Welcome to Module 20 of two dimensional dynamics. So for today's learning outcome, we're going to now extend these acceleration equations to a common mechanical part or device which is a wheel. We're going to look at the acceleration of the center of the wheel and the acceleration of the point of contact for a wheel rolling on a fixed straight surface. So here's a diagram, I've got my wheel, it's rolling on a flat surface. I've actually chosen my coordinates i to the right, j down, that way when I cross i would j k is into the board, or the rolling is positive in the clockwise direction. So let's go ahead and look at a demonstration. [SOUND] And so here is my wheel. I'm going to roll along here. And so, I'm going to start, I know the velocity of the point of contact for a no-slip wheel is going to be zero. So I am going to start from that known and go up to the unknown of the velocity of the geometric center of the wheel. And then I'm going to find the acceleration of the geometric center of the wheel. And then once I know that, I'm going to go from that known back down to find the acceleration of the point of contact for a no-slip wheel. So, even though the velocity of a no-slip wheel is equal to zero, we're going to find out that the acceleration of the point of contact is not zero. So, here is my my wheel. You'll recall the velocity equation, the relative velocity equation. And so I, I'm going to go from a known point, which is the ic for a no-slip wheel, where the velocity is equal to zero, up to the unknown point, which is the velocity of the center of the wheel. And so let's go ahead and write that. We've got the velocity of c now is equal to the velocity of the known point of contact, or the i c, plus theta dot k, crossed with r from the point of contact up to c is going to be a distance r, and it's going to be in the minus j direction. This is going to be minus r times j. And this, we said, was equal to 0. And so we get v sub c. Well, we know v sub c, we should find that v sub c is going to be only in the i direction, because it's rolling along a flat surface. So this is going to be v sub c in the i direction, k cross j is minus i. But I've got minus and minus, so that's going to be a plus. And so this is going to be r times theta dot in the i direction, or r omega in the i direction. And so that's the velocity of the center of the wheel. My question to you is how do I now get the acceleration of the center of the wheel? And your answer should be, let's just go ahead and take the derivative. Derivative of velocity is acceleration. So the acceleration is c is equal to I've got little r is constant, it's the radius of the wheel. i is constant in this case because the direction does not change, and so the only thing that changes is the theta dot. And so we're going to have, with respect to time, so we're going to have r theta double dot i. Or, we know that theta double dot is defined as the angular acceleration, so it's going to be r alpha in the i direction. And so now that's the acceleration of the geometric center of a wheel rolling without slip on a flat surface, or a fixed straight line. I did not label this c, and often I label the mass center as c, but this does not necessarily have to be the mass center. This is the geometric center. because, when I'm working with kinematics, I'm only working with the geometry of the motion, I'm not working with the kinetics. So I don't care where the mass center is. So this, this applies for the geometric center of the wheel. Okay, now let's take that acceleration of that center of the wheel and go down and find the acceleration of the point of contact which is unknown. So we're going to go from this new known to an unknown. You'll recall my equation for acceleration. And so I'm going to have the acceleration of the unknown, which is now the point of contact equals a of c which we know, we found, plus theta double dot of the wheel k crossed with r from known to unknown, minus theta dot squared, r from known to unknown. That should be i, not a very good looking i, but that's okay. And so I've got the acceleration of the point of contract, which I've labelled as the instantaneous center for zero velocity, is equal to now, I'm going to substitute in a sub c, which is r theta double dot i. And then I've got, plus theta double dot k, crossed with r from c to i now, is going to be in the positive j direction. So that's going to be r j, minus theta dot squared r in the j direction. r from c to a. And, if I cross k with j that's going to be minus i, so I'm going to get for this term, minus r theta double dot i. And, now you can see that this term and this term then cancel. And I end up with the acceleration of the point of contact equals just this last term minus theta dot squared r in the j direction. And so, that's the same as minus r omega squared in the j direction, because theta dot is the same as omega, the angular velocity. Or, since j is down, negative j would be up, so as an arrow, it would be r omega squared in the up direction, using an arrow. And so that's my acceleration of my point of contact for a rolling wheel on a, a, no-slip wheel on a flat, fixed straight line. And so even though the velocity of the point of contact is equal to zero, we see that the acceleration is not. And we'll come back next time and continue on.