《大學(xué)物理》英文課件9 Center of Mass and Linear Momentum

1、? ?大學(xué)物理大學(xué)物理? ?英文課件英文課件9 9 Center of Mass and Center of Mass and Linear MomentumLinear MomentumCenter of mass - a special pointThe center of mass - the point that moves as though (1) all of the mass were concentrated and (2) all external forces were applied there 1 1 1comcomcomzdmMzydmMyxdmMxVMdVdm 1

2、 1 1comcomcomzdVVzydVVyxdVVxSolid body321332211commmmxmxmxmx321332211commmmymymymySample Problem com of three particlescm 83comxcm 58comyProblem: three particles of masses m1 = 1.2kg, m2 = 2.5kg, and m3 = 3.4kg form an equilateral triangle of edge length a = 140cm. Where is the center of mass of thi

3、s system?Solution:m1 = 1.2kg, m2 = 2.5kg, m3 = 3.4kg, a = 140cmcm120,cm0,cm0,cm70,cm140,cm0321321yyyxxx0PSPPSSPSmmxmxmxRxmmxxPSSPSSP31areaareaSample Problem com of plate with missing pieceProblem: a uniform metal plate P of radius 2R from which a disk of radius R has been stamped out (removed) in an

4、 assembly line. Locate the center of mass comp of the remaining plate.Solution:3120,222RRRareaareammxRxPSPSPSSNewtons Second Law for a System of Particles nnrmrmrmrM 2211comnnvmvmvmvM 2211comnnamamamaM 2211comnet21comFFFFaMn comnetaMFF : external force; M: total mass of closed system; a: acceleratio

5、n of comzzyyxxMaFMaFMaFcom,net,com,net,com,net, Sample ProblemMFFFaaMF321comcomnet 2321321,comm/s 03. 1mmmFFFaxxxx2321321,comm/s 53. 0mmmFFFayyyy22,com2,comcomm/s2 . 1)()(yxaaao,com,com127tanxyaaProblem: the three particles are initially at rest. Each experiences an external force due to bodies outs

6、ide the three-particle system. The directions are indicated, and the magnitudes are F1 = 6.0N, F2 = 12N, and F3 = 14N. What is the acceleration of the com, and in what direction does it move?Solution:Linear MomentumvmpdtpddtvmdamF)(netm/skgunit th vector wia is -Linear momentum of a particle:Linear

7、Momentum of a system of particlescom22111vMvmvmvmPPnnnii dtPdFaMdtvdMdtPdnetcomcom Linear momentum of a system of particles:Closed and isolated system0 CnetFM0netdtPdFfiPPP constantfitt later time someat momentumlinear total timeinitial someat momentumlinear totalThe law of conservation of linear mo

8、mentum: If no net external force acts on a system of particles, the total linear momentum cannot change. Conservation of Linear MomentumHSMSfiivMvMPMvP)8 . 0()2 . 0( 500km/h and relrelHSMSMSrelHSvvvvvvv00km/h22 HSfivPPSample Problem one-dimensional explosionProblem: a space hauler and cargo module o

9、f total mass M travel along an x axis in deep space. They have initial velocity vi of magnitude 2100km/h, relative to the Sun. With a small explosion, the hauler ejects the cargo module of mass 0.20M. The hauler then travels 500km/h faster than the module along the x axis; that is, the relative spee

10、d vrel between the hauler and the module is 500km/h. What then is the velocity vHS of the hauler relative to the Sun? Solution:MSrelHSvvvSample Problem two-dimensional explosion0fyiyPPyfCyfByfAfypppP,o,50sin2 . 0 0fByfByfAMvppm/s 6 . 9fBvo,80sin3 . 0fCyfCMvpProblem: a firecracker placed inside a coc

11、onut of mass M, initially at rest on a frictionless floor, blows the coconut into three pieces that slide across the floor. Piece C with mass o.30M, has final speed vfC = 5.0m/s. (a) what is the speed of piece B with mass 0.20M? (b) What is the speed of piece A? Solution: (a)m/s5fCvSample Problem tw

12、o-dimensional explosiono,50cos2 . 0 5 . 0fBxfBfAxfAMvpMvpo,80cos3 . 0fCxfCMvp0,xfCxfBxfAixpppPm/s 0 . 3fAvProblem: a firecracker placed inside a coconut of mass M, initially at rest on a frictionless floor, blows the coconut into three pieces that slide across the floor. Piece C with mass o.30M, has

13、 final speed vfC = 5.0m/s. (a) what is the speed of piece B with mass 0.20M? (b) What is the speed of piece A? Solution: (b) Collision and impulse A collision is an isolated event in which two or more bodies exert relatively strong forces on each other for a relatively short time Collision does not

14、require “crash Single collision - impulse and Linear Momentum dttFpddtpdtF)( )(fifittppdttFpd)(fittdttFJ)(JppifImpulse-linear momentum theorem tFJavgImpulse:)(vmnpnJvtmvmtntvnmtJFavgvvvvifvvvvvvifif2Impulse on target: Series of Collisionsn number of projectiles that collide in a time interval t If t

15、he projectiles stop upon impact:If the projectiles bounce directly backward from the target with no change in speed:N 4 .3189avgtJFm/skg 2 . 6180coscosoifxmvmvJm/skg 5 . 10sinfymvJSample ProblemProblem: a 0.300kg baseball collides with a bat at interval 2.00ms. Before collision, its initial velocity

16、 is vi = 12.0m/s along opposite x direction. After collision, it travels at direction with angle 30o Solution:ofivvm30,m/s0 .10,m/s0 .12,kg300. 0m/skg4 . 622yxJJJCollision - Elastic collision (kinetic energy is conserved) - Inelastic collision (kinetic energy is not conserved) - Completely inelastic

17、 collision ( greatest loss in kinetic energy if the bodies stick together)Law of conservation of linear momentum - In a closed, isolated system, the linear momentum of each colliding body may change but the total linear momentum cannot change. Momentum and Kinetic Energy in CollisionsInelastic Colli

18、sions in One Dimension collision after themomentum totalcollision thebeforemomentum totalffiipppp2121ffiivmvmvmvm22112211Inelastic Collisions in One Dimension -completely inelastic collision Vmmvmvmii)(212211VmVmvmvmii212211212211mmvmvmViiVelocity of the center of mass:MPvppPvmmvMPcomiicomcom2121col

19、lision theduring conserved is momentumlinear TotalVMmmv)(ghMmVMm)()(221m/s 6302ghmMmvSample ProblemProblem: a large block of wood of mass M = 5.4kg is hung from two long cords. A bullet of mass m = 9.5g is fired into the block, coming quickly to rest. The block + bullet then swing upward , their cen

20、ter of mass rising a vertical distance h = 6.3cm before the pendulum comes momentarily to rest at the end of its arc. what is the speed of the bullet just prior to the collision?Solution:Elastic Collisions in One Dimension collision after theenergy kinetic totalcollision thebeforeenergy kinetic tota

21、lffiivmvmvmvm2211221122221211212222121121ffiivmvmvmvmElastic Collisions in One Dimension one object is at rest ffivmvmvm221111222212112121121ffivmvmvmififvmmmvvmmmmv121121212112Kinetic energy is conserved:Momentum is conserved:We get:ififvmmmvvmmmmv121121212112 iffvvvmm12121 0 ififvmmvvvmm121211122

22、ififvvvvmm1211212 Equal masses:A massive target:A massive projectile:Step 1: The descent of sphere 1 m/s252. 12111121121ghvghmvmiiStep 2: The two-sphere collision Sample ProblemProblem: two metal spheres, suspended by vertical cords, initially just touch. Sphere 1 with mass m1 = 30g is pulled to the

23、 left to height h1 = 8.0cm, and then released from rest. After swinging down, it undergoes an elastic collision with sphere 2 with mass m2 = 75g. What is the velocity v1f of sphere 1 just after the collision?Solution:m/s540 121211.-vmmmmvifCollisions in Two Dimensionsffiipppp2121ffiiKKKK212122211111coscosffivmvmvm222111sinsin0ffvmvm222212112121121ffivmvmvmKinetic e