Hi, this is module 18 of two dimensional-dynamics. Today's learning outcomes are to define angular acceleration, and then to derive what we'll call the Relative Acceleration Equation for a rigid body in planar or two-dimensional motion. And so first of all in defining the angular acceleration, we give it the symbol alpha. It's the time derivative of the angular velocity which is given the symbol omega, or the second time derivative of the angular displacement. So recall for velocity. Here's our body undergoing planer two dimensional motion. We have a body with two points in the body. We relate the linear velocity of those two points with the angular motion of the body. We found the absolute velocity of Q is equal to the absolute velocity of P, plus the angular velocity of the body crossed with the vector, the position vector from P to Q. Now to find the acceleration or the relative acceleration equation what do we do? And what you should say is we simply differentiate. And so I've got the velocity of Q dot, is equal to the velocity of P dot. And then we need to use the product rule here. So we're going to have theta double dot k crossed with rPQ plus theta dot k crossed with rPQ dot. So the derivative of the first crossed with the second plus the first crossed with the derivative of the second. And so, again, if you recall back to my relative velocity module, this is our fixed reference frame. It doesn't necessarily need to be inertial because were only talking about the geometrical aspects of motion. It would only need to be inertial for the kinetics where we applied Newton Euler's laws. But we say that the derivative of the velocity of Q, and the derivative of the velocity of P, since they're fixed in the reference frame, are called what I'll call the absolute acceleration. So this becomes the acceleration of Q Is equal to the acceleration of P plus again, theta double dot k, cross with rPQ, plus theta dot k, cross with rPQ dot. Okay, so here's the result we just came up with, absolute accelerations. We have theta double dot k, crossed with rPQ. On this term we're gonna need to be a little bit careful again, because we're taking the derivative of the position vector from P to Q, which is not fixed in the reference frame. And so we have to be careful on how we take that derivative, so let's look at this last term. And I've got theta dot K, or so I'm just looking at this term, crossed with rPQ dot is equal to theta dot k crossed with now rPQ dot, we did in an earlier lesson, when we were looking at relative velocities. We have, theta dot k, crossed with rPQ itself. And so this, this, Is from an earlier lesson. Okay, and so now let's simplify that some. And to do that, we're going to use a vector identity. This is a vector identity much like a trig identity. A x B x C = B(A dot C), -C (A dot B). And, so in this case, I'm going to call this my A vector, theta dot k. I'm gonna call this my B vector, theta dot k, and this my C vector, theta dot R from P to Q. And so that's A cross B cross C. And so I have now theta dot k crossed with rPQ dot is equal to. Okay I've got B, which is theta dot k, times A, which is theta dot k. Dotted with C, which is rPQ. And then I have minus C, which is rPQ times A, which is theta dot k, dotted with B which is theta dot K. And now I would like you to tell me what that term might be. This term right here. Do that on your own. And what you should say is okay, k is coming out of the board. RPQ is in the plane, the XY plane. So those are perpendicular vectors. And since they're perpendicular vectors when you take the dot product, that's going to be equal to zero. So we've got zero because k and rPQ are are perpendicular vectors. Okay so I just put are perpendicular. And so we end up with now, theta dot k crossed with rPQ dot is equal to theta dot k dotted with theta dot k is equal to what? And theta dot k dotted with theta dot k is just the square of the magnitude. So that's theta dot squared. And so I have minus theta dot squared times rPQ. And so, now I have an expression for this last term. And so this can go up into here. And that will then be what I call my relative acceleration equation. So I've got the acceleration of Q, is equal to the absolute acceleration of P plus theta, double dot k crossed with rPQ. It's that term. And then, this last term is simply, minus theta dot squared rPQ. And that's my relative acceleration equation. We call this my tangential acceleration term. And we call this my normal, or the normal, or radial acceleration term. And then we'll use that relative acceleration equation to look at the equation, the linear accelerations of two points on a body undergoing relative motion as related to the motion of the body itself. And we'll do that in a future module.