This is module 34 of two dimensional dynamics. And today let's start off by looking back at where we've come so far. The first part of the course was looking at particles, systems of particles, and the ki, kinematics and kinetics of those systems. And now we're into bodies in a planer rigid body two dimensional motion and we did the kinematics portion a looking at the geometry of the motion and now we're into the kinetics portion and we've already done the Newton/Euler equations. And the next step is to do work energy principle, we had done work energy principle for particle systems of particles. Now we're going to extend that to bodies in rigid planar two dimension motion. So the learning outcome for today is to develop the work and energy principle for bodies in 2D planar rigid body motion And, I want to start off by recalling, when we worked with particles, if I took any point in a body let's say it's a differential piece of mass out here, DM. Then its kinetic energy is going to be one half mv squared. One half DM, in this case for the differential piece of mass. Times its velocity squared. Now, if we want to integrate that over the entire body. We have T equals one half the integral of v dot v as a vector, dm and so, now we can relate the velocity of any point on the body to the velocity of the mass center. And so we can use the relative velocity equation, this was the relative velocity equation that we developed earlier, and, we can say okay this is any point here, DM. Remember, if this body is in rotational planar motion, that all the points along this line parallel to the Z axis is going to have the same, going to have the same velocity as the companion point here, in the plane of rotation which I'm going to call Q. And so we're going to relate the motion of Q to C and that will, that, that point Q will apply for any point along a line that's again parallel to the Z axis. So, I've got V of any point is V of C. And plus, how the body is rotating crossed with the position vector from c to q. So, we go out a distance x in the x direction and up y in the j direction. So, xi plus yj. And if I do that cross product, this is the result I get. And so, I now have my expression for kinetic energy for a body, and I have the through kinematics the velocity of any point on the body. And I'm going to have to dot that, because all these points are going to have to go into my equation for the kinetic energy. So I'm going to have to dot v of any point with v of any point. And so I've done the math here. I'll let you go through that on your own. And then I substitute that back in. [COUGH] Excuse me. My expression for kinetic energy. And this is what I arrive at. And let's see what we have for each term. We know that VC dot VC will be the magnitude of the velocity of the origin, of, excuse me, the velocity of the mass center of the body squared. The integral of DM, what is the integral of DM? Okay. What you should say is the integral of dm is just the mass of the body, and then I want to look at this term the integral of x squared plus y squared dm. You should recall from a previous module what that's equal to. And that's defined as the mass moment of inertia for rotation about the z axis. And then we have these two last terms, but we have the integral in, in these two terms, we have the integral of x dm and the integral of y dm. Well, if you go from the reference of the mass center and you go out a distance x to x, x. To each little piece of mass in the body, and you integrate over the bod, the entire body, that's going to be, by definition of the mass center, equal to zero. The same thing in the y direction. And so, what we are left with is then, these two terms dropping out at the end. And we have one half M, since we've integrated over the entire body. The velocity, the magnitude of velocity of the mass center squared plus one half I sub ZZ through the point at the mass center, a mega square. And so, I'll just abbreviate since we're only working with planer motion I'll just abbreviate this mass moment inertia is I or the mass moment inertia about point C and this is the translational kinetic energy which we've seen before and this term is now what we'll call the rotational kinetic energy. So, now a body that's able to not only translate but to also rotate, we'll have translational kinetic energy, and rotational kinetic energy. And you can see that the correlation of the analogy between these two terms. You have one half, one half M for translation, I sub c for mass moment of inertia, for the rotation, v sub c for the linear velocity and omega for the angle of the loss so there's a direct correlation there. And we'll go ahead and pick up at that point in the next module.