0:05

Continuous system, basically jello, blobs of jello, right?

Â This is kind of what we're starting out with.

Â So we have a dynamical system, that's this and our every self-respecting dynamics

Â class has to have these amorphous flying potatoes right.

Â It's just a general thing and we want to now compute certain properties of it.

Â Newton's Law is simply F = MA, written in this form.

Â 0:33

You can see.

Â And now, because I have a continuum, it's kind of hard to do it for the full thing,

Â a blob of Jello, right?

Â Different things will have different forces acting on it.

Â So what we want to do is track infinitesimal little elements here, and

Â that's my little box that's DM infinitesimal object.

Â Just like classic calculus that you've seen many many times before hopefully.

Â So df is the force that's acting on the differential element, dm.

Â R is my inertial position vector, so big R is the inertial position vector.

Â Little r is my position vector relative to the system's center of mass.

Â So whatever this bulb is you have an infinity of little DM's that all add up

Â to this thing.

Â They have in common at this instant that's where the center of mass is so

Â that's a classic location.

Â We'll be writing all our equations about that point so that's it.

Â That's basically F = MA times that.

Â So but for F = MA it was important that we have an inertia derivative so

Â these dots are derivatives as seen by the inertial frame,

Â and you have to write your position relative to the inertial origin.

Â That's why you wouldn't do little r double dot, you have to do big R double dot,

Â relative to the inertial frame then dF.

Â Now these forces, we can categorize them generally.

Â There's external stuff and internal stuff.

Â Who can give me an example of an external force that might be acting on

Â a spacecraft?

Â 1:58

>> Drag.

Â >> Drag, right, atmospheric drag is one of them definitely.

Â Gravity could be an external force acting on it.

Â You could have solar radiation pressure.

Â You may have your thrusters.

Â This is a man made object, maybe you have some way to control it.

Â Now you have to do some stuff.

Â That's all the external things, so

Â what are examples of internal forces acting on this blob of jello?

Â 2:24

>> Torquers?

Â >> Actually, good question.

Â So magnet torquers are actually objects that push off the external magnetic field.

Â I would see them as external [INAUDIBLE] influences acting on these elements.

Â The element might be inside the craft, but it's still pushing off an external field.

Â So what would be an internal force then that you might have?

Â >> Reactions.

Â >> Reactions is a good example.

Â because you have these fly disks that we put in,

Â sometimes called flywheels or reaction wheels or

Â control movement gyros that are all spinning objects inside our spacecraft,

Â and then they have motors attached that will spin them up or spin them down.

Â That motor is taking the wheel and it's pushing off the space craft.

Â So as a dynamical system it basically means some part of jello is pushing

Â off the other part of the jello and that's what cause the spinning.

Â This would be an internal force for the system all right?

Â If you do a free body diagram of just the reaction wheel then maybe we can treat it

Â as an external one on that system.

Â But here we have space craft jello could be fuel and all this stuff.

Â Fuel's another one.

Â Fuel slash backing back and forth.

Â Those are all internal forces.

Â And that will be key because we'll be talking about internal forces,

Â external forces, and how do things impact momentum and energy.

Â And there are some good high level things that are really useful as a spacecraft

Â analysts.

Â So we're going to break it off in two ways.

Â Now if I need a total force acting on jello,

Â I have it gazillion dm's, I have to integrate over them.

Â So if you look at chapter 2 we do a summation because we have a gazillion

Â of them.

Â And if you have a continuum you simply replace the summation

Â with an integral sign.

Â That's really, if you go back and look at calculus and how you did integrals,

Â you kind of broke it up into little bitty chunks and

Â then made those chunks infinitesimally small.

Â It's basically this big summation.

Â So if you're summing up over this, the claim here is the total force

Â acting on this blob is the sum of all the forces acting on the individual DMs.

Â Which is simply going to be the sum of the external forces.

Â What happens to the internal forces?

Â 4:27

>> It blots it, it cancels out.

Â >> Yeah, which law is that?

Â >> Newton's third law.

Â >> Yeah, it's one of Newton's laws, right?

Â So if you take a wall, and push against the wall, the wall is actually going to

Â push back with you exactly the same amount but in opposite directions.

Â So as I'm torquing on a spacecraft, applying forces to the wheel,

Â you're getting equal and opposite forces applied to the spacecraft.

Â When you sum them up all those things have to mutually cancel.

Â So internal forces never contribute to a net force on the system,

Â only the external ones do that.

Â 4:59

So I'm sure you've seen that before.

Â Now some basic properties of jello is any continuum, any dynamical system is

Â whatever the mass distribution is, it will have a center of mass location.

Â And there's many ways to define it.

Â I'm going to show you it's at least two here.

Â One of them is the center of mass is essentially your mass averaged location.

Â So Rc is the center of mass of this blob.

Â If you look at R times dm.

Â You really, you think of it you got one kilogram, half a meter out.

Â Two kilograms, a full meter out.

Â And five kilograms, a half a meter to the left, right?

Â You would add them up with a distances divide by a total mass.

Â That's going to be the center of mass location,

Â if you have this mass distribution on a system.

Â If it's continuum, instead of summing you do integrals of mass,

Â you know, the mass of the object times its location.

Â And then we divide by the total mass to get Rc.

Â So that's a classic definition of center of mass,

Â this works for any continuum dynamical system.

Â Now we can break this up because these are vectors, the inertial position vector.

Â So we can say R here has to be RC pass the r.

Â 6:12

Again, this is done in a purely vectorial way.

Â I'm not saying r has to be expressed in the body frame or in the inertial frame.

Â That's a book keeping element we keep to later.

Â Right now we're just adding vectors.

Â Vectors plus vectors and doing with that.

Â So if you do this you can see this will actually break up and

Â let's do that one quickly here.

Â We had that.

Â 6:41

Dm that's the sum of two vectors.

Â So I'm going to do this, and say okay.

Â That's going to be Rc dm + rdm.

Â And my little subscript b that notation is actually a triple integral.

Â I have to integrate over the entire body.

Â But instead of writing triple integrals with all the limits, this is a short kind.

Â I just do one integral with b that mean it understood to be integrating over

Â an entire, whenever this body is, all right?

Â So we can break these things up.

Â That's nice.

Â That should be a vector.

Â Then there's some simplifications that are going to happen.

Â We knew earlier, by definition, this was M times Rc right?

Â This term can somebody say how this is going to be refined?

Â 7:40

>> We get big M times R.

Â >> Yeah this is nothing but the sum of all the mass elements which is the total mass.

Â You can get there.

Â So now Kevin can you explain to me why it's okay to take

Â Rc varies with time and everything.

Â Why is it okay to take it outside the integral?

Â >> because we're only intergrating the phase.

Â >> Right. >> We're integrating over the body.

Â So when you see these B integrals, think of that, over this body,

Â 8:06

let's say my center of mass was right here by the microphone just easy right?

Â Stop I'm looking at this and going at this instant,

Â whatever the mass distribution is, my left hand is part of the system

Â the center of mass for the left hand is still down here, right, for the system.

Â That Rc location for the right hand is also going to be here.

Â Every element of my body, because it's a continuum,

Â it shares the same center of mass location.

Â So as far as the body integral goes, it's just a constant.

Â Just like integral of 3 times x.

Â You pick the 3 outside the integral and just do integral of x right?

Â And continue, that's how we can treat it.

Â So this is just a spatial integral.

Â Yes rt varies with time but

Â we're doing a spatial integration over the body volume right?

Â And that location is a fixed it doesn't change depending on if you're looking at

Â the nose of the spacecraft or the tail of the spacecraft.

Â It's one system and that system has one center of mass location.

Â So yes that's the reason, and we'll use this trick several times today.

Â So if it didn't quite make sense now hopefully it will sink in more and more.

Â So you can do all of this.

Â That gives me MRc Plus

Â little rdm = MRc.

Â These cancel and you will have little r times dm = 0.

Â This is the other way.

Â If you do your classic sophomore level center of mass stuff,

Â you have 5 kilograms here, 2 kilograms there, you add them up.

Â If all your positions are taken relative to the center of mass.

Â All these mass times positions have to sum up to be 0.

Â That means you've balanced it perfectly.

Â And that's what it looks like for generally speaking.

Â We have the little r dm's body integrals have to be equal to 0.

Â That's just a complete 3D version that works for any jello.

Â So good.

Â So that, from here, a few steps you can prove this.

Â These are the kind of things

Â I would definitely expect you to be able to do easily and quickly in an exam.

Â 10:06

So if you haven't done much of this, practice.

Â Now, what we're going to do next is we're going to differentiate this.

Â So we're going to take that definition that we just had,

Â the center of mass location, all this is good, and

Â I'm going to take two inertial time derivatives.

Â So that means on the left hand side, mass is constant.

Â I'm not losing or gaining mass in this dynamical system.

Â And RC, I just have to put two dots over it and I'm done.

Â This is kind of like homework 3.6 that you guys did that had inertial

Â derivatives and stuff.

Â In vectorial form this is dead simple, two dots and you're done.

Â The right hand side dm I guess doesn't change, we're not losing or

Â gaining mass with the system.

Â This is a body integral, as Kevin was saying.

Â And so the only thing that varies is r.

Â So we get big R and it has two dots over it.

Â But then you go back and look at Newton's equation that mass times inertial

Â acceleration is the force acting on that one.

Â And so this can be replaced with also the force.

Â The integral over the DF force and

Â we show that the integral of the DF force is nothing but the net external force.

Â You can ignore all the internal forces.

Â So you end up with an equation that should look pretty darn familiar.

Â 11:17

Mass acceleration equal to force.

Â But you've used it primarily I'm assuming on particles.

Â This is the particle, here's the mass,

Â here's the force this is where it's going to go.

Â This actually also works on a continuum.

Â Blobs in space.

Â 11:33

Anybody seen the Muppet Show?

Â There was pigs in space, you know, a lot of jiggling going on.

Â That's that kind of stuff.

Â This works for anything, even muppets.

Â Okay, so this is the key thing.

Â We call this the superparticle theorem.

Â What this basically says,

Â no matter what the dynamical system is, this complete closed system,

Â you can treat the center of mass of that system will act just like a particle.

Â 11:56

If I know there is two forces acting on it from a megatometer and

Â maybe from the atmospheric drag force that Daniel was talking about, that's it.

Â Then those are the only two forces I have to consider acting on the center of mass

Â and I will predict the translational motion perfectly.

Â Now that blob may pull apart.

Â It may do all kinds of weird stuff.

Â So if your external force depends on the shape, then it gets more complicated.

Â So if this blob would be pulling apart, then with gravity gradients the part's

Â closer to the planets would have different force than the parts top.

Â The net forces change because of the shape.

Â That makes it more complicated.

Â But we didn't have gravity.

Â We have just net thrusters you've got blobs in space.

Â Somebody fires a thruster.

Â And this thing is going to take off and go.

Â Just by looking at the force and

Â knowing the mass, I can predict what's going to happen to the center of mass.

Â I don't know what happens to the shape yet, but

Â I do know what happens to the center of mass of that system.

Â Which is a very powerful argument.

Â So that's what's called the super particle theorem.

Â Any blob in space.

Â If you just are concerned with the center of mass,

Â the center of mass will act like a particular.

Â F = Ma holds but F has to be the net external force.

Â M is the total mass and Rc double dots has to be the inertial second time derivative.

Â