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(
)
mb
2
+
c
2
mab
mac
1
1
1
3
4
4
(
)
2
2
=−
1
mab
1
m a
+
c
1
mbc
4
3
4
(
)
mac
mbc
m a
2
+
b
2
1
1
1
4
4
3
2
a bc
a bc
a bc
m
,,
E
,,
,,
.
3
222
222
222
After adding corresponding matrix entries, we have for the inertia tensor of a box
(with side lengths a , b , and c ) about its center of mass
(
)
mb
2
+
c
2
0
0
1
12
(
)
2
2
=
0
1
12
ma
+
c
0
.
(
)
0
0
ma
2
+
b
2
1
12
Example 14.7. Find the inertia tensor of a dome, having constant density ρ and
semiaxis lengths a , b , and c as shown in Figure 14.8, in a coordinate system
where center of mass coincides with the origin.
c
a
b
Figure 14.8. The dome used in Example 14.7.
Solution. The total mass of the dome is
, and the z coordinate of the
m ρπabc
=
2
center of mass can be calculated using the integral
2
2
2
11
w
1
v
w
π
 
2
2
mC ρabc
=
w du dv dw
=
abc
,
(14.100)
z
4
2
2
2
01
−− −−−
w
1
v
w
where we have made the substitutions
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