Linear Momentum. Center of Mass.

Transcription

1 Lecture 6 Chapter 9 Physcs I Lnear omentum. Center of ass. Course webste: Lecture Capture:

2 frequency 3 Exam Results verage 49./00 total grade dstrbuton 38 grade dstrbuton secton average grade Problem verage C 3.5 problem 6.9 problem problem 4 9. Total 49.

3 Outlne Chapter 9 -D collsons Systems of partcles (extended objects) Center of mass

4 Lnear momentum: Newton s nd law: If F ext 0 Revew p mv F m a dp, then 0 dt For an elastc head-on collson F, thus P const m v dp dt (Ths form s more general) m v m v m v m v m v v ( v v ) v Conservaton of momentum Conservaton of mechancal energy

5 Collsons n D You know how to fgure out the results of a collson between objects n D: -use conservaton of momentum- and, -f collson s ellastc, conservaton of mech. energy. You can contnue to use the same rules n D collsons as follows: If If ext F x ext F y 0, then momentum n x - drecton s conserved ntal fnal px px 0, then momentum n y - drecton s conserved ntal fnal p p ech. Energy s ( mv ) y ntal ( y conserved n elastc collsons (not n each dmenson) mv ) fnal

6 Example: Collsons n D projectle (m ) moves along the x-axs and hts a target (m ) at rest. fter the collson, the two objects go off at dfferent angles.

7 m Collsons n D: omentum Conservaton Snce net external forces n x and y drectons are zero, lnear momentum n x and y drectons are conserved X p Two equatons, can be solved for two unknowns ut, sometmes, these eq-ns aren t enough If collson s elastc, we get a thrd equaton (conservaton of mechancal energy) before x p m v' v before y p mv m m v Y 0 p sn ' m after x v ' after y 0 v' cos m v m ' v' mv sn ' ' cos m v Careful!!! n -D collsons v v v y m 0 m v m ( v v ) v x Three equatons, can be solved for three unknowns

8 Example: D collson all movng at 4 m/s strkes ball (of equal mass) at rest. fter the collson, ball travels forward at an angle of +45º, and ball travels forward at -45º. What are the fnal speeds of the two balls? conservaton of y-momentum 0 mv' sn 45 mv' sn( 45) m( 4 m ) mv' cos 45 mv' cos( 45) s 0 v' sn 45 v' sn 45 v' v' conservaton of x-momentum 4 ( )v' ( )v' m v v y m 0 m v m v x 4 v' v' v' 4 v' m/ s v' v'

9 Center of ass ()

10 Center of ass () We know how to address ths moton- knematc equatons How to descrbe motons lke these? The general moton of an object can be consdered as the sum of the translatonal moton of a certan pont, plus rotatonal moton about that pont. That pont s called the center of mass pont

11 Center of ass () Pure translatonal moton Translatonal plus rotatonal moton

12 How to fnd the center of mass? pont depends only on the mass dstrbuton of an object.

13 Center of ass: Defnton Poston vector of the : r m r r r mr mr m3r3 m m m m r 3 n mr ( x, y, z ) m m m3 Component form: n x m x y total mass of the system n m y r 3 m 3 z n m z

14 Center of ass Poston ( partcles) x=0 m m x-axs x x defnton x n m x poston for two partcles: Velocty of the : cceleraton of the : m x mx m m x mv v mv ma a ma where

15 Center of ass Velocty ( partcles) Consder velocty : For equal masses mv v v m m m m v X v m x v mv m( v) m 0 m m, v=0 v v/ X x v mv m(0) m v / For unequal masses m v v/3 X m x v mv m(0) m m v / 3

16 ConcepTest oton of Two equal-mass partcles ) t does not move ( and ) are located at ) t moves away from wth speed v some dstance from each other. Partcle s held C) t moves toward wth speed v statonary whle s D) t moves away from wth speed v moved away at speed v. E) t moves toward wth speed v What happens to the center of mass of the m v=? two-partcle system? v/ m v X m(0) m( v) Let s say that s at the orgn (x = 0) and s v v / m at some poston x. Then the center of mass s at x/ because and have the same mass. If v = dx/dt tells us how fast the poston of s changng, then the poston of the center of mv mv mass must be changng lke d(x/)/dt, whch s smply v. v

17 of a sold object Let s fnd of an extended body: efore, for many partcles we had r n mr r m Now, let s dvde mass nto smaller sectons m r r m mr lm 0 r mr rdm x xdm y ydm z zdm

18 coordnates (-5;4) m y of sold symmetrcal objects The easest trck s to use symmetry... m(5) m( 5)... x So, contrbutons to from x symmetrcal ponts cancel... m( 4) m(4)... each other and, as a result, m y the coordnates are (0;0) (5;-4) - coordnates If we break a symmetry, the wll be shfted Old Old

19 of Sold Objects (nce trck) How to deal wth objects lke ths? Dvde t nto symmetrcal objects m m m m of the orgnal object

20 ConcepTest Center of ass The dsk shown below n () clearly has ts center of mass at the center. Suppose a smaller dsk s cut out as shown n (). Where s the center of mass of () as compared to ()? ) hgher ) lower C) at the same place D) there s no defnable n ths case () () X

21 efore we used Newton s nd law for ponts (whch had masses but not szes) Now, Newton s nd law for a system of partcles (or extended bodes)

22 Newton s nd law for a system of partcles Poston vector of the : d r d t a r r r 3 r r n mr Snce, then acceleraton s : a F a F m m n F external ext n ma n a ma Where F forces actng on F F nternal ext net partcle However, nternal forces cancel each other ( N.3rd law) The of a system (total mass ) moves lke a sngle partcle of mass acted upon by the same net external force. n F

23 F pont descrbes translatonal moton of a system F It doesn t matter where you appled an ext. force, transl. moton of the system wll be the same F F

24 Newton s nd law for a system of partcles (II) a F net ext f ext F net 0, then a 0, and V const In the absence of external forces, the moton of the center of mass of a system of partcles (or an extended object) s unchanged.

25 Example: Center of ass ( partcles) What s the center of mass of pont masses (m = kg and m =3 kg), at two dfferent ponts: =(0,0) and =(,4)? y defnton: x y mx mx ( m m ) m y m y ( m m ) x y m = kg and m =3 kg ( 0) (3 ) 3 ( 0) (3 4) Or n a vector form: =(0,0) m =kg =(,4) m =3kg r.5î 3ĵ

26 Thank you See you on Wednesday

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