7.3 Solutions
305
Using (8) in (7) and (9) we obtain
0 = A + B
(10)
0 = ω(A − B) +
g
2ω
(11)
Solving (10) and (11) we get A = −
g
4ω 2 , B =
g
4ω 2
(12)
The complete solution for x is
x =
g
4ω 2 (e
−ωt
− e
ωt
+ 2 sin ωt)
7.10 Let the length of the cord be l. The Cartesian coordinates can be expressed in
terms of spherical polar coordinates (Fig. 7.17)
Fig. 7.17
x = l sin θ cos φ
y = l sin θ sin φ
z = −l cos θ
V = mgz = −mgl cos θ
(1)
v
2
= ˙
x
2
+ ˙
y
2
+ ˙
z
2
= l
2
( ˙
θ
2
+ sin
2
θ ˙
φ
2
)
T =
1
2
ml
2
( ˙
θ
2
+ sin
2
θ ˙
φ
2
)
(2)
L =
1
2
ml
2
( ˙
θ
2
+ sin
2
θ ˙
φ
2
) + mgl cos θ
(3)
∂ L
∂ ˙
θ
= ml
2 ˙
θ,
∂ L
∂θ
= ml
2 sin θ cos θ ˙
φ
2
− mgl sin θ
(4)
∂ L
∂ ˙
φ
= ml
2 sin
2
θ ˙
φ,
∂ L
∂φ
= 0
( 5 )