理论力学题库第一章 下载本文

?i????F??xFx?j??yFy:

?k??Fz?Fy????Fx?Fz????Fy?Fx????????i?????j??????k?z??y?z???z?x??x?y??FZ???22223333?(18abxz?18abxz)i??18abzy?18abzy?j??6abz?40bxy?6abz?40bxy?k?0

故力场为保守力场。

?U?6abz3y?20bx3y2??????(1)?x?U由 Fy???6abxz3?10bx4y??????(2)

?y?UFz???18abxyz2????????(3)?zFx??对(1)式积分得:U??6abz3yx?5bx4y2?f(y,z)?????(4) 对(4)式求偏导数得:

?U??f(y,z)???6abz3x?10bx4y???Fy??6abxz3?10bx4y ?y?y即

??f(y,z)??0上式得:f(y,z)?g(z) 代入(4)式得:

?yU??6abz3yx?5bx4y2?g(z)?????(5)

对(5)式求偏导数得:即

??g(z)??0 积分得:g(z)?c代入(5)式得: ?z?U??g(z)???18abz2xy???Fz??18abxyz2 ?z?zU??6abz3yx?5bx4y2?c??????(6)

取x?0,y?0,z?0,U?0 则c?0 所以势能函数为U?5bx4y2?6abxyz3

Fx?a11x?a12y?a13z已知作用于质点上的力为Fy?a21x?a22y?a23z

Fz?a31x?a32y?a33z式上系数aij(i,j?1,2,3)都是常数,问这些aij满足什么条件,才有势能存在?如

这些条件满足,试计算其势能。

解:要满足势能存在须使力场为保守力场,既力场的旋度为零,

所以??F??0

即?Fx?y?a?Fy12??x?a21 ?Fx?z?a??Fz13?x?a31 ?FyFz?z?a23???y?a32 即势能存在aij满足条件是:a12?a21 a13?a31 a23?a32

F?Vx?a11x?a12y?a13z???x.............(1)由Fa?Vy?a21x?22y?a23z???y............(2)

F?a?Vz?a31x32y?a33z???z.............(3)(1)式积分得V??12a211x?a12yx?a13zx?f(y,z)........(4)(4)式对y偏微分=(2)式得

?V?y??a?f(y,z)12x??y??a12x?a22y?a23z 即

?f(y,z)?y??a22y?a23z...........(5) (5)式积分得f(y,z)??12a22y2?a23zy?g(z)............(6)

(6)式代入(4V??12a2111x?a12yx?a13zx??2a22y2?a23zy?g(z)7.) (7)式对z偏微分=(3)式得 ?V?z??ax?a?g(z)1323y??z??a13x?a23y?a33z 式

..得

....即

?g(z)?z??a33z...........(8) (8)式积分得g(z)??12a33z2?c............(9)

(9)式代入(4)式得

V??12a2yx?a1111x?a1213zx??2a222y?a23zy?2a233z?c........(10)

取x?0,y?0,z?0,V?0 则c?0得势能为

V??1a2112211x?a12yx?a13zx??2a222y?a23zy?2a33z ??12(a211x?a222y?a33z2?2a12xy?2a23zy?2a31zx)15.某力场的力矢为F??x?i?y?j?zk?

试证明该力场是否为保守力场,若为保守力场,求出其势能函数。

?i?jk?解:由于??r???????Fz?Fy????Fx?Fz??x?y?z?????i???y?z?????z????Fy?x??j?????Fx?y??k??0FF??x??xFyz

故力场为保守力场

??F?x??V??x?x.............(1)由??F?Vy???y.............(2)

??y???Fz???V?z?z..............(3)积分(1)式得V??x22?f?y,z?????4? (4)式对y偏微分=(2)式得

?V?y??f?y,z??y??y (y,z)??y2f2?g?z?????5?

代(5)入(4)得V??x2y22?2?g?z?????6? 积分得 z2?V?g?z?(6)式对z偏微分=(3)式得???z 积分得g?z????c????7? 2?z?zx2y2z2???c 代(7)入(6)得V??222x2y2z2?? 取x?0,y?0,z?0,V?0 则c?0得势能函数为V??222???16.有一质点在xy平面上运动,质点受到的力为F?(x?y)i?(x?y)j,质点在

?平面上由点A(1,0)沿直线运动到点B(1,1),求力F所作的功

解法1:由功的定义计算

B?BB?W??F?dr??(Fxdx?Fydy)??(x?y)dx?(x?y)dy

AAA又x?1,dx?0

W??BA所以

??BB1F?dr??(Fxdx?Fydy)??(x?y)dx?(x?y)dy??(1?y)dyAA011?(y?y2)1?022解法2:由功的定义计算

B

??B(1,0)(1,1)112(1,1),0)W??F?dr??(Fxdx?Fydy)??(x?y)dx??(x?y)dy?(x2?xy)((1?(xy?y)(1,0)1,0)AA(1,0)(1,0)2211?1??22或

??BB(1,1)11W??F?dr??(Fxdx?Fydy)??(x?y)dx?(x?y)dy??d(x2?xy?y2)AAA(1,0)22 1212(1,1)1111?(x?xy?y)(1,0)?(?1?)??222222B解法3:由保守力性质计算

???ijk??Fz?Fy??????r??????x?y?z??y?zFxFyFz???(0?0)i?(0?0)j?(1?1)k?0?故力场F为保守力场

????Fx?Fz??i???z??x??????Fy?Fx???j????x??y??k ?????VF???x?y.............(1)?x?x??V?F???x?y.............(2) ?y?y??F???V?0..............(3)z??z?x2?xy?f?y?????4? 积分(1)式得V??2(4)式对y偏微分=(2)式得积分得f(y)??V?f?y???x???x?y ?y?y12y?c????5? 2x2y2??xy?c????6? 代(5)入(4)得V??22x2y2??xy 取x?0,y?0,z?0,V?0 则c?0得势能函数为V??22则由保守力与功的关系可知

11111111W??(V2?V1)?V1?V2?(?x2?y2?xy)(1,0)?(?x2?y2?xy)(1,1)???(???1)?22222222Fx?x?2y?z?5设作用于质点上的力场的力矢为Fy?2x?y?zFz?x?y?z?6

17. 求此质点沿螺旋线x?cos?,y?sin?,z?7?运行,自??0至??2?时力场所作的功

解:由保守力性质计算

???ijk??Fz?Fy??????r?????x?y?z??y?z?FxFyFz???(1?1)i?(1?1)j?(2?2)k?0?故力场F为保守力场

????Fx?Fz????Fy?Fx????i???z??x?j????x??y??k

?????