[BLANK_AUDIO]. Hi, and welcome to module two of Three-Dimensional Dynamics. Today's learning outcomes are to, first of all derive the derivative formula for taking a derivative of an arbitrary vector, expressed in one frame, what we'll call a moving frame, with respect to another frame. And to define angular velocity for three-dimensional motion. So in this course, we'll study bodies undergoing three-dimensional motion. To approach these problems, we're going to attach reference frames to the various bodies that are undergoing that 3D motion to describe it. To begin our study, let's review my previous course, and how we approach the situation of describing velocities, as expressed in two different reference frames for two-dimensional, or planar motion. And then we'll extend that to three-dimensional motion. And let's also look at what we call the derivation of the derivative formula. And we'll also expend, expand that to three-dimensional motion. 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. Let's go ahead and develop 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 go 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 sine theta in the big I direction. And we're going to go positive cosine theta in the big J direction. And now let's differentiate those vectors, the 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 of i with respect to time for the frame F is equal to i dot. Which is equal to now, derivative of cosine theta would be minus sine 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. So 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: dj dt in the F frame is equal to j dot. Which is equal to okay, the derivative of minus sine theta is minus cosine 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 stuff, 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 unit vectors and 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 could be any vector. We know vector quantities for kinematics 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 Ax dot i plus Ay 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 w hen 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.xi plus A.yj. So from the perspective of the little, of 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 in 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 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 A x j, plus excuse me, minus A y i. And 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 A x i plus A y j. And you can see that these are equivalent k cross i is j, so this A, x, theta dot j is this term and then k cross j is minus i so that becomes A, y, theta dot times i. And we can also realize that now A x i plus A y j 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 derivative 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 really important result, and we'll continue on, and use this theory in future modules to solve some real world problems. So this is the derivative formula we just came up with for two dimensional or planar motion. For taking the derivative in the B frame with respect to the F frame of an arbitrary vector. We applied it to problems in the previous course that were in two dimensions, here is a typical robot problem, planar robot problem or two dimensional robot problem. Where we have our frame F attached to one portion of the robotic arm, and frame B attached to another portion of the robotic arm. One thing you'll note in two-dimensional motion, is that the angular velocity here is only in the K direction by the right-hand rule. And in three dimensions now, this is going to be more generically omega, and it will be able, that's the angular velocity. And it will have components in not only the K direction, but also the I and the J direction. So, here's some clips of three dimensional robotic-type motion where you would attach an F and a B frame to different port, portions of this, these robots to analyze the, the kinematics of the motion. So again here I've extended my derivative formula to include a generic or a, a more general omega that can have i, j, and k components. I've did my body B and my body F, the generic bodies I show here now, have three coordinates instead of just the two planar coordinates x and y, now we have x, y, and z. Here's a typical robot, again, this is a robot that's been developed at Georgia Tech. I've got the F frame attached up here sort of like the base at the head of the robot, and then I've got a B frame, a moving frame, attached to this portion of the arm here, where the robot is pouring the drink into a glass. And so you see now that the angular velocity in, in the moving frame not only goes around the k axis but can have i, j, and k components. And so that's the definition of the angular velocity for frame B with respect to frame F having all three components. And that, that's it for this module, and we'll see you next time.
