复旦大学《大学物理》课件（英文）-第7、8章(1).pdf

Chapter 7 Systems of particles7-5 Conservation of momentum in a system of particles7-4 Center of mass of solid objects7-3 Many-particle system7-2 Two-particle system7-1 The motion of a complex objectWhen can the object studied be regarded as a mass point?Doing only translational motionTranslational+Rotational motionTranslational motion Rotational motion7-1 The motion of a complex objectWhen we projectile a rigid body,the motion of the body looks very complicated.We can consider the motion of the rigid body to be a combination of a parabolic trajectory of a“center of mass”(rotational motion is not considered)plus a rotation about“center of mass”(translational motion is not considered).How to find the center of mass(CM)of a particle system?7-2 Two-particle systemAs an example,we consider a baton consists of two particles,at its ends A and B,connected by a thin rigid rod of fixed length and negligible mass.122mm=1m2m1m2mABDo an experiment to find CM in a two-particle system.Clearly both and are accelerated,however,one point in the rod(point C)moves with constant velocity.2m1mWe give the rod a push along the frictionless horizontal surface and examine its motion.Snapshots of the locations of points A and B at successive intervals of time.If point c is regarded as a reference,A and B points rotate with a constant rotational speed.View the motion from the reference of point C.So point c is actually the center of mass.Fig 7-51m2m1rxyO212211mmrmrmrcm+=212211mmxmxmxcm+=212211mmymymycm+=cmr2rcmrCCBy building a Cartesian coordinate,position of point c is found at:(7-2)(7-1)or written as:1m2mABCFrom Eq.(7-1),the velocity and acceleration of the CM are:(7-4)(7-6)212211mmvmvmdtrdvcmcm+=212211mmamamdtvdacmcm+=Can we also find is zero from Eq.(7-6)?cma+=+rrFFamam2122110=ca=rrFF12How about the motion if the system has net external forces?Suppose there is an external force on each particle in above expt.,thenThis looks very like a particle of mass located at the center of mass.+=+=+rextrextFFFFFFamam2211212211021=+rrFF=+extextextFFF21,and if write+=cmextammF)(21=+extFamam2211+=212211mmamamacm21mm+Newtons second law for systems of particles7-3 Many-particle systemConsider a system consisting of N particles of masses .The total mass is(7-10)Each particle can be represented by its location ,velocity and its acceleration .The CM of the system can be defined by logical extension of Eq(7-1):1m2mNm=nmMnvnrnaIn terms of components,Eq(7-11)can be written as Taking the derivative of Eq(7-11)(7-13)Differentiating once again:(7-14)=nncmxmMx1=nncmymMy1=nncmzmMz1=nncmvmMv1=nncmamMa1 +=NnncmFFFamaM21Or(7-15)=+=nnNNNcmrmMmmmrmrmrmr1212211(7-11)(7-12),.By Newtons third law,in Eq(7-15)the vector sum of all the internal forces is cancelled,and Eq(7-15)reduces to(7-16)cmextaMF=cmaEq(7-16)is just the Newtons second law for the system of N particles treated as a single particle of mass M located at the center of mass(),experiencing .cmr=nncmamMa1=nncmrmMr1We can summarize this important result as follow:“The overall translational motion of asystem of particles can be analyze using Newtons law as if the mass were concentrated at the center of mass and the total external force were applied at that point.”These are general results that apply equally well to a solid object.Sample problem 7-3A projectile(射弹)of mass 9.8kg is launched from the ground with initial velocity of 12.4m/s at an angle of above the horizontal(Fig 7-11).At some time after its launch,an explosion splits the projectile into two pieces.One piece of mass 6.5kg,is observed at 1.42s after the launch at a height of 5.9m and a horizontal distance of 13.6m from the launch point.Find the location of the second fragment at that same time.540v0CM1m2mSolution:If the projectile had not exploded,the location of the projectile at t=1.42s should have beenIt is the location of the CM.mssmtvxx4.1042.1)/3.7(0=mssmssmgttvyy3.4)42.1()/80.9(21)42.1()/0.10(212220=212211mmxmxmxcm+=212211mmymymycm+=?By Eq(7-12)Fig 7-11mmymMyycm9.02112=mkgmkgmkgmxmMxxcm7.31.3)6.13()5.6()4.10()6.9(2112=0v0Cm1m2m7-4 Center of mass of solid objects1)If an object has symmetry,the CM must lie at the geometrical symmetrical center of the object.Suppose the mass is uniformly distributed.cm2)If the object has no symmetry,sometimes it is also easy to find its cm position:(a)(b)cmcmSample problem 7-4Fig 7-13 shows a circular metal plate of radius 2R from which a disk of radius R has been removed.Find the cm(x)of the plate.Fig 7-13xyC xDRSolution:Due to the mirror symmetry about the x axis,the cm must lie along the x axis.If the hole is filled with a disk of the same material of radius R,the cm of composite disk is at the origin of the coordinate system.RRRRRxmmxDxDx31)()2()(222=xDxxDDcmmxmxmx+=0CM for the big circular plate?3)If we encounter solid irregular objects,we can divide infinite small elements.And the sums of Eqs(7-12)transform into integrals:=xdmMmxMxnnmcm1lim10=ydmMmyMynnmcm1lim10=zdmMmzMznnmcm1lim10=dmrMrcm1(7-19)In vec