Hi and welcome to module 23 of two-dimensional dynamics. So far we've looked at the velocity and acceleration between two points on the same body. So it's all one body. Today we're going to look at situations where the velocity can be looked at of the same point expressed to two different reference frames or two different bodies in planar motion and later we're actually going to do accelerations as well. We're also going to derive something which is called the derivative formula, which is a very important tool that we'll be using for the remainder of the course, for taking the derivative of an arbitrary vector expressed in a moving frame with exp, with a respect to another frame. So, here is my generic situation of the velocities of the same point relative to two different frames or bodies. So let's take a point P out here, P. And I have a frame F, and I have a frame b. And we can look at the motion of P with respect to the frame or body B. Or the frame or the body F. And so I will often times talk about F as being what I'm going to call the fixed frame, and B as being the moving frame. But when I say fixed frame, it doesn't necessarily have to be fixed in an inertial reference frame. It only would have to be fixed in an inertial reference frame if we were talking about kinetics where we're going to apply Newton and Euler's laws. Right now we're only dealing with kinematics and so we're only looking at the geometry of the motion and so when I refer to F and B, those are just two different frames. In fact, If I had several bodies connected together, I could link them back together two frames at a time, and keep moving, moving the vectors, in, in different reference frames. And so, there are several situations that we're going to encounter where you have multiple bodies and you want to look at The kinematics of similar points in two different framers or bodies. here's, here's one example. As you know I'm a, I'm at Georgia Tech and our mascot is the Rambling Wreck or actually the yellow jacket and so he's singing our song, the Rambling Wreck from Georgia Tech here. Yellow jacket moving out on a vinyl record, and so it has some motion, and we could in this situation look at the, the motion or point P of the yellow jacket with respect the vinyl record, or we could look at it with respect to the frame F which I would have fixed to the earth. Here's another example of a mechanical device and we're actually going to solve this problem later on. But I can look at this motion, this point P is attached to bar R, and then there's a slider mechanism, that can slide up and down on bar B. And so we could call in this case, our frame F the ground. And we could call the moving frame, or body B, this vertical bar. Because P is going to move relative to that vertical bar as well. One more example this is labeled body C but I'm going to call it body B cause it's going to be my moving frame. I've got this cylinder. Here's a, here's a example of what I will call a fix frame but it's not fixed in an inertial frame it's, it's able to rotate. So here's my frame F, here's my frame B my moving frame and we can look at from the perspectives of both of those bodies or frames. This common point P and relate the velocities in that manner. And so we'll go ahead and do so now. Let's go ahead and develop the, the theory here. I've drawn my coordinate system for the frame F, capital I, capital J. And I've drawn the coordinate system for the frame B, which I'll call little I and little J. And we're going to relate the two, so I've got Little I is equal to, for little I I'm going to out a distance, cosine theta in the big I direction. And you should have, you should be really familiar with this, we've done it several times before. And then we're going to go a distance sine theta in the j direction. Big J direction, and then for little j. We're going to go minus sign theta in the big I direction. And we're going to go positive cosign theta in the big J direction. And, now let's differentiate those, Vectors, unit vectors in the F frame. Because I and J are going to move with respect to the big I big J frame. And so if I differentiate in the big frame we get, the derivative I respected time for the frame F. Is equal to I dot, which is equal to now the derivative of cosine theta will be minus sign, theta, theta dot. In the big I direction and plus sine. Derivative of sine is cosine theta Theta dot in the big J direction. And that equals, well let's look here, minus sine theta big I plus cosine theta big J is the same as little J. And so this becomes theta dot J. So the derivative of I with respect to T in the F frame is equal to theta dot J. We'll do the same thing for J. At D J, D T and the F frame is equal to J dot which is equal to, okay the derivative of minus sign theta is minus cosign theta. Big I times theta dot. And the derivative of cosine is minus sine theta. So this is minus sine theta big J. We can see that the stu, the information in parentheses is negative I, and so This becomes minus theta dot i. Okay, so we have the derivative of i and j with respect to the frame f. So the derivative of the univectors in the moving frame with respect to the fixed frame. And so, here are those results up here. So now let's look at an arbitrary vector expressed in my moving frame B. And we'll call that vector A. When I say arbitrary vector, it can be any vector, and we know vector quantities, for kinematics it could be position, velocity, acceleration. Any of those vectors could be represented by A. And so let's go ahead and express that vector A, in the moving frame. In the little I J components, and so I've got A, is equal to the X component of A, in the little I direction, plus the Y component of A in the little J direction. And now let's differentiate we'll be careful now. And differentiate that vector that's expressed in the moving frame with respect to the fixed frame. And so I've got A differentiated in the F frame is equal to. Okay we've got. A X dot I plus A Y dot J. And then we've got plus I also have to, by the product rule, take the derivatives of little I and little J because they, those unit vectors change with respect to time. When looking at them from the perspective of the F frame. So I've got plus A sub X times the derivative of I with respect to time, taken with the F frame. Plus A sub Y times D sub J DT. Taken with respect to the F frame so A sub, or A with respect to the F frame equals. Let's look specifically at these two terms. This says A dot x I plus A dot y j. So from the perspective of the little or the moving frame B if I was looking at. The vector A, from this moving frame B, so imagine yourself shrinking down onto that body B, looking at, the vector A. You can see that in that reference frame little I and little J do not change directions. The only thing that changes could be the magnitude of A, and both the X and Y Directions or components. And so this becomes, the derivative of A, with respect to the B frame, or my moving frame. Plus I can substitute now the, the derivatives of the unit vectors i and j in the moving frame that I found from the last slide and so I get plus Theta dot times AXJ, plus or excuse me, minus AYI. That's just substituting these into here. Let's see we'll substitute Di Dt into here, and Dj Dt into here. And that's the result I get. And then I can just make a mathematical manipulation which says that this is equal to A dot in the B frame, plus I can write this as theta dot K, crossed with AXI. Plus A, y, j. And you can see that these are equivalent. K cross i is j so this Ax theta dot j is this term, and then K cross j is minus i, so that becomes Ay theta dot times i. And we can also realize that now Ax. I plus AYJ is our original vector A. Expressed in the moving reference frame. And so our final result is A dot in the F frame is equal to A Dot in the B frame plus theta dot K crossed with A. And that's a really important, formula which i'm going to call the Directive Formula. And we'll use it a lot in the, in the, in the remainder of the course. What it says is, if I take the derivative of a vector in a moving frame, expressed in a moving frame with respect to another frame the derivative of the vector in the F frame is equal to the derivative of the vector in the moving frame plus theta dot K which is the angular velocity of the moving frame. Crossed with the vector expressed in the moving frame. So a, a really important result, and we'll continue on, and use this theory in future modules to solve some real word problems.
