第10章 動(dòng)力學(xué)方法選擇

1、laOE21laOAABBAaaa0nABalaaCABAB30coslaCACOCll23,30cosO30sin0ABBaa laCB21 CBBCyCxaaaa(b)30sinCBBCxaaalllaCCCCx41412130cosCBCyaalaCCy43lraBBB21CBmgGB2mgGGOClmamFCBBBI22I4121lmrmMCBBBlmmaFCCxCx41IlmmaFCCyCy43ICCmlM2I121lmlmmaFCOEO4321I22I6331lmmlMCOOlmFMrFMCBB21, 0, 0SISlmFFFFFFFCBBxBxBx23, 0, 0ISISmgFF

2、FGFFByByBy2, 0, 0NNlmFFFFFFFCBxCxAxCxBxAxx47, 0, 0IIlmmgFFGFFFFCByAyCCyByAyy43, 0, 0IlmFFMlFFlFFMCByAyCByAyAxBxC36343, 043)(4)(, 0IlmFFFFFCAxOxAxOxx47, 0, 0lmmgFFGFFFFCAyOyOAyOOyy43, 0, 0IlmmgFMlGlFMCAyOOAyO6321, 02, 0IlmmgFlmmgFCByCAy3631721,1831321lgC833mgFAx32321mgFAy165mgFBx1639mgFBy3233mgFOx323

3、21mgFOy3213mgF3297NmgF1633S，lgO169lgB833，r5BAvrv20101T222222121rmmvTAmgrrrmgW23)2(12mgrrmWTT2321,2221212rg32進(jìn)一步由動(dòng)能定理可以求出vA：nAaAannABABBAAaaaaa設(shè)桿OA的角加速度為2，A點(diǎn)的加速度為桿AB的角加速度為1，滑塊B的加速度為aB。在桿AB中以點(diǎn)B為基點(diǎn)，有 (a)graA322nraA20nABaraAB511n23,cosrgaaABArg231nAaAagararaaaBBABBA23,sin12graB232投影到nnABABBAAaaaaa0nCBag

4、raCB453251graB232gaaCBCx23cosgraaaaCyCBBCy43,sin2nCBCBBCyCxaaaaa (b)mgrJMCC851ImgmaFCxx23ImgrmmaFCyy432ImgG 043, 0, 02ImgmgrmGFFyyrg472 mgFFMrFrFMACAC850cos25cos25, 0NINmgFFFFFFAxAx23, 0, 0NINmgF167NmgFA1617， ；(a) (b)n2n1n1n2nrnrvvvvvve vvennvven2n1n1n2vvvve2112vvvve233233,uuvuuvmmmcoscos023uusinsin

5、233uuv1sin)cos(332vuuesin)cos(332vuu01tan2sin2cos1tantan, 01tantan1tan2tan1tan1222由以上三式中消去v3、u2、u3得到即212232CCBCCCaRBC224m/s575. 42,212gAHvmvAHmgvevevvvve54,nnnnvvv53) 1(48. 048. 048. 05453nevvevvvvx)916(04. 036. 064. 05354nevvevvvvyvm/s575. 4vm0246. 02)916(04. 0222gevgvKCy到達(dá)最高處的時(shí)間s071. 0)916(04. 0ge

6、vgvtym281. 0) 1(48. 0tevtvHKxm351. 045HKdm235. 043HKKCKJKCh)916(04. 0evvyyvhcosIJO0OyOxCIIIImvsin00cos0mdJhOlv1arveeravvvcosaevv cos,cos11lrlrcos)(31cos0101ImllIJJOOrJIIrJOO11,02120cos3cosOJmrmlr21201cos3cosOOJmrlmJI解：21amvLBBBJL221BBLL2mvaJB22232)22(61maammaJBav43（4）建立方程建立方程：物塊A在碰撞中對(duì)點(diǎn)B的動(dòng)量矩守恒。碰撞前物塊A

7、平移，它對(duì)點(diǎn)B的動(dòng)量矩為： 碰撞后A定軸轉(zhuǎn)動(dòng)，對(duì)軸B的動(dòng)量矩為： 令得 而 因此OAAvvvrvrvsin,sin nIIvvmmn)sin(sin,)cos(cosImvvmImvvmncoscosImvvm分別投影到?jīng)_量In與It方向上，有即sinsinImvvmrIJrIJJOOO)(,232mrJO)(32Imrcoscosnnvvvvk)cos2tan3(51tanvrk112lvvyBy222lvvyCyIlJJIvmvmOOyx20111111桿CD：IlJIvmvmOyx20022220BByCyvvve1212111122)()(22)2(2llvvllvlveyyyy021

8、xxvvmIvvyy211112OJlI222OJlI221121mlJJOO)1 (161,16111emlImlmlIe01xv)1 (16111elvy)35(8111e02xv)1 (16112elvy)1 (8312e1ImaF 121I52mrJMCC（3）受力分析受力分析： 碰撞過程中只考慮作用在球上的水平面的法向與切向約束碰撞沖量In與It； 碰撞后作用在球上的力有主動(dòng)力：重力G，約束力：作用在球與水平面接觸點(diǎn)P的法向約束力FN與摩擦力FS，以及球C的慣性力系向質(zhì)心C簡化的等效慣性力cosImvvmrIJJCC)cos52(52rrv121II52, 0, 0mrmrarFMMCP1152rvDrv1152)cos52(52rrvD)cos52(5211rrv11rv 7cos5210, 00 xxvvm)(,)(yyvvmIImvvmsin6,sin20mlIlIJCsin6lvvyACyxAvvvvsin2,sin20lvlvyy)sin31 (sin62lv2sin31mvIsin,cos2lylxBB2224lyxBB22226121mlJTA)sin31 (2sin32121222221mvJvmTCycos212lmgW1212WTTcos2)s