 Now, let's make things far more interesting. We're going to now change where we put O. Let's put it down here. Now nothing has changed. The physical system is still exactly the same. I have the Surat going up here, a little mechanism so that this can swing around. My point A is still there. Nothing has changed. I am just moved the point mathematically with respect to which I'm going to take angular momentum and torque. We still have our equations there for angular momentum and for torque. Things are going to change a bit though, because now I'm defining R, A of O as something different. And what would be an easy way? You've always got to think what would be an easy coordinate system to use. Well, in cases like this, if you just think of this particle moving at a constant height and just going around and around, a cylindrical coordinates really come to mind, because it's really just forming the cylinder. If it's cylindrical, now this was my position vector of A with respect to O was this. Now my position vector of A with respect to O is this up here, this vector here. And in cylindrical coordinates, it's still going to be R, R hat, R, R hat, which is this length here in R hat direction, which is the same as this, plus this. Now for that, let's call this height here, which remains very constant. It's a constant as two. Let's call it Z. And in my world frame, if uppercase K hat is up, here we'll have K hat in this direction as well. So that's just going to be plus Z K hat. So by moving my world frame, or the point with which respect to I'm taking the angular momentum and the torque, I am changing my position vector. Once again, I am going to need its first derivative and I am going to need its second derivative. Might as well get it going right from the start. Let's get R of A with respect to O, a dot, in other words, the velocity, that is still going to leave me with R dot R hat, plus this must really become second nature just to write this as R theta dot theta hat. It is the product of two vectors. That is a constant. That is not a constant. R hat is going to change all the time. We know that this term is going to fall away if R is a constant, which indeed it is. So this term is going to fall away. But now I've got to take the derivative of that as well. In other words, I am going to have Z dot K hat and Z d dt of K hat of K hat. So that's going to happen. Now think about it. If Z is a constant, this height stays exactly the same. This term is zero. And if we think of K hat, K hat stays exactly the same. It doesn't change. So its first derivative is zero as well. So all I'm left with is just this. For this problem, this is that if R was not a constant, fair enough, if Z was not a constant, fair enough, you can't throw those away for this problem. You can throw that away. So if we're going to second derivative of A with respect to O, all we now have to do is take the second derivative of that. Three terms, product rule, three times, three functions there. R dot, I'm not going to write that again. That first one's going to be zero. So I'm going to be left with R theta dot double dot theta hat minus R theta. This is ridiculous. Theta dot squared R hat. So that's what I'm going to be left with A as far as acceleration is concerned. R dot, the first term there was also just zero. So this is my velocity vector. This is my acceleration vector for this reference frame of point A. Now if I have to take the angular momentum of A with respect to O, I'm going to get R R hat plus ZK hat, this cross with MV, cross with P, which is going to be mass times this, cross M R theta dot theta hat. So you've got to do that whole, you've got to do this dot product, this cross product. So it's going to be this cross product, plus this cross product that. Let's do that. H of A with respect to O is going to be what? Let's do this cross product for this. We're going to have M, we have R squared theta dot. And what is R hat times theta dot, theta hat, that's K hat. And what are we going to have here? We're going to have an M, an R, a Z, a theta dot. And what is K hat times theta hat? That is negative minus R hat. Now if we think about this, this is a vector pointing in this direction up here. And that is now the angular momentum. That is the angular momentum. When you draw this vector, you'll see that it's perpendicular to this, always remain perpendicular to that. So there's going to be this angular momentum around K. And there's going to be this angular momentum around negative R. In other words, in this direction, R is pointing there, but it's in the negative R. So in this direction, you're going to get these two components. Now torque is going to be even more interesting of A with respect to O. It's going to be R R hat plus Z K hat. Now cross product with, cross product with F, which is M times this. So it's cross product. M R theta double dot theta hat minus M R theta dot squared theta hat. Oh, R hat. What am I doing? R hat. So this becomes a very interesting, this becomes a very interesting set of values here. So with respect to O, I'm going to get this times that. So I'm going to get M R squared, M R squared theta double dot theta double dot M R squared theta double dot R times theta. That's K hat. So that's one R hat times R hat. That's zero. Let's do this one. We're going to plus plus M R Z theta double dot K hat times theta hat is negative, negative R hat. And the last one, this one times that one is going to be a negative. I have M R Z theta dot squared and K hat times R hat, K times R is positive theta hat. So I have the three components keeping this torque, reducing this torque for me. One, two, three. And you can think about the physicalities of what is going, what is going on here, torque in this direction around the K direction. We're going to get torque in this direction, negative the R. And as far as theta is concerned, the torque down negative in the theta hat, well theta hat is, for us now is in that direction, so in this direction, so apologies, so in this direction. And that gives us an idea of the forces on this system, the forces on this arm with all the torques that are applied to it. Because remember, we can just work out what the forces are. It's just M times, M times this, there's going to be these two forces on the system there to give us, to give us this torque. Now if R was not a constant and Z was not a constant, obviously this would become even more complex. I could get rid of a lot of terms. All this is really, really simple, simple calculus. You've just got to take cognizance. And because we are, remember because we are using cylindrical coordinates R, we have R, we have theta, and we have K. So this is what we have. Just remember if it's not in this order, R cross theta gives you K. Theta cross K gives you, start in the beginning again, R. K cross R gives you theta. And if you turn those things around, if you go in the reverse, K times theta gives you negative R hat. Theta times R gives you negative K hat. I cross R gives you negative theta hat. So you just got to remember in which direction you do the cross product of these to give you a positive or negative. Otherwise it really isn't, especially if you do these two first, there really isn't a problem. Even so, even if this point starts to move, remember we had the fact that the sum of the torques was going to equal the DDT. DDT, let's just see if I can get something else to write with. We really have problems today. DDT of H of A with respect to O. And then we had this V cross P term there as well. And V of A with respect to O was zero. We could throw that term away. Or if these two were indeed, remember even if this was respect to some other point, if they moved in parallel, we could also throw that term away, which is in essence what is happening in this instance. So we can get, we are getting rid of that term. But if you do R cross F and you have this position vector in the mess so that you can just use F as M times this, then obviously just use this way. Otherwise, there's nothing difficult here to understand. Just remember, simple vector calculus.