Let's start our second example on using the transport theorem. The last one was mostly planar, but it showed you several ways of dealing with a series of different frames. Now in this one we're going to do some simple kinematics which actually takes us more into a three dimensional orientation and there are some particular challenges there. So let's start off with this dynamical system I have it in actual frame finds through and one, two and three the origin o. This piston is actually lined vertically we'll say lined up within three but the position is varying by L, so this piston is slowly moving across this space. And the L length can vary arbitrarily with time. n2 basically points kind of into the paper, so -n2, here's the reference out. On top of the piston there's a big disk that's spinning and the angle, how far it's rotated is theta, so the spin rate would just be theta dot. And, should be a dot there, and let's see, the angle theta is measured relative to the -n2 axis, so this tells you how far you've rotated. The radius of the disc is given through a R, and then from that disc, there's a rigid rod that's hanging down, and another disc is attached to it. And so that disc, it's spin axis is the same thing as this axis here, d1. I'm calling it d1 and spinning at a certain rate, the angle how far I've rotated is fi. So, this rotation rate would just be fi dot down here. And finally, the point I'm interested in is the point on this disk. So you can see this gets a little bit trickier. So we've got this moving baseline with a rotating disk, only to attach another bar with another rotating disk. And what's the inertial motion of this point p here? So we'll do the same steps we always do. The first one is, we have to write, I'm looking for the inertial velocity, so I have to write the inertial position of p relative to O in this case, back to the origin. That means I have to go from the origin first, at distance L, in the n1 direction. Then this piston is raising this disk and the current height is given through h that's going to be the n3 directions. So I have +hn3 that takes me from here all the way up to here. Now I need to get to the R to the edge of the disk that a distance R in the d1 direction Then I'm going to get lowered a constant distance. And this is vertically down, and vertical is n3, so = h n3 hat. And finally, I have to go a certain distance r, and I'm going to call the direction e r hat. So to draw this in here, this is hat and then a orthogonal direction, I would define as et hat. So I've got my position of vector, the next step is complete all the frames that you need. We have the n frame already as before, so that's n1, n2, n3. Okay, that's an n frame. Here I've got this disk frame. I only defined right now one axis direction, d1. So what I do want to do then is I want to call this other direction orthogonal to d1 and we'll call it d2 hat. And then to define the frame, I have d1 as the first one, d2 as the second one, and the third could just be n3. What did I write there? That didn't make sense. N3 hat, right, so that's our right handed coordinate frame for d, that works. Down here we've got another disk spinning and I'd like to attach a frame to that disk as well. And I'm just going to call, since I used e r and e fi, I'm just going to call that the e frame. The first direction is e r hat, the second direction is e fi hat and then finally, the third one, using the right hand rule would be nothing but d1 hat. So, we're sharing d1 across the two frames, it's a matter of convenience. You could give it a separate name and then, just say that this, whatever this name is, has to be equal to that. I typically just reuse the same variable. That way I know precisely where things are going. Okay, so now we have the frames. Third step is always write the angular velocities between these frames. We have the d frame is rotating, and this disc is rotating at a rate theta-dot about the n3 axis, so I can write that as theta.n3. Now, the other ones that are rotating is the e frame here, a second disk. How does it rotate relative to this system? And we're going to have omega e relative to d and that's going to be spinning, about fi. So that's going to be a magnitude fi dot. And the axis about which we're rotating is d1. All right, so as the disk rotates the spin axis here will rotate around as well, d1. Good, now we have two of them. If we have two we can get the third, just in case. If I needed e relative to n, it will be the sum of these two, so e relative to d + d relative to n. And if you add them up, that's a fi dot, nope, wait, I messed it up. These are different directions, I was thinking of the earlier example, delete this. So this is going to be fi.d1 + theta.n3. So we have two non orthogonal, or they are orthogonal, actually, vectors. I could have written here n3, it could also be d3, but I called it n3. So anyway, this works. I've got this e, now the last step is always differentiate. So this is a little bit more of a 3-D version, but very similar to what we did before. We're looking for p relative to o, its derivative, and the first term. This is already in the inertial frame, so this makes it very easy. We're just going to get an L.n1. The second one, h, is rising up in the n3 direction and h varies with time. Good, and let's see, and then we have the first two, the third one is given in the d frame. So we're going to choose to differentiate this as seen by the d frame, just to make our life easier. So rd1 hat, and then transport theorem. I need omega d relative to n across with the vector, that's rd1 here. Good, we've got this part. The next one, this is n3 again, so that's going to be easy. And actually h is a constant, I'll write it out but I need the inertial derivative, Of hn3. And h is a constant, and n3 is a constant as seen by the n frame, so this whole thing actually vanishes. And the last one, this is in the e frame, so I will choose to take an e frame derivative of the vector. With the right omega e relative to n crossed with the vector right to complete the transport theorem. So, good, we're getting close. Let's see, so these terms we can just combine. This fell out, so I have L.n1 hat + h.n3 hat. R is a constant, d1 is a constant as seen by the d frame, so this is going to vanish as well. I need omega d routed to n which is theta.n3 crossed with d1, so n3 crossed d1 those are all d frame vectors. That's simply going to give me d2. So I'm going to have + r theta.d2 hat. So that was nice, that one went easily. Now this next one is 0 as seen by the e frame, where r is a constant radius and hat is a constant as seen by the e frame. So this whole thing is nothing but constants and this derivative is going to go to 0 leaving us with this derivative. So I'm going to just write this one out, this is going to be phi dot d1 + theta dot n3, crossed with rer. So here's the challenge with this, if you go to 3D, very quickly all of a sudden, sometimes you end up with, well, is this, d1, I could figure out because d1 is orthogonal to that. That would work nicely and in fact, I can do that. So d1 cross let me write out start assembling the final answer. So d1 crossed this axis cross is going to give you + e fi, so that's going to be r fi e fi hat. But then you have here an n3 crossed with e r and that's a challenge, because n3 and e r are not orthogonal. So I can't just use regular base vectors to do this cross product. So here's a situation where we have to either map the e-frame components into d-frame components, remember n3 is part of the d-frame as well or have to map n3 into e frame components. I'm going to choose to do the first which is map into n frame components. So to do this, you want to draw a picture where you look down the last axis of rotation which here is d1 and it has d2 and n3 sticking up. So if I draw this picture, I've kind of got a unit I have to draw better. That will work better. Okay, just took it away, okay so let me draw this first. want to have a nice unit circle, good. And now here I can draw out more this distance is d2, I'm looking down the d1 access. The d2 would be to my right and n3 would be going up relative to this n3 axis now I am rotating the frame. So would be pointing here and e theta if I needed it, now e fi in this case would be orthogonal. The amount that we've rotated is fi, so you can draw your right handed triangle here. And you can see that hat ends up being a cosign, this is unit length, that it's a cosign of fi in the end three direction. And this length is going to be the sign of fi in the- d2 direction, there is a minus sign d2 hat. So now I'm going to use that result and bring this in to this product that I'm missing here. I got the first one already so the second one I'm going to have, take the scalars upfront r theta., I will have an n3 crossed with and is cosine fi and 3 fi- sine fi d2. If you do that n3 cross n3 that's going to drop out a zero and then you end up with n3 cross d2. Third cross the second gives you a minus, in this case e1. There's a minus sign there, so, This whole term simplifies to r theta.sin fi d1 hat, and that will be a plus, right? So that's going to go there, everything else, we found already at this stage. Okay, so you can see this is an example of more of a three dimensional stuff. But when we do even 3D at this stage, we tend to go from one frame to another via single axis rotation. To go from here to here that's really the same end frame axis work. Going from here to here, we're rotating about single axis theta about the n3 end frame and then to go from this frame to the other one were rotating about d1. And so this is a sequence all of one axis rotations that we're using and that's how we define all these frames. That was really the biggest trick, you do cross products hopefully most of them are between the same frame or lateral axes. If they're not, then and only then do we draw little circles and do the right cap-sines and cosines to map one base vector into another framed base vectors. And that's what we had to do here. So we only had one sine appearing despite some heavily three dimensional motion going on.