Solution 7.105
ππ
×
Solution 7.106
22
1,4
Solution 7.107
By symmetry, Ixy = Ixz = Iyz = 0 for whole propeller.
Thus for the three blades

=+ =

I
I
33
02
xx
II
()
ωωω
=−−
F
rom Eq. 7.21,
xxxx yyzzyz
MI I I
xy
3
30°
Solution 7.108
0;
ψψ
== From moment Eqs. 7.26,
2
From (c), p = const.
2
From (b),
2
zz
AB m p
b
θ
−+ = for
θ
=
π
/2 ………………… (d)
22
2
22
z
r
mrp
Bb
θθ

=−


y
x
Bz
p
Az
z
Z
mg
b
b
θ
ℓ
Solution 7.109
2
T
π
ω
== constant precessional rate about y-axis
M
z
M
M
x
ω
z
Ω
z
Solution 7.110
p
I
ψθ
=−
0
222
0
220 5 10 5
336
5105 2
xx
I I mr mh mr mh
r
== + = +


r
()
22
0
36
c,
hrI mr mr
== +
==


200 20.9 rad/s
60
p
39.1
=°
Body cone
Space cone
Ψ
·
h < r
–
2
pz
Ψ
·
Body cone
ω
ℓ
–
2
Solution 7.111
I = Izz = mr2
Solution 7.113
;O
MIp mk p
RR
=Ω =
2
O
mh
pmk
υ
= opposite direction to rotation of wheels
Solution 7.114
3
R
M
P
rotor
mass
mo
Rear views
Assume
right turn
Solution 7.115
()
()
()
2
0
83 6 4 16233 7.43in.sec
Ω× = − × − = − =
A
rikik jj
ωn
Solution 7.116
222
7
632
++
957 95
rad sec
n
ω
==
49
nik
x
3″2″4″
= 2 in./sec
υ
A
= 5 in./sec
Solution 7.117
2
kp
4
and from Table D/4, 2
10
Imr= so 22
10
kr=
Thus
2
24
rp
ππ
G
mg
z
h
p
θ
y
N
= 300
0.150 m
rev/min
Solution 7.118
With
ω
x =
ω
y = 0,
0xz z yz z zz z
Mass per unit of s is =
12 40 kg/m
0.150
3
40sin30 3
s
=− °×
π
ω
×
==
300 2 31.4 rad s
60
z
69.9
TJ
=