复旦大学《大学物理》课件（英文）-第10章 Angular momentum(1).pdf

Chapter 10 Angular momentum10-1 Angular momentum of a particle1.DefinitionConsider a particle of mass m and linear momentum at a position relative to the origin o of an inertial frame we define the“angular momentum”of the particle with respect to the origin o to be(10-1)PrL=PrLxzymPrIts magnitude is(10-2)where is the smaller angle between and,we also can write it as Note that,for convenience and are inxy plane.sinrpL=rpprLPrPr2.The relation between torque and angular momentum Differentiating Eq(10-1)we obtain(10-6)Here ,the =+=FrdtPdrPdtrddtLd=vdtrd0=PvPdtrdand Eq(10-6)states that“the net torque acting on a particle is equal to the time rate of change of its angular momentum.=FdtPdSample problem 10-1A particle of mass m is released from rest at point p(a)Find torque and angular momentum with respect to origin o(b)Show that the relation yield a correct result.boyxPrmmgdtLd=Solution:(a)(b is the moment arm)(b)=0zmgb=0zbmgtvmrL=0zbmgdtLd10-2 Systems of particles1.To calculate the total angular momentum of a system of particle about a given point,we must add vectorially the angular momenta of all the individual particles about this point.(10-8)As time goes on,may change.That is L=+=NnnNLLLLL121=+=nLdtdLdtddtLd21LTotal internal torque is zero because the torque resulting from each internal action-reaction force pair is zero.Thus(10-9)That is:“the net external torque acting on a system of particles is equal to the time rate of change of the total angular momentum of the system.”Note that:(1)the torque and the angular momentum must be calculated with respect to thesame origin of an inertia reference frame.dtLdextn=(2)Eq(10-9)holds for any rigid body.2.and Suppose a force acts on a particle which moves with momentum .We canresolve into two components,as shown in Fig 10-3:dtPdF=dtLd=FP/PPP/FF P+PPFFig 10-3The component gives a change in momentum ,which changes the magnitude of ;on the other hand,the gives an increment that changes the direction of ./F/pPF PP(a)(b)/LL/LL+LLWe once again resolve into two components and .The component changes the in magnitude but not in direction(Fig 10-4a).The component gives an increment ,which changes the direction of but not its magnitude(Fig10-4b).L/LLLLLFig 10-4The same analysis holds for the action of a torque ,as shown in Fig 10-4.In this case must be parallel to .tL=L/An example of the application of Eq(10-9)for rotational dynamics is shown in Fig 10-5.In Fig 10-5,a student pushes tangentially on the wheel with a force at its rim,in order to make it spin faster.The (/)due to increases the magnitude of .Fig 10-5aLLLffmgFNorfIn Fig 10-5b,we have release one support of the axis.There are two forces acting:a normal force at the supporting point o,which gives no torque about o,and the wheels weight acting downward at the Cm.Fig 10-5bmgOL LNNThe torque about point o due to the weight is perpendicular to and its effect is to change the direction of .Note that:(1).Eq(10-9)holds when and are measured with respect to the origin of an inertial reference frame.LLL(2).Eq(10-9)would not apply to an arbitrary pointwhich is moving in complicated way.However if the reference point is chosen to be the Cm of the system,even though this point may be accelerating,then Eq(10-9)does hold.10-3 Angular momentum and angular velocity1.Angular momentum and angular velocityFig10-6a shows a single particle of mass m attached to a rigid massless shaft by a rigid,massless arm of length perpendicular to the shaft.The particle moves at constant speed v in a circle of radius .Fig 10-6amyxrrrVWe imagine the experiment to be done in a region of negligible gravity,so that the only force acting on the particle is centripetal force exerted by the arm .The shaft is confined to the z axis by two thin ideal bearings(frictionless).Let the lower bearing define the origin o of our coordinate system.The upper bearing prevent the shaft from wobbling about the z axis.rThe angular velocity of the particle upwardalong the z axis no matter where the origin is chosen along the z axis.The angular momentum of the particle with respect to the origin o is(shown in Fig 10-6b),not parallel to.sinrvrv=PrLLIf we choose the origin o to lie in the plane of the circulating particle,then otherwise,it is not.Lxyo=/zLLrPLrFig 10-6b(10-10)Now is the rotational inertial of the particle about z axis.Thus(10-11)Note that the vector relation is not correct in this case.=ILILz=2mr2sinsinsinmrmvrrmvrpLLz=I2.Under what circumstance will and point in the same direction?Let us add another particle of the same mass at same location as the(first),but in the opposite direction.Fig 10-7oLL1m1m2m1P2P2L1LThe component due to this second particle will be equal and opposite to that of the first one,and the two vectors sum to zero.The two vectors in the same direction.Thus for this two-particles system,We can extend our system to a rigid body,made up of many particles.If the body is symmetric about the axis of rotation,which is called“axial symmetry”,and are pa