Physics 181. Particle Systems

Transcription

1 Physcs 181 Partcle Systems Overvew In these notes we dscuss the varables approprate to the descrpton of systems of partcles, ther defntons, ther relatons, and ther conservatons laws. We consder a system of dscrete partcles for smplcty, although a contnuous dstrbuton of matter has most of the same propertes; the dfference s manly that wth partcles we have sums over them, whle n the contnuous case we have ntegrals over the volume occuped by the materal. Some of the results we wll obtan have already been used n systems of two partcles. We wll not assume the system s closed, allowng for external forces. The nternal forces wll be assumed to obey the 3 rd law n the form F(1 on 2) =!F(2 on 1). Center of Mass and Momentum of the System We denote the ndvdual partcles by an ndex = 1,2,3,..., N.Ther postons at any tme wll be denoted by r and ther masses by m. The total mass s M =! m. The center of mass (CM) of the system s defned by specfyng ts locaton: r CM = 1 M There may or may not be any of the partcles at that pont. The velocty of the CM s the tme dervatve v CM = dr CM /dt, whch gves! m r Mv CM =! m!r. The rght sde s the total momentum of all the partcles, so we have a general result: p tot = Mv CM. Wth respect to momentum, the system behaves lke a sngle partcle wth the total mass M, movng wth the velocty of the CM. Another tme dervatve gves!p tot =! m!!r. 1

2 But for each partcle the 2 nd law says m!!r = F tot, so the sum on the rght s the sum of all the forces on the partcles. Ths ncludes forces arsng from outsde the system (external forces) as well as nteractons between the partcles (nternal forces). For a gven partcle we separate these types of forces: F tot = F ext + F nt, where each term on the rght represents the total of that type of force actng on ths partcle. But the nternal forces are mutual nteractons wth the other partcles, so F nt = " F( j on ). j! The total force on the whole system then becomes F tot ext nt =! F +! F = F tot + F( j on ) ext!!. The double sum conssts entrely of symmetrc pars, such as F(1 on 2) + F(2 on 1), all of whch add to zero by the 3 rd law. Therefore we have the smplfyng result F tot = F tot ext. Wth respect to the total force, only the external forces need to be counted. That s not generally true for the work done by the forces, as we wll see later. Returnng to the momentum, we have thus shown that!p tot = F tot ext. Of course we also have!p tot = Ma CM. That s, the center of mass moves lke a sngle partcle of mass M under the nfluence of the total external force. An mmedate consequence of ths s the law of conservaton of lnear momentum: If the total external force s zero, the total momentum s constant. Ths holds component by component, as was noted earler. The CM Reference Frame So far all the quanttes have been defned wth respect to some nertal reference frame, whch we wll call the lab frame. It s useful, as we wll see, to defne a reference frame wth ts orgn at the CM of the system; ths may not be an nertal frame. In the CM frame, denote the postons of the partcles by r!. Then we have r = r CM + r!. The veloctes obey!r = v CM +!r!. Clearly " m r! = 0, so we also have " m!r! = 0,.e., the total momentum s zero n the CM frame. That the total momentum s always zero n ths frame means that the total force n that frame must be zero, but one must nclude nertal forces f the reference frame s acceleratng. 2 j"

3 Angular Momentum of the System For a sngle partcle the angular momentum about a reference pont s L = r! p, where r s the vector from the reference pont to the partcle. For a gven reference pont (whch we wll take to be the orgn of the lab system) the total angular momentum s L tot = " r! m!r. (1) Usng r = r CM + r! and!r = v CM +!r!, we get four terms n the sum:! m r CM " v CM = Mr CM " v CM = r CM " p tot. Ths term s the same as f we had a sngle partcle wth the total momentum of the system, located at the CM. We call ths the angular momentum of the CM moton, L(of CM). " m r! # v CM. Ths s zero because " m r! = 0. r CM! # m!r ". Ths s zero because the sum (the total momentum n the CM frame) s zero. # r! " m!r!. Ths s the total angular momentum as measured n the CM frame, usng the CM as reference pont. We call ths the angular momentum about the CM, L(about CM). The net result s L tot = r CM! p tot + # r "! m!r ", or L tot = L(of CM) + L(about CM). Ths breakup of an mportant dynamcal varable nto ts value when only the CM moton s consdered, plus ts value as measured relatve to the CM, s a general feature of classcal mechancs. Now we consder how the angular momentum s changed by the forces actng on the partcles. We start from Eq (1) above and take a tme dervatve:!l tot = " m [!r!!r + r!!!r ]. The frst term n [ ] s zero of course. Now we nvoke the 2 nd law: m!!r = F tot, and as before separate external and nternal forces:!l tot ext = " r! F + " r! " F( j on ). j# 3

4 The frst term s the total torque on all the partcles due to external forces only, whch we denote by N tot ext. In the double sum we have pars agan, such as r 1! F(2 on 1) + r 2! F(1 on 2). Wrtng F(2 on 1) =!F(1 on 2) from the 3 rd law, ths becomes (r 2! r 1 ) " F(1 on 2). The vector r 2! r 1 s drected from partcle 1 toward partcle 2. If the nteracton between the partcles s along that lne (.e., an attracton or repulson) then the cross product s zero, and the whole double sum gves zero. Mcroscopc analyses usng quantum mechancs show that ths assumpton about the forces s vald. The result s a smple law: Ths s often called the rotatonal 2 nd law.!l tot = N tot ext. An mmedate consequence s the law of conservaton of angular momentum: If the total external torque s zero, the total angular momentum s conserved. Of course the torque and the angular momentum must have the same reference pont. Also, lke conservaton of lnear momentum, ths holds component by component. The total torque can be dvded nto two parts, just lke the angular momentum: ext " r! F = "(r CM + r #)! F ext = r CM! F tot ext + " r #! F ext. The frst term on the rght s the torque of the total external force appled at the CM. The other term s the total external torque about the CM. The former changes the angular momentum of the CM moton; the latter changes the angular momentum about the CM:!L(of CM) = r CM! F tot ext, L(about! CM) = N tot ext (about CM). The second of these holds even though the CM frame may be non-nertal. The reason s that any nertal forces act effectvely at the CM, so they produce no torque about that pont. Knetc Energy of the System Of course the total knetc energy s the sum of the ndvdual knetc energes, so Usng the fact that v = v CM + v!, we fnd T tot = 1 2 m 2! v. T tot = 1 m 2# [v 2 CM + v! 2 + 2v " v! ]. CM Because " m v! = 0 (the total momentum s zero n the CM frame) the last term above gves zero. So we have 4

5 T tot = 1 2 Mv CM " m v! 2. The frst term s called the knetc energy of CM moton. The second term s the total knetc energy measured n the CM frame. So we have T tot = T(of CM) + T(about CM). Ths breakup nto two parts s lke that of angular momentum. Total Mechancal Energy of the System Frst, consder the total work done by all forces, n nfntesmal form: dw tot = " F tot! dr = " m a! dr. Now durng tme dt the th partcle moves through dsplacement dr = v dt, and the acceleraton s a = dv /dt, so we have a! dr = (dv /dt)! v dt = v! dv. But also d(v 2 ) = 2v! dv, so we fnd m a! dr = 1 2 m d(v 2 ) = dt. So each term n the sum above s the (nfntesmal) change n that partcles knetc energy. The sum thus represents the change n the total knetc energy, and we have the work-energy theorem for the system: dw tot = dt tot. Now dvde the forces nto conservatve and non-conservatve. We fnd dw cons + dw non!cons = dt tot. But the conservatve forces are related to potental energy by dw cons =!du tot, where U tot represents the total potental energy for all conservatve forces (nternal and external). Rearrangng the above equaton, we have dw non!cons = dt tot + du tot = de tot. Here E tot = T tot + U tot s the total mechancal energy of the system. We see that ths quantty s changed only by work done by non-conservatve forces. If only conservatve do work, we have the law of conservaton of mechancal energy: If only conservatve forces do work, the total mechancal energy s conserved. Note however that any nternal forces that do work must be conservatve too. A note about gravty near earth s surface, where we use the approxmaton that g s unform. Ths feld produces a force m g on each partcle, so the total s Mg. But what about torques? Usng the above dvson of the total external torque nto that actng at the CM plus that about the CM, we note that the latter s " m r! # g. But " m r! = 0, so ths part of the torque vanshes. Ths gves us a smple rule: 5

6 For purposes of torques, gravty acts at the CM. In a non-unform gravtatonal feld ths argument fals. One result s tdal torques on planets and moons. As for the potental energy, any constant force s conservatve, wth potental energy gven by U(r) =!F "r + const., so we have (omttng addtve constants) U grav =!m g " r. The total potental energy s thus U grav =!# m g "r. But! m r = Mr CM, so U grav =!Mg " r CM. In a unform gravtatonal feld, the formula for the potental energy of a system s the same as for a sngle partcle wth the total mass located at the CM. (If one chooses the usual coordnate system wth the y-axs postve upward from the surface of the earth, then g! r CM = "gy CM and U grav = Mgy CM. Ths s the formula one uses n ntroductory courses.) 6