7: Rigid Body Kinetic Energy

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来自 University of Colorado Boulder 的课程

Kinetics: Studying Spacecraft Motion

23 个评分

University of Colorado Boulder

Kinetics: Studying Spacecraft Motion

23 个评分

课程 2（共 4 门，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.

从本节课中

Continuous Systems and Rigid Bodies

The dynamical equations of motion are developed using classical Eulerian and Newtonian mechanics. Emphasis is placed on rigid body angular momentum and kinetic energy expression that are shown in a coordinate frame agnostic manner. The development begins with deformable shapes (continuous systems) which are then frozen into rigid objects, and the associated equations are thus simplified.

5: Rigid Body Angular Momentum6:55

6: Rigid Body Inertia Tensor3:24

6.1: Rigid Body Inertia about Alternate Points3:02

6.2: Rigid Body Inertia about Alternate Body Axes6:05

7: Rigid Body Kinetic Energy6:52

8: Rigid Body Equations of Motion13:13

8.1: Integrating Rigid Body Equations of Motion1:54

8.2 Example: Slender Rod Falling19:59

(Tips for Solving Spring Particle Systems)5:47

Optional Review: Rigid Body Properties14:15

Optional Review: Rigid Body Equations of Motion19:47

与讲师见面

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

Kinetic energy, let's start here.

Now we did this for blobs in space.

And for blobs in space we were able to show kinetic energy was nothing but

the kinetic energy off the center of mass plus kinetic energy

about the center of mass.

But this second term contains both deformational energy,

if panels are flexing and doing or fuel is sloshing.

And the rotational part, right?

Now let's use the same assumption.

r dot for a rigid body, it doesn't go to zero, but

the r dot just became omega crossed r, the body relative derivative of zero.

So we're also just going to focus on the rotational side.

This is basically your translational energy which comes out of

your orbits class.

We're not going to focus on orbits in here, just the rotational dynamics, so

we're not going to cover that.

And this is where I plugged in r dot is omega cross r.

So you get a bunch of cross products, dot product.

And you know right away there must be some magic trick identity

that we can rearrange this.

That's kind of what I'm doing here, almost actually.

First, and I'm using trig here, this omega is a constant as seen of the integration.

If it's rigid, every point has to be rotating at the same rate.

So this omega I can just take outside.

There's a dot product.

Okay I did use a vector identity already to rewrite this into this form,

and now I could take your omega outside to the dot product.

Here I'm still left with omega in the middle but

we have the trick that a cross b is minus b cross a.

That puts it on the right hand side.

And you can take that one out as well.

In fact if you go back and look at Hc, this whole definition is nothing but Hc.

In fact this term becomes nothing but your inertia tense times omega.

And the inertia tense times omega was nothing but

angular momentum about that point c.

So it's nice, we now have, rotational energy can simply be written as one half

omega dotted with angular momentum.

Those are two vectors dotted together.

When you dot them,

mathematically make sure they're taken with respect to the same frame.

But here is written in a frame agnostic way.

So angle rotation, so that's the same thing, H is I omega,

you might remember as rotational energy being inertia over 2 times angular

rate squared for fixed axis rotation.

This is the 3D version of that.

You would have one-half omega transpose I omega.

And that's basically inertia omega squared over 2, essentially that's what it is.

Power.

We had earlier, if the time derivative from a blob,

the first part translation was just force times the inertial velocity, again,

that's the translational sign.

What we do care about here is T dot.

So if you differentiate this term, you will end up with this.

Now to get this form, you need to have the rotational equations of motion,

which we haven't derived yet.

But I will let you, this is something you should make a note in exam,

definitely go back and make sure you can go from here to here, but it simplifies.

But the beauty of this is, for translation,

we have force started with translational velocity.

With rotation, you have torque dotted with angular velocity.

It's really the same type of term that we're going to have.

And that's what makes, this is what gives you your power rate.

To the system.

When is T-dot going to be zero?

Let's ignore translation, we'll just focus on this.

When is my power equation going to be zero?

Sheila, what do you think?

Without reading more, I want to hear you think.

>> The angular velocity is zero or- >> You could, if you're not spinning.

>> Center of mass is acceleration is zero, the Rc.

>> Well we're going in translation right now.

>> Okay. >> Just focus on this one,

the rotational side.

What makes this second term go to zero?

Omega could be zero, yes.

It's kind of a trivial dynamical system then, yes,

if the energy is zero then the energy rate is going to be zero.

But that works, yep.

What else could you do?

>> [INAUDIBLE] Make Lc zero.

>> So what is Lc here?

[INAUDIBLE].

>> No?

What is LC here?

>> Torque.

>> It's a torque.

So this is a torque about point c.

And in fact when we get this torque,

this torque is due to external forces when we derived all this stuff.

So for this dynamical system, if your external torque on that

rigid body is zero, you won't dissipate energy.

This makes a great integration check.

So as you guys are writing numerical code, you've got something that's spinning and

there's no external torque on this system.

I know energy has to be preserved.

If we had H dot equal to L, that was the same thing.

The angle momentum as seen by the inertial observer,

that's what that dot meant right, inertial derivative.

As seen by inertial observer,

H has to be a constant vector if no external torque is acting on it.

These things become fantastic integration checks.

This is for a single rigid body.

But even if you do multi-body, complex panels flapping, reaction wheels,

gyros spinning, twisting, this stuff still has to hold, which is really nice.

And now if there is a torque acting on that system.

I can actually predict very quickly how my kinetic energy needs to change.

And this becomes a nice integration check, because I can compute kinetic energy from

the simulation numerically and then differentiate it to get T dot.

And compare,

I know I applied this gravity gradient torque, this is my spin rate at the time.

Energy should have changed by this many joules per second, right?

So these tools becomes very, very useful validation tools, but also for

analysis you will see.

So torque-free motion is when momentum is inertially preserved and

kinetic energy is preserved, because this is zero.

The third option actually could exist, in that the dot product goes to zero.

Maybe you have very particular torques being applied, such that no matter what

your spin rate is, right now the torques are orthogonal to that.

In that case you have a non-working force acting on it.

That's what they call it because it doesn't,

not doing work means it doesn't change my energy state.

So there's three ways, either omega zero, Lc zero, or

those two things are orthogonal.

Then T dot goes to zero.

And again over, if you have to work down between points and time, you can do this

with a time integral or you could also do it in the spatial integral.

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