A little bit more curious, angular momentum. With angular momentum we have to specify about which point we're taking moments. We take torques about a certain point and we'd measure angular momentum about the same point typically in a system. So here, I mean people often use center of mass, but I'm not that nice. I'm giving you a general point p. And so sigma is now my position vector of dm relative to this moment point. >> [COUGH] >> This is the hinge point. For some reason I want to know momentum about this on my body is tumbling out here. How does that behave? So we'll derive the general formulation and then you will see when it sympathize down to the down to the classy answers many of you are familiar with. The momentum here H about point P angular momentum. It's always position crossed linear momentum of that same quantity. So sigma, that was my position vector relative to P, so I have a dot, inertial derivative on that. And you plug it in. Sigma is R. RP would be the position vector going from here all the way up, right? So sigma is nothing but R- RP. You should start plugging this in. Let's see, we do it with differentiation so that we get a dot here and then we get a double dot on the second term. >> This first one, trivially, goes to 0, a vector crossed itself is always 0, that's kind of boring. But it's nice because this will vanish and we're left with this term. So, you're going to see okay. Now, what happens with this term? And we plugged instead of sigma double dot, we have R double dot and Rp double dot. And I plugged those in. Now, you notice R, I cannot take outside of the body integral because R depends on which element in the body am I pointing to. So that's going to be different for every dm. You just can't take it outside. Rp on the other hand that's a general hinge point that I have. It's not really connected to the body and keeping it very, very general. And so that one is just fixed Rp. Might be stationary, it might be moving, it might be spinning or who knows what it's doing. And whatever it's doing, if I differentiate it twice, I need its inertial acceleration. Its second order derivative. But that one I can take outside of my body integral. So good, we're left with those term. Now let's look at this one. If we focus just on what's the body integral of sigma dm. So sigma, again, is r minus rp. So I can break that up as the integral of Rdn minus the integral of Rpdn, but Rp I can take outside of the body integral, which allows me to just get a mass times this. And this part, if you recognize it, that was the right hand side of the center of mass definition. Right? n times Rc was equal to body integral big Rdm. So I'm using that definition from the other center of mass definition. So this just becomes M Rc, this one here becomes M and then a Rp. So this term here, I can rewrite as given by there. Okay, where's this going? If we look at torques, the torques about point B is basically the moment arm cross with the force acting on that particle. So, if there's a magnetic torque bar as mentioned earlier, it may have a particular torque, force acting. Once the moment of that force relative to point P. All right, so it's sigma crossed R double dot dm which is the same thing as the force acting on that element. If you write this, we can rearrange this, and so you can see this term actually appeared over here. Here, we've simplified that term to here, but this first term appears as being nothing but the torque around that point P. Minus this, well that was it, that is the talk about p. So I can use this, the inertial derivative of H about point p is equal to the torque about point p, plus this other stuff. And I'm leaving it as slightly greyed, because we typically, a lot of dynamics books don't talk about this, at least. Many people just remember the inertial derivative on momentum is torque which is true at times. So what must be true of point p to make this second term vanish? Tosh, how can you make this term go to zero? >> [INAUDIBLE] >> No because [INAUDIBLE] or not this doesn't matter actually. This only has Rc which is, this true for continuum as it is for rigid one and RP is just an arbitrary point outside of the body that I picked. And I could place it anywhere. CK. >> RP at RC. >> Yeah, if you set RP, it happens to be equal to RC. In that case, this one minus itself is zero and that will vanish. And then probably the most common use of h dot equal to l is they've assumed implicitly that is about a center of mass location, all moments are about Rc. And for spacecraft staff and if you keep spinning things, we will tend to use the center of mass as a point, a pivot point about which we take all of the moments. But it doesn't have to be. Sometimes you have a space craft and a CAD drawing, and for some reason what you got with the mass distribution is all the locations are not about that. It's like an off-balanced wheel. All of the sudden you have to account for that difference, and this is what you would have to do. What's the other way? There's a second way we can make this term go to zero. Evan. >> You can make the difference between Rc and Rp perpendicular to the acceleration of Rp. >> Okay, there's a third way you can make it go to zero. [LAUGH] That's kind of a hard way to retract that but you might come up with a condition that would do that. Good point. What's the second way? What do you guys think? Roda what is that? We have almost exhausted all the terms, what's left? >> RP is constant. A treat of. >> Okay if RP is constant the RP double both is zero. But that's true but I would say RP being constant is a little bit. >> RP dot is constant. >> Could also be constant right? So what do we call emotion that has, either it's fixed in space or it's only moving inertially at a constant rate? What do we call such motion? No not nonlinear, inertial, exactly. The definition for something being inertial, an inertial point typically is it's a non accelerating point, and that's what's required. The condition is really just RP double dot has to be 0. If that point is non-accelerating. So if it's moving in a constant line, we're fine. If something is spinning immediately it's centrifugal accelerations and stuff, you know, it's not going to to be inertial. But you could take a point that's not center of mass, but it's an inertial point about which you take moments and then H dot equal to L holds as well. So in fact, some of the homeworks we'll be getting into you need to use both. Sometimes you take Hs and Ls about center of mass. And sometimes you're finding this inertial point, o, and you can do it there. And both sets will give you some information that you can use to do that, all right? So we're not going to do much about general body fixed points or other different formulations of that that exists, you can look them up. Just be aware Hr is equal to l is not universally true. It's only true if you take your moments about center of mass or inertial points then we're guaranteed this is going to go to zero. Those are two conditions we'll use in class.