Chapter 4 Conservation laws for systems of particles

Transcription

1 Chapte 4 Consevaton laws fo systems of patcles In ths chapte, we shall ntoduce the followng geneal concepts:. The powe, o ate of wok done by a foce. The total wok done by a foce 3. The knetc enegy of a patcle 4. The powe-knetc enegy and wok-knetc enegy elatons fo a sngle patcle 5. The concepts of a consevatve foce and a consevatve system 6. The powe-total enegy and wok-total enegy elatons fo a consevatve system 7. Enegy consevaton fo a consevatve system 8. The lnea mpulse of a foce 9. The lnea momentum of a patcle (o system of patcles) 0. The lnea mpulse - lnea momentum elatons fo a sngle patcle. Lnea mpulse-momentum elatons fo a system of patcles. Consevaton of lnea momentum fo a system 3. Analyzng collsons between patcles usng lnea momentum 4. The angula mpulse of a foce 5. The angula momentum of a patcle 6. The angula mpulse angula momentum elaton fo a sngle patcle We wll also llustate how these concepts can be used n engneeng calculatons. As you wll see, to applyng these pncples to engneeng calculatons you wll need two thngs: () a thoough undestandng of the pncples themselves; and () Physcal nsght nto how engneeng systems behave, so you can see how to apply the theoy to pactce. The fst s easy. The second s had, but pactce wll help. 4. Wok, Powe, Potental Enegy and Knetc Enegy elatons fo patcles The concepts of wok, powe and enegy ae among the most poweful deas n the physcal scences. The most mpotant applcaton s n the feld of themodynamcs, whch descbes the exchange of enegy between nteactng systems. In addton, concepts of enegy cay ove to elatvstc systems and quantum mechancs, whee the classcal vesons of Newton s laws themselves no longe apply. In ths secton, we develop the basc defntons of mechancal wok and enegy, and show how they can be used to analyze moton of dynamcal systems. Futue couses wll expand on these concepts futhe. 4.. Defnton of the powe and wok done by a foce Suppose that a foce F acts on a patcle that moves wth speed v. k F(t) P By defnton: The Powe developed by the foce, (o the ate of wok done by the foce) s P = F v. O If both foce and velocty ae expessed n Catesan components, then P= Fv + Fv + Fv x x y y z z

2 Powe has unts of Nm/s, o `Watts n SI unts. The wok done by the foce dung a tme nteval t 0 t t s t W = F v dt t0 The wok done by the foce can also be calculated by ntegatng the foce vecto along the path taveled by the foce, as whee 0, W = F d 0 ae the ntal and fnal postons of the foce. 0 k O F(t) P Wok has unts of Nm n SI unts, o `Joules A movng foce can do wok on a patcle, o on any movng obect. Fo example, f a foce acts to stetch a spng, t s sad to do wok on the spng. 4.. Smple examples of powe and wok calculatons Example : An acaft wth mass kg flyng at 00 knots (0m/s) clmbs at 000ft/mn. Calculate the ate of wok done on the acaft by gavty. F D The gavtatonal foce s mg, and the velocty vecto of the acaft s v= vx+ vy. The ate of wok done on the acaft s theefoe P = F v = mgv y Substtutng numbes gves F T mg Example : Calculate a fomula fo the wok equed to stetch a spng wth stffness k and unstetched length L 0 fom length l 0 to length l. The fgue shows a spng that held fxed at A and s stetched n the hozontal decton by a foce F= Fx actng at B. At some nstant the spng A B has length l 0 < x< l. The spng foce law states that the foce actng on the spng at B s elated to the length of the spng x by F= Fx= kx ( L0 ) The poston vecto of the foce s = x, and theefoe the wok done s l k W = F d = k( x L0) dx = ( l L0) ( l0 L0) 0 l0 { } x F

3 Example 3: Calculate the wok done by gavty on a satellte that s launched fom the suface of the eath to an alttude of 50km (a typcal low eath obt). Assumptons. The eath s adus s km. The mass of a typcal satellte s 435kg - see, e.g. R The Gavtatonal paamete m = GM = km s (G= gavtatonal constant; M=mass of eath) 4. We wll assume that the satellte s launched along a staght lne path paallel to the decton, statng the eaths suface and extendng to the alttude of the obt. It tuns out that the wok done s ndependent of the path, but ths s not obvous wthout moe elaboate and sophstcated calculatons. Calculaton:. The gavtatonal foce on the satellte s. The wok done follows as GMm F= x R+ h GMm F= dx = GMm R h R R x Substtutng numbes gves J (be caeful wth unts f you wok wth klometes the wok done s n N-km nstead of SI unts Nm) h Example 4: A Fea Testaossa skds to a stop ove a dstance of 50ft. Calculate the total wok done on the ca by the fcton foces actng on ts wheels. Assumptons:. A Fea Testaossa has mass 506kg (see Testaossa.html). The coeffcent of fcton between wheels and oad s of ode 0.8 T x F N F mg 3. We assume the bakes ae locked so all wheels skd, and a esstance s neglected Calculaton The fgue shows a fee body dagam. The equaton of moton fo the ca s ( TF + TR) + ( NR + NF mg) = max. The vetcal component of the equaton of moton yelds NR + NF = mg. The fcton law shows that TR = mnr TF = mnf TR + TF = m( NR + NF ) = mmg 3. The poston vectos of the ca s font and ea wheels ae F = x R = ( x+ L). The wok done follows as. We suppose that the ea wheel stats at some pont x0 when the bakes ae appled and skds a total dstance d. x0 d x0 d+ L x0 d x0 d+ L FR R FF F R F R F x0 x0+ L x0 x0+ L W = d + d = T ( dx ) + T ( dx ) = ( T + T ) d L T R N R

4 4. The wok done follows as W = mmg. Substtutng numbes gves 5 W = 9 0 J. Example 5: The fgue shows a box that s pushed up a slope by a foce P. The box moves wth speed v. Fnd a fomula fo the ate of wok done by each of the foces actng on the box. The fgue shows a fee body dagam. The foce vectos ae. Appled foce Pcosα+ Psnα. Fcton Tcosα Tsnα 3. Nomal eacton Nsnα+ Tcosα 4. Weght mg The velocty vecto s vcosα+ vsnα Evaluatng the dot poducts F v fo each fomula, and ecallng that cos α + sn α = gves. Appled foce Pv. Fcton Tv 3. Nomal eacton 0 4. Weght mg snα P P α mg N T Example 6: The table lsts the expementally measued foce-v-daw data fo a longbow. Calculate the total wok done to daw the bow. In ths case we don t have a functon that specfes the foce as a functon of poston; nstead, we have a table of numecal values. We have to appoxmate the ntegal W = F d 0 numecally. To undestand how to do ths, emembe that ntegatng a functon can be vsualzed as computng the aea unde a cuve of the functon, as llustated n the fgue. Foce N Daw (cm) We can estmate the ntegal by dvdng the aea nto a sees of tapezods, as shown. Recall that the aea of a tapezod s (base x aveage heght), so the total aea of the functon s f+ f f + f3 f3 + f4 f4 + f5 F W = x x x3 x f = f f 3 f 4 f 5 x You could easly do ths calculaton by hand but fo lazy people lke me MATLAB has a convenent functon called `tapz that does ths calculaton automatcally. Hee s how to use t daw = [0,0,0,30,40,50,60]*0.0; foce = [0,40,90,40,80,0,70]; tapz(daw,foce) ans = So the soluton s 80.5J x x x 3 x 4

5 4..3 Defnton of the potental enegy of a consevatve foce F(t) Peamble: Textbooks nealy always defne the `potental enegy of a foce. P Stctly speakng, we cannot defne a potental enegy of a sngle foce 0 k nstead, we need to defne the potental enegy of a pa of foces. A foce can t exst by tself thee must always be an equal and opposte eacton foce actng on a second body. In all of the dscusson to be pesented n ths O secton, we mplctly assume that the eacton foce s actng on a second body, whch s fxed at the ogn. Ths smplfes calculatons, and makes the dscusson pesented hee look lke those gven n textbooks, but you should emembe that the potental enegy of a foce pa s always a functon of the elatve postons of the two foces. Wth that povso, consde a foce F actng on a patcle at some poston n space. Recall that the wok done by a foce that moves fom poston vecto 0 to poston vecto s W = F d 0 In geneal, the wok done by the foce depends on the path between 0 to. Fo some specal foces, howeve, the wok done s ndependent of the path. Such foces ae sad to be consevatve. Fo a foce to be consevatve: The foce must be a functon only of ts poston.e. t can t depend on the velocty of the foce, fo example. The foce vecto must satsfy cul( F) = 0 Examples of consevatve foces nclude gavty, electostatc foces, and the foces exeted by a spng. Examples of non-consevatve (o should that be lbeal?) foces nclude fcton, a esstance, and aeodynamc lft foces. The potental enegy of a consevatve foce s defned as the negatve of the wok done by the foce n movng fom some abtay ntal poston 0 to a new poston,.e. 0 V( ) = F d + constant The constant s abtay, and the negatve sgn s ntoduced by conventon (t makes sue that systems ty to mnmze the potental enegy). If thee s a pont whee the foce s zeo, t s usual to put 0 at ths pont, and take the constant to be zeo. Note that. The potental enegy s a scala valued functon. The potental enegy s a functon only of the poston of the foce. If we choose to descbe poston n tems of Catesan components = x+ y+ zk, then V() = V(, xyz,). 3. The elatonshp between potental enegy and foce can also be expessed n dffeental fom (whch s often moe useful fo actual calculatons) as F = gad( V ) If we choose to wok wth Catesan components, then

6 V V V Fx+ Fy+ Fzk = + + k x y z Occasonally, you mght have to calculate a potental enegy functon by ntegatng foces fo example, f you ae nteested n unnng a molecula dynamc smulaton of a collecton of atoms n a mateal, you wll need to descbe the nteatomc foces n some convenent way. The nteatomc foces can be estmated by dong quantum-mechancal calculatons, and the esults can be appoxmated by a sutable potental enegy functon. Hee ae a few examples showng how you can ntegate foces to calculate potental enegy Example : Potental enegy of foces exeted by a spng. A fee body dagam showng the foces exeted by a spng connectng two obects s shown n the fgue.. The foce exeted by a spng s F= kx ( L0 ). The poston vecto of the foce s = x 3. The potental enegy follows as L V ( ) = F d+ constant = k( x L0) dx = k( L L0) 0 L0 whee we have taken the constant to be zeo. x F Example : Potental enegy of electostatc foces exeted by chaged patcles. The fgue shows two chaged patcles a dstance x apat. To calculate the potental enegy of the foce actng on patcle, we place patcle at the ogn, and note that the foce actng on +Q +Q F patcle s QQ F= x 4πε x whee Q and Q ae the chages on the two patcles, and ε s a fundamental physcal constant known as the Pemttvty of the medum suoundng the patcles. Snce the foce s zeo when the patcles ae nfntely fa apat, we take 0 at nfnty. The potental enegy follows as x QQ QQ V () = dx = 4πε x 4πε x

7 Table of potental enegy elatons In pactce, howeve, we aely need to do the ntegals to calculate the potental enegy of a foce, because thee ae vey few dffeent knds of foce. Fo most engneeng calculatons the potental enegy fomulas lsted n the table below ae suffcent. Type of foce Foce vecto Potental enegy Gavty actng on a patcle nea eaths suface F= mg V = mgy m F y Gavtatonal foce exeted on mass m by mass M at the ogn GMm F= 3 V GMm = F m Foce exeted by a spng wth stffness k and unstetched length L 0 F = V = k( L0 ) F k ( L0 ) Foce actng between two chaged patcles QQ F= 4πε 3 V QQ = 4πε F +Q +Q Foce exeted by one molecule of a noble gas (e.g. He, A, etc) on anothe (Lennad Jones potental). a s the equlbum spacng between molecules, and E s the enegy of the bond. 3 7 E a a F = a 6 a a E F

8 4..4 Defnton of the Knetc Enegy of a patcle Consde a patcle wth mass m whch moves wth velocty v= v + v + v k. By defnton, ts knetc enegy s x y z ( x y z ) T= mv = mv + v + v 0 k O P v 4..5 Powe-Wok-knetc enegy elatons fo a sngle patcle Consde a patcle wth mass m that moves unde the acton of a foce F. Suppose that. At some tme t 0 the patcle has some ntal poston 0, velocty v 0 and knetc enegy T 0. At some late tme t the patcle has a new poston, velocty v and knetc enegy T. 3. Let P = F vdenote the ate of wok done by the foce t 4. Let W = Pdt = F d be the total wok done by the foce t 0 0 The Powe-knetc enegy elaton fo the patcle states that the ate of wok done by F s equal to the ate of change of knetc enegy of the patcle,.e. dt P = dt Poof: Ths s ust anothe way of wtng Newton s law fo the patcle: to see ths, note that we can take the dot poduct of both sdes of F=ma wth the patcle velocty dv d F v= ma v= m v= m ( v v ) dt dt To see the last step, do the devatve usng the Chan ule and note that a b= b a. The Wok-knetc enegy elaton fo a patcle says that the total wok done by the foce F on the patcle s equal to the change n the knetc enegy of the patcle. W = T T 0 Ths follows by ntegatng the powe-knetc enegy elaton wth espect to tme Examples of smple calculatons usng wok-powe-knetc enegy elatons Thee ae two man applcatons of the wok-powe-knetc enegy elatons. You can use them to calculate the dstance ove whch a foce must act n ode to poduce a gven change n velocty. You can also use them to estmate the enegy equed to make a patcle move n a patcula way, o the amount of enegy that can be extacted fom a collecton of movng patcles (e.g. usng a wnd tubne)

9 Example : The longest sngle-span escalato n the Westen hemsphee s located at the Washngton Meto staton n Montgomey County, Mayland. Some techncal specfcatons fo the escalato span can be found on Wkpeda (we have not checked ths data!). Addtonal nfomaton concenng escalato standads can be found hee. Calculate the knetc enegy of a sngle 80kg de standng on the escalato The KE s mv /; we have that the speed s 7m pe mnute, so KE s 8.J. Calculate the change n potental enegy of a sngle 80kg de who tavels the ente length of the escalato span. The change n PE s mgh, whee h = 35m, so PE s 7468J Assumng the escalato opeates at ts theoetcal capacty of 9000 passenges pe hou, estmate the powe equed to opeate the escalato. The powe s the numbe of passenges pe second multpled by the enegy change pe passenge. Ths gves P=68.7 kw Example : Estmate the mnmum dstance equed fo a 4 wheele that tavels at the RI speed-lmt to bake to a standstll. Is the dstance to stop any dffeent fo a Toyota Echo? Ths poblem can be solved by notng that, snce we know the ntal and fnal speed of the vehcle, we can calculate the change n knetc enegy as the vehcle stops. The change n knetc enegy must equal the wok done by the foces actng on the vehcle whch depends on the dstance sld. Hee ae the detals of the calculaton. Assumptons:. We assume that all the wheels ae locked and skd ove the gound (ths wll stop the vehcle n the shotest possble dstance). The contacts ae assumed to have fcton coeffcent µ = The vehcle s dealzed as a patcle. 4. A esstance wll be neglected. T A T B TC T D N A N B N C N D mg Calculaton:. The fgue shows a fee body dagam.. The equaton of moton fo the vehcle s ( TA + TB + TC + TD) + ( N A + NB + NC + ND mg ) = max The vetcal component of the equaton shows that NA + NB + NC + ND = mg. T + T + T + T = m N + N + N + N = mmg 3. The fcton foce follows as ( ) ( ) A B C D A B C D

10 4. If the vehcle skds fo a dstance d, the total wok done by the foces actng on the vehcle s d W = ( TA + TB + TC + TD) dx = ( TA + TB + TC + TD) d = mmgd 0 5. The wok-enegy elaton states that the total wok done on the patcle s equal to ts change n knetc enegy. When the bakes ae appled the vehcle s tavelng at the speed lmt, wth speed V; at the end of the skd ts speed s zeo. The change n knetc enegy s theefoe T = 0 mv /. The wok-enegy elaton shows that Substtutng numbes gves mgd = mv / d = V / mg Ths smple calculaton suggests that the bakng dstance fo a vehcle depends only on ts speed and the fcton coeffcent between wheels and tes. Ths s unlkely to vay much fom one vehcle to anothe. In pactce thee may be moe vaaton between vehcles than ths estmate suggests, patly because factos lke a esstance and aeodynamc lft foces wll nfluence the esults, and also because vehcles usually don t skd dung an emegency stop (f they do, the dve loses contol) the natue of the bakng system theefoe also may change the pedcton. Example 3: Compae the powe consumpton of a Fod Excuson to that of a Chevy Cobalt dung stopstat dvng n a taffc am. Dung stop-stat dvng, the vehcle must be epeatedly acceleated to some (low) velocty; and then baked to a stop. Powe s expended to acceleate the vehcle; ths powe s dsspated as heat n the bakes dung bakng. To calculate the enegy consumpton, we must estmate the enegy equed to acceleate the vehcle to ts maxmum speed, and estmate the fequency of ths event. Calculaton/Assumptons:. We assume that the speed n a taffc am s low enough that a esstance can be neglected.. The enegy to acceleate to speed V s mv /. 3. We assume that the vehcle acceleates and bakes wth constant acceleaton f so, ts aveage speed s V/. 4. If the vehcle tavels a dstance d between stops, the tme between two stops s d/v. 5. The aveage powe s theefoe mv /4d. 6. Takng V=5mph (7m/s) and d=00ft (6m) ae easonable values the powe s theefoe 0.03m, wth m n kg. A Fod Excuson weghs 900 lb (470 kg), equng 5 Watts (about that of a lght bulb) to keep movng. A Chevy Cobalt weghs 68lb (6kg) and eques only 36 Watts a vey substantal enegy savng. Reducng vehcle weght s the most effectve way of mpovng fuel effcency dung slow dvng, and also educes manufactung costs and mateal equements. Anothe, moe costly, appoach s to use a system that can ecove the enegy dung bakng ths s the man eason that hybd vehcles lke the Pus have bette fuel economy than conventonal vehcles.

11 Example 4: Estmate the powe that can be geneated by a wnd tubne. The fgue shows a wnd tubne. The tubne blades deflect the a flowng past them: ths changes the a speed and so exets a foce on the blades. If the blades move, the foce exeted by the a on the blades does wok ths wok s the powe geneated by the tubne. The ate of wok done by the a on the blades must equal the change n knetc enegy of the a as t flows past the blades. Consequently, we can estmate the powe geneated by the tubne by calculatng the change n knetc enegy of the a flowng though t. To do ths popely needs a vey sophstcated analyss of the a flow aound the tubne. Howeve, we can get a athe cude estmate of the powe by assumng that the tubne s able to extact all the enegy fom the a that flows though the ccula aea swept by the blades. Calculaton: Let V denote the wnd speed, and let ρ denote the densty of the a.. In a tme t, a cylndcal egon of a wth adus R and heght Vt passes though the fan.. The cylndcal egon has mass πr ρ Vt 3. The knetc enegy of the cylndcal egon of a s T = ( πr ρvt) V dt π 3 4. The ate of flow of knetc enegy though the fan s theefoe = R ρv dt 5. If all ths enegy could be used to do wok on the fan blades the powe geneated would be dt π 3 P= = R ρv dt Repesentatve numbes ae () A densty. kg / m ; () a speed 5mph ( m/sec); () Radus 30m Ths gves.8mw. Fo compason, a nuclea powe plant geneates about MW. A moe sophstcated calculaton (whch wll be coveed n EN80) shows that n pactce the maxmum possble amount of enegy that can be extacted fom the a s about 60% of ths estmate. On aveage, a typcal household uses about a kw of enegy; so a sngle tubne could povde enough powe fo about 5-0 houses. 3

12 4..7 Enegy elatons fo a consevatve system of patcles. The fgue shows a `system of patcles ths s ust a collecton of obects that we mght be nteested n, whch can be dealzed as patcles. Each patcle n the system can expeence foces appled by: Othe patcles n the system (e.g. due to gavty, electc chages on the patcles, o because the patcles ae physcally connected though spngs, o because the patcles collde). We call these ntenal foces actng n the system. We wll denote the ntenal foce exeted by the th patcle on the th patcle by R. Note that, because evey acton has an equal and opposte eacton, the foce exeted on the th patcle by the th patcle must be equal and opposte, to R,.e. R = R. m ext F m 4 R R 3 R 3 R m ext F R 3 R 3 F 3 ext m 3 Foces exeted on the patcles by the outsde wold (e.g. by extenally appled gavtatonal o electomagnetc felds, o because the patcles ae connected to the outsde wold though mechancal lnkages o spngs). We call these extenal foces actng on the system, and we wll denote the extenal ext foce on the th patcle by F () t We defne the ate of extenal wok (o extenal powe) done on the system as ext ext P = F () t v () t foces We defne the total extenal wok done on the system dung a tme nteval t 0 t t as the sum of the wok done by the extenal foces. t F v foces t0 ext ext W = () t () t dt The total wok done can also nclude a contbuton fom extenal moments actng on the system, but we wll woy about ths when we analyze gd body moton.. The system of patcles s consevatve f all the ntenal foces n the system ae consevatve. Ths means that the patcles must nteact though consevatve foces such as gavty, spngs, electostatc foces, and so on. The patcles can also be connected by gd lnks, o touch one anothe, but contacts between patcles must be fctonless. If ths s the case, we can defne the total potental enegy of the system V as the sum of potental eneges of all the ntenal foces. We usually compute the total potental enegy by summng up all the tems fom the table n Sect Mathematcally, the potental enegy depends n some complcated way on the dstances between all the patcles, and the esultant foce on the th patcle s elated to the total potental enegy by TOT V R = TOT

13 We also defne the total knetc enegy patcles. TOT T of the system as the sum of knetc eneges of all the The wok-enegy elaton fo the system of patcles can then be stated as follows. Suppose that. At some tme t 0 the system has and knetc enegy T 0 TOT TOT. At some late tme t the system has knetc enegy T. 3. Let V 0 TOT denote the potental enegy of the foce at tme t 0 4. Let V TOT denote the potental enegy of the foce at tme t ext 5. Let W denote the total wok done on the system between t 0 t t Powe Enegy Relaton: Ths law states that the ate of extenal wok done on the system s equal to the ate of change of total enegy of the system ext d TOT TOT P = ( T + V ) dt Wok Enegy Relaton: Ths law states that the extenal wok done on the system s equal to the change n total knetc and potental enegy of the system. ext TOT TOT TOT TOT W = T + V T + V ( ) 0 0 Enegy consevaton law Fo the specal case whee no extenal foces act on the system, the total enegy of the system s constant ext TOT TOT TOT TOT W = 0 T + V = T + V ( ) 0 0 It s woth makng one fnal emak befoe we tun to applcatons of these laws. We often nvoke the pncple of consevaton of enegy when analyzng the moton of an obect that s subected to the eath s gavtatonal feld. Fo example, the fst poblem we solve n the next secton nvolves the moton of a

14 poectle launched fom the eath s suface. We usually glbly say that `the sum of the potental and knetc eneges of the patcle ae constant and f you ve done physcs couses you ve pobably used ths knd of thnkng. It s not eally coect, although t leads to a moe o less coect soluton. Popely, we should consde the eath and the poectle togethe as a consevatve system. Ths means we must nclude the knetc enegy of the eath n the calculaton, whch changes by a small, but fnte, amount due to gavtatonal nteacton wth the poectle. Fotunately, the pncple of consevaton of lnea momentum (to be coveed late) can be used to show that the change n knetc enegy of the eath s neglgbly small compaed to that of the patcle. Poof of the enegy elaton Recall the powe-knetc enegy elaton fo a sngle patcle dt F v = dt The total foce on one patcle conssts of the extenal foce, plus the sum of all the foces exeted on the patcle by othe patcles. The powe-enegy elaton fo the th patcle theefoe becomes ext dt ( F ) + R v = dt Snce the ntenal foces ae consevatve, we can wte V R = Futhemoe, v = /, so that dt We can now sum ths ove all the patcles TOT TOT V R v = d dt TOT ext V d dt F v + = dt dt TOT We know that V depends only on the postons of the patcles, so the second tem on the left hand sde s a total dffeental TOT TOT V d dv = dt dt (by the chan ule). We also ecognze the fst tem as the total extenal powe, and the ght hand sde as the tme devatve of the total KE, so we see that TOT TOT ext dv dt P = dt dt Reaangng ths gves the powe-ke elaton fo the system, and ntegatng t wth espect to tme gves the wok-enegy elaton.

15 4..8 Examples of calculatons usng knetc and potental enegy n consevatve systems The knetc-potental enegy elatons can be used to quckly calculate elatonshps between the velocty and poston of an obect. Seveal examples ae povded below. Example : (Bong FE exam queston) A poectle wth mass m s launched fom the gound wth velocty V 0 at angle α. Calculate an expesson fo the maxmum heght eached by the poectle. If a esstance can be neglected, we can egad the eath and the poectle togethe as a consevatve system. We neglect the change n the eath s knetc enegy. In addton, snce the gavtatonal foce actng on the patcle s vetcal, the patcle s hozontal component of velocty must be constant. Calculaton:. Just afte launch, the velocty of the patcle s v= V0cosα+ V0snα. The knetc enegy of the patcle ust afte launch s mv 0 /. Its potental enegy s zeo. 3. At the peak of the taectoy the vetcal velocty s zeo. Snce the hozontal velocty emans constant, the velocty vecto at the peak of the taectoy s v= V 0 cosα. The knetc enegy at ths pont s theefoe mv 0 cos α / 4. Enegy s conseved, so mv /= mgh + mv cos α / 0 0 h= V0 ( cos α) / = V0 sn α V 0 α h Example : You ae asked to desgn the packagng fo a senstve nstument. The packagng wll be made fom an elastc foam, whch behaves lke a spng. The specfcatons estct the maxmum acceleaton of the nstument to 5g. Estmate the thckness of the packagng that you must use. Ths poblem can be solved by notng that () the max acceleaton occus when d the packagng (spng) s fully compessed and so exets the maxmum foce on the nstument; () The velocty of the nstument must be zeo at ths nstant, (because the heght s a mnmum, and the velocty s the devatve of the heght); and () The system s consevatve, and has zeo knetc enegy when the package s dopped, and zeo knetc enegy when the spng s fully compessed. Assumptons:. The package s dopped fom a heght of.5m. The effects of a esstance dung the fall ae neglected 3. The foam s dealzed as a lnea spng, whch can be fully compessed. Calculatons: Let h denote the dop heght; let d denote the foam thckness.. The potental enegy of the system ust befoe the package s dopped s mgh d h

16 . The potental enegy of the system at the nstant when the foam s compessed to ts maxmum extent s kd 3. The total enegy of the system s constant, so kd mgh = 4. The fgue shows a fee body dagam fo the nstument at the nstant of maxmum foam compesson. The esultant foce actng on the nstument s ( kd mg), so mg ts acceleaton follows as a= ( kd / m g). The acceleaton must not exceed 5g, so kd / m 6g kd 5. Dvdng (3) by (4) shows that d h/8 The thckness of the potectve foam must theefoe exceed 8.8cm. Example 3: The Chapy Impact Test s a way to measue the wok of factue of a mateal (.e. the wok pe unt aea equed to sepaate a mateal nto two peces). An example (fom s shown n the pctue. You can see one n Pnce Lab f you ae cuous. It conssts of a pendulum, whch swngs down fom a pescbed ntal angle to stke a specmen. The pendulum factues the specmen, and then contnues to swng to a new, smalle angle on the othe sde of the vetcal. The scale on the pendulum allows the ntal and fnal angles to be measued. The goal of ths example s to deduce a elatonshp between the angles and the wok of factue of the specmen. The fgue shows the pendulum befoe and afte t hts the specmen.. The potental enegy of the mass befoe t s eleased s V = mgl cosα. Its knetc enegy s zeo.. The potental enegy of the mass when t comes to est afte stkng the specmen s V = mgl cosα. l m The knetc enegy s agan zeo. α The wok of factue s equal to the change n potental enegy - m α W = V V = mgl cosα cosα F ( )

17 Example 4: Estmate the maxmum dstance that a long-bow can fe an aow. We can do ths calculaton by dealzng the bow as a spng, and estmatng the maxmum foce that a peson could apply to daw the bow. The enegy stoed n the bow can then be estmated, and enegy consevaton can be used to estmate the esultng velocty of the aow. Assumptons. The long-bow wll be dealzed as a lnea spng. The maxmum daw foce s lkely to be aound 60lbf (70N) 3. The daw length s about ft (0.6m) 4. Aows come wth vaous masses typcal ange s between gans (6-38 gams) 5. We wll neglect the mass of the bow (ths s not a vey ealstc assumpton) Calculaton: The calculaton needs two steps: () we stat by calculatng the velocty of the aow ust afte t s fed. Ths wll be done usng the enegy consevaton law; and () we then calculate the dstance taveled by the aow usng the poectle taectoy equatons deved n the pecedng chapte.. Just befoe the aow s eleased, the spng s stetched to ts maxmum length, and the aow s statonay. The total enegy of the system s T + V = kl, whee L s the daw length and k s 0 0 the stffness of the bow.. We can estmate values fo the spng stffness usng the daw foce: we have that F D = kl, so k = FD / L. Thus T0 + V0 = LF D. 3. Just afte the aow s fed, the spng etuns to ts un-stetched length, and the aow has velocty V. The total enegy of the system s T+ V = mv, whee m s the mass of the aow 4. The system s consevatve, theefoe T0 + V0 = T+ V LFD = mv V = LFD / m 5. We suppose that the aow s launched fom the ogn at an angle θ to the hozontal. The hozontal and vetcal components of velocty ae Vx = Vcosθ Vy = Vsnθ. The poston vecto of the aow can be calculated usng the method outlned n Secton 3.. the esult s = ( Vt cosθ) + Vt snθ gt We can calculate the dstance taveled by notng that ts poston vecto when t lands s d. Ths gves ( Vt cosθ) + Vt snθ gt = d whee t s the tme of flght. The and components of ths equaton can be solved fo t and d, wth the esult Vsnθ Vsnθ t = d = g gcosθ The aow tavels futhest when fed at an angle that maxmzes sn θ / cosθ -.e. 45 degees. The dstance follows as V LF d = = D g mg

18 6. Substtutng numbes gves 064m fo a 50 gan aow ove a mle! Of couse a esstance wll educe ths value, and n pactce the knetc enegy assocated wth the moton of the bow and bowstng (neglected hee) wll educe the dstance. Example 5: Fnd a fomula fo the escape velocty of a space vehcle as a functon of alttude above the eaths suface. The tem Escape velocty means that the space vehcle has a lage enough velocty to completely escape the eath s gavtatonal feld.e. the space vehcle wll neve stop afte beng launched. Assumptons. The space vehcle s ntally n obt at an alttude h above the eath s suface. The eath s adus s km h 3. Whle n obt, a ocket s buned on the vehcle to ncease ts speed to v (the escape velocty), placng t on a hypebolc taectoy that R wll eventually escape the eath s gavtatonal feld The Gavtatonal paamete m = GM = km s (G= gavtatonal constant; M=mass of eath) 5. Calculaton. Just afte the ocket s buned, the potental enegy of the system s V = GMm /( R + h), whle ts knetc enegy s T = mv /. When t escapes the eath s gavtatonal feld (at an nfnte heght above the eath s suface) the potental enegy s zeo. At the ctcal escape velocty, the velocty of the spacecaft at ths pont dops to zeo. The total enegy at escape s theefoe zeo. 3. Ths s a consevatve system, so T + V = mv / GMm /( R + h ) = 0 v = GM /( R + h) 4. A typcal low eath obt has alttude of 50km. Fo ths alttude the escape velocty s 0.9km/sec.

19 4. Lnea mpulse-momentum elatons 4.. Defnton of the lnea mpulse of a foce In most dynamc poblems, patcles ae subected to foces that vay wth tme. We can wte ths mathematcally by sayng that the foce s a vecto valued functon of tme F () t. If we expess the foce as components n a fxed bass {,, k }, then F() t = Fx() t + Fy() t + Fz() t k whee each component of the foce s a functon of tme. k O F(t) P The Lnea Impulse exeted by a foce dung a tme nteval t 0 t t s defned as t I= F () t dt The lnea mpulse s a vecto, and can be expessed as components n a bass I=I +I +Ik t0 x y z t t t I = F () t dt I = F () t dt I = F () t dt x x y y z z t0 t0 t0 If you know the foce as a functon of tme, you can calculate ts mpulse usng smple calculus. Fo example:. Fo a constant foce, wth vecto value F 0, the mpulse s I= F 0 t. Fo a hamonc foce of the fom F( t) = F 0 snωt the mpulse s I= F cos( ω( t + t)) cos ωt / ω [ ] It s athe ae n pactce to have to calculate the mpulse of a foce fom ts tme vaaton. 4.. Defnton of the lnea momentum of a patcle The lnea momentum of a patcle s smply the poduct of ts mass and velocty p= mv The lnea momentum s a vecto ts decton s paallel to the velocty of the patcle Impulse-momentum elatons fo a sngle patcle Consde a patcle that s subected to a foce F () t fo a tme nteval t 0 t t. Let I denote the mpulse exeted by F on the patcle Let p 0 denote the lnea momentum of the patcle at tme t 0. Let p denote the lnea momentum of the patcle at tme t.

20 The mpulse-momentum equaton can be expessed ethe n dffeental o ntegal fom.. In dffeental fom d F () t = p dt k Ths s clealy ust a dffeent way of wtng Newton s law F=ma.. In ntegal fom I= p Ths s the ntegal of Newton s law of moton wth espect to tme. x O y P z The mpulse-momentum elatons fo a sngle patcle ae useful f you need to calculate the change n velocty of an obect that s subected to a pescbed foce Examples usng mpulse-momentum elatons fo a sngle patcle Example : Estmate the tme that t takes fo a Fea Testaossa tavelng at the RI speed lmt to make an emegency stop. (Lke many textbook poblems ths one s totally unealstc nobody n a Fea s gong to tavel at the speed lmt!) We can do ths calculaton usng the mpulse-momentum elaton fo a sngle patcle. Assume that the ca has mass m, and tavels wth speed V befoe the bakes ae appled. Let t denote the tme equed to stop.. Stat by estmatng the foce actng on the ca dung an emegency stop. The fgue shows a fee body dagam fo the ca as t bakes to a standstll. We assume that the dve bakes had enough to lock the wheels, so that the ca skds ove the oad. The hozontal fcton foces must oppose the sldng, as shown n the pctue. F=ma fo the ca T F T R follows as N F N R ( TR + TF) + ( NF + NR mg) = max The vetcal component of the equaton of moton shows that NF + NR = mg. The fcton law then shows that TF + TR = m( NF + NR) = mmg. The foce actng on the ca s constant, so the mpulse that bngs the ca to a halt s I= ( TR + TF) t+ ( NF + NR mg) t=mmg 3. The lnea momentum of the ca befoe the bakes ae appled s p 0 = mv. The lnea momentum afte the ca stops s zeo. Theefoe, p= mv. 4. The lnea mpulse-momentum elaton shows that I= p mmg t = mv t = V /( mg)

21 5. We can take the fcton coeffcent as µ 0.8, and 65mph s 9m/s. We take the gavtatonal acceleaton g = 9.8. The tme follows as t 3.7s. Note that a TestaRossa can t stop any faste than a Honda Cvc, despte the pce dffeence 4..5 Impulse-momentum elaton fo a system of patcles Suppose we ae nteested n calculatng the moton of seveal ext patcles, sketched n the fgue. F m 4 ext F Total extenal mpulse on a system of patcles: Each m 3 patcle n the system can expeence foces appled by: R Othe patcles n the system (e.g. due to gavty, R 3 R 3 m R 3 electc chages on the patcles, o because the 3 patcles ae physcally connected though spngs, o R R 3 because the patcles collde). We call these ntenal m foces actng n the system. We wll denote the ext ntenal foce exeted by the th patcle on the th F patcle by R. Note that, because evey acton has an equal and opposte eacton, the foce exeted on the th patcle by the th patcle must be equal and opposte, to R,.e. R = R. Foces exeted on the patcles by the outsde wold (e.g. by extenally appled gavtatonal o electomagnetc felds, o because the patcles ae connected to the outsde wold though mechancal lnkages o spngs). We call these extenal foces actng on the system, and we wll ext denote the extenal foce on the th patcle by F () t We defne the total mpulse exeted on the system dung a tme nteval t 0 t t as the sum of all the mpulses on all the patcles. It s easy to see that the total mpulse due to the ntenal foces s zeo because the th and th patcles must exet equal and opposte mpulses on one anothe, and when you add them up they cancel out. So the total mpulse on the system s smply t TOT ext I = F patcles t0 () t dt Total lnea momentum of a system of patcles: The total lnea momentum of a system of patcles s smply the sum of the momenta of all the patcles,.e. p TOT = patcles m v

22 The mpulse-momentum equaton. In dffeental fom patcles F ext dp () t = dt TOT. In ntegal fom TOT TOT TOT I = p p 0 Ths s the ntegal of Newton s law of moton wth espect to tme. Consevaton of momentum: If no extenal foces act on a system of patcles, the total lnea momentum s conseved,.e. TOT TOT p = p 0 Devng the mpulse-momentum equaton We stat wth the mpulse-momentum elaton fo a sngle patcle n dffeental fom ext dp F + R = dt (The left hand sde s the total foce on the th patcle t ncludes the extenal foce, as well as all the foces exeted by all the othe patcles). Now sum ove all the patcles ext F + R = d dt p

23 Notce that snce R = R R = 0 (To see ths ust wte out the full sum evey ntenal foce on one patcle s equal and opposte to the foce on anothe, so the total has to cancel) Theefoe, evaluatng the sums TOT ext dp F () t = patcles dt We can ntegate ths expesson wth espect to tme to get the ntegal veson of the theoem Examples of applcatons of momentum consevaton fo systems of patcles The mpulse-momentum equatons fo systems of patcles ae patculaly useful fo () analyzng the ecol of a gun; and () analyzng ocket and et populson systems. In both these applcatons, the ntenal foces actng between the gun on the poectle, o the moto and popellant, ae much lage than any extenal foces, so the total momentum of the system s conseved. Example : Estmate the ecol velocty of a fle (youtube abounds wth ecol demonstatons see. e.g. fo samples. Be waned, howeve a lot of the vdeos ae tasteless and/o sexst ) The ecol velocty can be estmated by notng that the total momentum of bullet and fle togethe must be conseved. If we can estmate the mass of fle and bullet, and the bullet s velocty, the ecol velocty can be computed fom the momentum consevaton equaton. Assumptons: 4. The mass of a typcal 0. (.e. 0. damete) calbe fle bullet s about 7 0 kg (dealzng the bullet as a sphee, wth densty kg / m ). The muzzle velocty of a 0. s about 000 ft/sec (305 m/s) 3. A typcal fle weghs between 5 and 0 lb (.5-5 kg) Calculaton:. Let m b denote the bullet mass, and let m R denote the mass of the fle.. The fle and bullet ae dealzed as two patcles. Befoe fng, both ae at est. Afte fng, the bullet has velocty vb = v; the fle has velocty vr = V. 3. Extenal foces actng on the system can be neglected, so ptot ( t0 + t) = ptot ( t0) mb vb + mrvr = 0. mv b + mv R = 0 V= mv b / mr 4. Substtutng numbes gves V between 0.04 and 0.08 m/s (about 0.4 ft/sec) m g V m b v

24 Example : Deve a fomula that can be used to estmate the mass of a handgun equed to keep ts ecol wthn acceptable lmts. The pecedng example shows that the feam wll ecol wth a velocty that depends on the ato of the mass of the bullet to the feam. The feam must be bought to est by the peson holdng t. Assumptons:. We wll dealze a peson s hand holdng the gun as a spng, wth stffness k, fxed at one end. The end-pont stffness of a human hand has been extensvely studed see, e.g. Shadmeh et al J. Neuoscence, 3 () 45 (993). Typcal values of stffness dung quas-statc deflectons ae of ode 0. N/mm. Dung dynamc loadng stffnesses ae lkely to be lage than ths.. We dealze the handgun and bullet as patcles, wth mass m b and m g, espectvely. Calculaton.. The pecedng poblem shows that the feam wll ecol wth velocty V= mv b / mg. Enegy consevaton can be used to calculate the ecol dstance. We consde the feam and the hand holdng t a system. At tme t=0 t has zeo potental enegy; and has knetc enegy T= mv ( ) g = mv b / mg. At the end of the ecol, the gun s at est, and the spng s fully compessed the knetc enegy s zeo, and potental enegy s kd. Enegy consevaton gves kd = ( m v) / m b 3. The equed mass follows as ( ) g m = m v / kd. g b Example 4 Rocket populson equatons. Rocket motos and et engnes explot the momentum consevaton law n ode to poduce moton. They wok by expellng mass fom a vehcle at vey hgh velocty, n a decton opposte to the moton of the vehcle. The momentum of the expelled mass must be balanced by an equal and opposte change n the momentum of the vehcle; so the velocty of the vehcle nceases. Moto, mass m 0 Rocket, mass M Analyzng a ocket engne s qute complcated, because the popellant caed by the engne s usually a vey sgnfcant facton of the total mass of the vehcle. Consequently, t s mpotant to account fo the fact that the mass deceases as the popellant s used. Assumptons:. The fgue shows a ocket moto attached to a ocket wth mass M.. The ocket moto contans an ntal mass m 0 of popellant and expels popellant at ate dm = m0 (kg/sec) wth a velocty v0 = v0 elatve to the ocket. dt 3. We assume staght lne moton, and assume that no extenal foces act on the ocket o moto.

25 Calculatons: The fgue shows the ocket at an nstant of tme t, and then a vey Rocket, mass M shot tme nteval dt late.. At tme t, the ocket moves at speed v, and the system has momentum p= ( M+ mv ), whee m s the moto s mass. v. Dung the tme nteval dt a mass dm = mdt s expelled fom Moto, mass m the moto. The velocty of the expelled mass s v= ( v v0 ). v-v 0 Rocket, mass M 3. At tme t+dt the mass of the moto has deceased to m mdt. m dt 4. At tme t+dt, the ocket has velocty v= ( v + dv). Moto, mass m- mdt 5. The total momentum of the system at tme t+dt s v+dv theefoe p= ( M + m m0dt)( v + dv) + m0dt( v v0) 6. Momentum must be conseved, so ( M + m) v= ( M + m m0dt)( v + dv) + m0dt( v v0) 7. Multplyng ths out and smplfyng shows that ( M + m) dv mdtv0= 0 whee the tem µ 0 dtdv has been neglected. 8. Fnally, we see that ( M + m) dv = mv0 dt Ths esult s called the `ocket equaton. Specfc Impulse of a ocket moto: The pefomance of a ocket engne s usually specfed by ts `specfc mpulse. Confusngly, two dffeent defntons of specfc mpulse ae commonly used: I = v sp 0 v I 0 sp = g The fst defnton quantfes the mpulse exeted by the moto pe unt mass of popellant; the second s the mpulse pe unt weght of popellant. You can usually tell whch defnton s beng used fom the unts the fst defnton has unts of m/s; the second has unts of s. In tems of the specfc mpulse, the ocket equaton s dv ( M + m) = misp = mgisp dt Integated fom of the ocket equaton: If the moto expels popellant at constant ate, the equaton of moton can be ntegated. Assume that. The ocket s at the ogn at tme t=0;. The ocket has speed v 0 at tme t=0 3. The moto has mass m 0 at tme t=0; ths means that at tme t t has mass m0 mt Then the ocket s speed can be calculated as a functon of tme:

26 Smlaly, the poston follows as v t dv mispdt ( M + m0 mt) = misp dv dt = ( M + m mt) v M + m0 v v0 = Isp log M + m0 mt x t dx M + m0 M + m log 0 v = = Isp + v0 dx = Isp log + v0 dt dt M + m0 mt M m mt + Isp M + m x ( v 0 = 0 + Isp ) t ( M + m0 mt) log m M + m0 mt These calculatons assume that no extenal foces act on the ocket. It s qute staghtfowad to genealze them to account fo extenal foces as well the detals ae left as an execse. Example 5 Applcaton of ocket populson equaton: Calculate the maxmum payload that can be launched to escape velocty on the Aes I launch vehcle. Escape velocty means that afte the moto buns out, the space vehcle can escape the eath s gavtatonal feld see example 5 n Secton Assumptons. The specfcatons fo the Aes I ae at Relevant vaables ae lsted n the table below. Total mass (kg) Specfc mpulse (s) Popellant mass (kg) Stage I Stage II As an appoxmaton, we wll neglect the moton of the ocket dung the bun, and wll neglect aeodynamc foces. 3. We wll assume that the fst stage s ettsoned befoe bunng the second stage. 4. Note that the change n velocty due to bunng a stage can be expessed as M0 v v0 = Ispglog M0 m whee M 0 s the total mass befoe the bun, and m s the mass of popellant buned. 5. The eath s adus s km 5 6. The Gavtatonal paamete m = GM = km s (G= gavtatonal constant; M=mass of eath) 7. Escape velocty s fom the eaths suface s v = GM / R, whee R s the eath s adus. e

27 Calculaton:. Let m denote the payload mass; let m, m denote the total masses of stages I and II, let m0, m 0 denote the popellant masses of stages I and II; and let I sp, I sp denote the specfc mpulses of the two stages.. The ocket s at est befoe bunng the fst stage; and ts total mass s m+ m + m + m0 + m0. Afte bun, the mass s m+ m + m + m0. The velocty afte bunng the fst stage s theefoe m+ m + m + m0 + m0 v = Ispglog m+ m + m + m0 3. The fst stage s then ettsoned the mass befoe statng the second bun s m+ m + m0, and afte the second bun t s m+ m. The velocty afte the second bun s theefoe log m + m + m + m + m v Isp g Ispglog m + m + m = + m+ m + m + m0 m+ m 4. Substtutng numbes nto the escape velocty fomula gves v e =.8 km/sec. Substtutng numbes fo the masses shows that to each ths velocty, the payload mass must satsfy m m <.6log log m m+ 04 whee m s n kg. 5. We can use Matlab to solve fo the ctcal value of m fo equalty clea all syms m eq =.798==.6908*log((m+77096)/(m+65780)) *log((m+8395)/(m+04)) solve(eq,m) so the soluton s 8300kg a vey small mass compaed wth that of the launch vehcle, but you could pack n a lage numbe of people you would lke to launch nto oute space nonetheless (the ente faculty of the school of engneeng, f you wsh) Analyzng collsons between patcles: the esttuton coeffcent The momentum consevaton equatons ae patculaly helpful f you want to analyze collsons between two o moe obects. If the mpact occus ove a vey shot tme, the mpulse exeted by the contact foces actng at the pont of collson s huge compaed wth the mpulse exeted by any othe foces. If we consde the two colldng patcles as a system, the extenal mpulse exeted on the system can be taken to be zeo, and so the total momentum of the system s conseved. The momentum consevaton equaton can be used to elate the veloctes of the patcles befoe collson to those afte collson. These elatons ae not enough to be able to detemne the veloctes completely, howeve to do ths, we also need to be able to quantfy the enegy lost (o moe accuately, dsspated as heat) dung the collson. In pactce we don t dectly specfy the enegy loss dung a collson nstead, the elatve veloctes ae elated by a popety of the mpact called the coeffcent of esttuton.

28 Resttuton coeffcent fo staght lne moton Suppose that two colldng patcles A and B move n a staght lne paallel to a unt vecto. Let A0 A0 B0 B0 v = vx, v = vx denote the veloctes of A and B ust befoe the collson A A B B v = v, v = v denote the veloctes of A and B ust afte the collson. x x The veloctes befoe and afte mpact ae elated by A B A B0 A0 v v = ev ( v ) whee e s the esttuton coeffcent. The negatve sgn s needed because the patcles appoach one anothe befoe mpact, and sepaate aftewads. v x A0 A v x A * v x B0 v x B B B Resttuton coeffcent fo 3D fctonless collsons Fo a moe geneal contact, we defne A 0 B0 v v to denote veloctes of the patcles befoe collson A B v v to denote veloctes of the patcles afte collson A v A0 B v B0 In addton, we let n be a unt vecto nomal to the common tangent plane at the pont of contact (f the two colldng patcles ae sphees o dsks the vecto s paallel to the lne onng the centes). n The veloctes befoe and afte mpact ae elated by two vecto equatons: B A B0 A0 ( v v ) n= e( v v ) n B A B A B0 A0 B0 A0 ( ) ( ) ( ) ( ) v v = v v n n v v v v n n v A A B B v To ntepet these equatons, note that. The fst equaton states that the component of elatve velocty nomal to the contact plane s educed by a facto e (ust as fo D contacts). The second equaton states that the component of elatve velocty tangent to the contact plane s unchanged To undestand the second equaton, note that. Dung the collson, a vey lage foce acts on both A and B at the contact between them. Because the contact s fctonless, the decton of the foce must be paallel to n. Also, the foce on A must be equal and opposte to the foce on B. Thee s no foce actng on ethe A o B paallel to t. Ths means that momentum must be conseved n the t decton fo both A and B ndvdually. We can wte ths mathematcally as B B B0 B0 v v n n= v v n n A A A0 A0 v v n n= v v n n

29 (ths looks funny, but we have ust subtacted the nomal component of velocty fom the total velocty. Ths leaves ust the tangental component). Subtactng the second equaton fom the fst shows that ( v B A ) ( B A ) ( B 0 A 0 ) ( B 0 A 0 v = ) v v n n v v v v n n Combned 3D esttuton fomula The two equatons fo the nomal and tangental behavo can be combned (ust add them) nto a sngle vecto equaton elatng veloctes befoe and afte mpact ths fom s moe compact, and often moe useful, but moe dffcult to vsualze physcally ( v B A B0 A0 B0 A0 v ) = ( v v ) ( + e) ( ) v v n n Values of esttuton coeffcent The esttuton coeffcent almost always les n the ange 0 e. It can only be less than zeo f one obect can penetate and pass though the othe (e.g. a bullet); and can only be geate than f the collson geneates enegy somehow (e.g. eleasng a peloaded spng, o causng an exploson). If e=0, the two colldng obects stck togethe; f e= the collson s pefectly elastc, wth no enegy loss. The esttuton coeffcent s stongly senstve to the mateal popetes of the two colldng obects, and also vaes weakly wth the geomety and the velocty of mpact. The two latte effects ae usually gnoed. Collson between a patcle and a fxed gd suface. The collson fomulas can be appled to mpact between a gd fxed suface by takng the suface to be patcle B, and notng that the velocty of patcle B s then zeo both befoe and afte mpact. Fo staght lne moton, v = ev A A0 0 0 Fo collson wth an angled wall v A A A = v ( + e) v n n, whee n s a unt vecto pependcula to the wall. v x A0 A v x A A * A v A0 n v A A

30 4..8 Examples of collson poblems Example Suppose that a movng ca hts a statonay (paked) vehcle fom behnd. Deve a fomula that wll enable an accdent nvestgato to detemne the velocty of the movng ca fom the length of the skd maks left on the oad. Assumptons:. We wll assume both cas move n a staght lne. The movng and statonay cas wll be assumed to have masses m, m, espectvely 3. We wll assume the cas stck togethe afte the collson (.e. the esttuton coeffcent s zeo) 4. We wll assume that only the paked ca has bakes appled afte the collson Ths calculaton takes two steps: fst, we wll use wok-enegy to elate the dstance sld by the cas afte mpact to the velocty ust afte the mpact occus. Then, we wll use momentum and the esttuton fomula to wok out the velocty of the movng ca ust befoe mpact. Calculaton: Let V denote the velocty of the movng ca ust befoe mpact; let v denote the velocty of the two (connected) cas ust afte mpact, and let d denote the dstance sld.. The fgue shows a fee body dagam fo each of the two cas dung sldng afte the collson. Newton s law of moton fo each ca shows that R+ N + N mg = ma ( ) x ( ) ( ) R+ T3 + T4 + N3 + N4 mg = ma x N + N = mg; N3 + N4 mg.. The vetcal component of the equatons of moton gve 3. The paked ca has locked wheels and skds ove the oad; the fcton law gves the tangental foces at the contacts as T3 + T4 = m( N3 + N4) = mmg 4. We can calculate the velocty of the cas ust afte mpact usng the wok-knetc enegy elaton dung skddng. Fo ths pupose, we consde the two connected cas as a sngle patcle. The wok done on the patcle by the fcton foces s ( T3 + T4) d = mmgd. The wok done s equal to the change n knetc enegy of the patcle, so mmgd = 0 ( m+ m) v mmgd v = m+ m 5. Fnally, we can use momentum consevaton to calculate the velocty ust befoe mpact. The momentum afte mpact s h = ( m+ m) v, whle befoe mpact h0 = mv. Equatng the two gves ( m+ m) m( m+ m) mgd V = v = m m

31 0 0 Example : Two fctonless sphees wth adus R have ntal velocty A B v v. At some nstant of tme, the two patcles collde. At the pont of collson, the centes of the sphees have poston vectos A B y y. The esttuton coeffcent fo the contact s denoted by e. Fnd a fomula fo the veloctes of the sphees afte mpact. Hence, deduce an expesson fo the change n knetc enegy dung the mpact. Ths s a staghtfowad vecto algeba execse. We have two unknown velocty vectos: A B v v, and two vecto equatons momentum consevaton, and the esttuton coeffcent fomula. Calculaton. Note that a unt vecto nomal to the tangent plane can be calculated fom the poston vectos of the centes at the mpact as A B A B n= ( y y ) / y y. It doesn t matte whethe you choose to take n to pont fom A to B o the othe way aound the n fomula wll wok ethe way.. Momentum consevaton eques that A B B A B0 A0 mb v + mav = mb v + mav v A B v 3. The esttuton coeffcent fomula gves B A B0 A0 B0 A0 ( v v ) = ( v v ) ( + e) ( ) v v n n B 4. We can solve () and (3) fo v by multplyng (3) by m A and addng the equatons, whch gves B B0 A0 B0 A0 B0 A0 ( mb + ma ) v = mb v + mav + m A ( v v ) ( + e) ( ) v v n n B B0 ma B0 A0 v = v ( + e) ( ) mb + m v v n n A A 5. Smlaly, we can solve fo v by multplyng (3) by m B and subtactng (3) fom (), wth the esult v A A0 mb B0 A0 = v + ( + e) ( ) mb + m v v n n A 6. The change n knetc enegy dung the collson can be calculated as ma A A mb B B ma A0 A0 mb B0 B0 T = v v + v v v v + v v B A 7. Substtutng the esults of (4) and (5) fo v and v and smplfyng the esult gves A B( e ) B0 A0 ( ) ( ) m + m m m T = v v n A B Note that the enegy change s zeo f e= (pefectly elastc collsons) and s always negatve fo e< (.e. the knetc enegy afte collson s less than that befoe collson). A v A0 B v B0

32 Example 3: Ths s ust a bong example to help llustate the pactcal applcaton of the vecto fomulas n the pecedng example. In the fgue shown, dsk A has a vetcal velocty V at tme t=0, whle dsk B s statonay. The two dsks both have adus R, have the same mass, and the esttuton coeffcent between them s e. Gavty can be neglected. Calculate the velocty vecto of each dsk afte collson. V B Calculaton. Befoe mpact, the velocty vectos ae A0 B0 v = V v = 0. A unt vecto paallel to the lne onng the two centes s n= (3+ 7 ) / 4 (to see ths, apply Pythagoas theoem to the tangle shown n the fgue). 3. The veloctes afte mpact ae B B0 ma B0 A0 v = v ( + e) ( ) mb m v v n n + A A A0 mb B0 A0 v = v + ( + e) ( ) mb + m v v n n A Substtutng the vectos gves B + e v = ( + e) ( V ) (3 7 ) / 4 (3 7 ) / 4 (3 7 7 ) + + = V + 3 A 3 7 ( + e) 5 7e v = V+ ( + e) ( V ) (3 + 7 ) / 4 (3 + 7 ) / 4 = V + V 3 3 A R R/ 3R/ Example 4: How to play pool (o snooke, bllads, o you own favote ba game nvolvng balls, a stck, and a table ). The fgue shows a typcal poblem faced by a pool playe whee should the queue ball ht the eght ball to send t nto the pocket? Ths s easly solved the eght ball s statonay befoe mpact, and afte mpact has a velocty v B ( ) A0 = + e v n n Notce that the velocty s paallel to the unt vecto n. Ths vecto s paallel to a lne connectng the centes of the two balls at the nstant of mpact. So the coect thng to do s to vsualze an magnay ball ust touchng the eght ball, n lne wth the pocket, and am the queue ball at the magnay ball. Easy! The eal secet to beng a successful pool playe s not pottng the balls that pat s easy. It s contollng whee the queue ball goes afte mpact. You may have seen expets make a queue ball evese ts decton afte an mpact (appeang to bounce off the statonay ball); o make the queue ball follow the stuck ball afte the mpact. Accodng to the smple equatons developed hee, ths s mpossble but a pool table s moe complcated, because the balls otate, and ae n contact wth a table. By gvng the queue ball spn, an expet playe can move the queue ball aound at wll. To make the ball ebound, t must be stuck low down (below the cente of pecusson ) to gve t a evese spn; to

33 make t follow the stuck ball, t should be stuck hgh up, to make t oll towads the ball to be potted. Gvng the ball a sdeways spn can make t ebound n a contollable decton lateally as well. And t s even possble to make a queue ball tavel n a cuved path wth the ght spn. Neve let t be sad that you don t lean useful sklls n engneeng classes! 4.3 Angula mpulse-momentum elatons 4.3. Defnton of the angula mpulse of a foce The angula mpulse of a foce s the tme ntegal of the moment exeted by the foce. To make the concept pecse, consde a patcle that s subected to a tme vayng foce F () t, wth components n a fxed bass {,, k }, O x z then F() t = Fx() t + Fy() t + Fz() t k y Let () t = x() t + y() t + z() t k denote the poston vecto of the patcle, and M() t = () t F() t = ( ytf () z() t ztf () y() t) + ( ztf () x() t xtf () z() t) + ( xtf () y() t ytf () x() t) k the moment of the foce about the ogn. The Angula Impulse exeted by the foce about O dung a tme nteval t 0 t t s defned as t t A = M () t dt = () t F () t dt t0 t0 The angula mpulse s a vecto, and can be expessed as components n a bass A= A+ A + Ak x y z t ( () () () ()) A= ytf t ztf t dt x z y t0 t ( () () () ()) A = ztf t xtf t dt y x z t0 t ( () () () ()) A= xtf t ytf t dt z y x t0 If you know the moment as a functon of tme, you can calculate ts angula mpulse usng smple calculus. Fo example fo a constant moment, wth vecto value M 0, the mpulse s A= M 0( t t0) k (t) F(t)

34 4.3. Defnton of the angula momentum of a patcle The angula momentum of a patcle s smply the coss poduct of the patcle s poston vecto wth ts lnea momentum h= p= mv The angula momentum s a vecto the decton of the vecto s pependcula to ts velocty and ts poston vectos Angula mpulse Angula Momentum elaton fo a sngle patcle Consde a patcle that s subected to a foce F () t fo a tme nteval t 0 t t. Let () t denote the poston vecto of the patcle Let M() t = () t F () t denote the moment of F about the ogn Let A denote the angula mpulse exeted on the patcle Let h 0 denote the angula momentum at tme t 0 Let h denote the angula momentum at tme t x O y k P z The momentum consevaton equaton can be expessed ethe n dffeental o ntegal fom.. In dffeental fom d M = h dt. In ntegal fom A= h h 0 Poof: Although t s not obvous, these ae ust anothe way of wtng Newton s laws of moton. To show ths, we ll deve the dffeental fom. Stat wth Newton s law dv F= ma = m dt Take the coss poduct of both sdes wth dv F= m dt Note that v mv= 0 snce the coss poduct of two paallel vectos s zeo. We can add ths to the ght hand sde, whch shows that d d d d d F= m v + v mv = m v + mv = ( mv ) = h dt dt dt dt dt Ths yelds the equed elaton. Angula momentum consevaton: Fo the specal case whee the foce s paallel to, the moment of the foce actng on the patcle s zeo ( = F 0), and angula momentum s constant

35 d dt ( m ) v = Examples usng Angula Impulse Angula Momentum elatons fo a sngle patcle The angula mpulse-angula momentum equatons ae patculaly helpful when you need to solve poblems whee patcles ae subected to a sngle foce, whch acts though a fxed pont. They can also be used to analyze otatonal moton of a massless fame contanng one o moe patcles. Example : Obtal moton. A satellte s launched nto a geostatonay tansfe obt by the ARIANE V launch faclty. At ts pegee (the pont whee the satellte s closest to the eath) the satellte has speed 0.km/sec and alttude 50km. At apogee (the pont whee the satellte s futhest fom the eath) the satellte has alttude km. Calculate the speed of the satellte at apogee. v a p a v a Assumptons:. We assume that the only foce actng on the satellte s the gavtatonal attacton of the eath. The eath s adus s km Calculaton:. Snce the gavtatonal foce on the satellte always acts towads the cente of the eath, angula momentum about the eath s cente s conseved.. At both pegee and apogee, the velocty vecto of the satellte must be pependcula to ts poston vecto. To see ths, note that at the pont whee the satellte s closest and futhest fom the eath, the dstance to the eath s a max o mn, and so the devatve of the dstance of the satellte fom the eath must vansh,.e. d d d d ( ) = ( ) = 0 + = 0 v = 0 dt dt dt dt whee we have used the chan ule to evaluate the tme devatve of. Recall that f the dot poduct of two vectos vanshes, they ae mutually pependcula. We take a coodnate system wth and n the plane of the obt, and k pependcula to the obt. 3. We take the satellte obt to le n the, plane wth k pependcula to the obt. The angula momentum at apogee and pegee s then hp = p mvp = pmvpk ha = a mva = amvak whee a, p ae the dstance of the satellte fom the cente of the eath at apogee and pegee, and va, v p ae the coespondng speeds. 4. Snce angula momentum s conseved t follows that hp = h a pmvp = amva va = pv p / a 5. Substtutng numbes yelds.6 km/s

36 Example : Moe obtal moton. The obt fo a satellte s nomally specfed by a set of angles specfyng the nclnaton of the obt, and by quotng the dstance of the satellte fom the eaths cente at apogee and pegee a, p. It s possble to calculate the speed of the satellte at pegee and apogee fom ths nfomaton. Calculaton. Fom the pevous example, we know that the dstances and veloctes ae elated by pmvp = amva. The system s consevatve, so the total enegy of the system s conseved. 3. The potental eneges when the satellte s at pegee and apogee ae GMm GMm Vp = Va = p a 4. The knetc eneges of the satellte at pegee and apogee ae Tp = mvp Ta = mva 5. The knetc enegy of the eath can be assumed to be constant. Enegy consevaton theefoe shows that GMm GMm mvp = mva p a 6. The esults of () and (5) gve two equatons that can be solved fo va, v p n tems of known paametes. Fo example, () shows that va= v p p/ a- ths can be substtuted nto (5) to see that GMm p GMm mvp = mvp p a a a p GMa vp = GM vp = a p a p( a + p) Smlaly, at apogee GMp va = ( + ) a a p