南开大学：《力学》课程教学资源（PPT课件，英文讲稿）Chapter 4 Rigid Body Dynamics

Chapter.4Rigid Body Dynamics

Chapter 4 Rigid Body Dynamics

s 1.Law of conservation of angular momentum1-1 Angular momentumMomentum: Some otherquantity is conservedbeside mechanical energyQ0

Momentum: Some other quantity is conserved beside mechanical energy. §1.Law of conservation of angular momentum 1-1 Angular momentum  0 

History Keplerellipse无法显示该图片无法显示该图片KdsC三dt无法显示dsdrI1FXVXds=二xdr2dt2dt2x=vrsin=by

r v dt dr r dt ds     =  =  2 1 2 1 C dt ds = ds r dr   =  2 1 r v = vrsin = bv   History Kepler ellipse 

Notice:Fysi'sPhysics(2)Rotated90(3)Rotatedaboutdegreesaboutx.x,theny.(1)The book in itsoriginalorientation(4)Rotated90degrees abouty.(5)Rotatedabouty, then x

Notice:

1-2 theorem of angular momentum1. ParticleM=djiM=rxFJ==m政dt2. System of particlesdEJM-4ZM,=-dtM,=M"+M'Mex =rxFexMin =rxFinAdjMerZMin =rx(Fin +F)-E(F+r)xF, =0exdt

1-2 theorem of angular momentum in i ex Mi Mi M    = + dt dj M   = dt d J M i i i i   =   dt dJ Mex   = in i i in Mi r F    =  ex i i ex Mi r F    =  dt dJ M i i   = M r F    =  J r P r vm      =  =     ( + ) −( + ) ij = 0 i j i j in ji in i ij in Mi r F F r r F        2. System of particles 1. Particle

M= O J= constant (also fori=1)conservation of angular momentum(1).Isolated systemFex=0Mex=0(2).Central forcerotation×Fex =0exPulsarNeutron starT = 1.187911second

Mex = 0  J = constant (also for i =1)  conservation of angular momentum (1).Isolated system (2).Central force 0 ex Fi = 0 ex Mi = r  Fex = 0   Pulsar rotation Neutron star T =1.187911second

Example① V-?How is the work down by F?Express F by J ,M, R3Solve:Rmv, = mvRRyRF11州y-1>0mymy222mv2J23FmR3R

Example : V=? How is the work down by F? Express F by J ,M, R. Solve:       F 3 2 2 mR J R mv F = = R 0 mv 0 = mvR R R v v 0 0 = ( ) 1 0 2 1 2 1 2 1 2 0 0 2 2 0 2        − = − v R R mv mv m

Vellipse['FdR='dwRosunExamplej=rxmv=mvRk, =rxm,v, =myRkM =rxm g=-Rmgm,gm,gM, = rxm, g= Rmg

v sun ellipse  =  R w FdR dw R 0 0 m g m g 1 2 r r2 1 Example: → → →  J = r  m v = m v R k 1 1 1 1 1 1 → → →  J = r m v = m v Rk 2 2 2 2 2 2 M r m g Rm g 1 = 1 1 = − 1 → → → M r m g Rm g 2 = 2 2 = 2 → → →

Problem 4-11kk1(r):Center force:V(r)福1L=×m=bmvkRVmOb0

Problem 4-11 Center force: o m b v R v  = r r k f r 2 ( )  r k V (r) =  L = r vm = bmv k0   

k1I?mvomVb-212Yminbvo = rmin mVb一1bvo)?2+k一mom2-2km-mb=0mv1m22IminImin2k2b?=0Aminmy2k2k+ 4b2myomvoKk(k2+b2+mamn2m1>0(k+yk?+(bmy))minmvomn

min 2 min 2 0 0 ( ) 2 1 2 1 r k r bv mv = m + 2 0 2 0 2 min 2 min 2 mv0 r − k r −mb v = 0 2 2 2 0 2 min − − b = mv k r 2 2 2 0 2 0 2 2 2 0 2 0 min ( ) 2 ) 4 2 ( 2 b m v k m v k b m v k m v k r =  +  + = 0 r min  ( ( ) ) 1 2 2 0 2 2 0 min k k bmv mv r = + +      = = + b b bv r mv r k mv mv 0 min min 2 2 0 2 1 2 1