1、12.4 Theorem of motion of the center of massChapter 12:Theorem of Momentum12.1 The center of mass of a system of particles12.2 Momentum and impulse12.3 Theorem of momentum1.The center of mass.2.The center of mass of a system of particles is called center of mass.It is an important concept representi
2、ng the distribution of mass in any system of particles.)(imMiiCiiCrmrMMrmror got we,FromkzjyixrccccMzmzMymyMxmxiiCiiCiiC ,The position of thecenter of mass c is 12.1 The Center of Mass of a System of ParticlesExternal forces are the forces exerted on the members of a system by particles or bodies no
3、t belonging to the given system Internal forces are the forces of interaction between the members of the same system.As far as the whole system of particles is concerned,the geometrical sum(the principal vector)of all the internal forces of a system is zero.The sum of the moments(the principal momen
4、t)of all the internal forces of a system with respect to any center of axis is zero,too.。0)(or 0)(;0)()()(iixiiOiiFmFmF2.External forces and internal forces of a system of particles1)Momentum of a particle.The product of the mass of a particle and its velocity is called the momentum of a particle.It
5、 is a time-dependent vector with the same direction as the velocity,the unit of which is kgm/s.Momentum is a physical quantity measuring the intensity of the mechanical motion of a material body.For example,the velocity of a bullet is big but its mass is small.In the case of a boat it is just opposi
6、te.12.2 Momentum and Impulse1.Momentumiivmp 2)The momentum of a system of particles is defined as the vector equal to the geometric sum of the momenta of all the particles of the system:iiiiirmdtddtrdmvmP CiiiicrMrmMrmrThe momentum of a system is equal to the product of the mass of the whole system
7、and the velocity of its center of mass.iivmPCiivMvmPCCzzCCyyCCxxzMMvPyMMvPxMMvP ,3)Momentum of a system of rigid bodies:Assume that the mass and the velocity of the center of mass of the i-th rigid body are .For the whole system we get then ciivm,CiivmPCiiCizizCiiCiyiyCiiCixixzmvmPymvmPxmvmP.,.,In t
8、erms of projections on cartesian axes we have),252ABlPClvmC21,1llvmABC2525,2lvmC2,3In the mechanism shown in the figure,OA rotate with a constant angular velocity .Assume that OA=L,OB=L.The rods OA and AB are homogeneous and of mass m.The mass of the slide block at B is also m.Determine the momentum
9、 of the system when j j=45.Example For the rod OA mass=.For the slide block mass=.For the rod AB:mass=Solution:ivvvmvmvmvmPCCCCCC)cossin(321321j)sincos(21jvvCCj)101252221()2103252221()sin2545cos21()2cos2545sin21(jimljllilllmFdtFId 21ttdtFI2)Force is a variable vector(Include magnitude and direction.
10、)The elementary impulse is The impulse is)(12ttFI1)Force is a constant vector.F2.Impulse:The product of a force and the action time of the force is called impulse.Impulse is used to characterize the accumulated effect on a body of a force acting during a certain time interval.3)The impulse of a resu
11、ltant force is equal to the geometric sumof the impulses of all component forces:ittttttIdtFdtFdtRI212121m/s.kg s)m/s(kg sN2The unit of impulse is the same as that of momentum.212121 ,ttttttzzyyxxdtFIdtFIdtFI1.Theorem of momentum for one particle:FvmdtdFdtvdmam)(so BecauseThe derivative of the momen
12、tum of a particle with respect to time is equal to the force acting on the particle.This is the momentum theorem for one particle.In a certain time interval,the change of the momentum of a particle is equal to the impulse of the force during the same interval of time.IdtFvmvmtt2112 Integral form.The
13、 differential of the momentum of a particle equals the elementary impulse of the force acting on it.IddtFvmd)(Differential form:12.3 Theorem of momentum Projection form:xxFmvdtd)(yyFmvdtd)(zzFmvdtd)(2112ttxxxxdtFImvmv2112ttyyyydtFImvmv2112ttzzzzdtFImvmvIf,then is a constant vector and the particle i
14、s an inertial motion.0Fvm The law of conservation of the linear momentum of a particle02112ttxxxxdtFImvmvCmvmvmvxxx120)(FvmdtdCvm 0 xFxmvIf,then is a const and the motion of the particle along the axis X is an inertial motion.2.Theorem of momentum of a system of particles )()()(eiiiiiFFvmdtd get wea
15、nd 0But.)()()(iieiiiiiFFFvmdtd)(eiFdtPdthe momentum theorem of a system of particles.For the whole system of particles,we haveFor any particle i in the system,we have .The derivative of the linear momentum of a system of particles with respect to time is equal to the geometric sum of all the externa
16、l forces acting on the system.The differential of the linear momentum of a system of particles is equal to the geometric sum of the elementary impulses of all the external forces acting on the system.During a certain time interval,the change in the linear momentum of a system of particles is equal t
17、o the geometric sum of the impulses of all the external forces acting on the system during the same time interval.2)Integral form:1)Differential form:ieIdtFeiP)()(dd.)(12eiIPP3)Projection form:)(eixxFdtdP)(eiyyFdtdP)(eizzFdtdP21)()(12tteixexxdtFIixPP21)()(12tteiyeyydtFIiyPP21)()(12tteizezzdtFIizPP,0
18、)(eiFIf,0)(eixFIfCvmPtheniiCvmthenPixixOnly external forces can change the total momentum of a system of particles,while internal forces are incapable of changing it.They only may cause that the momenta of the particles or parts of it are exchanged between the particles.4)The conservation law of the
19、 linear momentum of a system of particles:A big triangular column of mass M is placed on a smooth horizontal plane,on the slope of it lying a small triangular column of mass m.Determine the displacement of the big triangular column when the small column slides down the shape to the end.Example Solut
20、ion:1 Choose the system composed of these two bodies as the object to be investigated.Analysis of force,0)(exFCPxrvvAssume that the velocity of the big triangular block is the velocity of the small triangular block with respect to the big triangular block is .reavvvAnalysis of motion2 From the conse
21、rvation law of the linear momentum in horizontal direction(the system being at rest at the initial moment)we obtain0)(axmvvM0)()(vvmvMrx).(bamMmSmMmSrx,AsmmMssmmMvvrxrx0 xF)m/s(and 21vvA fluid flow through a bend,the fluid velocities at sections A and B being respect.Determine the dynamic force(addi
22、tionally the dynamic reaction)exerted by the fluid on the bend.Assume the fluid is incompressible,the flow rate Q(m3/s)is constant and density of the the fluid is(kg/m3).Example)()(12aBAaBbaBABabPPPPPPP.,)()(for 1212vtQvtQPPPPPAaBbaBaB3 From the momentum theorem of a system of particles,)(lim21120RP
23、PWvvQtPdtPdt1 Choose the fluid between the sections A and B as the system of particles under consideration.2 The analysis of the force is shown in the figure below.Assume that the fluid AB moves to the position ab during the time t.Solution:Static reaction ,dynamic reaction)(21PPWR)(12vvQRWhen calcu
24、lating we often use the projection form R)(12xxxvvQR)(12yyyvvQR)()(1221vvQPPWRThe force contrary to the force R is just the dynamic force exerted by the fluid to the bend.,)(lim21120RPPWvvQtPdtPdtTo a system of particles,using the momentum theorem of a system of particles we obtain:CvMP.)()(eiCFvMdt
25、dIf the mass of the system does not change)(eiCFaM)(eiCFrM The formulas above are the theorem of motion of the center of mass(or the differential equations of the motion of the center of mass).The product of the acceleration of the center of mass of a system and the mass of the whole system is equal
26、 to the geometric sum of all external forces acting on the system,the principal vector of the external force system.12.4 Theorem of motion of the center of mass1.Projection forms:;,)()()(eizCCzeiyCCyeixCCxFzMMaFyMMaFxMMa ;0 ,)()(2)(eibeinCCneiCFFvMMaFdtdvMMa)(eixcxFMa2.System of rigid bodies:3.Denot
27、ing by the No.i the body of mass mi and velocity vci we have)(eiCiiFam)(eiCiiFrmor)(eiCFaM)(eiCFrorM)(eiycyFMa)(eizczFMa 3.The theorem of motion of the center of mass is an equivalent expression to the momentum theorem of a system,being similar in form to the differential equations of motion for one
28、 particle.For a system of particles in arbitrary motion,the center of mass of the system moves as if it would be a particle of mass equal to the mass of the whole system to which all the external forces acting on the system are applied.4.The conservation law of the motion of the center of mass:0)(ei
29、FIfCvoaCC and 00CvifCrC 0)(eixFIfCvaCxCx and 000CxvIfCxC2)Knowing the external forces acting on the system,determine the law of motion of the center of mass.5.The theorem of motion of the center of mass may solve two types of dynamical problems:1)Knowing the motion of the center of mass of a system,
30、determine the external forces acting on the system(including the reactions of constraints).A motor body is mounted on a horizontal foundation.The mass of the rotor is m2 and the mass of stator is m1.The shaft of the rotor passes through the center of mass O1 of stator,but there is a distance e from
31、the center of mass O2 of the rotor to O1 due to an inaccuracy in the production.Determine the reaction forces of the constraints exerted on the mount of the motor by the foundation when the rotor is rotating with the angular velocity .Example Solution 1 choose the motor as the system of particles to
32、 be investigated and the analysis of forces is shown in the figure 2 Analysis of motion:The acceleration a1 at the center of mass of the stator O1 is zero.The acceleration a2 at the center of mass of the rotor O2 is e 2(directed to O1).sin ,cos2222teateayxa1=0，a2=e2,3 According to the theorem of mot
33、ion of the center of mass we havexxeixCixiNtemamFamcos ,2222)(gmgmNtemamFamyyeiyCiyi212222)(sin ,temgmgmNtemNyxsin,cos222122The dynamic reaction caused by the deviation from the center is a periodic function of time.There is a floating hoisting boat of weight P1=200kN with the jib of weight P2=10kN
34、and length l=8m,The weight of the lifted load P3=20kN.Assume that at the initial moment the whole system at rest and the angle between the jib OA and the vertical was 1=60,Neglecting the resistance of the water determine the displacement of the boat at the instant when the jib OA makes an angle 2=30
35、 to the vertical.Example 321332211321332211mmmxmxmxmmmmxmxmxm1 Choose the system consisting of the hoisting boat with the jib and the load as the object under study.CxPCPxPxciiicic 2 The analysis of the force is shown in the figure.and because the whole system was initially at rest the coordinate of
36、 the center of mass of the whole system will remain at rest.0)(exF 0ciixmSolution:The displacement of the boat is x.The displacement of the jib is.2/)sin(sin2112lxxThe displacement of the weight is.)sin(sin2113lxx0/)sin(sin2/)sin(sin Therefore2113211211lxPlxPxP)sin(sin)(2221321321lPPPPPx)30sin60(sin8)2010200(220210m.318.0The negative sign shows that the boatmoves to the left.0 iixP 0Becauseiixm
