Environmental Engineering Reference
In-Depth Information
to the lab axes. Using Eq. (10.66) and Eq. (10.14) we obtain
L
=
I
ω
,
=
IA
cos
(t
+
φ)
e
1
+
IA
sin
(t
+
φ)
e
2
+
I
3
ω
t
e
3
.
(10.69)
Now
L
·
ω
is also a constant since, using Eq. (10.66) and Eq. (10.69), we
have that
IA
2
I
3
ω
t
L
·
ω
=
+
(10.70)
and the right-hand side is manifestly constant. Since Eq. (10.70) involves the scalar
product of two vectors it is independent of the co-ordinate system, and so must
also be true in the lab, even though we have calculated it in the body-fixed frame.
As an aside, we can show that
L
is also constant for free rigid bodies
even when they do not have an axis of symmetry. To prove this, we work in
the centre-of-mass frame and write the rotational kinetic energy of the particles
making up the rigid body as
·
ω
1
2
m
α
v
α
,
T
=
α
1
2
m
α
(
=
ω
×
r
α
)
·
v
α
.
(10.71)
α
Rearranging the triple scalar product we obtain
1
2
m
α
ω
·
T
=
(
r
α
×
v
α
),
α
1
2
ω
·
=
m
α
(
r
α
×
v
α
),
α
1
2
ω
·
=
L
.
(10.72)
If there is no net torque, there is no work done to rotate the body about its centre of
mass and the rotational kinetic energy must therefore be conserved. Hence,
L
·
ω
is constant.
Returning to the free symmetric top, we have already shown that the magnitudes
of
and
L
are both constant. Constant
T
implies that there must be a fixed angle
between
ω
ω
and
L
.Since
L
is a constant vector in the lab frame the most
ω
can do
in the lab is rotate about
L
maintaining a constant angle to it.
The result that there is a constant angle between
L
and
is not quite enough
to tell us exactly what the body is doing. What we really need to figure out is
what happens to the vector
e
3
(the symmetry axis of the body) in the lab frame.
Fortunately, we can show that
e
3
also lies in the plane defined by
L
and
ω
ω
by
constructing the vector
