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PROBLEM B.68
Given a homogeneous body of mass
m
and of arbitrary shape and three rectangular axes
x
,
y
, and
z
with
origin at
O
, prove that the sum
Ix
+
Iy
+
Iz
of the mass moments of inertia of the body cannot be smaller
than the similar sum computed for a sphere of the same mass and the same material centered at
O
.
Further, using the result of Problem 9.176, prove that if the body is a solid of revolution, where
x
is the
axis of revolution, its mass moment of inertia
Iy
about a transverse axis
y
cannot be smaller than 3
ma2
/10,
where
a
is the radius of the sphere of the same mass and the same material.
PROBLEM B.68 (Continued)
(ii) First note from Figure 9.28 that for a sphere
2
2
5
xyz
I
II ma
body 6
y
or 2
body 3
( ) Q.E.D.
10
y
Ima
PROBLEM B.69*
The homogeneous circular cylinder shown has a mass
m
, and
the diameter
OB
of its top surface forms 45 angles with the
x
and
z
axes. (
a
) Determine the principal mass moments of
inertia of the cylinder at the origin
O
. (
b
) Compute the angles
that the principal axes of inertia at
O
form with the coordinate
axes. (
c
) Sketch the cylinder, and show the orientation of the
principal axes of inertia relative to the
x
,
y
, and
z
axes.
2
1
2
22
zx z x aa
I I mz x m ma
Substituting into Equation (9.56)
322
13 3 13
12 2 12
KmaK
22
222
13 3 3 13 13 13 1 1 1 ()
12 2 2 12 12 12 2
22 22 ma K
22
223
13 3 13 13 1 3 1
12 2 12 12 2 2
22
13 1 1 1 1
2()0
12 2
22 22 22
ma
PROBLEM B.69* (Continued)
Simplifying and letting 2
Kma
yields
11 565 95 0
12
The principal moments of inertia are then
2
10.363Kma
2
21.583Kma
2
31.720Kma
() 0
xxxyyzxz
IK I I
(9.54a)
xyz
(Note: Since ,
x
yyz
I
I Equations (9.54a) and (9.54c) were chosen to simplify the “elimination” of
y
during the solution process.)
Substituting for the moments and products of inertia in Equations (9.54a) and (9.54c)
222
13 1 1 0
12 2
22
ma K ma ma
222
11130
212
22
ma ma ma K
and 11 13 0
xy z
12 2 12
PROBLEM B.69* (Continued)
Thus, a third independent equation will be needed when the direction cosines associated with 2
K
are determined. Then for 1
K and 3
K
12 2 2 12
or zx
12 2
22
12
xxx
or
28 1
(iii)
1
2 8 0.363383 ( ) 1
xz
and then 17
( ) 2 2 0.363383 (0.647249)
y
12 x
or 33
( ) ( ) 0.284726
xz
xz y
PROBLEM B.69* (Continued)
K
2
: For this case, the set of equations to be solved consists of Equations (9.54a), (9.54b),
and (9.57).
Now
0
xy z
ma ma K ma
13 1
2
22 22
Substituting the value of
2
into Eqs. (i) and (iv):
13 19 1 1
() () () 0
12 12 2
22
or
222
1
() () () 0
2
xyz
2
2
y
and then
22
() ()
yx
Substituting into Equation (9.57)
22 2
222
xyz
(
c
) Principal axes 1 and 3 lie in the vertical plane of symmetry passing through Points
O
and
B
.
Principal axis 2 lies in the
xz
plane.
0
PROBLEM B.70
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.
32
0.39460 10 kg m
xy
I
Eq. (9.55) then becomes
0
xxy
IK I
Substituting: 32
114.30368 10 kg mK
or 23 6
(34.54583 10 ) 287.4072 10 0KK
and 32
320.58145 10 kg mK
or 32
320.6 10 kg mK
PROBLEM B.70 (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
yz zx
111
xy x y y
11
(0.39460 10 )( ) [(20.55783 14.30368) 10 ]( ) 0
xy
Adding yields 11
() () 0
xy
xyz
K2: Begin with Eqs. (9.54b) and (9.54c) with 0.
yz zx
II
xy x y y
Substituting into Eq. (i):
33
2
(0.39460 10 )( ) [(20.55783 13.96438) 10 ]( ) 0
xz y
x
and 2
( ) 0.059740
y
22 2
( ) 3.4 ( ) 86.6 ( ) 90.0
xy z
33
()()0
zz
IK
0 0
PROBLEM B.70 (Continued)
Now
33
() 0
zz
IK
The differences in the above values are due to round-off errors.
(
c
) Principal axis 1 coincides with the
z
axis, while principal axes 2 and 3 lie in the
xy
plane.
PROBLEM B.71*
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.
x
xy
32
31.1726 10 kg m
y
I
32
4.0627 10 kg m
yz
I
2226
2
2
(2.5002) (4.0627) (8.8062) ](10 )
[(26.4325)(31.1726)(8.5773) (26.4325)(4.0627)
(31.1726)(8.8062) (8
K
2
.5773)(2.5002)
Solving yields
32
14.1443 10 kg mK
32
1
or 4.14 10 kg mK
332.2541 10 kg mK
3
