Environmental Engineering Reference
In-Depth Information
and that leads to a quadratic rather than cubic characteristic equation that is
easier to solve. However, for the square we do not need to do even that because the
diagonal of the square is an axis of reflection symmetry. We have seen that, if we
choose a basis such that one of the basis vectors lies along the symmetry axis then
the corresponding products of inertia vanish (e.g. see Example 10.2.2) and that fact
alone is sufficient to guarantee that the axis is also a principal axis. In the case
of the square plate it therefore follows that the diagonal is a principal axis, call it
e
1
. The third axis now comes for free, because it must be orthogonal to the other
two axes. In a right-handed co-ordinate system the three principal axes of a square
plate are thus as shown in Figure 10.9.
e
2
e
3
e
1
Figure 10.9 The principal axes of a square plate for rotations about a corner. The
e
3
vector
points out of the page.
10.4 FIXED-AXIS ROTATION IN THE LAB FRAME
All that was rather technical but now we are ready to start analysing the general
motion of rotating bodies. By the end of this chapter, we shall have succeeded in
understanding what happens to an object thrown through the air (it wobbles and
spins), how a gyroscope works and why some rotations of a tennis racquet are safer
than others! In this section we “warm up” by re-examining the simpler instance of
fixed-axis rotation.
First we shall consider the case where the fixed axis just happens to also be a
principal axis,
e
3
say. In this case we can write
L
=
I
3
ω
e
3
,
(10.39)
and crucially
L
is also parallel to the
x
3
axis and the problem maps onto the more
familiar one-dimensional one we met in Chapter 4, i.e.
L
=
I
3
ω,
(10.40)
where
d
Vρ(
r
)(x
1
+
x
2
).
I
3
=
(10.41)
V
Note that in order to maintain rotation only in the
e
3
direction any torque
τ
that
acts must be parallel to the
x
3
axis, then we can write
τ
=
τ
e
3
(10.42)
