[BLANK_AUDIO]. Hi, this is Module Nine of Three-Dimensional Dynamics. We just finished a block of instruction where we looked at velocities as expressed in moving frames of reference. And now we're going to derive the equation for acceleration expressed in moving frames of reference. And so here's our derivation for velocities. We said that the velocity of P with respect to a frame F is equal to the velocity of the origin of the moving frame B plus the relative velocity of P with respect to that moving frame B. Plus the angular velocity of B, the moving frame, with respect to F crossed with the position vector as expressed in the moving frame from O prime to P. We want to now go on and derive accelerations in those moving frames of reference. And so, what do I do next? And what you should say is I, I need to take the derivative. And so, I'm looking for the acceleration of P with respect to F. That's the same as the time derivative of the velocity of P with respect to F taken in the F frame. That's the derivative of the velocity of p with respect to f taken in the f frame that's going to equal now I'll go through and take the derivative of each of these terms so I've got the derivative of the velocity from o prime to f in the f frame plus the derivative of the velocity from p to b taken in the f frame plus now in this last term, I'm going to have to use the product rule. So I'm going to have the derivative of the first cross with the second plus the second cross with the derivative of the first. So I'm going to have plus the derivative. Of omega B with respect to F, taken in the F frame crossed with R from O prime to P, and then plus omega B with respect to F crossed with the derivative. Of the position vector from o prime to p taken in the f frame. Okay so we have the acceleration of p with respect to f is equal to now what is the velocity what is the derivative of the velocity of o prime with respect to f taken in the f frame. And what you should say is that's just going to be the absolute acceleration of O prime with respect to F. And then, plus now I've got the velocity of the derivative of the velocity of P with respect to B taken to the F frame. Here again I have to be careful because the velocity of P with respect to B is the relative velocity in the moving frame, yet I'm taking the derivative in the F frame. And so try that on your own for that term, and what you should come back with is, you're going to have to use that derivative formula, so we're going to have plus the derivative of the velocity of P with respect to B in the B frame. Plus now the angular velocity of B with respect to F, crossed with the velocity of self, P with respect to B. Okay. So that takes care of that term. All of this is equal to that term. Then I have plus now,. What is the derivative of the angular velocity of B with respect to F taken in the F frame? And what you should say is, okay that's the angular acceleration of B with respect to F. So we're going to say angular acceleration of B with respect to F. And then I'm going to take a little short hand here for my position vectors since it's in the moving fray. I'm just going to write it as r. And then I have, plus now, omega B with respect to F crossed with, again, I've gotta be careful. I've got the derivative of the position vector expressed in the moving frame, but the derivative's being taken in the B frame. And we've done that several times before using the derivative formula. That's going to be. The velocity of P with respect to B plus omega B with respect to F crossed with R itself. Okay, and so again, this term matches with. This. And so let's keep going and so now I have the acceleration of p with respect to f is equal to the acceleration of o prime with respect to f and what is the? What's another way to express. The, derivative of the velocity of P with respect to B taken in the B frame now. And what you should say now is, if I take the derivative of the velocity of P with respect to B with respect to the B frame, that's just going to be the acceleration of P with respect to the B frame. This is plus the acceleration of P. With respect to B. And so that takes care of that term and that term. Let's take care of this term here. So I've got now plus the angular velocity of B with respect to F crossed with R. And then plus now lets group a couple of these terms. I've got omega, B with respect to F, crossed with V of P with respect to B and then over here I also have omega, B with respect to F crossed with the velocity of P with respect to B. And so that gives me. 2-omega-B with respect to f crossed with, v of p with respect to B. And then the last term I'm left with is. Omega-b with respect to F. Crossed with omega B with respect with F crossed with R. So that's omega B with respect to F, crossed with omega B with respect to F, crossed with R. [SOUND] 'Kay, and so that's an expression for my acceleration in moving frames of reference. Okay, so let's circle that, because that's important. That's accelerations expressed in moving frames of reference. Let's now. Label each of those [COUGH] excuse me. Let's now label each of those terms. So this is the absolute acceleration of P and B. This is the. Acceleration of the origin of the moving frame. 'Kay, this point here. This now is the relative acceleration of B, P, with respect to my moving frame. So this is the relative, acceleration, of P and B. This now is this term is due to the angular acceleration of the moving frame B with respect to F. This is term is due to angular [NOISE] acceleration. Of B just say of B with respect to F, okay. This is called 2 omega F crossed with VP of B. And that's called the Coriolis express, acceleration. And we'll talk more about that in the next module. And then finally, this last term is what we call the centripetal acceleration. Centripetal. Acceleration, and if you look at it just from a scalar standpoint that's just going to be omega squared R. So we've seen something like that before, even back in particle acceleration. okay. We can write this as shorthand just like we've done before, this is my position vector. In the moving frame. We're going to call big R the position vector of the origin of my moving frame. And so I've got now the absolute acceleration of P and B, I'm just going to say is the absolute acceleration of P equals now the acceleration of the origin will just be R double dot. We'll call this the relative acceleration so this is a RLP is b plus now alpha crossed with R. That's just the angular acceleration of the moving frame crossed with the position vector in the moving frame plus now two omega. Cross with v rail, that's my coriolous expression for coriolous acceleration plus now finally omega cross omega cross R. And so that. Is a short hand for my acceleration equation again this is exactly the same equation I came up with for two dimensional dynamics except now my vectors have 3 exponents xyz rather than 2 exponents x and y we'll come back next time and solve an actual problem