Greenwood D.T. Advanced Dynamics
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409 Answers to problems
3.17.
˙
x =
(lω
0
− 6v
0
cos θ) sin θ
2(1 +3 cos
2
θ)
,
˙
θ =
lω
0
− 6v
0
cos θ
l(1 + 3 cos
2
θ)
ˆ
C
y
=
m
v
0
+
1
2
lω
0
cos θ
1 +3 cos
2
θ
3.18. (a) u
1
=−
2
√
2
ˆ
F
5m
, u
2
=
18
ˆ
F
5ml
(b)
ˆ
C =
2
√
2
5
ˆ
F
3.19. (a) ω =
g
R
2 −
l
R
, (b) ω =−
g
2R − l
3.20. (a) θ
max
= 65.07
◦
, (b)(ω
0
)
min
= 2
g
r
3.21. ω =
6v
0
7b
i +
6v
0
7a
j, v
c
=−
6
7
v
0
k
3.22. (a) v
1
= v
1t
e
t
+ v
1l
e
l
, v
2
= v
2t
e
t
+ v
2l
e
l
where v
1l
= v
2l
=
−
L
2
+r
2
v
0
4
5
r
2
+ 3
L
2
+r
2
v
1t
=
4
5
r
2
+ 4
L
2
+r
2
4
5
r
2
+ 3
L
2
+r
2
v
0
,v
2t
=
4
5
r
2
v
0
4
5
r
2
+ 3
L
2
+r
2
(b)
ˆ
M
1
=
4
5
r
2
L
2
+r
mv
0
4
5
r
2
+ 3
L
2
+r
2
,(c)
ˆ
P =
4
5
r
2
mv
0
4
5
r
2
+ 3
L
2
+r
2
3.23.
˙
θ
1
=
v
1
2
√
2l
−
3v
2
8l
,
˙
θ
2
=−
v
1
2l
+
5v
2
4
√
2l
˙
θ
3
=−
v
1
2
√
2l
−
5v
2
8l
3.24. (a)
ml
2
3
¨
θ sin
2
α −
µml
2
3
˙
θ
2
sin α cos α
=
1
2
mgl
(
sin θ sin α − µ cos θ cos α
)
(b) θ = 48.19
◦
3.25. (a) v
B
=
ˆ
F
m
(
−sin α i + 4 cos α j
)
,
m =
m
1 +3 cos
2
α
3.26.
m
1
=
5
4
m,
m
2
=
5
16
m
410 Appendix
3.27.
4
3
ml
2
¨
θ
1
−
3
2
ml
¨
x sin θ
1
+
3
2
ml
¨
y cos θ
1
+
1
2
ml
2
¨
θ
2
cos(θ
2
− θ
1
)
−
1
2
ml
2
˙
θ
2
2
sin(θ
2
− θ
1
) = 0
1
3
ml
2
¨
θ
2
−
1
2
ml
¨
x sin θ
2
+
1
2
ml
¨
y cos θ
2
+
1
2
ml
2
¨
θ
1
cos(θ
2
− θ
1
)
+
1
2
ml
2
˙
θ
2
1
sin(θ
2
− θ
1
) = 0
2m
¨
x sin θ
1
− 2 m
¨
y cos θ
1
−
3
2
ml
¨
θ
1
−
1
2
ml
¨
θ
2
cos(θ
2
− θ
1
)
+
1
2
ml
˙
θ
2
2
sin(θ
2
− θ
1
) = 0
Chapter 4
4.1. (a) T =
1
2
m(
˙
X
2
+
˙
Y
2
+l
2
˙
φ
2
sin
2
θ + l
2
˙
θ
2
+ 2l
˙
φ
˙
X cos φ sin θ
+ 2l
˙
φ
˙
Y sin φ sin θ + 2l
˙
θ
˙
X sin φ cos θ − 2l
˙
θ
˙
Y cos φ cos θ )
+
1
2
I
xx
(
˙
φ sin θ sin ψ +
˙
θ cos ψ)
2
+
1
2
I
yy
(
˙
φ sin θ cos ψ −
˙
θ sin ψ)
2
+
1
2
I
zz
(
˙
ψ +
˙
φ cos θ )
2
, V = mgl cos θ
(b) u
4
=−r
˙
θ sin φ + r
˙
ψ cos φ sin θ +
˙
X = 0
u
5
= r
˙
θ cos φ + r
˙
ψ sin φ sin θ +
˙
Y = 0
(c) Q
1
= 0, Q
2
= mgl sin θ, Q
3
= Q
4
= Q
5
= 0
ij
=
10 0
01 0
00 1
0 r sin φ −r cos φ sin θ
0 −r cos φ −r sin φ sin θ
4.2. (a) 2m ˙v
r
− 2mr
2
cos(φ − θ) − ml(
˙
φ + )
2
= 0
ml
2
¨
φ + mlv
r
˙
φ + 2mlv
r
+ mlr
2
sin(φ − θ ) = 0
(b)
˙
E =
˙
T
2
−
˙
T
0
= 0
4.3. I ˙ω
1
+ I
r
R −r
ω
2
ω
3
= 0
(I + mr
2
)˙ω
2
− I
r
R −r
ω
1
ω
3
= 0
(I + mr
2
)˙ω
3
= mgr sin θ
411 Answers to problems
4.4. (a) (I + mr
2
)˙ω
x
− mr ˙v cos α = mgr sin α
(I + mr
2
)˙ω
y
= 0, I ˙ω
z
= 0
(m + m
0
)˙v − mr ˙ω
x
cos α = 0
(b)
˙
x = rω
y
,
˙
y =−rω
x
4.5. (a) I ˙ω
r
+
r
R −r
I ω
φ
ω
z
= 0
(I + mr
2
)˙ω
φ
−
r
R −r
I ω
r
ω
z
=−mgr
(I + mr
2
)˙ω
z
= 0
(b)
˙
z = r ω
φ
,ω
φ
= A cos
I
I + mr
2
˙
φt + B sin
I
I + mr
2
˙
φt
4.6. (I
a
+ mr
2
sin
2
θ)˙ω
1
+ mr(l + r cos θ )˙ω
2
sin θ
+mr(l + r cos θ )ω
2
ω
3
cos θ + mr
2
ω
1
ω
3
sin θ cos θ = 0
[I
p
+ mr(2l +r cos θ) cos θ ]˙ω
2
+ mr(l + r cos θ )˙ω
1
sin θ
+[I
a
+ mr(l + r cos θ ) cos θ]ω
1
ω
3
+[I
p
+ mr(2l +r cos θ) cos θ ]ω
2
ω
3
cot θ = 0
(I
p
+ mr
2
+ 2mrl cos θ )˙ω
3
− (I
a
+ mr
2
+ mrl cos θ )ω
1
ω
2
−[(I
p
+ mr
2
+ 2mrl cos θ ) cot θ + mrl sin θ ]ω
2
2
− mrlω
2
3
sin θ
= mgl sin θ
4.7. (a) (I + mr
2
)˙ω
x
− I ω
y
=−mgr, I ˙ω
y
+ I ω
x
= 0,
(I + mr
2
)˙ω
z
+ mr
2
x = 0
(b)
˙
z = r ω
x
=
−mgr
2
I (I + mr
2
)
sin
I
I + mr
2
t
(c) v
0
>
1
2
r
4.8.
5
4
mr
2
+ m
0
(r + l sin ψ )
2
¨
θ − m
0
(r + l sin ψ )l cos ψ(˙ω
d
+
˙
θω
d
cot θ − 2
˙
θ) −
1
4
mr
2
+ m
0
(r + l sin ψ )l sin ψ
ω
2
d
cot θ
+
3
2
mr
2
+ m
0
(r + l sin ψ )(r + 2l sin ψ )
ω
d
=−
[
mr + m
0
(r + l sin ψ )
]
g cos θ
412 Appendix
1
4
mr
2
+ m
0
l
2
cos
2
ψ
˙ω
d
− m
0
(r + l sin ψ )l
¨
θ cos ψ
+
1
4
mr
2
(
˙
θω
d
cot θ − 2
˙
θ) +m
0
l cos ψ
lω
2
d
cot θ sin ψ
+l
˙
θω
d
cot θ cos ψ − 2l
˙
θcos ψ − (r +2l sin ψ)ω
d
= m
0
gl cos θ cos ψ
3
2
mr
2
+ m
0
(r
2
+l
2
+ 2rl sin ψ )
˙
−
mr
2
+ m
0
(r + l sin ψ )
2
− m
0
l
2
cos
2
ψ
˙
θω
d
+m
0
l cos ψ
(r + l sin ψ )
ω
2
d
−
˙
θ
2
+r(
2
− ω
d
cot θ)
=−m
0
gl sin θ cos ψ
4.9. (m
0
+ 3m)˙v + m
0
l
˙
θ
2
= 0
I
c
+ m
0
l
2
+ 3mb
2
+
1
2
mr
2
¨
θ − m
0
lv
˙
θ = 0
4.10. I ˙ω
1
+ Ibω
2
ω
3
cos α = 0
(I + mr
2
)
˙ω
2
− bω
2
3
sin α
− Ibω
1
ω
3
cos α =−mgr cos α
(I + mr
2
)( ˙ω
3
+ bω
2
ω
3
sin α) = 0
4.11. (5 + cos
2
φ)m ˙v + mL
¨
θ sinφ cos φ + mL
˙
θ
2
L
l
cos
3
φ − cos
2
φ−3
+
mv
2
l
sin
2
φ cos φ + mv
˙
θ sin φ cos φ
1 −
2L
l
cos φ
= F
mL
2
¨
θ(3 +sin
2
φ) +mL˙v sin φ cos φ + mLv
˙
θ
3 +sin
2
φ −
2L
l
sin
2
φ cos φ
+mL
2
˙
θ
2
sin φ cos φ
L
l
cos φ − 1
+
mL
l
v
2
sin
3
φ = 0
4.12. (a) m
˙v
θ
+
r
R
v
θ
ω
θ
= 0, I
˙ω
r
−
1
R
v
θ
ω
θ
= 0
(I + mr
2
)˙ω
θ
−
mr
R
v
2
θ
+
I
R
v
θ
ω
r
= 0, I ˙ω
z
= 0
(b) H
0
= (I ω
r
− mrv
θ
)e
r
+ (I + mr
2
)ω
θ
e
θ
+ (I ω
z
+ mRv
θ
)k
(c) Final values: θ = 113.58
◦
,v
θ
= 0,ω
r
= 1,ω
θ
= 0.6547
4.13. (a) 2m
1 +
l
2
4R
2
˙v − ml
1 −
l
2
4R
2
ω
2
3
= 2mg sin θ sin φ
ml
2
˙ω
3
+
1 −
l
2
4R
2
mlvω
3
= mgl sin θ cos φ
(b) ˙α =
1
R sin θ
v cos φ −
1
2
lω
3
sin φ
413 Answers to problems
˙
θ =−
1
R
v sin φ +
1
2
lω
3
cos φ
˙
φ =
1 +
l
2R
cot θ sin φ
ω
3
−
v
R
cot θ cos φ
4.14. (a) I
t
¨
φ sin θ + 2I
t
˙
φ
˙
θ cos θ − I
a
˙
θ = mlrω
2
0
cos(φ − ω
0
t)
I
t
¨
θ − I
t
˙
φ
2
sin θ cos θ + I
a
˙
φ sin θ
= mgl sin θ + mlrω
2
0
cos θ sin(φ − ω
0
t)
I
a
˙
= 0
(b)
˙
E = mlrω
2
0
[
˙
φ sin θ cos(φ − ω
0
t) +
˙
θ cos θ sin(φ − ω
0
t)]
(c)
˙
φ
1,2
=
I
a
tan θ
0
2(I
t
sin θ
0
+ mlr)
1 ±
1 −
4mgl(I
t
sin θ
0
+ mlr)
I
2
a
2
tan θ
0
4.15. (a) m ˙v − ml
¨
θ sin θ − ml
˙
θ
2
cos θ − ml
˙
φ
2
cos θ = 0
(I
xx
sin
2
θ + I
zz
cos
2
θ)
¨
φ + 2(I
xx
− I
zz
)
˙
φ
˙
θ sin θ cos θ + mlv
˙
φ cos θ = 0
I
yy
¨
θ − (I
xx
− I
zz
)
˙
φ
2
sin θ cos θ − ml ˙v sin θ =−mgl cos θ
(b)
˙
X =−v sin φ,
˙
Y = v cos φ
4.16. (a) (I + mr
2
)˙ω
1
+ (I + mr
2
)
r
R +r
ω
1
ω
2
cot θ
+ I
r
R +r
ω
2
ω
3
= 0
(I + mr
2
)˙ω
2
− (I + mr
2
)
r
R +r
ω
2
1
cot θ − I
r
R +r
ω
1
ω
3
= mgr sin θ
I ˙ω
3
= 0
(b) In terms of
˙
φ,
˙
θ,
¨
φ,
¨
θ, we obtain
(I + mr
2
)
R +r
r
¨
φ sin θ + 2(I + mr
2
)
R +r
r
˙
φ
˙
θ cos θ − I
˙
θ = 0
(I + mr
2
)
R +r
r
¨
θ − (I + mr
2
)
R +r
r
˙
φ
2
sin θ cos θ + I
˙
φ sin θ
= mgr sin θ
which are the top equations with correspondences
I
t
= (I + mr
2
)
R +r
r
, I
a
= I, l = r
4.17. m ˙v = F
x
cos β cos φ, mv
˙
φ cos β =−F
x
sin φ
4.18. (a) (I + mr
2
)˙ω
x
− mr
2
ω
y
+ mr
2
y = 0
(I + mr
2
)˙ω
y
+ mr
2
ω
x
− mr
2
x = 0
I ˙ω
z
= 0
414 Appendix
(b) x (t ) =
1
∗
(
˙
x
0
sin
∗
t +
˙
y
0
cos
∗
t) + x
0
−
˙
y
0
∗
y(t) =
1
∗
(
˙
y
0
sin
∗
t −
˙
x
0
cos
∗
t) + y
0
+
˙
x
0
∗
where
∗
=
I
I + mr
2
(c) H
0h
= [(I + mr
2
)ω
x
− mrx]i + [(I + mr
2
)ω
y
− mry]j = const
4.19. (a) (I + mr
2
)˙ω
x
=
mgrπ h
L
sin
2π y
L
(I + mr
2
)˙ω
y
−
2π
2
mr
3
h
L
2
ω
x
ω
z
cos
2π y
L
= 0
I ˙ω
z
= 0
(b)
˙
y =−rω
x
1 +
2π
2
hr
L
2
cos
2π y
L
4.20. (a)
1
4
mr
2
¨
θ(1 +4 cos
2
θ) −mr
2
˙
θ
2
sin θ cos θ −
1
4
mr
2
ω
2
d
cot θ
+
1
2
mr
2
ω
d
=−mgr cos θ
1
4
mr
2
˙ω
d
−
1
2
mr
2
˙
θ +
1
4
mr
2
˙
θω
d
cot θ = 0,
˙
= 0
(b) ω
d
=
2g
r
(
√
θ
2
+ 2θ − θ),
˙
φ =
ω
d
θ
Chapter 5
5.5. (a) 3m ˙v =−mg sin α cos φ,
¨
φ = 0
(b)
˙
x =−v
0
sin ω
0
t +
g
3ω
0
sin α sin
2
ω
0
t
˙
y = v
0
cos ω
0
t −
g
6ω
0
sin α sin 2ω
0
t
(
˙
x)
av
=
g
6ω
0
sin α, (
˙
y)
av
= 0
5.6. (a) 2m
¨
y + ml
¨
φ cos φ + 2m
˙
y
˙
φ tan φ =−F sin φ cos φ
ml
2
¨
φ cos φ + ml
¨
y + ml
˙
y
˙
φ tan φ =−Fl sin φ cos φ
(c) ml
2
¨
φ + Fl sin φ = 0
5.8. 2m
˙
u
1
(1 +cos
2
φ) +
2mu
2
1
l
sin
2
φ cos φ −
2mu
2
2
l
cos φ = 0
2m
˙
u
2
(1 +sin
2
φ) +
2mu
1
u
2
l
cos
3
φ = 0
415 Answers to problems
5.9. (a) T = 2m(
˙
x
2
+
˙
y
2
) +ml
2
(
˙
ψ
2
+
˙
φ
2
)
+2ml
˙
x(
˙
ψ sin ψ cos φ +
˙
φ cos ψ sin φ)
+2ml
˙
y(−
˙
ψ cos ψ cos φ +
˙
φ sin ψ sin φ)
T
0
= ml
2
˙
ψ
2
(2 tan
2
φ + 1) + ml
2
˙
φ
2
(2 cot
2
φ + 1)
(b) 2ml
2
(2 tan
2
φ + 1)
¨
ψ +
2tan
2
φ − 1
sin φ cos φ
˙
ψ
˙
φ
= 0
2ml
2
sin
2
φ
(cos
2
φ + 1)
¨
φ +
˙
ψ
2
sin φ
cos φ
− 2
˙
φ
2
cot φ
= 0
5.10. 3m ˙v − ml
¨
θ sin φ − ml
¨
φ sin φ − ml(
˙
θ +
˙
φ)
2
cos φ − mr
˙
θ
2
= 0
[I + m(r
2
+l
2
+ 2rl cos φ)]
¨
θ + ml(l + r cos φ)
¨
φ
−mrl(2
˙
θ +
˙
φ)
˙
φ sin φ − ml ˙v sin φ + m(r + l cos φ)v
˙
θ = 0
ml
2
¨
φ + ml(l +r cos φ)
¨
θ − ml ˙v sin φ + mlv
˙
θ cos φ + mrl
˙
θ
2
sin φ = 0
5.11. (a)
1
4
mr
2
¨
θ = 0,
1
2
mr
2
¨
ψ(1 + 2 cos
2
θ) −mr
2
˙
θ
˙
ψ sin θ cos θ = 0
m
¨
x = 0
(b)
˙
ψ
min
=
˙
ψ
0
,
˙
ψ
max
=
√
3
˙
ψ
0
v
min
=
˙
x
0
,v
max
=
˙
x
2
0
+r
2
˙
ψ
2
0
5.12.
4
3
ml
2
¨
θ
1
+
1
2
ml
2
¨
θ
2
cos(θ
2
− θ
1
) +
3
2
ml ˙v −
1
2
ml
2
˙
θ
2
2
sin(θ
2
− θ
1
) = 0
1
3
ml
2
¨
θ
2
+
1
2
ml
2
¨
θ
1
cos(θ
2
− θ
1
) +
1
2
ml ˙v cos(θ
2
− θ
1
)
+
1
2
ml
˙
θ
1
(v + l
˙
θ
1
) sin(θ
2
− θ
1
) = 0
2m ˙v +
3
2
ml
¨
θ
1
+
1
2
ml
¨
θ
2
cos(θ
2
− θ
1
) −
1
2
ml
2
˙
θ
2
2
sin(θ
2
− θ
1
) = 0
Chapter 6
6.1. (a) y
n+1
− y
∗
n+1
=−0.0001626, E
n+1
=−0.0001667
(b) ξ = 0.024813
6.2. (a) −
h
6
240
y
(6)
n
(b) ρ
2
+ bρ + 1 = 0 where b = 2
5h
2
ω
2
0
− 12
h
2
ω
2
0
+ 12
Unstable for h >
√
6
ω
0
416 Appendix
6.3. (a) a = 2, b =−3, c = 1
(b)
23
24
h
2
...
y
n
6.4. (a) ρ
2
+ (4h − 2)ρ + (2h
2
− 4h + 1) = 0
(b) ρ
1,2
= 1 − 2h ±
√
2h, stable for 0 ≤ h ≤ 0.5858
6.5. (a)
2
3
ml
2
¨
θ(1 +3 sin
2
θ) +2ml
2
˙
θ
2
sin θ cos θ − mgl cos θ = 0
(b)
t θω ˙ω
0.5 1.131533 2.774195 −0.762194
1.0 2.518895 2.913179 0.059869
1.5 2.914747 −2.396043 −9.159207
(c) ω
max
= 2.913254 at t = 0.319054,θ= 0.615481
θ =
π
2
at t = 0.660765
ω
max
= 2.913254 at t = 1.002475,θ= 2.526105
θ
max
= 3.141593 at t = 1.321531
6.6. Correct results:
t = 0.5 x =−0.345172 v
x
= 2.262982
y =−1.251733 v
y
= 1.142012
z = 2.596904 v
z
=−3.404994
t = 1.0 x =−0.066909 v
x
=−5.738694
y = 0.351428 v
y
= 1.211603
z = 0.715481 v
z
= 4.527092
t = 1.5 x =−1.331422 v
x
=−0.600386
y =−0.548644 v
y
=−2.141998
z = 2.880066 v
z
= 2.742384
z
min
= 0.585786 at t = 0.953950
z
max
= 3.414214 at t = 0 and t = 1.907900
6.7. (a)
˙
x = v
x
,
˙
y = v
y
,
˙
φ = ω
˙v
x
=
1
2
lω
2
cos φ − ω sin φ(v
x
cos φ + v
y
sin φ)
˙v
y
=
1
2
lω
2
sin φ + ω cos φ(v
x
cos φ + v
y
sin φ)
˙ω =−
ω
l
(v
x
cos φ + v
y
sin φ)
417 Answers to problems
(b)
t = 0.1 x = 0.004983 v
x
= 0.099336
y = 0.000332 v
y
= 0.009958
φ = 0.099917 ω = 0.997505
t = 1.0 x = 0.366104 v
x
= 0.517681
y = 0.265792 v
y
= 0.688060
φ = 0.925775 ω = 0.793278
t = 10.0 x =−4.964991 v
x
=−0.853882
y = 10.81251 v
y
= 1.127333
φ = 2.219039 ω = 0.001699
6.8. (a) ml
2
¨
θ + ml
2
2
sin θ cos θ − mgl cos θ = 0
(b)
t θω ˙ω
0.5 0.403383 1.264691 0.068683
1.0 0.883130 0.459049 −2.365847
1.5 0.814167 −0.727807 −2.245828
(c) θ
max
= 0.927295 rad at t = 1.192006 s
ω
max
= 1.264911 rad/s at t = 0.506431 s
6.9. (a) t = 0.2s
ω
x
= 14.983846
x
= 0.996466 W = 1100.8044 N·m
ω
y
= 0.880327
y
= 0.010208 α = 0.091516
ω
z
= 0.499509
z
= 0.044588 η = 0.070453
t = 0.4s
ω
x
= 14.969975
x
= 0.134382 W = 1500.8666 N·m
ω
y
= 1.199892
y
= 0.035882 α = 0.201343
ω
z
=−0.317395
z
= 0.093878 η =−0.985820
t = 0.6s
ω
x
= 14.999716
x
=−0.977615 W = 145.9451 N·m
ω
y
= 0.116755
y
= 0.060906 α = 0.136529
ω
z
=−0.297954
z
= 0.030712 η =−0.199038
t = 0.8s
ω
x
= 14.995077
x
=−0.266247 W = 607.6461 N·m
ω
y
= 0.486064
y
= 0.055357 α = 0.151824
ω
z
= 0.507382
z
=−0.051838 η = 0.960917
t = 1.0s
ω
x
= 14.962897
x
= 0.942681 W = 1668.4876 N·m
ω
y
= 1.333689
y
= 0.041312 α = 0.113436
ω
z
=−0.024909
z
=−0.038818 η = 0.328846
418 Appendix
(b) W
max
= 1669.4220 at t = 0.331667 and t = 0.995000.
(c) α
max
= 0.201447 at t = 0.405759
(d) t = 0.663333 s
6.10. (a)
¨
ψ =
2
˙
θ
cos θ
+
˙
ψ sin θ
¨
θ =−
1
5
˙
ψ
2
sin θ cos θ −
6
5
˙
ψcos θ + 16 sin θ
˙
=
2
3
˙
ψ
˙
θ cos θ
(b) t = 0
ψ = 0 θ =
π
6
= 0.523599 φ = 0 X = 0
˙
ψ = 4
˙
θ = 0 = 2 Y = 0
t = 1
ψ = 2.891062 θ = 0.096692 φ = 2.754937 X = 0.156089
˙
ψ = 1.726344
˙
θ =−0.678622 = 1.294624 Y =−0.919876
t = 1.5
ψ = 3.480268 θ =−0.595660 φ = 3.225671 X =−0.071231
˙
ψ = 0.534888
˙
θ =−2.748494 = 0.831908 Y =−0.910777
t = 2.0
ψ =−3.157916 θ =−0.517934 φ = 10.294732 X =−0.436413
˙
ψ = 0.633105
˙
θ = 2.483279 = 0.857619 Y =−0.856059
t = 2.5
ψ =−2.533888 θ = 0.116376 φ = 10.793919 X =−0.666352
˙
ψ = 1.785449
˙
θ = 0.645929 = 1.317532 Y =−0.784324
t = 3.0
ψ =−1.375078 θ = 0.380226 φ = 11.849700 X =−0.844784
˙
ψ = 2.927582
˙
θ = 0.476119 = 1.706236 Y =−0.328526
t = 3.5
ψ = 0.423844 θ = 0.522848 φ = 13.623335 X =−0.168626
˙
ψ = 3.993075
˙
θ =−0.050408 = 1.998268 Y = 0.013511
θ
min
=−1.542703 rad (−88.39
◦
)att = 1.735130 s
θ
max
=
π
6
at t = 0 and t = 3.470260 s when
˙
ψ,
˙
θ, and also
return to their initial values.
θ = 0att = 1.125585 s and 2.344675 s.