 All right, so we're doing exactly what we did with particle dynamics where we went from what? We went from kinematics then to kinetics when we looked at particles. And if you remember, when we got to the kinetics, we had three flavors of methods to solve kinetics problems. Remember what they were? Come on, take that. Yeah, yeah. The other two come right from that, so it's not like they're exclusive, but as I think we've seen, they did kind of lay themselves out for different types of problems. Bob just said the work energy method and then impulse momentum. That's even what we did in physics one, so it was kind of a review, but we did a lot more in-depth, especially since we took non-constant acceleration problems. And then we started with rigid body motion, rigid body dynamics. We just finished the kinematics of that. And now we're going to go to the kinetics, just like we did with the particles. We're going to see what it takes to actually get the accelerations and velocities that we had in the rigid body motion. But don't forget that we're also not doing just rotation. We also still have the possibility of exclusively translation or the combination of translation and rotation. So our kinematics involve all of those things. But the kinetics, a little bit more involved, we still will have this equation, but we're now also going to have to do something about the fact that forces can cause things to rotate. When we looked at our particle motion, there was no difference to us between a problem like that. When we looked at particle problems, there was no difference in those two situations. Yet you know that most certainly there is, that we can replace this second possibility with something that looks a little bit more like the first. But the fact that we've displaced this force by some distance, now we have to also put in a moment. Alex, since you jumped over static, you didn't hear the term moment really. We take it as torque and in physics one, so very same thing. But as you remember, these two systems for our need are equivalent systems. Particle motion does not take into account the fact that these first two are even different. So now we do. So we're going to have to come up with something where we also look at some of the moment's business. We'll set that up in a second. And then we are going to have, again, the work energy and the impulse momentum methods that will work for rigid bodies as well. All right, so that's where we're going here in the next couple of days. Well, we only have three weeks left. And one of the days is a test. And I think one of the days is review, just so the test wouldn't be on the last day. Give you kind of a break from the professors doing that. So let's see if we can improve our kinetics so that they reflect the kind of thing we need. All right, first and foremost, whatever the object is, there is some center of mass to it, some center of gravity. If it's a uniformly dense solid and a uniformly thick solid, then that center of mass is also at the area centroid, which is a simple fact we've used several times before. Now we've got this possibility that there are forces or a-force acting. Maybe that's the, well, we'll go ahead and call it the sum of the forces. And then that can cause some kind of acceleration. And I'll put a little G on there to remind us that, well, in particle motion we didn't need that, because that's all that would happen. There'd be some unbalanced force on some mass and it would accelerate. Now there's a little bit more going on as you'll see in a second. So we're going to have this subscript of G. But we know that the forces that are unbalanced will cause that center of gravity to accelerate somewhere. We have to add to it the possibility that there's going to be some kind of rotation or at least some kind of moment due to that force. So we have to take that into account too. So we have, if we sum the moments about G. If you remember, we set this up a little bit ago and that was the time rate of change of the angular momentum. These two equations were actually very similar because this one was time rate of change of the linear momentum. And this one was the time rate of change of the angular momentum. That looked a little bit familiar from a couple of weeks ago, I think. I hope trouble is, I don't know about you, my trouble is that I find the first one quite easy to work with. One, we're just more familiar with. And two, when we have forces on something and cause it to accelerate, it's kind of right there. You can see what it's going to do. You can see what it takes to do something. You look at this equation, I can understand what moments are. I can visualize them. I can't visualize that. That's not something that my mind can work with simply. So we can do a little bit more with that and get this to be somewhat more useful to us. Go something like this. A little bit of it has to do with what we did before. Remember this angular momentum, not the time rate of change of the angular momentum. Just the angular momentum itself was the fact that we were some distance from a reference point, which we can take for now to be this point G, because, I don't know how to erase this, but that was sort of our good reference point with our original situation where we had a force that was not acting through G, and that's the type of thing we're trying to investigate now. And then that was crossed with the linear momentum, that of the center of mass, which we kind of need to pay attention to now because the center of mass may have some velocity that other points on the body do not have, because it may be rotating. It can be doing all kinds of things. If it was just simple translation, every point on the body would have the same velocity, and I wouldn't need this. Now that we've got rigid body motion, we need to pay attention to the fact that the center of mass is going to be going some velocity, and, well, for the time being, God knows what the other points could be doing. So we're going to have to pay attention to them, get ready for them in a little bit. All right, so we have that little bit. That's, I don't know, a little bit more workable than this is, but let's see where we can go with this to make it even more useful to us in a second. If we're talking about multiple parts to the body, we can, or multiple components of some kind, we can sum all this business up over all of those pieces, and that will then represent the whole body, all the body pieces. So we can sum over all these little pieces doing this. Every little piece might have its own velocity. It doesn't seem like that's an easier step for us, but it will be in a little bit, because when we sum up what all the little pieces are doing, of course that's got to be what the whole body is going to do, so we should be okay with it. We're looking at 2D motion, so we can simplify that cross product. Those vectors will be mutually exclusive in some part, and we'll take our usual direction as positive, so this can become then the sum of r, m, v, k. Now we'll put it into our 2D only. It's not that it doesn't work in 3D, but things are going to get easier for us if we do this. So now we've got the magnitude and the direction of everything. Oh, skipped a little piece. That's in fact no shock. So we've got this r. Now that v we just had as omega cross r. That's the part we needed. Now we can go to the 2D step. Sorry about that. Now we go to the 2D step. Now we get the sum of r i squared omega. Remember for a rigid body, omega is constant for the entire piece, so that can come out of the summation. So redirect you with the magic arrow of algebraic redirection. And we get omega comes out of the summation. We get all of the rest left over, which is just a characteristic of the object itself. None of the dynamics is now in there. That's just whatever the object, that all has to do with object geometry. In fact, if we wanted to make dm small enough, these mass pieces small enough, this would then become just the integral of r squared dm. Either way, it's the same thing. And either way, I hope that this integral r squared dm or the summation of mr squared for all the pieces looks maybe a little bit familiar. No, not quite. You'll recognize it. It's the mass moment of inertia, which you remember is a purely geometric property. So this whole thing becomes i omega. If you think back to physics one, where we had a rotational analog for every linear component variable we had, we also had a rotational equivalent for mass, and that was the mass moment of inertia. Now, the reason this becomes so very useful to us is all the way back here, because now this time rate of change of the angular momentum, which we've now determined is, we'll have to put a g on this i, because as we know, moment of inertia is different about different axes. We now have that, that's the time rate of change of that vector. Our 2d analysis. Remember the moment of inertia is a characteristic of the solid itself no matter what it's doing. Omega is what it's doing, so i g comes out of the integration. We get d omega dt, which is alpha. And now we have our two equations governing rigid body motion. And I find i g alpha a lot easier to work with than h dot was. I don't know about you. I do. For our two-dimensional problems, oh, I guess I needed the vector sign in here, sorry. For our two-dimensional problems, this is three equations, because the sum of the forces is in x, y, and the sum of the moments is in k, so we have three equations. We can handle problems with three unknowns. Four two-dimensional problems. If we had three-dimensional problems, this would be a full six equations with three dimensions. Don't forget, though, that we also have any kinematic equations we used before. So we can solve problems of greater than three unknowns because we also can pull in, in fact, will pull in the kinematics equations. I don't understand this term here. This f sub i goes to zero, and that's how we go to r. No, as for this, if we're talking about a solid of discrete masses, so there's m1, there's m2, there's r2, there's r1, you know, like a satellite type thing might be, then this equation applies nicely. If we're talking about a continuous solid, then we break it into little elemental pieces. So this is dm, and it's located at some place, r, I don't know, what do I put? dm, and then we have to do that for the entire solid all over the place and for it to be accurate, dm has to go to very, very small pieces so that we do the continuous integral over the entire object. Kinetics equations. So obviously for problems now, that moment of inertia is going to become important. Let me just do a little bit of review for us on that. It's the integral over the entire solid, r squared dm, for a continuous solid rather than for a solid made up of discrete parts. We could even handle, if we need to, the possibility that the density changes with position. So we could do the same integral that way. However, as you can imagine, these integrals exist for regular solids so that we don't need to do the integration. That's not the point of this class. That's in appendix B4 in your book. Way, way at the back, have lots of regular solids, cylinders, hollow cylinders, parallel pipe beds, all kinds of stuff. The usual one, spheres, hollow spheres, conical sections, all kinds of things. And in there are the moment of inertia through the centroid but also through other parts of the object. Remember that moment of inertia through an axis through the center of gravity will be different than the moment of inertia about some other axis parallel to it and separated by a distance d. If you remember, that's the parallel axis theorem. And so if we have the moment of inertia about an axis through the centroid, through the center of mass, we can find the moment of inertia about an axis parallel to that. That's the parallel axis theorem. From this business, we also define then a term may or may not sound familiar. I don't think we used it in Physics I, radius of gyration. Have you heard that term before? Radius of gyration is some distance, we call k, that if all of the mass was collected at that distance in an infinitesimally thin ring but all the mass was there, that would have the same moment of inertia as whatever we're talking about in the problem. And they're related by that equation. There's not any great use to this other than the fact that a lot of times in some of these problems or in other problems in industry, I guess you're more likely to hear the radius of gyration rather than the moment of inertia and it's the moment of inertia that we need for this equation. It's also true that we can get the moment of inertia about another axis and the radius of gyration about another axis but the equation still is the same. So if we have some object and we took all of its mass and collected it at some radius such that those two objects had the exact same moment of inertia, that radius would be the radius of gyration. An abstract concept but one that is of common use in these problems. All right. Obviously, moment of inertia is going to be important in these problems. There are five ways you'll come up with the moment of inertia in a problem. It might just be given in a problem. This is common for more complex solids like satellites and bicycles and other things that we might need to look at. It's just given. It's also possible for you to look it up, checking that appendix D4. If the solid we're using is in that table, just use the moment of inertia from the table. Not often are our problems just simply objects out of the table but it might be composites of objects given in the table. So you can sum up all the moments of inertia of them individually, applying the parallel axis theorem that's necessary. It could also be that you're given the radius gyration instead and then just simply calculate the moment of inertia from that. The fourth possibility and a student favorite integrate it and figure out what it is. Figure it out yourself. That could be though if you add up the moments of inertia of regular solids then you're okay. The fifth possibility, one that sometimes sneaks up on students is that you just simply don't need it in a problem. If you remember back to physics one or even some of the problems we did in this class there were occasionally times when you thought the mass was important and it turned out it wasn't more important than the problem from the equations and that wasn't even necessary. Freefall problems were always that way. Remember the mass didn't apply but if you did freefall problems as a work energy problem the mass is in there and it doesn't cancel out to the end and then you remember all you have freefall problems we didn't need it. Then just a reminder the last little piece for composite bodies. Bodies made up of some of those regular solids that are in that table from method two. You can sum up all of the pieces applying the parallel axis theorem if necessary. All right, sounds like a horrible mess but it's not so we'll do it. Maybe I'll have to mic these for this class. Imagine a 747 looks like that's what they look like right there. Man, that's pretty darned. That's like a photograph. Where's Swing's? They're there. Can't draw them because they're swept back. See, it's got this big landing gear type thing if you remember plus some little nose thing there. That's what they look like. Exactly what they look like. I want to find out the reactions the forces on the wheels. Now as a statics problem not a big deal center of gravity right about there we'll put it let's say that it's 5 meters in front of the back landing gear and 22 meters behind the front landing gear 5 meters off the deck. See, when we get part of a motion none of that matters we didn't care where the forces were we just did what they were so we can figure out then from a statics problem what these reactions are and we'll call them A and B we can do that pretty quick we can do it oh I think we need to give it a mass too we need mega-grams favorite units. So we can figure out very quickly I hope what those reactions were for example we can sum up moments about one of the points one of the forces we pick one of them and we don't have to worry about it Alex we did this in physics one some of these simple type problems 5 meters must equal B times 27 meters is that right just practice quick practice quick warm up because that's just a statics problem just sitting there on the sitting there on the tarmac waiting to taxi out that's not good enough for us that's just a statics problem but I'd like to compare it to what happens to these type of problems during takeoff back that's why they have those multiple wheels there the front you've often seen guys just walk out they lift up the front and move the plane around you've seen that haven't seen them do that you have A and B go ahead and give them to me gee how about a few more figures in there 454 your A was somebody's got it Colin got it that's just a statics problem that's no good to us here because during takeoff now do we all draw on the wings the wings are maybe something like here with the engines below them that generates thrust that is a little bit below the center of gravity by about 2 meters how does that change then the reactions at the wheels that now becomes a statics problem because that's going to generate some moments that weren't there before cause some acceleration of the plane itself that's the point of turning on the engines so now let's figure out what the reactions of the wheels are to the dot notation just because we do have a subscript here if we are doing things in the x direction then we have another layer of subscript which I'm not really fond of so if we use the dot notation we only have the one subscript g forces in the x direction obviously the thrust there's a little bit of rolling friction at the wheels but not much certainly not much compared to the thrust so that's our x direction equation oh by the way we do have the thrust it's 700 kms so we can figure out the acceleration of the center of gravity of the plane itself down the tarmac and that problem is no different than what we would have done in particle motion no different than what we would have done in physics one so we'll take it a step further we still haven't found the reactions all that and then is give us the acceleration so we're going to have to sum the forces in the y direction as well we'll have already done that that's a plus b equals w because there is no acceleration in the y direction sum the moments remember about some point that's convenient it might be convenient to do some other points however don't because the only equation we have so far is the moment summed about g oops I didn't put the g in there we only have an equation for summing the moments about g we're going to see in a minute things change when we sum the moments about other points all right going to sum the moments about g what's the moment of inertia of a fully loaded 747 with respect to an axis through its center of gravity well go down that list of five ways to find it it's not given it's not in that table don't have the radius of gyration probably don't want to integrate it especially since we don't have any other dimensions than just these and we certainly don't have the weight distribution so the possibility is that we don't need it why don't we need it though it's right there plane is taking off still going down the runway well we don't even know if it's takeoff it's just a thrust of 700 km maxing but it is just running down the runway what's alpha it's not rotating it's only translating so we can sum the moments about g and we get essentially the same equation except we have the thrust in here where we didn't have any thrust there so we'll take all the since they all sum to zero counterclockwise thrust I mean moments so that's t times 2 meters a makes a clockwise moment about g b also does a counterclockwise moment about g so we'll add it over here of course that 22 meters b those are the two counterclockwise moments and that must equal sum to zero on this problem clockwise moments which is p we have two equations p plus b equals w and two unknowns static's problem when there wasn't any thrust this equation here the moment equation is a different equation than we had here the statics sum of the moments I didn't know this is a taxing situation so this thrust is very much smaller than takeoff thrust you don't get anything close to what these two are at together a plus b equals w is the same equation what was your what was the mass the mass was 250 mega grams so its weight was 9.8 what have we got get what you agree or don't agree don't agree with each other if you agree with me you're at path path's had a stellar day that's what you get something wow all kinds of different things common create the tie under one to one so if you agree with somebody that's not an accurate answer bobby what do you have for a no for b 439 all the numbers are right sometimes a transmute table everything looks okay 22 meters w forward what do you have pat for there I got 203,000 do we what do you have for a 2061 2061 Alex Colin 2048 Jake nothing that's a spirit in the plane Colin you were within round off 9.8 that's what I had contributing yet 402 they're mine 600 well look at the problem does it make sense that B would go down and A would go up the thrust is pushing under the center grab the center grab is the point about which it would tend to rotate if it could it can't of course because it's in contact with the pavement so the thrust pushing under the center gravity would try to turn the plane that way which would increase A and decrease B which is what several of us got but not everybody but I increased to exactly this 1800 1800 is less than 2000 well the only place to go wrong is algebra if what 9.1 where did that come from oh I thought you had 2300 well 2061 oh 2061 somebody else had 2300 alright anyway something close to this alright we got to take care of other possibilities though turns out that this moment equation with the moment summed about the center of gravity isn't always the easiest thing to use we have some situations where we might have some some lovely object of some kind with some center of gravity somewhere and a couple forces acting on it maybe there's a force there and a force there and a force there could be that we don't know what all those different little pieces are and the like or we don't know exactly how they act or something but it might be easier for us instead of summing the moments about G three components to that often times we've seen as we have before that there's some other point it might be more efficient or for us to sum the moments about it might be a little easier for us to sum the moments about point P most of our problems will be this oddly complex we know that a lot of the very regular programs or problems point P could be quite easy to find in a lot of type of problems we look at but the some of the moments about point P could indeed cause it to rotate however it'll be a center moment of inertia with respect to that point P itself same type of equation however it's about a different P a different point either equation is perfectly fine always make sure that these subscripts match the trouble is we might not know what IP is the moment of inertia with respect to an axis through P remember these are two dimensional problems so the axis is in the z direction the k direction often what the moment of inertia through the center of gravity is however when we apply the parallel axis theorem knowing that this will these forces will cause some kind of acceleration we can adjust things by using this form of some of the moments equations this just comes from the parallel axis theorem you can derive this in about three seconds using the deal that a equals r alpha actually in this case it'd be d instead of r our book happens to use d where d is the perpendicular distance from the point to the direction of acceleration whatever that might be so is d the same as r yes this is the form you're more familiar with d is what our book happens to use for the parallel axis theorem so we need another bracket there you go gives us an alternative way to sum the moments we have several possibilities now when we're summing the moments depending on what's easiest for a problem first one still applies it's not that it doesn't apply for other problems it just might not be the easiest one to use second one also applies but again may not be the easiest one to use the third one may apply but may not be the easiest one to use use whichever one of them is the easiest problem and see how this applies contemporary problem that applies right up to the very moment of our lives one of you has a little projector clicker you stole from Bobby you got a smart phone that has a projector app on it there we go see it's spring let's do a lawnmower problem alright what? draw up that's a photograph man it's a picture of my lawnmower and my grass let's see some of the details we can see center of gravity is 215 millimeters off the deck there's a wheel base with respect to center of gravity and well we're going to do this problem where it's a self-propelled lawnmower so take p as zero so we want to find that we want to find the forward acceleration for the self-propelled mower rear wheel drive without the operator actually having to do any pushing two little things you'll need that aren't up there let me make sure I got the right one is the coefficient of friction US and UK we want to determine what the acceleration is of course this acceleration is going to be parallel to the ground and forward kinda like the plane this was well the place the acceleration lives is in the x direction equation remember there's a p in the drawing but p is zero for this problem let's see what will be the forward acceleration if the wheels are engaged but the the operators just let go it's a phone ring or something maybe he knows that his daughter was playing out in front of the mower instead of turning off the mower and he ran around to the front together what yeah it's a lot more exciting than that alright are there any forward forces here there must be or there wouldn't be any forward acceleration we're trying to determine what a sub g is or x dot dot g same thing what forward forces are there remembering that p even though it's in the picture is zero well we can't sum the force if we don't have a decent free body diagram so let's finish that then any forces we want to jump in with easy ones so they get the rest of the act in and off good Alex the weight acting at the center of gravity we didn't necessarily care where we acted before other than it was acted down now we have to pay attention to the fact that there's a bunch of different positions in all of these problems and the center of gravity is most clearly one of them specifically listed as such what other forces normal force at each wheel b and c what else we still have no forward force friction as the back wheels tend to turn they produce friction in contact with the ground front wheels will just take to be simple rolling wheels with minimal resistance what else we don't have all the forces no sense doing the problem we're doing the wrong problem I want forces there's a free body diagram I did put the kinematics result of the forces in a different color what are the forces that's it that's all there are so we know that the friction times the normal force well, times that normal force b is going to cause it to accelerate we're looking for that acceleration oh, 50 kilograms by the way it doesn't matter if the friction is going quite in the wrong way it looks like it's going to speed speed it up it is we turn on the gas the wheels start to turn and it accelerates forward so the friction has got to be forward remember oh, okay the wheels are pushing back on the grass the wheels so that's our x-direction equation which coefficient of friction for greatest acceleration well, the greater this is the greater that's going to be so the static friction as you should know anyway you'll get better acceleration if the wheels aren't slipping you want them to remain in contact so the maximum acceleration will come at the maximum friction well, we don't know b the only way to find that is with the sum of the moments in the y-direction of course we don't want to accelerate now, remember in these type of vehicle type problems there's always the possibility maybe desirers of a wheelie in that case the center of gravity would have a y-direction acceleration even though the object was still in contact with the ground those are things we never had to consider before so d plus c equals w so we're a little bit stuck in that we don't have we've got two equations three unknowns so we still got to keep going sum the moments about what point do it about g if we knew the moment of inertia about g well, we wouldn't need the moment of inertia about g would we because there's no angular acceleration however that will load this equation with all of these three unknowns and things are still algebraically complicated if we sum the moments about say b two of the forces drop out of the problem but we will have this piece to it of that business that comes from the parallel axis so notice that the sum of the moments about g might be zero but not about b however this is pretty easy stuff in that it's all directly part of the problem the counterclockwise direction to be the positive direction that fits with our x and the y so let's see the c times the wheel base this is between b and c is a positive moment what's the wheel base for millimeters you agree that's a positive moment summing the moments about b which eliminates two of the forces from the problem so that's a negative we know what the mass is so we don't consider the weight an unknown its moment arm is 200 millimeters is that it for the forces now here's one place we do have to be very careful this acceleration is in a clockwise direction with respect to b so we have to put it in as a negative sign that is a tricky part about using this method 50 kilograms x double dot we're looking for what is d d is the minimum distance between b the point of summing the moments and the acceleration vector acceleration vector is a minimum distance of the 215 so watch out for that minus sign that's an easy one to easy one to skip alright so we've got all the pieces then it's a little bit easier probably have a debate whether that's easier but it's a pretty easy demonstration out there let's see what it's doing because I don't see your homework in here I guess I'm going to have to take a clear note of that you know when you get out of 747 because we have to have a race between the two that'd be kind of exciting equation there's only millimeters but they cancel because there's millimeters in every term so it doesn't matter what the units are in this term in this equation as long as they're all the same units because millimeters appears in every equation these ones didn't appear that doesn't in w but now that they've canceled out of the other terms it doesn't matter if you put in w that would just give you the units there so if you use meters per second squared here you get meters per second squared there 11.03 meters per second squared for 18 no I got it you may have to find x double dot you may have to find a and b sorry b and c on the way and then notice if p did exist that force exerted by the user all of this changes the acceleration no acceleration lottery colon you're just cruising here today you'll get out of mowing along I have 4.14 meters per second squared check those g here's 9.8 yeah I used 9.81 so that might be that b is 414 and c I got was 70 sessions