哈密頓正則方程應(yīng)用

1、分析約束分析約束, ,確定自由度確定自由度選好選好寫出系統(tǒng)的寫出系統(tǒng)的T,V,L 寫出寫出Attention:H=H(q,p,t).)3 , 2 , 1(qLqLH),(tqqpqLp廣義動(dòng)量),(tpqqq 正則方程解題步驟正則方程解題步驟代入正則方程求解代入正則方程求解哈密頓正則方程的應(yīng)用哈密頓正則方程的應(yīng)用例例1 1：雙原子分子哈密頓量：雙原子分子哈密頓量1m2mc1r2rr相對(duì)質(zhì)心位矢相對(duì)質(zhì)心位矢1r2r相位矢相位矢r02211rmrm21rrrrmmmr2121rmmmr2112雙原子雙原子分子運(yùn)動(dòng)分子運(yùn)動(dòng)質(zhì)心運(yùn)動(dòng)質(zhì)心運(yùn)動(dòng)相對(duì)質(zhì)心運(yùn)動(dòng)相對(duì)質(zhì)心運(yùn)動(dòng)),(zyx),(r相對(duì)質(zhì)心運(yùn)動(dòng)相對(duì)質(zhì)

2、心運(yùn)動(dòng)) 1 (1211frm )2(2122frm fffmm211212)2() 1 (frmmmm 21212121mmmm折合質(zhì)量折合質(zhì)量r相對(duì)質(zhì)心運(yùn)動(dòng)相對(duì)質(zhì)心運(yùn)動(dòng)fr 221rT相對(duì)質(zhì)心相對(duì)質(zhì)心運(yùn)動(dòng)動(dòng)能運(yùn)動(dòng)動(dòng)能)sin(21222222rrrT21222)(21mmmzyxmTc)(21222zyxmT)sin(21222222rrr0r)(rV)(rV)(0rV)(00rrrVr 2022)(210rrrVr簡諧近似簡諧近似r221)(rkrV0rrr)(21222zyxmVTL)sin(21222222rrr221rk)(21222zyxmVTL)sin(21222222rrr2

3、21rkxmxLpxymyLpyzmzLpzmpxxmpyympzz)(21222zyxmVTL)sin(21222222rrr221rkrrLpr2rLp22sinrLprpr 2rp22sinrpppprpzpypxLHrzyx)(21222zyxpppm)sin(21222ppI)(22rVpr2rI轉(zhuǎn)動(dòng)慣量轉(zhuǎn)動(dòng)慣量平動(dòng)平動(dòng)轉(zhuǎn)動(dòng)轉(zhuǎn)動(dòng)振動(dòng)振動(dòng)例例2 2一維諧振子一維諧振子解解:xxdtdxv221mvT 動(dòng)能221xm2xk21V勢(shì)能VTL222121kxxmkxxLxmxLxmxLdtd )()(xLdtdxL0kxxm xmkx demonstrationVTL222121kxxmxL

4、pxxmmpxxqLqLHxLxLxpxkxxm222121xxxpmpkxmpm2221)(21mpkxx22122mpkxHx22122mppHxxxkxxHpx根據(jù)正則方程根據(jù)正則方程mpxx xmkx ABxy例例3 3一半徑為一半徑為r,r,質(zhì)量為質(zhì)量為m m的實(shí)心圓的實(shí)心圓柱體在一半徑為柱體在一半徑為R R的大圓柱體的大圓柱體內(nèi)表面作純滾動(dòng)內(nèi)表面作純滾動(dòng), ,試用哈密頓試用哈密頓正則方程求其在平衡位置附近正則方程求其在平衡位置附近作微振動(dòng)的周期作微振動(dòng)的周期. .1o1oo分析分析ABABA)( rRrrR坐標(biāo)數(shù)坐標(biāo)數(shù)3約束數(shù)約束數(shù)2rrR oo1rR自由度自由度1demonstr

5、ation22)(43rRmVTLcos)(rRmg取取 為廣義坐標(biāo)為廣義坐標(biāo)Lp)(23rRm)(32rRmppLHprRmgrRmcos)()(4322cos)()(322rRmgrRmpHcos)()(322rRmgrRmpH根據(jù)正根據(jù)正則方程則方程pHHp sin)(rRmg2)(32rRmp 2)(32rRmp)(3sin2rRg)( 32rRg)( 32rRg例例4 4用哈密頓正則方程求自由質(zhì)點(diǎn)在球坐標(biāo)下加速用哈密頓正則方程求自由質(zhì)點(diǎn)在球坐標(biāo)下加速度的表達(dá)式度的表達(dá)式. .設(shè)其受力在設(shè)其受力在r,r, , , 三個(gè)方向的分量三個(gè)方向的分量分別為分別為F Fr, r, F F , ,

6、 F F r解解:廣義力廣義力非保守系拉氏方程非保守系拉氏方程必先求動(dòng)能必先求動(dòng)能)sin(21222222rrrmTrmrTprmprr2mrTp2mrpsin2mrTp22sinmrpppprTHr)sin(21222222rprppmr根據(jù)正則方程根據(jù)正則方程rrQrHprQmrpmrp23232sinQHpQmrp322sincos)sin(21222222rprppmHrQHpQrrFQ rFQ FrQsinrmprrmpr rrQmrpmrpp23232sinrFmrmrmrmrrm23222322sin)sin()( mFarr222sin rrr同理可得同理可得cos2sin2

7、sin rrrmFacossin22 rrrmFa222sin rrrmFarr例例5 5質(zhì)量為m的相同二質(zhì)點(diǎn)用一長為 的輕桿連接初始時(shí)直立靜止在光滑水平面上,以后任其倒下,試用正則方程求桿落地時(shí)的角速度.l分析分析坐標(biāo)數(shù)坐標(biāo)數(shù)3約束數(shù)約束數(shù)2自由度數(shù)自由度數(shù)1取如圖所示取如圖所示為廣義坐標(biāo)為廣義坐標(biāo)xy1m2m2122112212120)(xxxmxmlyxxxysin2lyccycos2lyc根據(jù)柯尼西定理根據(jù)柯尼西定理2221221ccIymT2221)2(2mllmIc)cos1 (41222mlTsinmglV )cos1 (41222mlVTLsinmglLpsin)cos1 (41222mglmlVTL)cos1 (2122ml)cos1 (222mlppLHsin)cos1 (41222mglmlpsin)cos1 (222mglmlpHsin)cos1 (222mglmlpH2222)cos1 (cossin2mlpHp cosmgl)cos1 (222mlppH)cos1 (222mlp 222)cos1 (sincos4mlp)cos1(2122mlp0cos2cossin)cos1 (22gll