PROBLEM 9.180
For the component described in Problem 9.165, determine
(a) the principal mass moments of inertia at the origin, (b) the
principal axes of inertia at the origin. Sketch the body and show
the orientation of the principal axes of inertia relative to the x, y,
and z axes.
SOLUTION
(a) From the solutions to Problems 9.141 and 9.165 we have
Problem 9.141: 32
32
32
13.98800 10 kg m
20.55783 10 kg m
14.30368 10 kg m
x
y
z
I
I
I
Problem 9.165:
32
0
0.39460 10 kg m
yz zx
xy
II
I
PROBLEM 9.180 (Continued)
(b) To determine the direction cosines ,,
x
yz
of each principal axis, use two of the equations of
Eq. (9.54) and Eq. (9.57). Then
K1: Begin with Eqs. (9.54a) and (9.54b) with 0.
yz zx
II
11 1
111
()()()0
() ( )() 0
xxxyy
xy x y y
IK I
IIK
PROBLEM 9.180 (Continued)
Now
33
() 0
zz
IK
Substituting into Eq. (ii):
33
33
(0.39460 10 )( ) [(20.55783 20.58145) 10 ]( ) 0
xy
or
33
( ) 16.70618( )
yx
Using Eq. (9.57):
222
333
( ) [ 16.70618( ) ] ( ) 1
xxz
0
PROBLEM 9.181*
For the component described in Problems 9.145 and 9.149,
determine (a) the principal mass moments of inertia at the
origin, (b) the principal axes of inertia at the origin. Sketch
the body and show the orientation of the principal axes of
inertia relative to the x, y, and z axes.
SOLUTION
(a) From the solutions to Problems 9.145 and 9.149, we have
Problem 9.145: 32
26.4325 10 kg m
x
I
Problem 9.149: 32
2.5002 10 kg m
xy
I
32
31.1726 10 kg m
y
I
32
4.0627 10 kg m
yz
I
32
8.5773 10 kg m
z
I
32
8.8062 10 kg m
zx
I
PROBLEM 9.181* (Continued)
(b) To determine the direction cosines , ,
x
yz
of each principal axis, use two of the equations of
Eqs. (9.54) and Eq. (9.57). Then
1:K Begin with Eqs. (9.54a) and (9.54b).
11 1 1
1111
()()()()0
() ( )() () 0
xxxyyzxz
xy x y y yz z
IK I I
IIKI
Substituting:
333
111
[(26.4325 4.1443)(10 )]( ) (2.5002 10 )( ) (8.8062 10 )( ) 0
xyz
333
111
(2.5002 10 )( ) [(31.1726 4.1443)(10 )]( ) (4.0627 10 )( ) 0
xyz
Simplifying
PROBLEM 9.181* (Continued)
Simplifying
22 2
22 2
1.3405( ) ( ) 3.5222( ) 0
1.8005( ) ( ) 2.9258( ) 0
xy z
xy z
Adding and solving for 2
():
z
22
( ) 0.48713( )
zx
and then 22
2
( ) [ 1.3405 3.5222( 0.48713)]( )
0.37527( )
yx
x
Now substitute into Eq. (9.57):
22 2
22 2
( ) [0.37527( ) ] [ 0.48713( ) ] 1
xx x
or 2
( ) 0.85184
x
PROBLEM 9.181* (Continued)
Now substitute into Eq. (9.57):
22 2
33 3
( ) [ 2.5795( ) ] [0.071276( ) ] 1
xx x
(i)
or
3
( ) 0.36134
x
and
3
( ) 0.93208
y
3
( ) 0.025755
z
PROBLEM 9.182*
For the component described in Problem 9.167, determine (a) the
principal mass moments of inertia at the origin, (b) the principal axes of
inertia at the origin. Sketch the body and show the orientation of the
principal axes of inertia relative to the x, y, and z axes.
SOLUTION
(a) From the solution of Problem 9.167, we have
22
22
22
11
24
1
8
53
68
xxy
yyz
zzx
WW
I
aI a
gg
WW
I
aIa
gg
WW
I
aIa
gg
PROBLEM 9.182* (Continued)
(b) To determine the direction cosines , ,
x
yz
of each principal axis, use two of the equations of
Eq. (9.54) and Eq. (9.57). Then
1:K Begin with Eqs. (9.54a) and (9.54b).
11 1 1
1211
()()()()0
() ( )() () 0
xxxyyzxz
xy x y y yz z
IK I I
IIKI
Substituting
222
111
113
0.163917 ( ) ( ) ( ) 0
248
xyz
WWW
aaa
ggg
PROBLEM 9.182* (Continued)
Substituting
222
222
113
1.05402 ( ) ( ) ( ) 0
248
xyz
WWW
aaa
ggg
Adding and solving for 2
()
z
22
( ) 2.96309( )
zx
and then 22
2
( ) [ 2.21608 1.5( 2.96309)]( )
2.22856( )
yx
x
Now substitute into Eq. (9.57):
3:K Begin with Eqs. (9.54a) and (9.54b):
33 3 3
3333
()()()()0
() ( )() () 0
xxxyyzxz
xy x y y yz z
IK I I
IIKI
Substituting
222
333
113
1.11539 ( ) ( ) ( ) 0
248
xyz
WWW
aaa
ggg
PROBLEM 9.182* (Continued)
Adding and solving for
3
()
z
33
( ) 0.707885( )
zx
and then
33
3
( ) [ 2.46156 1.5( 0.707885)]( )
1.39973( )
yx
x
Now substitute into Eq. (9.57):
22 2
33 3
( ) [ 1.39973( ) ] [ 0.707885( ) ] 1
xx x
(i)
PROBLEM 9.183*
For the component described in Problem 9.168, determine (a) the
principal mass moments of inertia at the origin, (b) the principal
axes of inertia at the origin. Sketch the body and show the
orientation of the principal axes of inertia relative to the x, y, and z
axes.
SOLUTION
(a) From the solution to Problem 9.168, we have
I
I
g
I
I
4
18.91335
x
t
I
a
g
4
1.33333
xy
t
I
a
g
PROBLEM 9.183* (Continued)
Solving yields
12 3
2.25890 17.27274 19.08046KK K
(b) To determine the direction cosines , ,
x
yz
of each principal axis, use two of the equations of
Eq. (9.54) and Eq. (9.57). Then
1:K Begin with Eqs. (9.54a) and (9.54b):
11 1 1
1211
()()()()0
() ( )() () 0
xxxyyzxz
xy x y y yz z
IK I I
IIKI
Substituting
444
111
(18.91335 2.25890) ( ) 1.33333 ( ) 0.66667 ( ) 0
xy z
ttt
aaa
gag
PROBLEM 9.183* (Continued)
2:K Begin with Eqs. (9.54a) and (9.54b):
22 2 2
2222
()()()()0
() ( )() () 0
xxxyyzxz
xy x y y yz z
IK I I
IIKI
Substituting
44 4
222
[(18.91335 17.27274) ( ) 1.33333 ( ) 0.66667 ( ) 0
xy z
tt t
aa a
gg g
2
x
Now substitute into Eq. (9.57):
22 2
22 2
( ) [4.02304( ) ] [ 5.58515( ) ] 1
xx x
or 2
( ) 0.14377
x
and 22
( ) 0.57839 ( ) 0.80298
yz
222
( ) 81.7 ( ) 54.7 ( ) 143.4
xyz
PROBLEM 9.183* (Continued)
Simplifying
33 3
33 3
0.12533( ) ( ) 0.5( ) 0
0.11705( ) ( ) 0.62695( ) 0
xy z
xy z
Adding and solving for
3
()
z
33
( ) 0.06522( )
zx
and then
33
3
( ) [ 0.12533 (0.5)(0.06522)]( )
0.15794( )
yx
x
Now substitute into Eq. (9.57):
PROBLEM 9.184*
For the component described in Problems 9.148 and 9.170,
determine (a) the principal mass moments of inertia at the origin,
(b) the principal axes of inertia at the origin. Sketch the body and
show the orientation of the principal axes of inertia relative to the
x, y, and z axes.
SOLUTION
(a) From the solutions to Problems 9.148 and 9.170. We have
22
2
0.32258 kg m 0.41933 kg m
0.096768 kg m 0
xyz
xy zx yz
III
II I
Substituting into Eq. (9.56) and using
0
y z xy zx yz
II I I I
PROBLEM 9.184* (Continued)
Substituting
11
11
(0.096768)( ) (0.41933 0.22583)( ) 0
(0.096768)( ) (0.41933 0.22583)( ) 0
xy
xz
Simplifying yields
11 1
( ) ( ) 0.50009( )
y
zx
2:K Begin with Eqs. (9.54a) and (9.54b)
22 2 2
()()()()0
xxxyyzxz
IK I I
2222
() ( )() () 0
xy x y y yz z
IIKI
Substituting
222
(0.32258 0.41920)( ) (0.096768)( ) (0.096768)( ) 0
xyz
(i)
22
(0.96768)( ) (0.41933 0.41920)( ) 0
xy
(ii)
0
PROBLEM 9.184* (Continued)
3
:K
Begin with Eqs. (9.54b) and (9.54c):
3333
() ( )() () 0
xy x y y yz z
IIKI
Simplifying yields
33 3
() () ()
yz x
Now substitute into Eq. (9.57):
22
33
() 2[()] 1
xx
(i)
or
3
1
() 3
x
0