 I think I left you with one. Did we think of one with the big slab or whatever was going down the rollers? We didn't finish that one? OK. The trickiest part of that one was realizing that the, because of the no slip condition, that's on both sides of the rollers, where the rollers contact the ground and where they slab contacts the rollers. And so you can relate the speed of the rollers themselves to the speed of the slab. And also for the potential energy, gravitational potential energy part, the slab actually goes twice as far as the rollers do. So good. I guess we're OK on that one. All right. So the last little part here that we've got to do is, again, the motion business that we're putting together. Only remember that just like we did with particle motion, we had three different ways to solve problems. The classic Newton's laws part, that's very good for problems that have either constant forces or constant acceleration. But it could work for forces that change, I guess. But it's not as easy as the other ways to do it. We had with that also the rotational equivalent of that, if you would. So that was one way to solve our general motion problems. Then the next one was the work energy equation. And that we got on Friday. That doesn't come in a completely different form. And mostly the concern was that for the rotational kinetic energy, it just goes in as one of the kinetic energy terms. And the rest of it, for the most part, is the same. And then we'll do the impulse momentum method now for both types of motion that we can then put together. So this will be for rigid body. So the translational part, of course, still holds. Then we take it a step farther, look at the time rate of change. And if we assume the masses are constant, then it's only the time rate of changes of velocity. But if the masses aren't constant, the time rate of change is the product of the two. This then, of course, is dV dt. And if we move things around a little bit, we get the differential form of the impulse momentum equation. But the integral form for us does a little bit better. So we can agree between any two times and any two velocities. And then we can get the impulse momentum form for translation. We had this in physics one. It's no different than what we had for the translational part of the particle mechanics. No difference with anything we've got here. And with constant mass, then it just comes out to be m w v. If we want to be, I guess, a little more complete, we can just say, change in the linear momentum. So that's just review. That was the particle motion part that we had before. But of course, we've got a rotational part to go with that now, because not only are these things moving, but they also have to turn. And it's both of those that we need to take care of. So it starts from very much the same place, only with angular momentum, rather than linear momentum. And then we pretty much go through the very same steps that we did there, take the time rate of change of that. And if we assume constant moment of inertia, then we get that term. Remember that like every part of the rotational part we've been doing, we generally, initially, refer to it around the center of mass. We can refer to it around other points. And this is true for this as well. But as long as these two subscripts match, you're always going to be OK with whichever point you generally prefer to use. And then we can go through the very same steps we did there. And we get then the differential form. And then integrate it. And we get then both a translational impulse momentum form and a rotational impulse momentum form. Both of those, at the same time, must be a part of whatever it is we're looking at. Remember that left-hand side is the area under the time diagrams, either of the momentums. We would graph the vector. However, we could graph the two separate directions independently, as we've done so many times before. I guess the only change, just to be a little more complete, is that this is actually the sum of all the forces, and this is the sum of all the moments being applied as well. Actually, that's not the forces. Where would I need the sum? The sum would have been down here. So we don't really want the sums in there. Is that what you're going to say, Chris? This was the h dot. Is that h plus v dot in the end of the top right? Here, yeah. OK, but we don't need a sum on any of those. Oh, actually, we could, because this part here, this is i g alpha, because it's the omega dot. But the acceleration is caused by moments, just like back here. Right here, we would have put it in the sum of the forces. So that's where I wanted the sums. This would be either one of those two parts. So this is actually, yeah, that'll do. All right, so that brings us up to a chance to do a couple problems. So we'll do a few without any particular translational part. And then we'll do one with translation and rotation. So our classic pulley and force problems. So we've got a 20-pound disc with a radius of 7.5 feet and an applied torque as well as that force being there. Applied torque of 4 foot-pounds. Starting from rest, fine then the angular speed, 2 seconds. So even though all the forces and the moments are constant, not a function of time, the fact that the problem has a time component in it can indicate that it might be worth using the impulse momentum method. In this case, it'll be pure rotation. If we actually had a weight here, we'd have to do a translational impulse momentum on it as well. But it'll suffice for us to just use the angular momentum part here. After 2 seconds, but that time rate of change of the angular momentum is that sum of the moments part. And so it's simply a matter then of putting in the moments and integrating those between two times. And that's going to cause then a change in the angular momentum. So got all the parts all we're looking for is this last little bit solved for the omega 2. And since it's a two-dimensional problem, we don't really need to consider the full vector form. It'll suffice just to put a plus or a minus on the direction of rotation and the direction of the moment. So take a second and knock that one out. That's the weight of the pulling, the weight of the wheel. What's the moment? What would you do as I was telling you? For a uniform disk, like we take that space since if you're given the radius of duration, it generally is telling you that it's not a uniform disk. It's a disk with maybe a hub and a rim or something. But for just a uniform disk, it's 1 1⁄2 MR squared. So then the only part that we don't have is the omega 2 that we're looking for. Just a bit of a warm up one. Got all the pieces. Just make sure you take into account the fact that there are two moments being applied to that, both the moment from probably some kind of motor and the moment due to the weight. Not currently. There's not enough. Now there is 10 pounds. And if that was also subject to time, you just have to complete the game, or we'll do one of those in a second. Good chance to get it. And they're all ready. And some of the moments, integrate them with time, which is easy. They're all constant in this. That's equal to i g omega 2. Some of the moments, they're both in the same direction in both constants. So that's a fairly easy interval to do. And then you can find an omega 2. Now what I got? Oh, I bet you weren't watching your units. Watch your units. It's time. I hope it's on this end of the military. See, unit, yeah, you may be going to see you don't even put them in at all. Doesn't depend. What you do, Travis? Yeah, well, don't just don't need to go to inches at all. There are no inches in the problem. That's in feet. That's in feet. So your problem's elsewhere, but it's with your units. I bet. Yep. Did you say you're stuck there? Well, I don't know. For i g? Why, what's wrong with, is that i g there? I don't know if I have it separately. I do. I have it, pounds, feet, second squared. However, you're given the weight of the disc, but here we need mass. And the easiest thing to do is just leave the units just like that, because they're gonna cancel out again. Anyway, don't change it to slugs or pound mass or any of those things, it makes no sense. And then that's the number that you need in over here. The 20 divided by 3, by 32, 2. Did you do that, Mark Travis? Didn't know. Of course you have something. So watch your units on these. As always, be on your headstone, Travis. It's finally caught up with her. Using that mass, I don't have that written down, but we have to have this, so you can check it. 1.75 pounds, feet, seconds, second squared. Second squared. And just leave it like that. Don't bother going to slugs or pounds of mass or anything else, just leave it like that because they're gonna come right out in the wash anyway when you put it back in the equation. Remember this big M is the moment, some applied moment of motor breaks or something. This would be great on the direction of this turn. Tom, okay. Alan? Boom five. 32. Use something else for G. Second squared. Which part? Omega. Okay. You're okay with these units? Yep. All right. That's the whole NMR squared. So the moment is four foot pounds. P times R is also foot pounds. So what is that? 7.5. So that's the moment. And then you're integrating that with respect to time. So there'll be one T in there and then evaluating between zero and two seconds. And then that should equal, this IG is the one half MR squared with those units. And then we're looking for omega two, which will be in radians per second. So this is feet pounds seconds. This is feet pounds seconds that cancels over here. We're left with just a second so that you're right then. So yeah, that's right then. Yeah, so we're okay. Yeah, because that'll leave a second here canceled by a second on the bottom here and then all the others cancel on both sides. So it's all right. They all agree. You tell me for minutes, right? Yeah. For omega two? You don't have to hold for the answer. Yeah. Yeah. Is 132. Oh, I got 131.4. Okay. I think that's what Travis got too. He's got a little bit of round off somewhere as well. All right. Tommy, okay. Phil. Yep. All right. David. Okay. All right. So we'll step it up a little bit just to make sure we're getting all the parts of it before we put the whole thing together. So, imagine a large gear here that is being started up by a smaller motor or a smaller gear on a motor. Starting from rest, we wanna find how long it takes to get the motor up to a steady state speed of 600 RPM. So, this is 10 kilograms as a radius of duration about its own center of 200 millimeters and a radius of millimeters. And that comes because it's not a uniform solid disc. It has some distribution to its mass somehow. And then the smaller gear to which the motor is attached is three kilograms, radius of duration, 80 millimeters, and a radius of its own, I don't need a G on those, a radius of its own of 100 millimeters. So, again, that's simply because it's not a uniform disc. And the applied moment, six Newton meters applied to the disc will put it in this direction. All right, with that set up, we wanna find the time to get the motor to 600 RPM. The motor's attached directly to the smaller gear there. So, again, generally the very same idea is is just that we have to take into account the changing momentum of both of the gears. Since they're gears that automatically makes it a no slip condition between the two, which of course is important to us. So, again, it's pretty much the same thing. Might as well start with the motor. If you want, we can label these that'll help or just do them separately on paper somehow. But again, it's the time component we're gonna find there because that'll integrate out as a, since the moments are constant. So, being this motor moment is constant. Okay, so you can set that up first for the little gear B remembering that it's trying to drag the big gear around to a start itself, does this tell you? So, you might wanna start that on the small gear itself because that's where we've got the omega remembering that it's trying to drag on the big motor so there's a force it's feeling from that. Take the moment to be constant, so that force there we can also take to be constant. Yeah, you don't need the R in there. It's just two pi radians per revolution. And then, yeah, in one minute or actually eight seconds. Maybe that'll give you less to work on. That 600 RPM is 62.8 radians per second. Not by factor of time, I had 6.28. Yeah, well you had the radius in there as well. So, starting up on B, get some of the moments. These are opposing here, so we're gonna need some direction. However, we'll take it to be constant so they integrate out easily. So, we put it back to the positive. Okay, so in the direction of the moment I am I'm gonna take this positive, so it'll be M minus F R B. We don't know what F is, but we do know what the moment is. We don't know what the radius is, of course. And these integrate out, so the time part just becomes delta T. So, just doing that, just doing the left hand side, we have two unknowns. The delta T, which we're supposed to find, and the force that one gear exerts on the other, which we're not asked to find, but we may need to. And then, that's going to be related to the startup of the small gear. IGB, we can find because we know the radius of gyration of the gear B. And the delta omega, it's starting from rest and needs to get up to this speed. So, we've got that side of it. So, we have one equation and the two unknowns, F and delta T. So, we need another equation. We've got two unknowns. We don't need to find F, but it is an unknown, so we still need another equation. Whether we find it or not, just depends on how we eliminated things from the system of equations that we'll get. Where's another equation? Notice, maybe that's one from yesterday. One equation, two unknowns. Yeah, because we're trying to get the motor up to 600 RPM from rest. So, it's got to go from zero to that speed. So, there's the delta omega for B right there. Won't be the same for the big disc, but it's the inverse of the radius. IGB uses inverse. And in general, it's that. So, where's the second equation? Yeah, you've got to go where the F is, one of the unknowns leads you, and that leads you to disc A. And so, it's got a moment on it caused by the very same force in the opposite direction. So, this force is just gear A on B or B on A, same either way. So, you can set up the very same thing for gear A. Just working that one out a little bit farther. Equation one. And so, do the same thing for A. That will also involve F and delta T as an interesting solution step you're gonna have to take that can really speed things up. If you don't take it, you should still solve for it. Just gets more difficult. The positive moments in the direction is gonna go against F, R, A, delta T. That's the applied impulse on gear A. Since the moments are, forces are constant, that integrates out to simply that, and then that's going to equal that. Remembering that, oh, we actually have four unknowns because of the maze or unknown, we're looking for those. That's actually the velocity of the contact point. Number two. Of course, that's assuming no slip, which is the point of gears can slip. So, what we'll do as often is a belt mine. The interesting part of the solution is that you can solve this part for F, delta T. And then that, you can stick directly into there and we don't need to separate F and delta T. When you put it in there, then you have delta T, and the final rotational velocity, angular velocity, and you can finish it. So, it just speeds up the solving of that system of equations. You can do it that way, just leave the F and delta T. It actually looks like it says FAT. Solve for that, and then just stick it right in there. You have that delta T as a term here. And then you go back to one and just solve it directly for delta T, that'll be your last unknown. That makes sense, Joe? Can solve separately, but we don't need F, so that just eliminates it right away. Just that quick to get rid of it. It's okay, as long as you get that. Phil, that makes sense? Yeah, see all those things we know. So, you can get a value for F delta T and just stick it right in there. There's F delta T, and then all that leaves is the unknown, is that one delta T, that's dope. That's just the easiest of the way you go through the algorithm. Get 40.2 Newton seconds for that, and the units should work out for those units. It's R A, omega A. Yeah, for omega 2A, I have 25.1. That's from the nose flip condition here. Phil, what do you got? What's that? See, omega 2B was given. We're gonna get it up to 600 RPM, which is this many radians per second. That is omega 2B. So there, use that to find omega 2A, which you need right here. Yeah, that's what I got. And then if you want, you can go back and find this course, but it wasn't asked for. Look at some of me and hold on to us. Use that as F delta T as a single, yeah. It's a lot simpler that way. 0.871 seconds, that's 40.2. That's the F delta T, which you have in here. So don't forget there's an R in there still. Tom, you got it? For the time? So man, I got it. 0.871 seconds, and so did Travis. David, did you come up with a number? That's 0.871, but that's an M. That was a coefficient version, which for properly mixed years is, yes, something good. Okay, there we go, good. All right, two more minutes, maybe. Alan, you stuck somewhere? Well, I kind of went off on a tangent point. You? No, I guess I have this. Bill, you're almost there. It's pretty easy to put the wrong radius in somewhere. I forgot to multiply by this radius. Yeah, that's easy to do too, because all we have here is F delta T from the second equation. But he's got the radius in there. You know what you got, Chris? Same thing? You're squaring everything that needs squaring. There's a couple in his hand. You just square the radius and the duration. You have to do that too. Small amount, there are all your numbers, so it could go into it. All right. Well, take a little time if you need it, double check that if you're still struggling whether to come see me. But you should get about 0.871 seconds. Did you get that too, Tom? All right, so we've got several less than that. I've gotten that. Okay, Chris, find your mistake. All right, easy enough since those don't actually translate. So let's go to one that translates and rotates and put the whole thing together. One of these compound wheel pulley things. Looks like it's going downhill. No. Force being applied to it, it varies with time. T is in seconds and then there's got to be some conversion thing in there and doing this per second just to make the units work. Mass of 100 kilograms, individual radiuses, that's 0.4, 0.75 meters. And the radius of the gyration about its center, 0.35. So I've got this time-varying force being applied. T is in seconds, assuming no slip as usual. Find the angular speed after five seconds. Yeah, it's like a same general idea. This time it's not pure rotation. There's also a translation with it. This one's got a little bit more to it. You can't do that because that's what the force is at the end. But it varies with time. At zero seconds, it's only 10 and then it goes up to 15. But it varies with time. So remember we're doing intervals with respect to time. Mass of the whole thing more. Yeah, mass of the whole thing. You've got the radius of gyration. So you launch it the same way, I guess. Start with what we've got that gets us closest to omega. And that's the interval that some of the moments with respect to time is going to get this thing rolling. Starts from rest. So we're just trying to find the angular speed after five seconds. It's called omega 2. So sum of the torques. If there aren't any unknowns in there, you can finish it right through. If there aren't any unknowns. Oh, I hate when you say that. Just be careful when you sum the torques. If I get too much farther, let me help you with this one. Obviously, it's got a torque applied to it due to the varying force p. It varies with time. But there's also a torque being applied to it by friction along the bottom. If that wasn't there, then it wouldn't rotate nearly as much as it does. It would slide and rotate a little bit. The thing is, this also varies with time. So you're going to have to integrate that as well. So the sum of the torques over five seconds is p that varies with time times the little r. We call one r little r bigger one big r, I guess. And then the torque due to the friction in the opposite direction. Oh, actually in the same direction, aren't they? With respect to g. Sorry, that's a plus there. And the f friction force, of course, has to do with the moment. But in this case, the moment is just equal to the weight. If there is an angle to that p, then the normal force would vary with time as well. Sorry? Do you need a coefficient of friction? Well, maybe. If you need to find that, yeah, I'm going to need a coefficient of friction. And occasionally, man, I do forget some things, but not this time. You don't coefficient of friction, which means you can find this, solve this without it. So is the friction. So it has to be integrated as well. Well, you've got it already? Come on. I know you don't. It's probably not, right? No. No, no, no. You can't do that. You have to integrate it. No, no, no, no. You have to integrate the whole thing, which is the t plus 10 when you integrate. Oh, you've got the squared in there. Yeah, so when you set this up, 0 to 5 seconds, t plus 10 times little r, 0.4 meters, so this will be meters, meters seconds for that bit there so far, plus the friction, which you do not know, but you know the radius at which it acts, 0.75, and then all of that gets integrated. I did it for the initial start using just 10, yeah. This you're going to need to do very much the same thing we just did. This first part, you can integrate the 0.4 comes out. Remember, that's the radius. That's in meters, t squared over 2 plus 10 t, 0 to 5 seconds. That's the first part of that integral. And that you can do, you can finish that bit there. Then you have this back piece, 0.75. That's the big radius comes out. And you've got the integral f dt, which you can't integrate because we don't know just how the force varies with time. It's not equal to p. If that was the case, the wheel wouldn't accelerate. It's a function of time, but we just don't know what the function of time is, so we can't do that integral. So then what? We have to do a lot the same as we did on the last one. Remember, we found f delta t as a separate unit and swapped it out in the first equation. This time, we do the same thing, only it's the whole integral. We need to find some other source for exactly that. We can swap it out. Where is that one other source? This is the impulse momentum equation for rotation. This wheel is not just in rotation. It's also in translation. So do the impulse momentum equation for translation. And one part of that is going to be the integral f dt. And so you don't need to actually solve the integral. You can substitute it and eliminate it. Of course, all of this equals omega 2 since omega 1 is 0. Normally, it equals i g delta omega, but of omega 1, 0. It just equals omega 2. I'm pretty sure you're struggling, because John's not here to do better than that. Or you got it already. What did you do about that? Well, is this no slip or not? Yeah, no slip. So I don't see how if you think about that it's not slip. You don't see what? How can you not go out if it's not slip? Because if p equal f, it wouldn't accelerate. It's slipping, but there it's not slipping, but it is turning. So if p equal f, and it started from rest, it stayed at rest. So there's, I guess, a portion during which p would equal f, but then? Well, no, there's no slip. But I'd always show the radiuses to each other. p to f is just the ratio of the radiuses. Maybe, I don't know, you don't need it. You can eliminate it from the translational impulse momentum equation. Well, that's the rotational part. The translational also integrates 0 to 5 seconds. It's just p. This time it's minus f because they're in opposite directions, dt. And then right from there, you can get the integral f dt. And solve for it directly, put it in here, and then find omega 2. Equals then the linear momentum, and v2, because again v1 is 0. p and f were equal. That integral would just be 0, and the delta v would be 0. Makes sense, Joe? You'll need one last little bit, but that's the no slip condition on the velocity and the angular velocity. Big R. Second equation will have v2 in it. The first equation has omega 2 in it, so you'll also need that to tie the two together. I know I've validated. If the integral of f dt is equal to ikw2, it's not. R me 0.75. No, that's not saying this is equal to that. This whole thing is equal to that. This is the impulse. This is the angular impulse. We'll cause a change in angular momentum. I guess what I'm doing is trying to figure out the substitution down there. Well, it's almost done for you. You can separate these two integrals by doing that, right? M r omega 2, just using the no slip condition. Oh, that's not supposed to be squared. It's supposed to be 2. And then this you can solve for that as a whole unit, and then put it in there. You can do all this part. Well, it'll be a function omega 2, but then you solve that last bit for omega 2. Because this is just a constant when you finish it. That's a constant, but omega 2 is unknown. So when we put it in here, we'll put it in a constant, plus a term of omega 2, and then you solve for omega 2. The angular velocity. For which part? For this right here? That no slip condition is based on that. Velocity is equal to r omega. So it's got to be the distance between the velocity and center mass and this contact point. So that's the big r, the 0.75 for this part right here. Well, for something that wasn't slipping, we could put the r in there. The P of t was 1. It's not slipping either. If you wanted to relate the speed of the rope to the angular speed, but the speed of the rope is greater than is the speed of the wheel itself. For that, you'd need both radiuses, because it's that total distance down to the contact point. Got it? Yeah, that's how I got to it. Well, you got it? Yep. Oh, that's get out of place. Question? I don't have time for anything more. So pack up and hit the road. Then do FWD separately. You should be able to get that this part here is 62.5. Yep. Minus that integral part equals, and then everything else is a conscious 75 omega 2. And you should be able to find that omega 2 by readings per second. Be careful with the minus signs. The units all work out. As long as you don't forget an r or somewhere. The little radius only came into the problem once at all. And I wish back here that everything else was based on the big radius.