[BLANK_AUDIO] Welcome to Module 15 of Two Dimensional Dynamics. Today's learning outcome is to solve for the velocities of a planar rigid body in motion using the relative velocity equation that we developed in the previous modules. So here's a picture of what we developed before. We have a, a generic body, two points on the body, and the relative velocity equation relates the linear velocity of those two points on the body to the angular velocity of the body, the motion of the body, and its geometry. So, here's a worksheet of an actual problem. This is the piston cylinder mechanism that I used as a demo before. And this problem is is just a mirror image of this piston cylinder. I have a crank arm here, which body B 1. Which just is a, a portion of my wheel. But it's the same body. It's got a fixed point at O. So it, it's rotating about a fixed axis. I've got a, a connector arm here, B2, connected to a piston that undergoing linear translation. And I know in the problem that the, the crank arm has a angular velocity of ten radians per second clockwise, but in my, my model here, since it's a mirror image, it's counter-clockwise. And we want to find at this given instant, what the velocity is, the linear velocity is of the piston itself. Okay, so let's go ahead and talk about a strategy on how to solve this problem. Generally, I, I, I, I would think the best approach is to pick a body one of the bodies in the problem, and go from knowns to unknowns. And in, so, in this case the body I know something about is body B1, the crank arm. I know that it has a ten radian per second clockwise angular velocity. And I also know that it's velocity at point O, since it's a fixed pin, is zero. So let's go between O and A and write our relative velocity equation. So I have velocity at A, which is unknown is equal to the velocity at O plus theta dot from O to A in the k direction, crossed with r from O to A. And so in this case I know that the velocity of O is 0. I'm given that theta dot, or the angular velocity of the crank arm B1 is 10 radians per second clockwise. And so if I go 10 radians per second clockwise by the right hand rule, that is into the border in the minus k direction. So I've got plus, minus 10 K crossed with r from O to A in going, the position vector in going from O to A is minus 3 in the i direction and plus 4 in the j direction. So, that's going to be minus 3 i, plus 4 j. And so the velocity of A then is equal to k cross j is minus i, but I got minus 40 times minus i so that's plus 40 i. And k cross i is j minus 3 times minus 10 plus 30. So we get plus 30 j inches per second. Okay, so I, I, I now know the velocity of this point on body B1, the angular velocity of body B1, and the linear velocity of point A. I want to keep going toward the unknown that I want to solve for, which is the velocity of, of pin B, and so, let's continue on. And so here's my results so far. Now I'm going to go from what is now know as the velocity of A to velocity of B using the body B2. And so, now I have the velocity of B is equal to the velocity of A. Plus theta dot AB in the k direction crossed with r from A to B. And that's just an application of the relative velocity equation. We know what A is, we don't know what VB is, we don't know what theta dot AB is. We, we can find r AB from the geometry. Do we know anything? We want to reduce, as much as possible, the number of unknowns of a, a, as we can. So do we know anything special of the velocity of point B that will allow us to reduce some of the unknowns. And what you should have thought about and discussed, or thought about and, and arrived at is the velocity of B is in the i direction only. So, the only unknown for the velocity of B is the magnitude, not the direction. So I can then write, the velocity of B in the i direction is equal to the velocity of A, which we found to be 40i plus 30j, plus theta dot AB in the k direction, the angular velocity of body B2, cross with r from A to B. We're going to have to use a little geometry here. And so I know by Pythagrom's theorem rise is 4. The hypotenuse is 13 so this distance is 12. And so I'm going to go, and going from A to B I'm going to go 12.4 in the i direction and down, 4, so we're going to have cross with 12.4 i minus 4 j. 'Kay so let's continue on again. Here's the result that I have. So, I've got, let's do the cross products out here. We've got V B i equals 40i plus 30j. Then I've got k cross i is j, so I'm going to get plus 12.4, theta dot AB j and then k cross j is minus i. But I've got minus i times minus 4 times theta dot AB or plus 4 theta dot AB i. And now I'd like to think, you to think about, you know what do I do next. How can I finish up the problem? I want to find, I know the direction of the velocity of B. I need to find the magnitude of it. And what you should arrive at is that you should use match components like we have in, in, in, past courses and, past modules. So I'm going to match the j components to start, on both sides of the equation. And, on the left-hand side of the equation, I don't have any j components, so that's 0. Equals, on the right-hand side, I've got 30 plus 12.4 times theta dot AB. And what that allows us to arrive at is that theta dot AB, if I solve for it, equals minus 2.42, or theta dot AB, the angular velocity of bar B2 is a vector is -2.42K radians per second. And so what that means is, this bar B2 here, is turning with an angular velocity of 2.42k radians per seconds. Since it's negative K it's a clockwise rotation. Finally, we need to solve for the velocity of B. How do I do that? The magnitude of the velocity of B? And what you should say is, match the i components. And when I do that, on the left hand side i have VB. On the right hand side I've got 40 plus, j, j, i component four times theta dot AB. And I know that theta dot AB, I just solved for it, is minus 2.42. So, I arrive at VB equal to 30.3 That's the magnitude of VB, I know it's direction, so my final answer is VB, as a vector, is equal 30.3 inches per second in the i direction. And I've solved my problem. And I want to go ahead and give you a chance to solve a similar problem. So here is a worksheet for you to do on your own. I've included the solution in the module handouts, and I'll see you next time.
