y
CD
Solution 7.74
Solution 7.75
Solution 7.76
0, ,
0, 0
so Eqs. 7.23 become
xy z
xyz yz
I
ωω ωω
ωωω
== =
=== =
33
2
bc
x
z
ω
Ox
pgc
Oz
θ
c
–
2
b
–
2
Solution 7.77
2;cossin
8
Solution 7.78
From sample Prob. 7.7:
π
22
B
y
m
1
m
2
r
Solution 7.79
For parallel-plane motion with 0
ω
= and 0
xz
I=,
Equations 7.23 give
12
z
y
z
M
x
Solution 7.81
ω
==
3g
a
nd for 0, 2
y
M
Solution 7.82
0, , 0, 0
xy z z xz
I
ωω ωωω
== = = =
ππ
π
2
x
ω
mg
ℓ
—
z
Solution 7.83
xz c c
Ixzdm
′′ ′′
=
Solution 7.84
=Δ +Δ +Δ
g
2
1mg 3
e
UTVV
ω
=
2
,
zz
xyzz
MI
22
32
xy
AAA ab
=+=
+
Solution 7.85
:MI pMiI j
ρ
=Ω× − =Ω×
y
Ay
By
Solution 7.86
B
Direction of bearing forces acting
on airframe
P
Ω
M
P
= M1
Ω
Solution 7.88
υ
Ω
600 5.03
b
Solution 7.89
()( )
0.8 9.81 0.180
MI
b
ψρ
=
−
Solution 7.90
2
1
MI
MM mk r
ρ
υ
=Ω
== Ω
y
ψ
⋅ = 0.2 rad
150 mm
A
Forward
D
Solution 7.91
Solution 7.92
==
20,000 2094 rad s
p
P
Ω
M
Solution 7.93
M
As viewed by passenger
P
z
Solution 7.94
G
p
z
M
1
M
2
Ω
2
Solution 7.95
M needed on structure of ship to counteract zroll to port (left).
60
M
p
–
Solution 7.96
()()
4 9.81 0.08 3.14 N m
A
M
==⋅
b = 400 mm
320 mm
yy
O
Ω
0.08 m mg
Solution 7.97
For rotor,
() ()
23600 2
4 0.12 2 43.4 N m
xzz
MI p
π
=Ω= = ⋅
yyy
yx
z
y
M
x
M
x
A320 mm
mg mg
x
1
–
Solution 7.98
Solution 7.100
60 sec
51.5
Solution 7.101
For zero moment Eq. 7.30 is
ψ
==
2
p
Ip
1.5
Solution 7.102
00M gives 89sin70M=× °
67.7 lb-in.Mi=
M
M
Z
Y
O
O
Y
P
G
X, x
ΩΩ8 lb
8 lb
8 lb
K
Z
9″
dH = Mdt
F
n
Z
z
z
b/2
Solution 7.103
Solution 7.104
From Eq. 7.30,
p
Ip
ψθθ
==
10