Physcs 207: Lecture 20 Today s Agenda Homework for Monday Recap: Systems of Partcles Center of mass Velocty and acceleraton of the center of mass Dynamcs of the center of mass Lnear Momentum Example problems Momentum Conservaton Inelastc collsons n one dmenson 1 System of Partcles: Center of Mass For partcles, the components of R are: mx my ( X, Y, Z ) =,, M M mz M For a contnuous sold, we have to do an ntegral. N M = = 1 m y x dm r R rdm rdm = = dm M where dm s an nfntesmal mass element. 2 Page 1
System of Partcles: Center of Mass We can use ntuton to fnd the locaton of the center of mass for symmetrc objects that have unform densty: It wll smply be at the geometrcal center! 3 System of Partcles: Center of Mass The center of mass for a combnaton of objects s the average center of mass locaton of the objects: m 2 R N mr = 1 = N m = 1 R 2 -R 1 m 1 R 1 R R 2 y x so f we have two objects: R m1r1 m2r2 = m m 1 m2 = R1 M 2 ( R R ) 2 1 4 Page 2
Lecture 20, Act 1 Center of Mass The dsk shown below (1) clearly has ts at the center. Suppose the dsk s cut n half and the peces arranged as shown n (2): Where s the of (2) as compared to (1)? (a) hgher (b) lower (c) same X (1) (2) 5 Lecture 20, Act 1 Soluton The of each half-dsk wll be closer to the fat end than to the thn end (thnk of where t would balance). The of the compound object wll be halfway between the s of the two halves. Ths s hgher than the of the dsk X X X X (1) (2) 6 Page 3
System of Partcles: Center of Mass The center of mass () of an object s where we can freely pvot that object. pvot Gravty acts on the of an object (show later) If we pvot the object somewhere else, t wll orent tself so that the s drectly below the pvot. Ths fact can be used to fnd the of oddly-shaped objects. mg pvot pvot 7 System of Partcles: Center of Mass Hang the object from several pvots and see where the vertcal lnes through each pvot ntersect! pvot pvot pvot The ntersecton pont must be at the. 8 Page 4
Lecture 20, Act 2 Center of Mass An object wth three prongs of equal mass s balanced on a wre (equal angles between prongs). What knd of equlbrum s ths poston? a) stable b) neutral c) unstable 9 Lecture 20, Act 2 Soluton The center of mass of the object s at ts center and s ntally drectly over the wre If the object s pushed slghtly to the left or rght, ts center of mass wll not be above the wre and gravty wll make the object fall off mg mg (front vew) 10 Page 5
Lecture 20, Act 2 Soluton Consder also the case n whch the two lower prongs have balls of equal mass attached to them: mg mg In ths case, the center of mass of the object s below the wre When the object s pushed slghtly, gravty provdes a restorng force, creatng a stable equlbrum 11 Velocty and Acceleraton of the Center of Mass If ts partcles are movng, the of a system can also move. Suppose we know the poston r of every partcle n the system as a functon of tme. R 1 N r M m = = 1 N M = = 1 m So: V = dr N r N v dt = 1 M m d dt = 1 M m = 1 = 1 And: A = dv N v N a dt = 1 M m d dt = 1 M m = 1 = 1 The velocty and acceleraton of the s just the weghted average velocty and acceleraton of all the partcles. 12 Page 6
Lnear Momentum: Defnton: For a sngle partcle, the momentum p s defned as: (p s a vector snce v s a p = mv vector). So p x = mv x etc. Newton s 2nd Law: F = ma = m dv dt d = ( mv) dt F d p = dt Unts of lnear momentum are kg m/s. 13 Lnear Momentum: For a system of partcles the total momentum P s the vector sum of the ndvdual partcle momenta: N N m = 1 = 1 P = p = v But we just showed that N mv = MV = 1 V 1 = M N m = 1 v So P = MV 14 Page 7
Lnear Momentum: So the total momentum of a system of partcles s just the total mass tmes the velocty of the center of mass. P = MV Observe: dp M d V = = MA = ma = F, dt dt net dp We are nterested n so we need to fgure out dt F net, 15 Lnear Momentum: Suppose we have a system of three partcles as shown. Each partcle nteracts wth every other, and n addton there s an external force pushng on partcle 1. F = F, NET 1, EXT = ( F F F ) 13 12 1, EXT ( F F ) 21 23 ( F F ) 31 32 (snce the other forces cancel n pars...newton s 3rd Law) All of the nternal forces cancel!! Only the external force matters!! F 13 m 1 F 31 F 12 F 1,EXT m 3 F32 F 23 F 21 m 2 16 Page 8
Lnear Momentum: Only the total external force matters! dp dt = F, = F EXT NET, EXT m 3 Whch s the same as: dp F = = MA NET, EXT dt m 1 F 1,EXT m 2 Newton s 2nd law appled to systems! 17 Center of Mass Moton: Recap (really know ths!) We have the followng law for moton: F EXT dp = =MA dt Ths has several nterestng mplcatons: It tells us that the of an extended object behaves lke a smple pont mass under the nfluence of external forces: We can use t to relate F and A lke we are used to dong. It tells us that f F EXT = 0, the total momentum of the system can not change. The total momentum of a system s conserved f there are no external forces actng. 18 Page 9
Example: Astronauts & Rope Two astronauts at rest n outer space are connected by a lght rope. They begn to pull towards each other. Where do they meet? M = 1.5m m 19 Example: Astronauts & Rope... They start at rest, so V = 0. V remans zero because there are no external forces. So, the does not move! They wll meet at the. M = 1.5m m L x=0 x=l Fndng the : If we take the astronaut on the left to be at x = 0: M m(l m(l x cm = ( 0 ) ) ) 2 = = M m 25. m 5 L 20 Page 10
Lecture 20, Act 3 Center of Mass Moton A man weghs exactly as much as hs 20 foot long canoe. Intally he stands n the center of the motonless canoe, a dstance of 20 feet from shore. Next he walks toward the shore untl he gets to the end of the canoe. What s hs new dstance from the shore. (There no horzontal force on the canoe by the water). 20 ft? ft 20 ft before (a) 10 ft (b) 15 ft (c) 16.7 ft after 21 Lecture 20, Act 3 Soluton Snce the man and the canoe have the same mass, the of the man-canoe system wll be halfway between the of the man and the of the canoe. Intally the of the system s 20 ft from shore. 20 ft X X of system x 22 Page 11
Lecture 20, Act 3 Soluton Snce there s no force actng on the canoe n the x-drecton, the locaton of the of the system can t change! Therefore, the man ends up 5 ft to the left of the system, and the center of the canoe ends up 5 ft to the rght. He ends up movng 5 ft toward the shore (15 ft away). 15 ft 20 ft X 10 ft 5 ft X x of system 23 Center of Mass Moton: Revew We have the followng law for moton: F EXT dp = =MA dt Ths has several nterestng mplcatons: It tells us that the of an extended object behaves lke a smple pont mass under the nfluence of external forces: We can use t to relate F and A lke we are used to dong. It tells us that f F EXT = 0, the total momentum of the system does not change. The total momentum of a system s conserved f there are no external forces actng. 24 Page 12
Lecture 20, Act 4 Center of Mass Moton Two pucks of equal mass are beng pulled at dfferent ponts wth equal forces. Whch experences the bgger acceleraton? (a) A 1 > A 2 (b) A 1 < A 2 (c) A 1 = A 2 A 1 (1) M T A 2 F (2) M T 25 Lecture 20, Act 4 Soluton We have just shown that MA = F EXT Acceleraton depends only on external force, not on where t s appled! Expect that A 1 and A 2 wll be the same snce F 1 = F 2 = T = F / 2 The answer s (c) A 1 = A 2. So the fnal veloctes should be the same! 26 Page 13
Lecture 20, Act 4 Soluton The fnal velocty of the of each puck s the same! Notce, however, that the moton of the partcles n each of the pucks s dfferent (one s spnnng). V Ths one has more knetc energy (rotaton) ω V 27 Momentum Conservaton F EXT dp = dt dp = 0 F EXT = 0 dt The concept of momentum conservaton s one of the most fundamental prncples n physcs. Ths s a component (vector) equaton. We can apply t to any drecton n whch there s no external force appled. You wll see that we often have momentum conservaton even when energy s not conserved. 28 Page 14
Elastc vs. Inelastc Collsons A collson s sad to be elastc when knetc energy as well as momentum s conserved before and after the collson. K before = K after Carts colldng wth a sprng n between, bllard balls, etc. v A collson s sad to be nelastc when knetc energy s not conserved before and after the collson, but momentum s conserved. K before K after Car crashes, collsons where objects stck together, etc. 29 Inelastc collson n 1-D: Example 1 A block of mass M s ntally at rest on a frctonless horzontal surface. A bullet of mass m s fred at the block wth a muzzle velocty (speed) v. The bullet lodges n the block, and the block ends up wth a speed V. In terms of m, M, and V : What s the ntal speed of the bullet v? What s the ntal energy of the system? What s the fnal energy of the system? Is knetc energy conserved? x v V before after 30 Page 15
Example 1... Consder the bullet & block as a system. After the bullet s shot, there are no external forces actng on the system n the x-drecton. Momentum s conserved n the x drecton! P x, = P x, f mv = (Mm)V v = M m m V v V x ntal fnal 31 v = M m m V Example 1... Now consder the knetc energy of the system before and after: Before: E mv m M m 2 M m B = 1 2 = 1 1 V = ( M m) V 2 2 m 2 m After: EA = 1 ( M 2 mv ) 2 2 2 So E A m = M m E B Knetc energy s NOT conserved! (frcton stopped the bullet) However, momentum was conserved, and ths was useful. 32 Page 16
Inelastc Collson n 1-D: Example 2 M m V v = 0 ce (no frcton) M m v =? 33 Example 2... Use conservaton of momentum to fnd v after the collson. Before the collson: After the collson: P = MV m( 0 ) Pf = ( M m) v Conservaton of momentum: P = P f MV = ( M m) v M v = V vector equaton (M m ) 34 Page 17
v = M m m V Example 2... Now consder the K.E. of the system before and after: Before: E MV M M m 2 M m BUS = 1 2 = 1 1 v = ( M m) v 2 2 M 2 M After: EA = 1 ( M m) v 2 2 2 2 So E A M = M m E B Knetc energy s NOT conserved n an nelastc collson! 35 Lecture 20, Act 5 Momentum Conservaton Two balls of equal mass are thrown horzontally wth the same ntal velocty. They ht dentcal statonary boxes restng on a frctonless horzontal surface. The ball httng box 1 bounces back, whle the ball httng box 2 gets stuck. Whch box ends up movng faster? (a) Box 1 (b) Box 2 (c) same 1 2 36 Page 18
Lecture 20, Act 5 Momentum Conservaton Snce the total external force n the x-drecton s zero, momentum s conserved along the x-axs. In both cases the ntal momentum s the same (mv of ball). In case 1 the ball has negatve momentum after the collson, hence the box must have more postve momentum f the total s to be conserved. The speed of the box n case 1 s bggest! V 1 V 2 1 2 x 37 Lecture 20, Act 5 Momentum Conservaton mv nt = MV 1 -mv fn V 1 = (mv nt mv fn ) / M mv nt = (Mm)V 2 V 2 = mv nt / (Mm) V 1 numerator s bgger and ts denomnator s smaller than that of V 2. V 1 > V 2 x V 1 V 2 1 2 38 Page 19
Exploson (nelastc un-collson) Before the exploson: M After the exploson: v 1 v 2 m 1 m 2 39 Exploson... No external forces, so P s conserved. Intally: P = 0 Fnally: P = m 1 v 1 m 2 v 2 = 0 m 1 v 1 = - m 2 v 2 M v 1 v 2 m 1 m 2 40 Page 20