A
Yy
υ
rel
Solution 7.42
Angular velocity of x–y–z axes is 12
iJ
ωω
Ω=− +
υυ υ υ
==+Ω× +
ωω ω ωω
⋅
Ω=− + =− Ω× =
2
12 1 12
B
iJ i k
Solution 7.43
From sample problem 7.2,
()
22
43 9 33 radsec
ij k
π
=++
Y
y
θ
R
Solution 7.44
rr
Angle d
φ
measured in x–y–z turned by wheel in time dt is
xyz xyz
Noting Ω is constant in XYZ and xyz.
2
Thus 0
XYZ
dRpRp
kk
dt r r
Rp Rp
ω
α
==+Ω×Ω×=Ω×
rr
Solution 7.45
ωγω γ γω γ
Ω
=+=− +
11
111
00 sin cos
XYZ xyz
xyz xyz
dt dt
dd jk
dt dt
Solution 7.46
0.1 0.2 0.2
Axyz
−
Expand and equate coefficients to set
()
0
0
0
.2 0.2 0.32 1
yz
ωω
−=
Solution 7.48
With
ω
x =
ω
y = 0,
ω
z = –
ω
,
6
b
Solution 7.49
x–y–z are principal axes so
=− − ⋅
2
2.70 744 kg m s
A
A
Hjk
θ = π = 0.524
30
rad
.
Solution 7.50
z
z
33
bb2
y
3
2
b
Solution 7.51
From Eq. 7.14 using O for A,
()
3
15 1 26
35 5
bb
rm b i j k b i j i j k
ωρω
υρ
×= ++ × −+= −−+
Solution 7.52
0,
xy z
ωω ωω
== =
2
41
0, 0 ,
322
so
xz yz zz
yz z zz z
rb
II m cImr
HI jIk
π
ωω
==+ + =
=− +
3
z
12
Solution 7.53
0;,,0
xyz
HHrG p
ωωω ω
=+× = = =
ω
=+
22
11
w
here 12 4
x
Hmbmr
ω
=++ +
22
22
0
1
34 2
br
Hhmimrpj
Solution 7.54
About G,
3
2
xx
I
Hmb
=+ Ω
Ω=Ω +Ω +Ω
where
xy z
ijk
Solution 7.55
()
cos sin
pk i N j k
ωγ γγ
=−+ +
0
Solution 7.56
From Eq. 7.11, with 0
xy
ωω
==
,
HIiIjIk
ωω ω
=− − +
O
I
xz
= 0
z
x
Solution 7.57
Introduce axes xyz
′′′
−−
()
22
11
0, 2
xy yy
I
Imama
′′ ′′
== =
222
11
yy y zz z
ωω
′′ ′ ′′ ′
′
()
1
1
sin 45 cos 45 2
kj k jk
′=°+°=+
′=− °+ °= − +
()
()
222
13
so 2 20 0.04 0.06
66
22
Hmbjabk j k
ωπ
=−++= −+
Z
b
O
b
x’
𝜔 = 20π rad/s
z
z’
Solution 7.58
2
1
2
zz
Imr
′′
=
Equation 7.11 yields
1
cos 4.96 for 10
Hk
H
βα
−
⋅
==°=°
Solution 7.59
,0,
xyz
p
ωωω
=Ω = =
2
2
222
02
131 1
2104 2
h
THmr p
r
ω
=⋅= +Ω+
Solution 7.60
Let Ω= angular velocity of x–y–z about z0
()
22 2
2sincos
z
Gzzz
G
Hmfk ik kmkpk
πθθ
′
=− + +
Solution 7.61
12
,,
xyz
p
ω
ωω ωω
=− = =
()
222
12
2
E
quation 7.14, , ,
,,, 0
442
111
Equation 7.11, 442
xx yy zz xy xz yz
B
HHOBGOBbiGm mbk
I mrI mrI mrI I I
Hmr imrjmrpk
υω
ωω
=+× = = =−
======
=−+ +
4
r
11
F
rom Eq. 7.15, 22
B
Tm H
υυ ω
=⋅+⋅
p
ω
Solution 7.62
xyz
′′′
−− are principal axes of inertia
222
2
111
So 4222
232.2 12 2
ω
ω
′
=+= +
O
H mr i mr p j mr i p j
2
111 1
10 40 10
4 32.2 12 2 2 32.2 12
84.29 63.85
148.1 ft-lb
ω
ππ π
=++
=+
=
Solution 7.63
With 0
xy
ωω
==
and z
ωω
=, the components of angular momentum become
6
Sphere: 0
yz
I=
22222
653
121
sin 2 cos 2
653
O
Hm c j r c bk
ωβ β
=+++
x
z
bbx2c cosβ
b
y
b
z
z
dℓ
b
O
1
x
MW
z
b
Solution 7.64
Let
ρ
= mass per unit of panel area
b
22
22
4cos
33
cb
bc a ac
ρθ
=+++
633
By symmetry, principal axes are 0–1, 0–2, 0–y
()
22
2
3
3
cb
Imb
+
Solution 7.65
ω
=
2
2
2
2
mb
M
ω
Solution 7.66
xy z z
2
33
yz
LL
ImRmR
=−
2
2
2
2
2
,
33
0, 0
axes :
,
33
0, 0
xyzz
yy
yx
xyzz
yy
yx
mLR mR
BL B
MB
xy M I
mLR mR
AL A
MA
ω
ω
ω
ω
ω
′′
′
−=− =
==
′′
−=
−
==−
==
Solution 7.67
2
cos
2
x
mb
Bc
ωθ
=
and
()
22
cos sin ,
22
mb mb
BijBB
cc
ωω
θθ
=+==
A
x
x’
z
Rm
A
y
L/3
L/3
x
x
A
Solution 7.68
From Eq.
7
.23, with 0
z
ω
=
x
z
z0
Solution 7.70
Let p = mass per unit length
2
MI
ω
=−
0
22 2
cos cos 2 2
4
22
so or
rrmr
Mmr Mmrj
ρθ θρ
π
ωω
ππ
=−− = =
=− =−
Solution 7.71
2
22 2
13
22
oo
zz z z
mr
II mr
ρ
π
=
=+
ωππ
π
=
=− = =− =−
==−
2
2
200
2
0
23
24
2
:33
4
0, so 3
xz
zzzz O zz
xxzz x
y
Imr
MM
MI MM mr
mr
M
MM i
z
ω
Solution 7.72
zz
MI
α
=
where Iz is given by Eq. B.10 with
22
0
cos sin
II
αθθ
=+
Solution 7.73
z