So good, analytic one, this is pretty straightforward. Actually, you do something on this in the homeworks, as well. So if we have torque free motion, we've assumed the principal frame. These are the three equations you have. The only thing I've dropped was the l1, 2, 3, there's no external torque. So this is a pure spin, you take an object, and you spin it up, what happens? But in particular now, we want to look at an axisymmetric case. So here we're going to assume this diagonal still. And if we assume it's axisymmetric, like a cylinder, rocket bodies. Many spacecraft kind of have axisymmetry just to coming out of the launch vehicle, to have to get through the atmosphere. So it is a common shape assumption. We picked b1, b2 to be a transverse and then b3 is the axis of symmetry. So I1 and I2 have to be equal, I'm just calling it It, right? Those are the two transverse axes. If you do that, and plug it in, the last one that was I2 minus I1, or I1 minus I2, that's going to go to 0. So immediately we have an omega 3 dot. Whatever spin you give it about the axis of symmetry, that spin rate just stays a flat line. Doesn't change. The other two will actually change and they are the function of the inertia about the symmetry axis and the transverse inertia, and of course, initial conditions. So now we have two equations and two unknowns. This one falls out because omega 3 dot, that makes omega 3 a constant. That means up here omega 3 is simply the initial omega 3. It's no longer not known, it's a known quantity by initial conditions. So, we're left with omega 1, and omega 2 and we've got two first order differential equations. How do we solve this? This is one of the cases when you're trying to solve these things, you try to get rid of the dots. And the easiest path is not to get rid of the dots is actually to add more dots and that gives you an answer that we can find in the end. If I differentiate this equation this is shouldn't be bold actually, unbold. But just take the dots. Inertia is a constant. Omega 2, omega 3's a constant, so there's no omega 3 dot, but omega 2 could vary. You can see omega 1 only depends on omega 2, and if I go back a slide, the omega 2 dot only depends on omega 1, again, treating omega 3 as a constant. So if I differentiate 1, I now get an omega 2 dot, which depends on omega 1. And you can plug that in and you end up with this equation, that's omega 1 double dot plus something times omega 1 equal to 0. Which for engineers, like yes, I knew there'd be a spring-mass system somewhere, right. This is a typical spring-mass system. You know how to find these answers, and they will give you the typical sines and cosine results. So, x double dock + kx can you solve that differential equation? That's what it boils down to, for an axis symmetric case. You can do the same things for omega 2, it gives you the same kind of a constant Inertia's a constant, omega 3 is a constant, so all of this is nothing but k in the end, right? If you did a spring-mass analogy. So for axi-symmetric cases, it's easy to do. Then, I think in the homework you do this yourself. Show me that you remember how to solve simple first order differential equations. You plug in the conditions, come up with these coefficients. And this one here is called omega p. I can rewrite it, and so now it has to do with this. What's interesting here is is I3 is for example less than IT. That's this kind of stuff. The inertia about the transverse on this pen is much bigger than the inertia about the symmetry axis all right. So in that case this is a ratio less than 1 minus 1, it's going to give you omega 3 which is positive. That's going to give you now a negative omega p. We saw on the pole hold plus does momentum spheres and energye lipsoids if I'm doing a spin near the axis of least inertia, that was the i-3 axis, so those wobbles up there. We would be orbiting in a negative sense, and that comes from the sine right here. If we have I3 being the axis of largest inertia, so we have just a plate. Think of a big, flat plate basically spinning, right, and you're off slightly. Now I3 is actually, so it's oproid body, I3's bigger than the transverse inertia, and this becomes more than 1, then -1 is still positive. That's why there we saw those curves orbiting in a positive sense. But this is the classic answer that you have for a rigid body, no torques, but it's axis symmetric. So we make that one assumption, and this is just a classic result. Make sure you can derive this. This is pretty straightforward. Here are some answers. If you integrate the equations, you would see, just using those differential equations we talked, i omega dot plus omega ~ i omega equal to 0, in this case. And you integrate this stuff, you would get these curves, which then show you yes, as predicted, I get a sine Like curve with the face shift, or a cosine in some cases. And you get [INAUDIBLE] in omega. That's for the rates. If you do the attitude motion, this is the same thing. I'm just showing here all your angles. Again, people, try to be diverse. Something non-MRP. We'll get to those a lot more later. So what happens is, the attitude motion is very difficult to predict. So for torque-free motions, the parts that are easy are the spin parts with this equation. And we can predict spin stability, but we don't necessarily have pointing stability. That's a whole nother thing. And we'll get to that in three-axis control. But if you need spin stability, that's where these kinds of analysis really help.
