 Alright, let's go back to where we were before we step into something new with remember we were doing constrained motion. Sorry, unconstrained motion. Constrained motion is coming up. So we're doing unconstrained motion where we're looking at the relative motion of two objects, the motion of one object relative to another. So to bring us back up to speed let's look at another traffic problem that worked very well for the things we're working on. So a nice 90 degree corner with two cars in it, each of course with their own velocity and their own acceleration. Come up with the relative acceleration. We'll call that one A and that one B. I only want to come up with something about the relative acceleration, velocity acceleration between the two. And as you probably already anticipate it has also to do with the geometry of the corner itself because of the acceleration effects we have to take into account there. So it's on the A cars on the corner of radius row with some velocity and each may as well also have some kind of acceleration. Anticipating forthcoming arguments we've already ascertained that the acceleration is given for the car in the corner is the tangential acceleration. Just the very thing that you would get a crude estimate of if you were actually looking at your speedometer or needle and timing how it changed with certain values. So we want to find the velocity of one with respect to the other and same for acceleration. So if you remember we used that terminology, that symbolism, for our relative velocities and relative accelerations. And hopefully you know just what those equal to and how we define the relative motion relation. Simple as that. If you do it like this, which certainly works for me in my small brain, it's just the two velocities in the same order as written, AB, AB, that is just the two velocities subtracted and the same thing for the acceleration because it's really just the time derivative equation of the equation above. So it was a matter of putting each of these into their particular forms. The two velocities, well the velocity A is a little bit, we've got to put a little bit to it. I don't have a specific angle here, just some generic angle theta. And so where it is in the corner is going to determine what its velocity looks like. We know it to be ahead of time in the tangential direction only. However, that in a little bit itself is a challenge in that the tangential direction, easiest place to put these, the tangential and normal directions change with the position of the car around the corner itself. They are theta dependent. However it's not too big a deal for us to cast those unit vectors in i, j unit vectors themselves. So we can do that, let's see, I believe that's also theta. So the tangential unit vector is a small component in the i direction and a small component in the j direction. It'll hardly draw, let's see if a larger drawn would help. So there's the normal direction, the perpendicular to it, the tangential direction. And the angle theta is that angle, is it not? I believe so, yeah. As that angle closes then this comes down, that angle closes too. I also think that angle is theta. Alright, so we can cast the tangential direction into i, j components. There's the i component of it and here's the j component of it. So the tangential direction is sine theta, sine theta, sine i minus cosine theta, j. That's nice to do because the other car is purely in the i direction, so it'd be nice if we can cast both of these problems into the either make both of them i, j, notation, then we can handle any theta or specify a theta and then that's what we could do in either notation. But to keep it generic with an open theta we want the possibility to handle either one. So the normal direction is cosine theta i plus sine theta j, if i, j are in the usual directions. That look okay? That's just, again, the normal vector, the normal unit vector into i, j notation. And so V a becomes, well, whatever it is, whatever the magnitude, whatever the speed is of car a in the tangential direction which we can now put in i, j notation. And then it's no problem. It's no problem combining the velocity of a and the velocity of b because b is in purely i direction itself. So the velocity of a relative to b is that we just had minus then the velocity of b which is V b i. And then whatever numbers those are, whatever speeds it has, whatever angle it might be, we now know the velocity of a relative to b. For any angle any place around the corner. So again, as I remember telling you when we talked about the normal and tangential directions, coordinate system, often times it's very useful to us to put it also in the i, j notation. Most people are a little more comfortable with i, j, x, y. This concept of a moving coordinate system is a little bit different because we can do the same kind of thing with the acceleration vector. The acceleration of a, however, is a little bit problematic in which way. And we saw on Friday when we looked at the acceleration vector of one of the cars in particular. In this case it would be this one. Not that it's always changing. There's not only the tangential component of acceleration, there's always a normal component of acceleration because of the fact it's on a curve. So a becomes the acceleration of a in the tangential direction plus the acceleration of a in the normal direction. That's the centripetal part. Then from that we subtract the acceleration of the, which again is no problem because it's already going to be in the i direction by definition of the problem. So if it does indeed have some tangential component of acceleration, that again is in the tangential direction. So we have to use the tangential unit vector that we've got there, sine theta i minus cosine theta j. And then what about the normal component? Its magnitude remember is v squared over r. So this is plus actually is minus v squared, this is v of v a, v squared over rho times the unit vector in the normal direction. Which we've already determined from up there. That whole thing, oops I don't want to vector sine over that. That whole thing there is the acceleration of car a, which is the first part of our relative velocity equation. So we've got the tangential portion and the normal portion. So make sure you've got that down. I want to ask you an important question about it. That question being why minus sine here? Because this is the tangential portion, this is the normal portion and I have a plus sign right there. Why do I have a minus sign there? And it's deliberate, it's not a mistake. Remember I even started to put a plus there and I deliberately changed it back to a minus. So why is there a minus sign there? Maybe that should be the whole test on Friday. What? Yes, it's exactly it. The normal unit vector is out but we know the centripetal acceleration to be in. So this is this normal component, the centripetal acceleration is opposite the direction of the normal, the unit vector in the normal direction. What part? For this, for v a, for v a, it has no normal component. You just wrote it up there for reference. Oh, here? Because I was going to use it. I needed it for a here, didn't use the normal one but I knew it was coming up so I just went ahead and established it while I still had that little drawing that was there. And then we take off of that the acceleration that v might have if any and it would be something in the i direction. If it's slowing down that would be a minus making the whole terminal plus. So there we have it for generic values of any possible configuration. So just to see if we got it all down right, let's give it some values. So a's got a speed of 10 meters per second, v is 18.5, the acceleration of a, let's say, oh, a's, a's the one on the corner. So this will be the tangential component anticipating a call from your lawyer plus 5 meters per second squared v, which is on the straight way so it has no tangential component, no tangential or normal component, 2 meters per second squared. Let's put it on a 100 meter corner and figure it at an angle of 45 degrees. Just to run through the numbers, make sure we got all the pieces right and then see if we can at least visualize what that looks like when we get all those pieces and just double check everything I've got there. Okay, so we've got all those pieces in. Take a second, just run through the numbers and then you can also sketch them on to a simple picture. You might want to redraw it since this one's got an awful lot of stuff in it. Notice that it does not depend where v is as long as v isn't in the corner yet. Of course, both cars are red. And as usual, notice how important the minus signs are. Important is any part of it up there. Is it okay? Did he help you, Joe? Or make it worse? If he helped you, make sure you give him a tip. Like someone were coming in the same, same type of number that I had. Which is good, we're doing the same problem. Combine like terms and let's see what we've got. Do you guys agree? Alright, anybody ready with the velocity of A relative to B? Alan, you have it? You've got negative 11.4 I minus 7.1 minus 7.1 J wrong, meters per second. Thank you. Is that what you got? Travis, are you okay with that, Phil? Alright? Okay, that's after collecting the I and the J terms. We have two I terms, one J terms, so it's just a matter of collecting a little cleaning things up a little bit. How about the, oh let's see what that looks like. So it's a little bit of a negative I, a little bit in the negative J. So A relative, oh wait, I have two. Velocity of A relative to B is something like that. About the same, a little more in the I direction. 32 degrees. 32 degrees, yeah. Something like that. If you were in car B, it would appear car A is going sort of to the side of you, which it is. If you're looking out the front, you're seeing it go more and more and more to your right field of view. And that's all that says. And then the acceleration, what's the I component? Travis, you have that? Alright, just what I got. And the J component, Joey, you got that one? Who does, Ken? Anybody? John? Huh? Is that what others had? What I had? Is it? I have a negative 4.2. Alright, let's go with that because I hate being wrong as often as it happens. Alright, I have a 4.2 as well. Okay, so double check, double check that, Ken, to see what was the difference. And remember, that one's not nearly as intuitive as the velocity is. It's just much harder to actually see acceleration calculate on. We can even estimate velocity on the fly. Everybody can do that with just paying attention to car travel. Somehow you can make some estimates there. Acceleration is very much different than that. Phil, alright, confirm that. Alright, any questions before we clean up and go on to the next phase of relative motion that we're going to look at? That's unconstrained. Remember, that's only because the two objects in the problem are independent of each other. They're not necessarily independent of the path as in a traffic problem, but they are independent of each other. They can each have their own velocity, their own acceleration, and be in their own position on the path. I guess with the traffic problem, they can be at the same point on the path, but that's for CSI Queensbury. Oh, what can you do? Recharge, occupy the same time-space coordinate. Okay, constrain motion is some kind of motion where the velocity and acceleration of the two objects are in some way linked to each other. Probably the simplest example I can come up with is two objects connected by a roper cable. Clearly, as one moves, the other's motion is going to be affected. Alright, so we can figure out how one works with the other. First thing we can do is establish some kind of system for locating the two. So let's say, for example, that I say object A is that distance below the roof. That way, if I find any time rate of change in that, I'll know the velocity of A. If I find any second order time differential of that, then I'll know the velocity and the acceleration of it. And as well for B. Here's a genius. I couldn't have picked anybody better. I don't know how many extra credit points you're going to get on an upcoming exam. It's going to be a lot. Alright, here's what we do. We need to link not just those positions with some reference, but we need to link them to each other. So what we do is determine the length of the cable or rope that's connecting the two. Because that is literally and figuratively the connection between these in this constrained relative motion. And we write down the length in terms, the length of the cable connecting in terms of what we have here. So let's see, that would be xA minus whatever this distance is. For that side of the rope, let's see, what is that distance? That's, well, however long this link is. Well, just however long the link is. That will do because that takes us right to this point. So we'll call that length some length h. So that's this section of the rope that goes from A to first contact with the pulley. Then there's a little bit of the rope that goes over the top half of the pulley. So that is 1 half 2 pi r or just pi r, where r is the radius of the pulley. That takes care of the part from A up to the pulley over the top of the pulley. And then the last little bit is B minus h. That's the total length of that rope as reference to an arbitrarily chosen origin. Or datum line if you will. Does that look right? The part hanging down to A up to the pulley, top over the pulley, and then B from B to first contact with the pulley. Up to here? Do. Go ahead. That's still xA minus h plus a quarter of 2 pi r plus then another quarter of 2 pi r, which is what I have there is pi r. And then down here anyway. What do you mean to put the reference line to the top of the pulley? Yeah, it could do that too. As you're going to see in a second, it doesn't matter. So we collect terms a little bit. Remember we're trying to get how A moves with respect to B. So we put those two together, then combine all the other things pi r minus 2h. Does that look like good algebra? I think? Alright. And this business depends upon this rope or cable being inextensible. That this length is a constant. Which it must be. If this rope stretches, then as one moves, it's going to be very hard to figure out how the other one's moving because of that stretch. It's going to depend upon how much there is there, what the spring constant is, all kinds of things that would be troublesome with it. So this is, we're assuming a constant length connection between the two. So here's what we're going to do. Get the time rate of change of the length of the rope. DLDT or L dot, if you will, well there's our equation for L. So we'll just take the time derivative of this equation. So that becomes xA dot plus xB dot plus the time rate of change of these terms. Which is the rate of change of those terms. I already did the first two, xA dot, xB dot. Now we've got the velocity of the two. What's the time rate of change of this term? Zero. These are all constants. That's why it doesn't matter where we put the measurement because they all drop out. The length of the rope is constant as well. So L dot is zero. We now have the direct correlation between the motion of one and the motion of the other. And it's as straightforward as that. And it even matches what we would have known anyway. However fast A is going up, B is going down by the same speed. If A goes up, B goes in the opposite direction with the same magnitude. Which you would have known anyway. But now with that simple setup, we can do much more difficult problems. The trick becomes, as much as anything, just figuring out some place from which to measure things. And you can do it almost anywhere. As John even anticipated, we could have done it from here. And then H wouldn't have been any part of it. And we wouldn't have had to worry about the top, the part of the cable that goes over. Because that's also a constant anyway. That's what these two terms are. We wouldn't care what they were. So that's it. Just where you put your reference line is immaterial. However, I do have two suggestions that makes it a bit easier. As you decide where to put your reference line. And we'll go through a couple problems and practice them. And you can even choose your own and put them in different places to see if we all should get the same answer. The first thing I suggest is that it's not a moving reference. Which means don't attach the reference point, the datum line, to one of the objects because they're going to move. Also, as we get more involved, rope and pulley systems, some of the pulleys move, some don't. Don't attach it to the actual moving object. You can do it. It's just an awful lot more difficult. The other suggestion I give you is don't put it between the objects we're concerned with. Which means A and B. We're not that concerned with how the pulleys are moving. We are concerned with how A and B are moving. Don't put the reference line between the two. The second reason I suggest you don't do that, again, it's still doable. It doesn't change the physics. Remember these origins are arbitrary. Which you can choose them in places and makes the problem a lot easier. If you put the reference line between the two, then the plus signs and the minus signs become much more confusing in terms of what they mean. In this case, we know that if XA is getting longer, which means it's positive, B is going to do the opposite. Which means it's getting shorter. And the minus signs carry direct reference if you avoid putting your datum line in between the two objects. Because then they're both going in different ways and the minus signs become much more difficult to understand. To anticipate just what they mean. But the whole thing depends upon the simple notion that the length of the cable connecting the two is constant so the time-rated change of that length is zero. But you've done that. There's not much left. What about the acceleration of one relative to the acceleration of the other one? We take the derivative, the time-rated change of this equation and we're all done. That's all we needed to do. Now, we can add other parts to it if we want to do the velocity of one relative to the other. But we've got all the pieces that would go into it. So there's not a whole lot to be added to it. Yeah, I'll point out that you wrote an A there so that we wouldn't get that over. Oh, yeah. X double dot. Yeah. I got to stop talking when I write. That's why I won't let you talk during tests because you put down a bunch of stupid stuff like I do. There we go. X double dot. Not that the acceleration wouldn't have a time-rated change. And that equation is still true, I guess. The time-rated change of acceleration would be X triple dot. We can still keep doing that forever. Alright. A little more complicated problem, which means more pulleys, more links, more ropes. Okay. So maybe a tow truck type of situation. We have one object here, maybe a car that needs to be towed out. So the tow truck operator sets up this pulley system, puts a pulley on the front of A, attaches another pulley to a guard rail, and then runs a rope through it all from the center of A on the pulley back around that pulley and then out to his tow truck. Okay. I drew the rope a little loose around the pulleys just so you can see it. But as we've already seen how much is wrapped around the pulleys doesn't matter as long as it doesn't change. It's just one of the constants that goes into the pot. Want to find out for some velocity in B what is A going to do, both its direction and its magnitude. Alright. So let's pick a reference that's not moving and not between the two objects. We could put a reference here on the guard rail, but it's between the two objects. It's still doable. It just makes the minus signs harder to understand. And they give us enough trouble as it is. So we'll put it over here. Maybe that's where my mother-in-law stands. She's a little drove it into the ditch. Sorry, I'm just kidding. Alright. So now we know where A is. And from the same reference we know where B is as well. And remember we're treating these objects as particles, as points. So just where we go to on the piece probably doesn't matter. We could have a problem where it does. But for these ones just to know where the front of the car is is enough to know where the entire car is. Okay. So figure out in those terms what's the length of the rope. Well, let's see. We've got this big piece at the top here. I guess that would be XB minus XA plus some constants in there. That's the top piece of the cable. XB minus XA plus some constants. And we know they're going to drop out anyway. So we're not real concerned about those. What? Don't we need to know where the guard rail is? Well, that's for that top part of the rope. For the middle part of the rope, yeah, we're going to need to know where the guard rail is. Because wherever it is determines how long this piece is and how long this piece is below, because they're both the same length. So let's, I'm not sure what we're going to do with that. Well, of course I am, but you're not. Let's locate the guard rail. It's at some position XC. So we've got the top of the rope. Maybe we'll call this piece one. Now we've got piece two that we're adding on. Piece two is what? It's XB minus XC. Would that give us piece two plus some constants for wrap around the pulleys? Would that give us piece two XB minus XC? No, no. That would be this. We want just this part, which is XC minus XA. Yeah, XC minus XA. Yeah, sorry. XC minus XA plus some constants. We know enough now not even care what they are. So that's XC minus XA plus a little bit of constants here. Oh, in fact, we have another one of those. Yeah, times two. Because section three is just the same. So we put a plus two times two in the front. Clean things up a little bit. Let's see. We've got a minus XA plus a two minus XA. So that's a minus three XA plus an XB plus a two XC plus a bunch of constants. We'll just throw them all in the same pot because we don't care what they are. Does that look like good algebra there? Did I do okay? We have a minus XA and two more minus XAs. We have one XB and then we have an XC and a bunch of constants. Look all right? Algebra looks okay. So L dot, which of course is zero because the rest of the rope is constant, is minus three XA dot plus XB dot plus XC dot, two XC dot, which is zero. The guard rail we presume isn't going anywhere. So XC itself is a constant. So we don't care just where the guard rail is as long as it's in this iteration between the two. We have a bunch of constants. The whole thing there is all constant and so that whole extra piece there is zero. We get that X dot B over three equals X dot A. So if the rail truck, which is B, moves 12 meters per second, the car itself is going to move four meters per second, but in which direction? Since there are no minus signs in here, they are going to move in the same direction. They are both going to be pluses. That would not have been the obvious, as obvious as a result, if we put the origin somewhere in between. Since those have the same sign and our origin is outside of the two, then we know they are both moving in the same direction at all times. If B moves to the right, A will also move to the right at one-third the speed. If B moves to the left, what happens? Nothing. You can't push with a rope. You guys know that. It won't work at all. So it can only go to the right. I almost tricked you there. What about the acceleration? Same thing. Just take the time rate and change that. You get X double dot B over three equals X A double dot. Simple as that. I'm going to do another reset here. Thanks, Tom. All right. What for you to do? I'm working way too hard. So just so we get everything lined up, well it's easier if you draw it this way first. First mask, then this goes down to another pulley up to the board. I mean the ceiling. And then from that one hangs B. OK, pretty simple. Pretty simple setup. Determine the relative velocities. If we know the velocity of one, what's the velocity of the other? Not necessarily the relative velocities using what we had last section, but the related velocities. First thing you need to do is pick some reference point. Turn the length of your line with regard to that. And the time rate of change of that length of line will be zero because it's a constant length row. Inextensible. That's obit from Earl at A's hardware. Once you've done L dot zero, you'll have X and A, X dot A and X dot B. And of course, if you took the L equation, you also know how much each one moves in relation to the other. So pick a reference line, measure everything from there. I recommend you don't attach it to something moving and you don't put it between the two. But if you put it to something moving, it's going to have its own velocity. And if you put it between the two, you get extra negative signs or fewer negative signs when you need exactly the right number on the negative signs. You'll draw X dot A is going from your reference down to the object, not both ways. Because we need to know whether that distance is increasing or decreasing to really give a full picture of the resulting velocity. As usual, your first time, we're getting some pretty interesting interpretations. Generally, students tend to over-configure these things. Joe, what are you doing? You're doing okay. There's something that might be simpler than others, but they should come up the same in the end. The choice of reference point should not affect the physics. It just affects the algebra. You got it, David? Got something. We'll see. Okay, I see a lot of people doing the same type of thing. We'll see how it comes out. Once you've sat on a reference point, it looks like most everybody picked the ceiling. That's fine. It's not moving and it's not between the two. It's as good as any. Once you've picked the two, then establish where the two objects are and the length of the rope and those turtles. Interesting interpretations of the coordinate. Did you two guys agree? Looks like it. Is that comforting or disquieting? We'll honor the other of them. Let's see. Most people picked the ceiling as the reference line. So if you do that, then that might locate A, most certainly, and locate B. However, notice that wherever this pulley goes, so does B go because of this connection here as a constant. So you can go down to here and then you'd have to add on that constant or subtract it off, or just say B is going to go wherever this is and just stop it there. Either one, it's just a little bit simpler if you just bring B down to there because those two are connected. It just reduces one constant and in fact you could have even had your reference line through this upper pulley because it's not moving either and then you wouldn't have had to worry about that length. But those are all just constants and they end anyway. So if I do it from the ceiling where most of you did, and maybe we'll call that H there, if you want to, you can make this one H1 and this one H2. If you don't want to bring it down to here, if you want to bring it down to here, you'll need that. Either way, it all comes out in the same because those are all constants and they all disappear. So the length of the rope XA minus H plus a little bit over the top. So XA plus some constants. I don't care what they are, I don't care how big the pulleys are, I don't care how long this link is, they never change. This little bit here is XB minus H. Is that right? So plus XB minus H, I'll just throw it in the pot for the constants. I don't care what it is, I don't care if it's a minus, I don't care if it's an H or a G or whatever, it doesn't matter. And then this third rope is just simply XB. If you had your line all the way down to here, it's XB minus H, it's just something else that dumps into the constant pot. So another XB and another thing to throw in the pot for the constants. L dot of course is zero because the length of the rope does not change and we're all done. X dot B plus two X dot B or as one goes up, the other goes down by a factor of two. They go in opposite directions at all times. Is that what it would look like if you imagine actually having this thing strung up and A was going down would B go up? Yes, we're pulling more rope above A, there's going to be less rope above B, B is going to have to go up. However fast B goes up, A goes half the speed in the opposite direction and the acceleration comes right out of that as well. So David, that's what you had. You and Chris had that. Anybody else get that too? Some of you, I think somebody had this distance in there. That's troublesome line. It's going to be changing. That distance itself is going to be changing. You would have another constant in there. But it wasn't important because we can get rid of how that changes by incorporating it in with the other constants or the other distances and some constants. So as simple as that, all I can do now is keep adding more and more pulleys. So let's add more and more pulleys then. Everybody done with this one? Any questions? Can you get that one okay? Oh, as a bit of a side note, let's see. What happens we had this one here, I think. What happens if we've got some angle between the pulleys like this? Does that affect things or is that always just something that we don't have to worry about much if the pulleys aren't moving? No, of course they're moving. Well, in this case they're not. Whatever the drawing was we just had. Well, I guess we didn't have this one strapped up. This one was hanging here and we had that there. So what if we had something like that? I guess it depends on just how far apart these are. The farther apart they are, the more that this is vertical and really has no angular component to it. The closer they get together, the steeper this is and that may be a factor in it. So the ones we're looking at, we're always going to keep these things lined up unless some very specific things are said otherwise. So that's part of why it kind of takes some time putting a sketch up for you so you can get those things all nicely lined up. So here's another one. Pulleys there with a mass there. It comes down to another pulley there. The cable goes around that one up to another pulley over that one. Attacks that to the roof or we're going to have a mess. Okay. So we've got that kind of system. B is B we already moved. So we stored it from the last problem. I know why I called that C. Call it B if you wish. C. I know why it's C. It's a coffin. Yeah, there we go. We're moving a coffin. All right, so same thing. Find the velocity and accelerations of one relative to another. In fact, go ahead and put some numbers on here. Six feet per seven per second up. What is A going to do? Two feet per second squared down. What is A going to do? And there is no B in this one. Really I want to test your flexibility now by not having a B in this problem. We're kind of a major curveball actually with this problem. So on the switch. So remember you can put your reference line anywhere even against my suggestions of not moving and not being between the two. You can do otherwise if you wish. Works pretty well to do that. I took those two pulleys. I was nice enough to line them up. But you don't have to. You can do it from the roof. The distance down there is constant anyway so it's not going to matter. Keep it as simple as you can. As few of those variables as possible. If we can do it just in terms of XA and XC plus a bunch of constants, then a hard part is done. You take the rest of the day off and leave it for your resistance. Like everybody's got a pretty fair picture of XA, XC or you could even go down to the middle pulley. Because wherever it goes that's exactly what C is going to do. And we're showing C stays level. So if you want to make it a little bit simpler. Just take C down to there. Because whatever that pulley does, the block's going to do as well. So we have 1 XA and 3 XCs. And then some constants for some wrap here and there. 1 XA plus 3 XCs. So is it what you said it was? You said 2 to 3. Famous last words. Maybe that was your mother-in-law in there. Since the time rate of change of that length is 0, then we're all done. Now let's see. This says if A is C, there is a B in the problem. See it was B for box or C for constant, whichever you prefer. As A moves up, actually we have C moving up. A moving to move is going to move down because of the minus sign. If we didn't have the reference line outside, if we had it between the two, we wouldn't have that minus sign. And it wouldn't be as obvious they move in opposite directions. You'd have to pay more attention to it. But you could still get the same answer. And of course then also the acceleration. So there's some numbers you can throw them in. I think what you need is either get out of class question for the day or get back into class question for Friday. So here's the setup. And a winch there that is 4.2 centimeters in radius. Maybe it's a red herring. The size hasn't mattered before. And it's turning at 47.6 radians per second. It's winching up the following. This line A is coming from a pulley. One pulley there. That line goes back up to a pulley hanging from the ceiling. That line comes back down to another pulley. Goes around that pulley, back up to another pulley that's connected to that one. And then they both come down to your body. Okay. Is that viable? Yeah. Everything is strapped in. We don't have anything floating in space. There you go. Find the velocity of C. This one's a little bit different. Before we had two blocks moving. One made the other move. Now we have a solitary stationary winch moving the system. Yeah, that's a solid link between those two. That's a bright pink line. You need to know more than that. A little bag. You might be right. It's just a bag what three means, what two means. So it'll be more specific. Plus I was looking for a velocity of C directly. So you can put your records line, your data points. If you can, the lower you put it, the less constants that are in the problem. Every time you lower it to something, another constant drops out of the whole problem. Not that we're worrying about any of them anyway. Just put them all in the constant pot. You could put it right there. That's the lowest possible non-moving point. That would certainly work and down to the highest possible and that'll give us the least amount of cable in the problem. So we don't have to worry about that link. We don't have to worry about that link. We don't have to worry over here because the winch isn't moving. So whatever point right there, we do have this link here. We can't ignore that. But it is a constant so we're not going to be worried about it. Give it any letter you want. We're done with it. So what you need to do is work the speed of the winch into it. No, no, no. That says that the rope is getting longer by 200 centimeters per second. L is the length of the rope so L dots the rate at which the rope is lengthening. Oh, it's what you're saying that it has to be. But negative is not shortening, is that true? No, the length of the rope is constant. The rope is being pulled up into here but we're not losing any rope. Are we? Are we losing or gaining rope? Just kind of use the thought of the purposes of this. You could say the rope is getting shorter by pulling the poles together. I guess you could. You could say that the rope is disappearing at a certain rate. I'm not doing that way. I guess you can. But you don't need to still do this as a constant length rope. One thing that might help is if you just pick some point on the cable going up into the winch and that can then serve as the second object in the problem you can relate to that velocity because you know the rate at which that point is going up into the winch. Or I guess indeed you could say shortening. Eldon is shortening by the rate at which rope is going up into the winch. Do you want us to find out the velocity of the seam? Oh. Yeah. Do what I say. Now what I write. Especially if they're in opposition to each other. We're at the end so I'll give you this one. The rope going up into the winch should be about 200 centimeters per second. That's just r omega. The velocity of A equals r omega. And you should get a velocity for c of minus 0.667. The minus being that x dot c is decreasing. It's going up. Which would make sense if we're taking rope up into the winch then of course the block is going to rise. No, that's 200 centimeters per second. I think that should be meters. But that one should be meters. But that's right, isn't it? 200. All right. That's a wrap.