At the end of Part 1 we had obtained 2 coupled differential equations:
I1 w1' + (I3 - I2 ) w3w2 = 0
I2 w2' + (I1 - I3 )w1w3 = 0
Where: b = (I3 - I2 )/ I1 = - (I1 - I3 )/ I2
And:
w1 = A1 e pt
w2 = A2 e pt
Substituting:
1)' I1 A1 p e pt + (I3 - I2 ) w3 (I3 - I2 ) w3 = 0
2)' I2 A 2 pe pt - (I1 - I3 ) w3 A1 p e pt = 0
Then take:
I2 A 2 x (1)' and: I1 A 1 x (1)'
and subtract:
Þ
A22 I2 p e pt (I3 - I2 ) w3 + (I1 - I3 ) I1 A22 w3 = 0
Now, if I1 = I2
Þ
A22 b + A12 b = 0
Or:
A 2 = Ö - A12 = + iA1
And: P = -(I3 - I2 )/I1 { A2 / A1 }
Þ
P = - b w3 + i = + i b w3
Þ
w1 = A1 exp [ + i b w3 t ]
w2 = A2 exp [ + i b w3 t ]
Where:
w1Re = A cos (b w3 t + f)
w2Re = A sin (b w3 t + f)
Þ
A cos (b w3 t + f)e1^ + A sin (b w3 t + f) e2^ + w3 e3^
Pictorially w precesses about the w3 axis as shown in sketch of cone:
For the space cone half angle:
as = = cos a s · (w) w3
cos a s = /w w3
tan a s = A /w3
Thus, for Earth, the angular momentum vector is a constant in space, i.e. 0 =
For the half angle of the precession body cone:
cos ab =
ab =
We need to obtain space cone half angle ( as ) in terms of: ab and w . Note for the inertia tensor:
w = w1 e1^ + w2 e2^ + w3 e3^ = w N
(w · L) =
After much algebra:
as =
= I1 w 2 (1 + b cos 2 g) / (wI1 )(1 + 2b cos 2 g + b 2cos 2 g) 1/2
= I1 w 2 (1 + b cos 2 ab ) /(wI1)(1 + 2b cos 2 ab +b 2cos 2 ab) 1/2
Note here that b = (I3 - I1 )/ I1 can be greater or less than zero. For b > 0, space cone is inside the body cone, then:
L = w I (n^ + b cos ab e3^)


No comments:
Post a Comment