Hi and welcome to module 25 of two dimensional dynamics. Today's learning outcome is to solve an actual engineering problem for the velocity of the same point relative to two different bodies or reference frames, moving in, in planar two-dimensional motion. And so this is where we left off last time, we developed this equation for the velocity between two [COUGH] of the same point in two different reference frames. and, now we're going to look at a a typical engineering mechanism here. In this problem, I'm going to call my fixed frame F the Earth. And I'm going to actually call the moving frame. I'm going to attach my moving frame or weld my moving frame to this vertical bar B. And so that corresponds to the situation here, and the common point that I'm going to look at same point I'm going to look relative to these two frames is going to be this point P. So here is our our problem. We've got this collar C pinned to bar R. And it's free to slide along the vertical bar, which I'm going to call, my moving frame B. And we're given the angular velocity of r to be 0.2 radians per second clockwise, at the instant shown, snapshot in time. We want to find the angular velocity of the vertical bar at that time, and we want to determine, the relative velocity of P with respect to the moving frame, B. And so, this is a rather difficult concept, so you may have to, review this module a couple of times, but we're going to have our fixed frame F, fixed to the ground, that's going to be capital X, capital Y. And then, I'm going to have a frame, a moving frame, welded to the bar B. And so it can move with the bar B and it's going to be little x, little y. And so, the green frame moves with respect to the blue frame which is fixed to the Earth. Now, as this system moves to the right, if I were looking at this from the perspective of the moving frame, if I shrank myself down, and I looked, I was standing on this bar B and I looked at this sliding collar, with point P, from my perspective, P is only going to move In the j direction. And so that's what I mean by relative velocity. So let's go ahead and use our relative, our, our velocity equation that we developed last time. And we've got V of P, now is equal to V of the origin of the moving frame, which is O prime plus V REL. So V REL again, that's the relative velocity of P in my moving frame which, from the perspective of bar B, can only be in the j direction. So this is going to be R REL j plus omega B which is the, angular velocity of the moving frame and I don't know what that is. So that's going to be an unknown omega B in the k direction crossed with R in the moving frame. That's this distance from O to P. You're given this distance is 0.5, you're given this angle of 30 degrees. And you can find by geometry that this is 0.866 in the little j direction. So this is going to be time, or cross 0.866 in the little j direction, or, the velocity of P, is equal to well, V [COUGH] O prime is zero because the origin of my moving frame is fixed to the ground at this pin. And, then we have, V REL j plus, k cross j is minus i so this becomes minus, 0.866 omega B in the i direction. And so that's, that's an application of that that equation. So let's continue on. Okay. Here I've shown my fixed frame, fixed to the Earth, my moving frame, X Y fixed to body, this is body B now. And, I want to go ahead now and look at point P from the perspective of bar r. Now, very important point here, from the perspective of bar R, this is between two points on the same body. Okay? So there's no relative velocity in this case. So we go back to our, our normal relative velocity equation between two points on the same body. And so, for member R, I've got, V of P is equal to, and, here's the relative velocity equation for two points in the same body. The known, the unknown point is going to be V of P in this case. The known point is V of O. Which is a fixed point. because this is a RAFA body now. Rotation about a fixed axis. Plus Omega of R with respect to k. Well, we're given that omega R is, R is rotating clockwise. So that's going to be in the minus k direction. This is minus omega R in the k direction, crossed with R from O to P. And so we've got V of P, equals, again, V of O is 0. So I get minus, 0.2k radians per second, that's the angular velocity of bar O, R crossed with R from O to P. We go minus 0.5 meters in, the little i direction. And plus 0.866 because this distance can be found by geometry to be 0.866 in the j direction. So this is minus0.5i plus 0.866j. And if you do that cross product, you get the velocity of P, from the perspective of member R, is equal to, 0.1j plus, 0.1732i. Okay. So, let's continue on. I now have two equations for velocity of P. And so, I can equate those. [BLANK_AUDIO] And I'm going to get V REL j minus 0.866 omega B i equals 0.1j plus 0.1732i. And my question to you is, how do I finish up this problem? And what you should say is we equate components, like we have in the past, so let's equate the i components. On the left-hand side, we got, we have for the i components is, -0.866 omega B. On the right-hand side we have 0.1732. And so that gives us omega B equals minus 0.2 or the angular velocity of bar B is minus 0.2 in the k direction, radians per second. So what we find is, and that's one of our answers. What we find is, the angular velocity of bar B is actually the same, this is bar B, is actually the same as the angular velocity of bar R. And so that's, that's, one of our questions that was asked. To find the velocity of point P relative to the vertical bar, we just equate the j components, and in this case we have, V REL on the left-hand side, equals 0.1, on the right-hand side. And so V REL, as a vector, is equal to 0.1, and as we said before from the perspective of the vertical frame, or vertical bar, or, or the, moving frame B, it's only moving in the j direction. And its units are going to be meters per second. And so that concludes the problem. Now as I mentioned these are fairly difficult problems. I will I'll be honest with you. And so it takes a lot of practice, and to, to do these problems. And so what I've done is, I've included a couple of more worksheets for you to do. And I've also included the solutions. they're, they're real world type engineering mechanisms and so I'll let you go through, take your time, go a step at a time, try not to look at my full solution before you try it on your own. Kind of step yourself along because if you look at the solution, you'll say oh, I can understand how that goes, but if you try and do it without looking at the solution. You'll, you'll, you'll, you'll see where you, where you get stuck and, and kind of help yourself along. So that's one problem. Then I have another problem here. And those will be great, but if, if you're able to do these two problems then you'll pretty much have in control, the concept of looking at the velocity of the same point, with respect to two different bodies, or two different reference frames in 2D motion. And with that, I'll see you next module.
