 Maybe we need to go to a conference. Any questions? Friday we looked at what I hope was mostly review. And we looked at pure rotation. Very, very similar to what we did in physics one. The only possibility really is that physics one we pretty much stuck to constant acceleration. But here we would not necessarily be stuck to that. But the rotational business is so similar to the particle business that there you go, it was worth one day, one day only. So any questions before we take our next step beyond just rotation? So now we're going to look at general plane motion. General means that most anything can happen. There's lots of possibilities. It's not pure translation any longer. It's not pure rotation any longer. Could be combinations of the both. Plane means that whatever objects we're talking about will stay a 2D plane. So the easiest example of something like that is simply a wheel rolling along the ground. If we have some reference line on it, some little bit of time later, that reference line will have slipped through some angle. But that's not pure rotation, of course, because there's no point on that wheel about, there's no point on that wheel that didn't move somewhere. Everything moved at least some distance. Certainly the center moved that far. And whatever other points that we could look at, that was just some random point A. Then A would move that far. I guess we should put a displacement on those, rather than a. And all possibilities of the types of things could have happened with that. We're also going to be concerned with the angular acceleration of the wheel, which may or may not be constant, of course, and the associated velocity at any one point in time, remembering that not only is there a velocity relative to the center, but there's going to be some velocity of the center itself. And it's the combination of those things that's going to give us the true velocity of that point. And we're going to be able to do this for any point as we go through these. So we're going to look at this in two different ways. First, we're going to start with absolute motion analysis. And then tomorrow we'll look at relative motion analysis. And we'll stick more with that, but allow us to do some more complicated things. Absolute motion analysis is very good for single things like wheels, mildly compound things like a two-gear system or a one-gear and a linkage, that type of thing, whereas the relative motion analysis will let us do a lot more complex things as we go through it. So our absolute motion analysis. And we'll stick with this general example of a wheel turning. It's one of the easiest to do. We need to sort of set up a procedure for looking at these. And then we'll go through a couple problems where we do just that. If possible, locate some point that travels a prescribed path. If it's a straight line, that's about as good as you can do. For example, the center of the wheel travels a straight line. So that's a nice place to start the analysis. For very general motion, this might not be as easy a step as it might seem. But for our time here to get started, most of the ones won't be able to do something like that. Second, sort of make sense, align the coordinate system with that path. If this wheel later plays, we know that the center takes a path like that. And so we'll line up our coordinate system with that. And that's just what we'll do in a second for more of the analysis. That'll help with the linear position. But then from that, after that, we're going to have to determine angular position. And it may be just as simple as picking some arbitrary reference line and then seeing what that does as the wheel moves. Once we've got those two, essentially two coordinate systems, the linear translational reference frame and then the angular one, then we need to relate those two. So we want to relate S to theta. And then from there, we'll be able to relate V to omega and A to alpha. And we'll be able to put the whole problem together. Now, sometimes that steps a little bit easier said than done. But for most of the problems that we're going to do, they're fairly straightforward. Right, for example, this problem, if we allow it to turn through some angle and we let it do so, expect it to do so without slipping, things are much different if it does slip, but we'll let it do so without slipping. We've already got the coordinate system laid out. Now we've got an angular position. And hopefully it's not too big a stretch for you to remember. Then the distance it traveled will actually be how much arc length rolls by. And then those two are related just that way. So that in itself is one of our no slip conditions. This would not be true if the wheel was slipping on the ground. That's going to be important to us. I don't think we'll do any problem. I don't remember any problems coming by that have any kind of slipping with them. So if we're talking about gears of some kind, then that's definitely a no slip situation. Because that's the intent of the hearing. You see the arc length? The arcs of the radius are the radius of the wheel. So that's OK? Remember this only works if theta is in radians. This does not hold if you have theta in degrees. And then it's easy enough to continue it from there. We know that the velocity is the time rate of change of the position. R is a constant. So this becomes then the velocity equivalent of the no slip condition. Again, now omega is in radians per second. And then the acceleration. Where we remember, I hope, well, this is, I guess, without saying, this is the position of the loss in acceleration of the center points. And that's where we've pegged everything. And then we're going to have to, if we look at particular points, we're going to have more than just this acceleration. Because not only are these points translating, the center is in simple translation, where that may or may not be constant. However, every other point is not only in translation, but also in rotation about the center as the center itself translates. And so we're going to have to put that together with it in a more complex analysis. So these are all of the center point, I guess we could call it g, sort of per center gravity type point. And then for each one of those three things to hold, that either requires or confirms the no slip condition, which is going to be real important to us as we go through this. All right, so we're going to redo that analysis in a little bit more detail now. Because that's simple enough. They're just the center. But for any other point on this object, we need to know more than just this. This is the velocity, position, acceleration of the center. We're going to need to know more. Not too bad, too bad for a hand-drawn circular. All right, so there's the initial position of our wheel. We're going to use the same coordinate system and let it roll. I'm not going to draw the floor necessarily. But then we're going to let there be a reference line that looks something like this. I'll label that point A, that point B, give it some angular velocity that will cause it to roll along that surface. Nice and big. I hope you're drawing it nice and big. All right, it's some little bit of time later. It will have rolled to here and gotten all flat and lumpy. That's a little better. That'll do. And that reference line, let's say it has rolled to something like this. So we'll call that A prime and B prime just to help us with the notes as we're writing this. So something like that, it's rolled that distance. OK, so some radius are some angular speed omega. Some little bit of time later, it's gotten to there. See what we've got. The center has traveled from there to there. I guess we can call that G to G prime. So let's see. We'll let that be our distance, x. So x equals G to G prime. What else does that equal? It equals this arc length B all the way down to this bottom contact point, which of course is r theta. So it's traveled that far. What else can we make out of it? I think that's all we need for there. All right, but let's put some other. That's no more than we had over here, but let's do a little bit more. Let's see. Let's figure out where point A is now, or A prime, as it's called later. But it's the same point, just different spots and times. Let's see. It's a distance x, which is just what the center did. But there's a little bit more distance added on to that, which is what? r cosine theta. Is that right? r theta rather than the x. r theta plus r in the x direction. Is that where point A is now? And it's in the y direction. Happens to have gone down. That's minus r sine theta. Theta happened to be less than 90. We'd actually want this to be a plus, wouldn't we? Because it would be above r point. That would then be theta. Right, that's cosine theta, isn't it? Yeah. Got these backwards. It only rolls to here. g prime plus r sine theta. Cosine theta, so we do want that to be a plus. And then if theta happens to go greater than 190, then this turns minus automatically, and it's still OK. That make sense? Help us out here, Alex. You went to a math conference. Is that about right? That's a little nice. That's right. That's about it. That's beautiful. That's what that is. OK, so I think we're all right with that. OK, so that's where point A is at any time, or not any time, we don't have time on here yet. But that's where it is at any position. So now we want to know something about the velocity of A. That's the time derivative of that position vector we just established. But that's got a couple things going on in it, though. So that's going to be drA d theta, d theta dt, because r is not a function of t. It's a function of theta. Not too big a deal, because what's this little beast right there, d theta dt? First is our angular velocity, omega. All right, so we need to do that. All right, first piece is r, which is constant, then theta dot, same as r omega. Second piece, r, again, is constant. So it's derivative of this respect to theta, which is cosine. That's dr d theta of this first little part, d theta dt, which, again, is omega. OK, we're all right. And then we need to do the derivative of the j part, r's constant, so this is just sine theta d theta dt. Because we took the rd theta, no, took the derivative of this, dr d theta, this part, memory, when you take derivatives, you can do each part as you go. So that is just this one. r sine theta, the derivative of that is cosine theta. And then d theta dt of the inside part, because it also varies with time. It's a product rule. I think that's what you call it, chain rule? Oh, yeah, chain rule. I thought that this would be nice to take the derivative of respect to theta, and then you can just multiply them that entire thing by. That's the derivative of respect to theta, and then that's d theta dt, because that's the chain rule. Why don't you just take the derivative of respect to theta and then multiply that entire thing by d theta dt. In order to get almost the same thing, but then r omega to v r omega squared. That's what's confusing me. Why don't I do that? Because it's not right. Where would, where does the extra omega come from? There's only one. Answer will be the same. I do it this way, because that's what's right. That's the, I mean, chain rule. Anyway, I can factor out omega, r omega, one plus time theta. That one's negative sign. Okay. Oh, yeah, good. So we have a general expression for the velocity of some point a. And a general expression for the acceleration of some point. We're gonna take a look and see if any of what's contained in there is stuff we, stuff we need and we're, stuff we need to remember. All right. So we need to do the derivative of that piece there. Omega dt, which is r alpha. This part plus cosine theta. That's gonna be a minus. Omega, because that's there. Plus another omega, so that's an omega squared and theta. This part taken as constant, just that part, which is plus r in the i direction. Let me check. I mean, theta. This is r cosine alpha. r and omega is constant in the derivative of this cosine. That's the middle part. Then there's another d theta dt. Then the third part was we take this as constant and just that. Oh yeah, yeah. So that's not omega, that's alpha. That's what you're saying, do we? Yeah. Yeah, you're right. So that's the, that's the i part, the j part. It's gonna be a little simpler. We've got r sine theta cosine theta d theta dt, which is another omega. Plus sine theta, yeah. Plus omega squared cosine theta. Yeah, okay. All right, I think we've got all the pieces. Now, are we okay with that? Now we've got the three, the three possibilities. So let's check them at something we either already know or something we're gonna find as very, very useful. So at theta equals 180 degrees on a wheel, that's useful to us because that means now we're gonna know the velocity and acceleration of the contact point. That'd be really useful if this wheel was a gear. Do you know what a rack is? There are some gears that run on racks, which are sort of linear gears. So that the rack can move and oftentimes the gear is fixed and now we can associate the acceleration of the rack with this contact point because the contact point is always, of course, at the bottom. It's also where our no slip condition is. So let's check it. R, the position of point A, one plus cosine of 180, minus one, right? Minus one, so that cancels one minus one in the I direction. Find the velocity of point A. No, the location of point A. Just a second, something's not right here because that says it hasn't moved at all. If the I position is zero. So what's wrong? What? This goes into 180 is zero. No, goes into 180 is negative one, right? All right, let's look at it. Are we using that equation instead of the one to your left? Oh yeah, that's the velocity equation. There's the trouble. Phew, that makes, yeah, let's use the right equation. Sometimes that helps. So we've got, what's 180? That's pi. So this is pi R. That's this part. Sine theta of 180 is zero. So that disappears, so it's pi R I. And this is minus one. Cosine of 180 is minus one, so that's minus R in the J direction. Is that where point A is? It's a distance, here, let's use this picture. It's a distance pi R, right? Half of circumference at roll. So that makes sense. And then it's down a distance minus R. So that makes sense. So that worked okay. That's no surprise, we knew that's where it was anyway. But we got some more, so a little bit more to learn with these next two pieces. Velocity of point A when theta equals 180 degrees. So it's R omega times one plus minus one in the I direction minus R sine theta of 180 degrees is also zero. So this is plus zero J. What then is the velocity of point A at the instant when theta equals 180 degrees? The contact point at any instant is not moving. A moment later, the wheel has moved somewhere else, but that's a new contact point. You can also see it as, well, if the wheel's in contact with the ground and not slipping and the ground's not moving, then of course the contact point has no velocity. So that's something we're gonna be able to use, but you've got to remember that's gonna come back to us, we're gonna need that. Not so much, I don't think we need it so much again today, but we are gonna need it in starting Wednesday, or Friday with the relative motion. All right, the acceleration of point A, when point A is at the very bottom, that instant it's there, we've got R alpha, sine 180 is zero and cosine is minus one so that's minus R alpha. So there's zero x direction acceleration of the contact point. I think that makes as much sense as it's not having any velocity. Let's see what happens next. Oh, we've got a minus, sine theta, that's zero. R omega squared cosine theta of 180 is minus, so we have two minuses, that's a plus R omega squared j, cancels. The acceleration of point A is R omega squared j. So here's the contact point plus j, it's got an acceleration of R omega squared. That's what? No, there's another name for it than that. Centripetal. That's the centripetal acceleration. Remember centripetal acceleration was R omega squared. So that's the centripetal acceleration. It's, yeah, it's normal to the surface, but that's where the center of that circle is. So at that instant, that point A is in a simple circular motion about the center. An instant later, that point doesn't really exist anymore. There's a new contact point, so don't think that since the contact point has zero velocity, the wheel's not moving. That's not true. At that instant, the contact point has no velocity. Well, let's see, let's see how we can use that. How that's going to look to us. At any instant in time, this point has zero velocity. The center has velocity of V, whatever the velocity of the wheel is. If that was your car, that's what we'd see as the velocity of the car. What velocity does that point have? Let's see, that's when theta equals zero, so we could put all those pieces in here. Cosine of zero is what? One, so there's actually an r omega plus an r omega. That's two r omega, and that part's zero. So it has some velocity, two r omega. Is the velocity of the center with respect to r and omega? It's r omega, so the top point is going twice the velocity of the middle point, and in fact, we can even draw a velocity profile, and we can figure out the instantaneous velocity of any point along that diameter at that instant. That's useful if we need to hook something to those. If that's some kind of gear or a can or something, we need to hook something there. Now we know the velocity and the acceleration of any of those points, so that's useful. What do I want to do? So that was our absolute motion analysis since we knew that the center moved along a prescribed layout. We also had the no slip condition, and now we know the position, velocity, and center of the contact point at any instant in time. Okay, so let's do another problem. Here we have a cam. It's a circular cam. However, it doesn't rotate about its center. It rotates about a point right there. Contact with that at a point level to the mounting center. The spring loaded rod, spring loaded just so that it'll follow, stay in contact with that wheel. Like an indicator might, or something you'd need so that you get a periodic motion perhaps in a manufacturing, I'm wondering. Those rollers just hold that thing nice and level. So as this cam turns, remember it rotates about this point, not about its own center, as that cam turns, this will go in or out depending upon what its contact does with the cam itself. We'll also give it the possibility of some acceleration too. You can imagine we're gonna need to know the velocity and the acceleration of the rod itself, which of course will be the velocity acceleration of that contact point as well. So let's see, this point will make a good basis for our coordinate system. That rod's not gonna go anywhere in the y direction, so that's not a great important, but it is gonna have some velocity here in the x direction. It's a cam, an eccentric cam, off-center cam, with a follower. As this wheel turns, as this cam turns about this point, this rod's going to go in or out depending on where it is. So at some time it might be like that, a little bit later, it'll move to there, and the rod will have moved to there, possibly it comes back again when we get the rod to oscillate in and out. So the position here, call it point B, call it center point A, the position of point B is two R theta if that angle is theta. Because we have an isosceles triangle there and twice the base will be two R cosine theta. Y is by definition of the setup, zero, and it won't change, so we don't need to mess with that. That's some nice animation. Sounds like a pain to do. So the velocity of point B, which is the velocity of the rod itself will be the time derivative of that piece before. Two R's constant, derivative of cosine is minus the sine, and then again, d theta dt, which is omega, because the X position here's theta, so there is cosine theta. That could work. Yeah, now theta's right here, the line drawn to the center makes, and the acceleration of point B with dvv, a little more involved, omega, cosine theta, and then we need the d theta dt, which will give us a omega squared, taking that as constant, we get, not to say that alpha. So we have the velocity and the acceleration of the rod, because that's the same as the velocity and acceleration of point B. Good job, about a while ago. So you're not even writing any of it down. Oh, I wrote that, I'll follow up on this later, but I'm kind of lost. I think the more examples we get, the better. That's a three line problem. They don't come any shorter than that. Right. Except two line problems. Except, well, I got it now. All right, well, another problem then. A hydraulic actuator opens up this door, this window. You okay with that one? Yeah, so much so good. Yeah, attached right in the middle, got the picture and acceleration of the door back next week and then entered question a week after. Yeah, was, oh my God, I'll be right back. Somebody hit him? Is everybody else okay with this picture? Okay. Yeah, I think he was asking where. Don't have it there yet. Oh, I guess I better now that I said it. That's angle theta. Essentially what we need to know is, if we can find the speed and acceleration of point B, we're gonna be able to relate that to the speed and acceleration of the door. I'll get you started here with the position of the door. We'll call this, we'll call the length of the actuator at any instant S. So we can do the law of cosines, final O and that will give us then the position S. We take the derivative of that, we'll get the velocity of the actuator, which is a known and it'll be a function of, then the position of the door or theta will be with respect to that. Then we can find alpha as well. We'll turn to the law of cosines. Because that will relate S to theta. S squared equals, because the S is opposite the angle we have in the interest, so that's how the law of cosines sets up. Then what? The two sides, two other sides adjacent to that angle, squared, so that's O B squared plus O A squared. Well we know both of those distance, one's a meter, one's two meters. What's the rest of the law of cosines, remember? O B, O A, cosines. Good, so there's the, that relates the position, the length of the actuator as a function of theta. Reduces, because we know some of those things to five minus nine theta. That would be a meter squared. One, that's two squared, so that's five, that's okay. That's one, that's two, two cosines theta. Where'd I get it for? From my notes, that's where. The velocity of the actuator is S dot, so you take over from here. This, when you do it, will give you omega as a function of both the velocity and the angle. And then solve for omega at 30 degrees. And then acceleration, no, it's a distance, it's a length, so you don't want to warm the velocity. I know it is four, right? You didn't remember the law of cosines properly? It was two. There's a two in there, yeah. Thanks, Jake, for your calculator. Go on the cosine, because then we have theta and the omega we're looking for, necessarily showed that the units work out velocity and acceleration. Each of this will have, as part of it, omega. So that's how we're gonna find omega. And then this part of it will have alpha. And when you put in values, then you can solve for each of them, power by one and all that stuff. One half is outside the front season, So that happens for a minute. Oh, I just didn't write it in. Oh, yeah, yeah. A reform too, that's fine. Fond of the form of omega at 30 degrees. Omega's got to be meters, radians per second to be the meters per second. So omega's got to be in meters per second. Yeah, I think so, if I remember what you wrote. Do you know what S dot is? Because that was given, and you can solve for omega. S dot, and S dot, we already have. Let's read it here again. You have to take it as is, because if you put in theta equals 30, this becomes a constant. Which, when you take the derivative, it's not a constant. I'm sorry, I don't know what that is. Collin, you got it? Is that a hand up? I don't want to put the, to do this, you have to take the derivative that without the 30 degrees in there. I think that's, it's a little bit bigger than that. Oh, okay. That's that. Oh, you're solving. But we have two unknowns. What? It's gotta be, it's in radians per second. Omega's in radians per second. Omega at 30 degrees, you can solve for alpha at 30 degrees. Because there's three things that vary, omega, S, and theta. So when you take the derivative of this, you're gonna have three terms in the chain rule. So this first one is those kept constant, take the derivative of that. Which is it's two omega assigned theta constant, take the derivative of S to the minus one, help everything constant, but the omega. Omega dot, or alpha. It's not zero. A is zero. A is zero, but alpha isn't. Alpha's the angular speed of the door, whatever that is. And that's not zero. Take a frank or take the rest of the day off. You can solve for, like the four omega sent in over two times that one. Different answer for both, these two. So if you put 30 degrees in here, you get a different answer than if you put 30 degrees in here. How do you do it? Okay, well, I don't have a calculator here. I only did this one, so I don't know. I didn't check that one. At least get the velocity and then just check on the speed. That's what I did.