0:09

So basically, think of this rod.

Â You sign in this up on a frictionless surface.

Â Think of an ice rink, a very slick ice rink and he's standing it up and for

Â some reason this is perturbed.

Â There's no friction force here holding your lower end in place.

Â So once you slide off you're just going to slide out.

Â So it's basically just going to fall down and slide down like this.

Â That's what's going to happen.

Â And what we're going to try to find here are our equations of motion.

Â And we're going to use these very principles, super particle theorem,

Â HL equal to L.

Â The inertia tenser definitions and so forth to derive all the right quantities.

Â 0:46

So I'm giving you here from a material statics book you can find easily

Â if you have a slender rod then the inertia about it's center of mass is mass over 12.

Â L squared, L's length of the rod.

Â So slender means we're ignoring the other dimensions.

Â It's just a homogeneous, very, very thin mass distribution, right.

Â So that's what you get there.

Â So we're going, okay, so we have that.

Â 1:13

Now let's look at the angular momentum.

Â If we want to write Hc, the angle momentum above the center of mass,

Â we needed the inertia tensor times the angular velocity of this stuff.

Â And you can see, I've got a body fixed point,

Â a body fixed frame, E, that I'm defining here.

Â So EL, E theta, where is E3 pointing?

Â Let me just make sure you're looking at the picture, right.

Â 1:42

Into the board or out of the board?

Â >> Into. >> Into the board, right?

Â eL, e theta, e3, third finger points into the board,

Â that's where it's pointing, okay?

Â So the angular velocity as we're tilting in 1d,

Â it's going to be theta dot about e3, right.

Â So that's what this part right here, those two terms,

Â that's omega of e relative to n for this body.

Â Now I need the inertia tensor about this one, so let's just write that one out.

Â 3:01

In different components,

Â which is equivalent to that I3 theta dot E3.

Â And in this problem I3 is the N over 12 L squared.

Â That's the inertia about that, the third axis that we have, in and

Â out of the board.

Â All right, we didn't need the inertias about the long axis or

Â anything else like that.

Â That's where, so when you've, this is something you want to practice.

Â Once it clicks, you go how was this ever confusing?

Â But you want to be able to go from the inertia stuff to this quick other

Â formulations especially with the plane of rotation.

Â It should always give you equivalent results.

Â 3:35

So good, so that's how we get this part.

Â We have to angle momentum about C.

Â Now to get equations of motion,

Â one way we often do with rotational side is we use Euler's equation,

Â which means H dot = L and taking moments about center of mass, H dot = L holds.

Â And I need to have L above the center of mass.

Â So let's see what do we have here?

Â 4:04

What torch is being applied?

Â The only torch that's being applied here is this normal force,

Â somethings holding that tip from sliding beneath the I surface, right?

Â So the I is still pushing up.

Â It's just not pushing laterally.

Â So it's giving it lateral stability, but

Â it is holding that point in place, all right?

Â Why does gravity's acting on this?

Â Why doesn't gravity cause any torques?

Â >> Ax on the center of mass.

Â >> Right, gravity ax on the center of mass and therefore the moment on for

Â the gravity force.

Â That's when you apply it, it would actually vanish.

Â Very good, okay.

Â So, we need the moment arm.

Â What is the point of force relative to the center of mass then that's this vector.

Â So, this length is L, so

Â if going from here to here is minus L over 2 in the eL direction, right?

Â Back to week one, vectors are distances times direction.

Â The distance is eL over 2.

Â The direction is minus eL hat.

Â That's the moment arm crossed with the force,

Â which is the normal force on the surface, right?

Â And you now have an eL crossed in n2.

Â Well you're going to have to map one into the other and that's what you get,

Â a sign term in here in terms of e3.

Â So we can write this out.

Â So good, we're looking for equations of motion for theta, but

Â all of a sudden we have to introduce an extra unknown.

Â We don't know what this force is on the surface.

Â So we know right away H equal to L is not going to be enough.

Â You can't have two unknowns in one equation.

Â People try, they try really hard, it just doesn't work.

Â 6:03

It's fixed, actually, right.

Â It's the other two that are rotating.

Â So e3 is actually N1, it's the same thing as minus N3.

Â And N3's definitely an inertial frame so that's why taken the intertial

Â derivative of this, I don't have to treat e3 as a rotating axis.

Â It's actually a fixed axis.

Â So this inertia is fixed.

Â I'm not changing the mass distribution of the rod.

Â The only thing that varies is theta.

Â So you very quickly get the inertia times theta double dot.

Â And here I've dropped e3.

Â Because in the end, everything's along the e3 axis.

Â So it gives me my one thing.

Â So H dot equal to L.

Â I brought L over to the left hand side.

Â And that's how that comes in.

Â So this is one equation.

Â I'm looking for always second order differential equations for

Â equations of motion.

Â And so acceleration, that's my theta double dot acceleration equation.

Â But it have an extra unknown.

Â So H dot equal to L was not enough, what else do we have?

Â 7:21

An inertia point that was another approach.

Â And you can try H equal to L about center of mass or about the inertial point and

Â they can give you different information.

Â And that might be a way to get a second equation.

Â But what we're going to do here is actually use Newton's equation.

Â Basically, super particle theorem, right?

Â That said, well whatever the sum of the external forces is,

Â mass times the inertial acceleration to center of mass,

Â has to be equal to that force, right?

Â 8:09

That's my big M times rc double dot that we have to

Â review the particle theorem, right.

Â Now that has to be equal to the sum of the forces acting on the system, and

Â this system only has two forces.

Â There is a force on the surface that's pushing back against the pen tip,

Â keeping it just in place.

Â And there's gravity acting on the slot and that's why we're falling.

Â Otherwise, if you have slippery surface on the space station and

Â without any other human slip.

Â So with gravity +N it's in the N2 direction and

Â NG is in the -N2 direction so this gives you your differential equation.

Â 9:16

Spencer, are those two independent coordinates of the dynamical system?

Â Or another question is, this dynamical system has how many degrees of freedom?

Â >> One.

Â >> One, right, there's only one degree of freedom.

Â There's only one angle, right.

Â Now with that angle, we can prescribe what that height has to be.

Â So there must be a geometric relationship between the y coordinate and

Â the theta coordinates, right.

Â Because you have y and theta, this didn't turn it into a two degree freedom problem.

Â It's still a one degree of freedom problem.

Â So what is the relationship?

Â And that's what we do here.

Â It's very simple, this line would be L over 2 cosine, that gets you to heighth y.

Â That's what I have here, all right.

Â So if I have theta, I know y.

Â And now here, I need theta double dot,

Â that means you simply take 2 times derivative of this y expression,

Â chain rule of this cosine, so you get some sines and cosines and other stuff.

Â But that's how your y double dots relate to your thetas.

Â Theta dots and theta double dots, right.

Â So those are not, we had two kinematic variables that was convenient.

Â This y was very convenient when writing f = ma.

Â But in the end, we have to pick one of them and

Â write everything in terms of one of the coordinates.

Â So if you do this, you plug this in here.

Â You can now, actually this depends on y and

Â n, this depends on y and theta, I can now write n in terms of theta.

Â 10:40

So I get rid of one of the variables, y.

Â And now the last step is this, this has to go back up in here and

Â now we can write our equations of motion for

Â this slender rod falling in place, slipping down like this.

Â Looks a little bit more complicated.

Â It's amazing how life gets complicated quickly, even with simple examples.

Â Now with this differential equation, is this a linear differential equation or

Â a non-linear differential equation?

Â Non-linear.

Â There's sines, cosines of the states, right?

Â There's even theta dot squares, sine squared of stuff, definitely not linear.

Â So who thinks they can quickly solve this differential equation?

Â 11:20

Darn, I always hope somebody raises their hand because then I want to talk to you

Â about a PhD program.

Â because I can't solve this.

Â Maybe a mathematician can solve this.

Â But it's tricky, right?

Â So how do we solve this?

Â Well, we go back to integration, right?

Â You put this in state space form.

Â x1 is equal to theta.

Â x2 is equal to theta dot.

Â Then you get your sets of equations.

Â You can integrate.

Â So numerically, no problem.

Â But analytically, this can be a challenge.

Â And very often in these problems we want to have at least

Â integration of one time.

Â 11:53

And it happens, you've done this, high school physics problems.

Â Bouncing balls, perfectly elastic bouncing ball, right?

Â The total energy has to be preserved.

Â It's a conservative system.

Â So therefore, we have our potential energy of the ball, and

Â we have the kinetic energy.

Â And as you're up, you have a lot of potential energy.

Â No kinetic energy.

Â Let go, it trades potential energy for

Â kinetic energies, but the sum has to be the same.

Â Potential energy always depends on position.

Â Never rates.

Â 12:20

Kinetic energy, you see the definition.

Â It's always ex dots, theta dots, it's all the rate of the coordinates.

Â So if you're looking for a relationship, and one of the homeworks asks you for

Â this in part B in particularly,

Â what's the relationship between some states and rates?

Â Your first step shouldn't necessarily be to try to integrate this equation once.

Â That might be quite hard.

Â The trick is you're going to look for is the system conservative, and

Â if yes, now we can look at the energy balance.

Â So people was talking earlier about energy,

Â this is typically where I use energy systems and I can come up with directly

Â in analytic, at this angle what is my rate going to have to be?

Â It's good conservative.

Â Now in this system there's no lateral friction.

Â There's nothing doing work on the system it's just a conservative one.

Â And this point here is not doing work because this point is not moving

Â up and down.

Â Work is typically force times distance moved and

Â this point is not moving in this direction.

Â So this force end also.

Â This system is conservative.

Â And we can use conservation.

Â So this part should hopefully be boring.

Â Very simple, mgh, right?

Â Or this case, y, we have it as in terms of the angle theta.

Â I know what the potential energy is of this rod.

Â I need the total mass of this rod, the moment arm,

Â the height of the surface, I've got that.

Â The kinetic energy, now remember how do I get this?

Â Kinetic energy, like momentum, I can break up as the kinetic energy of the center of

Â mass plus the kinetic energy about the center of mass.

Â For weird complicated tumbling things like this I really recommend that approach.

Â Sometimes there are shortcuts with other stuff that you can quickly do but

Â this is a way that's very rigorous, you won't double count terms all of a sudden.

Â So the kinetic energy, all the center of mass is mass

Â over 2 times the only velocity we'll have is y dot.

Â And we know how to relate y dot to theta dot right,

Â by using that earlier geometry expression and differentiating.

Â And then here is the moment over 2 times theta dot squared.

Â That's the one-half omega transpose i omega but

Â put back into everything's about E3.

Â And it all simplifies down to the simple scalar equation.

Â So we've got that.

Â So you plug in that kinematic relation between y dot and

Â theta theta dot simplify.

Â This is what you end up with.

Â So you do something very similar in the homework to this.

Â So now we have kinetic energy and potential energy and

Â if you sum the two up, I'm seeing initially this rod is standing up,

Â and yes, something has to disturb it but

Â the disturb it's been treating us infinitesimal.

Â So the total energy is basically just the initial potential.

Â And then something turns it ever so slightly and it will start to slip, right?

Â 16:04

So when you solve this equations, and there's a few in the homework,

Â if you want to look for h dot equal to l, but apply it about points that make sense.

Â Either center of mass, that's often a very good place to start.

Â Or you might find that inertial point as well.

Â And that might, both of those equations might give

Â you all the information you need.

Â The other one that we have is F = m a of the center of mass,

Â that's the super particle theorem.

Â Between the combinations of that you should always get all the states,

Â everything that you need.

Â 17:09

Well let's do a quick thought experiment.

Â So now what we'll move next into is torque free motion.

Â Daniel already mentioned earlier that if its torque free motion certain special

Â things happen.

Â And in fact, spacecraft once deployed from a rocket,

Â the initial stage is always tumbling in one way or a manner.

Â And then you have to recover, maybe we use magnetic torque bars or

Â we use thrusters or reaction wheels, or something.

Â And we want to go positive.

Â Or these all these different challenges.

Â And even otherwise, you don't have a situation typically where you're

Â continuously controlling attitude because you just run out of resources,

Â even with reaction that only use electrical energy.

Â It's a lot of effort.

Â You try to exploit as much you can the natural dynamics, and

Â the natural dynamics is no torque, it's just there.

Â So can we come up with spacecraft that spin?

Â And people do that quite often.

Â They spin stabilize a single rigid body.

Â Or we'll be looking at dual spinners that means we have multiple parts that spin,

Â but without any feedback.

Â And then multiple parts that's basically a stepping stone into reaction wheels,

Â momentum exchange devices, which we'll derive afterwards and

Â stay up there as well.

Â So that's kind of the motivation for why do we want to look at torque free motions?

Â Really a classic problem, even with three axis control these days you still want to

Â understand the different modes.

Â Why are some motions stable?

Â Why are some motion unstable?

Â And one of the equilibria spin conditions.

Â And if I put it in a particular spin, will it just stay there or

Â will it wobble out of that spin and go and get in the way?

Â 18:51

But this first cut at it though is very complicated because

Â if I do everything in the inertia frame, I will get three components.

Â They all have to ce zero.

Â That's great but your momentum in the body frame, if I have a principal

Â coordinate frame, is simply i 1 omega 1, i 2 omega 2, i 3 omega 3, right?

Â We mentioned principal coordinate frame that inertia tension that's just diagonal.

Â So it becomes a very simple product.

Â h is equal to i omega.

Â But then we have to map this into the inertial frame.

Â And here I'm using [INAUDIBLE] in terms of three to one Euler angles.

Â And immediately, you get all these relationships.

Â But you can see there's tons of sines and cosines and omegas and

Â everything varies with time.

Â And true, that's a constraint, but

Â it's very had to wrap your brain around that and get some analytic insight.

Â So next time when we start off we're going to start here and

Â I'm going to show you a different approach which leads to the pole hold methods.

Â Those of you who have taken 3200 have seen a little of this stuff already.

Â But it's a classic approach in how we can discuss stability, but

Â we'll take it all a step further then from that as well.

Â