 If you remember what we were talking about on Wednesday, we're now looking at rotation, objects that rotate about a stationary center. So, as easy as any, just draw something around doing that. And we started to look at the kinematics of rotation. Now we've done kinematics of translation, or kinematics of the straight line motion that we were looking at, even the two-dimensional motion. What is it that was very, very useful in terms of how much new material we needed to cover on Wednesday? What was very, very useful to what we were looking at? And I'm not that useful. Well, Phil's holding up the constant acceleration equations, but those aren't for rotation. Those are the constant acceleration equations for translation, an object moving along some kind of path, usually straight line path, but not always. So, I don't understand why you held that up, Phil, if that was not, those aren't the constant acceleration equations for rotation. Well, because the, everything that we picked up in translation in the first half of the term still applies here. We just apply it to a slightly different problem, but everything was pretty much the same. We have position defined by some angle from some reference line. That's not very much different than position was from some arbitrary reference spot, some origin to that place. So, position. Now, that was, we took that to be a vector. I mean, it is a vector. We're not gonna need any more than plus or minus on this, but the fact that this angle could be through any one of the three dimensions. We're gonna do it so that it's an angle that happens to lay right in the plane of the board, or as you take notes, it's right in the plane of your paper. And that's all we're gonna do with it. But angles could go in any direction. Real-life things really do that. So that's why we need a three-dimensional aspect to it, even if we don't quite get to it much in this class. So we have this angular position idea. What were the units on that? Radians, the same angles, the same angle measurement that you used when you took trigonometry, where 45 degrees is what, pi over eight, I think pi over four is 45 degrees, 90 degrees is pi over two, that those are just angles measured in radians. We can talk among ourselves using degrees, but in the equations, in those constant acceleration equations, the unit of degrees does not work in those equations. Then we talked about how rapidly this line could be moving some angular velocity. Do you remember the symbol we gave to that? It looks like a kind of a fat round bottom W, but it's an omega. And it also is a vector because an angular velocity could go clockwise or counterclockwise. Which direction clockwise or counterclockwise is our usual positive direction? Linearly, we usually say that's x direction positive and that's y direction positive. What do we usually say for angular measurements and velocities as positive? Counterclockwise, that goes with the right hand rule where if we put our fingers in the direction of curl, that puts our thumb in an automatic orientation of straight out of the board, which is the plus z direction for a right hand coordinate system, x, y, z. So all that went together. So that's our angular velocity. Units of radians per second. I can't remember ever in any of these classes talking about degrees per second, but we could. We could say that something is rotating at so many degrees per second, but then you just have to make sure that it's radians, which we'll double check in a second here. And then we could have with that all angular acceleration and that's in radians per second squared, radians per second per second. And then all of the equations were exactly the same. I didn't really have to define angular position velocity and acceleration because they all kind of made sense like the other ones did, but all the equations that related all these were exactly the same. Even the constant acceleration equations that are on that sheet and the constant acceleration problems are solved in exactly the same way. And so we started one, we'll take it a little bit farther now. We had a CD that was spinning from rest up to, what was it, 500 RPM in 5.5 seconds. Wanted to find the average acceleration. Now we have a page of equations, the one Phil held up, there were constant acceleration equations, but this I asked for the average acceleration. So what's the deal with that? What do you do? No, no, what equations are you gonna use? We got constant acceleration equations, but what if I asked for the average acceleration? What are you gonna do? Cross out constant and put average? You could do exactly that. If acceleration's constant, isn't that the same as its average? If you have a number that never changes, what's the average of that number? Exactly the same thing. So if we say find the average acceleration, that's the same thing as saying, assume the acceleration's constant and find out what it would be. All right, remember, constant acceleration problems all happen in the same way. What was the deal? Says it on that sheet, right? How to solve constant acceleration problems? There's four variables in the problem, and three of them you know and one of them you don't. So what's one of the variables? What's any order? We don't care what order they come in, it's gotta have all four and the three that you know. Joey speaks up, gets the easy one. Initial velocity, zero. So on these constant acceleration problems, write down the things you know so you know what four variables have gotten and which ones you're looking to find. What else? Don't talk with your mouth full, Alan. I have to call your mom and tell her. Yeah, of course, we got the time in here. One more that we know. We sort of know omega two. What do I mean by sort of? This is an RPM revolutions per minute that does not mean radians. That means revolutions per minute. We need it in radians per second. Believe me, that will come up again. Half of you here will forget that, use the number 500 right in the equation. Who is that half? When you're volunteering, land volunteering, that's nice. Yeah, take the bullet for the class. Yeah, thank you guys, thank you. All right, any time though, here's how you make the conversion. Revolutions, this even works for degrees slightly different. Well, we can do it, we'll do it. What I'll do is I'll put this into degrees and then put it into radians so you can use it however you want. In one revolution, there are 360 degrees so it's as simple as that. To change this into degrees. Now we know how many degrees per minute that thing is spinning through. To do radians, we can do the very same thing. 60 degrees is once around, how many radians is that? Two pot, remember we're doing this on a unit circle so the circumference is two pi r. If r is one, then it's two pot and that's all you need. And the reason why I didn't need to go to degrees first is because it just cancels out anyway. Two pi radians per 360 degrees or two pi radians per revolution is all you need. And then we gotta change minutes to seconds. It's on the bottom and it's on the top. 500 times two times pi divided by 60, just like it says, what's that come out to be? 55, what? Oh my God, don't make me ill. 60. We're engineers, we don't think like 55 pi over, whatever you said, I have no idea what that number means. Give me a number, huh? There, now I know. Now I know which one's going faster. 60.7. 60.2.4. It's got 52.4, that's what it is. It is accurate, that's very accurate. What it isn't is exact. And we don't have exact things in the real world and that's where we work. That's wrong, that's right. What was theirs? 52.14.14. Math teachers, did you know service by saying angles have to be measured in exact units? There is no such thing in the real world but that's not where the math teachers live. They don't live in the real world. We do, we live in the real world, we make real measurements, we design real things. They've been saying that for years. We got work to do, huh? They've been saying that for years? No, I haven't, but it's very different. Yeah, well they don't listen to us either. But we've been saying it for years too. All right, now we did this already. So we're gonna do something else now. We did this on Wednesday. All you do is pick out a equation. So if instead of me asking you for the acceleration, what if I asked you for the number of revolutions? How many times did it actually go around starting from rest, getting up to 500 revolutions per minute and it took 5.5 seconds to do that. How many revolutions did it actually go through? Asking for there. Distance, well, kinda, yeah, I guess. In terms of the kinematic things we have, really what we're looking for is change in position, which if we were talking about linear motion, that would be delta S, that would be the distance traveled. What I need to find is delta theta here. We can find it in radians and then change it into revolutions and we'll have the answer. How do we do that? Well, how do you know to do that? Is that one of the constant acceleration equations? Yes, third one. Still three things we know. One we're supposed to find. One equation does that for us. What equation was it, Glenn? The final minus the initial change in omega. Yeah, I was wondering what plan or review work. We have the change in velocity. We've got the time that it took. We can find out the number of radians that transpired in that time and then we can convert it into revolutions. So you do that real quick. No different than the kinda things we were doing, man, about the third week or so we were doing this stuff, weren't we? Remember when you were a little physicist just this high? All doing the same problem. So we should get the same numbers. We can check that when we get there. Boy, that's not delta omega. That's not listening to Glenn. But I should stop listening to you. That was the big point. Yeah, if you were to let me look at the key. That was what gonna be right. Just wrote down what Glenn told after I fixed it. Change in omega times the change in time. It's not changing omega over change in time. Change in time is changing time. What is it we need? Since maybe a, do we have an equation that has these two, three variables plus this fourth variable? Which is? The initial velocity times time. We don't have h to square. No, no. We don't have alpha written down. I don't have it. We already figured out the last step. Some people did, some people weren't here. Oh, we'll come use it or not. You can use it to check your answer, but we should be able to do it both ways. We're not gonna come up with two answers, are we? What's the equation that works with just these things we have? These three things. And the fourth we're looking for. We're gonna have to be omega squared. We're gonna have to be omega squared to a, but no. Omega squared. What is it? And that's essentially what we had. This is average velocity, but you're right. Velocity is changing. The average velocity isn't changing. The instantaneous velocity is changing. What does the average velocity equal? Velocity equals the average of two numbers. Both of which we have. So that you can do then and solve for delta t. And then make that into revolution. If you have a, and use a different equation, you should get the same answer. It shouldn't change just because we have more information. As long as the more information we have is good. In fact, if you've got the acceleration, you can use any of the equations that have delta theta in or delta s on your paper. Use the easiest one, and I don't know what's much easier than this equation anyway. Keep going. Maybe I should mind my own business. What do you think, Patrick? Put it down, Joe. Are you possessed? That was the, that was the, I'll make it by that time. She put the average for me. So you're only gonna get a little too much. That's not what I wrote. I don't know why. Is it? I don't know, I don't know. Come on, let's be easy. Delta theta equals, average velocity. Ooh. Times the times. We have all those numbers, don't we? Anybody agree yet on the answers? Anybody get an answer that you could check with somebody and agree? That's fine. Okay. It's about to be as easy as the problem we can do. It's not what I have. Oh. We'll give you an answer in radians for the questions. Oh, radians? Yeah. I asked for it in revolutions. That's what went on the end. But you can check it in radians before you go any farther if you want. And you know, in no sense do an extra work if what you had is wrong. Yeah. Remember, this is delta theta of WP is average. That's why she's a number of girls. Flossy. Agree on radians? Finally, Mike, what do you have in radians? I buy two personal ones. Oh, radians, 144 radians. Does that sound right to people? Yes. 144 radians. That's delta theta in its proper units. I asked for a number of revolutions. So what'd you do? Two pi radians in a revolution. So there's your rotation, if you'd rather. There's your unit conversion right there. And that comes out to be 23, 22, 29, 23. So go around almost 23 times starting from rest, heading up to 500 RPM in that five seconds. So the last question for this problem. How far does a point on the edge of the CD actually travel? And we'll say the radius of six centimeters. So now this is the actual linear distance it travels. And that has to do with the wear on the CD head and how often it actually has to be reading and moving. I want the answer in, yes, meters. We're going to go what distance, then, if we unrolled that and laid it out. Or took a CD, I guess, rolled it along the floor for 23 rotations. And you can measure it. You could do that, run out into the hall and do it. Hang on until then, would you? Three weeks. And it goes very fast. If you're behind, you're going to just get behind or real quick. Don't count on taking an angel plea at the end of the term. Nine times out of 10, they never get completed. They turn automatically into an F. Would you check with Allen? Boycotting everybody? So filled with hatred, Allen. Actually, that hatred only flows this way in that class. But you guys got it going side to side. It's getting natured after 360 degrees. Or how many degrees? Got it? How'd we do it? Len, did you do it? Well, you got it? Mike? Bill, check your hat if you need to. Going to help. Bill can see it and he's already done. If you turn your head and hat around, it would slow fill down. But I'd speed up. The idea here is the very same idea you've had before, that an angle is going to represent some, if that angle or that change in angle is in rate. And that came out to be, now you got it, Mike? No, Bill, 8.6 meters. Simply the six centimeters times the 144 radians. And then just the conversion from centimeters to meters. Should have worked. Other questions before I give you another one? This is from an old test. A lot of these things we've been finding, there may be more than one path through the problem. All right, in this one, an electric motor is turned off and its angular speed increases from 500 RPM to 200 RPM. I want you to find the average, the number rotations continue to a complete stop if it has the same acceleration as before. 200, find the average acceleration, total revolutions made in that same four seconds and then how much extra time to come to a complete stop. You need to know that because if you go in there and try to grab it and it's still running, it's going to hurt. For that third part of the acceleration, I don't think you have any use. I'm still using them all the time. I'm a school teacher like me. Good morning, I'd like to hear that every day for the rest of your life. That's about right there at the college. Now, once you find the first one, the acceleration, there may be more than one way to find the second one, the distance, rotational distance traveled, but if there's more than one way, it's a way for you to check. I'll make sure to agree. Hopefully where I'm going is right in front of me somewhere. I'll make it. Where's somebody? What's up? What's up? Wait, I thought, I thought it was going to be fun. Yeah. She's not eating. Not eating, not eating. Is anybody else? No, I thought it was going to be fun. It's as easy as you are. Somebody's going to be asking. Yeah, I feel. It's always a nice guy. I feel like I'm going to fall in love with him. Hey Bill. Hey Bill. Yeah. It's me and Mike. I'm sorry. Can I ask you something? What is that? What are you asking? Would you have 30,000? 7.35. Okay. I didn't do anything afraid of you. Who's in the video? I love you. It's on iTunes. You can watch it on your phone. It's on iTunes. You can watch it on your phone. You're a problem writer. But. Anybody done that? Have to ask Mel? Can I see the phone number? You buddy. You're on the screen. So it's on the screen. Is that you done that? I would just tell you. I would just tell you. Have you done that? Checked out our video. Because if you can't read this on the full screen, you're just going to have to read it out there. No. Did you agree? Find him with Bill? It's right. He can sit down. No, we just agreed. I'm sorry. I didn't do anything. I was asked. Well, I would have said they changed the policy. But, well. It's fine. Bill. You okay? Stop. I screwed up the acceleration. Why go any further? Why go that? So. Are you going to do some contouring? Yeah, I guess. I guess I can do it. Just watch this. Does that mean I have to redo all of this? I don't know. I'm going to put it like that. That's a new romance. Ah, what? What is this? Is this 20-9? You have a risk of getting into this problem? 20-9 changes? Yeah. But I have 21. And I want you to have a gloss. I need 2.1. Which is actually a change. So, just some of the reading and conversion. And then changing it. You can do the three-year-old. Yeah. Take that right four. So it's not deflecting, it's growing. Did I know how to do it, but I didn't know how to do it. So I'm just trying to make up for that. Can I have you dealing off of what? Is that the electron in the 2 pi r? I did a rotation across the square. What do you think? The equations are pretty the same. I'm just saying the other two are a little bit different. The term is a little bit different, but the equations are the same. Does that help? I know. You should be able to call it RBI. That means everything's fine. That's all you have to tell me? Okay, that's all I wanted to know. See Samantha, you can be helpful. You can be pleasant. I'm in the anger management class. Bill, how are you doing? How are you and Glenn doing? I'm fine. Are you? You guys can check. We're going a long way. We look busy. Do we at least have some agreement? Some agreement? I'll give you a joke. Not that he's giving me an answer. Wait, so now we're waiting for the radians and we're not in radians? No. So you change the data in the years that you want to implement some radians? What? Say that again. What about it? There's change in angular position, but then there's change in linear position. This is changing in angular position. So do you want that on us in radians? This would be in radians. And let's set otherwise, because our unit for theta is radian. Oh wait, no. I love this at one point. Yeah. We're about ready to check this. I jumped the gun on this. Ten radians. We get some good agreement finally over here? No. We got acceleration. Alright, let's see. How'd you do acceleration? What were you doing for acceleration? Omega. Omega. Acceleration is dealt with over delta t. Because we had just constant acceleration equations. That's the one that fit. Omega two. Well, what did you put in for omega two? 52.4 radians per second. Not 200 RPM? Oh, omega two is about 20.9 radians per second. What are you talking about? The figure is small. Omega two is 2A, because it decreases to 200 RPM. Anybody just put in? Huh? Did anybody just put in RPM's here? What? We totally just went over that. Maybe you repeated what I said. He's like, it's 21. I wasn't listening anyway. 200 RPM, or revolutions per minute. Which is in most situations, the base unit for rotational speed is not the same thing as angular velocity, though. What does this come out to be? And radians per second. Yeah, but I just want to make sure that if this number is wrong, then no sense going to the equation. So this is what? 20.9 radians per second. Is that what I'm hearing? And omega one, which was 500 RPM, just multiply that by two and a half. Is that right? It would work. Comes out to be 52.4. All right, so now we can do four seconds and what that comes out to be. There you go. Because it's decelerating 7.9 radians per second squared. All right, everybody agreed on that one, though, evidently, but then things started to fall apart as a group. After that, Bill seems to be together with whoever gets an answer. What did you answer? Yeah, that's what I got. Yeah, that's what I got. All right, now we're looking for delta theta. Which equation do you use? Of the four constant acceleration equations, which do you use? Do you have to use that one? Because we certainly had to use this one. None of the ones would have worked. But what one works for delta theta? Any of the ones with delta theta and them should work. Use whichever one is the easiest. In fact, you can use any one of several and they should get the same answer. So what one do you think was easiest? Len, you said which one? The average velocity. Use the average velocity one. That's the one we just used a minute ago. That one, because we have the two velocities, we can take the average and it should work well. Anybody use one of the other ones? Any of the ones with delta s in it that you convert to delta theta should work. What did you get for that? 46.6 degrees. 147 radians, or however many revolutions that is. So what, 20? 23.3. But you could use any of the other equations that had delta s in it, which you convert into delta theta would work just fine. It's a matter of time. How about the last part? It's time to complete stock. How did you approach that one? There's a couple different ways to do this one as well. You have 200 rpms, is that it? Uh-huh. That's 300. You could take two thirds or four seconds. Something you can explain easily that I can understand. You set it up as a proportionality. What's proportional to what? 147 radians, or over 300 rpms. Proportional to 200 rpms. The change in radians over the rpms. Do it in a way you can explain to me that I'd understand? A philosophy over time. Well, you can use several equations. The idea being that you essentially have a new constant acceleration problem where now you have a different beginning velocity, which is omega 2, and a different final velocity, omega 3, which is zero. You're looking for delta t now, a new delta t. So maybe we'll want to call that delta t1. And this would be delta t2. It's a constant acceleration problem. There's two things you know. One thing you don't. You need three things you know. Assuming the acceleration is the same as it was before, then you've got three things you know, and one thing you don't. However, well, I'll let you finish that, and then I'll take up somebody's question that was in there for a second. Would you get a bill? And here's one thousandth of a second. You were given this to a tenth of a second, so don't go too much beyond the tenth of a second. 6.7 seconds? No. 6.7. 6.7 seconds? No. How about total time? I said the additional time to stop. The additional time? 2.6 seconds. 2.6 seconds. All right. Let me ask you this. Angular speed. We know that at zero time it was going 500 RPM, or what was that? 52.4 radians per second. Four seconds later, 20.9 radians per second. Can you just say, oh, I figure it'll do that, and I know when it'll come to a stop at zero speed, which is what I asked, the time to complete stop. Could you do this? This will take a vote, unless somebody shouts out their vote ahead of time. Could you do this? Draw a straight line, extrapolate it down to here, and see where it hits. Patrick says yes. Samantha says, I should have watched the video. Alec, Alan says, there's a yes, a no. I ain't going to play. You're not playing. You said no, you cannot do that. Well, I drew it, yeah, but is that valid? Just because I can sketch it at the board, because maybe that's not the curve that fits these two points. Maybe it goes like that. Maybe it goes like this. Maybe it goes like that. Answer the question. Everybody else did. Will you call in a lawyer and say, I'm sorry, I would like to petition the court for more information? You can't. I'll hold you in contempt. What did we say? No, no. I said, can I do this and get the answer? I'm referring to something you said earlier in the semester. You said, if I draw something out the board, it's right. It's in pink. Yes. It's very light and it's a different color. That should tell you everything. Phil. No, no, no. No, yes, yes. No. Sure. Did you wake up long enough to catch the question? Yes. Yes, did you answer? Yes. Yes? Phil says yes. So that's three yeses, three noes. Fourth yes. Yes. I don't know. You're worried more about them. Because they're far away. John? Yes. So we get a lot of yeses. Three noes and a little island. Did I change to a yes? No. I'll do it then. Yes, you can, but you lose your deposit. What made you change then, Len? Well, he did tell me, but then they recognized that the acceleration was constant. Some of the velocity is just going to be a straight line. What's the slope of the velocity curve? This is the velocity curve. What's the slope of the velocity curve? Yes, it's delta omega. In fact, there's delta omega over delta t. That would be the slope of the velocity curve. It's acceleration. I said the acceleration is the slope of the velocity curve. Acceleration, I said the acceleration is constant. Is that a constant line? Constant slope line? It sure is with slope alpha. So if the acceleration is constant, you most certainly can do this. And it should come out to an additional 2.64532691 seconds. So yes, you can do that because the velocity is constant. All right. Now we're ready for our next big step in this angular motion business. What a wise guy. But you're going to apologize to me if you're going to say, wow, that was a big step. So far, everything we've done with this business of rotational motion has had a perfect equivalent from the rotational motion we started with. And all we had to do was swap the symbols. Didn't have to really do anything more. We could use the exact same equations. We have a position or distance or whatever it was, however it went, we just swap out that S for theta and the equation was fine. If we had velocity, we put in omega, we had acceleration, we put in, oops, we had acceleration, we put in alpha. It worked so well. I didn't even give you a new sheet of constant acceleration equations. You're welcome to print up your own if you want. We just did those on math type. You don't use math type. But there's no real need to. The equations are so similar. What about kinetic energy? Let's see, what was kinetic energy? One-half mv squared. So kinetic energy is kinetic energy. So I don't know that we could swap out a K. We don't have angular kinetic energy, I guess. The one-half, that doesn't swap for anything else. The v swaps, we already got that on the page there. What about the mass? Well, there's one very easy way to tell. There's one very easy way to tell. Can we just put m in here and we'll be fine. There's one very easy way to tell. The suspicion is maybe, yeah, that would kind of make sense. We just put m in there. We've got this massive thing that's going at some rotational speed. Sure, that's going to have some kinetic energy. That kind of makes sense. It's going to take a while to slow down. It's going to take a while to speed up all those kind of things we learned from kinetic energy earlier. It might seem like maybe we can just put mass in there and we're okay. How can we check to see if that might be okay? Or can we do something we could check to see that's not going to work at all. We're going to need some other idea. What? Units. Check the units. The units for kinetic energy are jewels or Newton meters. So check to see if we put mass in here, will we have units in Newton meters? If we do, it might be okay. We'll just put m in there and just keep going. So check the units. I'll do the units on the one half for you. I'm very serious about this. Bill hasn't even ready turned his hat around. No way I'm that far along. I didn't know they ever made just a black hat. Nothing written on it? What's the last time they made a hat with nothing written on it? Doesn't even say, you know, Gooch's feed store. Gooch's feed and see. Check the units. You have an answer for me? Will it work? If we just put mass in here and keep going. Will it work? Joey? Joey says yeah. Bill's sitting next to him so he says yeah too. Does it work? Just put mass in? Phil? Yes or no? It's simple. No extensions? I know. No what? Why not? I know. I want to make sure he, so he just answered yesterday. He said what you said, I want to make sure he knew what he was answering, because his eyes were kind of blank in half. They're useless. Len? Yes. Yes we can do that. Tyler? Yes. No. Yes. Yes we can. Samantha? No. Allen? About even now. I don't know. You don't know? I answered it can't be yes or no. It's yes and no. How can it be yes and no? Well if you put mass in there you'd have to do radians. Remember the radians is kind of a magical unit when we didn't need it to disappear, when we didn't need it to come back, when we did r times omega, which was meters times radians per second, the radians just disappeared and we had velocity, meters per second. So we don't have to worry about the radians part, so this will be just her seconds squared. Just looking at the units. Are these equal? No they're not. So we can't put mass in there. We need some rotational equivalent of mass. So here's how, here's how, here's one way we can see it. So there's, there's something rotating about a fixed point. That's what we're talking about this last week or two. Let's, let's imagine we've got a little piece of it right there. Some little piece of mass and right there, this thing has some angular speed, it's rotating, I'll pick that direction. What, what's the velocity of this point at this second? Can we come up with some, some idea of what that is? Not by number, but just how we can draw where it's moving at that second? Yeah, this, anything moving in a circle is always, always has a velocity tangential to that circle or perpendicular to the radius at any time. So in fact let's call this little piece of mass one. It's at radius one and its velocity is v one. Now here's a little piece of mass m two out, let's say twice as far, just to make it easy. It's at a distance twice as far, its distance is r two, which is twice r one. How fast is it moving? And in what direction? At that instant. John says this piece is moving twice as fast as this piece. Is that true? Same what? And in the same direction? So twice as fast and parallel. So would I draw s one, m one? Is it going twice as fast as what? Well let's think about it. These two are going to make one circle at the same instant because they're, they're on the same object. Which one's going farther? Two or one? Two's going a lot farther in the same amount of time it must be moving faster. In fact it's going how much farther? Twice as far. Twice as far, twice as radius, twice as circumference, twice as velocity. Is that right? Joey you buy that? Here's the last point out on the rim. Just for argument's sake let's say that's three times r one. R three equals three r one. Just to make it a nice easy drawing. So those r's aren't the additional distance, it's the total distance from the center. How fast is that one moving? If this one's moving v one, how fast is this one moving? Number point three. It's three times farther out. Has to go three times the distance in the same amount of time it's got to have three times velocity. In fact couldn't we just draw a nice proportionality similar triangles just like that? Because they're all on the same object. If they were on the same object that wouldn't be true. But since they're all on the same object that velocity relationship holds. Alright let's see. The kinetic energy of little piece of mass one is one half m one v one squared. What's the kinetic energy of little piece of mass two? m two v two squared, isn't it? That little piece has that kinetic energy at that instant. In fact we could say that for the third piece and however many pieces we wanted to make in fact we could do it for the entire object. Just make the object bunch of little pieces of mass all going in a circle. We know the velocity of each one of those these little pieces of mass because we know where they are from the center. In fact we could rewrite this one half m one v one we could take out and put in r one omega. Isn't that true? Isn't that v one? Remember the arc length is equal to r theta. The velocity is equal to r omega. It's just a time derivative of the equation before it. And then acceleration on the single point is r alpha. This didn't work. That's why we're doing this. So I could do this with the second one m two r two omega squared and however many points by one I could do a bazillion points an infinite number of points if I made the little masses small enough I could do that. So this is really just the summation from i equals one to n where n is the number of little pieces of mass I want to cut this thing into I chose a bazillion but maybe you want a bigger number one half m one r one squared omega squared oh no sorry not m one m i because we're doing the summation now, right? That follow, right? Everything's okay? Now if you remember the summation well they're just things we're adding together we can pull any common terms out certainly the one half is a common term so I'll pull it out of the summation because it appears in every single little piece what else does? Omega does too, remember that's the angular speed of whatever this object is all of these pieces are the same object they have the same omega so that's not part of the summation either so I'll pull that out actually let me pull it out by being a proctologist on it and I'll pull it out the backing Alan you didn't even laugh at that I laughed inside What's Amanda? How do you go from your r i omega squared to r squared omega squared How do I go from here to here? Just the square of two numbers multiplied if we have A B to the nth that equals A to the nth full of exponents I don't know that but I guess we factor it out I haven't factored it out yet I'm not done factoring I've factored that out the back end and this out the front end That's good because here's here's what I didn't factor out well whatever's left m i r i squared is that right? oh yeah sorry sorry you're right I did lose the summation you're right I think it's a little late it really looks bad yeah alright there we go that okay? check the units on that in meters so it's meters squared and omega n radians for this what are those units? what can we reduce those to? these two Newton meters for kinetic energy so this must have something to do with the way in which mass rotates we'll wrap that up okay yeah we'll be okay wanted to get through that but didn't quite what? why radians just turn the work yeah because radians for those other equations sometimes the radians unit has to disappear because it doesn't fit in units on velocity meters per second units on radius meters units on omega is radians per second but we don't talk about meters radians per second equal so the radian unit just has to disappear I don't know any other way to explain it other than