复旦大学《大学物理》课件（英文）-第9章Rotational dynamics(1).pdf

Chapter 9 Rotational dynamics9-1 Torque1.TorqueIn this chapter we will consider only case in which the rotational axis is fixed in z direction.Fig 9-2 shows an arbitrary rigid body that is free to rotate about the z axis.PzOMFrdM*A force is applied at point P,which is located a perpendicular distance r from the axis of rotation.and lie in x-y plane,and make an angle .FFrThe radial componenthas no effect on rotation of the body about z axis.Only the tangential componentproduces a rotation about the z axis.cosFFR=sinFF=The angular acceleration also depends on the magnitude of rThe rotational quantity “torque”is defined as(9-1)The unit of torque is the Newton-meter ()sinrF=mNIf r=0-that is the force is applied at or through the axis of rotation;If or ,that is the force is applied in the radial direction;If =0.1800=When =0?sinrF=F2.Torque as a vectorIn terms of the cross product,the torque is expressed as (9-3)=FrsinrF=Magnitude of :Direction of :using righ-hand rulesinrFPzOMFrdM*+=kyFxFjxFzFizFyFFFFzyxkjiFrxyzxyzzyx)()()(+=kzjyixr+=kFjFiFFzyxComponents of torque:SoSample problem 9-1Fig 9-5 shows a pendulum.The magnitude of the torque due to gravity about the point o is it has the opposite direction.Fig 9-5sinLmg=mgmgx.9-2 Rotational inertia and Newtons second law1.Rotational inertia of a single particleFig 9-7 shows a single particle of mass m which is attached by a thin rod of length r and of negligible mass,and free rotates about the z axis.oxyrmsinFFFig 9-7A force is applied to the particle in a direction at an angle with the rod(in x-y plane).Newtons Second law applied the tangential motion of the particle gives and ,we obtainFsinFmaFtt=rmFz=sinrazt=rmsinFz=rrzzmr2=amF2mrII=,zzWe define to be the“rotational inertia”Iof the particle about point o.(9-6)The rotational inertia depends on the mass of the particle and on the perpendicular distance between the particle and the axis of rotation.2mrI=2mr2.Newtons second law for rotation Fig 9-82m1m0 xyt 1F1FRF12FRF2t 2F(For system of particles)2m1mPrT1rT21T2T1rxy02rThat is(9-7)For each particle Substituting them into Eq(9-7),we obtain221121)()(rFrFttzzz+=+=ttamF111=ttamF222=222211zz222211z222z21122t211t122t11tzrmrmII)rmr(mrmrmramram)rF()rF(+=+=+=+=+=,(9-8)The are the same for both particles,the total rotational inertia of this two-particle system:(9-9)The obvious extension to a rigid object consisting of N particles rotating about the same axis is(9-10）z222211rmrmI+=22222211nnNNrmrmrmrmI=+=Torque of Internal Forces:2211frfrM+=In21ff=112121frf)rr(M=In=01f2f1r2r12rTorque of Internal Forces is zero!For system of particles,what kinds of forceinduce torque?2m1mPrT1rT21T2T1rxy02rFig.9-8Tensions and have also no torque about o.1T2TThus the torque about o is due only to the external force .Thus we can rewrite the Eq(9-8)as(9-11)This is the rotational form of Newtons Second law.zzextI=,PNotes:,I,must be calculated about same axis.For rotations about a single axis,I is scalar.If many external forces act on the system,we add up the torques due to all the external forces about that same axis.zext,zSample problem 9-2Three particles of masses =2.3kg,=3.2kg and =1.5kg are connected by thin rods of negligible mass,so that they lie at the vertices of 3-4-5 right triangle in the x-y plane.Fig 9-9341m2m3mX(m)y(m)c1r2r3rF=4.5N302m3m1mQuestion:(a)Find the rotational inertia about each of the three axes perpendicular to the x-y plane and passing through one of the particles.(b)A force of magnitude 4.5N is applied to m2in the xy plane and makes an angles of 300with the horizontal.Find the angular acceleration about oz axis.(c)Find the rotational inertial about an axis perpendicular to the xy plane and passing through the center of mass of the systemSolution:(a)consider first axis through Similarly for the axis through ,we have For the axis through 22222153)4()5.1()0.3()2.3()0()3.2(mkgmkgmkgmkgrmInn=+=2222258)0.5()5.1()0()2.3()0.3()3.2(mkgmkgmkgmkgI=+=23117mkgI=1m2m3m341m2m3mX(m)y(m)c1r2r3rF=4.5N30(b)Using Eq(9-2)(c)Rotational inertial about an axis passing through the cm.(3)(4.5)cos30zrFmN=21(3)(4.5)cos3053zzmNIkg m=mmymynnncm37.1=mmxmxnnncm86.0=2222162.2myxrcmcm=+=22222240.3)(myyxrcmcm=+=22232374.11)(myxxrcmcm=+=2222235)74.11()5.1()40.3()2.3()62.2()3.2(mkgmkgmkgmkgrmInncm=+=IcmI1,I2,I33.The parallel-axis theoremThe result of the previous sample problem leads us to an important general result,the parallel-axis theorem:“The rotational inertia of any body about an arbitrary axis equals the rotational inertial about a parallel axis through the center of mass plus the total mass timesthe squared distance between two axes”.2MhIIcm+=4.Proof of Parallel-Axis theoremFig 9-10 shows a thin slab in the x-y plane,the rotational inertia about oz is Fig 9-10oxXhcyycmxcmyPncmnxxx+=ncmnyyy+=)(222nnnnnyxmrmI+=Substituting these transformations,we have Regrouping the terms,we can write this as)()(22ncmncmnyyxxmI+=+=ncmcmnncmnncmnnnmyxymyxmxyxmI)(22)(22222cmIMh=+0=cmnnMxxm0=cmnnMyym9-3 Rotational inertia o