Environmental Engineering Reference
In-Depth Information
substituting the eigenvalue back into the eigenvalue equation. So, if we denote an
eigenvector by
ˆ
β
then
−
3
−
3
−
3
β
1
β
2
β
3
=
−
3
−
3
−
3
0
,
−
3
−
3
−
3
which yields the equation
β
1
+
β
2
+
β
3
=
0
.
(10.37)
This is the equation of a plane that lies perpendicular to the direction of the first prin-
cipal axis. To see this notice that we can write Eq. (10.37) as
0
. Eq. (10.37)
admits an infinity of solutions, all corresponding to vectors that lie in the plane
perpendicular to
ˆ
β
·
α
=
ˆ
. However, only two vectors are needed in order to form a basis
in the plane (i.e. any other vector can be written as a linear combination of the
original two). We are free to choose any such pair of vectors and, rather arbitrarily,
we pick the first to be
α
1
−
1
√
2
11
12
Mb
2
.
ˆ
;
β
=
1
0
I
β
=
This vector clearly has components that satisfy Eq. (10.37) and we have set the
length to unity through the choice of the factor of
1
√
2
. The third principal axis is
now fixed (up to an overall sign) by the requirement that it is perpendicular to the
other two (in order that the principal axes should form an orthonormal basis). You
should be able to show that
1
1
−
1
√
6
11
12
Mb
2
.
;
γ
=
ˆ
I
γ
=
2
,
ˆ
To finish off we can write the moment of inertia tensor in the basis (
ˆ
α
β
,
ˆ
γ
) as
12
Mb
2
20 0
011 0
0011
1
.
I
=
The directions of the principal axis vectors are shown in Figure 10.8.
Notice that when we considered the moment of inertia of the cube about its
centre, we showed that the moment of inertia tensor, Eq. (10.27), was diagonal in
a co-ordinate system aligned with the edges of the cube. Thus the basis vectors
e
1
,
e
2
and
e
3
already defined a set of principal axes and we left the matter there. On
the other hand, when we considered rotation about a corner, one principal axis lay
along the diagonal and the other two lay anywhere in the plane perpendicular to
the diagonal. But one ought really to recognize that for the cube rotating about its
