Problem: Space and Body Cones for the torque free top
(a)
One way to do this is as follows. Designate the body frame unit vectors
as
i
,
j
,and
k
, and assume they lie along the principal axes of the body. In the
body principal axis frame we can write the angular momentum vector
L
as
L
=
I
1
ω
1
i
+
I
2
ω
2
j
+
I
3
ω
3
k
,
(1)
so that
L
·
k
=
L
3
=
I
3
ω
3
=
L
cos
θ,
(2)
and

L
×
k

=
L
sin
θ
=

L
2
i
−
L
1
j

=
I
1
q
ω
2
1
+
ω
2
2
=
I
1
A,
(3)
where
θ
is the Euler angle between the space and body
z
axes and the compo
nents of the angular velocity vector are
ω
1
=
A
cos
Ω
t
and
ω
2
=
A
sin
Ω
t
.T
h
e
s
e
two results give us
tan
θ
=
I
1
A/
(
I
3
ω
3
)
.
(4)
If we repeat the same operations with the vectors
ω
and
k
,we
f
nd
ω
·
k
=
ω
cos
θ
00
=
ω
3
,
(5)
and

ω
×
k

=
ω
sin
θ
00
=

ω
2
i
−
ω
1
j

=
q
ω
2
1
+
ω
2
2
=
A,
(6)
from which we have
tan
θ
00
=
A/ω
3
.
(7)
From the ratio of Eqs.(4) and (7), we
f
nd the desried result
tan
θ
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 Fall '10
 Wilemski
 mechanics, Angular Momentum, Body Cone, space cone

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