00:03
So we know that w3 is a constant.
00:12
So the other euler equations are w1 is equal to negative i3 minus i over i, all times w3, all of that times w2.
00:28
And then we know that w2 is equal to negative i3 minus i over i times w3, all of that times w1.
00:38
So solving these and by already making the substitution because we are dealing with constants, we get i3 minus i over i is equal to w, nope, sorry, times w3.
01:01
So w1 plus i, w2 minus i w2 minus i w1 plus i w2 is equal to zero and so we need to let or be known that q is equal to w1 plus iw2 so that means q with a dot is equal sorry minus i q that equals zero and so that will equal qt is equal to a, e to the i, t powers.
01:56
So then we know that w1 plus i, w2 is equal to a cosine plus ia sine of the same thing.
02:12
So w1 of t is a cosine of the sky, and then w2t is equal to a sign of the same thing.
02:28
So the x3 axis is the symmetry axis of the body.
02:33
So the angular velocity vector processes about the body x3 axis with the constant angular velocity.
02:39
So that is shown as over iw3.
02:48
So that moves to i3w3 over i1 cosine of theta.
03:12
So in order to prove this, we have to look at the two cone figure angular moment and components, where l is directed along the vertical space axis, and theta is the angle between the space and body vertical axis.
03:28
So l1 is 0, and l2 is l sine of theta, and then l3 is l cosine of theta.
03:40
So if a is the angle between w and the vertical body axis, then we have w1 is also 0, and w2 is w sine of a, and w3 is w cosine of a.
04:06
So they're essentially the same intertwined components.
04:13
So the angular momentum components in terms of a are l1 is i1w1 equals 0 to where l2 is i1, w2 is equal to i1, w2 is equal to i1, w sign of and l3 is 3w3 equals i3w cosine of a.
04:42
So using the euler angles in the body frame, we find that w2 is equal to this sign of theta cosine, not good at drawing greek letters, minus theta sine of that guy.
05:05
That w2 is equal to this sign of theta.
05:11
So this is essentially w2 over sine of theta, which equals w sign a over sign of theta, which also equals w times l, sorry, i2 over iw, l over l2, l over l2 equals l over i1.
05:43
So the simplest way to show would be sign of theta prime equals this over this times sine of theta times the derivative two times may be constructed by using the cross product of w times l and w times x cubed...