0:06

Let's start our second example on using the transport theorem.

Â The last one was mostly planar, but

Â it showed you several ways of dealing with a series of different frames.

Â Now in this one we're going to do some simple kinematics which actually

Â takes us more into a three dimensional orientation and

Â there are some particular challenges there.

Â 0:31

This piston is actually lined vertically we'll say lined up within three but

Â the position is varying by L, so this piston is slowly moving across this space.

Â And the L length can vary arbitrarily with time.

Â n2 basically points kind of into the paper, so -n2, here's the reference out.

Â On top of the piston there's a big disk that's spinning and the angle,

Â how far it's rotated is theta, so the spin rate would just be theta dot.

Â And, should be a dot there, and

Â let's see, the angle theta is measured relative to the -n2 axis, so

Â this tells you how far you've rotated.

Â The radius of the disc is given through a R, and then from that disc,

Â there's a rigid rod that's hanging down, and another disc is attached to it.

Â And so that disc, it's spin axis is the same thing as this axis here, d1.

Â I'm calling it d1 and

Â spinning at a certain rate, the angle how far I've rotated is fi.

Â So, this rotation rate would just be fi dot down here.

Â And finally, the point I'm interested in is the point on this disk.

Â So you can see this gets a little bit trickier.

Â So we've got this moving baseline with a rotating disk,

Â only to attach another bar with another rotating disk.

Â And what's the inertial motion of this point p here?

Â So we'll do the same steps we always do.

Â The first one is, we have to write, I'm looking for the inertial velocity, so

Â I have to write the inertial position of p relative to O in this case,

Â back to the origin.

Â That means I have to go from the origin first, at distance L, in the n1 direction.

Â 2:17

Then this piston is raising this disk and

Â the current height is given through h that's going to be the n3 directions.

Â So I have +hn3 that takes me from here all the way up to here.

Â Now I need to get to the R to the edge of the disk that a distance R in the d1

Â direction Then I'm going to get lowered a constant distance.

Â And this is vertically down, and vertical is n3, so = h n3 hat.

Â And finally, I have to go a certain distance r, and

Â I'm going to call the direction e r hat.

Â So to draw this in here, this is hat and

Â then a orthogonal direction, I would define as et hat.

Â So I've got my position of vector,

Â the next step is complete all the frames that you need.

Â We have the n frame already as before, so that's n1, n2, n3.

Â Okay, that's an n frame.

Â Here I've got this disk frame.

Â I only defined right now one axis direction, d1.

Â 3:33

So what I do want to do then is I want to call this other

Â direction orthogonal to d1 and we'll call it d2 hat.

Â And then to define the frame, I have d1 as the first one,

Â d2 as the second one, and the third could just be n3.

Â What did I write there?

Â That didn't make sense.

Â 3:59

N3 hat, right, so that's our right handed coordinate frame for d, that works.

Â Down here we've got another disk spinning and

Â I'd like to attach a frame to that disk as well.

Â And I'm just going to call,

Â since I used e r and e fi, I'm just going to call that the e frame.

Â 4:32

So, we're sharing d1 across the two frames, it's a matter of convenience.

Â You could give it a separate name and then,

Â just say that this, whatever this name is, has to be equal to that.

Â I typically just reuse the same variable.

Â That way I know precisely where things are going.

Â Okay, so now we have the frames.

Â Third step is always write the angular velocities between these frames.

Â We have the d frame is rotating, and

Â this disc is rotating at a rate theta-dot about the n3 axis, so

Â I can write that as theta.n3.

Â Now, the other ones that are rotating is the e frame here, a second disk.

Â How does it rotate relative to this system?

Â And we're going to have omega e relative to d

Â 5:23

and that's going to be spinning, about fi.

Â So that's going to be a magnitude fi dot.

Â And the axis about which we're rotating is d1.

Â All right, so as the disk rotates the spin axis here will rotate around as well, d1.

Â 5:41

Good, now we have two of them.

Â If we have two we can get the third, just in case.

Â If I needed e relative to n, it will be the sum of these two, so

Â e relative to d + d relative to n.

Â And if you add them up, that's a fi dot, nope, wait, I messed it up.

Â These are different directions, I was thinking of the earlier example,

Â delete this.

Â So this is going to be

Â fi.d1 + theta.n3.

Â So we have two non orthogonal, or they are orthogonal, actually, vectors.

Â I could have written here n3, it could also be d3, but I called it n3.

Â So anyway, this works.

Â I've got this e, now the last step is always differentiate.

Â So this is a little bit more of a 3-D version, but

Â very similar to what we did before.

Â We're looking for p relative to o, its derivative, and the first term.

Â This is already in the inertial frame, so this makes it very easy.

Â 7:07

Good, and let's see, and then we have the first two,

Â the third one is given in the d frame.

Â So we're going to choose to differentiate this as seen by the d frame,

Â just to make our life easier.

Â 7:25

So rd1 hat, and then transport theorem.

Â I need omega d relative to n across with the vector, that's rd1 here.

Â Good, we've got this part.

Â The next one, this is n3 again, so that's going to be easy.

Â And actually h is a constant, I'll write it out but I need the inertial derivative,

Â 7:50

Of hn3.

Â And h is a constant, and n3 is a constant as seen by the n frame, so

Â this whole thing actually vanishes.

Â And the last one, this is in the e frame, so I will choose to take an e frame

Â derivative of the vector.

Â With the right omega e relative to

Â n crossed with the vector right to complete the transport theorem.

Â 8:18

So, good, we're getting close.

Â Let's see, so these terms we can just combine.

Â This fell out, so

Â I have L.n1 hat + h.n3 hat.

Â R is a constant, d1 is a constant as seen by the d frame,

Â so this is going to vanish as well.

Â I need omega d routed to n which is theta.n3 crossed with d1,

Â so n3 crossed d1 those are all d frame vectors.

Â That's simply going to give me d2.

Â So I'm going to have + r theta.d2 hat.

Â So that was nice, that one went easily.

Â Now this next one is 0 as seen by the e frame,

Â where r is a constant radius and hat is a constant as seen by the e frame.

Â So this whole thing is nothing but constants and

Â this derivative is going to go to 0 leaving us with this derivative.

Â So I'm going to just write this one out,

Â this is going to be phi dot d1 + theta dot n3, crossed with rer.

Â 9:30

So here's the challenge with this, if you go to 3D,

Â very quickly all of a sudden, sometimes you end up with, well,

Â is this, d1, I could figure out because d1 is orthogonal to that.

Â That would work nicely and in fact, I can do that.

Â 10:03

So d1 crossed this axis cross

Â is going to give you + e fi, so

Â that's going to be r fi e fi hat.

Â But then you have here an n3 crossed with e r and

Â 10:24

that's a challenge, because n3 and e r are not orthogonal.

Â So I can't just use regular base vectors to do this cross product.

Â So here's a situation where we have to either map the e-frame components into

Â d-frame components, remember n3 is part of the d-frame as well or

Â have to map n3 into e frame components.

Â I'm going to choose to do the first which is map into n frame components.

Â So to do this, you want to draw a picture where you look down the last axis of

Â rotation which here is d1 and it has d2 and n3 sticking up.

Â 10:59

So if I draw this picture,

Â I've kind of got a unit

Â I have to draw better.

Â That will work better.

Â Okay, just took it away, okay so let me draw this first.

Â want to have a nice unit circle, good.

Â And now here I can draw out more this distance is d2,

Â I'm looking down the d1 access.

Â The d2 would be to my right and n3 would be going up

Â relative to this n3 axis now I am rotating the frame.

Â So would be pointing here and

Â e theta if I needed it, now e fi in this case would be orthogonal.

Â 12:02

The amount that we've rotated is fi, so

Â you can draw your right handed triangle here.

Â And you can see that hat ends up being a cosign,

Â this is unit length, that it's a cosign of fi in the end three direction.

Â And this length is going to be the sign of fi in

Â the- d2 direction, there is a minus sign d2 hat.

Â So now I'm going to use that result and

Â bring this in to this product that I'm missing here.

Â I got the first one already so the second one I'm going to have,

Â take the scalars upfront r theta.,

Â I will have an n3 crossed with and

Â is cosine fi and 3 fi- sine fi d2.

Â 13:00

If you do that n3 cross n3 that's going to drop out a zero and

Â then you end up with n3 cross d2.

Â Third cross the second gives you a minus, in this case e1.

Â There's a minus sign there, so,

Â 13:26

This whole term simplifies to r theta.sin fi d1 hat,

Â and that will be a plus, right?

Â So that's going to go there, everything else, we found already at this stage.

Â 13:45

Okay, so you can see this is an example of more of a three dimensional stuff.

Â But when we do even 3D at this stage,

Â we tend to go from one frame to another via single axis rotation.

Â To go from here to here that's really the same end frame axis work.

Â Going from here to here, we're rotating about single axis theta about the n3

Â end frame and then to go from this frame to the other one were rotating about d1.

Â 14:12

And so this is a sequence all of one axis rotations that we're using and

Â that's how we define all these frames.

Â That was really the biggest trick, you do cross products hopefully most of them

Â are between the same frame or lateral axes.

Â If they're not, then and only then do we draw little circles and do the right

Â cap-sines and cosines to map one base vector into another framed base vectors.

Â And that's what we had to do here.

Â So we only had one sine appearing despite some heavily three dimensional motion

Â going on.

Â