PH 221-2A Fall 2014 Center of Mass and Lnear Momentum Lectures 14-15 Chapter 9 (Hallday/Resnck/Walker, Fundamentals of Physcs 9 th edton) 1
Chapter 9 Center of Mass and Lnear Momentum In ths chapter we wll ntroduce the followng new concepts: -Center of mass (com) for a system of partcles -The velocty and acceleraton of the center of mass -Lnear momentum for a sngle partcle and a system of partcles We wll derve the equaton of moton for the center of mass, and dscuss the prncple of conservaton of lnear momentum Fnally we wll use the conservaton of lnear momentum to study collsons n one and two dmensons and derve the equaton of moton for rockets 2
The Center of Mass: Consder a system of two partcles of masses m and m 1 1 1 2 2 1 2 2 1 2 at postons x and x, respectvely. We defne the poston of the center of mass (com) as follows: x com mx m mx m We can generalze the above defnton for a system of n partcles as follows: x mxmx mx... mx mxmx mx... mx 1 n 1 1 2 2 3 3 n n 1 1 2 2 3 3 n n com mx m1m2 m3... mn M M 1 Here M s the total mass of all the partcles M m m m... m 1 2 3 We can further generalze the defnton for the center of mass of a system of partcles n three dmensonal space. We assume that the the -th partcle ( mass m ) has poston vector r r com 1 n M 1 mr n 3
The poston vector for the center of mass s gven by the equaton: The poston vector can be wrtten as: r ˆ ˆ ˆ com xcom ycom jzcomk The components of r are gven by the equatons: com n n n 1 1 1 x mx y m y z mz com com com M 1 M 1 M 1 r com 1 M n 1 mr The center of mass has been defned usng the equatons gven above so that t has the followng property: The center of mass of a system of partcles moves as though all the system's mass were concetrated there, and that the vector sum of all the external forces were appled there The above statement wll be proved later. An example s gven n the fgure. A baseball bat s flpped nto the ar and moves under the nfluence of the gravtaton force. The center of mass s ndcated by the black dot. It follows a parabolc path as dscussed n Chapter 4 (projectle moton) All the other ponts of the bat follow more complcated paths 4
The Center of Mass for Sold Bodes Sold bodes can be consdered as systems wth contnuous dstrbuton of matter The sums that are used for the calculaton of the center of mass of systems wth dscrete dstrbuton of mass become ntegrals: 1 1 1 xcom xdm ycom ydm zcom zdm M M M The ntegrals above are rather complcated. A smpler specal case s that of dm M unform objects n whch the mass densty s constant and equal to dv V 1 1 1 xcom xdv ycom ydv zcom zdv V V V In objects wth symetry elements (symmetry pont, symmetry lne, symmetry plane) t s not necessary to eveluate the ntegrals. The center of mass les on the symmetry element. For example the com of a unform sphere concdes wth the sphere center In a unform rectanglular object the com les at the ntersecton of the dagonals C. C 5
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F 1 m 2 F 2 x m 1 O z F 2 m 3 y Newton's Second Law for a System of Partcles Consder a system of n partcles of masses m1, m2, m3,..., m and poston vectors r, r, r,..., r, respectvely. 1 2 3 The poston vector of the center of mass s gven by: n n Mr m r m r m r... m r We take the tme dervatve of both sdes com 11 2 2 33 n n d d d d d M rcom m1 r1m2 r2 m3 r3... mn rn dt dt dt dt dt Mvcom m1v 1m2v2 m3v3... mnvn Here vcom s the velocty of the com and v s the velocty of the -th partcle. We take the tme dervatve once more d d d d d M vcom m1 v1m2 v2 m3 v3... mn vn dt dt dt dt dt Macom m1a 1m2a2 m3a3... mnan Here acom s the acceleraton of the com and a s the acceleraton of the -th partcle 7
m 2 z Macom m1a 1m2a2 m3a3... mnan F m 1 1 F 2 We apply Newton's second law for the -th partcle: F 2 m 3 ma F Here F s the net force on the -th partcle O x y Macom F1F 2 F3... Fn The force F can be decomposed nto two components: appled and nternal app nt F F F The above equaton takes the form: app nt app nt app nt app nt Macom F1 F1 F2 F2 F3 F3... Fn Fn app app app app Ma nt nt nt nt com F1 F2 F3... Fn F1 F2 F3... Fn The sum n the frst parenthess on the RHS of the equaton above s just Fnet The sum n the second parethess on the RHS vanshes by vrtue of Newton's thrd law. The equaton of moton for the center of mass becomes: Macom Fnet In terms of components we have: F Ma F Ma F net, x com, x net, y com, y net, z Ma com, z 8
Ma com F net F F F Ma net, x com, x Ma net, y com, y Ma net, z com, z The equatons above show that the center of mass of a system of partcles moves as though all the system's mass were concetrated there, and that the vector sum of all the external forces were appled there. A dramatc example s gven n the fgure. In a freworks dsplay a rocket s launched and moves under the nfluence of gravty on a parabolc path (projectle moton). At a certan pont the rocket explodes nto fragments. If the exploson had not occured, the rocket would have contnued to move on the parabolc trajectory (dashed lne). The forces of the exploson, even though large, are all nternal and as such cancel out. The only external force s that of gravty and ths remans the same before and after the exploson. Ths means that the center of mass of the fragments follows the same parabolc trajectory that the rocket would have followed had t not exploded 9
m p v mv Lnear Momentum p Lnear momentum p of a partcle of mass m and velocty v s defned as: p mv The SI unt for lneal momentum s the kg.m/s Below we wll prove the followng statement: The tme rate of change of the lnear momentum of a partcle s equal to the magntude of net force actng on the partcle and has the drecton of the force dp In equaton form: Fnet We wll prove ths equaton usng dt Newton's second law dp d dv p mv mvm ma Fnet dt dt dt Ths equaton s statng that the lnear momentum of a partcle can be changed only by an external force. If the net external force s zero, the lnear momentum cannot change F net dp dt 10
z The Lnear Momentum of a System of Partcles p 1 m 1 p 3 In ths secton we wll extedend the defnton of m 2 x p 2 O m 3 y lnear momentum to a system of partcles. The -th partcle has mass m, velocty v, and lnear momentum p We defne the lnear momentum of a system of n partcles as follows: P p1 p2 p3... pn mv 1 1m2v2 m3v3... mnvn Mvcom The lnear momentum of a system of partcles s equal to the product of the total mass M of the system and the velocty vcom of the center of mass dp d The tme rate of change of P s: Mvcom Macom Fnet dt dt The lnear momentum P of a system of partcles can be changed only by a net external force F. If the net external force F s zero P cannot change net P p p p p Mv 1 2 3... n com dp Fnet dt net 11
Example. Moton of the Center of Mass 2. Isolated system means that Pconst; But PP Mv 0 v 0; x 0; they wll meet at x com com com f com 12
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Collson and Impulse We have seen n the prevous dscusson that the momentum of an object can change f there s a non-zero external force actng on the object. Such forces exst durng the collson of two objects. These forces act for a bref tme nterval, they are large, and they are responsble for the changes n the lnear momentum of the colldng objects. Consder the collson of a baseball wth a baseball bat The collson starts at tme t when the ball touches the bat and ends at t f when the two objects separate The ball s acted upon by a force F ( t) durng the collson The magntude Ft ( ) of the force s plotted versus tn fg.a The force s non-zero only for the tme nterval t t t dp Ft ( ) Here ps the lnear momentum of the ball dt tf tf dp Ftdt () dp Ftdt () t t 14 f
p J J Fave t dp F( t) dt dp p p p = change n momentum t t t t f f f t t t t f Ftdt ( ) s known as the mpulse Jof the collson t f J F( t) dt The magntude of J s equal to the area t f under the F versus t plot of fg.a p J In many stuatons we do not know how the force changes wth tme but we know the average magntude Fave of the collson force. The magntude of the mpulse s gven by: J F t where t t t ave f Geometrcally ths means that the the area under the F versus t plot (fg.a) s equal to the area under the F versus t plot (fg.b) ave 15
Collsons. Impulse and Momentum 16
The Impulse-Momentum Theorem 17
F ave Seres of Collsons Consder a target whch colldes wth a steady stream of dentcal partcles of mass m and velocty v along the x-axs A number n of the partcles colldes wth the target durng a tme nterval t. Each partcle undergoes a change p n momentum due to the collson wth the target. Durng each collson a momentum change p s mparted on the target. The Impulse on the target durng the tme nterval t s: J np The average force on the target s: J np n Fave mv Here v s the change n the velocty t t t of each partcle along the x-axs due to the collson wth the target m m Fave v Here s the rate at whch mass colldes wth the target t t If the partcles stop after the collson then v 0 v v If the partcles bounce backwards then v vv 2v 18
p m 1 1 p 3 m 2 p 2 x O z m 3 y Conservaton of Lnear Momentum Consder a system of partcles for whch Fnet 0 dp F net 0 P Constant dt If no net external force acts on a system of partcles the total lnear momentum cannot change total lnear momentum total lnear momentum at some ntal tme t at some later tme t f The conservaton of lnear momentum s an mportan prncple n physcs. It also provdes a powerful rule we can use to solve problems n mechancs such as collsons. Note 1: In systems n whch F net 0 we can always apply conservaton of lnear momentum even when the nternal forces are very large as n the case of colldng objects Note 2: We wll encounter problems (e.g. nelastc collsons) n whch the energy s not conserved but the lnear momentum s 19 P
respectvely If the system s solated.e. the net force F Momentum and Knetc Energy n Collsons Consder two colldng objects wth masses m1 and m2, ntal veloctes v and v and fnal veloctes v and v, 1 2 1f 2 f net 0 lnear momentum s conserved The conervaton of lnear momentum s true regardless of the the collson type Ths s a powerful rule that allows us to determne the results of a collson wthout knowng the detals. Collsons are dvded nto two broad classes: elastc and nelastc. A collson s elastc f there s no loss of knetc energy.e. A collson s nelastc f knetc energy s lost durng the collson due to converson nto other forms of energy. In ths case we have: K f K K K A specal case of nelastc collsons are known as completely nelastc. In these collsons the two colldng objects stck together and they move as a sngle body. In these collsons the loss of knetc energy s maxmum f 20
One Dmensonal Inelastc Collsons In these collsons the lnear momentum of the colldng objects s conserved p1 p2 p1f p2 f mv mv mv mv 1 1 1 2 1 1f 1 2 f One Dmensonal Completely Inelastc Collsons In these collsons the two colldng objects stck together and move as a sngle body. In the fgure to the left we show a specal case n whch v 0. mv mvmv V m1 m m 1 2 v 1 2 1 2 1 2 1 2 1 1 1 2 The velocty of the center of mass n ths collson P p1 p2 mv 1 1 s vcom m m m m m m In the pcture to the left we show some freeze-frames of a totally nelastc collson 21
Inelastc Collson 22
The Prncple of Conservaton of Lnear Momentum 23
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One-Dmensonal Elastc Collsons Consder two colldng objects wth masses m1 and m2, ntal veloctes v and v and fnal veloctes v and v, respectvely 1 2 1f 2 f Both lnear momentum and knetc energy are conserved. Lnear momentum conservaton: mv mv mv mv (eqs.1) 1 1 1 2 1 1f 1 2 f 2 2 2 2 mv 1 1 mv mv 1 2 1 1f mv 2 2 f Knetc energy conservaton: (eqs.2) 2 2 2 2 We have two equatons and two unknowns, v and v 1 2 2 1f 1 2 m1m2 m1m2 1 2 1 2 f 1 2 m1m2 m1 m2 1f 2 f 1f 2 f If we solve equatons 1 and 2 for v and v we get the followng solutons: m m 2m v v v 2m m m v v v 25
Specal Case of elastc Collsons-Statonary Target v 0 The substtute v 0 n the two solutons for 2 1f 2 f v and 2 v m m 2m v v v v 1 2 2 1f 1 2 m1m2 m1m2 2m m m v v v v 1 2 1 2 f 1 2 m1 m2 m1 m2 1 2 1f 1 m1 m2 1 2 f 1 m1 m2 Below we examne several specal cases for whch we know the outcome of the collson from experence 1. Equal masses 1 2 m m mm v v v v m m 1 2 1f 1 1 m1m2 mm m 2m 2m v v v 1 2 f 1 1 1 m1 m2 mm 0 m m 2m The two colldng objects have exchanged veloctes v v v 1 v 2 = 0 m m v 1f = 0 m m v 2f x x 26
v 1 m 1 m 2 v 2 = 0 x v 2f v 1f m 1 m 2 1 2. A mass ve target m2 m1 <1 m2 m1 1 m m m v v v v 1 m 1 2 2 1f 1 1 1 m1 m m 2 1 m 1 2 2m m 1 2 m 1 v2 f v1 v1 2 v1 m m 1 m2 1 1 m2 m2 Body 1 (small mass) bounces back along the ncomng path wth ts speed practcally unchanged. Body 2 (large mass) moves forward wth a very small m m 1 speed because <1 2 x 2 m 27
v 1 v 2 = 0 m 2 v m 1f 1 v 2f m 1 m 2 x x 2. A massve projectle 1 2 m2 1 m m m v v v v 1 m 1 2 1 1f 1 1 1 m1 m m 2 2 v 2m v 2 v 2v 1 m 1 2 f 1 1 1 m1 m m 2 2 1 1 m m m m Body 1 (large mass) keeps on gong scarcely slowed by the collson. Body 2 (small mass) charges ahead at twce the speed of body 1 2 1 <1 28
Elastc Collson 29
Collsons n Two Dmensons In ths secton we wll remove the restrcton that the colldng objects move along one axs. Instead we assume that the two bodes that partcpate n the collson move n the xy-plane. Ther masses are m1 and m2 The lnear momentum of the sytem s conserved: p p p p If the system s elastc the knetc energy s also conserved: 2 1 2 1f 2 f K K K K 1 2 1f 2 f We assume that m s statonary and that after the collson partcle 1 and partcle 2 move at angles and wth the ntal drecton of moton of m axs: 1 1 1 1f 1 mv 2 2 f 2 1 2 1 In ths case the conservaton of momentum and knetc energy take the form: x mv mv cos cos (eqs.1) yaxs: 0 mv sn m v sn (eqs.2) 1 1f 1 2 2 f 2 1 2 1 2 1 2 mv 1 1 mv 1 2 f mv 2 2 f (eqs.3) We have three equatons and seven varables: 2 2 2 Two masses: m, m three speeds: v, v, v and two angles:,. If we know 1 2 1 1f 2 f the values of four of these parameters we can calculate the remanng three 1 2 30
Problem 72. Two 2.0 kg bodes, A and B collde. The veloctes before the collson are ˆ ˆ v (15 30 ) m/s and ( 10 ˆ 5.0 ˆ A j vb j) m/s. After the collson, v ( 5.0ˆ 20 ˆj) m/s. What are (a) the fnal velocty of B and (b) the change ' A n the total knetc energy (ncludng sgn)? (a) Conservaton of lnear momentum mples mv mv mv' mv'. A A B B A A B B Snce m A = m B = m = 2.0 kg, the masses dvde out and we obtan v (15 ˆ 30ˆj)m/s ( 10ˆ 5j)m/s ˆ ( 5ˆ 20ˆ B va vb va j)m/s (10 ˆ 15 ˆj) m/s. (b) The fnal and ntal knetc energes are c 1 2 1 2 1 2 2 2 2 2 Kf mv' A mv' B ( 20. ) ( 5) 20 10 15 80. 10 J 2 2 2 1 2 1 2 1 2 2 2 2 3 K mva mvb ( 2. 0) c15 30 ( 10) 5 h 13. 10 J. 2 2 2 The change knetc energy s then K = 5.0 10 2 J (that s, 500 J of the ntal knetc energy s lost). 31 h
Systems wth Varyng Mass: The Rocket A rocket of mass M and speed v ejects mass backwards dm at a constant rate. The ejected materal s expelled at a dt constant speed vrel relatve to the rocket. Thus the rocket loses mass and accelerates forward. We wll use the conservaton of lnear momentum to determne the speed v of the rocket In fgures (a) and (b) we show the rocket at tmes t and t dt. If we assume that there are no external forces actng on the rocket, lnear momentum s conserved p( t) p t dt Mv dmu M dm v dv (eqs.1) Here dm s a negatve number because the rocket's mass decreases wth tme t U s the velocty of the ejected gases wth respect to the nertal reference frame n whch we measure the rocket's speed v. We use the transformaton equaton for veloctes (Chapter 4) to express U n terms of vrel whch s measured wth respect to the rocket. U vdvvrel We substtute U n equaton 1 and we get: 32 Mdv dmvrel
Usng the conservaton of lnear momentum we derved the equaton of moton for the rocket Mdv dmv rel (eqs.2) We assume that materal s ejected from the rocret's nozzle at a constant rate dm R (eqs.3) Here R s a constant postve number, dt the postve mass rate of fuel consumpron. dv dm We devde both sdes of eqs.(2) by dt M vrel Rvr el Ma Rvre l dt dt (Frst rocket equaton) Here a s the rocket's acceleraton, Rv the thrust of the rocket engne. We use equaton 2 to determne the rocket's speed as functon of tme f f dm dm dv vrel We ntegrate both sdes dv vrel M M f v M M M f rel M rel rel M f M f v v v ln M ln v ln v v v f rel ln M M f (Second rocket equaton) rel v M v M v f M v O 33 M /M f
Problem 78. A 6090 kg space probe movng nose-frst toward Jupter at 105 m/s relatve to the Sun fres ts rocket engne, ejectng 80.0 kg of exhaust at a speed of 253 m/s relatve to the space probe. What s the fnal velocty of the probe? M 6090 kg vf v vrel ln 105 m/s (253 m/s) ln 108 m/s. M f 6010 kg 34