Environmental Engineering Reference
In-Depth Information
d
L
d
t
body
+
ω
×
d
L
d
t
=
L
=
τ
(10.47)
and also
d
L
d
t
d
(
I
I
d
ω
)
d
t
body
=
body
=
,
(10.48)
d
t
body
since in this frame any matrix representation of
I
is time-independent. The
time-derivative of the angular velocity is actually the same vector in both frames,
because
d
d
body
=
ω
d
d
t
d
t
d
t
=
body
+
ω
×
ω
=
.
(10.49)
We now have all the elements we need to write the equations of motion in the
body-fixed frame. Using Eq. (10.46) - (10.49) we arrive at
τ
=
ω
+
ω
×
ω
=
ω
+
ω
×
I
(
I
)
I
L
.
(10.50)
Even though we have dropped our explicit denotation of the frame of reference,
it is crucial to remember that this equation is generally valid only in a body-fixed
frame since
I
must be independent of time.
Note that for the special case of rotation at constant angular velocity, i.e.
ω
=
0,
Eq. (10.50) gives
τ
=
ω
×
L
.
(10.51)
If the body is also rotating about a principal axis then we have seen that
L
and
ω
will be parallel, in which case
0 and so no torque is required (which confirms
the result from the last section). With Eq. (10.51) we now have a more direct
method to address problems like the one posed in Example 10.4.1. You might like
to check that you can obtain Eq. (10.45) by computing
τ
=
L
.
At this point we are still free to choose a set of coordinate axes in the body-fixed
frame. Taking the basis vectors to be along the principal axes we have
ω
×
e
1
e
2
e
3
ω
1
ω
2
ω
3
I
1
ω
1
I
2
ω
2
I
3
ω
3
ω
×
L
=
.
(10.52)
Expanding the determinant and taking components of Eq. (10.50) gives us the three
equations:
I
1
˙
ω
1
+
(I
3
−
I
2
)ω
2
ω
3
=
τ
1
,
I
2
˙
ω
2
+
(I
1
−
I
3
)ω
3
ω
1
=
τ
2
,
I
3
˙
ω
3
+
(I
2
−
I
1
)ω
1
ω
2
=
τ
3
.
(10.53)
