Environmental Engineering Reference
In-Depth Information
moment of inertia tensor
I
and write
=
ω
L
I
,
(10.14)
which is now an expression completely independent of our choice of co-ordinate
system. The tensor
I
may be thought of as a geometrical object that is able to map
the vector
on to the vector
L
. The operation that
I
performs generally changes
both the length and the direction of the vector
ω
ω
on which it is operating. That
I
can change the direction of
means that
I
itself has directional properties, e.g. it
tells us how the
e
1
component of
L
depends upon the
e
2
component of
ω
, through
I
12
. The moment of inertia tensor has two spatial indices and it is accordingly
known as a rank 2 tensor. Incidentally, scalars and vectors may also be thought
of as tensors: a vector has one spatial index and is a rank 1 tensor; a scalar has
no directional dependence and is a rank 0 tensor. Tensors of higher rank also exist
(and are used in General Relativity) but in this chapter we will need nothing higher
than rank 2. Note that we do not use any special typesetting to distinguish
I
as a
rank 2 tensor, but that should not cause any confusion. It is important to realize that
both the angular momentum and the moment of inertia tensor are defined
relative
to an origin
and so we should always speak of “the moment of inertia about a
point”. This is in contrast to the simpler treatment in Chapter 4, where we were
only ever interested in the moment of inertia for rotations about some axis, and the
component of the angular momentum about the same axis. As we shall very soon
see, the moment of inertia about an axis is something that lives within the moment
of inertia tensor - the latter being the more general object.
As soon as we choose a co-ordinate system with associated basis vectors we can
express the tensor
I
as a matrix. Thus the moment of inertia tensor
I
is represented
in the basis
(
e
1
,
e
2
,
e
3
)
as the matrix
ω
I
11
I
12
I
13
I
21
I
22
I
23
I
31
I
32
I
33
.
I
=
(10.15)
In the same basis you might like to check that we can express Eq. (10.13) as
multiplication of the column vector
ω
by the matrix
I
to give the column vector
L
:
L
1
L
2
L
3
I
11
I
12
I
13
I
21
I
22
I
23
I
31
I
32
I
33
ω
1
ω
2
ω
3
=
.
(10.16)
Let us for a moment consider the diagonal elements of the moment of inertia
tensor. The first of these is
m
α
r
α
δ
11
−
r
α
1
.
I
11
=
(10.17)
α
By Pythagoras' Theorem,
r
α
=
r
α
1
+
r
α
2
+
r
α
3
,also
δ
11
=
1, and we can write
m
α
r
α
2
+
r
α
3
.
I
11
=
(10.18)
α
