 All right. Any other questions as far as we get going? Appreciate other questions as you download the professors, because that's awesome. The question on the second homework also. What one was that? I was not quite sure. You have a question on the second homework problem? Yeah. Okay. I want to see a sweet diagram on this one. Oh, that is a sweet diagram. What acceleration of the collar, and if I assume it's in a crack and only to that, would give a 15 degree deflection? Make sure you wrote that down on the end of the sentence. What problem is that? I'll say it right now. That's not going to mess this problem. That's not going to mess this problem. Oh man, that broke me. I'm not going to stick here. That's not. No, that's not mustard. That's the color of those peeps, the marshmallow chickens. Is that your Easter set that you got? Kind of is, yeah. I bought it out for Easter. I've got to admit, the board in the first class was awful pretty. Hey, there's a lot more right there. What problem is this? Oh, there it is, number three. Oh, I see. I see. I see what's, I got it. All right. I didn't understand. All right. So that problem is you have this collar that can ride on a rail, and there's a thing hanging from it, arm of a little mass there, and the arm and the, and there have the same mass, and then this is given some acceleration along there, which of course will cause the arm to swing back at some angle. So the question is, what is the acceleration required to put that at 15 degrees? Any slower, of course, will be less, any greater, it will be more. So what'd you do with it? I tried some things like some new moments about peas, where's pea? Point pea is where it's connected to. Point pea's up there. Okay. I've gotten much farther. I'll sleep in the middle of it. Does the collar have a mass too? The collar. The particle, you can just think of something. You don't have to say so. Okay. Do you guys do this? What do you do with it? I don't start yet. I figure I'd save it until the day before it's good. I tried some of the forces on it and stuff. I'm just like a little confused about how they all went up. So it's some of the forces in the abstraction and some of the moments about point pea. It could be somewhere, but I haven't gotten to the bottom. All right. Well, this is, remember what we were looking at on whatever it was, Monday. This is a translation problem. We're looking at just the kinetics of translation. So what you need to look at, I guess, are a couple things. This will have the same acceleration for all, I guess those parents could have been the same length. All three of those must have the same acceleration. And each of those has a w on weight. And so some of the moments about pea that didn't help in it, what do you do next? Some of the moments about pea equals zero. Equals the difference. I didn't split the ball and the arm up. I used this. It's like the, well, where'd you put it? Yeah, that would also make it too difficult because I want to know where the center of mass is between the sphere at the end and the arm. But I didn't know where it is on the arm itself. So that would probably have helped. Probably have helped. You could do a force of X to Y and put a force like a bar. Some force causing the acceleration? Yeah. The bottom and the middle of the bar? The middle of the bar? Yeah, if you did a free body diagram of each of those, then that would put some force on that piece and some force on that. And it's the component of those that's causing the acceleration. But remember, there was a caution about summing the moments I think we've just gotten to. In fact, we're going to hit it again today in some details. Remember what the deal with that was? Very little access there. Yeah, that if you sum the moments about the center of gravity, then it's ideal. If you sum the moments about somewhere else, we'll do several problems today that highlight that. It's got to be about the moment of inertia, about that same point, which we don't always have, but we may have the center of the moment of inertia about G for regular shapes, but then we have to take into account the parallel axis there. Which if you put that in, you'll get just that with it, with the additional fact that A equals R alpha. Actually, R is the same as D in these cases. So if you sum the moments about P, just said they were zero, you didn't take into account this part because you're summing the moments about somewhere other than the center of gravity. So that one's zero, but that one's not. So try it again, see if that works. They're not due today, right? Yeah, so what are you going to do all weekend? Just eat peeps. We're doing kind of the... Well, we're sort of doing an extension of what we were doing before. Remember, what we're doing now is kinetics of rigid buttons. Translation, translation problem. Now, yeah, the bob hanging down had to rotate to that position, but then when you were to do the problem, it had already gone through that rotation. And so it wasn't really a translation, or it wasn't really a rotation problem anymore. Now we're going to do rotation problems, and then next week we'll put them together and do general motion problem. And that's the type of problem of gears running along a rack or something, those kind of problems. All right, so the idea is that we might have some rigid body here, maybe some kind of physical pendulum. A physical pendulum is a term slightly different than the usual type of pendulum that we looked at in physics class where the string has no mass or negligible mass compared to the bob itself, where in this case we have an object where the entire thing has some mass, and so it has a center of gravity that's somewhere other than at the very end length of the string that makes up the pendulum. So when we've got these type of problems we're looking, this is a pure rotation one, because that's all it's doing is about some point. So we might have then angular velocity and or angular acceleration at any one time, and because of that we need to look at the forces involved in holding it there and or other things that may happen. Because of the rotation at any one second it's best probably to use our normal and tangential coordinate systems. And then we can sum the moments about the pivot point, we can sum the moments about G, whatever it is we want to do as long as we deal with it appropriately and we'll look at both possibilities. In fact, every problem we've got today, I think every problem, has alternate solutions to it so you can do it however you best see fit. So for our problems we've then got kinetics equations we can use as necessary. That of course is two equations because of the two dimensions of it for our two dimensional problems. We can sum the moment about some spot and find the angular acceleration from that or maybe we have the angular acceleration and we want to find the forces that are causing the moments that give us that acceleration. That would be a third equation for us. No, remember the stipulation is that you can do it about any point as long as those two subscripts match. So one possibility is that the point you do it about is the center of gravity. You would have to use a parallel axis there. Yes. If you want to sum the moments about some point other than G and you don't have this then you have to use the parallel axis there which then gives you the alternate form I, G, alpha plus M, A, G where then D is the distance between those two axes. Actually just in a two dimensional problem just the distance between those two points the minimum distance between those two points. So either way that's one more equation and in terms of the kinetics that's all we've got. So after that you have to use the kinematic equation if you need more equations. The kinematics equations are the acceleration type equation that we just had a little bit ago. Oh, sorry. That one goes on the other side. So let's make it with respect to G and label this sum point P. Actually it could be again any point that we've got. I know that would have to be P. We have a relative acceleration equation but the acceleration of point P being the pivot point is zero. So a relative acceleration then becomes what we've had before R alpha in the tangential direction and squared in the normal direction. So those equations are possibility are possible equations and then it need be even the velocity relationship between them and that would be only in the tangential direction because we don't have in rotation motion any velocity at any point other than tangential. That bit there is kind of a reminder. Yeah, these colors are awesome for an Easter. You just get in the mood, aren't you? Explain the last part. Velocity of G is just R w tangential direction only. Yeah, where R is the distance from the pivot out to the center of gravity I guess we could call it R g if you want to and then times the angular motion itself. So for the picture shown I have to pick that is rotating that way so at that instant it's velocity is that and it's related to the angular acceleration. I'm sorry angular velocity in that way. Let's do a problem here. Kind of like that one. Imagine a slender rod pivoted about one end at the instruments in the horizontal position there's a moment applied to it. We'll call that end where it's pinned point A. Moment is 60 Newton meters. Also at that moment it happens to have an angular velocity of 5 radians per second. I think I'm just going to stop talking and just draw. It's so pretty. And the length of it is 3 meters and its mass is 20 kilograms. We want to find two things. The angular acceleration it'll have that instant to that moment as it passes horizontal and the reactions at that pin that's holding it to the wall. The blue arrow is a angular acceleration. No an applied moment. It will cause an angular acceleration. Alright we're going to have three solutions for this. All of you can give the same answer and God you pick which one you like the best and study only that one then. But we'll do all three. Alright, free body diagram. Let's see. Of course it's got some weight there. It's got some A there. Since its rotation we'll use the button there. I'm not going to need that spot. I guess I don't want this drawing but we'll put it here like this. That will be the normal direction. That's the tangential direction. Since it's accelerating in that remember normal tangential coordinate systems are defined at the instant by the motion itself. So we've got that. And then there's some component of the fours over here we need to find. So that's tangential direction that's the normal direction right? And then in terms of the load we also have that applied moment. So that's the free body diagram I think. We've got all the pieces. Anything missing? Any forces missing? No, we've got the reaction and the weight and the moment being applied. Why does the reaction force point that way to the opposite direction? Just because that puts it in the positive normal direction. If we're wrong, when we solve for it we'll get a minus. So we're not too worried. Now it helps in these problems especially with the general problems coming up if you also draw a kinetic diagram which is the motion that will result from those forces being applied. Remember in this class the force is going to sum to zero. So we're going to have some angular acceleration alpha that we're looking for. There will be some acceleration of the center of gravity in a tangential direction acceleration in the normal direction. So that's our kinetic diagram. Times to make sure we've got those because when you have to do that MAGD part it will really help if at times you've got that actually written down because you've got to know what that D is in fact you've got to recognize that it's even there. Sum the forces in what we've got to do in all directions I guess. We've got three unknowns so we're going to need three equations. Let's see, tangential direction forces are unbalanced that will cause an acceleration in the tangential direction. Well we can go ahead and put in a T there. We'll fix it in a second. Alright so what forces do we have in the tangential direction? Positives down for tangential direction because that's the way it's accelerating at the moment. So what have we got? W minus A tangential. We know W I don't actually have it calculated but we know M so we don't consider that an unknown. We have W so then that equals the acceleration which will be M times R times alpha because it's the acceleration of the center of mass where R will take just to be the half distance there. That will be the tangential acceleration of the center of mass. This part right here. Why is it minus A T? Because W's down and I drew A T up and that's why arbitrarily chose well not arbitrarily chosen positive direction. Do you need some more colors? I can go get green. Same thing in the normal direction at the normal acceleration and that's our positive direction there. So just to remind ourselves what do we have? An but it's in the positive direction so it doesn't even minus sign. That's it. Equals M A G which is R A squared. In fact we've got all those pieces knowing because we know all those numbers to bet. Vibrations per second. So we have 25 and 70 sorry? So that's the reaction and that's also what's our positive direction. What weight? What weight? Here? Oh that's the minus sign. So we still have two unknowns A T and alpha so we're going to need another equation that relates those so we'll sum the moments about what do they want to do for my first choice here. We'll do it about G M A it's not strictly about G itself but remember the moment is a floating vector. As long as you don't change the direction its location is the same for anywhere along the rigid body. Alex do you have another hand up? If there's some moments about G that we're given moments about A but I see we're down. G's at the center there but moments moments if you have the moment the moment applies anywhere for the whole piece. It's a floating vector whereas forces are sliding vectors. I can't take the weight and put it anywhere I want without changing the problem but I could draw it pushing instead of pulling and there's no difference in the problem. Vectors can slide along their length but moment vectors can float anywhere you need them not anywhere you need them but anywhere. So these two forces cause no moment because they go through point A so only have W causing any moment WR and that's going to cause the object to accelerate. What's IG? Remember there's five ways to handle moment of inertia. Find what? Independence? Yeah, that's IG for a slender rod is the moment caused by the weight about point we're doing the moment about G so actually we want AT we want ATR about G AT is going that's our chosen positive direction so no, it's a plus so that comes out to be meter second so we've got two unknowns again cause MA is known that's the applied AT we don't know it's a moment arm on AT IG is 15 alpha is unknown so we have two equations now two unknowns do I have some of the other numbers for that equation? Yeah, the weight is 196 times R is two unknowns now we can solve it get the last little pieces we already have that to skip the algebra I'll give it to you AT is 19 newtons and alpha 11.3 radians oh sorry, not 11.3 5.9 nothing too tricky there there's new stuff there's things to pay attention to gotta be very careful with the directions so specify that set them down and stick to it remember that moments are floating vectors so even though that's a moment about A, it's a moment from the entire rigid body close to three solutions three ways to solve the same problem there's one of alternate solutions you got all this? I'll tell you something right now alternate solution and it doesn't matter which which everyone makes you more comfortable we sum the moments about G but we could sum the moments about A let's see what that does for us let's see, we've still got MA there