作業(yè)及詳解第四章習(xí)題講解

1、第4章力學(xué)習(xí)題講解1. 在直角坐標系、柱坐標系和球坐標系中分別寫出質(zhì)點在勢場V (r) 中運動的解：在直角坐標系里，函數(shù).質(zhì)點的拉氏函數(shù)可表示為= L= L= L= mx ，p= my ，p= mz ，pxx yy zz H = piq i L p2p2p2p2p2p2yy=+V (x,y, z)x mzxzmm2m2m2m1 （p2 + p2 + p2）+V (x,y, z)=xyz2m1L = m (x 2 + y 2 + z 2 ) V (x,y, z) 2p= Lq 在柱坐標系里，質(zhì)點的拉氏函數(shù)可表示為= L= L= L= m2 , p= m , p= mz ,p zz H = 1 m

2、 ( 2+ 2 2 + z 2 ) +V (, , z )2p21() r 222=p+ p+V, , zz2m 在球坐標系里，質(zhì)點的拉氏函數(shù)可表示為= L= L= L= mr 2 , psin2 ,= mr 2= mr , pp rr H = 1 m (r 2sin2 2 ) +V (r, , )+ r 2 2 + r 22p2p21 +V (r, , )p2=+ +2m rr 2r 2sin2 2L = 1 m (r 2 + r 2 2 + r 2 sin2 2 ) V (r, , ) 2L = 1 m ( 2 + 2 2 + z 2 ) V (, , z) 22.已知系統(tǒng)的約束是穩(wěn)定的，

3、其函數(shù)為12 sinq1 試求相應(yīng)的日函數(shù)。p2= p ,=解:得： q q 由112sin2 q1p = q ,pnq1121= q p + q p1122 H1 2p2sin2 q2q p+ 2 + a cosq2 2111sin2 q11 ()22q sin q+ a cosq .211232L = q p H=1q= Hp3.試用方程導(dǎo)出單擺的運動方程。L = T V = 1 ml 2 2 mgl (1 cos )2= ml 2 = Lp =由 p解出 ml 2cos )2mlHp =p =2pmlml 2p H= mgl sin p = = mgl sin 5o sin 4 + g

4、= 0l + g sin = 0l解：如圖建立坐標系，取 x 為廣義坐標，以桌面為零勢能面。mgx 21L = T V =mx+222lx = px= L= mx 由 p解出xx mmgx 22mgx 2p12H = mx 2pxmx 2+=x2m2l2lHx =glpxxx mxx dx =xdxHxgx 2mgx lgl0l 2p = =xx = l122=x x =3gll45mx = mgxl解：的動能和勢能分別為T = m (x 2 + y 2 ) + m x 22V = mgy22xxy =y以x 為廣義坐標4a2am 22 x 2mgx 2x L = T V =1 + x2224

5、a4aLx 2x 2=x = mx 1 + x = pxm 1 + 4px2 24a a 61 p212 2mg2H = px x L =x m x+x2m 1 + x 2 / 4a2 24ax 2L=x =mx 1 + px24a x 2x= m 1 + m2p x x x224a 2ap2112mgH = px x L =x m x+222x2m 1 + x 2 / 4a2 4aHp2x / 2a2x= =+ m x mg2p xx x2m (1 + x 2/ 4a2)22amx=+= 0mg4a2 4a22a7Gkz pzGix pxGjy pyGGL =GGL = r pLz= xpy

6、 ypx證:Lz + Lz + Lz = LzLzLz , L z p x p y py pzpxpzzzxxyyLzx= p , Lz= y, Lz= p , Lz= x, Lz= 0, Lz= 0pxypyzpzyx = (2x + p ), = (2y + p = (2z + p ),),xpxypyzpzxyz= (2p+ x),= (2p+ y),= (2p+ z)xyz,Lz = 08 = (x 2 + y2 + z 2 + p2 + p2 + p2 + xp+ yp+ zp )xyzxyz度為1，選 為廣義坐標，則解：系統(tǒng)的yA = l cos x= l sin BT = 1 m

7、x 2 + 1 my 2= 1 ml 2 22V = mgy22BA+2 1 kx 222AB= 1 ml 2 22L = T V2p ml 2= L= ml 2 =p = 1 ml 2 2 + mgl cos + kl 2 sin2 + pH = L + p29p ml 2 H = =p pHH = mgl sin 2kl 2=sin cos p pp = (mgl sin 2kl 2 sin cos )/ ml 2ml 2 = g sin 2 k sin cos ml當 為小角時，sin , cos 12k g ,(2kl mgl ) + (2k g ) = 0 =mmllmlT = 22

8、kl mg10p222H = + mgl cos + klsin 2ml 28. 用原理求一質(zhì)點在平方反比有心引力作用下的運動微分方程。解：質(zhì)點在萬有引力場中運動，其拉氏函數(shù)為：mk 2122 2L =m(r 2+ r ) +r2 t()1mk r s = 2 2+ r 22+m r dtAdB = d(AB) B d A2t1t2mk 2 2 = mrr + mr r + mr2 r dt r 2t1= 011變分和積分交換順序r = d (r)dt由于 r= r= 0, t=t1= t=t2= 0,上式可變?yōu)椋簍=t1t=t2t22 ()mk r 2s = mr 2 mr + r m 2r

9、r + r dt =20t1mk 2 m (r r) + 2 r + m(2rr + r2) = 0r 2m (r r 2 ) = mk 2r 212 d (mr 2 ) = 0dt9. 一半徑為r, 質(zhì)量為m 的實心圓柱體在一半徑為R 的大圓柱體內(nèi)表面作純滾動, 試用作微振動的周期.解：對實心圓柱體，由于作純滾動，故正原理求其在平衡位置附近 = R r R = r( + )vc = (R r ) rr = (R r ) 取 為廣義坐標，體系的動能為= 3 m(R r)2 24T = 1 mv2 + 1 I 2 = 1 mv2 + 1 mr 22ccc2224V = mg(R r)cos 故體

10、系的拉氏函數(shù)為：L = T V = 3 m(R r)2 2 + mg(R r)cos 413L = T V = 3 m(R r)2 2 + mg(R r)cos 4t2 3 m(R r)2 mg(R r)sin dt = 0 2t1將上積分式中的第一項變?yōu)? m(R r)2 = d 3 (R r)2 3 (R r)2 dt 2 22 3 m(R r)2 + mg(R r)sin = 0 2當 為小角時，sin , cos 12gT = 2 = 23(R r) + 03(R r )2g14度為1，選x 為廣義坐標，則解：系統(tǒng)的x = 2a ,= a x c其中 為圓柱體所轉(zhuǎn)過的角度.T = T+Tcylinderboard= 21 mx 211 Mx 2 +I2 + 22c2= 1 Mx 2 + 3 mx 22V = Fx815L = T V = 1 Mx 2 + 3 mx 2 + Fx28I= 1 ma2axis2t2t2s = Ldt 4M + 3m x 2= + Fx