[BLANK_AUDIO]. Hi this is module eight of three dimensional dynamics, and today's learning outcome is to go ahead and use the development that we did last time to solve for the velocities of bodies undergoing three dimensional motion and expressed in moving frames of reference. And so here's the the equation from the velocities that we came up with last time. You'll recall from this video of an earlier, module, that, we've been looking at, a landing gear, retracting up into an aircraft. I had a simplified, or a less complex, system than that. But it still can be analogous to a, a landing gear, retracting up into an aircraft. We have the wheel itself spinning, and then the landing gear itself pulling up into the body of the aircraft and the body of the aircraft itself. And so we do have here a situation with several different moving frames. We have our ground if you will it's analogous to the Earth but in this case it's actually the airplane itself, the fixed fram attached to the aircraft. And now I want to find the velocity of a point on the landing wheel. And I've got, got it located here, point P with respect to the aircraft itself. And to do that, we're going to have to move through the a, a, a couple of frames here. Because we have a moving frame that's attached to the landing gear itself. It's moving with respect to the aircraft and then we also have the wheels spinning. And those are spinning with constant angular velocity. So omega-1 down here and omega-2 retracting up into the aircraft. So we're going to approach this problem a step at a time since it's involving multiple moving frames of reference. And to begin, I'm going to attach what I'm going to call my moving frame to the wheel, and I'll call that frame C. And for this, for this portion of the problem, I'm going to call B my fixed frame. And so I'm going to first find the velocity of P with respect to the B frame. And then I will then, from there, transform to find the velocity of P in the F frame, which is our ultimate answer that we're looking for. And so here's the generic situation that I have. Now I'm using frame B over here and my moving frame is frame C. Here's the expression for the velocities. The velocity of P in B now is equal to the velocity of the origin of the moving frame C plus the velocity of P with respect to C plus, the angular velocity of C, since its now considered my moving frame, crossed with the position vector in the moving frame which is R from O double prime to P. And so, lets look at these terms one at a time. The first term I want you to tell me is what is, what is the velocity of O double prime with respect to the B frame? And what you should say is, okay that's going to be o. If I shrink myself down and I'm on the point, or excuse me, the frame B, and I'm looking at this point O double prime, as the frame B turns from this vantage point of the landing gear, that point O double prime is not moving. So this is 0. Next, I want you to find the velocity of point P with respect to the C frame. And again, it can be a difficult concept, but if I shrink myself down and I'm standing on the frame C, the wheel, and, and I turn with the wheel, as I turn with the wheel and I look at point P from my vantage point, if I have blinders on, I can't see anything else around me. Then, to me, point P is just standing still. And so that relative velocity is equal to 0. Next I want you to find the angular velocity of the moving frame C with respect to B. And in this case by the right hand rule the moving frame C by the right hand rule has an angular velocity omega-1 in the I prime direction but you see that's also in the I direction. I'm going to put all of my vectors in terms of the B frame. And so I've got omega-1 in the I direction, and then crossed with V R from double, R from O double prime to P is just the little distance R, and again at this instance in time, we're solving for the velocity at this snapshot in time. That's going to be little R in the J direction if I put everything into the coordinates of the B frame. Okay, so now i can do that mathematics and I get the velocity of P with respect to B is equal to I cross J is K, so I'm going to get R omega-1 K. Okay. Let's go on from there. [BLANK_AUDIO]. So now, I have found the velocity of B, or excuse me P, with respect to the B frame. Now I want to move it back to the velocity in the F frame since that's the question that I'm asked. And so, here again is the generic situation. Now my fixed frame is F. Now my moving frame is frame B, attached to the landing arm. And here's my expression for the the velocities. The velocity of P now in F is equal to the velocity of the moving frame B origin, which is O-prime plus V of P with respect to B or the relative velocity, plus the angular velocity of B with respect to F crossed with the position vector now in this moving frame, which is from O-prime to P. And so let's go ahead and apply that theory to our problem. And my first question is, what is V of O prime to F? And again, if I'm standing on F in the airplane and I'm looking at O prime, from my vantage point, O prime does not move. It's got a zero velocity. Oops, so that's zero. Now let's find the velocity of P with respect to B. Well, we don't have to worry much about that because we just did that. That's R omega-1 K. And then I want you to write the angular velocity of the moving frame of B with respect to F. And if you do that. Okay, we've got omega-2 by the right hand rule. We going to put everything into coordinates for the B frame. So it's in the little j direction. So this is omega-2 J and then crossed with R from O prime to P in our moving frame when B is considered the moving frame. So in that case I'm going to go a distance little r in the J direction and then big R in the K direction. So this is little rj plus big RK. And as I said, I'm putting this all in the coordinate system for frame B and we're looking at this as a, as a snapshot in time, but realize and I've done this before in other modules, that I can always change from one set of coordinate system to another set of coordinate systems. And I even did that back in my two dimensional dynamics class and so. If you, if you want to change to and you can even express the velocity for all times in, in, in, in, in, in, in the fixed frame or how ever you want to do it. Just for convenience in this problem I'm going to put it into the B frame, to express our answer. And so let's go ahead and finalize this problem. We've got V of P now with respect to F, is equal to, okay, we've got R, omega-1 in the K direction, that's that term, and then I've got J cross J is going to be zero, J cross K is I, so this is going to be plus capital R omega-2 in the I, oops, I direction that should be an I. Okay and so that's my answer. I'd like you to go ahead and try a problem of your own, here's a typical crane problem. Here's a video of a crane not exactly the same as this crane, but same sort of of concept destroying a building here. And I'm also going to show here some photos of other cranes. again, each of them have a little different configuration but they're similar to the crane I show here, in fact, the last picture here I show is a rather, old crane back in the days, the dark days when maybe I was born. And so this is a problem, a real-world problem, I like to approach these as real-world problems. We want to find now the angular velocity of frame or body B with respect to G and we also want to find the velocity of point P, which is up here at the end of the boom with respect to G. We're given that the crane itself, the crane body rotates about the vertical axis with a angular velocity 0.2 radiants per second, that's constant. Simultaneously the boom B is being elevated at an increasing rate of 0.1t radiants per second. And the boom was along the Y axis when T was equal to zero, and you want to move these points and these velocities all the way back to the ground which is G. And once you complete that I've got the solution in the module handouts, and if you can do this you should have a pretty good handle on solving velocity problems, for velocities expressed in several moving frames of reference. [BLANK_AUDIO]
