Environmental Engineering Reference
In-Depth Information
Solution 10.3.1
We have already calculated the moment of inertia tensor for a cube
about a corner. It is
12
Mb
2
−
−
8
3
3
1
.
−
−
38
3
−
−
3
38
So that the characteristic equation for the eigenvalues may be written
I
8
−
−
3
−
3
I
−
38
−
−
3
=
0
.
(10.35)
I
−
3
−
38
−
I
We have simplified notation with the substitution I
=
12
Mb
2
,whereI denotes an
eigenvalue of
I
. Expansion of the determinant in Eq. (10.35) gives
1
I
)
[
(
8
I
)
2
I
)
I
)
]
(
8
−
−
−
9]
+
3[
−
3
(
8
−
−
9]
−
3[9
+
3
(
8
−
=
0
.
This cubic equation can be put into the form
I
)(
11
I
)(
11
I
)
(
2
−
−
−
=
0
from which we have I
=
2
or 11 (twice). The principal moments of inertia are
6
Mb
2
and
1
12
Mb
2
. Our next task is to figure out the corresponding eigenvectors.
We start with the eigenvalue equation for I
=
1
I
=
2
:
8
−
2
−
3
−
3
α
1
α
2
α
3
=
−
38
−
2
−
3
0
−
3
−
38
−
2
from which we obtain two independent equations:
6
α
1
−
3
α
2
−
3
α
3
=
0
,
−
3
α
1
+
6
α
2
−
3
α
3
=
0
.
(10.36)
Note that we need three equations to completely determine the eigenvector and
we have only two. That, however, is not surprising since any eigenvector can be
multiplied by an arbitrary constant and it will remain a solution to the eigenvalue
equation. Thus we are free to fix the overall normalization of the eigenvectors. It is
not difficult to show that the solution to Eq. (10.36) must satisfy α
1
=
α
2
=
α
3
.Now
is a unit vector then α
1
+
α
2
+
α
3
=
if we insist that
ˆ
α
1
and we can determine the
direction of one of the principal axes, i.e.
1
1
1
1
√
3
1
6
Mb
2
.
;
α
=
ˆ
I
α
=
We must now address the solutions corresponding to I
=
11
. The fact that this
solution occurs twice results in only one independent
linear equation upon
