 Alright, we're looking at general motion now, which is the possibility of both translation and rotation. We're going to look at some things where some stuff's in translation, some parts are in rotate. It just depends on what all the parts are. We had the general kinetics that we did on Monday, which was that an object, if subject to unbalanced forces, can't accelerate. We did that in physics one. We also looked at the possibility that if the moments are unbalanced, then there too might be angular acceleration. The general motion is the possibility, the likelihood that both of those apply. So that was the approach we took to problems on Monday. It works very well if the forces are constant, the accelerations are constant, but it is perfectly applicable if they're not. It also works well if all the problems concerned with is accelerations, either that's given or that's asked for. If we have other types of problems, just like we did in other parts of things we looked at, if we have other problems that either involve have a heavy dependence on position, or doesn't specifically concern itself with the acceleration, but rather the velocity, then the work energy equation tends to be much more useful. That's what we'll look at today. There's a few minor changes to it where its general original form is unchanged, but we're going to have to fill with it a little bit as we get to some of the things we're doing. This part here is work by non-conservative forces, meaning certainly not including gravity, because that's over on the other side anyway. We don't want to count for it twice, but gravity is also a conservative force, as are the elastic forces due to springs, at least the way we deal with them here. Of course, the big one here that we might be concerned with is friction. That's certainly the best example, but if we have anything in there like a rope or somebody grabbing in and reaching in and grabbing in and pushing or pulling or something, those are non-conservative forces, but friction is the big one. However, for most of our problems, it is actually a very little concern. If we look at a rolling wheel on a surface, naturally there is friction at that contact point, because if there weren't, depending on which way this is moving, if there weren't friction there, that wheel wouldn't turn. It's the friction that actually supplies the torque that makes this wheel turn. If there's no friction, then there would be no reason for the wheels to turn. If you have a car on an icy surface, not only icy, but greasy ice, ice with coal tar spread on it, which I found out from one of my other classes is the slipperiest thing there that you can, that's what I was told in the other classes, whatever would have this wheel on it, maybe your little toy red wagon would just slip down there and the wheels wouldn't turn because there would be nothing to make them turn. However, for most of our problems like this, where we have a rolling wheel and friction at the bottom, we observe the no slip situation then, which means we can directly calculate how much work is being done by the friction forces on a rolling wheel that is not slipping. We can calculate that right now for all problems where there's no slip. If we remember one little thing we hit a couple weeks ago, what's uniquely interesting about that contact point, if you remember, for a wheel that's rolling without slipping, the velocity, that contact point is not moving. If that contact point is not moving, then the work done by the friction force is zero. The friction in these situations, when we have the no slip condition, the friction is doing no work because it's acting on a point that does not move and to calculate the work done by a force, we need the force to move through a certain distance with that object and it doesn't, it's acting on a point that does not move. So we don't need to consider the friction if we have the no slip condition. If the wheel is slipping, which on some of the problems we've done, I think we did, do we do that lawnmower problem where the wheels skid at first? I can't remember if we did that or not. Oh, we did, we did what, a motorcycle or something, we did something where the wheels were skidding on startup and that, in that case we'd have to calculate how much work is being done by the friction forces. But if there are any of these non-conservative outside forces, we calculate them in the usual way, which is the integral of f dot ds if we have a force like that. But we can also have couples doing some kind of work as well, wherein for our simple problems that would be c dot ds, yeah I guess that'd be, I don't know if we can make the dot product out of that as well. So if we have a couple, a motor for instance doing work on something, it's going to depend upon, sorry that shouldn't be ds, that should be d theta. We can also determine the work done by that and so if we have, perhaps if we have some slipping or skidding or we have a rope pulling, we can figure out the distance through which it works. If we have some applied couple, some applied torque to the object, we can also figure out the work done by it depending upon the distance, the angular distance through which that couple works. So that's one change we're going to have to make now that we're adding both the possibility of translation of motion and rotational motion. The other place this could really apply is with the kinetic energy, and if you remember in this class our author happens to use a t for that. We have to look into the possibility that there's both rotational and translational kinetic energies. Translationally it's as, just as we've done it before, rotationally it's very similar to that, you can add to it. However if we've got a rolling wheel where it has both translational kinetic energy and rotational kinetic energy, I guess I drew that the wrong way, then we know we can take these and we can take out this velocity v, I guess you put a g on there, the velocity of the center, we know that that's related to omega at that instant because of the no slip condition of ours and so we can take out vg put in r omega squared plus the rotational part of it and then if we do a little more algebra with it, let me bring it up here just so I can finish it in some space I've got, I'm going to pull out the 1 halves, I can pull out omega squared leaving behind and, oh wait, sorry this is, omega's not squared quantity squared, so this is r omega quantity squared, so I can pull out the omega squared that leaves behind then m r squared and ig and that should also look familiar maybe if I turn around the other way, this is ia if a is a label from the contact point so we could figure out the, that's for a wheel, if we have other parts in the problem that aren't rotating in that way then we have to treat any translational and rotational components separately but this is in particular for a wheel in no slip and by wheel I pull out the omega squared and that could also be a pulley or a gear of some kind just as long as there's a no slip condition in the contact point that's not terribly different than what we've seen before and then the change in height in a gravitational field and any change in elastic potential energy are unchanged and there's no rotational difference between what we're doing and what might be on there otherwise, so we'll do a problem or two and then maybe see if we don't have a get out of class question for John as he wants to leave early. Alright so here's a situation, maybe some kind of automated door, some situation might come across here, so the upper part travels on wheels in a track, so does the lower part, so we have this, maybe a track door of some kind or other that needs to be controlled and we'll attach a spring to that bottom end, okay so that's the base picture, the basic setup, this is four tenths of a meter long and has a mass of 10 kilograms, we'll take the wheels themselves for this first part of it as we look at this to be a negligible mass, if they weren't then we have to add into it the problem of not only do we have to get those things moving, we have to get them rolling, so we'll deal with that in a little bit. The spring has a spring constant of 800 and the initial angle is 30 degrees where it's released from rest, so that'll be 0.1, when it reaches horizontal at that moment it'll have some angular velocity, we want to figure out what must be a bar, the door, whatever it is, reaches horizontal, oh what other thing you need, at horizontal the spring is unstretched, so that's when the bar, the spring is hooked up when it's at, when it's horizontal then it's pulled down to 0.1, take a deep breath and then let it go, I want to find the angular speed when it comes back to horizontal, this is one depends upon position, also one that does not concern itself with acceleration, there is acceleration but it's neither asked for nor is it given, so that's generally a good indicator to use the work energy equation. Alright there's the first step right down there, work energy equation, second step, no, well you might need it, but second step remember is to determine if these are 0 and the problem already gets smaller, what? 0, the first one is 0, by first point you mean this one or this one, this one, yeah there's negligible friction on the wheels even so they have the no slip addition the work done by that friction would be 0 anyway, so there isn't really anything else to, oh well there are other forces certainly and so that's another type of force that will not cause any work to be done, we looked at the friction force acting on a non-moving point but there's also the normal forces in any one of those places, perpendicular forces do no work anyway because f dot ds of those type of forces is automatically 0. So we're starting from rest but we're not finishing there so there's a non-zero delta T, we've got a portion of the mass at least that's increasing in height in the gravitational field, so the second third term delta BG is also non-zero and of course we've got a spring in the problem that finishes, starts and finishes at different lengths which automatically means delta VE is non-zero, so delta T is going to be T2 minus T1, starting from rest so T1 itself is 0 and so we need the kinetic energy of the bar, we need the wheels to, if mass of those was a concern but I said they are of negligible mass, so we have certainly the factor that this, this, this bar changes its angular orientation which is going to be reflected in the fact that it's going to have some angular rotational energy but it's also true that just its center of mass alone will also be moving, once it gets to that point the center of mass will have a velocity of 2, both of those have contributions to the angular velocity of the, of the setup, however this reduces to 1 half at i A omega square if A is that point, oh by the way these are all 2's on air because at the moment it reaches horizontal at that instant it is in the center in rotation about that point A, because that point A will move to the left and come to a momentary stop when the bar is horizontal then if the bar went past that point A would start moving back in, so at the moment the bar reaches horizontal then this roller point A is actually not moving and so the bar is at a different rotation about that point and we can use just an i A if we have it, which for a slender bar we do, so we can put those pieces in, 1 half, 1 third m l squared, that's the moment of inertia of a slender rod about one end and at the instant of concert point 2 point A is not moving and so it's in rotation at that point, so we've got enough pieces there we can claim it up from 1 sixth m was 10 kilograms l is 0.4 meters, the unknown omega 2 that we're looking for, so are the units right on that? if not we're missing something, we'll have kilogram meters per second squared and another meter, so it'll be Newton meters, so the units are okay and that reduces 0.267, omega 2 squared. Then the other two parts are simply as they normally would have been, there's no rotational change to those pieces, we just have some mass that's changing height and gravitational field, remember though that we take the mass of the bar to the edit center point, so it's only traveling up that little bit, but we can figure out what that is, 10 kilograms, g is 9.81 meters per second, delta h is increasing in height and that distance is 0.2 sine 30, that right? That change in height there, 0.2 sine 30 and it's positive because it's increasing height in the gravitational field, so it's increasing its gravitational potential energy, so that figures out to be 9.81 Newton meters, sine of 30 I think is one half, so that's where that came from, just coincidence. And delta v e, you set that up now and let's see if we get the same setup for it, just to make sure fk del 2 squared minus del 1 squared. No rotational concern with that, generally springs are so light we don't worry about their mass anyway, if it was a very heavy spring or all the other masses were very tiny in this problem, we'd worry about that. Yeah, here it's at rest. Yeah, the system's released from rest here and then the spring pulls that up to horizontal, actually if the spring's strong enough, if you can't get a solution to this then it means the spring's not strong enough. And there aren't any unknowns in this, so we should have a number for this one, the unknown is in the potential energy, the kinetic energy term, and then once you get that, it's up in the equation back together. Putting it all in there with all the units checked, all the signs checked positive because they're both increasing, everything's in Newton meters, so we should be able to find an angular velocity. Okay, looks like most everybody's got it. At the second point, remember the spring is unstretched, so that first delta term is zero. The second delta term is the amount of stretch that's here, the extra distance that's stretched out, which I think is point two. That gives this term then a negative sign which it should have because it's got a lot of energy stored in it when it's stretched here. By the time it gets horizontal, it's got no potential energy left in it, and so we can figure out that term and it comes out to be minus 16. The minus sign makes sense, the units work out, so as long as our simple math there's right, we're okay. And now, all you've got is a very simple equation with one unknown, and you can solve for omega two, an angular velocity of the bar in this situation. Got something? Where'd you get Travis? Yeah, that's what I had. John, check something. Yeah, it'll give up now. It should get 4.81 in the direction shown. David's 4.81, you got that too? Chris, you did, you're having a good day today? It's what? Yeah, it's kind of like we have a net average day. He goes down, we go up, everybody else stays pretty much average. Joe, okay? What's the number of just the units I'm going to use on it? No, no, no, we'll make a square there. That has kilograms per meter. How's that? Remember, this will be in radians per second, but it's squared, so we have a kilogram, one meter per second squared, that's a Newton times the other meter, that's Newton meters. It's got radians in there, but remember that's the magical mystical unit that disappears when we don't want it, when we're going through the calculations. So does that work okay then? Okay, questions before we clear up and do another one? We're trying to get to a get out of class question. This one, right? Yeah, what? I guess the get out of class question you want is, do you want to get out of class? Halfway through to class. We are, we're half way through to see we're having fun. We're not. We're more than. No, not quite. So let's get back to work. We don't have to, you guys are paying for this, not me. I don't imagine my bosses are watching these tapes, so I can do whatever I want. Alright, so here's another setup, kind of similar. So here's a wall and a floor and we have another long slender bar pinned at the bottom, vertical with a concentrated mass at the top. What I mean by concentrated mass is we don't have to concern ourselves with its radius. Its radius is small compared to all the rest. If it wasn't that way then as this thing twists we have to account for the fact that the mass itself is also turning. So right at the midpoint of the bar we have a spring. So that's, and the bar is two meters so that's one meter of bar above and below that. It's not to scale so the length of the spring at this point is 0.1 meters. So picture is not to scale. Very small spring kind of crammed in there against the wall. The bar is 9 kilograms and the mass whose radius is much less than all the other measurements in here so we really only need its mass as 4 kilograms. And at the moment shown it has an initial angular velocity of three radians per second. The spring has a rest length of 0.25 meters. So we've got it up at this point. The spring has squished a little bit there and it has an initial angular velocity. So we want to know how big the spring should be. By that I mean what's the spring constant? Find k. The spring constant such that as the bar reaches horizontal it just comes to a stop. To me, turn and go back up but the spring's got to be strong enough so as this bar comes down it has zero velocity right at that point. The spring's too weak. It's going to crash into the floor there. If the spring's too strong it won't actually make it down all that far. So find k such that omega with an angular velocity at horizontal is 0. Again, it's got quite a bit of dependence on position. It's got a spring in the problem. It's not asking for anything it has to do with acceleration. So it sounds like a good candidate for the work energy equation. Take a cut through there, see if any of the parts disappear. It's not a wheel, it's a mass. It'll look like a wheel. See, I didn't draw a little axle for it there. It should be square. Oh yeah, no, it looks good. Probably might stand. I hope I still have it. We'll do that on the get-out class on Monday when you're really in a bad mood. We do have those too. You have to figure it out. Yeah, you have. Well, you can do it as a single object if you know the moments of inertia of each as the same as treating them separately anyway. All right. Any parts of the work energy equation disappear? Not always do they, but sometimes do they. No outside work going on here. We're assuming there's no friction in this little bearing down here. If there were, we'd figure out how much work it does by how big it is and the fact it goes through 45 degrees. We've got two things moving here. So each of them is going to possibly have a kinetic energy. However, it's finishing at rest. So we don't need that. But we do need the initial kinetic energy of the system. Now the rod is in pure rotation about one end. So might as well just start with that fact because the moment of inertia with respect to that end is so easy to find. It's very well known. But then that's mass, the minus on the outside. The mass of course has some velocity, but it also has an angular momentum component to it that we need to figure. So we need to do one half a omega squared from it as well. But that's going to turn out to be very simple. And it's a mass, it's not a wheel. Oops, oops, oops. No, this would be cheap. All these subscripts have to match. This becomes one half m, the mass of just the sphere part i and omega one squared. If you want to do it that way, it doesn't matter either way. Works out in the wash anyway. Let's see what's the easiest way to put that together again. If we do this, then we can pull out omega. Remember the subscripts have to always match. So if we have this and this, or we could have one half m a squared but a's not moving. So it just becomes one half i a omega. Folks, that's not supposed to be a two. That's supposed to be a one. But that whole equation can reduce down to that part. So the moment of inertia for the rod is just to make sure you use the appropriate mass. We know that it's one third mL squared about one n. And this is clearly, the rod's clearly in rotation. For the mass itself, if we take it as a concentrated mass, it's just its moment of inertia with respect to that's elsewhere. It's just a mass of a particular distance from the point of rotation. So you can just figure that out straight away. And we've got all those pieces. We've got omega one was given the mass of each. Make sure you put the right mass on the right one. The mass of the rod is nine. The mass of the sphere at the end is four kilograms. And then the L is the full two meters distance. Because it's either the entire length of the rod for the first part or the distance the mass is from the point of rotation for the second part, which is the same. And the minus sign because it's losing kinetic energy starts with quite a bit, finishes with none. So put those numbers in, see if you get the same thing I got on that one. And then we've got the other ones to do as well. Is that your idea? Yeah. Yeah. Remember, if we do it this way about one end or some other point besides the center of gravity, that's all we have. If we do it about that point but with the numbers respect to the center of gravity, these substitutes have to both match. So what that means is the one half Mv A part is zero. That's a g. No, there's no delay in this problem. That's a g. Okay, should get minus 126 Newton meters on that one. You got what? So Chris, are you going to screw up on this one then? She's gotten down first, see, and then that would have forced John to screw up. Minus makes sense. It's losing kinetic energy. 126 makes sense because that's what the calculator said. We always believe it. And then we've got two other terms that we can... What? We have, we get in touch at the end of the term. We have two masses, each going through their own change in gravitational potential energy. Let's move that together. Because you need to know where the center of mass is then, of the two together. I think we'd probably be up here somewhere. I don't know if that's negligible. I mean you can declare it negligible. Let's see how that floats. The g comes out and the mass of the rod is 9. It falls, the center of mass of the rod falls one meter and then four kilogram falls two meters. And it should be negative as well because both of them are losing height in the gravitational field. And we've got all those numbers. So this should be minus 167. Doing it my way was 1.6. All right. Well, if you want to do it your way, that's fine. But you have to come here to the front of the class and sing, my way, okay? Keep that tape running. The pressure's on. It's crumpled. I sing all the time. Of course, I don't know if it's the way to edit that out. Oh, one. See that? That's what happened on the element, John. The original 167 was correct. We only have one spring in the problem. We're looking for its size. So there's our unknown. All we need then is del 2, del 1. Remembering that those are the extra stretch and squish in the problems. Not the length of the spring, but the change in the length of the spring from rest. So del 1 is easy. It's a 0.25 meter spring, but we've got a squish down to 0.1. That's minus 0.15 meters. So what's the rest of 0.25 meters? It's down to 0.1, so that's a change in length of 0.15. And then del 2, you just have to figure out from the triangle that the spring stretches out to how much longer that is. Remember, this is non-directional, which makes it pretty easy to use. We just care about the length of the spring, not which direction it's pulling. It makes it a lot easier. The work energy equation is not a vector equation. And this should be a positive number because there's a little bit of change in length, and we end up with a lot of change in length. If that wasn't the case, then the spring wouldn't stop the system from twirling down to its point. So are you ready to call or call? Place an order. We'll see. Where are you stuck? Remember, del is defined as L minus L0. So del 1 is L1 minus L0. So I think del 2 comes out to be 1.24 meters. It originally was squished, 0.15. Now it's going to be stretched out to, I think it's 1.49, something like that. 1.39 maybe it is. I'm just going to write that down. Okay? No? Tommy? Where are you stuck? 2, right there. All right, let's see. Del 1 is L1 minus L0. L1 is 0.1 meters. L0 is the rest length 0.25. So that's the easy one. L2 minus L0. L2 is this length. So you have the right triangle that's 1 meter there. And what was the length? Oh, 2 meters plus 0.1. So you have a right triangle. It's 1 meter plus 0.1. So this is 1.1 meters. This is 1 meter. And so now the final length of the spring is that. And then you subtract off the rest length of the spring. And that should come out to be 1.49 meters. 1.49 minus 0.25. And that gives us the 1.24 over here. And so you should get from the piece of that 758, and that's the last little bit, last little bit of our problem that we can now solve for K. And then call her all quarter of the spring. Feel that makes sense? Okay with it? And they should get 387 newton meters. And all the units work out as long as we put those lengths in here. Any parts sticking? Doesn't quite work? I had an issue with the kinetic energy, but I worked that out. It turns out I was right. The first time I did, I was right. Well, there's your... Ooh, then don't redo the problem. And then you get that same thing. And then that unit's can be right. All right, now you want to get out of class question. Wow, wait. Which means it's got to be a really hard one. So maybe I don't want that one. I didn't want this one. Since it's a Friday, we'll give you a get out of class question. Everybody all done? David, you okay? Yep. Tom, comfortable with that? Simple problem. One, the ancient Egyptians had to approach. So on an incline here, of angle three, is a massive concrete block on rollers. As if, you know, just a nicely rounded block. They're not wheels, but rollers. We'll give this the symbol big M for its mass. And each of the rollers is of little mass and radius R. But we'll say that the mass of one roller is equal to half the mass of the piece itself. So both rollers weigh the same as the, as the, what if that object is probably not a slab of granite. Because I don't know if the two logs weigh the same as the slab of granite. But one, the Egyptians didn't have any trees. It's Egypt. And two, they didn't have any granite. So it doesn't matter. It's academic. All right. So I think that's all the pieces you need there. Release from rest. Release from rest. Find the velocity of the box or the crate or whatever it is. We'll give that a big V because the wheels will have some velocity of their own. Maybe a little V would make sense to go through the lamp. Find big V after the rollers have gone through 10 revolutions. So the rollers will have rolled around 10. Now ignore the fact that you make, this thing's long enough that it doesn't come off of a back roller. It's not getting out of class that quick. And also assume no slip. In fact, without that, you can't do this. If things are slipping, we don't know how much if we're able to find out any of those parts. All right. It has a lot to do with position. It doesn't ask for anything about acceleration, just velocity. It might make sense then for, it's just some variable r. So your answer will be a function of r, m, and wheat. And anything else that you think applies. No outside forces and no spring, which is opposed to no winter, which we just had. Certainly this thing's picking up speed. So delta t will be positive, delta vg is dropping in height. That's where the speed's coming from. Are we assuming these rollers are attached to that? What, that these wheels are attached there? No. No, it's this big slab on these rollers, just like the Egyptians used to do to roll these big slabs. We said they're not flying out. No. It's kind of like one knee. That makes sense. Good. So we'll start like that a little bit later. It'll have rolled down some. And the wheels will be back a little bit farther. They'll have... Are they stationary though? No, they're rollers. How could they be stationary? We've got lots of bearings. No. No. No, it's... Here, here's the ramp we're going up. And the first thing you do is you put down these rollers. And they won't go anywhere because they're not supposed to yet. And then you put on top of that this big slab. No bearings, no attachment, no nothing. And then you let go. And so the slab starts to roll down. So not only does it go down farther, but the wheels also go down farther because they're rolling forward as the slab rolls on it. In fact, you should be able to relate the velocity of those two to each other since there's no slip. How do we have the picture? The delta T term is probably the more problematic. The delta VG term shouldn't be quite as problematic. You get the picture, John? These guys were asking about bearings and all kinds of stuff. How about bearings? Yeah. So the delta T term is going to take some thinking. We've got two things going. One's pretty easy. It's just simply in translational motion. The big slab thing, whatever it is, maybe it's a Pharaoh's coffin or something exciting. Using big M and a big V for it. But the wheels are going to be a little bit different because they are indeed translating. And it starts from rest. We'll have the T1 terms of no concern. So the wheels are translating but they're also rotating. There's two of them. So just multiply that by two because whatever one does, it's a good weekend film. A lot of those terms are related to each other so you can combine them. For instance, the mass of the rollers is half the mass of the slab or whatever it is itself. This looks suspiciously like that business we've done before. So that whole thing could be made into one-half I, A omega squared, where A would be which point? We'll remember for a rolling body. It doesn't matter if it's going downhill or not. It's in rotation about that point right at that minute. So about the contact point. That's the no-slip condition. But there's also a relationship between the big V, the speed of the slab, and little V, the speed of the wheels. I guess I put a G on that. So with that, you can even eliminate more of the constants, get it down to a fairly simple relationship. Leave in the big M's of big V's because we're looking for the big V and you should be able to get this all the way down to MV squared with a constant in front because we have one-half MV squared there. So this can be related. We can relate the masses using this and the V's can be related to each other. The V and omega can be related to each other and then that can be related to the velocity of the slab itself if we apply the no-slip condition. What? These two are equal. No. You can see there's no equal sign between them. They get to each other, but I'm leaving that to you for a couple minutes. Of course, we can also use in here that VG equals r omega squared. Take out the omega, put in the VG, because then when you relate little V to big V, you won't have anything left, the big M and the big V. Oh, yeah. You would have caught that when you checked your units. You want to jump in 10 minutes and then don't forget there's also a delta VG term, but that's fairly straightforward. Well, I'll leave that to you too for the slab and for the rollers, not forgetting that there's two of them. In that term, I think you can also get down to and that'll have r big M and G in it. The r just because it has to do with the number of revolutions. Delta H for the rotor is half delta H for the slab. Right. The distance the slab drops is twice the distance the rollers drop, but clearly it's not a scale drawn. But that also has to do with the relationship between the speed of the slab and the speed of the wheels themselves, the rollers themselves. Has anybody put that together yet? Remember, this is no slip. So the contact point between the rollers and the slab is opposite that point A. And so that will be the speed of the slab. How is that related to VG? Little V. Can you see it? This is the contact point for the slab. So however fast that point is moving, that's how fast the slab is moving. Pardon me? Yeah? Or just two times VG, VG, little V. The slab's going to be moving twice as fast as the rollers are. That's absolute speed relative to us standing here at the outside. That, again, is the no slip condition. If we were slipping at either point, either where the rollers are on the plane or the slab is on the rollers, if either point was slipping, then we couldn't finish that. What are you using for your I? Just leave it here and you don't need it. Because you can just use IG. Because VG is related to omega, so we can eliminate omega and just have VG. IG is just a roller, so whatever moment of inertia is about its center. And then once you've got this in VG, you can eliminate VG using this whole thing in terms of big M and V, and it just comes out to be a simple constant times MB. So IG, if that's what you need for the roller, if it's one half MR or IA, but you don't need, because if you use that equation, it's in the parentheses part, but you can get it directly. So I think we only have enough time. If you can come up with that constant, I'll give it to you. Only need for the kinetic energy? Changing kinetic energy. So you can eliminate the omega with this, VG plus R omega, and then you can eliminate VG from the two of them using M, and so you get down to nothing but MB squared. Again, big B squared. And that would be the energy for the entire thing, even though it's just in terms of the slab. Easier said than done, usually. It's getting a little simpler as you go, isn't it? You got that constant, David? Don't say that out loud. Just wait. Ah, darn. Oh, yeah, that's what I meant. So if you want, David, you can go add two minutes and 45 seconds of your weekend. Take almost two minutes and 45 seconds to pack up and get to the door. Alright, some people are getting closer. You don't have to touch that first part. The two comes together and cancels everything. VG squared plus IG, that's what, that's one-half MR squared, and omega is R squared VG squared over R squared. So the R's cancel, we get one-half MVG squared plus MVG squared. So that's been, what, three-halves little MVG squared. Is that right? What? It's one-half MR squared. Yeah, that's one-half MR squared is the moment of inertia of a cylinder. It's the roller power. Swap out the mass, and that's, we've already put the two in there, so that's the mass of one roller, because we already had the two in there. This factor of two rollers is already there. And VG is one-half MVG for that squared. So that puts two times two is four, times four on the bottom is 16 on the bottom. Yeah, so this becomes three-sixteenths, plus eight-sixteenths, and so you can add them together to get what David confirmed, eleven-sixteenths. That's just delta VG, so let me give you a B because now we're cutting into our weekend. Let me just give this to you so you can find it. Once you gather all the constants, like eleven-sixteenths and two pi for the revolutions, the M should cancel, and you should get 42.4 times the square root of R sign. So you should get that, and if you get that, you get a dominant. No early weekend in this week, maybe next week.