Momentum 1. MOMENTUM. An object of mass m traveling at velocity v has a linear momentum (or just B. momentum) p, given by. p m B v. (6.

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1 Mmentum 6 In this chapter we begin ur study f mre realistic systems in which the bjects are n lnger pint particles but have extensin in space. Up until nw we ve generally limited urselves t the dynamics f pint masses, first in ne dimensin and then generalized t tw and three dimensins. Indeed, nt all f the prblems we studied were limited t pint masses, but the bject s size and shape were nt relevant in the prblem and s were nt cnsidered. Fr such bjects we ve learned hw t describe and predict translatinal mtin using Newtn s laws, sme f the cmplicatins due t frictinal frces, and the imprtant cncept f energy. In general we can divide the mtin f real extended bdies int tw parts: translatinal mtin, described by fllwing a particular average crdinate f the bject, knwn as its center f mass as it mves abut, and all ther mtins with respect t this pint. This chapter fcuses n translatinal mtin f systems, r cllectins f bjects, and the fllwing chapter deals with rtatinal mtin. We begin this chapter by intrducing the imprtant cncept f mmentum. As we ve seen, all frces cme in pairwise interactins. When studying the interactins between different bjects, it turns ut that we can re-frmulate Newtn s secnd law in terms f mmentum. If the system we are studying is islated meaning that it des nt interact with the utside wrld then ur refrmulatin is particularly simple and leads t a new fundamental law, the law f cnservatin f mmentum. After seeing this fr a system f tw particles, we next define and learn hw t cmpute the center f mass f a system, that special average pint f a system at which all its mass appears t be cncentrated in rder t explain the net translatinal mtin f the system. The last sectin f the chapter shws hw t refrmulate the dynamics f translatinal mtin f any system in terms f the center f mass mmentum. Here we als see the general frmulatin f cnservatin f mmentum. 1. MOMENTUM Thus far in ur discussins f dynamics we have fcused n frces as the rigin f mtin accrding t Newtn s laws. There is a very useful alternative apprach based n mmentum that we wish t develp in this chapter. Very ften this alternative apprach is t be preferred because it des nt hinge n the specific frces r interactins between bjects, which are usually unknwn r nly incmpletely understd. In this sectin, we first intrduce mmentum, the basic quantity used in this apprach, fr a particle. Then we refrmulate Newtn s secnd law using mmentum and shw hw this leads t the cnservatin f mmentum principle fr a cllectin f particles. Later in this chapter we generalize this apprach t arbitrary cllectins f extended bjects. An bject f mass m traveling at velcity v has a linear mmentum (r just mmentum) p, given by p m v. (6.1) J. Newman, Physics f the Life Sciences, DOI: / _6, Springer Science+usiness Media, LLC 2008 M OMENTUM 139

2 Newtn s secnd law fr an bject can be written in terms f mmentum as F net dp dt Using the definitin f p (Equatin (6.1)) and the prduct rule fr derivatives, we can write this as v) F net d(m m d v. dt dt v dm dt In the case when the mass is nt changing the last term vanishes and using the definitin f acceleratin we get the usual frm f F net ma. In cases where the mass is changing (e.g., a rcket ejecting substantial amunts f fuel), the full expressin is needed and this frm f Newtn s secnd law is the crrect expressin.. Nte that mmentum is a vectr quantity, defined as the prduct f the mass, an intrinsic prperty f the bject, and its velcity, a quantity depending n its mtin. It has units f kg-m/s, which have n ther special name. Clearly, based n Newtn s first law, a particle with n net frce n it will maintain a cnstant mmentum. When the particle feels a net frce, due t sme interactin, its mmentum will change with time. Als clearly, based n Newtn s secnd law, the larger the interactin (frce) acting n the particle, the greater will be the change in its mmentum. Hw des the mmentum f a particle cntrast with its velcity? First, we nte that bth f these quantities are vectrs, in fact with the same directin. If we cmpare tw particles f different mass traveling at the same velcity, the ne with larger mass will als have prprtinally larger mmentum. Fr example, a truck with fur times the mass f a car, bth traveling at the same speed alng a highway, has fur times the mmentum f the car, in accrd with ur cllquial usage f the wrd mmentum. On the ther hand, if the same truck is traveling at nly 1/4 the velcity f the car, then bth vehicles have the same mmentum. Hw des the mmentum f a particle cntrast with its kinetic energy? Nw, nte that these are very different quantities, with kinetic energy a scalar and mmentum a vectr. A particle with a given mass will have its mmentum dubled if its velcity dubles, but will have its kinetic energy quadrupled in that case. Kinetic energy is prduced by ding wrk n a particle, as we ve seen in the wrk kinetic energy therem. Hw is mmentum prduced? Well, clearly they are related, but the direct answer is that mmentum is prduced by frces acting n the particle as we nw shw. Newtn s secnd law fr an bject can be written in terms f its mmentum by nting that ma is defined as and because m is cnstant, we can further write that We therefre find that m lim t : 0 v t, mv p m a lim lim t :0 t t :0 t. p F net lim t :0 t. (6.2) This is actually the frm that Newtn prpsed fr the secnd law and is mre general than the frm F net ma, because it allws fr cases in which the mass f an bject may change with time. Such a situatin might arise when mass is either being added r remved frm the bject ver time (see the bxed discussin). Fr example, a rcket burns fuel and decreases its mass by ejecting the waste gases r the mass f a bat that is drifting by a pier may suddenly increase when yu jump int it. In bth f these cases ur riginal frm f F ma des nt apply because the mass is changing. Example 6.1 An E. cli bacterium f mass m kg is initially swimming at a cnstant velcity f 8 m/s tward the east. One ms later it is fund t be swimming at 10 m/s tward the nrth. Find the change in the E. cli s mmentum and the average external frce acting n the bacteria during the 1 ms time interval. 140 M OMENTUM

3 Slutin: It is very tempting t write that the change in the bacterium s mmentum is the prduct f its mass and the change in its speed (10 8) 2 m/s. This temptatin must be strngly resisted because it is the change in the velcity vectr that is apprpriate and this is nt a ne-dimensinal prblem. Figure 6.1 shws a vectr diagram fr the initial and final mmenta and the change in mmentum f the bacterium. Frm the figure it is clear that the change in mmentum is fund frm the hyptenuse f the triangle frmed s that mv fin mv ini mv ini θ mv fin FIGURE 6.1 Vectr subtractin fr Example 6.1. p m1 vini 2 v 2 fin The directin f this mmentum change is given by where the angle is measured west f nrth as shwn in the figure. The average frce acting ver this interval f time is then given by Equatin (6.2) (withut the limit) and is fund t be F p t kg # m/s in the same directin as the mmentum change. u tan 1 a 8 b 39, N Example 6.2 The fastest passenger elevatr in the wrld (in a 70-stry building in Ykhama, Japan) attains a maximum speed f 12.5 m/s (28 mph) taking passengers frm the grund t the tp flr in 40 s. Find the maximum change in yur mmentum if yu were t ride in this elevatr. What is the net change in yur mmentum fr the entire trip? Slutin: The maximum change in yur mmentum wuld ccur during the acceleratin r deceleratin phase f the ride. Assuming yur mass t be 80 kg, during the acceleratin phase yur mmentum wuld increase frm zer t p (80 kg) (12.5 m/s) 1000 kg m/s, s that yur maximum change in mmentum wuld just be 1000 kg m/s. Fr the entire trip t the 70th flr yur net change in mmentum is zer because bth yur starting and ending mmentum are zer. Suppse that tw therwise islated pint particles underg a cllisin. We wuld like t understand what ccurs and be able t predict the utcme. When far enugh apart, the tw particles mve independently and d nt interact. They will each have sme mmentum and if they are t cllide must be mving alng a line cnnecting them; let s call this the x-axis and we see that this prblem fr tw-pint particles is really ne-dimensinal. Mmentum is a particularly useful cncept in this situatin, as we shw. Suppse that particle #1 has mmentum p 1 and particle #2 has mmentum p 2, bth directed alng the x-axis. Fr them t cllide they must be mving tward each ther, but they might bth be mving in the same directin with ne M OMENTUM 141

4 particle catching up t the ther, s let s label the mmenta as bth psitive fr this discussin. If we write Equatin (6.2) fr each f the particles we have (6.3) where the nly frce n each particle is frm the ther ne. These frces need nt be cntact frces acting nly during a (macrscpic) cntact between the tw particles, but can als be lng-range frces acting ver lng distances. Nw, using Newtn s third law, we knw that these tw frces are reactin-pair frces and are always equal and ppsite t each ther. We can cnclude then that because the vectr sum f the tw frces always adds t zer, we must have at all times that p 1 t p 2 t F 2 n 1 p 1 t 0 r and F 1 n 2 p 2 t, ( p 1 p 2 ) 0 r ( p t 1 p 2 ) 0. (6.4) Fr this t be true it must be that the net mmentum f the tw particles remains cnstant with time. We say that, in this situatin, mmentum is cnserved. We have specifically written these last few steps using vectrs and in a general way t shw the pwer f the law f cnservatin f mmentum, even thugh ur current example is ne-dimensinal. All we have used in this derivatin are Newtn s laws (specifically the secnd law written in terms f mmentum and the third law) and the fact that the particles were therwise islated, nt interacting with any ther bjects. Thus, we really have prven that any tw islated bjects, nt necessarily pint particles, that cllide will have a ttal mmentum that remains cnstant (Figure 6.2). Furthermre, even if there are external frces acting n the tw particles, as lng as there are n external frces acting alng the directin f their mtin, mmentum will still be cnserved. Fr example, mmentum will be cnserved fr hrizntal (frictinless) mtin f tw clliding bjects even thugh gravity may act vertically. We shw this in a cuple f examples just belw. What des this tell us abut the interactins between the tw particles and the utcme f the cllisin? The beauty f this frmulatin is that the utcme is independent f the interactins; we d nt need t knw anything abut the details f the interactin in rder t predict smething abut the utcme. All we need t knw is cntained in Equatins (6.3) and (6.4). During the cllisin the tw bjects will exert equal and ppsite frces n each ther fr sme perid f time. If the cllisin invlves shrt-range frces, s that the cllisin time t is shrt, then the prduct f the (typically) large frce n ne particle frm the ther and the shrt cllisin time is called the impulse, Impulse F t p p final p initial. (6.5) FIGURE 6.2 Even tw clliding galaxies cnserve mmentum. 142 M OMENTUM

5 The impulse represents the net effect f a cllisin between tw bjects. It lumps tgether the acting frce and its duratin int a single parameter that is able t predict the change in mmentum f the particle due t the cllisin. Figure 6.3 shws a plt f a typical interactin frce n a particle as a functin f time during a cllisin. The impulse represents the area under this curve, equal t the average frce acting multiplied by the duratin t. Suppse, fr example, the tw bjects are identical, with the same mass m, and are traveling tward each ther at the same speed v. Then, althugh each bject has mmentum mv, the net mmentum befre the cllisin is, in fact, zer. D yu see why? (Remember that mmentum is a vectr!) In this case, cnservatin f mmentum predicts that the final mmentum must be zer as well. There are tw pssible final situatins fr which the final mmentum can be zer. In ne case the tw particles stick tgether and cme t rest, whereas in the ther case they bunce ff each ther and g ff in ppsite directins with the same magnitude f mmentum that they had, and thus at the same speed. Althugh bth f these situatins cnserve mmentum, they differ in whether they cnserve kinetic energy. The tw particles that stick tgether and cme t rest clearly have lst their kinetic energy, giving it up t ther frms f energy such as sund and heat, because we knw that ultimately energy must be cnserved. In mre cmplex situatins with tw unequal mass bjects traveling at different speeds, the algebra becmes a bit mre invlved and the pssible utcmes will depend n whether kinetic energy is cnserved. We d nt dwell n these situatins in detail, but simply pint ut that cnservatin f mmentum ffers a majr additinal tl in their study. In the third sectin f this chapter we generalize ur frmulatin f cnservatin f mmentum t mre cmplex systems. A few examples shuld help yu t appreciate the pwer f this new cnservatin law. F F Δt FIGURE 6.3 Typical frce acting n a particle during a cllisin. Usually the frce is large and shrt-lived. The area under the curve equals the impulse, which is als equal t the prduct f the average frce and the cllisin duratin because the area under the rectangle equals that under the frce curve. t Example 6.3 A 60 kg by dives hrizntally with a speed f 2 m/s frm a 100 kg rwbat at rest in a lake. Ignring the frictinal frces f the water, what is the recil velcity f the bat? Slutin: Since there are n external hrizntal frces acting (we have ignred the frictinal resistance frce f the water here), mmentum is cnserved as the by dives ff the bat. ecause the initial mmentum f the (by bat) system is zer, the ttal mmentum immediately after the by dives ff the bat must als be zer s that the by and the bat must have equal, but ppsitely directed, mmenta. Nte that it is the mmenta that must be equal and ppsite, nt the velcities. In equatin frm P ini P fin, with P ini 0 and P fin P by P bat. We therefre have that 0 (60 kg) (2 m/s) (100 kg) v bat, s that the bat s recil velcity is fund t be 1.2 m/s in the directin ppsite t the by s velcity. Example 6.4 Tw ice skaters, bth traveling at a speed f 5 m/s and heading straight tward each ther, cllide and lck arms tgether. If their masses are 80 g and 50 kg, find the velcity with which they mve tgether after the cllisin. (Cntinued) M OMENTUM 143

6 Slutin: There are n hrizntal external frces acting, s therefre mmentum is cnserved and we knw that the sum f the skaters tw initial mmenta is equal t their cmbined final mmentum. Initially their mmentum is P ini (80 kg) (5 m/s) (50 kg) (5 m/s) 150 kg-m/s in the directin the 80 kg skater is traveling. When they lck tgether, their cmbined mass is 130 kg and we must have that P fin (130 kg) v fin P ini 150 kg-m/s, s that v fin 1.2 m/s in the directin the heavier skater was traveling. In Example 6.3 we ignred the fluid medium and its frictinal frce. The surrunding fluid medium is ften f primary imprtance. Let s turn ur attentin t the prblem f animal lcmtin and, in particular, the mtin f sea creatures such as the squid r jellyfish. These creatures, and indeed all animals that swim r fly thrugh a fluid medium, mve by virtue f reactin frces prvided by the surrunding fluid medium. The jellyfish prpels itself by jet prpulsin, ejecting a vlume f water in a jet that prvides a thrust frce in the ppsite directin. Fish and birds generate thrust in a mre cntinuus fashin by pushing back n the fluid medium with fins r wings (Figure 6.4). In any case, we can analyze such lcmtin in either f tw ways: a difficult methd using the detailed reactin frces r much mre easily using mmentum. Let s discuss the jet prpulsin f a jellyfish in rder t derive an expressin fr the thrust prpelling it. We can mdel the jellyfish as a balln that fills with water and then cllapses driving water ut in a jet (Figure 6.5). Let the initial mass f water cntained within the balln be m 0 and suppse that the cllapse results in a unifrm rate f decrease f the mass, m/ t. Then the rate at which mmentum is ejected frm the balln will be p t m t v, where we assume a cnstant velcity fr the jet f water expelled. y Newtn s secnd law, the rate f mmentum ejectin prvides a net frce, knwn in this cntext as the thrust. If we take the initial vlume f the jellyfish t be that f a 0.1 m radius sphere filled with water f density 1000 kg/m 3, then m 0 (vlume) (density) 4/3pr 3 r 4.2 kg f water. If this water is ejected in 1 s thrugh a 1 cm radius circular FIGURE 6.4 ird, rcket, r fish, prpulsin is by thrust, a reactin frce, that cnserves mmentum. 144 M OMENTUM

7 aperture, then we can first calculate the velcity f water flw. This is fund by assuming that the ttal vlume f water flws ut in a cylindrical jet whse length is prprtinal t the velcity; that is, AL m 0, where L vt. Knwing the mass m 0, density, crss-sectinal area A (.01 m) 2, and time t 1 s, we can calculate the water velcity t be v 13 m/s first, and then F m t v t find abut 55 N f thrust is generated. This is actually a greater frce than the weight f the initial water cntained in the balln. A similar analysis can explain the thrust f a rcket r that f a bird, but a realistic analysis will be mre cmplex because f the nncnstant velcities invlved in the prblem. FIGURE 6.5 Mdel f a jellyfish as a balln. 2. CENTER OF MASS The simplest systems are cmpsed f a single pint particle intrduced in the previus chapters. Here we begin ur systematic study f increasingly mre cmplicated systems f tw particles, f many particles, r f a single extended bject such as a persn. Fr any system there is a well-defined pint, the center f mass, at which the entire mass f the system can be cnsidered t be cncentrated in rder t understand its translatinal mtin. A rugh analgy t finding the center f mass can be made t lcating the ppulatin center f the United States. Rather than weighting lcatins by their mass, they are weighted, in this case, by their lcal ppulatins. This tw-dimensinal prblem n a map culd be attacked in a number f apprximate ways, ne f which we illustrate. Using census figures fr the state ppulatins and chsing sme apprpriate lcatin as the ppulatin center within each state (e.g., by specifying latitude and lngitude f its largest city) ne culd find the U.S. ppulatin center by separately averaging the latitude and lngitude f the states, weighting each by its ppulatin. Thus Califrnia, Texas, and New Yrk, tgether with mre than ne third f the U.S. ppulatin, dminate in the calculatins and we expect t find a ppulatin center smewhere in the Midwest, even thugh the Midwest ppulatin is nt particularly large. This example illustrates the ntin f weighting lcatins by a lcal prperty r characteristic. T intrduce the definitin f center f mass cnsider tw particles f masses m 1 and m 2 attached by a light (massless) rd f length L as shwn in Figure 6.6. If this system were tssed int the air it wuld translate and rtate abut befre landing n the grund. One special pint, the center f mass f the system, wuld travel in the same trajectry as a single particle f mass (m 1 m 2 ) launched with the same initial velcity (we shw this in the next sectin). Qualitatively this pint can be imagined t be determined by finding the balance pint alng the rd. That is, imagine mving yur finger alng the rd until yu can balance the rd with its masses n either end. That pint is als the center f mass. Fr example, if m 1 m 2 the center f mass wuld be lcated in the center f the rd at a distance f L/2 frm either end. If m 1 m 2, then the balance pint wuld be clser t m 1, but hw much clser? ecause the balance pint in Figure 6.6 will be clser t the mre massive particle, we want t define the center f mass as an average psitin f the tw particles, with mre massive particles cunting mre in the averaging prcess. We therefre define the center f mass alng ne dimensin, x cm, relative t an arbitrary rigin, as L m 1 m 2 FIGURE 6.6 Tw masses separated by a light rd. C ENTER OF M ASS 145

8 m 1 m 2 x cm a bx m 1 m 1 a b x 2 m 1 m 2, 2 (6.6) where x 1 and x 2 are the x-crdinates f masses m 1 and m 2, respectively. Example 6.5 Find the center f mass f the Earth mn system given that the mean radius f the Earth is m, the mean radius f the mn is m, the Earth mn mean separatin distance is m, and that the Earth is 81.5 times mre massive than the mn. Slutin: The Earth mn separatin is s much larger than the radius f either; therefre we can treat bth bdies as pint masses fr the purpses f this calculatin. With an rigin at the center f the Earth (see Figure 6.7), we can write M e L M m x FIGURE 6.7 The Earth mn system. x cm M e 102 M m (L) M e M m 1 1 M L L 4.63 * 10 6 m. e M m Thus, the center f mass f the Earth mn system actually lies within the Earth. We can generalize this definitin in a straightfrward way t systems f mre than tw particles by simply adding terms fr additinal pint masses in bth the numeratr and denminatr in Equatin (6.6). Hwever, with mre than tw particles, the system need nt be ne-dimensinal if the masses are nt c-linear. In this case we can als define the y- (and z-, if needed) cmpnents f the center f mass in a similar way and cmbine them by writing r cm (x. Using the cm, y cm, [z cm ]) summatin ntatin that indicates t sum ver all particles in the system, we can write (with a similar equatin fr z cm ) x cm a a m i M bx i and y cm a a m i M by i, (6.7) where M is the ttal mass f the system. The subscript i dentes a particular numbered particle and the summatin sign indicates that i is t be varied frm number 1 t the ttal number f particles in the system while perfrming the additins indicated. Example 6.6 Find the center f mass f a water mlecule using the fllwing data (Figure 6.8): radius f O 0.14 nm, radius f H 0.12 nm, bnd length f O H bnd nm, and H H angle subtended at O M OMENTUM

9 y y x x Slutin: We slve this prblem in tw different ways using tw different crdinate rigins t see that the answer is independent f the chsen rigin. (1): In the first slutin we set the rigin n the O center and use the axes shwn n the left. In this case the atms have their centers lcated at: O (0,0); H( cs 52.3, sin 52.3 ) ( 0.059, 0.077), where the O H bnd is the center-t-center distance and we have fund the x- and y-cmpnents f the H centers. We slve fr the x- and y-crdinates f the center f mass (taking the masses f O and H as 16 and 1) by writing and where the zer value fr y cm shuld be expected frm the fact that the tw H atms are symmetrically situated abve and belw the x-axis. (2): Using the crdinates shwn n the right in the figure the atms have their centers lcated at: O (0.097 cs 52.3, sin 52.3 ) (0.059, 0.077), H (0, 0) and H (0, 2 # sin 52.3 ) (0, 0.15). Using the same basic relatins we write and FIGURE 6.8 Space-filling mdels f a water mlecule with tw different crdinate system rigins x cm nm, y cm 0, x cm nm, y cm nm. 18 Althugh these tw answers appear at first glance t be different, the shift in rigins must be accunted fr in cmparing them. The rigin n the right is lcated at the pint ( 0.059, 0.077) with respect t the rigin n the left and if we cmpare the actual spatial lcatin f the center f mass in bth parts (by, e.g., adding the rigin crdinates n the right with respect t thse n the left (Cntinued) C ENTER OF M ASS 147

10 t the answers in part 2), we find identical results, indicating a unique spatial lcatin fr the center f mass. A different apprach t the prblem might be t first recgnize that by symmetry the y cm must lie alng the x-axis using the left set f crdinates in Figure 6.8 and then t set up the prblem as a tw-mass system with the ttal H mass lcated at x nm and the O mass at the rigin. Try it. In the case f a slid extended bject, if it is unifrm thrughut and has sme symmetry we can ften determine its center f mass by inspectin, based n the ntin f a balance pint that was qualitatively intrduced with Figure 6.6. Fr example, a unifrm slid rd will have its balance pint, r center f mass, at its gemetric center. Even if an bject has multiple parts, each f which is unifrm thrughut and has sme symmetry, we can reduce the prblem t finding the center f mass f a cllectin f particles, ne fr each part f the bject with the mass f each part lcated at the center f mass f that part. This is illustrated in the fllwing example. Example 6.7 The slid bjects shwn in the three figures belw are all made frm the same unifrm material and have the same thickness. In part (c) there is a small hle in the larger circular plate. Find the center f mass f each bject using the crdinate system shwn. Take R 0.1 m and L 0.05 m. 2R y 2R R y L 3L y R R 2L L R 2R x 3L 2L a L x c b FIGURE 6.9 Unifrm slid bjects fr Example 6.7. x Slutin: ecause all the bjects are made f a unifrm material and have the same thickness, their masses are simply prprtinal t their areas. This is true because the mass m is equal t the prduct f the density f the material, its thickness t, and its area A, r m r t A. ecause bth the density and thickness are cnstants, m r A, and furthermre, because in Equatin (6.6) nly the rati f masses appears, we d nt need t knw the thickness r density f the materials as they cancel and d nt appear in the final result. In what fllws we therefre set the prprtinality cnstant simply equal t 1 and numerically equate masses and areas. We prceed by replacing each regular shape with a pint mass having the same ttal mass as that prtin f the entire bject and lcated at its center f mass (these are fund by inspectin because the shapes are highly symmetric). Each prblem then reduces t a set f pint masses, all lcated in the same plane. In part (c) we use a trick: let the hle have a negative mass accrding t its size and superimpse the larger slid circular plate with the 148 M OMENTUM

11 smaller circular plate f negative mass, thus canceling the mass within the hle regin! In (a) we have the fllwing three bjects: M 4R lcated at (R, 3R) (0.1, 0.3); M R at (R, R) (0.1, 0.1); and M R at (3R, 3R) (0.3, 0.3). We then find that and In (b) we have three pint masses: M 2L at (L/2, L) (0.025, 0.05); M 4L at (2L, 2.5L) (0.1, 0.125); and M 3L at (2L, 4.5L) (0.1, 0.225). The center f mass is given by and In (c), using the trick mentined abve, we have tw pint masses: M (2R) at (2R, 2R) (0.2, 0.2); and M (R/2) at (2R, 3R) (0.2, 0.3). Using the same methd we find as expected, and x cm 4R2 (R) pr 2 (R 3R) 4R 2 2pR m, y cm 4R2 (3R) pr 2 (R 3R) 4R 2 2pR m. x cm 2L2 (L/2) 4L 2 (2L) 3L 2 (2L) 9L m, y cm 2L2 (L) 4L 2 (2.5L) 3L 2 (4.5L) 9L m. x cm y cm 4pR 2 (2R) - p R2 4 (2R) 3.75pR m, 4pR 2 (2R) - p R2 4 (3R) 3.75pR m. This last number seems reasnable because with a hle cut ut we expect y cm t be smewhat less than 2R 0.2 m. Yu shuld g thrugh each answer, fllwing all the steps, and see that the center f mass psitin makes qualitative sense. In the general case f nnunifrm and/r nnsymmetric bjects, the center f mass can always be fund experimentally by finding the balance pint alng three mutually perpendicular axes, if pssible, r by suspending the bject separately frm three different pints, drawing vertical lines frm thse pints, and lking fr the intersectin f the three lines (Figure 6.10). The reasn why this latter methd wrks becmes clearer after we have discussed rtatinal mtin, but the center f mass must lie suspended vertically under the suspensin pint. Alternatively, ne can use mre cmplex mathematics t calculate the center f mass psitin. We shw a mre C ENTER OF M ASS 149

12 FIGURE 6.10 Nde I, a part f the Internatinal Space Statin, being readied (left) and having its center f mass determined by suspending it frm abve. FIGURE 6.11 (tp) A gd high jumper has a center f mass that actually ges under the bar in welldefined tw-dimensinal free-fall mtin while his flexible bdy ges ver it. (bttm) An unmanned Titan rcket expldes shrtly after takeff in Despite fragmenting int many pieces the center f mass cntinues in a well-defined trajectry (see Example 6.8). direct experimental apprach t finding the center f mass in the next sectin. There we shw that the center f mass translates abut in space as if all external frces act directly n the entire mass f the system lcated at its center f mass. 3. CENTER OF MASS MOTION: NEWTON S SECOND LAW AND CONSERVATION OF MOMENTUM In the last sectin we learned hw, in principle, t find the center f mass f any bject, and in practice, t find that pint fr a cllectin f particle masses r symmetric bjects. Here we shw that the translatinal mtin f a system f particles r an extended bject is fully described by knwledge f the center f mass mtin. The main gals f this sectin are t generalize Newtn s secnd law fr a particle t a very similar result fr the center f mass f a system and t generalize the law f cnservatin f mmentum. The derivatin f the generalizatin f Newtn s secnd law t a system f extended bjects is straightfrward using sme calculus (see bx n the next page), but therwise is cumbersme. The resulting Newtn s secnd law fr a system is F net, ext g F ext M a cm, (6.8) where the sum is ver all f the external frces acting n the system, M is the ttal mass f the system (assumed cnstant; a system that des nt exchange mass with its surrundings is knwn as a clsed system), and a cm is the acceleratin f the center f mass. In this expressin the nly frces that prduce an acceleratin f the center f mass are frces exerted n the system by bjects that are external t the system, s-called external frces. All f the internal frces between the particles f the system cancel pairwise because they are equal and ppsite accrding t Newtn s third law. This equatin als applies t extended bjects because they can be cnsidered t be built up frm particles. We cnclude that the translatinal mtin f a system can be cmpletely described by replacing the entire system by a pint mass with ttal mass M lcated at the system s center f mass,, with nly external frces acting (Figure 6.11). r cm 150 M OMENTUM

13 As a byprduct f the derivatin f Equatin (6.8), we shw (in the bx) that the ttal mmentum f the system, the vectr sum f the individual particle mmenta, is equal t the mmentum f the center f mass, r the prduct f the ttal mass M and the center f mass velcity, v cm P cm M v cm g p i. (6.9) Thus an alternative way t write Equatin (6.8), in terms f the center f mass mmentum, is t :0 F net, ext lim P cm (6.10) We see that the center f mass mves as if all the mass f the system is lcated there and experiences the net external frce n the entire system. S, n matter whether the system is cmpsed f a single extended bject (such as the high jumper f Figure 6.11) r many independent parts (such as the explded rcket in that figure), the center f mass f the system mves in a well-defined trajectry based n the ttal mass and the net external frce n the system. Written in this frm we can deduce a very imprtant cnsequence: t. A derivatin f Equatin (6.8): Starting frm a rewriting f Equatin (6.6) fr the x-cmpnent f the center f mass M x cm m i, we can differentiate bth sides f the equatin with respect t time t find r P cm p i x i M v cm m i P cm p i, where and are the mmentum f the center f mass and the individual particles. Thus we see that the center f mass mmentum is equal t the ttal mmentum f the system f particles. If we further differentiate this equatin with respect t time we have Ma cm dp cm /dt d p, i /dt F i, net, v i In the absence f a net external frce n a system, the center f mass mmentum, r ttal mmentum f the system, des nt change with time, and is said t be cnserved. This is a statement f the principle f cnservatin f mmentum, a very pwerful and general result, which hlds fr all islated systems, thse with n net external frces applied. It is a fundamental principle that hlds n every scale f distance: n the atmic r nuclear scale as well as n the scale f the size f the universe. We saw a preliminary versin f this in the first sectin f this chapter fr cllisins between tw particles, but the principle is much mre general than we saw there. Cnservatin f mmentum is the secnd f a handful f cnservatin laws that we study in this bk. We have already learned the cnservatin f energy principle and seen its tremendus value as a tl in understanding mtin. Later n we demnstrate its value in all ther areas f physics that we study. Energy and mmentum cnservatin are tw f the crnerstnes f physics. ecause the mmentum f an islated system is cnstant, if we cmpute the ttal mmentum at any time, its value at any ther time will be the same vectr, namely the same value and in the same directin. Just as with energy cnservatin, we can use ur knwledge f the situatin at ne instant f time t find the ttal mmentum, which will remain cnstant as lng as there are n external frces acting. On the ther hand, unlike energy cnservatin, mmentum is a vectr and therefre a directin as well as a magnitude is fixed in time. We nte that the kinetic energy f a particle KE 1 2mv 2 can in fact be rewritten in terms f the particle s mmentum in place f its velcity. Using the definitin f the magnitude f the mmentum p mv, we have that KE p 2 /2m. Yu need t keep in mind that althugh the kinetic energy, a scalar, can be written in terms f the square f the particle s mmentum, the cnservatin laws f energy and f mmentum are tw different laws that keep different quantities cnstant. Fr a clsed system (ne with n exchange f mass with its surrundings) with n external frces acting, the ttal, r center f mass, mmentum will be cnserved as will the ttal mechanical energy. Hwever, the kinetic energy f the system may change because it can be exchanged fr ptential where we have used Newtn s secnd law fr each particle, assumed that the ttal mass f the system is cnstant, and F is i,net the net frce n the ith particle. T cmplete the derivatin f Equatin (6.8), we nte that the frces n particle i are f tw types: external, arising frm bjects utside the system, and internal, arising frm ther particles in the system. These latter internal frces cancel pairwise in the summatin because a frce n particle 2 frm particle 3 is equal and ppsite t the frce n particle 3 frm particle 2 and all pssible pairs f frces will be summed. The final result is Equatin (6.8). Newtn s law can be generalized still further t a system f extended bjects with n change t Equatin (6.8). C ENTER OF M ASS M OTION 151

14 energy within the system. In the first sectin f this chapter we saw a few examples f the applicatin f mmentum cnservatin t the cllisin between tw bjects. Tw quite different examples shuld help t prvide an appreciatin fr the pwer f the cnservatin f mmentum principle. Example 6.8 A rcket f mass M expldes int three pieces at the tp f its trajectry where it had been traveling hrizntally at a speed v 10 m/s at the mment f the explsin. If ne fragment f mass 0.25 M falls vertically at a speed f v m/s, a secnd fragment f mass 0.5 M cntinues in the riginal directin, and the third fragment exits in the frward directin at a 45 angle abve the hrizntal (see Figure 6.12), find the final velcities f the secnd and third fragments. Als cmpare the initial and final kinetic energies t see hw much was lst r gained. M v 1 v 2 v 3 FIGURE 6.12 The rcket befre (tp) and just after (bttm) the explsin f Example 6.8. v Slutin: Althugh the rcket is nt an islated system, the frces in the explsin are assumed t be s much greater than the weight f the rcket that we can neglect gravity at the mment f the explsin. This situatin is very similar t that f any cllisin in which tw bjects interact very strngly fr a very shrt time as, fr example, when a tennis racket hits a ball. In all such cases we can neglect gravity during the cllisin and treat the system as islated. Therefre the initial mmentum f the rcket P ini Mv in the hrizntal directin must be cnserved during the explsin and the sum f the mmenta f the three fragments must add up t exactly this same P ini value. Using vectr additin, we can write that cnservatin f mmentum in the hrizntal directin implies Mv 1 2 Mv Mv 3 cs 45, where the velcities are labeled as in the figure. Nte that the first fragment falls vertically and des nt cntribute t this equatin fr the hrizntal mmenta. Cnservatin f mmentum in the vertical directin gives a secnd equatin Mv Mv 3 sin 45. Substituting that v m/s, we find first, frm the secnd equatin abve after canceling the cmmn factr M, that v m/s, sin 45 and then, n substitutin int the first equatin abve, that v # 1.7cs 45, s that v 2 19 m/s. The initial kinetic energy is KE i 1 2mv 2 50 M, and the ttal final kinetic energy is KE f 1 2 (M/4)(1.2) (M/2)(19) (M/4)(1.7) M, bth measured in J with M in kg. The kinetic energy has increased by ver 80% with the 152 M OMENTUM

15 excess cming frm the chemical energy released in the explsin. Kinetic energy alne is nt cnserved in this example, but mmentum is. On the ther hand, the general principle f cnservatin f energy is beyed with the ttal mechanical, chemical, and ther surces f energy remaining cnstant fr the rcket. In the abve example we ve seen hw in an explsin we can use cnservatin f mmentum t learn abut the mtin f the final pieces. Similarly in a cllisin between bjects we can use cnservatin f mmentum during the cllisin t learn abut the final mtins f the bjects after cllisin. Fr micrscpic bjects that interact during a cllisin, f say atms, the frces are all cnservative and the cllisins, aside frm cnserving mmentum, tend t be elastic, cnserving energy as well. In mst cases f macrscpic bjects clliding, the cllisins tend t be inelastic, s that energy is lst (r gained in the explsin f the last example) even thugh mmentum is cnserved during the cllisin. The next tw examples illustrate sme f these pssibilities. Example 6.9 A hckey puck f 0.5 kg mass traveling at a speed f 5 m/s cllides with an identical statinary puck in a glancing (nt head-n) cllisin. If the first puck is deflected by 30 and travels with a final speed f 3 m/s, find the final velcity f the puck that was hit if it mves ff at a 45 angle as shwn. Ignre any frictin between the ice and pucks. befre after Slutin: ecause there are n external hrizntal frces acting, mmentum is cnserved. With the initial directin f mtin chsen as the x-axis, the initial mmentum is nly p ix m v 0 (0.5)(5) 2.5 kg m/s. After the cllisin, bth pucks have x mmentum (cmpnents f their mmentum vectrs) that must add up t the initial mmentum as 2.5 kg m/s (0.5 kg)(3 cs 30 m/s) (0.5 kg)( v cs 45), where v is the final velcity f the secnd puck. Slving fr v we find v 3.4 m/s. Ntice that energy is nt cnserved in this cllisin because the initial KE 1/2 (0.5)(5) J, whereas the sum f the final KE 1/2 (0.5)(3) 2 1/2 (0.5)(3.4) 2 5.1, amunting t a lss f abut 18% f the initial KE. Example 6.10 Suppse ne prtn mving with a speed v 0 cllides with a secnd prtn initially at rest. If ne f the prtns emerges at a given angle frm the incident directin, find the speeds f bth after the cllisin and the angle (Cntinued) C ENTER OF M ASS M OTION 153

16 at which the secnd prtn emerges frm the cllisin. Wrk this ut in general and then take v m/s and 30. y v 2 v Slutin: We are searching fr three unknwn quantities, and s require three independent equatins. We wrk this prblem ut withut substituting in numbers s that we can learn abut the general case. These equatins can be btained frm cnservatin f mmentum (tw equatins, ne fr the incident directin, say x and ne fr the directin perpendicular t that, say y) and cnservatin f energy. Cnservatin f mmentum in the x-directin gives p 0x mv 0 p fx mv 2 cs u mv 1 cs f, (1) where m is the prtn mass, and v 1 and v 2 are the prtns final velcities. Cnservatin f mmentum in the y-directin gives θ φ v 1 x p 0y 0 m v 1 sin f m v 2 sin u. (2) Energy cnservatin gives us the equatin 1 (3) 2 mv mv mv 2 2. We nw have ur three equatins in three unknwns this was the physics part f the prblem and the remainder f the prblem is t slve fr them algebraically. This is a bit cmplicated, s fllw clsely. First we can cancel all the m s in all three equatins and if we then slve fr v 2 cs and v 2 sin in Equatins (1) and (2) we have v 2 cs u v 0 v 1 cs f, and v 2 sin u v 1 sin f. (4) We can then square each f these and add them tgether t find, using sin 2 cs 2 1, that but frm Equatin (3), after canceling 1/2 m frm each term, we have that v 0 2 v 1 2 v 2 2 v 1 2 (v 0 v 1 cs f) 2 (v 1 sin f) 2. Expanding ut the terms in parentheses and cmbining again we have Simplifying this, we have v 2 2 (v 0 v 1 cs f 2 ) (v 1 sin f) 2, v v 1 2 v v 0 v 1 cs f. 2v 1 (v 1 v 0 cs f) 0, which has the slutins v 1 0 r v 1 v 0 cs. The slutin v 1 0 gives v 2 v 0 indicating a head-n slutin in which ne prtn stps and the ther ges n in the frward directin (we must reject the negative slutin fr v 2 as unphysical.) The ther slutin, frm Equatin (3), gives v 2 v 0 sin. In that case t find, after substitutin fr v 1 in Equatins (4) we have that 154 M OMENTUM

17 v 2 cs u v 0 v 0 cs 2 f v 0 (1 cs 2 f) v 0 sin 2 f and v 2 sin u v 0 sin f cs f. Dividing these equatins we find that tan u 1/tan f. Therefre, given an angle fr the first prtn, the secnd emerges such that 90. In ur case, if v and 30, we find v m/s, v m/s and 60. Yu can check these by direct substitutin int Equatins (1) (3), after canceling m. In this chapter we have learned hw t describe the translatinal mtin f a system f extended bjects using the center f mass and mmentum cnservatin. In general such systems will have tw ther types f mtin: verall rtatinal mtin and internal mtins. Internal mtins include all relative mtins f prtins f the system ther than verall rtatinal tumbling, including shape changes as well as vibratinal mtins. We cme back t this tpic much later in the bk in discussins n the structure f matter. Rtatinal mtin is taken up in detail in the next chapter. CHAPTER SUMMARY The mmentum f a particle f mass m is defined as x cm g a m i M bx i and y cm g a m i M by i. (6.7) p mv. (6.1) Using this definitin, we can write Newtn s secnd law fr the particle in terms f its mmentum as Then fr a system f such masses, Newtn s secnd law can be shwn t be F net, ext g F ext M a cm, (6.8) p F net lim t :0 t. (6.2) r written in term f mmentum, as If tw particles interact in the absence f any external frces, then their ttal mmentum is cnserved, meaning that it will remain a cnstant in time. A useful cncept in discussing cllisins is the impulse, defined by the prduct f the cllisin frce and its duratin, and shwn t equal the change in mmentum f the bject: Impulse F t p p final p initial. (6.5) Fr a cllectin f masses m i, each lcated at (x i, y i ), with ttal mass M, we define the center f mass t be lcated at the pint t :0 F net, ext lim P cm (6.10) In the case f an islated system, with n external frces acting, the center f mass mmentum, equal t the ttal mmentum f the system, is cnserved: P ttal cnstant. This is a vectr equatin and, in general, stands fr the three independent equatins fr which each cmpnent (x, y, and z) f mmentum remains cnstant. t. C HAPTER S UMMARY 155

18 QUESTIONS 1. What are the differences and similarities between mmentum and velcity? etween mmentum and kinetic energy? 2. Is it pssible fr the center f mass f a slid bject t lie physically utside the bject? Give an example r tw t supprt yur assertin. 3. Fr unifrm (cnstant density) bjects, is it true that the center f mass must lie alng a symmetry axis, if there is ne? Give sme examples. 4. Explain, in yur wn wrds, why nly external frces result in a change in the center f mass mmentum f a system f interacting particles r an extended bject. 5. Carefully define an islated system. Give sme examples and explain why it is that mmentum is nly cnserved fr an islated system. 6. Is a rcket traveling in uter space an example f an islated system? If s, hw can the rcket change its mmentum if it is t be cnserved? 7. Tw identical twins f equal mass are ice skating tward each ther at the same speed. What happens when they cllide? What happened t their mmentum? 8. In a cllisin f a tennis ball with a racket, why shuld the tensin in the strings f the racket be made as large as pssible? 9. When a cllisin between tw bjects ccurs and there is a net change in the mmentum f ne bject there are very large frces acting fr a very shrt time. The prduct f the average frce n the bject during the cllisin and the duratin f the cllisin is called the impulse. If a tennis ball f mass m and velcity v bunces ff a wall and rebunds with the same speed, what is the impulse n the ball? Why des a new tennis ball bunce higher than an lder tennis ball when drpped frm the same height? 4. The magnitude f the average frce n the car during this time is (a) 720 N, (b) 73 N, (c) 420 N, (d) 210 N. 5. The directin f the average frce n the car during this time is (a) 38 S f W, (b) 38 N f W, (c) 52 S f W, (d) 52 N f W. 6. An 80 kg man and a 40 kg girl are skating n smth level ice. Initially, they are in cntact and at rest. The man pushes the girl away frm him with a frce f 30 N. Immediately after they are n lnger in cntact the girl s speed is 2 m/s. At the same instant the man s speed (a) must be zer, (b) must be 2 m/s als, (c) must be 1 m/s, (d) depends n hw much frce the girl exerts n the man. 7. A 40 kg by is standing n a 5 kg skatebard at rest. If he jumps ff with a hrizntal velcity f 1 m/s, neglecting frictin the recil velcity f the skatebard is (a) 1 m/s, (b) 0.1 m/s, (c) 0.03 m/s, (d) 8 m/s. 8. A 50 kg astrnaut in rbit can give a 10 kg wrench a speed f 10 m/s by thrwing it. The speed the astrnaut will recil with after ding s will be (a) 0 m/s, (b) 2 m/s, (c) 10 m/s, (d) 50 m/s. 9. In a head-n cllisin between a seagull and a jet airplane (a) The mmentum f the airplane is exactly cnserved. (b) The ttal kinetic energy is exactly cnserved. (c) The magnitude f the change in mmentum f the seagull divided by the cllisin time equals the magnitude f the average frce n the jet. (d) The ttal mmentum is zer. (e) Nne f the abve is true. 10. A 0.1 kg meter stick has tw masses attached: 0.3 kg at 20 cm and a 0.4 kg at 100 cm. The center f mass f the system lies at the fllwing indicatr n the meter stick. (a) 57.5 cm, (b) 63.8 cm, (c) 65.7 cm, (d) 70 cm, (e) nne f the abve. MULTIPLE CHOICE QUESTIONS 1. A 3 kg mass has psitin crdinates ( 2, 2 m) and a 1 kg mass has psitin crdinates (3, 0 m). The center f mass f this system has crdinates (a) (1, 2 m), (b) ( 3, 6 m), (c) ( 0.75, 1.5 m) (d) (0, 0 m). 2. A 2 kg mass is at x 0 m, y 2 m, and a 3 kg mass is at x 2 m, y 0 m. The x- and y- crdinates, respectively, f the center f mass f this system are (a) 6/5 m, 4/5 m, (b) 2/5 m, 2/5 m, (c) 0, 0 m, (d) 2 m, 2 m. 3. A 5 kg bwling ball with a center f mass velcity f 4 m/s strikes the padded end f the bwling lane and cmes t rest in 0.01 s. The average frce exerted n the ball is (a) 400 N, (b) 2000 N, (c) zer, (d) 500 N. Questins 4 and 5 refer t a car weighing 900 N that is heading nrth at 14 m/s. It makes a sharp turn and heads west at 18 m/s. During the turn, a gd luck charm hanging frm the rear view mirrr is angled frm the vertical fr a ttal f 5 s. PROLEMS 1. Find the center f mass f the fllwing sets f pint masses. (a) A 2 kg mass at x 5 cm and a 5 kg mass at x 2cm (b) A 1 kg mass at y 0 and a 4 kg mass at y 10 cm (c) Three small bjects each f the same mass m, lcated at the fllwing pints (0,0), (0,10 cm), (10 cm, 0) (d) Pint mass m at (0,0), pint mass 3m at (0, 5 cm), pint mass 5m at (5 cm, 0) and pint mass m at (5, 5 cm) 2. Using Table 6.1, find the center f mass f (a) A persn standing upright with hands at sides (b) An utstretched arm and an arm bent upward at the elbw by a right angle (c) A persn bent ver s that there is a right angle between her straight legs and upper bdy/head and between her upper bdy and straight arms 156 M OMENTUM

19 Table 6.1 Distances and Masses f Prtins f the Typical Human dy (Expressed as % f Ttal Height and Mass) Hinge Pints Center f Mass (frm flr) (frm flr) Mass Neck 91.2 Head Shulder 81.2 Trunk/neck Elbw 67.2 Upper arms Hip 52.1 Lwer arms Wrist 46.2 Hands Knee 28.5 Upper legs Ankle 4.0 Lwer legs Feet Frm the text discussin, yu knw that the center f mass can be fund thrugh balancing methds, that is, suspending an bject frm a pint. This prcedure indicates that fr three equal masses situated at the vertices f an equilateral triangle, the center f mass will be at the intersectin f the three angle bisectrs f the triangle. Frm elementary gemetry therems, it is knwn that the three angle bisectr segments intersect at a pint that is 2/3 f a segment length away frm its angle vertex. Calculate the center f mass fr the mass arrangement shwn and cmpare its psitin t the intersectin f the angle bisectrs. Nte that because the height f the triangle is a 3/2, the methd is a physical manifestatin f the therem that the bisectrs f angles f an equilateral triangle intersect at the center f mass f the triangle (usually called the centrid by mathematicians). This is true whether the physical triangle is cnstructed f sides nly, f similar and unifrm crss-sectin, r if the triangle is a unifrm plate. a a 4. Calculate the center f mass fr three equal masses situated at the vertices f a right triangle. 5. Calculate the center f mass fr the arrangement f three masses als situated at the vertices f a right triangle, but where the masses are in the rati 3:4:5, with the largest ppsite the hyptenuse and the smallest ppsite the shrtest side. Cmpare the result with the previus prblem. 6. y symmetry, the center f mass f a unifrm regular plygn is at its center. Similarly, this is true fr an arrangement f equal masses situated at the vertices f such a plygn. Where is the center f mass fr each f the arrangements shwn, where nly a subset f the vertices f the plygn is ccupied by masses? Make a x L sure yu arrange the crdinate framewrk t take advantage f any remaining symmetry. (Hint: Missing masses can be represented as M M 0 mass, s that the sum f a negative mass can be added t the situatin with a full cmplement f masses at the vertices.) L 7. Cnsider the three spherical masses shwn. Hw far t the right f m 2 shuld m 3 be s that the center f mass f the entire arrangement is lcated exactly at the psitin f m 2? x 1 x 2 m 1 m 2 m 3 8. Cnsider a unifrm linear arrangement f ten masses, that is, with equal spacing between, ranging frm 1 t 10 kg, each 1 kg mre than the previus. Where is the center f mass f the assemblage? 9. Three unifrm spheres f radii R, 2R, and 3R lie in cntact with each ther frm left t right in the rder given with their centers alng the x-axis. Remembering that the vlume f a sphere is given by (4/3) r 3, find the psitin f the center f mass f the three spheres as measured frm the left edge f the smallest sphere. 10. Find the center f mass f a screwdriver with the fllwing characteristics: a wden cylindrical handle (density f wd kg/m 3 ; cylinder length and diameter 10 and 2 cm) and a steel cylindrical rd (density f steel kg/m 3 ; 15 cm lng and 0.5 cm in diameter, with an additinal 3 cm flat unifrmly tapered head with a triangular crss-sectin). 11. Three unifrm rds (identical except fr their lengths) frm the right triangle shwn with crdinates measured in meters. y (0,3) (0,0) (4,0) x Q UESTIONS/PROLEMS 157