Readng: Chapter 9 The Center ass Center ass and mentum See anmatn An Object Tssed Alng a Parablc Path The center mass a bdy r a system bdes s the pnt that mves as thugh all the mass were cncentrated there and all external rces were appled there
Fr partcles, Fr n partcles, m x m mx m m x mx x x m x mnx n In general, x y z n n n m x,, m z m y In vectr rm, r n m r
Sld Bdes x y z xdm, ydm, zdm I the bject has unrm densty, dm dv V Rewrtng dm = dv and m = V, we btan x y z V V V xdv, ydv, zdv Nte: [] I the bject has a pnt symmetry, then the center mass les at that pnt I the bject has a lne symmetry, then the center mass les n that lne 3
I the bject has a plane symmetry, then the center mass les n that plane [] The center mass an bject need nt le wthn the bject eg a dughnut Examples 9- Three partcles masses m = kg, m = 5 kg, m 3 = 34 kg are lcated at the crners an equlateral trangle edge a = 40 Where s the center mass? m = x = 0 y = 0 m = 5 x = 40 y = 0 m 3 = 34 x 3 = 40cs60 y 3 = 40sn60 m x mx m3x3 x ( )(0) (5)(40) (34)(70) = 83 (ans) 7 m y m y m3 y3 y ( )(0) (5)(0) (34)() = 58 (ans) 7 4
9- A unrm crcular metal plate P radus R has a dsk radus R remved rm t Lcate ts centre mass lyng n the x axs Let = densty, t = thckness ass: Object P: m S = (R) t R t = 3R t Object S: m P = R t Object C: m C = (R) t = 4R t Center mass: Object P: x P =? Object S: x S = R Object C: x C = 0 Snce mpxp ms xs x C 0, mp ms ms xp xs m P R t ( R) 3R t R (ans) 3 5
Newtn s Secnd Law r a System Partcles Fnet a In terms cmpnents, F F F net, x net, y net, z a a a, x, y, z,, See Yutube Ze Ballet Grand Jeté 6
Lnear mentum Fr a sngle partcle, the lnear mmentum s p mv Newtn s law: F net dp dt Fr a system partcles, the ttal lnear mmentum s P p p m v m v n n n Derentatng the pstn the center mass, v m v m v n n P v The lnear mmentum a system partcles s equal t the prduct the ttal mass the system and the velcty the center mass Newtn s law r a system partcles: dp dv a dt dt Hence F net dp dt 7
Cllsn and Impulse Newtn s law: F dp dp F( t) dt dt Integratng rm just bere the cllsn t just aterwards, p p t d p F( t) dt t Change n lnear mmentum durng the cllsn: t p p p F( t) dt The ntegral s a measure bth the strength and the duratn the cllsn rce It s called the mpulse J the cllsn: t J F( t) dt t t 8
Impulse-lnear mmentum therem: p p p J In terms cmpnents, p p p x y z p p p x y z p p p x y z J J J x z y,, Average rce: J F t, where t s the duratn the cllsn 9
Seres Cllsns n partcles cllde wth R n tme nterval t Ttal change n mmentum = np Accrdng t Newtn s law, average rce actng n the partcles = rate change mmentum n n p t t m v Hence the average rce actng n bdy R s: F n t m v I the clldng partcles stp upn mpact, v = v I the clldng partcles bunds elastcally upn mpact, v = v 0
Examples 9-4 Fg 9- shws the typcal acceleratn a male bghrn sheep when he runs head-rst nt anther male Assume that the sheep s mass s 900 kg What are the magntudes the mpulse and average rce due t the cllsn? Snce dv a, dt t v( t) a( t') dt' 0 Hence the change n velcty s gven by the area enclsed by the a-t curve v (07)(34) 459 ms Usng the mpulse-mmentum therem, J p mv (90)( 459) 43 kg ms The magntude the mpulse s 43 kg ms (ans) The magntude the average rce s J 43 F av 530 N (ans) t 07 Remark: The cllsn tme s prlnged by the lexblty the hrns I the sheep were t ht skull-t-skull r skullt-hrn, the cllsn duratn wuld be /0 what we used, and the average rce wuld be 0 tmes what we calculated!
9-5 Race-car wall cllsn A race car clldes wth a racetrack wall at speeds v = 70 ms and v = 50 ms bere and ater cllsn respectvely (Fg 9-a) Hs mass m s 80 kg (a) What s the mpulse J n the drver due t the cllsn? (b) The cllsn lasts r 4 ms What s the magntude the average rce n the drver durng the cllsn? (a) Usng the mpulsemmentum therem, J p p mv mv m v v ) ( x cmpnent: J m( v v ) x x x (80)(50cs0 70cs30 ) = 90 kg ms y cmpnent: ( ) J m v vy (80)( 50sn0 70sn 30 ) y y = 3495 kg ms The mpulse s then agntude: J J ( 90ˆ 3500 ˆ) j kg ms J x J y 366 kg ms (ans) 3600 kg ms J y Drectn: tan 754 80 05 (ans) J x J (b) 366 5 F 583 0 N 6 0 5 av N (ans) t 004 Remark: The drver s average acceleratn s 583 0 5 /80 30 ms = 39g atal cllsn!
Cnservatn Lnear mentum I the system partcles s slated (e there are n external rces) and clsed (e n partcles leave r enter the system), then P cnstant Law cnservatn lnear mmentum: Examples P P 9-7 Imagne a spaceshp and carg mdule, ttal mass, travellng n deep space wth velcty v = 00 km/h relatve t the Sun Wth a small explsn, the shp ejects the carg mdule, mass 00 The shp then travels 500 km/h aster than the mdule; that s, the relatve speed v rel between the mdule and the shp s 500 km/h What then s the velcty v the shp relatve t the Sun? Usng the cnservatn lnear mmentum, P P v 0 ( v vrel) 0 8v v v 0vrel v v 0vrel = 00 + (0)(500) = 00 km/h (ans) 3
9-8 A recracker placed nsde a ccnut mass, ntally at rest n a rctnless lr, blws the rut nt three peces and sends them sldng acrss the lr An verhead vew s shwn n the gure Pece C, wth mass 030, has nal speed v c =50ms - (a) What s the speed pece B, wth mass 00? (b) What s the speed pece A? (a) Usng the cnservatn lnear mmentum, Px P x P y P y m v cs80 m v cs50 m v 0 () C C B B m Cv C sn 80 mbv B sn 50 0 () m A = 05, m B = 0, m C = 03 (): 03v sn 80 0v sn 50 0 C 03sn 80 v 5 964 ms B 96 ms (ans) 0sn 50 (b) (): 03v cs80 0v cs50 0 5v C B A ( 03)(5)cs80 (0)(964)cs50 B v A 30 ms (ans) 05 A A 4
Inelastc Cllsns n One Dmensn In an nelastc cllsn, the knetc energy the system clldng bdes s nt cnserved In a cmpletely nelastc cllsn, the clldng bdes stck tgether ater the cllsn Hwever, the cnservatn lnear mmentum stll hlds m v ( m m ) V, r V m m m v 5
Example 9-9 The ballstc pendulum was used t measure the speeds bullets bere electrnc tmng devces were develped Here t cnssts a large blck wd mass = 54 kg, hangng rm tw lng crds A bullet mass m = 95 g s red nt the blck, cmng quckly t rest The blck + bullet then swng upward, ther center mass rsng a vertcal dstance h = 63 bere the pendulum cmes mmentarly t rest at the end ts arc What was the speed v the bullet just prr t the cllsn? Usng the cnservatn mmentum durng cllsn, mv ( m) V () Usng the cnservatn energy ater cllsn, ( m) V ( m) gh () V gh m m (): v V gh m m 54 00095 ()(98)(0063) 630 ms (ans) 00095 6
9-0 Cnsder the cllsn cars and wth ntal velctes v = +5 ms and v = 5 ms respectvely Let each car carry ne drver The ttal mass cars and are m = 400 kg and m = 400 kg respectvely (a) What are the changes v and v durng ther head-n and cmpletely nelastc cllsn? (b) Repeat the calculatn wth an 80 kg passenger n car (a) Usng the cnservatn mmentum, mv mv mv mv Snce the cllsn s cmpletely nelastc, v = v = V mv mv ( mm ) V mv mv V mm (400)( 5) (400)( 5) 0 400 400 v v v 0 ( 5) 5 ms (ans) v v v 0 ( 5) 5 ms (ans) (b) In ths case, m s replaced by 480 kg ( 480)( 5) (400)( 5) V 0694 ms 480 400 v v v 0694 ( 5) 43 ms (ans) v v v 0694 ( 5) 57 ms (ans) Remark: The rsk atalty t a drver s less that drver has a passenger n the car! 7
Elastc Cllsns n One Dmensn Statnary Target In an elastc cllsn, the knetc energy each clldng bdy can change, but the ttal knetc energy the system des nt change In a clsed, slated system, the lnear mmentum each clldng bdy can change, but the net lnear mmentum cannt change, regardless whether the cllsn s elastc Cnservatn lnear mmentum: m v m v m v Cnservatn knetc energy: v m v mv Rewrtng these equatns as m 8
Dvdng, m ( v v ) mv m ( v v ) m v, v v v We have tw lnear equatns r v and v Slutn: v v m m m m m m m v v, Specal stuatns: Equal masses: I m = m, then v = 0 and v = v (pl player s result) A massve target:i m >> m, then m and v v v m v v The lght ncdent partcle bunces back and the heavy target barely mves 3 A prjectle:i m >> m, then v v and v v The ncdent partcle s scarcely slwed by the cllsn 9
Example 9- Tw metal spheres, suspended by vertcal crds, ntally just tuch Sphere, wth mass m = 3 g, s pulled t the let t heght h = 80, and then released Ater swngng dwn, t underges an elastc cllsn wth sphere, whse mass m = 75 g What s the velcty v sphere just ater the cllsn? Usng the cnservatn energy, mv m gh gh v ()(98)(008) 5 ms Usng the cnservatn mmentum, mv mv mv Fr elastc cllsns, mv mv mv ( v v m v () m ) m ( v v ) mv Dvdng, v v v m ( v v ) m ( v v () m m 003 0075 v v 5 m m 003 0075 = 0537 ms 054 ms (ans) ) 0
Cllsns n Tw Dmensns Cnservatn lnear mmentum: x cmpnent: mv mv cs mv cs, y cmpnent: m v sn m v sn Cnservatn knetc energy: m v 0 v mv m Typcally, we knw m, m, v and Then we can slve r v, v and
Systems wth Varyng ass: A Rcket Assume n gravty Cnservatn lnear mmentum: P P Intal mmentum = v Fnal mmentum the exhaust = (d)u Fnal mmentum the rcket = ( + d)(v + dv) v ( d ) U ( d )( v dv) Suppse the rcket ejects the exhaust at a velcty v rel U v dv v rel Substtutng and dvdng by dt, dvrel dv, d dv vrel dt dt Snce the rate uel cnsumptn s the rst rcket equatn: Rv rel a R d dt, we have
T Rv rel s called the thrust the rcket engne Newtn s secnd law emerges T nd the velcty, Integratng, dv v rel d, v v d dv vrel, v v vrel ln (secnd rcket equatn) Remark: ultstage rckets are used t reduce n stages Example A rcket wth ntal mass = 850 kg cnsumes uel at the rate R = 3 kgs The speed v rel the exhaust gases relatve t the rcket engne s 800 ms What thrust des the rcket engne prvde? What s the ntal acceleratn the rcket? T = Rv rel = (3)(800) = 6440 N 6400 N (ans) T a 6440 76 ms (ans) 850 Remark: Snce a < g, the rcket cannt be launched rm Earth s surace 3