[BLANK_AUDIO]. Welcome to module ten of Three Dimensional Dynamics. Today we're going to go ahead and solve for the velocity and acceleration of bodies undergoing three dimensional motion. Expressed in moving frames of reference. And so here's the theory we came up with from the last several modules. We did the velocities in moving reference frame. That's the equation. We did the accelerations in moving reference frame. This is my generic example with frame F fixed, frame B the moving frame, and a point P. We applied it to a number of different situations, and now we're going to apply it to another situation. You know, music's come a long way these days, and so now we have these MP3s and digital music. Prior to that, if you're as old as I am, we had CDs, in fact, there's still some CDs around. Prior to that, we had things like 4-track cassette tapes and prior to that, we had things like, 8-track tapes. Some of you may not even have heard of 8-track tapes, but I remember when I was back in school, in my younger days. I was very proud when I got an 8 tat, track tape player. And even prior to that we had vinyl records. And here's a a picture of a vinyl record, and we're going to solve a problem using this, this rotating vinyl record. In fact vinyl records, I guess, are coming back now some. So let's go ahead now and solve a problem, using this as a kind of framework. A real world problem. So here, here is a, a picture of a vinyl record player. It's says here I'm a rambling wreck from Georgia Tech. That's the music being played. And I've got a little yellow jacket here. It doesn't look much like a yellow jacket, but that point, p. That's a little yellow jacket. So the, fr, the frame B, or body B, is going to be the vinyl record itself. Frame F is going to be the ground or the record player. And I'm given that the vinyl record is spinning with an angular velocity of mega, in the k direction. Which is the k direction is pointed up and its constant. And the velocity, of the yellow jacket. The relative velocity with respect to that of moving frame. Or body b is going to be some value u, could be you know, whatever meters per second or inch per second in this case. And it's going to be in the I direction and it's going to be constant. And what I'm going to find is what is the velocity of P with the respect to the ground. And what is the acceleration of P with respect to the ground. I'm going to have my fixed frame F which is fixed to the ground. And I'm going to put right over top of it now my moving frame B which is welded to the record itself. And so, it's going to turn with the record whereas F is going to just stay put. It's my fixed frame. Okay, so for the velocity of P with respect to B, here is my expression. And so, I've got, why don't you go ahead and try it on your own, and then come on back and we'll, we'll do it together. And so we've got the velocity of P with respect to F is equal to the velocity of O prime. Which is the velocity of the origin of the moving frame with respect to F. And so if I'm on the ground looking at the origin, the center of that record it doesn't move. So this is going to be 0. And then we had the relative velocity or the velocity of P with respect to B. Now, if I'm standing on the record and I'm turning with the record. And I'm looking at the yellow jacket point P, it's velocity now. Is given as being U in the I direction attached to my reference frame. So it's given as being U in the I direction as it walks out. And then, plus we're given that the moving frame is turning with an angular velocity of mega K crossed with R from O prime to P. And the moving frame that's a little distance R in the I direction. And so my velocity now, of P with respect to F, is simply ui plus k cross i is j. And so that's going to be r omega j. And that's my answer for my velocity of P with respect to F. 'Kay, let's now move on to accelerations. So here's my expression for acceleration. Again I'd like you to try it on your own, then come back and we'll do it together. And so, we have the acceleration of P with respect to F. Which we want to find is equal to the acceleration of O prime with respect to F. And that, if I look at O prime with respect to F. Again, I'm on the ground I'm looking at the center of the record, and from my perspective. It does not move and so it's got got 0 acceleration. Then I have got the acceleration of P with respect to B. If I'm on the record and I look at point P or I look at the yellow jacket. It is moving out, what is it's acceleration? What is it's relative acceleration? What you should say is 0 since the relative velocity was constant. So its going to be plus 0. And then how about this term. What's it going to be equal to? And if you look at the angular acceleration of my record with respect to the ground, frame b with respect to the ground. Well since the angular velocity was constant, this is going to be zero for the angular acceleration crossed with r and so that's also a zero term. Now we have my Coriolis term which is 2 omega B with respect to F is given as a mega K crossed with V of P with respect to B. We said was u of i. And then we've got plus omega B with respect to F is, again, omega K, crossed with omega K, crossed with r, was little r in the i direction. Okay, so there's my expression. So I get the acceleration of P with respect to F is equal to K cross I is J, so this is going to be 2 omega u in the j direction. Plus omega K crossed with okay K cross I is J so that's going to be R omega J. And so finally I get a of P with respect to F is equal to two omega u, j. And then I've got k cross j is minus i so this is going to be minus r omega squared i. And that's my acceleration of my yellow jacket with respect to the ground and so we see that it's two terms here. There's this Coriolis acceleration term. And there's this normal, or centripetal acceleration term. So let's look at these two terms more closely. And let's have a discussion of something that can be a little difficult troubling but you can grasp it here if you follow along with me. And that's the Coriolis acceleration. And so here's, here's my acceleration of P with respect to B. Let first of all look that situation. Here's, here's my top view of the record, so here's the center of the record, here's point P. Let's first of all look at what goes on. If P is just sanding still, if there was no, if the yellow jacket just standing still and there was no relative velocity. In that case, this point P would go round and round with a record. And the only acceleration term I would get then would be my so there's no u here, there would be no relative velocity. And so all I'd have was the centripetal or normal acceleration term. So that would be an acceleration that's directly railing inward. And that's omega squared R would be its magnitude. So, if the yellow jacket wasn't moving we'd only get centripetal or normal acceleration. But now when we do have both angular velocity of my moving frame. And we have relative velocity we're going to get this Coriolis acceleration term as well. So again, we're going to look at a top view. As the yellow jacket moves out. After some period of time as it's walking out it's going to end up over here. But as it walks out because the frame is turning or the record is turning. If it has shoes on, the yellow jackets walking out and has shoes on that have friction with the record it's going to be pushed along. With the the, the, the record itself. And there's going to be an acceleration corresponding to that force and that's our Coriolis exper acceleration. If it wasn't there. If say the, the yellow jacket had what we'll say is friction less shoes. What would happen is that the moving frame would slip out from underneath? The yellow jacket and move out with the velocity of u and i and its friction on shoes it would just end up moving straight out. And the moving reference train would slip out from underneath it. And so if the yellow jacket has relative velocity. And it is in a moving reference frame, we have this Coriolis acceleration term. And so let's look at a couple of other exam, I'm going to go through a couple other thought experiments with you. To see if we can again, grasp this Coriolis expression. Let's say that we have, and we're just doing an imaginary thought experiment here, let's say we had a really, really long. A bowling alley, from maybe the north pole down to, the equator. And when, say we rolled a gutterball down that, equator, or excuse me, down that, that bowling alley. So we had a ball rolling down the gutter. And, if it was rolling a flat surface with no, with no friction it would, the moving reference frame would turn out from underneath it. But, with the gutter it would go straight out. And there would be a force. Being applied to it by the gutter as, as, as the reference frame turns. And the force is accelerated, is excuse me, the force is associated with this turn we call the Coriolis acceleration. One other example, let's say we're flying in a plane from Alaska down to California. And in so doing we have to account for this Coriolis acceleration to end up in California at the end of our journey. Because as we take off, the earth is turning and California would end up over here. So if I kept on the same line. Is what I started off with I, I miss California and we land in probably the ocean. And so, think about that, read a little bit about Coriolis acceleration on the internet and see if you can grasp that and And understand that it occurs again when you have a relative velocity term and you have a moving reference frame. So, that's the theory. I'd like you to go and we did a problem together. I'd like you to also do a problem on your own. This that crane problem again. You'll recall that we solved for the velocities. And now I want you to solve for the ac, angular acceleration of the boom B with respect to the ground. And the linear acceleration of point P with respect to the ground. And I've got the solution in the handouts, and that's that's it for this module. [BLANK_AUDIO]