Physics 141. Lecture 14. Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 14, Page 1


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1 Physcs 141. Lecture 14. Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 1
2 Physcs 141. Lecture 14. Course Informaton: Lab report # 3. Exam # 2. MultPartcle Systems (Chapter 9): The center of mass of a multpartcle system. The momentum prncple for multpartcle systems. The energy prncple for multplepartcle systems. The moton of the center of mass and varable mass systems. Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 2
3 Course nformaton. Labs: Lab report # 3 wll be due on Frday October 28 at noon. Lab # 4 wll take place on Monday October 31 n B&L 407. Exam # 2: Date: Tuesday October 25 between 8 am and 9.30 am. Locaton: Hoyt. Chapters covered: 5, 6, 7, 8, and 9. Note: some sectons n Chapter 9 (those dealng wth rotatonal moton, moment of nerta, etc.) wll not be covered untl later n the course and wll appear on exam # 3. Detals wll be provded durng the exam revew. Format: same as exam # 1 (and yes, there wll be a nonphyscs queston,.., but you know the answer). Revew: Frday 10/21 between 4 pm and 6 pm n B&L 106. Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 3
4 Rocket propulson. Observed n Hoyt. Notce the shadow of the exhaust on the screen. Projectle Moton. The launch Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 4
5 Rocket propulson. Observed n Hoyt. The collson Remanng exhaust. Frank L. H. Wolfs The destroyed rocket. Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 5
6 Locaton of the center of mass? In two or three dmensons the calculaton of the center of mass s very smlar, except that we need to use vectors. If we are not dealng wth dscreet pont masses we need to replace the sum wth an ntegral. r cm = 1 M V rdm Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 6
7 Moton of the center of mass. Nonrelatvstc lmt. Ma cm can be rewrtten n terms of the forces on the ndvdual components: M a cm = d dt ( Mv cm ) = d P cm dt = F = F net,ext The moton of the center of mass s thus only determned by the external forces. Forces exerted by one part of the system on other parts of the system are nternal forces and the sum of all nternal forces cancels. Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 7
8 Moton of the center of mass. Lnear momentum. Now consder the specal case where there are no external forces actng on the system: d P tot dt = 0 Ths equatons tells us that the total lnear momentum of the system s constant. In the case of an extended object, we fnd the total lnear momentum by addng the lnear momenta of all of ts components: P tot = M v cm = m v = p Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 8
9 Systems wth varable mass. Rocket moton s an example of a system wth a varable mass: M ( t + dt) = M ( t) + dm As a result of dumpng the exhaust, the rocket wll ncrease ts velocty: v ( t + dt) = v ( t) + dv Mass of exhaust. dm < 0 kg Snce ths s an solated system, lnear momentum must be conserved. The ntal momentum s equal to p = M ( t)v ( t) Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 9
10 Systems wth varable mass. The fnal lnear momentum of the system s gven by p f = M ( t) + dm ( ) + dv ( )U ( )( v t ) + dm where U s the velocty of the exhaust. Conservaton of lnear momentum therefore requres that M ( t)v t ( ) = M ( t) + dm ( ) + dv ( )( v t ) + dm The exhaust has a fxed velocty U 0 wth respect to the engne. U 0 and U are related n the followng way: U U 0 = v ( t) + dv ( )U Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 10
11 Systems wth varable mass. Conservaton of lnear momentum can now be rewrtten as M ( t)v ( t) = M ( t) + dm ( ) + dv or We conclude ( )( v t ) + ( dm ) v t M ( t)v ( t) = M ( t) v t ( dm )U 0 = M ( t) ( dv) ( ( ) + dv +U 0 ) ( )+ dv ( dm )U 0 Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 11
12 Systems wth varable mass. The prevous equaton can be rewrtten as dm dt U 0 = M t ( ) dv ( t ) In ths equaton: dm/dt = R where R s the rate of fuel consumpton. U 0 = u where u s the (postve) velocty of the exhaust gasses relatve to the rocket. dv/dt s the acceleraton of the rocket. Ths equaton can be rewrtten as RU 0 = Ma rocket whch s called the frst rocket equaton. Ths equaton can be used to fnd the velocty of the rocket (second rocket equaton): v f = v + u ln M M f dt Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 12
13 Rocket propulson at the U or R. Introducng the Nmbus A unque present. See Mke Culver for orders. Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 13
14 3 Mnute 52 Second Intermsson Snce payng attenton for 1 hour and 15 mnutes s hard when the topc s physcs, let s take a 3 mnute 52 second ntermsson. You can: Stretch out. Talk to your neghbors. Ask me a quck queston. Enjoy the fantastc musc. Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 14
15 Energy of multpartcle systems. Knetc and potental energy. In order to determne the (mechancal) energy of a multplepartcle system we need to determne both ts knetc energy and ts potental energy: The knetc energy of the system wll be the sum of the knetc energy of the centerofmass and the knetc energy of the moton of the partcles wth respect to the center of mass. The potental energy of the system may or may not depend on the poston of the center of mass: The gravtatonal potental energy can be expressed n terms of the poston of the center of mass. The electrostatc potental energy depends on the poston of charges, not on the poston of mass, and does not depend on the poston of the center of mass. Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 15
16 Energy of multpartcle systems. Knetc energy. The knetc energy of a multplepartcle system wll have two components: The translatonal component: the knetc energy assocated wth the moton of the center of mass. The relatve component: the knetc energy assocated wth the moton of the partcles wth respect to the center of mass. Ths type of moton an be vbratonal, rotatonal, a combnaton of these two, etc. The decomposton of the knetc energy nto ts components can be carred out wth a bt of effort. Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 16
17 Energy of multpartcle systems. Knetc energy. Consder a multpartcle system for whch we have specfed the poston of the center of mass r cm, and the poston of the partcle wth respect to the center of mass r. The knetc energy of partcle s equal to K = 1 2 m = 1 2 m d dt ( r cm + r ) 2 = 1 2 m v cm + v 2 ( v cm + v )( v cm + v ) = 1 2 m { v 2 cm + 2v cm v 2 + v } The total knetc energy of the system s thus equal to K = 1 2 m vcm2 + v cm m v m v 2 Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 17
18 Energy of multpartcle systems. Knetc energy. K = 1 2 m vcm2 + v cm m v m v 2 v cm m v = v cm d dt m r = v cm d dt ( 0) = m vcm2 = 1 2 Mvcm2 = K translatonal Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 18
19 Energy of multpartcle systems. Knetc energy. We thus see that K = 1 2 Mv cm 2 m v 2 = K translatonal + K relatve The relatve knetc energy (relatve to the center of mass of the system) can be equal to The vbratonal knetc energy (e.g. the vbratonal moton of atoms n an element). The rotatonal knetc energy (e.g. energy assocated wth the rotaton of a wheel)... Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 19
20 Energy of multpartcle systems. Potental energy. Consder a multpartcle system located close to the surface of the earth. The gravtatonal potental energy of ths system s equal to U = m gy = g m y = = gmy cm The gravtatonal potental energy thus depends on the vertcal poston of the center of mass of the system. Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 20
21 Conservaton of lnear momentum. An example. Two blocks wth mass m 1 and mass m 2 are connected by a sprng and are free to slde on a frctonless horzontal surface. The blocks are pulled apart and then released from rest. What fracton of the total knetc energy wll each block have at any later tme? m1 v1 v2 m2 Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 21
22 Conservaton of lnear momentum. An example. The system of the blocks and the sprng s a closed system, and the horzontal component of the external force s 0 N. The horzontal component of the lnear momentum s thus conserved. Intally the masses are at rest, and the total lnear momentum s thus 0 kg m/s. At any pont n tme, the veloctes of block 1 and block 2 are related: m1 v1 v2 m2 v 2 = m 1 m 2 v 1 Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 22
23 Conservaton of lnear momentum. An example. The knetc energes of mass m 1 and m 2 are thus equal to m1 v1 v2 m2 K 1 = 1 2 m v K 2 = 1 2 m v 2 = ( ) 2 = 1 2 m 2 m 2 v 2 m 1 ( ) 2 m 1 v 1 = m 1 K m 2 m 1 m 1 2 The fracton of the total knetc energy carred away by block 1 s equal to f 1 = K 1 K t = K 1 K 1 + K 2 = K 1 K 1 + m 1 m 2 K 1 = m 2 m 1 + m 2 Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 23
24 Next lecture: collsons. Conservaton of Lnear Momentum at RHIC (STAR Experment). Frank L. H. Wolfs Department of Physcs and Astronomy, Unversty of Rochester, Lecture 14, Page 24
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