[BLANK_AUDIO] Hi, this is module 28 of Two Dimensional Dynamics. And I'd like to take some time now and talk about the what, what I refer to as the forest versus the trees. Sometimes we get lost in the trees when we're doing some specific problems and doing some real specialized stuff. So I want to pull back for a second and look at the forest of, of, of where we're at in, in, in the course and where does this all fit in. And so we're talking about engineering mechanics which ties together the basic math and basic phy, physics to engineering applications like mechanical engineering, civil, aero, all the different engineering applications. In engineering mechanics we are looking at rigid bodies in the study I have done so far I, in my previous courses we looked at static equilibrium and that was my courses in Introduction Engineering Mechanics and Applications in Engineering Mechanics. In this course, we are looking at dynamics, and so and we broke dynamics into talking about the kinematics or the geometrical aspects of motion relating things like position, velocity, acceleration and time. And kinetics, where we now put forces on that body, and we look at the motion of the body which we can describe with the kinematics, and so here is an overview of the topics from the course. We, in the first part of the course looked at particles and systems of particles, and now we are into the planar 2D rigid body motion portion of the course. We have just now completed, as of the last module, all of the kinematics. We can completely describe the motion of bodies in planar 2D motion, for all the geometrical aspects. We took, talked, talked about relative velocity and relative acceleration between two points in the same body. We looked at the instantaneous center of zero velocity concept. And then we did the velocity and accelerations for the same point between two different reference frames. And now we're going to use of all that information and apply it by putting forces on bodies and seeing what kind of motion results. And we're going to use the same types of kinetic methods that we did for particles, systems of particles. Newton, Euler equations, then the work-energy principle and finally impulse and momentum. So, today we are going to do for our learning outcome of the equation for the kinetics of a rigid body, and first of all we are going to look at a rigid body that simply translate. And so, this is a review, I want you to recall our development of Euler's second Law or what's called the Moment Equation, which you'll find in an earlier module. We started with Newton's second law, Newton's Law applied to particles, we said that a particle has some linear momentum, mass times velocity, this is mi vi for the ith particle. You take the time derivative of that momentum and you get the it has to, it, it, it balanced with the forces that are acting on that particle, the external forces acting on that particle. And then we made that we took a system of particles now. We took several particles together and we crossed, we took and we crossed the, both sides of this equation with a position vector from some point P to each of these particles. And we crossed it on both sides and we came up with r cross F we said was defined as the moment about this point P. And then, on right-hand side, it was the time derivative of the summation of r cross mv for each of the particles. Or we said that was the moment of momentum, or what's often called the angular momentum. And then we can extend that even further, we can go ahead and take our system of particles and add so many particles in that this becomes a continuous body. And then our summation signs become an integral. So I've got the integral of r crossed with, and if I take the time derivative of v, m is constant, and we get r crossed with a, the time derivative of v, or the acceleration dm. Okay. So, that's the development, that's a review of the moment equation. So from there now let's look a the translational motion of a body. If we have external forces acting on it and it translates. Then for translation, all points on the body are going to have the same acceleration. An acceleration would be, since acceleration is the same for every point on the body, we can factor it out of the integral. And I'll move the dm over here and then I have the integral of rdm and you should recall back. In fact I'll let you answer that on your own. What, what does that equal to? And what you should say is, okay, that's the same as if I take r to each little piece of mass and integrate, that's going to be the total mass times r from p to the mass center. That's by the definition of the mass center. And so this is the result I get Let's just do one more step, which is, if we look at a very particular point, point C where P is equal to C, so r from P to C is equal to zero, then we find out for, a body in translation that the sum of the moments about the mass center has to be equal to zero. And so we will pickup next time and do a problem.