CHAPTER 8 SYSTES OF PARTICLES
CHAPTER 8 COLLISIONS 45 8. CENTER OF ASS The ceter of mass of a system of particles or a rigid body is the poit at which all of the mass are cosidered to be cocetrated there ad all eteral forces were applied there. I this sectio we wat to kow how to determie the ceter of mass of a system. y cm m m cm Figure 8. A system of two particles m ad m. The poit labeled cm is the positio of the ceter of mass of the system. Cosider a system of two masses m ad m located alog the -ais as show i Figure 8.. The positio cm of the ceter of mass of these two masses is defied to be cm m + m =. (8.) m + m d If m = m, = 0, ad = d, we fid that cm =, i.e., the ceter of mass lies midway betwee the two masses. For a system of -particles m, m,, m, the ceter of mass is cm m + m + Lm = (8.) m + m + Lm
46 cm = mi i (8.3) i= Where = m + m + L+ m is the total mass of the system. If the system of particles is distributed o three dimesio, the y ad the z coordiates of the ceter of mass are similarly defied by ad y cm = mi yi, (8.4) i= z cm = mi zi. (8.5) i= I vector otatio, the positio vector of the ceter of mass r cm ca be epressed as rcm = cmi + ycmj + zcmk (8.6) Or rcm = miri (8.7) i= To fid the ceter mass of a rigid body (cotiuous mass distributio) we treat the body as cosistig of so large umber of small elemets dm such that the sums of Equatios 8.3-8.5 become itegrals ad the coordiates of the ceter of mass become
CHAPTER 8 COLLISIONS 47 cm = dm, (8.8) ycm = ydm, (8.9) zcm = zdm, (8.0) The vector positio of the ceter of mass of a rigid body is epressed as r cm = dm r. (8.) The itegrals are to be evaluated over all the mass distributio of the object. Eample 8. Three particles of masses m = kg, m = kg, ad m 3 = 3 kg are located as show i Figure 8.0. Fid the ceter of mass of this system. Solutio The -compoet of the ceter of mass is y (m) kg m + m + m3 3 cm = m + m + m3 0 + + 3 4 4 = = =.33 m. + + 3 6 kg 3 kg (m) 4 Figure 8. Eample 8..
48 The y-compoet is m y + m y + m3 y y 3 cm = m + m + m3 + 0 + 3 0 = + + 3 = = 0.33m 6 The positio of the ceter of mass is therefore rcm =.33i + 0. 33j. m Eample 8. Show that the ceter of mass of a uiform rod of mass ad legth L lies midway betwee its eds. Solutio Let the rod be located alog the -ais as show i Figure 8.3. By symmetry it is obvious that y cm = z cm = 0. Let us take a small elemet of mass dm ad legth d. From Equatio 8.8, we have cm = dm y L dm d Figure 8.3 Eample 8.. To solve the itegral we eed to fid a relatio betwee the mass dm ad the variable. To fid such a relatio we defie the liear dm mass desity λ ( mass per uit legth), as λ = =. Now the d L above equatio becomes
CHAPTER 8 COLLISIONS 49 cm λ L = d = 0 λ L 0 λl = Substitutig for λ =, we get L L L cm = =. L 8. DYNAICS OF A SYSTE OF PARTICLES From Equatio 8.7 we have r cm = m i r i i= Differetiatig the above equatio with respect to time gives d( r ) d( = i ) cm r mi i= (8.) d(r Kowig that cm ) is the velocity of the ceter of mass ad d ( r i ) is the velocity of the i th particle, Equatio 8. becomes v cm = m i v i (8.3) i= Differetiatig Equatio 8.3 with respect to time leads to
50 Where a cm = m i a i (8.4) i= d (v cm ) is the acceleratio of the ceter of mass ad d ( v i ) is the acceleratio of the i th particle. From Newto's secod law we kow that m i a i represets the resultat force F i that acts o the i th particle. Thus we ca write Equatio 8.4 as a cm = F i (8.5) i= Remember that F i is the vector sum of the eteral forces actig o the i th particle ad the iteral forces resultig from the other particles of the system. From Newto's third law the iteral forces form actio-reactio pairs so that they cacel out i the sum of Equatio 8.5. So, the right had side of Equatio 8.5 is the vector sum of all the eteral forces F et that act o the system. Equatio 8.5 the reduces to a cm = Fet (8.6) Equatio 8.6, like ay vector equatio ca writte as three equatios correspodig to the compoets of a cm ad F et alog the coordiates aes, that is,
CHAPTER 8 COLLISIONS 5 F F F et et y et z = a = a = a cm cm y cm z (8.7) Equatio 8.6 tells that if o et eteral force actig o a system, the acceleratio of its ceter of mass is zero ad thus the velocity of the ceter of mass of the system remais uchaged. Eample 8.3 A shell is fired with a iitial speed v 0 at a agle of θ above the horizotal. At the top of the trajectory, the shell eplodes ito two equal fragmets. Oe fragmet, whose speed immediately after eplosio is zero, falls vertically dow, as show i Figure 8.4. How far from the iitial poit does the other fragmet lad. Solutio As the forces due to v o the θ eplosio is cm iteral, they do ot affect Figure 8.4 Eample 8.3. the motio of the ceter of mass. Sice the oly eteral force actig o the system is the force of gravity, the ceter of mass follows a parabolic path ( the dotted path show i Figure 8.4) as the projectile did ot eplode. From Eample 3. we obtai
5 v o si θ cm = g But from Equatio 8.3 we have m m + cm = Kowig that = cmad m m = = we get or cm = cm + 3 = cm = 3 vo si θ g Eample 8.4 A railroad car of mass ca move alog a smooth horizotal track. A ma of mass m is, iitially stadig at oe ed of the car, which is iitially at rest, as show i Figure 8.5. If the ma starts to walk toward the other ed of the car, describe the motio of the car. Solutio There is o eteral force actig o the ma-car system alog the horizotal directio. This meas that the velocity of the ceter of mass of the system will ot chage ad must remai zero, ad so the positio of the ceter of mass of the ma-car system is the same before ad after the ma starts to walk. To maitai the positio of the ceter of mass uchaged, the car will move i a directio opposite to the directio of the walkig ma, as it clear from Figure 8.5.
CHAPTER 8 COLLISIONS 53 g mg cm L mg g cm Figure 8.5 Eample 8.4. Let us ow study the motio of the car aalytically. Let be the positio of right ed of the car (at which the ma is iitially stads) relative to a fied ais, ad is the ew positio of the same ed. Assumig that the car is uiform ad the ma walks a distace L betwee its eds, the positio of the ceter of mass of the ma-car system whe the ma is at the right ed is ( + L) m + cm = m+ Whe the ma is ow at the left ed of the car the positio of the ceter of mass is
54 ( + L) + ( + L) m cm = m + Equatig the above two equatios we obtai m = L + m The last equatio tells that if the ma moves a distace L to the left, ml + m. the car will move to the right a distace ( ) 8.3 LINEAR OENTU The liear mometum p of a particle of mass m movig with velocity v is defied as p = mv (8.8) I SI uit system the mometum has a uit of kg.m/s. The compoets of mometum p i three dimesios are p = mv p y = mvy p z = mvz. Let us ow fid the relatioship betwee the liear mometum ad the force. From Newto's secod law we have dv d( mv) F = ma = m =, (8.9) where m is costat. Newto's secod law thus ca be writte as
CHAPTER 8 COLLISIONS 55 dp F = (8.0) For a system of particles, each with its ow mass ad velocity, the liear mometum P of the system as a whole is the vector sum of the liear mometa of each particle idividually, that is P = p + p + L + p = m v + m v + L + m v = mi vi (8.) i= Comparig this equatio with Equatio 8.3 we get P = v cm (8.) From Equatio 8. we defie the liear mometum of a system of particles as the product of the total mass of the system ad the velocity of its ceter of mass. Differetiatig Equatio 8. with respect to time we fid dp dv = cm = acm (8.3) Comparig Equatios 8.6 ad 8.3 we write dp F et = (8.4) Which is the geeralizatio of Newto's secod law to a system of particles.
56 8.4 CONSERVATION OF LINEAR OENTU If a system is isolated, that is the resultat eteral force actig o the system is zero, Equatio 8.4 gives or dp F = = 0, P = costat. (8.5) This importat relatio is called the coservatio of liear mometum priciple, which states that, if o et eteral force acts o a system, the the total mometum of the system remais costat. Equatio 8.5 ca be writte as P i = P f (8.6) This meas that, for a isolated system, the liear mometum of the system at some iitial poit i is equal to the liear mometum at some fial poit f. It is worth to metio that Equatios 8.5 ad 8.6 are vector equatios ad so both are equivalet to three separately equatios correspodig the three perpedicular aes. Depedig o the eteral force actig o the system, the liear mometum might be coserved i oe or two directios, but ot ecessary i all directios. I aother word, if oly oe compoet of the et eteral force actig o a system alog a ais is zero, the the compoet of the liear mometum of the system alog that ais is costat oly. The other two compoets i this case is ot costat.
CHAPTER 8 COLLISIONS 57 Eample 8.5 A cao of mass 000 kg rests o a smooth, horizotal surface as show i Figure 8.6. The cao fires, horizotally, a ball of mass 5 kg with a speed of 75 m/s relative to the earth. What is the velocity of the cao just after it fires the ball? 75 m/s Figure 8.6 Eample 8.5. Solutio We take our system to cosists of the cao ad the caoball. The two eteral forces, the force of gravity ad the ormal force, are both perpedicular to the motio of the system. Therefore, the -compoet of the liear mometum of the system is coserved. Before firig the liear mometum of the system P i is zero, while the liear mometum of the system just after firig P f is P f = mcvc + mb vb Where c ad b refer, respectively, to the cao ad the caoball. Coservatio of liear mometum i the horizotal directio requires that m cvc + mb vb = 0 Solvig for v c yields v c m = b vb mc = 5 000 75 = 0.94 m/s The egative sig idicates that the cao recoils to the right, i the directio opposite to the motio of the ball.
58 Eample 8.6 Two blocks of masses m = kg ad m = kg v k m m v are coected by a sprig of force costat k= 00 N/m. The two blocks are free to slide alog a Figure 8.7 Eample 8.6. frictioless horizotal surface, as show i Figure 8.7. The blocks are pushed i opposite directio compressig the sprig a distace of cm, ad the released from rest. Fid the velocities of the two blocks whe the sprig returs to its equilibrium state. Solutio Our system is the two blocks ad the sprig. Therefore, the total mometum i the horizotal directio is coserved. Kowig that the system is iitially at rest we obtai So we get 0 = m v + mv m v = v m Which gives the relatio betwee the two velocities at ay istat of the motio. Now applyig the coservatio of mechaical eergy priciple we ca write k = m v + m v Substitutig for v from the previous equatio we get m v + m v k m m =
CHAPTER 8 COLLISIONS 59 Solvig for v we obtai km (00)()(0.) = = = m + m + m v 4.0 m/s Agai usig the relatio betwee the two velocities we get v m = (4.0) = = v m.0 m/s
60 PROBLES 8. Two particles are located alog the -ais. The first particle with mass m =4 kg has the coordiates (m,0,0), while the other particle with mass m =8 kg has the coordiates (4m,0,0). Fid the coordiates of the ceter of mass of the system. 8. A particle of mass kg is located o = - m, ad a particle of mass 3 kg is o = 4 m. Fid the positio of the ceter of mass. m 3 y(m) m 4 3m (m) Figure 8.8 Problem 8.3. 8.3 Three masses are located as show i Figure 8.8. Fid the ceter of mass of the system. 8.4 Three thi rods each of legth L is arraged as show i Figure 8.9. The two vertical rod have equal mass, ad the horizotal rod has a mass. Fid the ceter of mass of the system. L L L Figure 8.9 Problem 8.4.
CHAPTER 8 COLLISIONS 6 8.5 Show that the ceter of mass for a rectagular plate of sides a ad b is at the ceter of the plate. y 8.6 A lamia i the shape of a right triagle, with dimesios a ad b as show i Figure 8.5, has a b uiform mass per uit area. Fid a the coordiates of the ceter of O mass of the lamia. Figure 8.0 Problem 8.5. 8.7 A square piece of side 0 cm is cut out of a square plate of side 30 cm. Fid the coordiates of the ceter of mass of the plate. 8.8 A 6 kg particle moves alog the -ais with a speed of 4 m/s. y(cm) 30 0 0 (cm) 0 0 30 Figure 8. Problem 8.7. A aother particle of mass 4 kg moves alog the -ais with a speed of 8 m/s. Calculate the velocity of the ceter of mass. 8.9 Two boys oe of mass 45 kg ad the other of mass 35 kg, stad o a frictioless, horizotal surface holdig a rigid rope of legth 5 m. Startig from the eds of the rope, the boys pull themselves alog the rope util they meet. How far will each boy move? 8.0 Two particles are iitially at rest ad.5 m apart. The two particles attract each other with a costat force of mn,
6 which is the oly force actig o the system. If the masses of the particles are kg ad 0.5 kg, fid a) the speed of the ceter of mass, b) the distace from the first particle's positio at which they collide. 8. At the istat the traffic light turs gree, a car with mass 500 kg starts from rest with costat acceleratio of m/s. At the same istat a truck of mass 3000 kg, travelig with costat speed of 0 m/s overtakes ad passes the car. a) How far beyod the traffic light is the ceter of mass of the car-truck system at t= s? b) What is the speed of the ceter of mass at that istat? 8. Cosider a body of mass -kg ad velocity of (i-3j) m/s. a) Fid the ad y compoets of mometum. b) The magitude of its total mometum. 8.3 Calculate the magitude of the liear mometum for the followig cases: a) a proto of mass.67 0-7 kg movig with a speed of 5 06 m/s, b) a 5-g bullet movig with a speed of 500 m/s, c) a 75-kg ma ruig at a speed of 0 m/s, ad d) the earth of mass 5.98 06 kg, movig with a orbital speed of.98 0 4 m/s. 8.4 Two blocks of masses m = 3 kg ad m = 6 kg are coected by a sprig of force costat k= 800 N/m. The two blocks are free to slide alog a frictioless horizotal surface, as show i Figure 8.. The blocks are pulled i opposite directio stretchig the sprig a distace of 0 cm, ad the released from rest.
CHAPTER 8 COLLISIONS 63 a) Fid the velocities of the two blocks whe the sprig is at its equilibrium state. b) What is the maimum compressig distace of the sprig? m v v k m v v k m m Figure 8. problem 8.4. Figure 8.3 problem 8.5. 8.5 A block of mass m = kg moves with speed of v =0 m/s to the right o a frictioless surface collides with a mass m=8 kg movig with velocity of v =3 m/s to the right. A light sprig of sprig costat k=000 N/m is coected to the block of mass m, as i Figure 8.3. Whe the blocks collide, what is the maimum compressio of the sprig? 8.6 A 60-kg studet is stadig o a cart of mass 0 kg. The cart, origially at rest, is free to slide o a smooth, horizotal surface. The studet begis to walk alog the cart at a costat velocity of 0.8 m/s relative to the cart. a) What is the studet's velocity relative to the groud? b) What is the velocity of the cart relative to the groud? 8.7 A block of mass 6 kg slidig o a frictioless surface eplodes ito two equal pieces. Oe piece goes south at 4 m/s, ad the other piece goes 30 o orth of west at 5 m/s. What was the origial velocity of the block? 8.8 A block of mass 0-kg iitially at rest eplodes ito three pieces. A 4.5-kg piece goes orth at 0 m/s, ad a -kg piece moves eastward at 60 m/s. a) Determie the magitude ad directio of the velocity of the third piece.
64 b) Fid the eergy of the eplosio. 8.9 A cart of mass 40 kg is travelig at a speed of 3 m/s alog a horizotal smooth surface. A 90-kg ma iitially ridig o the cart jumps off with zero horizotal speed. What is the fial speed of the cart? 8.0 A cao car of mass 5000 kg that rests o a smooth, horizotal surface is attached to a post by a sprig of force costat of 4 0 4 N/m, as show i Figure 8.4. The cao fires a ball of mass 50 kg with a speed of 00 m/s, directed 30 o above the horizotal. a) What is the velocity of the cao just after it fires the ball? b) Fid the maimum etesio of the sprig. 00 m/s 30 o Figure 8.4 Eample 8.0.