Dr. Carissa Rowe


Dr. Carissa Rowe
GPA 3.97


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Class Notes
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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.


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Date Created: 10/20/15
Lecture 2 Outline Restating the obvious 0 Work and gradients of Potentials Sections 11 12 o Constraints and Generalised Coordinates Section 13 0 Some examples Recap 0 Conservation of total linear momentum given 17 26 0 assumes weak law of action and reaction is valid for internal forces Fm Fjl 0 Conservation of total angular momentum given external Torque A71 0 requires that strong law ie central internal forces Fl F a 77 77 aft J 31 l j lj Single Particle Work De nition Work done by an external Force upon 2 particle between points 1 and 2 is W12 Fd 1 eg d dx dy dz Udt and 13 Fag7 Fy F2 so 1305 dex Fydy dez and thus dz m d 77102 dS m dt 11gt dt 2 dtv dt 2 constant mass and dt v11 div vdt 2d 11 771013 11 T2 T1 in terms of kinetic energy W12 Conservatives and Friction 0 De nition A force eld is conservative if W12 is the same for all paths ie 5612 1305 0 0 De nition A dissipative force eg friction strictly has 13d gt 0 thus 56f 13d gt 0 Single Particle Potential I o For W12 to be independent of path a necessary and suf cient condition is that 13 VF 0 ie the gradient of some scalar function best seen as 2 2 W12 Fd VVd 1 1 2 V 2 a V0l93 V0lg7L3 0lz dVV1 V2 1 3x 33 dz 1 0 Thus W12 V1 V2 T2 T1 and T V is conserved o This is assuming that V07 is constant in time Single Particle Potential II c To be able to express 13 VF the necessary and suf cient condition is really that the curl 6 X 13 6 0 Note also that the zero value of V07 is arbitrary and 8V 8V 8V 8V Fd d d d F 8 3x x3y y3z z 38gtds sds System of Particles Doing Work I 0 Consider all forces on each particle W12 f12 0 Where 1 and 2 denote con gurations Cg the total work 2 2 dv 2 1 W12 Eds mid tZ Zdt d 5mm i i 1 Z 1 1 c From equations of motion E ml 0 total work done is just the change in total kinetic energy 1 1 1 1 2 T E mwivi E 2 1 2 W12 d T2 T1 1 1 1 System of Particles Doing Work II 0 Further consider total kinetic energy T 7711171471 0 we also want the energy equations of the centre of mass 1 1 1 T E nil17 mum 17 5M1 E nilum 1 1 d E 7711112 E mivil2 0 Now the force on each particle 7 16 Zj so W12 ff Reds 392 2m ff 0 Assuming the external forces 17 16 V f iVi Wie2Z12F d ZA261Wd 2 ZWE Internal Forces and Work 0 Work so far W12 ff 0 Assuming that the internal forces are conservative ie that both and 17 17 depends on the interparticle separation a mm wa m am iii 6in jvm Fh Where ViW l Fjl 737 0 ie satisfy the strong law of action and reaction 0 if Vij depended on other vectors eg instrinsic angular momentum forces might not be along Energy Conservation System of Particles 0 Consider the pairwise interaction terms of j ff fixUda jvmdgj figMs o where 617 is the gradient with respect to 7 o and d j df df j gt d3 d j dis g thus 2 2 2 1 1 Z Fjidsi Z Visz39jdmj Z de39j m 1 m 1 m 1 0 Factor of two since we have summation over 239 j and ji o Sotota1 energy T V is conserved 2 1 2 W12 Z1 ij1Vg V1 ETg Tl i 39J o and total potential energy V Vl Vij Rigid Body System of particle Where the distances are xed and cannot vary with time The vectors d j must be perpendicular to 77m Thus the internal forces do no work ijlijd j Rigid bodies have constant internal potential energy 1 V E Z Vij constant M Constraints To solve mechanics problems just equation of motion OFF mi dt2 Fie LZFJ i j holonomic constraints are those that can be written f771772t 0 eg the rigid body has cm 0 First problem Constraints couple 772 s and diff eqn s but usually there are nonholonomic constraints Second problem Forces of constraints are unknowns eg usually speci ed by effect on motion Generalised Coordinates With no constraints N particles 3N Cartesian co ords With constraints is eqns 771772 t 0 leaves 3N 16 degrees of freedom Introduce indep variables q1q2 q3Nk eg 2 D double pendulum has 4 co ords 2 variables 0 Useful in systems Without constraints eg 7 6 gb Constraints Vertical Disk nonholonomic Constraints Pendulum holonomic


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