 All right, if you remember now, looking at rigid body motions, specifically we're working on the kinetics of rigid body motion and as such, we had to break it into a couple parts. So we actually can break this into, well, we're going to break it into three parts. The first part we're on, but it's also broken into three parts, so the first part that we're working on now is the F equals MA, part of the kinetics, not just acceleration, but it's how we cause that acceleration, but we also now are including in there this general business of we're not just strictly accelerating things, but we're also rotating things. This we're also breaking into three parts. We did the first part Monday and that was pure translational motion, however, for the first time really, well, I guess for the first time really in the work we've done over the last couple of years the actual size of the object really mattered in the whole part. We did some things with some friction business where we wanted to see if an object would tip over if we pushed it and then we cared about how big it was, but for the most part those were acceleration problems, those weren't dynamics problems, so that required us to pay attention to the size of the body, and of course that had to do with the moment of inertia of the object. Remember, however, in the pure translational motion there was a bit of a trap and it came with this particular equation here or its alternate form where instead of summing the moments about the center of gravity sometimes it made more sense to sum the moments about some other point, just as it had in statics all those times. Sometimes problems get a little bit easier if you pick a different moment about which to make your sums. Sometimes the forces disappear from the equation, so the solution is just easier. If you do that, don't forget that these subscripts, you can use any form of this equation as long as these subscripts match each other, and that's where students sometimes get into some trouble, because for translational motion naturally there's no angular acceleration. If there was, then we have a rotational component, and we don't when we're talking about pure translation. In that case, then this alpha equals zero, but this one does not, because remember that we had to add on to it. This is actually an angular momentum term, and now this alpha here equals zero, this term left over does not equal zero, which shows that this alpha is not equal zero. As much as you may be tempted to make it zero, it's not. It's a very easy trap to fall into, which is my subtle way of saying I've fallen into it, so I've found over the years it's just best to really make a belabor point of being careful with this. Today we're going to look at pure rotation as part of the problems, and in this case now there will be some angular acceleration, and this doesn't become as much of a problem, because alpha isn't zero in any of the parts of it, so you're not going to get trapped with the difference between these two things. If we need to sum them all at a different place, we'll just do it. It is still true that this part must be in here if we're not summing about the center of gravity, but we'll just add it in, and we'll do a couple problems with that, and hopefully it'll make some good sense. And then Monday we're going to put these two back together to get the general motion of rigid bodies, general motion being both translation and rotation at the same time. Again, the simplest example of just that type of motion is a car tire. As the car moves along, the tire rotates and translates, then we've got to put those two motions together to get a good feel of what we're doing. Alright, before we go any further I want to make sure that you know how to get this moment of inertia when needed. So there's five ways in this class to get the moment of inertia. Now remember if you need the moment of inertia about another point, use the parallel axis theorem. But five ways to get the moment of inertia for these problems. Not in any particular order they are one. It might just simply be given. You'll notice in some of the problems when you need moment of inertia, it just simply says so right on there what the moment of inertia is. Pay attention to the moment of inertia about which point or axis. Two, you might need to look it up. If we're working with a regular shaped body, then it's just in the tables and you've seen those before. We have 2D tables. Those are tables I think it's 3D and 3D tables 3D objects is table 40. So if it's just an ordinary regular shape, go there for the moment of inertia in particular simple shapes. If it's neither one of those, you might need to calculate it. And that can take two forms. One you might need to integrate it directly right from the definition of the moment of inertia. Integral of bar squared dm. That will work for semi regular solids because you're going to need the functional relationship between the shapes to do the integration. But as we established in the earlier class this morning we still love integration as much as anything we ever do. So that's a possibility. It's also a possibility to remember to take various breakthroughs from the tables and combine them into a compound shape to get the moment of inertia you might need. Again this is no particular order. It's just maybe the thought, the steps you go through until you find it. You can also get it from the definition of the radius of gyration which is the, we use the little letter K. It also is about a specific axis. So again these two subscripts must match each other. You can have the radius of gyration about a different axis just as you can have the moment of inertia about a different axis. This is an artificial construct. It's as if we've taken whatever object we're talking about and we know whatever its moment of inertia is. We move all of the mass out to some distance from the center of rotation K, the radius of gyration and it's as if we put all of the mass right there. So if we've got some wheel of some kind and it's got a rim on it that might be thicker than the inner part or even could be a spoke wheel of some kind. Whatever its moment of inertia is, we can get an idea and a relationship between that and the radius of gyration. If we imagine an artificial object with all of the mass located at this distance, the radius of gyration, those two things, if they have the same moment of inertia then the radius of gyration is the place where all the mass would be concentrated in an artificially infinitesimally thin ring at that distance. It's just another way of representing what the radius of gyration is, what the moment of inertia is and then the fifth way. Anybody know what that one might be? No, that's not on there. Asking me it's not on there. Actually it might be worth it on tests. It's really a type of test to forget to put something in. I never do that but I've heard of it in the past. There are problems where you just don't need it. You might think you do but as you work farther through the problem you'll find out that it later cancels. If you remember we had some problems over the years where we needed the mass but the mass ends up cancelling and we don't need it in the end anyway. So there might be some problems where you think you need it but you don't and you might not need that and you might not know that until you get a little farther. So if you're doing a problem and you don't have the moment of inertia and you think you need it, take the problem a little bit farther and it may just simply cancel out and you're fine without it. Then you don't have to bother me. Which at this time of year you do it your own peril. That is on the tape. Alright, so we're ready to look at a couple of rotation problems. Alright, the general setup of what we're doing can illustrate with a simple picture. We've got something that can rotate in this case what's called a physical pendulum. It'll have some place where its center of gravity exists and it may have some angular acceleration and it may have some angular velocity. I'll just arbitrarily choose them in that direction mostly because when we do so that establishes our tangential and normal directions if we so choose to use those. The tangential direction is always in the direction of the motion at the moment as shown by the omega in that direction and then the normal direction of motion, normal coordinate direction is always towards the center of rotation. So that just simply sets up our general method. Some of the problems, those coordinate directions happen to fall in the typical direction of the horizontal vertical direction we're used to so often and so you can just go ahead and use the IJK there if you want. But we've got a couple things that we can use for our typical two dimensional problems as this one is. We know that if the forces are unbalanced on it in any way then the center of gravity will accelerate. So for our purposes that's two equations because that's in the x and the y direction that we could do that and then we can sum the moments about some point and it may be the point g it might not be and for our two dimensional problems and that's one equation. If we're doing three dimensional problems then there are six of these available to us. Oops, I need the m in there. That makes more sense. There we go. And these are our kinetics equations of rigid body motions. And then of course we've got any possibility, any available kinematics equations that we can bring in as needed. For example the acceleration of the center of gravity will be relative to the acceleration of point a if point a is the pivot plus the acceleration of g relative to a. But of course in pure rotational motion like we're looking at today that pivot point doesn't accelerate. It will when we look at general motion on Monday but today it doesn't. We're looking at just pure rotation. So the center of rotation itself does not accelerate. So we can count that as, well let's see, we can do more with it. Remember then that that is r alpha in the tangential direction plus r omega squared in the normal direction. And typically for us that, well depending on the problem I could count as either one or two equations. And then the other kinematic piece we can bring in has to do with the velocity which is also dependent upon the velocity of point a and the velocity of g relative to a just as we looked at the week before. But in pure rotational motion point a is not moving. So we've got then just the piece r omega in the tangential direction because there is no normal component to the velocity of g. g cannot move anywhere closer to point a or farther away from point a because it's a rigid body. So there's two more equations there and the kinematics that we can bring in and any others that might be appropriate. But that pretty much covers it all for the ones we want to use. So that's the general setup of where we're going today. We'll do a problem now to help illustrate it. So imagine we've got a slender rigid bar at that instant. It happens to have a slender rigid bar already some angular velocity to it. It's 3 meters long, has a mass of 20 kilograms and just at that instant as it's rotating past the horizontal a moment is applied about this point rather than certainly rather than g because g is at the center. The type of thing you do, you know this is some kind of hatch coming down and you need to apply a breaking moment. You'd apply one in the opposite direction and it might not occur or tell some particular point in the motion. This one we happen to have it as an aiding moment but could be virtually anything as we go through this. So we'll give that a magnitude of 60 Newton meters and omega zero we'll say at that instant is 5 radians per second. And so we want to find two things in particular. One, once it passes that point and once it has the moment applied what's the acceleration now and what's the reaction at a? We'll find that it's a little bit different than we might have thought if we'd approached this with the nae and the te we would have had doing it in statics. Two things we generally need to do. Generally to find forces we need a free body diagram and to find something about the particular motion we'll need a kinetic diagram. You're welcome if you're careful enough to do these on the same diagram but just be careful with it. If you're drawing acceleration vectors on a free body diagram make sure that they don't look like force vectors and you then put them into the particular solution. So free body diagram of course we have r that has some particular weight to it. We don't take that as an unknown since we know what the mass is. There's also some reaction at a so we'll break that into components. Let's establish a coordinate system. We'll just use normal and tangential coordinates tangential down because that's the way it happens to be moving at the time. That will give us a positive on some of the velocities and accelerations we need. And the normal is always towards the center of rotation. So that will be a in the tangential direction and then a in the normal direction as well. Is that it? A free body diagram we also have to draw any applied moments. So we have this applied moment ma as well. Anything that's any of the kinetics that's affecting the kinematics of the object goes on the free body diagram. Alright then the kinetic diagram we'll put what we expect to be the result of any of these forces and moments as applied. So naturally expect the kinetic that the center of gravity will accelerate and certainly at that point we expect it to celebrate. Well that's our tangential direction so let's just leave it as that. We will sum the forces in the moments to determine what makes up that bit. We also expect some angular acceleration of course but we're trying to find that not only is it falling past that point and would probably be accelerating in some way and I believe that was a physics one problem that we did. It also has this moment applied to it at this instant that will contribute to the acceleration. What else? There's one other piece and this is the piece that's easier of any of them to forget and it's also the piece that will show us that this normal component on the reaction is not zero. Certainly our temptation would be say oh this has got to be zero but it's not. These are not static problems. David? Centripetal acceleration. The centripetal acceleration and we'll put that as acceleration of the center of gravity in the normal direction and that's always defined as straight towards the center. So what we've drawn in the free body diagram causes what we expect to see in the kinetic diagram itself and so we do what we've always done when we're looking for accelerations and reactions the length of some of the forces and the moment. So some of the forces in the normal direction I guess tell us what that normal acceleration is and then we'll do the same thing in the tangential direction and we'll do the same thing with the moments. So this part, this is, well we've done this kind of thing before so let's see. A and that's in the positive direction of how we've established, well because the normal coordinate is always towards the center of rotation. A t of course doesn't figure, ma isn't a force anyway and so that's all there is but again that's not equal to zero because that's the centripetal, that's a result of the centripetal acceleration. And that we know to be, just like it always is, r omega squared. And so I think that's I think we've got all those pieces so we can actually calculate this one. Mass of the objects 20 kilograms. What do we mean by r? Is that the whole length of the bar or this distance out to the center? You said something, John? Yes, it's the distance from the center of rotation to the center of gravity because it's the center of gravity we're talking about right here is the acceleration so that's how we have to peg it. So that r will be 1.5 meters omega. At that instant remember it has an angular velocity of 5 radians per second. So that will give us, let's see, kilogram meters per second squared. Remember when we use the kinematics and the kinetics the radians disappear as a unit. So the units work out and that comes out to be 750. Is that right? 750. Well that one was easy, just kind of a warm up. You know, get the juices flowing here now. So we're having a tangential direction. Some forces, if they don't balance and we certainly wouldn't expect them to we'll have some tangential acceleration. So let's see. Positive is down, that's the tangential direction so it's w minus a t. Those are the two vertical forces. Remember the moment's not a force so it doesn't contribute to this. And that's going to cause the center of gravity to have some tangential acceleration but that tangential acceleration is going to be r alpha. We have m and r but we don't have alpha. In fact that's one of the things we're supposed to find. So there's the first appearance of r unknown. So we've got to keep going. We have two unknowns. Alpha and a t are in here. We need to figure out more. We'll sum the moments about the center of gravity and that should give us some idea of what the angular acceleration is. We'll sum about the center of gravity because for a slender rod that's easy to find. It's one of those ones that's in the book. IG equals 112 ML square and we have all those. We know the mass is 20 kilograms and we know the length is three meters so we can figure that out. As requested this morning we want me to do the hardest stuff so you don't have to. So this is 15 kilogram in our square. Alright so let's set up. We'll take the, we expect the acceleration to be clockwise so we'll take that as our positive direction. It doesn't hurt to indicate that just so you don't mess up. So sum the moments about G. MA is a moment on the entire piece so it goes in there just like that. It's all we need to do with it. AN goes through G so it has no moment that it contributes. AT has a positive direction to it which would be, its moment would be ATR and W goes right through G so we don't put it in there either. So that's good. We haven't introduced any more unknowns. I think we only have two unknowns anyway so we're coming up on the two equations so we need to solve it. And that's going to be equal IG alpha and we know what, we know what I is and we're just looking to find alpha. So there we go. There's two equations, two unknowns. Two unknowns alpha and AT are two equations that, two equations we just came up with. So as you go through these be careful what we're doing here in dynamic is different than what we are doing in statics. Don't try to apply your statics equations unless you know for sure that in a particular direction there's no acceleration. It would be very, very tempting to have this point said AN equals zero. But it doesn't. As we saw, it's not only not zero. It's, we're 750 atoms. Alright, two equations, two unknowns. Then we can solve it. It's just algebra left so I won't belabor that. We have alpha 5.9 radians per second squared and we only need the tangential component of a reaction force because we already have the normal component and this is 19 N. So in this particular situation it's that normal component you might have thought was zero that is significantly larger than is the tangential component. That, at least to me, was not at all intuitive. So you've got to trust the equations. You've got to be careful with them. You've got to avoid the temptation to just start knocking some stuff off as zero because it should be. It will make the problem easier. As calling a bunch of stuff that's not zero, zero always makes the problem easier. It just also makes the problem wrong. So be careful with it. Alright, questions on that before I clean up? Pretty straightforward. Just be careful. Take your time through it. Make sure you get your minus signs right. Make sure you don't miss any of the parts but that's always been the case. Certainly ever since you started with me in Physics 1 we've been watching minus signs and making sure we're doing the complete problem. So any questions before I clean up? And we do an alternate solution to this? Okay, alternate solution. Well, there's no easier way to have found that A n. So we're not going to find an alternate solution to that. That only appeared in one equation and so there's only going to be one way to solve for it and that's summing the forces in the normal direction that we already did. But we have an alternate possibility. So, alternate solution. Let's sum the moments, not about G but about A. We have two forces that go through A. One of them is unknown. So let's sum the moments about A and see if that isn't a little bit easier. Remember though that we need the moment of inertia about A. But for a slender rod that's no big deal. That's one of the ones in the book. You might even remember that for a slender rod it's just four times what it was about its center of gravity. That's only true for a slender rod. It's not true for other pieces. It's just true for this for a slender rod. And by slender rod we mean that the diameter is much, much less than the length. So we're not even concerned with the diameter. Okay, so let's give this maybe a try. Let's see. MA is still there. That's just one of the givens. That's the kinetic set of driving the whole problem. And W causes a moment in the same direction. It's moment arm is R. They happen to be perfectly, perfectly a moment so that's easy. And that's all there is. The other two forces go through the point A. So it's then I, A, alpha. And the only unknown in that problem is alpha. And so it is a little bit easier to solve for. Much to R relief you get the same thing as you did before. Still though you need to solve for the other component of the force. So you have to then sum the forces in the tangential direction and actually get the tangential component then with the force there. So it doesn't cut down on the number of equations but it does mean that each of the equations independently solves for one of the unknowns and you don't have to use a system of equations. So it's a little bit easier in that regard. Questions there because we need an alternate, alternate solution. Just when you were hoping things were easier to decide about, they aren't necessarily. So we'll have an alternate, alternate solution. And can't do anything with that but what we can do is just look at the summation a little bit differently. It could be that we didn't have something as simple as a slender rod to use and thus maybe it would be a pain to find the moment of inertia about a different point. So we can go back and do it about the original point, the center of gravity. But we need to take into account the fact that that's not a point where we're summing the moments themselves. So you have to put it together like that. And taking the clockwise direction as positive as before because that's where our alpha goes. That looks like ma plus wr just like before because we're still summing the moments about a just as we did before. It's just now things are a little bit different because we use ig instead of ia. The mass is the mass of the bar. Ag with respect to point a is r alpha. So we only have one unknown in this equation but it does appear twice. And then d is the distance between the two points which for us is that r. That's ag. Specifically ag in the tangential direction. Then r is just that part d. So the numbers come out to be exactly the same as they were before. I have a question for you. This ag, I've got a bt taken care of. How do we handle agn? Because that is not zero. That took care of agt, not agn. I don't like to think about it a little bit because I need to go get a new piece of chalk. I'm working way harder than I am. It's my chalk sponsor. Chalk does come off the board. Oh come on. I don't generate chalk does do I? I'm placing it there for a couple million bucks. Alright. What did you decide? How do we handle the normal component here? Because this is not specific to either direction. This is just the moments about a and the kinematics they cause. Who's got an idea? He's great over a million bucks. What? He's great over r. Well we can certainly do, we can find that. It's r omega squared. We have both of those. So I could certainly put it in. In fact, just for the sake of argument, I will. So there's r omega squared. That is ag in the normal direction. That's normal direction. David? I was thinking it didn't count because it was parallel to d. No. You're close. But you need to be even more specific than that. What is d for this component? It's not r. It's zero because this component goes straight through the point a and it's the minimum distance between that acceleration and the point of interest that is d. So it has a d on it, but for that component of the acceleration, it's zero. So I told you this was going to be advanced physics one. Some of these things are pretty subtle. You need to trust the physics, trust the equations, and be careful about jumping to conclusions just because some things might fit. They don't necessarily. So one equation, one unknown alpha, it just happens in two places so it's a little bit more complex. It also has this potential trap in it. But that's why it's the second alternate solution and not the first alternate solution. But you get the same answer. You get a little shell shock from the dynamics here. Yeah? Oh, let's see how we're doing. It just gets hard and hard. Now we'll do it in two pages, I promise. It need to be a bi-ish paper. Okay. Another problem. Harking back a little bit to physics one for you, those of you that had it, not those of you who had it with me. Those of you who had it with me. All right. Imagine a pulley centers mounted somewhere to, you know, hanging from a skyhook or something. And we have a mass on it. All right. Starting from rest, find the angular acceleration of the pulley. All right. Some of the details, the radius of the pulley, 0.4 meters. The mass of the pulley, m sub p, we'll call it, 60 kilograms. And the radius of gyration. What this really tells you when we have that instead of the moment of inertia is that it's not a uniform density solid. Certainly it has mass distributed in different places. Maybe more at the rim or something. Maybe there's some spokes or something that helps it turn on the axle, whatever it is. Doesn't matter because that leads to it directly to the moment of inertia if that's what we need. And that is, it's a regular length. It's the radius at which we can concentrate all the mass and still have the same moment of inertia. And then the little mass there itself called the mass of the block is 20 kilograms. All right. So find the angular acceleration of the pulley. So I'll leave that to you a bit. We see what you can put together. Start with a free body diagram. And separately if you wish, or you can combine a magnetic diagram as your purpose. He has no way to draw circles and just keep going. There's no stop in Travis. Do not get between Travis and the solution. All right. So set up your free body diagram. Do not assume forces are zero because we just showed a problem where might have been very tempting to say certain forces were zero and they're not. And remember every force is caused by something real. There are no forces caused by the motion. The motion causes reporting can also have two fugues. So get just the right number of equations under the forces and the free body diagram and proceed. A few minutes on the free body diagrams. Some of you are getting them. Some of you are throwing in a couple extra just for fun. Yes, John, what are you doing? I'm throwing arrows in all over the place. Notice everybody's got the basis here for a free body diagram. Of course the pulley weighs something. So there's some force down. It's counter by reaction. We'll call it A. We'll call it AY. That thing's not gentlemen any longer so we might as well use XY ordinance. And then AX. I happen to pick in this direction of course the tension in the line. That's actually what's causing it to turn. That's the applied torque as far as the drums. And then we've got the mass with the same tension up and its own weight of a WP for the pulley and WB for the block. So that should be your free body diagram. Any major disputes? The kinetic diagram is the cause of those forces. It's going to cause the wheel, the pulley, to rotate. It's not going to cause the pulley to do anything else. That's all it's going to do is it's going to start to rotate. It's going to require some angular acceleration. Nothing else. The block will have some translational acceleration. And of course those two are related. Alpha where R is the radius of the pulley is going to be the acceleration of the block. How can I say that? That's a bit of an assumption but I talked about it the other day. Remember what I called that? That is only true if the rope is not slipping on the pulley. If that's slipping on the pulley, if the rope is slipping on the pulley not wrapped around well enough, then I can't say that. That's equivalent to saying that this point on the edge of the rim will have the same acceleration as the block at that instant. That point also has a normal component because it's in a circular path but that has nothing to do with the block. The block is not going to have a normal component to the acceleration because that point will go in but the contact point with the rope will not stay where it is. So now we've got enough of the pieces and you can now start to solve the problem. So we've got one, two, three, looks like four unknowns. However, two of them, the two reaction forces go away if we sum the moments about the center point. We're not asked for those so that would be a good place to start. Plus the radius and the gyration that we have is about the center. So I don't know if it's worth calling that A or calling it G. It doesn't matter. It's just semantics. The center of gravity are the same. Okay, pick the positive direction as the direction we expect the positive acceleration just makes things a little bit easier. Less hassle with minus signs. So the moments about point G or A, the center, is just the tension at a particular radius. And that is positive in the direction specified. And that will equal KG squared MP alpha. Two unknowns, one equation so we're not ready to solve yet. We've got to do something else. It'd be nice if the two equations didn't bring any more unknowns into it. If we can do an equation without AX and AY, we'll be that one's closer to the solution. If we bring in that equation as A and X, Y, we've got to keep going because we've got it in an unknown. Right now we have two unknowns. If we can keep it at that one more equation, we'll do it. Well, we sum the forces in the vertical direction, that brings AY into it. What? R alpha is equal to AB, where AB is the absolute block times acceleration of gravity. Well, if we sum the forces on the block, it'll give us that acceleration, but that acceleration we know to be directly related to alpha. So technically we would bring in another unknown, but it's very easily taken out, perhaps our simplest equations. So sum the forces in the Y direction on the block will cause the block to accelerate. And we've got T up, WB down. We've got to cause the block to have an acceleration that we know to be related to R alpha. So there's a second equation with just as many unknowns as before. So it's a matter of a high school algebra to solve that, and you'll have to do that because I didn't write it down. Oh, wait, there it is. 11.3 for the angular acceleration. We should guess ahead of time how it's kind of an office pool on what these numbers should be, something like a little bit of cash before we head out for summer vacation. Alright, there's another way we could have solved this one, and it eliminates an unknown by taking the block and the pulley as a single system, and not separating them like we did here. We didn't need to find T, so it may not have been the most prudent thing to do to bring it into the problem. Still, we need a free body diagram. So we've got the weight of the pulley there, and the weight of the block there, and the two reaction forces, neither of which we were asked for, and we didn't bother finding them the first time. Since we don't want those, it makes sense again to sum the moments about that point, where this is of the entire system. We can make it a little bit simpler if we use I, G. Remember point G and point A are the same anyway, but we need to add on the effect of the acceleration of the block itself. We don't know the moment of inertia of that whole system with respect to point A, because the block itself changes that, but we can split that out as a separate part then. And now we don't need to know the tension, nor do we need to know the reaction forces. So it comes out to be very similar to what it was before. None of those have anything to do with it. We have w, b at a distance r. It's the only moment applied about point A, and then this is KG squared m alpha, mp alpha, and I had that over there design. That's just the moment of inertia from the radius of gyration. And this is, by the way, that's the block, and it's the point of it being off center. So that's mvr, alpha, or this is the radius of the pulley, that again is our no slip condition. And then d is the radius of the pulley yet again. That little piece is the acceleration of the block, and this little piece is d, the distance of the, actually the line of action of that acceleration with respect to the point picked, which is point A. Again, it's a trade off, there's only one unknown in there, just the alpha. Everything else is given. However, getting to that point is a lot more subtle than I think this was. That's a lot more straightforward. So, maybe it's an academic exercise for you, which one you want to do, but it's your choice. Even if you're not going to use something, I think it's helpful to see how it works, even without using it. So, oh, I think we need to get out of class question at the end of the weekend. What? Yeah, but then I got to add lid for 12 minutes, and that's not pretty. We've all seen the kind of things I say when I'm winging it up. So, here's a problem for you. It was actually a quiz problem from a couple of years ago. Pulling a crate up a ramp using a winch, and the details are, like, it's 20 degrees. It is coefficient of kinetic friction between the block and the slope is .4. The block is 45 kilograms, and the winch is radius of 15 centimeters, a moment of inertia about the pulley part, 4 kilograms, 5 meters per second, I'm sorry, for meter squared, kilogram meter squared, and the moment applied by the motor to the pulley is 50 Newton meters. That's the motor turning the pulley itself. So, not only does the motor have to haul up the mass, it also has to get the pulley itself started. Let's see, find the, well, just set up the equations. Finding the acceleration or something is just a matter of solving the system of equations. That's just down front. So let's get the physics set up. Set up the equations, however you choose, and then then it's just a matter of, I've got the equations right, then it's just algebra after that, and that's not all we're here for. So, free body diagram. The kinetic diagram's pretty simple in some of these problems. It might not be worth it. It could be worth it in more complex problems. It might be worth it, not worth it in these, especially in small, some of you draw. Okay, so set up the free body diagram. Set up the equations. You will need all three forces in the x direction and y. It probably makes sense to incline them with the slope. If you're looking for the acceleration of the crate, if you incline your axes with the unknown, it usually just makes the solution a little bit simpler. But you can do it any way you want with it. Free body diagram of the crate. Now, some of that is statics. The dynamics is just the t, a little bit of w down, and the f down. But when you sum the forces in the different directions, it comes out in the wash anyway, that's how we get the normal force that we need for the friction. And then for the winch, we've got the tension. We've got reactions, and it's weight, but the weight wasn't given, and the reactions weren't asked for. So all we're really concerned with is the dynamics of the situation, which is just the tension that's causing some angular acceleration. Oh, it's got the moment on it too, of course. But all that's going to result is that alpha. So we can wrap everything up with some of the forces in the x, y direction on the block, and some of the moments on the winch should be able to relate all of those. So I already said that the box is a crater, whatever, coffin. Not up, we're talking about murder history. We're talking about murder history. We're talking about patient stand-hold. I'm going to find out until later, while north of here, so it's a long drive. The mass? Nope. Well, yeah, it does, but not given, because it exists. But those are static forces. This is the dynamic force that causes the acceleration. So we don't need the mass on the pulley, the mass of the pulley. If you sum the moments about the center of gravity of the pulley, those don't come into it, because they go right through there. So I'm not concerned. If you sum the moments about g on the pulley, all that's in there is t and m, and n is given t we don't have, but that's what's going to cause the acceleration of the block up the table. So we'll be able to find that. We're overseeing the acceleration of the block. So just set up the three equations, and we'll see what we've got if we all compare. On the block, we sum the forces in the x direction. That's what's going to cause the acceleration we're looking for. That'll have in it the friction force. We need the normal force before we can solve for the friction force. So we'll also have to sum the forces in the y direction, but it is in a static balance in the y direction. So the dynamic balance in the x direction, but not the y direction. So that'll give us n, which will give us f. So we'll have two unknowns, the tension in the line and the acceleration of the block. And then the tension in the line we can solve with the third equation. Sum the moments about g for the pulley. That way we don't care what the weight of the pulley is. We don't care what the reaction forces are supporting the pulley on the motor. That will leave us with three unknowns. From this one, we're going to have t up, f down, which is mu n minus w sine 20. Is that right? And that's all the x components. And so that's going to cause the block to accelerate in the x direction. We're looking for that. So that should be the equation. And then we also need, of course, that f equals mu k n, which we find in the y direction. So that's mu k w cosine 20. The sum of the forces in the y direction is solely going to find the normal force for us that we need for the friction. So how we do it? t is unknown. f will take as known because it's just a matter of filling all those little pieces in. And the acceleration of the block is unknown. So some of the moments about the center of the pulley taking the positive direction to be the direction of the acceleration. Just, you don't have to, but it makes things a little easier. So we've got the tension kind of turn up one way. The moment exerted by the motor trying to turn it the other way. And the difference between those two is going to cause, actually I do the acceleration in the wrong direction. So that's a matter of coming out now as a minus sign so we knew we suffered during the wrong direction. So now how many unknowns? The tension, the acceleration, the tension's already there. R's given, M's given. Alpha's not given. Three unknowns, two equations. This doesn't count as a second equation because we can use that just to find n. So essentially two equations, three unknowns. Not treating that friction force as an unknown. Just because it's a matter of filling stuff in. What's the third equation? Forty-five seconds into your weekend. Third equation. Who's going to rescue the class and send us all home? Travis? The no-slip condition. Assuming the cable's not slipping on the pulley. There's your last equation and now it's just a matter of the algebra to solve it. So there we go. We'll do general motion on Monday.