 Alright, just to get us back up to speed, as we solve these dynamics problems, especially in terms of kinetics that we're working with now. Remember that we started with the kinematics, which is just where things are, how fast they're moving, what's the acceleration at certain times. And then the kinetics is how do we get those accelerations, how do we get those blocks, etc. And we have three ways now that we solve these problems. The first, using f equals ma, works very good for general problems, works great for constant force, constant mass, and that's constant acceleration problems, general problems, those that either need or have the acceleration in them, because it's right there, and then especially easy if the forces are constant because then the acceleration is constant. We just finished up a little bit ago with the work energy equation. If you remember its general form, where it takes into account that any work done on the system is going to change the mechanical energy of the system, and that's what these three terms here are. The kinetics, the gravitational, and the elastic potential energies in there, those are all mechanical energy terms as opposed to thermal, nuclear, chemical type energies, so sometimes given delta E. This was really good if you remember for position dependent problems, because the work term have a distance in it, the distance over which it force acts. The gravitational and elastic potential energy terms both had position components to them, so very, very good for position dependent problems. Those that have some distance as part of them are asked for those distance. Now, one thing I didn't point out, I don't believe, it's very important you remember about this, especially in contrast to the two other methods that we've looked at, this F equals MA, and then the impulse momentum method, is that this one is not a vector equation. The kinetic energy when we square the velocity in there, that has no directional components, the velocity that that object is moving in any direction, whatever the direction, it's just simply one half MV squared. There's no directional component to that, and that's true of all the other terms. Well, there's a bit of a directional component in this one, because you have been here the gravitational field direction, and that's the movement in that gravitational field, not a cross movement, but still it's not a vector equation. And then this elastic potential energy didn't care which direction the springs were pulling, it just cared out much that they were stretched or squished. Nothing else in it mattered. The third one we had was the impulse momentum equation, very good for time dependent problems, and it is a vector equation. This one does have a definite dependence in it upon direction in the problem. And with that F equals MA and this one, what we usually do with those is we split them into their component directions and use the equation in exactly that form. Remember that this G is the linear momentum, very, very good for time dependent problems. Problems that have some component in it of time, either the force varies with time, in that case the integral takes us the area under that graph. So those are the three new things we've had so far. We're going to add to this one a little bit today as we look at angular momentum. It's a little bit non-intuitive, especially for those cases where something's not going along some circular path, but it does hold and have its usefulness in certain general type problems, not just those where something is going in an angular fashion. So set that up now. All right. For those of you that who took statics, remember from that we had our second of the big tools that we used that the sum of the moments with respect to any particular point O was the position vector from point O to whatever object we're talking about cross with the net force. And that can be either done as R cross F for the individual forces or you can add up all the forces and do R cross F on the residual forces. Either way, it didn't matter. In statics, this would equal zero. In dynamics, it's not necessarily going to equal zero anymore. But we did set this up for that basic piece from statics. All right, we're going to do a little bit more with it now. Let's see, R zero cross MV dot because the sum of the forces is MA, of course. This could also be written if we wished as R zero cross G dot, the time creative change of the linear momentum. So this is in a way the rate at which the linear momentum is changing as the object goes past some other point. If it's going towards the point, then the cross product diminishes. So it is mostly taking into account how fast that object might be going past some arbitrary, often arbitrary point. So it might look something like this. If we have some object of mass and moving along some path, then at any point there, it's got a momentum and the sum of the moments about that point are going to be equal to R cross MA as we go about that point. So that's the kind of thing we're looking at even if we're not necessarily going in a circular path of any kind. So let's see. What's not clear at the moment is what that thing there is, this new part to it we just got. So let's see if we can illuminate on that a little bit. Let's do this first and then we're going to see that this will come out of it. So if we take the time-rated change of R cross MV, that's going to require what I think is the product rule. So we'll have R dot cross MV plus the dot using the product rule. Look about right? Now this first one here, let's look at that. We've got R dot cross MV, but R dot is itself V because however the position vector changes with time is the velocity. So I've got MV cross MV. Anybody want to do that cross product for me? Why is it zero? These two vectors are parallel to each other, not the same vector, but they're parallel to each other. It's just the vector multiplied by a positive constant, those two vectors are parallel to each other. This is then zero. And you can make up some numbers and put them in and work out the cross product and you'll see exactly that. So this first part of it is identically zero and the second part of it is what we had right here. So that we know then that this is the sum of the moments is the part here that we started with. That part we started with, this part right here is defined as the angular moment of an object with respect to some point, oh whatever that point may be, it might be the origin, it might be some other part of the problem. Our book in its wisdom uses a capital H for angular momentum. So the angular momentum is defined as R zero cross. And of course remember that the cross product is a result in a vector itself. So what we've got now is the sum of the moments, whatever moments, whatever torques if you will are applied in some problems is equal to the time rate of change of the angular moment. A different creature just as moments were different than the forces, but as always these things are very closely related. The sum of the forces was the time rate of change of the linear momentum. So these two are closely related, together they are the time rate of change of the momentum. That's a plurinable momentum. I think it is now those two things together completely define the dynamics of a particle of space. That's all we've been looking at is particle dynamics for the first part of the course. We'll go to rigid body dynamics very very shortly here, but those two together will completely define the dynamics whatever the force or whatever the object is doing and what the forces are that are causing it to do exactly that. Alright let's double check as we usually do in here. The units on angular momentum, let's see of course r is in meters, mass is in kilograms, and velocity is in meters per second. A kilogram meter per second is equal to a Newton second and we have a meter left over, so this is equal to a Newton meter second. Let's see if that makes sense. The time rate of change of that will have seconds on the bottom, canceling the seconds is exactly what the units of moment are. So the time rate of change of the momentum, angular momentum will give us units that work out just right. Just like the linear momentum equation also does. Parts all work out well. H o is r cross m v in matrix form is r x r y r z and then m v y or x dot y dot if you wish and then remind yourselves a little bit on how to do matrix, a simple matrix I got a simple cross product. Interesting to look at for 2d problems which is where we spend the bulk of our time for 2d problems, then there is no z component for either term which automatically makes the other two components of the cross product zero and leaves you then with just the k component is r x m v y minus, we're going to do the cross product with the minus in between r y m v x in the k direction and we can even draw that until we take a second to copy that down. Make sure I got all the little notation things right. We've got a lot of going in here. There's dots coming and going and time rate of changes coming and going and going. Put that picture we just had up there again. Here's some mass going with some velocity that gives it a linear momentum m v. It's got a position r zero. A 2d problem lies right there on the board. If we look at what those two components are, let's see the x component times the y component of the linear momentum. So the x component, there's r x times the y component of the momentum. There's m v y. So we get r x times m v y. Let's see if we can figure out where the minus sign comes from. If we look at the other component, here's r y r y. That's the moment r times m v x. There's m v x there. Notice that both of those have a moment in the same direction. Both of those r x times m v y is doing a counterclockwise moment. m v x times r y is also doing a counterclockwise moment. But the minus sign comes because v x is in the negative direction. It's the negative direction of v x that gives a similar angular momentum to that piece. And using our right hand rule, both of those then are positive in the k direction. So v x would be negative, making this positive, the two add together. For this particular picture, for other r's, other v's, it's still going to work out that they contribute to each other in the same way. So for that particular example, it's the easiest form to see. Alright, so one other piece that we get to that is, let's see, we've got some of the moments, d dt of r zero cross m v. That whole term there on the right hand side is the h zero dot, the time right of change of the angular momentum. But we can do what we've done before to do a little bit of simple differential algebra, d t over, technically let me just put h zero in there. It's going to be simple, I don't need that. And then of course we can integrate that over time. And we get then that if the moment is time variant, then we've got the area under that curve, the moment time graph. And we can do it in a separate component direction, do it in the x direction, do it in the y direction. It's a little difficult to draw a vector time graph into the missions. And then of course this piece over here is delta h zero. So it looks very, very much like our linear impulse momentum equation where the impulse caused a change in the momentum. So we have those two very same equations again related both. It's happening at the same time on these problems. It's not that one happens or the other happens. Unless the object has a velocity straight through the origin and then it would be normal. But that would show up in the cross product game zero as well. All right, so we're the principle of angular momentum, impulse momentum, and linear impulse momentum. All right, we'll work through a couple now. Easier said than done. Usually the things get a bit simpler. Then we'll start working on some things. So let's imagine a problem. We have a circular track, one quarter of that circular track, and a crate is sliding down that track. Something like UPS or FedEx. We have to worry about it if you ever end up designing some of their equipment. So at some particular point, at some angle theta, it's got a velocity v down the path. And we want to figure out what's the angular momentum and want to find at that time what its acceleration is. Which would be the tangential acceleration. Because the normal acceleration is just v squared over r. And if we're assuming that we know some velocity at a particular angle, there's nothing there to calculate that. So we're not going to bother with that one. H zero is r cross mv. Now we can do this in two ways. We can do the cross product itself, or we can use a little bit simpler way to do it, especially for two dimensional problems as this one is. If we remember what the magnitude of the cross product is and the direction we can get from simple observation. So here's our two dimensional problem. We've got some object at some point, r, moving with some velocity that gives it a momentum mv. The magnitude of the cross product for a two dimensional problem like this is, but we don't actually have to do the cross product all the time. The magnitude, and you can write it whichever way you want. I just simply leave off the vector sign to get magnitude. Is the magnitude of each of the components of it, r and v times sine phi, where phi is the angle between the two vectors. It's that angle there. And then for that particular picture, it would be obvious that the direction is in this particular instance in the plus k direction. If that's x and y there, then positive c is out. So we can save ourselves a little bit of trouble and do that. The angle phi is the angle between these two. That's the velocity. So it's also the direction of the linear momentum mv and the angle between them is simply 90 degrees. So this is nothing more than r mv. So the angular momentum, then in this case, we've got the magnitude. All we need to do is apply some direction to it if we use the same coordinate system we had before. This time, you can see it's going into the board. So it would be a minus r mvk, or in other words, what we typically call clockwise. So that can just come from inspection. For visualizing other cross products, remember r cross mv, we need to visualize the two vectors as being tail to tail before we do the cross product. So I can take these vectors, put them like this, then you do r cross mv. And for that picture, it brings the momentum vector out of the board. r cross mv. It's got to be into the board for that example. So perhaps a simpler way to do some of those things. Okay. So that was easy. We're happy with that, I think. I hope you are too. We've got to find then the tangential velocity, which we also know as v dot. Mind from some of the moments, because v dot is right in there. Just a piece of that time rate of change of the linear momentum. Alright. Let's see. The time rate of change of the linear momentum we've got. It's just that piece over there. So that's not going to be any great difficulty, because we can just take the time rate of change of that part of the vector. So there's our v dot right there. What may not be as apparent is what is the sum of the moments. Don't forget that this box has some weight, mg. It also is experiencing a normal force from the track itself. The sum of the moments is the sum, those two moments, those two forces cause with respect to our main point o. The normal force causes no moment, because it goes right through o. So just to make things a little clearer, I'm going to erase that one, because we've got then the two components of the weight. This one was equal and opposite to the normal force anyway. Goes also through the origin. We're not concerned at that point. We're concerned only with this component here that is perpendicular to r. It is of magnitude mg sine flea, because there's flea there. So I can then say, let's see, the sum of the moments is mg sine, oh, not flea, I guess it's theta, sorry. And that problem I have to draw theta. That's why I should take notes in chalk. And that is also in the minus k direction. And so we can solve for v dot equals mg sine theta, whatever angle it happens to be at where it has the velocity v divided by rm, the m's cancel, and we're left with g sine theta over r. No, hang on, hang on, this is just the force, but what I don't have in there is the moment arm. So I need an r in here as well. If you notice the units wouldn't have worked with that. So I need a minus r in here. Now the r's cancel, the m's cancel, we get just g sine theta. So now the units work, because they both have units of acceleration. I forgot the moment arm working on this perpendicular component of the weight. What? Okay. Okay. So the two dimensional problems get easy, especially when we notice when we've got components that go right through O and components that are perpendicular to O. Had you? Well, at remember that in curvilinear components, the tangential velocity, the object only has tangential velocity. Therefore, v dot entirely gives us the tangential acceleration. Remember that the normal acceleration is a centripetal acceleration. That was supplied by the centripetal force n. Okay. I know angular momentum is not always too terribly intuitive at times, but we're not going to be doing three dimensional problems. We're not going to be doing terribly complex ones. So it'll grow on you. Let's try another problem. So here's a circular track with a car going around the track. It's a 100 meter radius corner. At this instant, we'll call it 1, it's got a speed of 5 meters per second. And at that moment, driver hits the gas, gives it some force that accelerates the car along the track. And the force is a function of time 150 t squared, where t is in seconds. And so we want to find v2 five seconds later after this acceleration starts, where it hits the gas so it gets a forward force from the tires. Find v2 five seconds. All right. We can use angular momentum as a two dimensional problem, two dimensions as seen there, which is nice. It allows us to ignore a couple of the other forces in the problem. One is that there's a normal force on the car that is up from the bottom of the track. It's tough to draw. I guess I could try if the car is here. Then it's got this force there driving it forward, but it's also got a normal force up and its own weight down. It's also got a frictional force directed radially inward. That's actually what's supplying the centripetal force. But we don't need to worry about any of those three forces because they have nothing to do with the direction of motion. The n and the w are not, they're a two dimensional problem with r, but not in the direction of motion. The friction force goes through the origin so it contributes no moment. So we only need to consider the force f that's doing the acceleration in terms of what change in the motion it'll cause. All right. Time to end the problem. We can use our impulse momentum of equation and we'll use curvilinear components. Look at this only in the tangential correction. It's a little bit easier. We don't have to worry about x and y changing. And so the moment then becomes an integral from 0 to 5 seconds by 0 seconds is when the acceleration started. The moment is the moment caused by the force doing the acceleration which is r times f. We only need their magnitudes because they are by definition perpendicular r cross f and the r cross mv is also perpendicular so we can use r and then it'll be not the right thing there. This is just delta h, not h dot. Sorry. That makes more sense because now this is the delta v we're looking for. All right. So take a couple seconds, finish that up. We know what r is. It's constant. f is given as 150 t squared. Then m, sorry, the car has a mass of 1.5 megagrams. The standard unit of weight for x is 100. F is given so you can do the integral. Then you've got rm and the first v you can find the second v. Watch the units as all. If I made it back equally. Alan, doing okay with that? It should be. Is it? The hardest place with the units is the megagrams just because it's not something we're taking to do. Two-dimensional problems with lots of 90 degrees are inherently easier. Yeah, that's a force. This is going to have newtons per meter squared. No, second square, not meters. Was that your question? Yeah. Sorry about that. Who would have done traps with a couple of units? And what you do, you're coming around as far as constant, but 150 is constant. Both of those come out. Integrate t squared in five seconds. And then the other side's just the rmv. So actually, again, r cancels. As a matter of what size track we're going on with this acceleration. 1.5 megagrams is 1500 kilograms. And then we're looking for v2. And v1 is 5. All that's left is the integration. Now a little bit of algebra. Chris, you got something? Thank goodness. I'm going to go forward with a drawing. Check something. t squared integrates to t cubed over 3. And then you can solve for v2. You can get something like 9.17. You got it, Travis? No? Phil, did you get that? Alan? Oh, yeah. Plus that's delta v, is that? Yeah, you got it. Tom, all right? Do the integral right. t squared integrates to t cubed over 3. I'm not going to show you that. Two limits. Anthony, okay? Got it, you think? Close to it? What, Tom? All right? Do you think I'm out? All right. Okay, another problem. Kind of a miniature satellite problem. Imagine we have a pulley that runs on a shaft on which we apply a force. Back on the shaft somewhere are four that are in the same plane as is the pulley. So if we happen to look at it down the shaft, you'll see the pulley with a force on it that's turning this. And that could be, you know, a simple propulsion thing of some kind. And then behind it, you see each of these, each of those parts there. That seems pretty clear, doesn't it? Travis, okay? Come with me with that. All right, put a couple numbers to it. These arms are very slender, 400 millimeters long. The radius on that front disc is 100 millimeters. Right? The radius? Yeah, the radius is 100 millimeters. And each of these little pieces here are 3 kilograms. All right, so that's the basic setup. Or picture? Picture? Oh, one other little piece. The force is 20 millimeters. All right, using angular momentum, figure out that if its initial angular velocity is zero and its final angular velocity needs to be 150 rpm, find how long that force must be applied. If this was a little satellite, you're trying to spin up to some speed, how long would you have to spin it before you could release it in the moment? It means some spinning rate would be too low. Stabilize it before it's released. All right, do this with angular momentum. That's a radius of 100. What? The radius of this is a disc where the force is applied. All right, again, watch your units on this one. Fairly straightforward problem. Each of these things will have some final velocity that we're trying to find. Well, not that we're trying to find that we have. Sort of. So that's the angular momentum side and then the impulse side has to do with the force being applied for a certain amount of time. Just be careful with your units. Use the impulse momentum. Angular impulse momentum. Move on this one. It's time dependent because I'm asking for time, but the forces aren't time dependent. All of these with respect to some central point, which I think would be the shaft. Pretty obvious down that axis. The dimensional problem is drawn in the second drawing. Well, there's subtle pieces to it, though. You need to watch out for. Watch your units. Got an awesome drawing to start with. That always helps, doesn't it? Careful, there's two radii in here. Not yet. Let's do these with pieces of time. For example, impulse side. Zero to some time later. You want to figure out how long that is. The moment is, of course, the force times the moment arm. Those are both constant, but you just have to watch your units. Since they're constant, they come out and integrates to delta t and there's the delta t we'll be looking for. So that's the impulse side unless I made a goof. I don't think so. Never, ever do. And then we're looking for, we've got both of the momentum sides. All we have to do is figure them out. Since the velocity is always perpendicular to the radius of that object's shaft, then each of these becomes a lot simpler. For each is r across mv, but since they're perpendicular, they just become rmv. And since r and m are constant, they're minus v1. However, what's the mass on this part? It's not one of these. It's all four of them. Four out in front. And what are the velocities? These are not the velocities. These are angular velocities, angular velocities. What we need are linear velocities in there, but maybe you remember from physics one of these was r omega. When r and v are perpendicular, as they are here. So starting from rest, getting up to a rotational speed of 150 rpm. Travis, okay? Travis, this question was where did the four come from? Four of these three kilogram masses. Each one of them contributes to the angular momentum. So there's only one other little trap in here. One trap left. r's no trouble. That's the 400 millimeters. This r was 100 for the momentum of the force. m is three kilograms, but there's four of them. That's the four. But to find the velocity, you need omega. It's very easy when it's zero, but you can't put 150 rpm in here and get the right number. You've got to make this into radians per second to get the velocity to work. And that's 15.7 radians per second. So now you can figure out the two velocities, then all you've got left is to equate these and solve for delta t. How to get this? 150 revolutions per minute times one minute is 60 seconds. So the momentum's canceled. And then 2 pi radians for every revolution. So the revolutions cancel then. And we left it with radians per second. That should be 15.7. Was it? Yeah. Oh, great. Yeah. Sorry, radians per second. So that gives a total change in momentum. Does anybody have that number? I don't even have it. Yeah, well, it starts with all we need is h2, because h1 is zero. v1 is zero. And I've got 30.1 kilogram meters squared per second, which is a 50 meter second. Then all you have to do is solve for delta t. And you get 15.1 seconds. Sounds right. Now we're starting another problem. So any questions, fill? Do that one perfectly. Got the radians per second.