Hi, this is module 44 of two-dimensional dynamics. Today's learning outcome is use the Impulse-Momentum Theory that we developed last time to solve a problem for a two-dimensional body of motion. And so we left off last time with the linear impulse equaling the change in linear momentum and the angular impulse equaling the change in angular momentum. I can now add those two equations, and I'll pull my initial momentums over on the left hand side. So I have my initial linear momentum, my initial angular momentum plus, on the left hand side, my linear impulse and my angular impulse equals my final linear momentum and my final angular momentum. And I like to use graphical tools as a I solve these problems as I have before and so I'm gonna take the initial momentum plus the impulse and equal it to the final momentum on what I'm gonna call impulse momentum diagrams. And so here is my initial momentum. My linear momentum is a mass times velocity vector. I've broken it into two convenient orthogonal directions in this case, mv at point c in the x direction and mv at point c in the y direction plus my angular momentum. And then I have my impulse which is integrated over some period of time. I've got all of my forces in moments acting on my body. And this is an American football. I'm restricting the body that I'm using to planar two-dimensional motion. Okay, we know that a football actually is in 3D motion, and we'll study 3D motion in my following course, Advanced Engineering Dynamics. But for now we're just gonna stay in the plane, or two-dimensional motion. That's gonna equal my final momentum, and here I've shown my final momentum vectors. Okay, so let's use that theory and this graphical tool to go ahead and solve a problem. I've got a cylinder with an inner drum with a string attached, where I'm pulling on the cylinder with a time-dependent force. And we're told that the force is in the shape of a half sinusoid here, and the cylinder's initially at rest. It rolls without slipping, and we're asked to find the angular velocity after this load application. And so here's my initial system. I'm gonna add the impulse, add my final momentum. And so my first question to you is, what is the initial momentum of the system? And you should say that it's zero because the system starts at rest, so this is zero. For the impulse, I want you to go ahead and draw the forces and moments acting on the body. This is just simply the same as doing a free-body diagram. And if you did that, you should have came up with, okay, you have this time of variant force here. We have the weight, w, we have the normal force, n, and we have a friction force acting at the point of contact, f. And that's a good free-body diagram, so integrating over that gives us our impulse. And then finally for the final momentum, again, I always like to start with three momentum vectors, and then I can zero out ones that are not present in the problem. So in this case I do have a vc final, so I'm gonna have a vm at point c in the x direction final. As far as linear momentum in the y direction we can see that that's zero because the cylinder doesn't go up and down, but I also have an i sub zzc omega final for the final angular momentum of this cylinder. And so that's a good impulse momentum diagram. Here I show it again. And so my question now is, where would be a smart point to sum my Impulse-Momentum vectors about if I want to kind of get rid of as many unknowns that I'm not interested in as possible? And what you should say is, well, let's just go ahead and sum momentum and impulse vectors about point A because I've got these three vectors which I'm not interested in. I'm interested in the velocities here. And so I'm gonna sum my impulse momentum vectors around point A. I'll go ahead and choose counterclockwise positive, and we'll say about point A. Whoops, and I've got, here we go, zero for my initial momentum because there's no momentum there. Then I have, let's see, the impulse vector here. There's forces to the right, tendency to cause a clockwise rotation. That's opposite my sine convention, and so it's gonna be negative, and the integral is going to be from, okay t initial is zero and t final is t sub naught. And I've got p times its moment arm, which is R plus one half R or 3R over 2 dt equals now, I've got my angular momentum vector which is causing a clockwise rotation, so that's gonna be negative in accordance with my sign convention. I sub zz c times omega final. And then I also have my moment linear momentum vector, and I wanna take the moment of that momentum, so I'm gonna multiply it by its moment arm. It's also gonna tend to cause a clockwise rotation opposite my sign convention, so it's gonna be minus m V sub cx final times its moment arm, which is R. We can substitute in I sub zzc for a cylinder. You can look it up from any reference again, but it's 1/2 MR squared. And I've got all minuses here. So I'm gonna make this a plus, this a plus and this a plus. And I will end up with the integral from 0 to t sub 0, 3/2 R times P. And P is given as (W times sin of (pi t / t0)) dt =1/2 MR squared omega f + M V cxf times R. I can substitute in for M, which is going to be the weight over acceleration due to gravity. And then I can cancel some common terms out here. I'll cancel the R here, the R here, and this is an R squared so it becomes R. I can cancel the Ws. And so I'm left with the integral from 0 to t0, 3/2 sin of ((pi t) over t0) dt = 1/2 R omega f + 1, v sub c, in the x direction final. And so this is what that looks like. Let's go ahead and do the integral. Well, before we do that, I've got one equation, and I've got two unknowns. Is there any way I can take these two unknowns and relate them to reduce it so that I have one equation and one unknown? And hopefully you'll remember your kinematics and say oh yeah, v sub f, the v of the geometric center for wheel rolling without slip is going to be omega R, or R omega. So this is going to be R times omega final for no slip wheel kinematics. And so I end up with, I'll pull the 3/2 out. And then I'm going to do this integral. The integral of sin is minus cosine, but I've also got to pull out the t0 over pi. So I'm going to have -t0/pi times cos of pi t/t naught, all evaluated between 0 and t0 equals. Now I've got 1/2 R + R, + 1R, which that's gonna be 3/2 R, omega f. And then I can get rid of the 3/2 on both sides. And I've got, let's pull out the t0/pi, times -cos of pi now because t0 and t0 cancelled, cos of pi is -1. So minus a minus is gonna give me a +1 minus this integral evaluated here at zero. Cos of 0 is going to be 1, but then I have a negative here so it's going to be- a -1 = R omega sub f. Or I can solve for my final angular velocity which I'm required to find, omega sub f. And I'll express it as a vector, final, is equal to 2gt naught/R pi, and we see that it's in the clockwise direction. And so this is because 1- a -1 is 2, so that's where I get my 2 from. The g comes over t naught / pi R, and that's my answer. And so we've done an impulse momentum problem. And so the next thing is for you to practice and do an impulse momentum problem on your own. And so I've got a nice little worksheet here for you. And when you finish it, I've got the solution in the hand outs. And I'll see you next time.