Hi and welcome to Module 43 of Two Dimensional Dynamics. And I wanna take a look at the overview of the course cuz we're getting near the end. We did the section on particles and systems in particles. We're in the section on bodies in rigid planar two-dimensional motion, and we went through all the kinematics. And we're in the kinetics section, we did Newton-Euler equations. Last time we finished up the work-energy principle by doing a problem in work-energy. And we're gonna do the last block of instruction now, which is applying impulse and momentum to bodies in 2D motion. And so we'll get started. Before we do so, however, I just wanted to make a little observation about how to approach these problems, because it can be difficult as far as, how do you start these dynamics problems? And so I've made some observations here that are really oversimplifications. But they can be at least a good starting point when deciding how to approach a dynamics problem. So in looking at the kinetics methods, when you're using the Newton-Euler method, free body diagram equals kinetic diagram, often the problem is given in terms of some forces and some accelerations. Remember, I do the FBD for my forces and moments. My accelerations are on the kinetic diagram, my motion vectors, and the interval in time is an instant in time. We usually look at a snapshot in time. And so, that's a good way, if you get a problem in these terms, that might be the way to go. With the work-energy principle, the problem is usually in terms of forces and velocities, because we're talking about energy, kinetic energy in terms of velocities. And the interval is some displacement between two positions. And then finally, we're gonna work on the principle of impulse-momentum which we also did with particle systems of particles. In those type problems, you're often dealing with forces and velocities again. But we're always looking over a period of time, cuz an impulse is forces acting on a body over a period of time. And so the interval is time. And again, these are oversimplifications. Obviously, if you get the acceleration, you can integrate to get velocities, or you can go back and forth, but these are a nice little kinda rule of thumb, if you will. Okay, so, for today's learning outcome, we're going to go ahead and develop the principle of impulse-momentum for 2D motion. And so we'll start off recalling the definition of linear momentum for a system of particles. Here I have a system of particles. Each one of those particles has a little mass times its velocity. I can sum them up over the entire body, and that's the entire linear momentum. We then extended that to a body of continuously distributed mass. And we saw that then the linear momentum is mass times the velocity of a very particular point, the mass center. If we take the derivative of momentum, we get for constant mass, classical mechanics, m dv dt, or m times the acceleration of the mass center, which we know is also the sum of the forces. And so we can also write the sum of the forces is the time derivative of the linear momentum for a continuously distributed body of mass. And so, we'll use that relationship to now develop the impulse-momentum by integrating both sides over some period of time. So I'm going from t1 to t2. I integrate the forces acting on the body. I get a change in linear momentum, so I have mv final- mv initial. And this is called the linear impulse, forces acting over some period of time. And this is the change in linear momentum. And so, similarly, for a rigid body of continuously distributed mass for rotation, we have moments acting on the body. We saw that the sum of the moments was equal to the time rate of change of the angular momentum about the mass center. And that form was good for two points, the mass center or about a point fixed in space. But you can see that it's analogous, there's an analogy here, forces and moments, time derivative of linear momentum and time derivative of angular momentum. And so with that relationship for angular momentum and moments, we can integrate both sides between two periods of time again. And we have the angular impulse, or the moments due to couples acting on a body over some period of time, about point C equals the change in angular momentum of the body about point C. And so as a recap, here's the linear impulse equals change in linear momentum. The angular impulse equals the change in angular momentum. And remember in general, for two-dimensional motion, this is the expression we came up with for the angular momentum. This was a product of inertia, angular velocity, product of inertia, angular velocity, mass moment of inertia, and angular velocity. But for symmetric bodies, we had the products of inertia vanish so that we stayed in planar motion. And so we just had the angular momentum was equal to the mass moment of inertia about the z axis through point C times the angular velocity, all in the k direction. And we can take that now, substitute it up into our expression for impulse-momentum for angular impulse and angular momentum, and we get this result. And so now we have developed and can put together next module and solve a problem, the impulses on the left-hand side, the momentum, changing momentum on the right-hand side. So we'll get started next time, and I look forward to seeing you then.