0:05

Torque free motion.

Â Momentum has to be constant.

Â The momentum magnitude has to be constant,

Â which is what I'm writing here in the first equation.

Â And kinetic energy has to be constant.

Â So we're not considering actual mechanical energy loss right now.

Â Like, before, like the Paul will plots, let's assume T and

Â H are perfectly preserved.

Â You can write the momentum at energy constraints in terms of your omegas.

Â H was I omega, so that for the magnitude is the component squared.

Â That's easy, magnitude squared.

Â And kinetic energy was 1/2, so the 2 comes to the left-hand side.

Â This one was just to make a transpose I Omega, that I is a diagonal.

Â Again, makes it very easy.

Â This is the form that you would have.

Â Now these are two constraints.

Â And we have three variables.

Â And quickly this illustrates again, that the omega curve that you're

Â going to get isn't going to be a plane or a volume, the solution space.

Â It's going to be one dimensional curve in three dimensional omega space, right?

Â That's what we're looking for.

Â We have three unknowns, omega one, two and three and two constraints.

Â So I can always use these two constraints and that's what I'm laying out here.

Â I say, okay well let me just pick to write everything in terms over omega 1.

Â Why? because it's the first thing one that

Â appears.

Â So I can use this and solve for omega 2 squared and

Â that's what you have in this equation.

Â And I can solve this for omega 3 squared which comes up here.

Â So you've got two equations and two terms, omega one and

Â two that I'm solving for, so just rule any algebra and you reduce it.

Â So, now because of these constrains if I have an omega one and I know my momentum

Â and I know my energy added at the beginning, I can quickly tell you what

Â the magnitude squared of omega two has and the same thing for omega three, all right?

Â So it's just the way of re-writing momentum and

Â energy constrains exclusively in omega terms.

Â 1:58

Now what makes omega 1 special?

Â Nothing really.

Â There's nothing special about omega 1.

Â You could've just as easily solved everything in terms of omega 2 or

Â you could've used in two equations and say, well,

Â I'm going to solve everything for omega 1 and 2 in terms of omega 3, right?

Â You can solve this any which way you wish.

Â And substitute them again.

Â I will use all of them.

Â This was in the paper by Junkins, Jacobson,

Â Blanton that you've got a reference down here.

Â It's actually from 1973.

Â Most of you weren't even born.

Â 2:30

But it's an eloquent result here and you will see,

Â this has some analogy to the axisymmetric cases.

Â But it gives you a more general formulation that you don't often see.

Â So this page, all we've done is we've taken momentum energy equations and

Â we've used it to express the omegas in terms of momentum and energy and

Â one of the other omegas, that's it.

Â So let's go back to the original problem.

Â This is the one we're trying to solve.

Â We've got these differential equations.

Â There's no L's on the right-hand side because there's no torque.

Â How do we solve this?

Â Before with axis symmetry, one of them was 0.

Â So we could say omega 3 is fixed and therefore we took a derivative of this,

Â substitute into the others, right?

Â And came up with nicely decoupled ones.

Â Now everything varies with time and I can no longer do that trick directly.

Â 3:18

But there's still similar steps.

Â We're going to start out with the same kind of a step.

Â We have three fully coupled and now non-linear differential equations,

Â whereas before I had two coupled linear differential equations.

Â I'm still going to take a derivative just because it worked so well for the axis and

Â metric let's try it here, right.

Â And you do this, and then with chain rule, Omega is Omega 2's 3's Omega 1, everything

Â varies with time, and here you go, to make a 1, 2, 3 double outs divided by inertia.

Â That's basically it.

Â So you're going okay, great, where is this going?

Â So you took a derivative again.

Â Now similar as to with the axis symmetric case, omega 1 dot for

Â the axis symmetric only depended on omega 2 because omega 3 was constant.

Â Well, what we're going to do now is we plug in the second and

Â third differential equation into here.

Â 4:11

And rewrite this all and

Â you end up with this expression here that you going to have.

Â So omega 1.., is equal to omega 1 omega 2 squared.

Â And omega 1 omega 3 squared.

Â As you can see there's still quite coupled.

Â There's still non-linear Just I went from three non linear first order differential

Â equations to three non linear second order differential equations.

Â Seems like we're going backwards, we're not making much progress.

Â But here's the elegant thing now, from the prior slide,

Â 4:42

I have for example omega 2 squared In terms of just a constant and

Â something times omega 1.

Â And the same thing for omega 3 squared.

Â So if you look at the second order differential equations,

Â here is omega 1, omega 1.

Â I can write omega 2 squared in terms of a constant and omega 1.

Â And the same thing for omega 3 squared.

Â So by using the momentum and the kinetic energy constraints,

Â I can substitute those in and instead of having time varying quantities,

Â I have everything in terms of constants of energy and consular momentum magnitude.

Â And what you can do for one, you can follow the same process here.

Â We use here the expressions where omega 1 and 3 are written in terms of omega 2.

Â And here we use the third set that writes omega 1 and 2 in terms of omega 3, right.

Â There's three different ways we could take two constraints and solve for

Â two to three variables.

Â If you do that and plug them in, so that's the process,

Â those equations we found, and you plug them in here.

Â And then you do the same thing for the other two look what happens in the end.

Â This is omega 1, just a bunch of constants.

Â Omega 1 times constant something times omega 1 squared.

Â So you're going to get omega 1 term and

Â then something times omega 1 squared will give you omega 1 cubed essentially.

Â So the final form for everyone of these omegas, I can write it as a,

Â what's called a homogeneous undamped Duffing equations.

Â Homogeneous just means the right hand side is 0.

Â 6:19

Undamped, there's no omega dot.

Â Just omega double dots and omegas.

Â And the Duffing equation, is this classic, this would be a spring mass equation.

Â The Duffing equation is a spring mass with a cubic Stiffness term added to it, so

Â like a cubic spring instead of a linear spring.

Â It's a common modeling thing to see as well.

Â So, all three of the omegas actually have an equation of this form.

Â But you would have these As and Bs which depend on omega 1, 2 and 3, and

Â you can see, the inertias are computed differently, and

Â when you plug it in, energy and momentum come in with different terms so

Â we have formulas for the As and Bs have to be.

Â So this is actually very elegant because we often see systems that are,

Â we end up with x double dot plus kx, right?

Â 7:09

If it's x double dot plus k equal to 0,

Â it should be trivial to talk about linear stability, at least.

Â You've linearized the system for small motions locally.

Â If k is positive, we know we'll have a restoring force let me just draw

Â that out because we're about to mute that.

Â So if you, If I have M.

Â Forget M, we'll just do normalized.

Â 7:49

And it's a linear relationship.

Â So if you're too far ahead, the force is going to restore you back to the origin.

Â If you're too far behind, and x is negative,

Â -k be the positive value that then makes it positive, you can catch up.

Â This gives you the typical oscillator equation that you'd have, but

Â it's a stable motion.

Â 8:06

So that works.

Â But this result is typically due to a linearization that you get.

Â Here have we linearized anything when we derived this?

Â No, we just used full nonlinear differential equations,

Â differentiated them, substituted in energy momentum, and

Â you end up with three equations, we've a first order term and a cubic term.

Â The cubic term 2, if you do structures, you often see this cubic term.

Â But it's the result of an approximation.

Â Once you go beyond the first order stiffness,

Â the next term people typically include is the cubic term.

Â If the fractions get really big like strain hardening on a spring or something.

Â You get this extra stiffness that comes in, that's the cubic term.

Â But again,

Â in structures, that's an approximation of a much more complex behavior.

Â It's like the next order term that's included.

Â Here this is a perfectly rigorous equation.

Â So in structures, this is often found as an approximation.

Â Here there's no approximation.

Â Is it exact thing for tumbling general inertia, motions.

Â So good, so we want to study this now.

Â 9:16

A little bit more detail.

Â We have three equations.

Â When we only had these parts,

Â that was the axisymmetric case it was easy, spring mass.

Â It was a sine and a cosine response, right?

Â Now here it's going to get a little more complicated.

Â Unfortunately I don't have analytic answers as we have with the axisymmetric

Â codes but there are some arguments we can talk about like global stability of these

Â motions and departures that you can write it in this form is kind of a nice thing.

Â So yeah, then I'll lead departure from Hooke's law.

Â That's the linear part, this is an exact differential equation.

Â So we can do this.

Â 9:54

Are these three equations decoupled?

Â >> [COUGH] >> Sorry, what was your name?

Â >> Mark. >> Mark, are they decoupled?

Â >> Yes.

Â >> Okay. Are they independent?

Â >> Yes.

Â Well, not because of the constants.

Â >> You said it with such conviction you almost had me convinced.

Â Why are they not independent?

Â 10:18

Everybody agrees they're decoupled.

Â If you look at this each equation has only omega 1s in there.

Â And the next term only has omega 2's and the other one has only omega 3's there's

Â no omega 1 times omega 3's as we had originally.

Â So we definitely made them independent.

Â Once you have all the initial conditions and these parameters you can solve three

Â independent differential equations numerically and get the answer.

Â 10:43

But independent is something different that means,

Â lets say I have a spin about omega 1 only and

Â the next simulation is a spin about omega 1 and tumble about B2 or something.

Â It is the same omega 1 but it has a different B2.

Â What will be different in these differential equations?

Â because you're right, they aren't independent, Matt?

Â >> They are energy momentum terms?

Â >> Exactly.

Â The energy momentums are actually what cross couples them, so

Â they're not independent.

Â I can't just go well, if omega 1 is one radium per second,

Â I have a fixed answer that's not true.

Â Because the coefficients that go here depend on momentum and energy.

Â And so it's a little bit more complicated if you also tumbling about another access,

Â your energy state is different, momentum states are different, and

Â that will give you different A's and B's.

Â So you have to solve it for that particular case.

Â So there are three uncoupled equations but not independent equations.

Â Is that make sense?

Â They're coupled through this coefficients through energy and momentum, precisely.

Â 11:50

So, the three uncoupled oscillators and that's there.

Â If I write them out to As and Bs, this is what they are, as you can see, T and

Â H has appeared.

Â The inertia's always the same, it's the same with your body.

Â But as your math was saying, if H is different and

Â T is different, I would have different coefficients, and all of a sudden and

Â then we get a different response with these omegas, right?

Â So that's where it does matter.

Â But it's very elegant so for the general case you can always write this as

Â a harmonious, homogenous on them to Duffling equation.

Â Now with these parameters we'll look at this a little bit more carefully but

Â you can see, for

Â example, the cubic stiffness terms Bs, there's differences of inertias again.

Â So we know right away if it's axis symmetric,

Â some of these things will go away, some insight you get.

Â But at the same time, I'd say if it's a cube or

Â a sphere, all of these terms go away.

Â For a sphere, none of that matters.

Â Even in here there's always differences of inertia so for sphere again, As and

Â Bs go away which we talked about.

Â A sphere you can tumble about any axis and it'll just continue to tumble,

Â your mega 1, 2, 3s are all flat lines, always, or for a cube the same response.

Â But for general bodies, if I1 is your largest inertia.

Â You can see this is going to be a positive term times a positive term,

Â this is going to be a positive cubic stiffness.

Â Down here again, if I1 is the largest, so this is going to be negative,

Â that's negative, negative times negative is positive again.

Â Again we'll have a positive stiffness whereas here

Â you'll have a negative stiffness.

Â 13:20

So is positive stiffness good or bad for a stability for a cubic one?

Â >> Good.

Â >> It's good.

Â It's pretty much analogous.

Â Instead of a liner response, now you have something that's kind of cubic.

Â But you need the same kind of a sign, right?

Â If you're positive, you want something pulling you back to the origin.

Â The origin is the equilibrium that we'll be studying here.

Â And if you're negative, you want something positive to pull you back.

Â It's just got to be more aggressive for big departures, but

Â way less aggressive for small departures.

Â So there's a difference in the cubic terms.

Â That's why for a linearization, this is the linear part it's going to dominate.

Â But if you go really big departures, the cubic one is always going to win.

Â At some point, x cubed is going to be bigger than x,

Â regardless of what these constant coefficients are, these As and Bs.

Â So good.

Â So you can see just looking at the signs of this we can get

Â some insight into what must be happening and

Â we'll cover this a little bit more in another slide as we go through it.

Â