Solution 7.119
22
2
Rr Rr
π
−
O’
z
o
Solution 7.120
OR
r
θ
A
ry
θ
z
zO
Ω
Solution 7.121
Using Eqs. 7.6
22
33
τ
=−−+−−
A
aRrjrk
RrR
G
θ = 15°
Solution 7.122
()
Ω≈ =
2
2
9.81 0.080
g
r
z
z′
20°
p
Solution 7.123
Equation 7.14 becomes 0,,
CC
HHrmrOC
υυυ
=+× = =
ππ
××
4
With 0
x
ω
= and principal axes xy z−−
′′
, Eq. 7.13 gives
2 32.2 2
11.30 ft-lb
=
y
‘
4″
z
z’
β
β = 20°
C
p
Solution 7.124
Equation 7.14 becomes 0,,
CC
HHrmrOC
υυυ
=+× = =
For disk, 120 2 12.57 rad sec
x
π
ωβ
×
== =
22
10.01380 lb-ft-sec
Imr
10 0.833 ft
ri i
==
232.2 2
11.85 ft-lb
=
3″
x
Solution 7.125
MI
ω
=
22
79.6 137.9 159.3 lb
AB
RR
== + =
Solution 7.126
0.0435
32.2 ft
==
()()
2
2
22
22 2
336
0.0435 125.7 257 lb-ft
2212
xz
yz
Mmb
ω
=− =− =−
( ) () () ()
2
1
200
yz
bb
Im bmmmb mb
=− + + + =−
xy
44
1
33
221
b
bb
b
zz
M
M
Solution 7.127
Let m = mass of each plate mass per unit area = m/(
π
R2/4)
22
9.42 9.42 13.33 N m
M
=+= ⋅
Solution 7.128
With 0
ωωω
===
and
ω
=2
200 rad sec ,
22
1.910 1.910 2.70 N m
M
=+=⋅