4: Torque Free Motion Integrals of Motion

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Do curso por University of Colorado Boulder

Kinetics: Studying Spacecraft Motion

16 classificações

University of Colorado Boulder

Kinetics: Studying Spacecraft Motion

16 classificações

Curso 2 de 4 em Specialization Spacecraft Dynamics and Control

As they tumble through space, objects like spacecraft move in dynamical ways. Understanding and predicting the equations that represent that motion is critical to the safety and efficacy of spacecraft mission development. Kinetics: Modeling the Motions of Spacecraft trains your skills in topics like rigid body angular momentum and kinetic energy expression shown in a coordinate frame agnostic manner, single and dual rigid body systems tumbling without the forces of external torque, how differential gravity across a rigid body is approximated to the first order to study disturbances in both the attitude and orbital motion, and how these systems change when general momentum exchange devices are introduced. After this course, you will be able to... *Derive from basic angular momentum formulation the rotational equations of motion and predict and determine torque-free motion equilibria and associated stabilities * Develop equations of motion for a rigid body with multiple spinning components and derive and apply the gravity gradient torque * Apply the static stability conditions of a dual-spinner configuration and predict changes as momentum exchange devices are introduced * Derive equations of motion for systems in which various momentum exchange devices are present Please note: this is an advanced course, best suited for working engineers or students with college-level knowledge in mathematics and physics.

Na lição

Torque Free Motion

The motion of a single or dual rigid body system is explored when no external torques are acting on it. Large scale tumbling motions are studied through polhode plots, while analytical rate solutions are explored for axi-symmetric and general spacecraft shapes. Finally, the dual-spinner dynamical system illustrates how the associated gyroscopics can be exploited to stabilize any principal axis spin.

Module 2 Introduction1:22

1: Torque Free Motion Polhode Plots33:27

1.1 Example: Special Polhode Plots3:00

2: Torque Free Motion Axisymmetric Solution5:51

3: Torque Free Motion General Inertia Case14:25

4: Torque Free Motion Integrals of Motion6:16

5: Torque Free Motion Phase Space Plots9:07

5 Example: Phase Space Plots for Varying Energy Levels4:58

6: Torque Free Motion Attitude Precession11:43

6 Example: Phase Space Plot of Duffing Equation4:35

Optional Review: Torque Free Motion10:11

Conheça os instrutores

Hanspeter Schaub
Glenn L. Murphy Chair of Engineering, Professor
Department of Aerospace Engineering Sciences

There's an integral, every one of those equations we can actually integrate and

come up with these conserved quantities.

You can think I'm calling him K here,

this is not kinetic energy, it's something different.

But it's like energy, so it's an integral of motion basically.

So it leaves us a waypoint to go to a set of slides.

So let's say we have our classic mx double dot

= -kx.

Anybody know what the integral of motion of this one is?

What's the preserved quantity of this differential equation?

Spring mass system.

What's constant?

You did it in the homework with two particles in the spring mass.

Part B, what did you have to use to get that relationship?

Energy, right.

Basically.

We know out of this, this is nothing but a spring mass system.

So we know that that kinetic energy and

the spring potential energy summed up will give us a constant, right.

But we know that.

Can we mathematically prove it?

So I'm going to show you quickly how to integrate these system.

And this is a simple linear system.

With this other term there, the cubic stiffness has an x cubed instead of an x.

But even x cubed is trivial to integrate.

I expect you guys to know how to integrate x cubed.

In the next homework you will have other terms in there.

There was one homework where you end up with a sine theta,

but even sine theta isn't that evil.

I know you guys know how to integrate sine theta, hopefully.

So you can come up with these integrals of motions, all right.

So let's see how we can do this.

And there's some little tricks.

You'd have to write them down to remember all the little steps.

So I've got the second ordered derivative of x equals to minus kx.

This works nicely.

Now I'm going to do a substitution where I'm saying v is simply x dot, right?

It's commonly done, like you do in second order system,

we put in first order forms and integrate it.

That's typically what we do.

So this side can be rewritten as dv/dt = -kx.

And you can can also say, well,

let me multiply this dt Over to the other side.

So mass times differential velocities equal to -kx times differential time.

This phone, this side is easy to integrate.

That's just mdv.

The integral of that is just going to a give you v plus the constant.

But the right hand side, x depends on time and we don't have an answer for

it in this form yet.

Is not as easy, so once you get to the stage breaking these dts and dvs up,

the trick is you want to multiply both sides times v

which is of course dx/dt, right that's what .x means.

So if I multiply both sides over with this.

Here on the left hand side, I'm going to have mvdv, that's easy.

So, right hand side -kx times v and

v is dx/dt.dt.

These dt's cancel and

you end up with nothing but

-kxdx, that's it.

Now we've got a set of equations we can easily integrate and I'm not doing

in between two specific points in time I'm just going to get the indefinite integral.

So, if you do the indefinite integral of this,

the integral v with respect to v is just going to be v squared over 2 right.

So you end up with mass velocity squared over two which is nothing but

kinetic energy.

The right hand side,

the integral of x with respect to x is just going to be -kx squared over 2.

And then out of this whole thing you have to do plus a constant.

What's this constant physically going to be in this problem?

If you look at it the constant mv squared over 2 plus kx squared over 2,

what is that?

Total energy exactly that's it.

That's because it has a physical thing but in a more abstract sense when you have

differential equations we can often do this step but

you don't get something that's actually in terms of energy it's just

an equation constant of whatever these other things are right?

So here happens to be total energy but if you go back to these differential

equations and do the same steps bring this to the right hand side.

Instead of just having a minus A omega I also Also have minus B omega cubed,

I did the integrals of those in the end then I'll have an x to the 4th over 4 and

an x squared over 2 with the right constants, and

an integration constant, right?

And that's what we found here, so if you do that step,

and I encourage you to practice that once because in the homework you have to

use it anyway for one of the solve the homework for, you can get this.

So this is the integration constant, you can see this is not mass over velocity

squared the units are wrong this is actually velocity rates,

this is an acceleration squared, this can't be energy but

it's something energy like in that term and this is a preserve quantity.

If I get my resulting tumbling motion out of your integration, and every

integration times that if you're computing on mega .2 integrated and you have

your current omega and you go through this math, you should always get the same k.

All right?

So this becomes the great integration check for talk free motion as well.

And I think one of the homeworks, in the next homeworks said has you numerically

stimulate this and validate that this actually, you stimulate this numerically,

you compute this analytically with the T's and the H's.

And this math should always give you back that same value

that everything make sense.

That's a good numerical check of what's going on.

So this k1, 2's and 3's are now written out in terms of the energies,

these your integration constants for the system.

So I'm giving you the complete answer here at this stage.

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