 All right, any questions? We do an exam Friday, open book, open notes. We'll be over, I think, the first three chapters, which takes us, no, first two chapters is off. So we've got the rectilinear motion, curvilinear motion, couple of coordinates. Won't be a big deal. Now, in any of those, I'm not trying to trick you on any of this stuff. Just see if we got the basics well enough to keep moving. Because we do accumulate a lot of what we're getting here. Have a couple choices of what we can do today. Got another constrained motion problem. I also have some of the problems from last year's exam. So we can look at any one of those you want. What? What was the name of last year's exam? Last year's exam. Okay. You may have lost, I had a dream about you. Right. It was in physics class, so it was exam day, you started taking an exam, and then you started talking to someone. But you took your exam away, and then you started crying, because you said the reason you were talking was because somebody died, you just got a text message. I decided, well, do it. He's getting text messages here in an exam. Should I let that go? Or, so, ended up just taking the test away for 10 minutes, cried for a while, and I woke up. So this is the paper. This is what? One of the homework questions? Yeah. Was it? Okay. Well, that probably got us in from the same place. So, we don't need that one. That's just a rectilinear motion on one, deep, and only go up or down. So you need to assume constant accelerations for other things. This one was, well, it was an illustration, kind of an illustration, to show that different coordinate systems can be used on the very same problem. Because with the northeast coordinate system, some of the students chose to do X, Y, I, J notation. Some others went ahead and put in the normal tangential coordinate system and answered it in that way. And I accepted either one, because I didn't make a specification of how it should be. Well, that one should look pretty straightforward, is it? Like, coordinates, coordinates to use. All right? What? All right. No. My question is super. That's fine. That's okay, because that's what I just said. This one was to illustrate that on some problems, either one can use. That's how it can work. In fact, even the polar coordinates would work on this one as well. So on this one, you were given a car traveling in a circular radius path at a certain point, at a speed of 45 miles per hour, 30 degrees north of east. Didn't actually label it, but I guess I expected the students to come up with that, was 30 degrees. And it's speed is increasing in a rate of five feet per second. You were to determine the acceleration. Remember, acceleration is a vector, so I needed some representation that would give me a full vector solution for this. I also said I wanted it to be feet per second squared. Okay? Let's tell you to read it. 450 foot radius on that term. For a car, actually, it's a fairly tight term. Now, this is one I think would be easier in normal tangential coordinates. Why is that? Because Alex asked that very thing, he said that I'm sometimes confused in which coordinate systems I should use. Well, I'm very good. Kind of the normal tangential. We have the normal component that's directed right towards the center and the tangential. Oh, wait, we used E. Some books use just the N and the D. Subscript C for C triple or C triple, or C triple. On that, I'd rather have the normal because it is the normal component. It's not a centripetal component. All right, I suppose you'd have to read it. Actually, yeah, now that you bring it up, I think our book uses the, yeah, I know they use a normal, but I think they had the normal component. No, they do have it to the inside of the circle. All right, so then the question remains and to help Alex with this, why would this be easier to do it than normal tangential components for this problem? Right, another question here? Yeah. Yeah, Jay? So do you have the tangential acceleration? We already have the tangential acceleration. In fact, remember this acceleration is made up of a piece in each of the two directions. And from the wording, this speed is increasing at five feet per second. That is the tangential acceleration. But not only that, what else is true? The normal acceleration is the centripetal acceleration, which is why Jake asked, can I call it that? And we know that the D squared are actually for this we call it D squared over row, but it's the same thing. And I would certainly wouldn't comment off for that. Alex Trebek would have put on, oh, by the way, big night tonight, the game shows. Jeopardy is gonna have a computer, is one of the contestants, an IBM computer. They've been working, I think, five years. Got through hundreds of test matches and now they're, tonight for three nights, the computer goes up against the two biggest Jeopardy champions ever. And on cash cab debuts in Chicago tonight, cash cab Chicago, one of the people getting on is a friend of mine from elementary and junior high school. Yeah, yeah, so I'm famous tonight. So extra credit for watching both and then writing more important. So the computer, Jeopardy's going home three nights. Three nights this week, so that'll be pretty interesting. All right, so what deals remain? What VDU's, since it's accelerated? Yeah, even though the velocity's changing, it's the acceleration, it's the velocity at that instant shown gives you an acceleration at that instant shown. So if you choose to just do it in the normal tangential components, I think the hardest thing left is you gotta make sure the 45 miles per hour is in feet per second. That's about the hardest part of the whole thing, was for normal tangential components, we tend to use for the radius of the curvature at that point. That could change as well as could the center of the curvature because very few curves are actually truly circular for more than a little bit. What? I mean, is it hard? Yeah. Do you like seeing that? Now you've got columns, not a missile disease. Spreading. That would go back to what you wish. Is someone saying that's not spreading? No, no, no. Is someone saying that's not spreading? Is someone? Yeah. Yeah, I'm happy to do it if you'd like, but. Happy to do it if you're stumped and homework's due Friday and so probably sometime Thursday I'll post the solutions. Perfect homework support this week. Remember, I'm not great at the content. I'm great at the performance. See if you do a good job putting them together and then you check and see if you're doing it right. Oh, why don't homework you wanna look at or other than these? What are you guys and gal that you need? Are we gonna have one that we have to use more than this? No. It. There may be one where I would hope you would. Remember, the ones that are good for that are the cases where the center point is fixed and so you have, especially some kind of tracking satellites type thing where the information from the satellite would be exactly the things you need for the pull up ordinance and it would be just a matter of filling it in and you have R, R dot and R double dot. That's very easy data. In fact, most of you I think have used those, those TI calculator physics things, CVLs I think they're called. Calculator based laboratory things. You might have used those in high school. We've got a full set of them. We just can't figure out how to use them. It gives exactly that kind of stuff and then also theta, theta dot and theta double dot and once you have any of those, it's a matter of just plugging them into the polar coordinate equations. I just know some of that homework is like really like there's a lot of substituting like the variables and the gauss and the gauss. Well, that's okay because we'd find you. We'd track you on radar. Okay. We'd just take a swell and fill in the equations come get you. So, let's take your little pup, Tim. Where do I get you? Simple as that, that one's pretty straight forward. No tricks I don't think. Like I said, the hardest thing is just getting that into the, you've already been given that. I think getting the end of the beat per second. So, just a matter of putting it in, multiplying it, keeping it new and straight. What students were allowed to do if they wanted to, though I can't imagine why they'd want to, is they were certainly allowed to put it into X and Y for it. Certainly allowed to, and it's not wrong if you do so, it's just remember every time you take a step like this when you go from one place to another place, it's a chance for an error to happen. Making that simple conversion. It's just a simple trick problem, but it's easy enough under the pressure test to just screw that up and so if you take something that's correct, it has here and move it to here but make a mistake, I can't let that ride, you goofed up. Would that just ignore the simple trick? No. Oh, yeah, I had the velocity in it. Well, no, what you'd end up getting, actually I guess I drew it on the wrong vector, you have some acceleration like that and then that one would break into XY components. Some students are very tied to XY proportions. So that one, I've written down 9.7, I thought you said 9.8, I was hoping for a void argument, but third one, they're actually four because we've covered more material on that one. There was the third one, it's a problem, we've been looking at those on Friday that actually not unlike the kind of thing you do if you ever had to winch a fallen climber out of a crevasse, just the kind of thing they set up. So given that block A is moving to the right, possibly an acceleration of block A. I'm going to look at this actually. For what I said is advice for a coordinate system, I think I gave it two particular pieces of advice for coordinate systems. Keep it fixed, wherever you put it, don't attach it to one of the moving objects. You can, it just makes them more difficult. But the other thing was not between the two objects, now this is where the curve ball is here because we have two directions, I'm going to talk to them, they have been retreating. I'm on I am with my lawyer, you have to pay attention. Remember what it is, the basic thing you need to solve these problems. The premise on this whole thing is what? Yeah, the length of the rope. So however you can find that, I don't know what that means, huh? Yeah, that's a magic pulley. There we go, put the thing on. We're using it for a coordinate system, still kind of thinking. Share it with Alex and he didn't like it. Actually I don't want to write anything this way. Write anything down when it's wrong, you can erase it. I don't want to waste my time here. Wait, you could have put your diagram and pen then your attempts and pencil and you could have just erased them off and your diagram would still be okay. And sketch them. That one's in there. That one's full of talent. I'm in a pencil group. I try to use a pencil and it's like a snap when I'm out of there. What'd you do? Here if you do that enough times. These three lines? Yeah. Well, no, they're not even right here. They're not pulling. You're the top pulleys, Well, it doesn't matter because that pulley moves and is as it does, the length of that will change, that will change, that will change. Once you've found the length of the rope, remember what the next step was? Oh, and you didn't watch the video? I didn't. It's not that long. All right, the whole deal is if we can find the length of the rope, then we take the length of the rope will be some function of the position of the two objects plus some other stuff that we call constants. That's things like the amount of cord wrapped around a pulley. That's just never gonna change. No matter where the object is, that doesn't change. So there are a bunch of little things like that that we just call constants. So it all had to do with where A is and where B is and then all the other stuff that never changes. And then we take the time rate of change of the length of the rope. What should that be? The rope doesn't change length. So now we have this as a function of the velocity of A and B because they integrate as well. And then what's the time rate of change of constants? Zero. So all that's left is some functional relationship between the velocity one and the velocity of the other. Which is exactly what I asked for. Gave you the velocity of A, asked for the velocity of B. Then you take the derivative again and you have a function that relates the acceleration, which again was the last part that I asked. So the is really the hardest step because these were very simple derivatives to take, which is all done. What'd you do? Did you even look over her shoulder? I would at least see what do we've got because maybe you should have made a little mistake and it's pretty easy to learn this. Well, where'd you put your origin? Here? Like where B and A, like? Where they intersect, sort of? Yeah. So you did something like that? Yeah. Okay. And then measured from there. That would be the distance down to B. How do you handle A? Does that work? Maybe it would rest more easily with your I if you even wrote that. Would that change anything? Not really, no, not at all. All this motion's in one direction, all that motion's in the other. They're linked, of course. So let's see then how to do the length of the cables in that regard. So we have this piece of cable, which is YB minus whatever that piece is. I call that constant. That's a constant, all right? So we've got YB constant. We'll just dump in there. So that little bit there above that line is a constant. We'll just chop it off, dump it over there. We've got the YB minus whatever that is. We've dumped it over there. What about this one? The second cable, the easiest way to do it, make sure you don't want to skip any parts. What about this cable? It's also YB minus that part, which is a constant, and minus this part, which is a constant. So it's another YB, put a 2 in front of there. I don't know how many we get. Now we've got two things dumped over there in constants. What about this cable? It's YB minus that amount, minus this little amount. But those two are constants, aren't they, right? All right, so that's a third YB plus a third little bit dumped into the constant bucket, where you have trouble. Yeah, I did. All right, so we've taken care of those three areas. They've been comfortable with those three. How about it's xA minus that little bit, and minus that little bit, plus xA plus another little twig thrown into the jump bucket. Well, that was this one. Now we have that one. We've got this one. OK, so that's xA minus that little bit, just more constant. YB plus 2xA, dj, that makes sense? Do we, that's what you've got up there? Pat, come figure that out? Yeah, you take the time derivative of that, 3y dot b, the velocity of b, well, we're asked to find that, plus 2x dot a, which we're given. Because everything else is a constant. Both the length is a constant, and of course, the constants are constant, so they all have to reverberate to zero. And b, which we are asked to find, is what, minus 2 third, whatever x dot a is. Don't even care. But we are given that, minus 10 thirds. What's that, 3.3? What's the minus mean? Yeah. It's like you just know the answer. Yeah. So this means that y dot b is decreasing with time, which means it's going up. As it would, you'd expect if a was going to the right. So if x dot a is positive, y dot b is negative, advice person. If a is getting bigger, b is getting smaller, advice person. So minus sign in terms of what's happening to the coordinate we set up. We could have done y b down to be negative, I guess. It's just a distance. So that's the best left positive, I guess. We can just take the second grid over the bad equation. Straight off. I'll give you another one of those. We've got that, and try to clean the board. Put a little bit more practical application to it. A winch set up here, for some reason named C. Oh, this is the winch, 4.2 centimeters. Angular velocity of the winch, 47.6 radians per second. Should have given it an r, d, m. Find the velocity of C. After a conference around there, we actually had that in there, pi r, for each of the different polies. Then quickly realized it's a constant. So we don't care about the details of the constants. Yeah, that's right. That's the easy part. Moving and between objects. Drew it there, but that's not having to measure anything. So we drew an origin, then ignored it. We used a y. Doesn't matter. So the length of the rope. In those terms, it might be more cables wrapping around that. It's always the second person to go is the first hit me. Let's say it's winching up. It wouldn't matter. The equation's all going to be the same, whether you have a negative sign on it or not. You do it? No. 3yc, you're getting that a lot of the answer. You know something's wrong. Because if you've got 3yc, and you take the time to rid of that, yeah, so that's not it. That's not going to work. Unless, well, if we turn the system off and go home early, I guess that handles it. Somehow you've got to get this in there, the fact that the cable's moving. No, no, non-stretchable cable. I bought it from Ace, got it from Earl of Ace. He was pretty dumb to buy a stretchy cable, so I imagine in especially large crane design, they have to take that into account. Luckily, we're not there yet. Don't just save this for Friday. We're on negative sign, which should give us a hint. What do you do? It's all up there. Give you that hint. Let's see. Maybe we can. We've got to get the fact this is turning in there somehow. A little mark on there maybe called it A. It's like A. It's the speed with which that rope is moving up that determines the speed of the whole rest of the system. This system, you could take out this winch, run this up to something else, and it wouldn't be any different than any problem we've already done. Because you know the speed A is moving. That point A, it's just the speed of the winch. Minus, well, there's some constants or something up there we're not really concerned with. We want to do that whole length of this cable, I guess. So its length is yc minus, oh, we have to worry about all those things. I'm going to manage this. Not worry about that. That's a constant. That's a constant. So that's not a big deal. That's a constant. What was the other dot there? Where? Here? Length c. This length c? Well, it's a little confusing. Length b. Make this length b, but that's changing. Let's do this. Let's start from the other side. So the length of this is yc plus a bunch of constants. Because that's a constant. That's a constant. The amount around each pole is a constant. How about this length? Well, y a only if it's perfectly level with that. But as this is going up, that's going to go down. You think, aren't these two lengths, what if we took those together? All we care about is we take track of the entire cable length. So these two together is yc minus this amount, minus that amount, and minus that amount, which are all constants, plus yc plus some more constants. Oh, sorry. You could, I guess, but you'd have to know you just need to locate some other things. You need to locate this. I would guess you'd have to locate that fully. And then, well, that one's a fixed distance away from that one, so that wouldn't be a concern. So it would come out outright to be the same thing, the point where we know it's moving. Plus a bunch of constants. It's yc minus ya, which isn't constant, minus that, which is. Length of the cable, these two cables taken together. Total distance is yc, top to bottom, minus this amount, which is constant, so we're not worried about it. Minus this amount here, yc minus that amount, it's as if we now took this off. We just have now yc minus ya, which is the point. This is the length to this point because we know that velocity from the speed of the winch. So that's yc minus ya minus a bunch of other constants that we don't care about. We don't care if they're minus, plus or minus. There are a bunch of constants that make this constant. But we don't care if it's all yc. So we want to pass those specific constants. That doesn't matter. And we're going to go with yc minus the ya. And minus a constant. That's it with you, Pat. Oh, yeah. Did he help, Alex? Is that a good explanation? I'm just too lost. All right, here. Maybe there's just a different way we can draw this. So let me try this. I don't know if this will work. Let's try if this will work. Here's a length of cable that's yc long. If I had a piece of cable like that, it would reach down to C from the top. Now, I'm going to take out of it that amount, which is no trouble because that's a constant. I think I just got it. And then I'll take out this amount, but that's ya. So I have yc minus a constant, minus ya. So yc minus ya plus some stuff in the constant. I realize that we're just the second pulley was also moving upwards. This pulley? Yeah. These two? Isn't that what's causing this? Yeah. Well, if cable A is going up, point A on the cable is going up, then that pulley is going to go up as well. Cable doesn't change to that song 0. Or we just had yc up there, or free yc for L. Why only put the time variable to 0? Well, when you originally had L equals 3yc. Yeah, why did it go to 0? Because the length of the cable is constant, supposedly. And that's what this is supposed to represent. Plus constants. So it can take the time derivative of it. That's 0. But the time rate of change of the length of the cable is also 0. That doesn't mathematically fall from here. It's a restriction we have on the problem that this is a non-stretching cable. The length of the cable itself is a constant. So if you set that to 0, then you don't have anything left. There's no way the velocity of the pulley, the winch, sorry, was in there. I mean, when I first looked at that problem, I automatically just set it up like this. Why is that? Here's what Jake had, and he asked the question, why is this wrong? This is how he set it up originally. He had called that w, called that distance w, and then that distance c. And you said what the length of the cable then is w plus c for this first piece? It did it like a what? Like a track, like starting at the winch and falling down. So you said that this is 1, 2, 3w, right? Plus 1, 2, 3c minus a couple constants with a k for constants, because you have a c in there. And then l dot equals 3w dot plus 3c dot equals 0. It doesn't make sense because the boys aren't doing anything good. Everybody see trouble? It's doable, but there's a big, huge extra layer of complication in here. Just end up having too many variables to build a software, right? I think it's great to just don't know what w dot or c dot is. You don't know what w or c dot is, so you can't, we need c dot. Well, actually what you need is w plus c dot. Plus this point right here to which he's measuring c is moving. This point right here is moving. So he's essentially put his origin on a moving object, this point, and between the two objects doing the moving. So he was able to violate both of my recommendations on where to put the origin. I didn't know I was doing that. Well, you can get there. It's all screwed up by the fact that that point there is moving. Now if you can figure out how fast it's moving, then you can account for that in these different images. But I wouldn't even want to touch that one. The velocity of a, just put that in there. Velocity of a, I think it was two meters per second. Do you want another one? Because you like drawing? You know, if you could bring in colored pencils yourself, I remember when I started doing the first professor I had who used colored pencils. I said, boy, that works great. Took me about half a semester to take into my notes that same way. All right, so here's, I'll leave her being part of a crane or something, maybe an overhead crane here, unfixed pulling below it. And that's where our load hangs. This is the second. This pulley here, so I'll put some little rollers on it because it's attached to a piston that pulls it in and out. All right, you got that? Before I put the cable in, pay attention to what the cable does. Around those two pulleys, I think there's slack in the line. The piston actuator must move to raise the load two feet. You're a slacker. Is that what you meant? Let's label everything here. If you want about any one of the pulleys, make sure you know which ones move and which ones don't. Now there's no slack along, and I just got excited taking the corners as a good call. We could use a chain here. And then some of these pulleys could be like bicycle derailleurs or change gear, change the radius of the pulley. The radius of the pulley isn't a constant as we've been using it. All right, I guess we have to make this a get-up class question. This is going to go up two feet. How far is that one got to go? Gotta make sure the piston's big enough. The actuator. Set up your measurements in terms of what's moving from something that's fixed. For example, we could do and we'd know where F is. Point E isn't moving. Same as the actuator. So the length of the cable. Let's see. XC plus XC plus some constants. Plus this piece here and that piece there. You're okay with these two cables? This cable. Well, that's a constant because this isn't moving, that end isn't moving and this is a constant. So all we're left over with is just the two vertical parts here. Which is two YF. So for F to go up two feet, the end of the actuator arm C has got to go which way? XC increases. Because these two are all together accounts. So if YF is decreasing, XC is increasing. It's got to go that way. Which would certainly make sense because if you take more cable up at the top you're going to have less down here. So that system is really too many things for mechanical advantage? No. Other than whatever was doing in this direction is now doing in this direction. Which itself may be an advantage. Maybe you need to get your mother in law out of a crevax. Sorry, Nana. Just kidding.