 if you so wish. All right, we're changing gears a little bit now. We have been looking at the kinematics and kinetics, in other words, the dynamics of particle motion. Now we're gonna do rigid body motion, kind of like we did in physics one. We also started there with particle motion. Went through very much some of the same stuff that we went through here, what we've done here, I hope was worthy of being a year farther along in your studies, two more in depth. Was it, Alex? Yeah, definitely. And then once we got to about the same point, then we stepped into rigid body motion, just like we're gonna do now. Though, of course, we'll take it a little bit farther. In physics one, when we got to rigid body motion, all we looked at was simple, pure rotation. We're going to revisit that good portion of this review, but then we're gonna take it a little bit farther. We've looked so far at translation of particles and the translation of rigid bodies is very much the same. That came in both linear and curvilinear fashion. In other words, 1D and 2D type motions, 3D motions. We looked at a little bit. We won't really visit that much. And now we're going to look at the rotation of rigid bodies. That's what we'll spend most of today doing. And then we're also going to look at the two of them together which we'll call general plane motion. The easiest type of example to see of that that you'd be most familiar with is a car tire. As the tire's rolling along the road, it's in rotation about a central point. However, that point itself is in translation, especially a car tire going over the Adirondack Hills like we do where it's going up and down. Very good example of rotation and curvilinear translation together of rigid body being very much general plane motion. In fact, that would be actually be general 3D motion. If we actually look at all the possible things you can do driving around the roads and like. All right, the surface first. We'll take a quick look at our definition of a rigid body because certainly we can get any kind of an object but we need some strict definition to allow us to say, yes, that's a rigid body as opposed to some other thing. We will look at things like linkages which are individual rigid bodies put together in such a way that together they're not a rigid body. So we'll get to that study very, very shortly. But a rigid body for our purposes is any object where we can put any three non-colonial points on that object that then of course describes a triangle. So any arbitrary triangle can be inscribed on a rigid body and no matter what that rigid body does that triangle will not change. So any side remains the same length. Any angle remains the same. Not just similar, the triangles don't change in some way that they remain similar triangles. It's always the same triangle. Any arbitrary triangle we can inscribe on that object, whether we do it, actually do it or just virtually do it, will not change in shape or form. Can however of course change in orientation. You can't imagine if we're gonna have something rotating about the central point that triangle is gonna change in its orientation but no aspect of that triangle will change and that's our definition then of a rigid body. So I guess if we wanna formalize it, we'd say something like A, B and C are, well I don't know what it means to say, oh yeah that's the sides, not the points. So A, B and C are constant in magnitude and the angles that go with them are also constant. However anything we use to locate these points, whatever we might call them, I guess we can give them capital letters and then say that the location vectors of the three points may or may not be constant. We're gonna have rotation about a central point and that central point happens to be A, B or C then the position vector of that point would be constant but then as all the other points rotate around it, none of the other location vectors would be constant and also for the velocity vectors, they might be constant, they might not. And of course the same for the acceleration vector. Depends on what the points are doing, what the points are and what the piece is doing. Simple enough, I hope it makes sense. Now you see it's something you probably could have come up with, you can imagine something committee, something sitting down somewhere and coming up with this type of thing. So we'll start our look in rigid body motion just like we did with physics one where we'll look at pure rotational motion first. The translation, we don't really need to revisit too much though we will a little bit but it's the pure rotational motion that generally needs a bit of review for most students. So it's rotation about fixed point or axis so the simplest thing we can do is imagine it's a disc, a rigid body will be a disc though it need not be and rotating about some fixed axis. If it is rotating about an axis as anything would be, I don't know how it rotates by the single point alone I guess, then at any point on it we can inscribe a reference line and then our first idea about position is any change of that reference line with time. Changes in position will be simply an angle made from the original reference direction, whatever that might be. And remember the units on this are radians. We might talk about degrees because I don't know about you somebody says some angle and radians I don't necessarily quickly understand what they're talking about I have to think about the transformation but then I make the transformation in my mind to degrees anyway because that's what we're most familiar with it's kind of like the way people can talk to us in a metric system but then in our head we're still converting things over to the English system because that's what we're used to in America. All right so we'll look at changes in that position with time and that of course will give us then angular velocity just like it did in physics one. We use the symbol omega which is kind of a kind of a fat round W. The average angular velocity is the change in angle with time and then just like we did in physics we define then the differential as time goes to zero giving us the instantaneous velocity. The very very same type of thing we did in physics one. And if you remember and I hope you do everything we did in terms of particle motion especially translation has a direct and direct and intimate allegory in terms of the rotational motion. All you have to do is take out any equation we had in translational particle motion swap out the position velocity and the acceleration symbols and you've got the same equations. We'll amplify on that a little bit because we do need to know a little bit more complete things about it than we did in physics one. And then we also add the instantaneous acceleration. It's the average acceleration as that time period goes to zero becomes an instant limit. And then not all the other choice symbols we could put with it. So hopefully that looks somewhat familiar as it should. So remember that any equations we had before we now have in translation we now have exactly the same thing in rotational motion. You just need to swap out the variables. So for example we had the translational we had the translational acceleration equation and now we have the equivalent rotational motion equation just the definition of the acceleration there. Obviously if the acceleration is constant the average acceleration and the instantaneous acceleration are always equal. What other constant acceleration equations do we have? Delta S equals 1 1 2 A T squared plus VI T. Swap out position, put in theta, swap out acceleration, put in alpha, swap out velocity and we get the exact same form for rotational motion. So our constant acceleration equations don't really change. You don't need to get a new tattoo. Your old one will suffice. All you have to do is swap out the symbols and you get additional forms straight away. Remember we had four constant acceleration equations. What was the fourth? Check your tattoo. These changes are supposed to be five. That, what was that again? It's constant. Average velocity is delta S over delta T. Of course in constant acceleration the velocity changes linearly and the way you find the average of any number is by adding them together and dividing by the number you have. So it's just a straight arithmetic average and we get the last of our rotational constant acceleration equations. So we have four constant acceleration equations for either translation or rotation. At this point remember we're not doing both at once. We're only doing one or the other. Right now we're focusing on the rotation but we still have the very same four equations and the like. All right, for the special case of constant acceleration however in this class we're not going to stay strictly with that so we have a little bit more we can look at. If you remember, if you remember we did have one equation look something like this in translation. Remember that one? That came about from the definition of velocity and the definition of acceleration. Combining those two equations and eliminating the DT we got that equation. That served us very well for those times when we had an acceleration that was a function of position. So that was one of our most useful non-constant acceleration equations. If acceleration is constant and you go ahead and integrate that you end up with the third form that we have there anyway. So we have a rotational equivalent then that we can use for those times when alpha is a function of beta or the angular acceleration is a function of the angular position. All right, now we're going to expand things a little bit, take them a little bit farther than we really did in physics one. So, oh, one other thing, one other thing I want to do with this. Remember that angular acceleration, angular velocity and angular position are all vectors themselves where the vector direction whether it's theta, omega, or velocity is all done with the right hand rule. Put your hand in the direction of the turn, your right hand, and the thumb will give the vector direction of that change. So it's sort of like a quasi two-dimensional. The objects we're looking at might be planar and turning in a plane. However, the vector that represents those, we allow to be in the third dimension for a way for us to describe this. So for example, I've illustrated what the positive direction might look like, or not necessarily positive, but what the vector representing those motions would look like with the right hand rule for a particular example shown in rotation. All right, so let's open things up a little bit. Imagine we have some objects rotating about some fixed axis, and we want to know something about the nature of motion of a position, a point on that object. So it might have some angular position relative to some reference. So that's easy enough to understand, I guess, at some certain time, that angle P that describes the position of that point P certainly could be a function of time. We'd expect it to be, as a matter of fact, otherwise things aren't moving. But some other things that are more interesting to us possibly are things like what the velocity of that point might be. I'm trying to draw it so that vector is lying right on that disc because if that disc is turning in its own plane, then the point P must always remain in that plane as well. And then we know the magnitude of the velocity of that point is the distance it is from the axis of rotation times the angular speed of the object itself. That's generally pretty easy if that R P, the minimum distance the point is from the axis is known. If that's not known, we're going to need something a little bit more robust. We can then find the velocity of that point at any time by taking the cross product, I knew it, which we would have caught in seconds when we do that. I don't want to do that. Cross product's the other way where this R P vector is from anywhere on the axis of rotation two point P. That might in some problems be a little bit easier. We might know some point down here or even up there and it's just for whatever reasons, a lot easier to know what that vector is from that point, rather than knowing exactly where the center of rotation is so that cross product form then could help us. And you can check it. It's a little difficult because now we're working in three dimensions, but if the axis of rotation is as shown then there's our omega vector straight up. Rotation and this is something students forget a lot. This vector, the rotational motion vector is a floating vector, which means it's the same anywhere on that object. So you might see the cross product a little bit better if we draw it like this, then an omega crossed into R is going to give you the velocity vector of that point. So remember the cross product is also done with the right hand rule where you put your fingers in that direction of that vector, move them the shortest distance to the direction of that vector that orients your thumb then in the direction of the vector you're trying to find. Omega crossed into R will put your thumb in the direction of the velocity at that point. Just to clarify, the omega that's the curved one, that's the angular velocity and then the one that's going straight up that we were just talking about. They're one and the same. They're one and the same. It's just, this is a more of a, just a, I don't know what to call it, kind of like a graphical demonstration of what the angular velocity is. This is our vector representation of that that we use for any calculations, especially cross product. This kind of thing is not a true vector and we can't put it into a cross product. This is a vector and can work in cross product. Very seldom in this class are we're going to have to actually do these cross products but there are times when we're going to have to imagine them. Most often we do know the perpendicular vector and the cross product itself is a perpendicular result from that. But we have it if we need it. And in some cases we will but more often than not we just need to use it to conceive the direction because most of our problems are going to be two dimensional problems rather than three dimensional. All right, the other possibility we need to look at, of course, is that this point P of interest, whatever that might be, maybe we've got something attached there. We're going to be spinning a lot more things than just discs here. We might also be concerned with the fact that there could be some angular acceleration of this object causing point P to have a tangential acceleration. But if you remember from points going in circular motion there's also a normal component of the acceleration as well. And so we have to have both of those. Acceleration of point P is that component in the tangential direction and that component in the tangential direction, point P in the normal direction. And both of those, we remember, let's just not bother them for point P because it's just the tangential. We're going to remember a point going in a circle as point P must. A point going in a circle will only have a tangential velocity, so the tangential acceleration is any change in that velocity. And it's related to the angular acceleration depending upon its particular distance from that axis. If we need it in a more complete way we can also do that as a cross product where again that r position vector rp locates that point from any point along the axis of rotation whether it's the center point of rotation or just some place that might be more convenient or anywhere along the axis of rotation. Doesn't really matter. And the normal acceleration if you remember, what was that? r omega squared v squared over r. Remember r is not the radius of the object, the radius of the circle the point P is on. And that also can be done as a cross product. Location of point P relative to anywhere from that axis. This last little quantity that omega cross r, well that is dp. So this is also then omega cross vp. We can check that. Remember this normal acceleration for an object going in a circle as point P is is always directed towards the center. And if we do omega cross vp, omega cross to vp, our thumb is going right towards the center of the rotation just as it should be. So this does indeed give us the centripetal acceleration that we'd expect. I guess we could write it as r omega squared vn and a minus meaning it's towards the center. Remember the normal direction is typically taken as out. So we've got all the pieces there we can do. We can do all kinds of damage now. However again, we don't actually have to do those cross products very often. Mostly though they're useful for visualizing to get an idea of the direction you're considering as appropriate. These are racist, such a pretty picture. Any more questions? Normal acceleration. Why is this negative r omega squared? Can anybody tell them? Because it's supposed to go outwards instead of outwards? The normal direction is typically outwards. Remember as we set this up when we did our normal and tangential components. So an object moving along some path has a velocity along that path. That sets the tangential direction. And then the normal direction was perpendicular to that away from the center of curvature. And of course as that object moves anywhere along that path then the normal tangential coordinate system moves with it. That's not a big deal when something's moving in a circle because that change in these is so predictable. It's a little more difficult if you're going along some general curvilinear path. Okay, so let's do some problems just to warm up on this. Remember we don't very often have to do cross products. So imagine we have a wheel rotating about its axle. The speed increases uniformly from 25 radians per second. No sorry, from 10 radians per second to 25, 10 seconds. AT and AN at five seconds of some point P, one foot from the center. I normally assume that if it's not said, it's going counterclockwise. Yeah, if it doesn't say you can assume that or just sketch it on there if you want. It's the same thing that we had a particle motion. We said it's moving with 10 meters per second. You can draw it moving to the right or to the left or up or down because it doesn't really matter. So feel free to just designate that in the picture. We can say though as a class policy, it must state it otherwise in the positive direction which would be counterclockwise. Give us a vector out of the board. Also find the total distance that point P travels. So just a bit of a warm up. Getting used to again this rotational motion business. Let's get you warmed up a little bit. Get your rotational juices flow on another thing. Tangential and normal acceleration is a little distance that P travels. Is this here right? Yeah, point P is one foot from the center of rotation by, I guess, change problems of changing radius. We believe that for a bit that angular motion, angular momentum handled that kind of thing fairly well. This is funny, did you laugh at that? Is that what he's laughing at? What's funny about that? That's not funny. This is serious stuff, but it's too big. What's the size limit on that? Sometimes last year, I just written that on the board one time. 175 gradients, you feel much more comfortable right now. No, but to say, you've never seen something come up before, sorry. That must be then what Bob's laughing at. I don't check with your classmates. They're still willing to talk to you, Bob and I aren't. So in this one, all those cross products are pretty easy. Remember, the cross product of perpendicular vectors has a magnitude of the product of the magnitude of vectors. Is this a constant acceleration problem? Did I say that? So this is a constant acceleration problem because I said the speed increases uniformly. I did write it down, but I said it's a problem and Bob's going to say it's a problem. So any of the constant acceleration equations will work. If it caught that, if it looked at your units, remember the radian units is kind of magical. It comes and goes when we need it. Oh, God, yeah, we're going to do this again. Just like that. Or put it back in. You got to remember to put things away when you're done. Just get something to this thing. It comes and goes. 175 gradients. Do you guys agree finally? I agree. I don't agree with that. Is this how you do your strength? That's how you do it. Oh, yeah, you do. It's a good test. I'm sure we got you done it. Look at that. I did it. I did it for a moment. I'm glad I'm praying. I was looking at that. Oh, thank you. I'm sitting at 0.7. I'm just glad you got that. Yeah, around. We're good. We're good. I'll do the other one. Yeah, distance graphing. I'm going to do a r-r-r-r-r-r. I'm going to do a r-r-r-r-r-r-r. I'm going to do a r-r-r-r-r-r. The total distance of travel is 10 seconds. That's a total of me. Don't you just multiply it by 2? Instead of doing one remultification, because if you have the equation for 5 seconds, you just change the numbers you put in. Are you sure it's a little bit by 2 because you have a tenth circle? All right. April 1st. Just like last night, you guys took me to the top of the rain. And by total distance, I mean the number of... I don't know what units were in meter. Oh, we're in feet. So the total feet that point feet your apples. That's so big. It's the biggest number I've ever seen. Since you do the total distance for 5 seconds, will that weigh in on the number that you bring with me? Yeah. Okay. How far in feet does point 2 travel? Figure it out. Delta theta equals that. Equals that. These can all be equal. Completely different things. That's delta theta, that's delta x. How can they be equal? Of course they are. I don't want to write delta x alone. I don't want to be correct. I just want to be lazy. I don't really care if the plane crashes that I design. As long as I got the rest of the day off. We get the angular acceleration. It's the... Oops. Alpha is what? 15 per second in 10 seconds. And AT are not equal. Just because you're multiplying by one foot doesn't mean they're equal. Does this one change at all? For 5 seconds. How do we do the fact that it's that at 5 seconds? Constant acceleration. Alpha is constant. Is this then constant? How do we find the normal acceleration? The centripetal acceleration. Omega squared. R is constant certainly. But omega is not. The angular acceleration is actually increasing. How do you find omega? Remember, this is at 5 seconds. Plus the change. Yeah, or if you notice since it's a linear increase and we're at 5 seconds at the midway point it'd be just the average velocities. Either way it comes out to be the same thing. So what's the average of that? 17 and a half. Don't forget it's squared. That's a very easy thing to forget in these things we're doing here. So what'd you get for that then? 306 per second squared. And then the total distance traveled delta s is r delta theta. And delta theta we can find from constant acceleration equations. It comes out to be an absolutely gigantic number that no one's ever seen before. A number that big. Good thing you're sitting down, aren't you? So the t is the 10 seconds for the total distance, alpha we have from above. Omega one was given. Do you want to avert your eyes while I write this down, Jay? This is 175 gradients. So how many times a round is that? We divide that by what? Two by. So divide that by about six since r is one foot. We got just that. Warm up. Simple one. Could have done that back in physics. Physics one. So we'll do a little bit more involved problem now. This will be your get out of class question. So we have a winch system with a small motor here driving a larger wheel over there. And the winch, the lifting cable is actually wrapped around the center there. And we have a load attached to that. Now, unstated here but something we're going to have to pay a lot of attention to. It's going to be very, very important to us because if it's not true then all of this for the next couple weeks becomes a lot harder. You can imagine that this would work best if these were geared wheels in connection with each other. I'm not going to draw all those teeth so you have to just assume they're there. But what that means to us is that there's no slip between those two. That as one turns it's intimately connected with the other one and will cause it to turn perfectly in response. So we'll call that wheel A this is wheel B No. Other ones. C just to keep them straight is what I have. The radius of A is 50 millimeters. The radius of B is 100 200. The motor at A starts from rest begins to turn clockwise at an angular acceleration of 0.2t radians per second squared. A starts to turn clockwise that will cause C to turn counterclockwise causing the load to start to rise. You're defined after 10 seconds these two things. How far does the load travel? How much does it rise? And the velocity of the load after 5 seconds. Again nice planar motion so we don't actually need any of the cross products here. Sorry, was there a question? No questions. Perfectly understood. There's your get out of class question. You start the weekend, call them the homies down to get the beer on ice. We'll be out of here early today. Guys, 300. They were playing the lottery every week so it cost them like 8 bucks. Think through this a little bit. Think where you're going to go with this. Get kind of a game plan for it. Of course B and C have the same angular accelerations, same angular velocity. They're stuck together. Delta H get Delta V B We'll have a big sale. We had to work big sale. Homies text back yet? Lots of busy people. What do you got? Let's see, Alpha A. Do you know how to get from Alpha A to Alpha B? Alright, let's look at that. Here's kind of a blown up picture. There's A and here's B connected to it. Any point at that instant where they're connected Oh yeah, that's sorry, that's C, not B. Hopefully everybody understands Alpha B and Alpha C are the same because they're connected whatever acceleration one has, the other has. That point where they're connected those teeth at that instant that point will have some linear acceleration and that linear acceleration is the same for both. So you can use that in the angular acceleration. It's going to be the ratio of their radii. That will give you this acceleration as a function of t because that was a function of t. So you'll have the acceleration of B as a function of t. So with that then you can find the changing speed from rest that will give you velocity of any particular time t. That's the same type. Omega is a function of t. Right? Well then what do you do with this? Since this velocity is not constant how are you going to get this distance turned from that? I realized once I got to that part I had to take the derivative of the derivative of what? The derivative of angular velocity. If you take the derivative of it you're going to get the acceleration. Yeah. I said I had to go bad but I didn't actually go bad. It's two integrations. If this integrates to some time t and then integrates that to get the distance in ten seconds then that will give you you're already ready. You snapped into action let's get out of class question. Nope. Not what I got. The velocity looks okay. Did you check mine? I know one of these is at least going to be incredibly wrong. That makes no sense. It makes our velocity and delta S be down. That's no, look how it rotates. Jesus, it's got to go up. For a lot of time the sense is just from whatever is in the picture. Okay. Don't put in minus signs just screw it up. Nope. Not what I got. What's the solution to this problem? We'll do the solution in a couple hours. We'll get tired, hungry, sleepy. You know, you're texting the boys to come get you. Sprinkly loose. Yeah, that's what I got. That's the point? Yeah. It's zero in front of those decimal points so they don't disappear. It's not wrong that you don't. But there's some day you're not going to see that decimal point. Like you did to put a line over it or your photocopy enough times it disappears. Just a precaution. Just one of those professional techniques you should use. Nope. Let's do it for Bob. Let's see. So Alpha B here is going to be rA over rB alpha A. Right? And what's that ratio? I don't know if that's a C. We need a C in there. That's a B. But not on the radius. The ratio of the radii is 1 fourth. So the angular acceleration is 0.05 T. Is that right? I think. So if we integrate that that will give us the change in angle but the original angle is zero so it will give us then the speed of B at any time T since we're only going to integrate up to any time T. So what's that integral become? Zero zero one? Zero zero two five. Power comes down. Raise the power by one and zero zero two five T squared. That's the speed of B at any time T because we integrated from zero to T the initial speed was zero at T zero. So the integral becomes the speed of B at any time T. So we can use that then to get the change in angle of B by now integrating zero to ten seconds. And that change in angle times the radius of B will tell us how much we raised point L. Because delta theta is the integral of omega DT. That's the point zero five. I saw you had a pair. Last year you said you didn't have a pair. Better? Divide it by three TQ evaluated from zero ten seconds delta H so I just multiply by the radius B and what was the other piece I asked you to find? The velocity of L VL equals RB omega. We know omega is a function of B as a function of T so we just put in the ten seconds. Makes sense? Text those homies, Bob. You're coming. Colin, what was wrong? Don't use constant acceleration problems for non-constant acceleration. Once we know how we just multiply by the radius of B okay? Got the last piece, Jake? Got the first piece, Jake? Now the velocity is the question. We have the angular velocity of B with time so you can just put it in there and evaluate the whole thing at the ten seconds. Because we have that, right? Yeah. Ten seconds. Are the sentences in it last time? Oh, millimeters. Yeah. Come on, Alex. Here's your problem right here. T squared. That's why you're off by a factor of 10.