 Okay, I believe we just had the impulse momentum equation. Is that right? I will put that on board. Oh, dang it. I knew it. They put tape on the board. Okay. It makes it harder to write on it. This is the third of the three ways we solve kinetics problems, if you remember. This one, like the others, comes directly from Newton's second law anyway. So it's not like there's different. They're just sort of recastings of the same thing that gives us a chance to solve different problems in slightly different ways. If I remember, this class uses g as momentum, linear momentum. And remember, this is a full vector equation. So don't go throwing vector signs around in a half-pazard way. It's very definitely a vector equation. And I believe we did a problem where we actually took this and then split it into the two component directions, the x-direction and the y-direction. And it works just like our others have for that, where, for example, that we sum the forces in the x-direction and that will give us a change in x-momentum. So it works very much like so many of our other vector equations. So we're going to take a little bit of a step beyond that, sort of add to it in much the same way. If I remember, hang on, we did, I guess we had an alternate form of this that I'll put down just for completeness. I remember, yeah, this was also took this one with the sum of the forces, which is equal to the time rate of change of that linear momentum. And, in fact, just divide it through by dt and then integrate, and you get that one up there at the top. So they're really very much one of the same. This is really nothing than our first one, which is f equals ma. So we're going to go a little bit beyond that now. I add to it some. From statics, if you remember, for Alex, you don't, but you'll remember at least the first take on this from physics one. We were a little more complete with it in statics. But the sum of the momentum in physics one, we call that torque and use the symbol tau. It's just a slightly different notation. Remember, with respect to some particular point, remember what the sum of the torques was in statics? You fell as you did that? How we calculated all the torques? Yeah, it wouldn't be good. Well, yeah, in the more complete form, it was actually a cross product. R cross and then the sum of the forces, R cross the net force on an object. Where we remember this R was the location of whatever force it is we're talking about. Bless you. Maybe the force is something to give us that. And we had some point O and R is the position vector locating that with respect to that position O. So we had that as a full cross product. And if you remember the angle between those two, maybe we'll call it theta, then this has a magnitude of R. Yeah, let me write it that way. Just trying to think of which is the clearest way to write it. The magnitude R times the sum of the forces R times the sign of that angle between them. And then since that's a full vector up there, this better be a vector here. So typically we, at least for the drawing given, would use the right hand rule then for that. And so we have that as the magnitude of the sum of forces, sorry, sum of the moments. And then the direction is out of the board for, at least for that bit of a picture there. Maybe you also remember that the cross product actually gives you the area enclosed by those two vectors as well. I don't know if you remember that or not. That comes to be very, very useful in some of the other more advanced business classes and engineering classes you're going to get to shortly. All right, so that's a little bit of a review. Alex, when we did this almost every single time these two vectors were perpendicular to each other, which made this 90 degrees, which meant that sign of 90 degrees was one. And so it was just our distance times force, force times moment arm, which is what TJ gave us a second ago. So it's not terribly different from what we did because it's just a little bit more advanced, I guess. All right, hopefully that's just a quick reminder because what we want to do with it is this. Let's see, so that's R cross. And then that's some of the forces. Let's see, that's MA, so that's MV dot. Same thing, that's just F equals MA, pulling that in there. Now let's see what that second part means. It's actually, we sort of have to come to it in a roundabout way. Let's see, let's do this. Let's take R cross MV and we'll take the time rate of change of that. Take that through and then we'll connect these two. We'll see what this business really is that we've got here. This takes two parts. It's doing the derivative of the product rule. One part is R dot cross MV. That's DET of the R part and then leave in the second part alone as you do in the product rule. And then we leave the first part, oops, that's a R vector. Then we'll leave the first part alone and take DET of the second part. That's good because that thing right there is that thing right there. That's why I did that. So we've got an idea then of what the second part is, this R cross MA, if you will. Look at that first part, though. The second part we're going to use over there. But the first part, what's that? Anybody have any idea? What if I wrote it this way? R dot, we know to be V. So this is R cross MV. Well, remember that the magnitude of the cross product, so let's see, V cross MV, those two vectors are parallel to each other because M is just a multiplier. And so the cross product of two parallel vectors is Tj? Zero. The cross product of two parallel vectors is zero. If they're perpendicular, then the cross product is just simply their magnitudes because then the angle between them is 90. The time the angle between them is zero. The sign of zero is zero. So this whole first part here is zero itself because that's the cross product of two parallel vectors. That means that this piece here is the time rate of change of R cross MV. Let's see if we can get just a picture of that. Here's some useful coordinate system. Some point that serves as our origin. We've got some path here that our object can follow. And there it happens. No, let me just let the vectors lay a little bit better. So there's our mass MV. It's momentum vector. And here's R for that instance. We have R cross MV. Let's see, that's the momentum as this object is moving past a particular point in some way depending on where it is along that path. But as it moves past that point, that's its momentum with respect to that point. That is actually what we call the angular momentum. It is with respect to a particular point which makes it a little less intuitive than linear momentum is. Linear momentum is just what it is. It doesn't matter where it is or where any origin is. But the angular momentum, a little less intuitive because it's different with respect to different places. But it's that cross product, that quantity R cross MV. We then have the sum of the moments on an object is the time rate of change of the angular moment. And that looks very, very much like the sum of the forces we just had which was the time rate of change of the linear moment. These two equations completely define the dynamics of a particle moving in space. The moments of the time rate of change of the angular momentum, sum of the forces, time rate of change of the linear momentum. So they're both very much the same type of ideas, same type of thing. The units, meters, kilogram, meters per second. So what is that? Kilogram, something like that, or Newton, meter. Remember, with respect to some particular point. For our two-dimensional problems, this will become rather easy for the most part. We don't have to do the full three-dimensional cross product time we were here. That's really, remember, just ethical ZMA. We got from that last time the impulse momentum equation, where the amount of time of forces being applied. Remember, that's the area under the force time diagram. Will be the change in the angular momentum. And we also have now another part to this that we just developed. And it, too, leads directly to the angular impulse momentum equation. So the amount of time that moments are applied, forces offset at some moment arm, will cause a change in the angular momentum. So now we have the impulse angular momentum equation form. Remember, just the same, same, they spring right from F equals MA, and some of the torque, some of the moments as you call it in this class. It's as intuitive as F equals MA is, but we'll also keep it a little more straightforward. So let's do a little piece of it. Here's a circular piece of track sliding down that track, such that when it gets to a certain point there, that will define with which angle? This angle. This is a circular track of radius R. When it gets to a certain point theta, it's got a velocity V at that instant. Just to get used to what we're doing here, find angular momentum at that instant. The acceleration at that instant as well, tangential acceleration. The angular acceleration is just going to be, the radial acceleration is just going to be v squared over R. So at some point theta, it's got some velocity V. Find then the angular momentum. It hasn't changed, so that's rather easy. In fact, because of the perpendicular nature of these two quantities, then this is just R and V, so this is a vector. So I guess taking our usual convention that clockwise into the board is negative. This would be minus R and Vk. Nothing more than that. So that's the angular momentum about point O. It's acceleration at that instant. Let's see. Tangential acceleration, if you remember, is the time rate of change of the velocity. Is it not? Because the velocity itself is always tangential. There's no normal component to the velocity at any time. For V dot, well, we have, well, we almost have a part of it. Most of it comes from a new form of this equation. Remember where this was? R up right there. So there's the V dot we're looking for. Two dimensions. This is a lot easier. We don't even really need the vector because it's either into or out of the board. We automatically know which. H dot zero of this we've already got. From here, we just need to take the dot product of it. We've got the minus sign implying that it's clockwise, which is our normal convention. Some of the moments, that's a little bit different. Well, just in that, we don't have it up there yet. So let's put it up there. There's the weight of the object. There's the normal force exerted by the track, which is, of course, perpendicular to the track and V itself. And it's the sum of the moments caused by these two forces. Those are the only two forces in the problem. We're assuming a smooth track, no friction. How much moment exerted by the normal force? The normal force goes right through the origin, so it's not going to exert any moment about the origin. So we only have to concern ourselves with that part of the weight that's exerting any moment. And so let's see. That's mg is the weight. This moment arm here, we move that R just so it's not confusing. That's, of course, the radius. This is the moment arm. Maybe we'll call it D. I think fairly typical in physics one. What's that length? Both these angles are theta. So that's R sine theta. So this is mgR sine theta, but that's also in the negative direction. Those two parts are equal. Minus sine's cancel. R cancels. M cancels. That level's not a vector, because we know automatically from inspection that it's clockwise. Minus M, R all cancel. G sine theta is all this way in the impulse momentum form of that very same business. So let's say we've got a little gizmo here, a little pulley there on an axle. We're going to pull on that pulley as the axle causes a like, so we're going to pull on that force causing that to spin. This is a very same type of thing that is done when a satellite is spun up to speed so that it can be released from the space station in the space shuttle. They give it a spin, an angular spin to help stabilize it. So this is a very type of thing that might do spin it up to speed. Probably something a little more complex than one of the astronauts pulling on the string there. But that's the idea. So put some numbers to this. Each of those pieces is three kilograms. The arms are very slender, 400 millimeters. The radius there is 100 millimeters and the force T is 20 newtons. All right, so that's the basic setup. Your spring break-addled brains can see that. Jake? No, Jake's just thinking 20 minutes after the endless buffet to put on anyway. Here's the other information. Starts from rest. There's some amount of time later that you need to find to such that its final angular speed is 150 rpm. The speed we wanted on it. That's the time we release it. Since time's involved, maybe you're thinking that application impulse angular impulse momentum equation might work. The impulse side is fairly straightforward. In fact, even more so if we look straight on to this pulley. If we see the pulley like that with this 20 newtons force at some moment arm, and you see that the impulse, 20 newtons, a radius of some unknown time T that you're defined. Angular momentum, because it's a two-dimensional problem when you look right down the axle itself. Finding then the change in angular momentum. I'll do half of it for you. Colin, you want me to do more than that? What's the initial angular momentum? Nothing's moving. There's no velocity, angular or otherwise. So there's my contribution. Colin at first didn't think it'd be enough knowing he realizes that was a huge chunk of the problem. He's willing to take off the rest. So if we can find this second angular momentum, then it'll be a straightforward matter of finding then the amount of time T that we need to apply this 20 newtons force. By more than that gets more speed, less than that will have less speed. We want it to have 150 rpm at release. Your duty is to find that final angular momentum. I know that this pulley has insignificant mass compared to the rest of the system. That, too, and the axle as well needs to be spun up to speed as well. But we're worried mostly about the contraption at the back. Momentum of this system with very slender axle, very slender, nearly massless arms, and a small pulley at the front. In the class we'll take into account the size of all of those things. Right now we're still looking at particle dynamics. Regular momentum. You think we're using for this reference, oh, that's right down the shaft because everything's turning around that. What that is, R. R would be the 400 millimeters. Down? Well, you've got that. What about V? Angular velocity. V, which is not angular velocity. That's linear velocity. This little piece right here itself is the linear momentum. Yeah? If you're again making revolutions per second. Any instant you've got these little masses at some distance R, that way such that it'll have some velocity. That velocity is directly related to the angular speed omega. It's there. Spring brake withdrawal. Go back and go to bed. You've been waiting this kind of the day for a week and a half. I know, I got a son about your age. I got a half-time son about your age because he doesn't get up till about now. I'm not sure about the connection between omega, R, and V. And there clearly is, because the greater R is, the greater V would be. The less omega is, the less V would be. All those things are clearly interconnected. What then is the connection between V and R and omega? R omega. That helps a little bit. However, it's not in vector form. What's that cross product become? It's since those two vectors, R and V are perpendicular. It's just the magnitude, R and V. However, what were you saying? About the masses? There's four of them. But each one contributes. So you have to multiply that total times four. This is the total angular momentum. You put a summation sign in front of it there. And then V, we now know to be... I give you omega, what I gave you is operand frequency, which is not the same as omega. Omega remembers angular velocity in radians per second. Did you okay? Call him. You got it? No? Did you hear what he had to do? Did you agree? Because he's more well rested than the rest of us, because we've been shoveling ice for two weeks. Pat, that helped? A little bit. Bobby? Talked somewhere? You're what? That's a good one. Find some help. Alex? He's talked somewhere? I didn't understand his question. You didn't understand his question? Bobby is stuck with omega. Because this is not omega, is it? Oh, he said omega is one. Oh, okay. Okay, cool. Bobby omega is for that operating frequency. However, when we use omega, do you remember what the units were for it? Yeah, radians per second. So we need to turn revolutions into radians. There's two pi radians for every revolution. Once around, there's two pi radians, and we need to make that into seconds. Yeah, 15.7 radians per second. So we've got r, and now the four, remember, is because there are four of them. And you should get them what for the final angular momentum. Since this is, I have it written in vector form. People, I may have it separately. Units are what? Three meters squared. Yeah, three meters squared per second. Then you can, that goes in here. And you solve for t. The total time it takes to bring this thing up to speed from rest, that's what you've got, right? Alex, you've got that, too. Good job. What would make that time longer? To what? Bigger masses. Longer arms. Both. More masses. Any time they add something to a satellite, all this calculation has to be done. All that has to be done again to get things just right. The sizes and the masses change for these things. Smaller force. What? Smaller force. Smaller radius, or both. Okay, maybe another satellite type problem. Some kind of slender linkage thing here. Pivoting on the center. And there's a little hook on each end there. Yeah, L for the length of the arms. L coming in here is a single mass moving at some velocity 3V down here is a double mass, 2M at just a single V, whatever that quantity is. So this is going to be terms of quantities. M, V, L, find the final angular velocity assuming it starts from rest. Things here may be a little more subtle but we'll come to why as we discuss it and see what you can do with it first. Where's omega F? Yeah, if we had the final velocity of those objects, because once they hit and stick to that arm, they're going to have the same velocity as each other and they're certainly going to affect each other. Once we find that angular velocity we can find the final angular velocity of the system itself velocity there once those stick on. Where's that going to come from? Angular momentum equation which we got a little bit ago today. So there's sort of a solution map that we can start with so we can find this final angular momentum and it'll be again in terms of V and R or not R, L and M even bigger. Angular momentum, I start moments and like respect to O, hopefully it's pretty obvious that should be this pivot point there. Moments half here, that's what we're looking for. H O, I won't bother putting, sorry H I, I won't bother putting O. It's about the only point where it's rotating. What's H I? These masses are moving with respect to that point so they're going to have an angular momentum with respect to that point of R cross Mv. So we'll add it together. Obviously it's going to be counterclockwise. What's that cross product going to be? At the instant those things hit this then the momentum vector and the R vector are perpendicular so it'll be just the product of the two. So the first one, R is really L here, M3v. That's the angular momentum of this first little piece as it hits the arm. This momentum might be different some other time before it hits there but we only care about once the instant it makes contact with the arm because that's when things start to change. What's the angular momentum of this piece? 2LMV. Initial angular momentum. I guess to make it a vector we'll say is a vector equation or results in a vector. So H I into is 5. That's the initial angular momentum of this. Well what about the arms itself? Not worrying about those. They're slender, they're light as always. Later in the term we'll take into account that that may not be the case get all of this thing spinning up to speed. From right now we're leaving those as very lightweight. So there's part of it. There's the initial angular momentum. Then what? What we need is the final angular momentum from that we'll get the velocity of any one piece of them and from that we'll get the angular velocity. So what's next? You think they're the same? They feel the same. Well what's going from there? If those are the same then the difference is zero and that's got to be zero then. Is that the case? Let's see if this doesn't help remind us. We have the angular or we have the impulse momentum equation for linear linear momentum. It looked like that. What was this quantity? Some of what forces? External forces. Applied forces. When somebody reaches in and pushes it or shoves it or pulls on that rope. What moments are these? The moments caused by those very same forces. The applied forces causing some applied moment. What applied moment do we have here? There are none. There are no applied external forces there are no applied external moments. There would be if the bearings here were rusty dragged a lot or if we had a motor attached there that was helping spin the thing or we had some brakes on there that tried to slow it down or something then we'd have some kind of applied moment but in this case we have no applied moment. So the impulse remember that's this side the angular impulse is zero. So we have conservation of angular momentum the change in momentum itself the zero. So in conservation of angular momentum no applied torques no applied moments no angular acceleration. This is that quantity V when they hit will they be moving with that same velocity? No, they'll have some other velocity when they hit. So we have to figure out what that is. The first piece which has mass m is a distance l so r m v will be r l m and then whatever the second velocity is we don't know what that is yet. So this will be l m v2 but that v2 I guess vf vf is what we need to find final angular velocity consistent with an f in there What's the angular velocity in this piece once it's stuck to the arm? 2m vf No, once it hits and sticks and is now moving with velocity vf both going to have to move at the same velocity because they're both stuck to the same arm At that instant its momentum is r m v The fact that this was a 3 is going to have an effect on this one you figure that momentum is going to transfer from one to the other not sure which one is well the this one has more angular momentum than this one does In the end though we're left with the same so what's that 3l m we can solve for vf because these two are equal m's cancel l's cancel we're left with 5 third certain v velocity original velocity factor this one is moving v the units work out here's per second, 5 by meters is per seconds, radius works out we have conservation of angular momentum when there are no external moments just like we had conservation of linear momentum when there were no external forces questions on that one you ready for a get out of class question any questions before I raise that one okay with that Alex comfortable with that now like I said this angular momentum is a lot more subtle not nearly as intuitive as linear momentum is it's easy to think at times there's no angular momentum just because nothing's turning at the time and that's not the case alright so here's here's another system we have this shaft rotating at 20 radians per second of course slender and nearly massless on that is an arm a collar this slide there's a stomp thing there at the instant that collar is at point A which is 3 inches from the center angular velocity 20 radians per second and the collar is allowed to slide then to the end of the arm the lightweight arm which is the angular speed of the system when the collar reaches point B initial speed at which point this collar is released because there's insufficient centripetal force it slides outward to point B what then is the angular velocity of the system when it gets out to point B I'm not sure get out of the class question picture long spring brake diet changes I'm changing omega I guess when you get it angular momentum will come full with David Carradine modern history ancient history full momentum equation again about which all this is calculated would be the shaft changing angular velocity since I practically solved it for you now in radians per second before anybody else did so you hold it up yep all of a sudden column happens to be magic when that's over it's exactly the same thing this is one we can actually solve if I've done it I could have been out of it Pat, you stuck?