19.11
19.10
19.15
βsin
23
)(2
,g
QP
QP
a
+
+
=
19.14
121
21
2
1
2
1
2
2
2
1
0
4
1
8
7
78
216
)(0
8
1
16
7
2
1
khvgvmavmavm
mm
khghm
v
khghmvmvm
AAAAAA
A
AA
=+?+?
+
=
=+?+
）式两边求导对（
19.18
1
28 khgm
a
A
=得
21
78 mm +
lmglk
l
mg =+=+=
1
2
10
,)31(
2
1
2
1
60sin
2
2 λλ
o
转向为顺时针
（
,
2
)735(3
2
1
)32(,
2
1
30sin
2
2
,
3
4
2
1
2
1
2
1
1100
2
2
21
22222
1
g
l
VTVT
lk
l
mgV
r
l
mlJJJT
AB
ABCBCABABC
C
PBC
BC
PAB
AB
A
=?+=+
=+=
===++=
ω
λλ
ωωωωωωωω
Q
o
19.19 解 取θ =0
o
为零势点
T
0
=0,V
(1) θ=30
o
时
(2) θ =0
o
时
转向为顺时针,
2
)13(3
2
1
0
)00(
3
1
2
1
2
1
2
1
2200
2
22222
2
g
l
VTVT
V
vmlJJJT
AB
CCABC
C
PBC
BC
PAB
AB
A
+
=?+=+
=
=?==++=
ω
ωωωωω
Q
Q
ω
C
ω
AB
P
C
ω
BC
P
BC
