Environmental Engineering Reference
In-Depth Information
moment of inertia about that axis, then
L
=
I
ω
=
ω
I
ˆ
α
=
I
α
ω
ˆ
α
,
(10.30)
i.e.
I
ˆ
α
=
I
α
ˆ
α
.
(10.31)
So, the vector ˆ
by a constant
but doesn't alter its direction. Eq. (10.31) is known as an eigenvalue equation and
ˆ
α
is special in that operation on it by
I
multiplies ˆ
α
α
is said to be an eigenvector of
I
.
I
α
is the principal moment of inertia about
the axis ˆ
α
, and it is the corresponding eigenvalue. In matrix form Eq. (10.31) is
written
I
11
I
12
I
13
I
21
I
22
I
23
I
31
I
32
I
33
α
1
α
2
α
3
α
1
α
2
α
3
=
,
I
α
(10.32)
where
(α
1
,α
2
,α
3
)
are the components of the
α
. Rearranging Eq. (10.32) gives us
=
I
11
−
I
α
I
12
I
13
α
1
α
2
α
3
I
21
I
22
−
I
α
I
23
0
.
(10.33)
I
31
I
32
I
33
−
I
α
A non-trivial
1
solution to this equation exists only if the matrix multiplying the
column vector has no inverse, that is if
I
11
−
I
α
I
12
I
13
I
21
I
22
−
I
α
I
23
=
0
.
(10.34)
I
31
I
32
I
33
−
I
α
Equating the determinant to zero
2
generally gives rise to a cubic equation in
I
α
,
known as the characteristic equation, which can be solved to obtain three possible
values of
I
α
. The solutions to the characteristic equation are the principal moments
of inertia. Then, each solution for
I
α
may be substituted in turn into Eq. (10.33)
to obtain simultaneous linear equations that can be solved to give the eigenvector
components
α
1
,α
2
,α
3
. Let us now look at an example of this procedure.
Example 10.3.1
Determine the principal axes of a solid cube of side b and mass
M for rotations about a corner.
1
We can get what is called a trivial solution by setting ˆ
α
=
0
but this does not determine a direction
for a principal axis.
2
We are assuming a certain familiarity with linear algebra and refer to any number of mathematics
textbooks for further details.
