理论力学课后习题解答附答案
?，则在固定坐标系中的速度：5.21解 取在转动坐标系的速度为广义速度v??qv?v??Ω?r，自由质点的动能T?12mv，设质点势能为V，则质点的拉氏函数 212mv?V 2L?T?V?根据定义：
?1???mv2?2?T1??v??Ω?r? 2??p???m?mv??q?v?2?v?在转动坐标系中：
???L?mv?v?-H??p??qi?1s121mv?V?mv?v-Ω?r??mv2?V 22?121212mv?mv(Ω?r)?V?p?(Ω?r)?p?V?p?Ω??r?p??V 22m2m上式中r为质点的位矢，p?mv,v为质点相对于固定坐标系的速度。
5.22解：取在广义坐标q1??,q2??,q3??.根据教材（3.9.21）和（3.9.19）式得动能：
T?1?2???2sin2??I3???cos?????2 I1?2????势能：mglcos?,l为原点距离
H?T?V? 1?2???2sin2??I3???cos?????2?mglcos?I1?2????根据定义式
p???T?T?,有p???I1?????q???sin2??I3cos????cos?????p??I1??cos?????p??I3??
1
H?故
?因为
p?12?p??p?cos??p????mglcos?22I12I32I1sin?221?12112?2??p?p?p?p???mglcos?????2?I1I3I1sin2??
dH?H??0 ???H,H???dt?t所以H?C1为第一积分.又
?p??p??0,?p??p????1,?0,?p??p??0,?p????sin?,??I3??p????0,?p?p?H?H??0?0,???????p?I
故
?p得p??C2为第二个第一积分. 同理
?,H??0
???0 p?p??p??p???即
?0,?p??p??0,?p??p??1,?p????0,?p????0,?0,?H?H?H??0;p,H?0?0,?0,??????p??p???
???0 p得p??C3为第三个第一积分.
2
5.23解如题5.23.1图,
yMa?
xC?tO题5.23.1图由5.6题解得小球的动能
T?1?22???2?m?4a?cos2?4a2???cos2?a2??① 2?22?根据定义
p???T??2ma2?cos2?② ?ma2??2??得
???根据哈密顿函数的定义
p?2??2?cos③ 22ma??L?p??H?p????T?V 222?22?2?2???1???p??4a?cos?4a??cos?a??V??2m?22?代入③式后可求得:
p?212?222H??2?pcos?ma?sin?④ ?2222ma由正则方程得:
???p??H???p?sin??ma2?2sin?cos?⑤ ?????代入⑤得
p?H???2?2?cos2⑥ ?p?ma222??psin??ma?sin?cos??????sin????2sin? ????ma整理得
????2sin??0 ?
3
5.24 如题5.24.1图,
bO?c? 题5.24.1图 ⑴小球的位置可由?确定,故自由度s?1
????. ⑵选广义坐标q??,广义速度q⑶小球动能
T?121mv1?I0?2 22又由
?,??v1?c?b??,代入①式得 v??c?b??cc1127?2?c?b??2?2?c?b?2 T?mv12?a?b????mc2????m?22510?c?2设小球势能为V,取固定圆球中心O为零势点,则
V?mg?c?b?cos?
小球拉氏函数
L?T?V?7?27?22?c?b?2?mg?c?b?cos?① m??c?b??V=m?1010根据定义
4
?T7?2???m?c?b?25??
p5????7m?c?b?2p??有
??LH?p??2
?5?7?5p??p?2???p???m?c?b?mg?c?b?cos??2?2?710?m?c?b???7m?c?b??
25p???mg?c?b?cos?214m?c?b??4?根据正则方程
???p?H?mg?c?b?sin???④
???对式两边求时间导得：
??????5p7m?c?b?5p??H?⑤ ?p?7m?c?b?22?5mg?c?b?sin?7m?c?b?2?5gsin?
7?c?b?故小球球心切向加速度
???5gsin? a??c?b??7
5.25解根据第二章§2.3的公式有：
??i????yipiz?zipiy???i?ziyJx??mi?yizi?1i?1?nn??i?xiz?i????zipix?xipiz??① Jy??mi?zixi?1i?1?nn??i?yix?i????xipiy?yipix??Jz??mi?xiyi?1i?1?根据泊松括号的定义：
5
nn
理论力学课后习题解答附答案
