08刚体的平面运洞葱炉

12345.结论56 7BAABvvv,tanuvB,sinuvBAABsin1lulvBAAB8ABBAvvv,tanuvB,sinuvBAsin1lulvBAAB910COOCvvv0COOCvvvRvRvOCO11RvRvOCO,2ivOAvOAvv2,jivOOBvvOBvv2：,jivOODvvODvv21260603013606030CAACvvvACvv14606030BCCBvvvACBvvv260cos15606030CCBCvtgvv360 CBBCCBvABCCBvlCBv316。17186030cosBAvv0vB=090rvvAB33230cosrvvAB19，306020解：ABvv30cossm 30 cosOA30 vvAB/231.0cossm vCDCBvvBBD/693.03306021DEvv30 cos3060sm vE/8.0sm vD/693.022AvAC 2324252627RvOOAvRv22OBvRv22：ODvRv2228 29 cot cos sinulluBCvvABB sinluACuvAB3060603031606030ACvv32lCBCCv3330 tan2AvCCBvlCCv32AACBvBvvllBCv2332260603033。nBAtBAABaaaa ABatBA2 ABanBA3435ntMOMOOMaaaaRvO36RatvRvdtdtOOOddRdd1)(22ntRvRaaRaoMOOMO37M3M1M4M2OaOaOaOaOaO基点的加速度基点的加速度M1M3M4M2Ot3OMat4OMat1OMat2OMan4OMan3OMan2OMan1OMa相对基点的加速度相对基点的加速度M1M3M4M2O1Ma2Ma4Ma3Ma合成加速度合成加速度38RvaM2012422tr2nr2)(2)(RvRvaaaaaaOOOOOM2242tr2nr34)(OOOMaRvaaaa242224)(2)(RvRvaaaaaaOOOOtrOnrMM1M3M4M2O1Ma2Ma4Ma3Ma39RvaM201M1M3M4M2O1Ma2Ma4Ma3Ma4041OArrv)(21,)(21tOArra221n)(OArra42OArrrrv22122 OOrrrttrrrtt22122122ddd)(d43,2122tOCArrra2222122n)(OCArrrrantntCACAAACaaaaa2212122122221nn)()()(OOOACACrrrrrrrrraaaM44ntntMAMAAAMaaaaaoArr)(21ta221n)(oArra222nrMAa22trMAa45ntntMAMAAAMaaaaaOOOMAAMxrrrrrrrraaa)(2)(21221221tt22212122212221nn2)()()(OOOMAAMyrrrrrrrrrrraaa464222122122)2(4)(OOMyMxMrrrrraaa2212)2(2 tanOOMyMxrrraa47曲柄滚轮机构曲柄滚轮机构 滚子半径滚子半径R=OA=15cm,n=60 r/min求：当求：当 =60时时（OA AB），滚轮的），滚轮的，rad/s 230/6030/n48解解：一、运动分析一、运动分析：rad/s 3245/30/1APvAABcm/s 30 OAvAP为其速度瞬心ABBBPv1P2P1vBP2为轮速度瞬心：定轴转动：定轴转动：OA,二、研究二、研究AB：平面运动平面运动:AB杆、轮杆、轮B，AB三、研究三、研究B轮轮：)(cm/s 32032330rad/s25.715/320/2BPvBB49解解：一、运动分析一、运动分析：rad/s 3245/30/1APvAABcm/s 30 OAvAP为其速度瞬心ABBBPv1P2P1vBP2为轮速度瞬心：定轴转动：定轴转动：OA,二、研究二、研究AB：平面运动平面运动:AB杆、轮杆、轮B，AB三、研究三、研究B轮轮：)(cm/s 32032330rad/s25.715/320/2BPvBB 50四、求点的加速度：四、求点的加速度：222cm/s60 OAaAnBAtBAABaaaa2ABnBAABa大小？方向 将上式向将上式向BA线上投影线上投影nBABaa0030cos22rad/s77.815/5.131/BPaBB基点：基点：A点，点，OABBP2nAanBAanAaBa动点：轮心点，动点：轮心点，Ba，方位水平方位水平作加速度矢量图作加速度矢量图由由tBAa)(cm/s5.13130cos/2nBABaaBAanBA沿,ABtBAABaBAanBA,51解：(a)AB作平动，),(,nBnAtBtABABAaaaaaavvBOAOBOaAOaBOvAOvtBtABA2122112211 ;/,/;/,/而又.;2121 已知O1A=O2B,图示瞬时 O1A/O2B 试问(a),(b)两种情况下1和 2，1和2是否相等？(a)(b)(a)(b)52(b)AB作平面运动作平面运动,图示瞬时作瞬图示瞬时作瞬 时平动时平动,此时此时BAABvv ,021221121,/,/,BOvAOvBOAOBAnBAtBAABaaaasin:tBAnAnBaaay,ctg2212112tBAnAtAnBtBaaaaatBAatAanAa加速度分析：加速度分析：cos:tBAtAtBaaaxsin/)(nAnBtBAaaasin/22latBABAaaAB作瞬时平动时并由此看出,53 542nRaaAA基点基点 的的202n)(vMAACRAMAMa基点基点的的 AMaMAtvvAACRACv0ntMAMAAMaaaa55选选 A点为基点，则点为基点，则B点的加点的加速度速度ntBABAABaaaan)cos(cosBAABaaa20RaA202n)(vBAACRlABacos coslACvcos1coscos)cos(2220 RaB56t)sin(sinBAABaaacos)sin(tBABAaaatancoscos)cos()sin(2220tlRlRABaBAntBABAABaaaa572220ncoscos2)cos()cos(lRRaaaMAAMt)sin(MAAMaaa21tAMaMAtancoscos)cos()sin(22220lRRaMntMAMAAMaaaa58BOAC3059BOAC30Avva2elOAvABBAvvvreavvv60 e30sinvvvABBlvvvBAB)(2er30cosvvABlv23rrevvvvABBABvABABBOAC3061BOAC30Aaa a0tealABaABAB22nlOAa22nelva2rC32ntABABBAaaaaCrteneaaaaaa62BOAC30CrnentaaaaaABABCnt30cos30sinaaaABABlaAB2t332t33ABaABAB63OBDllOA45A64OBDllOA45AlvOEvBEvOBvBEBntBEBEEBaaaa65n45 cosBEBaalvaaBEB2n245 coslvBEaBEBE22n2ntBEBEEBaaaaOBDllOA45A66OBDllOA45Areavvv,eavv,0rvvvvBelvOBvOAe67OBDllOA45AlvOBaOA22neBaa ateaaa rneteaaaaalvaaB2te222te2lvOBaOA