PHYS 705: Classical Mechanics. Newtonian Mechanics


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1 1 PHYS 705: Classcal Mechancs Newtonan Mechancs
2 Quck Revew of Newtonan Mechancs Basc Descrpton: An dealzed pont partcle or a system of pont partcles n an nertal reference frame [Rgd bodes (ch. 5 later)] r s a poston vector of the partcle relatve to O We are n 3D Eucldean space,.e., d v r dstance bet r1 and r r1 r  s ts velocty vector 3 1 O r partcle (m) Defne the lnear momentum vector as the product of the partcle mass m and v: p mv
3 3 Newtonan Mechancs: Basc Descrpton f 1 f The nfluence of the external world on ths partcle s encoded as forces (vectors) f actng on t r O The dynamcs of the partcle s descrbe by the Newton s second law of moton: F dp p where F f s the net sum (vector sum) of all forces actng on the partcle If m s constant n tme, d p d m m d v F v ma If m s not constant n tme, dfferentaton wll nvolve m e.g., rocket problem
4 4 Conservaton Theorems dp 1. If F0, then 0 pconstant Conservaton of Lnear Momentum Defne angular momentum by: Lrp Now defne torque as: NrF Recall that vector cross product s not commutatve,.e., ABBA In terms of these angular varables, the nd law can be wrtten as: dp dp d F rfr r m v d Nr m v
5 5 Conservaton Theorems Notce that: dl d d dr rmvr mv mv d r m v v mv Plug ths nto the prevous equaton for the nd law, we have: zero! d d N r mv m r v N d L nd law for rotaton dl. If N0, then 0 Lconstant Conservaton of Angular Momentum
6 6 Conservaton Theorems Defnton of work: W 1 F d s 1 1 s the work done by a force F along a partcular path between 1 to and t can be +, , or zero. We have taken F as beng the net force actng on the partcle, ds F And, now expandng the two separate terms n the ntegral, d dv F mvm ds v for a constant mass m
7 7 Conservaton Theorems Ths gves: dv W1 m v 1 Sde note: d d d d v v v v v vv 1 d dv v v So, we can wrte the above eq as: m d v m W1 dv 1 1 W mv v 1 1 1
8 8 Conservaton Theorems Now, defne T 1 mv as the knetc energy, ds Then, we have, 1 F W T T 1 1 Net work done KE The WorkEnergy Theorem A conservatve force s one such that the work done n gong from 1 to depends only on the end ponts and not on the partcular path taken. Equvalent statement: work done around a closed path wll be zero!  as an example, fcton s not a conservatve force whle gravty s.
9 9 Conservaton Theorems Mathematcally, f we have a conservatve F on a closed path F d s 0 Recall from vector calculus (Stoke s Theorem): F d s F d n surface where the surface s defned wth the path as ts boundary and dn s the normal on the surface. dn The rght hand sde s zero f there exst a scalar functon U such that, F U ( note : U 0) ds U s called the potental or potental energy. Note that one can add an arbtrary constant to U wthout affectng ths result. zero level for U s not mportant and only s physcally relevant. U
10 10 Conservaton Theorems So, for any conservatve systems, we can wrte, W Fds du U U From our prevous dscusson, we also know that the net work from ths conservatve force s also equal to, W 1 So, W T T 1 1 T T1 U1U Then, rewrtng, 3. T1U1 T U Conservaton of Mechancal Energy
11 11 Mechancs of a System of Partcles  For a system of partcles, one needs to dstngush between : external forces actng on the entre system and nternal forces actng wthn the system r  nd law for the th partcle s: O F F p nternal force on from j j ( ) j net external force on We further assume the weak form of Newton s 3 rd law apples such that, F j F j Note: there s a stronger form about the drecton of forces (later)
12 1 Mechancs of a System of Partcles Summng over all partcles (LHS): F j, j j F ( e) F ext tot = 0 by 3 rd law Summng over all partcles (RHS): (also assume m are constant n tme) dp dv d dr d p m m m r We now defne the center of mass vector R as, R mr m mr M wth M m r weghted avg of by the partcle s mass
13 13 Mechancs of a System of Partcles Then, LHS = RHS gves, F ( e) p F ext tot d d d R m M M r R d t The center of mass moves as f t was a pont partcle wth the total mass of the entre system and all the external forces actng drectly on t. F ( e) 1 ext F tot O r F ( e) O R M
14 14 Mechancs of a System of Partcles Now calculate the total lnear momentum : dr d dr p p m mr M tot Puttng ths n our prevous red equaton for the Center of Mass, F ext tot d R M gves F ext tot p tot Conservaton of Lnear Momentum for a system of partcles: If total external force, then the total lnear momentum conserved. F ext tot 0 ptot s
15 15 Mechancs of a System of Partcles Smlarly for angular momentum, startng agan wth the nd law, F F p j ( ) j for the th partcle Crossng r on both sdes and summng over all partcles, m m r F F r p d d r mv ( e) j j ( e) r Fj r F r m v j, j d d r v r v v mv N N ext tot d L L tot
16 16 Mechancs of a System of Partcles Lookng back at the frst term on the LHS: rfj Rewrte the sum n terms of pars of partcles: r Fj rfj rjfj j, j pars pars r r F r pars j j j F j By defnng, r r r j j j, j (by the weak form of the 3 rd law) O r r j r j
17 17 Mechancs of a System of Partcles Now, assume the strong verson of the 3 rd law as follows: r F j F j r j In adon to beng equal and opposte, F j also le along the lne connectng the th and j th partcle Then, the prevous sum term wll vansh, r Fj rj Fj j, j pars O r j 0 Puttng everythng together, we then have, dl tot N ext tot The system of partcles agan acts lke a sngle partcle Conservaton of Angular Momentum for a system of partcles: f total external torque = 0, then the total angular momentum s conserved.
18 18 Notes on the 3 rd law For a system of partcles, 1. Conservaton of Lnear Momentum weak form of the 3 rd law. Conservaton of Angular Momentum strong form of the 3 rd law Let dgress n consderng a movng charges n an EM feld, Recall BotSavert law: a pont charge q movng n the drecton gven by v gves rse to B at a poston r away, Br v r 3 4 r 0 q B (out of page) r RHR q v
19 19 Notes on the 3 rd law Recall also the Lorenz force equaton: a charge q wth velocty v n an E + B feld feels a force F gven by, q F E vb Now, we wll gve two examples where the 3 rd law does not hold (footnote on p.8): a) two +charges movng n a plane Consder the magnetc forces actng on each others : v 1 B 1 r 1 F 1 v The two nternal forces F 1 F 1 B and F 1 are equal and opposte (weak) but are not drected along r 1 jonng them (not strong).
20 0 Notes on the 3 rd law. two +charges movng n a plane B=0 F 1 =0 v v 1 F 1 B The two nternal forces F 1 and F 1 are not even equal and opposte here (not weak nor strong). 1
21 1 General Moton of a System of Partcles We have shown that for a system of partcles, we generally have: For Lnear translatonal moton of a system of partcles can be descrbed as a sngle partcle at CM wth ts total mass M concentrated at that pont. p tot dr M p tot = (total mass) (velocty of CM) And, for rotatonal moton of a system of partcles, a smlar statement n relaton to the CM can be made: tot ang mom L tot = (ang mom of M as f concentrated at CM) + (ang mom of rotaton about the CM)
22 General Moton of a System of Partcles To show the rotatonal moton clam, consder : O r r ' CM R Rr ' r Vv ' v Total angular momentum L for all partcles wth respect to O s: r L r p m v R r ' m V v ' R mv R mv ' r ' mv r ' mv ' RmVr' mv' Rmv' mr' V
23 3 General Moton of a System of Partcles Notce that the last two terms are zero by def. of CM: ' MR mr m Rr MR mr m r ' 0 ' Last term mr' V 0 Takng tme dervatve of above, d mr' mv ' 0 3 rd term R mv' 0 So, we have: R R m r ' m ' M r ' m v ' L V v V ang mom of CM ang mom of rot about the CM
24 4 General Moton of a System of Partcles Next, we can obtan a smlar separaton for KE: KE of a rotatng system = (KE of CM) + (KE of rotaton about the CM) To see ths, start wth, T 1 mv we also have, r Rr v Vv ' ' Notce that ' ' V V v ' v ' V v ' v v v Vv Vv ' V v ' V v
25 5 General Moton of a System of Partcles Puttng ths nto the expresson for T, we have, 1 1 T mv m v m V v ' ' V So, ths s our desred separaton, 1 1 T MV v m ' m v ' 0 (def of CM) KE of sys as f KE of moton about the CM
26 6 General Moton of a System of Partcles Lastly, ths separaton can also be appled to potental energy of a system of partcles, Assumptons: All forces are dervable from a potental functon (conservatve) Internal forces also satsfy the strong form of the 3 rd law 1 U U U j, j j (the ½ s here snce the sum doublecounts pars) Total PE of system = (external PE) + (nternal PE) Note: n order for the nternal forces to satsfy the strong 3 rd law, PE (U j ) dervable from nternal forces must also be a functon of r j dstance between partcles only.
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