We're doing a review of what we covered last time. We had dual spinners. So with a dual spinner, I had some spacecraft, didn't necessarily have to be symmetric. But this spacecraft has b1, b2, b3 and you'd have some panel attached to it. So now, Nathan. What do we do with the dual spinner? What's the set up that we do here? >> Well you have one section spinning to maintain your angular momentum. Or, so it's spinning about its axis of minimum energy, so maximum inertia. >> Okay. >> And then the other section is not spinning essentially. >> Is that required? Do we have to spin about an axis of a minimum of inertia? Does the second section have to be despun? >> I guess not. >> No, in fact that's what I'm going to show today through an example. We can stabilize which is the beauty of dual spinners. I can make any of the axis stable. Now if you have energy loss the minimum inertia one you have to kind of be a little bit more careful there's some power equations you can find on this stuff. But conceptually, we know without energy loss this is a stable spin and this is a stable spin. But a spin about the access of intermediate act inertia tends to flip around and be unstable, all right? So, with a dual spinner in this mathematics, we have a fly wheel. How do we line up, Brett, how do we line up this fly wheel in our dual spinner? >> One of the body axis? >> Yeah, one of the principal ones. So, if we wanted to have a nominal spin here, Omega1, that's my equilibrium spin. I am going to add a wheel, which has a spin axis, going to call it gs-hat, that's my wheel spin axis, which is the same thing as b1. Those are lined up. Good, then we went through, how do we find equations of motion? We still use H equal to L. But H is now due to two rigid bodies. And there's different ways to derive it. We went through that last time. There's angular momentum of the space craft and the wheel that was I, that was a summed I times omega plus the wheel momentum routed to the spacecraft. Or you can treat them as two separate ones, I showed you that as well and you can combine again. That's a great problem that I've used before in the final so make sure you go through that. It's a good simple exercise. Later in class, you can do spacecraft within variable speed CMGs. It's the same process but about 12 pages more algebra. So this is where you want to practice the process so you don't get silly sign errors. Then the rest of it just kind of becomes mechanical. I'm going to do this, I'm going to do this, and you can use computer tools as well to do that. So practice this. So we went through this. Now I labeled this as an equilibrium spin omega 1E, all right. What does it mean Mandar to be in equilibrium spin? What does that mean? [COUGH] >> All Omega dots go to 0. >> Exactly, any dynamical system, any flow process or whatever you have, if you have x dot equal to f, a nonlinear thing, for [INAUDIBLE] xe is x dot equal to 0. And the planar pendulum has one equilibrium here, one equilibrium here. Does this have anything to do with stability, finding equilibrius? No, which just it's a convenient point about which we tend to discuss stability. But stability is a separate concept from finding equilibrius, all right. So we found the equilibrius and in particularly for the single rigid body there were three equilibrium spins we discussed earlier. Spinning about axis of least inertia, maximum inertia, but also intermediate inertia. Those are equilibrium spins. If you put it perfectly on there it would tumble that way. But if you bump it, if you perturb it, that little delta that we were talking about earlier, do you stay close? What happens to my spin rates? Right, do I wobble just a little bit and stay there, which is true for this kind of a thing? But if I throw it this way you can see it did several gyrations right away, that's an unstable motion, all right? That's what we're after now. And so good, so we identified while the single rigid body could have equilibrium spins about all three axis. For the duel spinner the equilibrium spins we end up using is primarily the one about the GS axis. I say primarily, let's look at that equation again. [SOUND] Here, we argue that if I have a non-0 Omega 1 and Omega 2's and 3's are 0, I always have all my dots. Omega dots are going to be 0 like Mandar was saying earlier. Perfect, but you can also have Omega E1 be non-0. Omega E2 is 0. Which makes this 0 and this 0. You could pick particular inertias, a particular E1 at a particular wheel speed, such that this combination with a non-0 omega E3 actually goes to 0 as well. That's also an equilibrium, but it's not one we typically fly, because now all of a sudden you have a very particular solution. You have to have exactly this wheel speed. If you're a little bit fast, or little bit lower, you're out of that equilibrium part. Whereas this other spin we had if this wheel speed is supposed to be 10 rpm and I put in 11 rpm it's still an equilibrium actually and still going to hold the stair. And so, we did that, here we just added what is the equilibrium plus delta and then dropped tie over terms, right? And you end up with these differential equations. That it's stable, if you can see here, again. If instead of putting in that 10 rpm but I gave it something slightly faster and perturbed, this can hold that speed because my angular acceleration about the one axis, the spin axis of the wheel, is just 0. So you just stay there. The other two, tends to couple and we have two first order differential equations that are coupled and the way we solve these was, we differentiate again, substitute in. So we replaced two coupled differential equations with one second order decoupled differential equation, and in the end, it's all about these two bracketed terms. This is your effective stiffness. X doubled up plus K X, K has to be positive right? And either we make this positive or this positive or we make both bracketed terms negative. Different ways we can write it? We'll have in that example today, you will see omega hat that's a non-dimensionalized. We take our wheel speed, 100 RPM, divide by the spacecraft speed, 2 RPM, so megahat would be 50 in this case. It's a 50 multiplier factor between the spacecraft spin and the wheel spin. And then it just breaks down to these inertias that you need. And you have to pick spin speeds such that both terms are positive, or both terms are negative. If you put in 0 spin speed we recover the case of having a single rigid body. And that's how we quickly found that this stiffness is only going to be positive if my inertia is either the maximum inertia, that means we're doing the flat spin we discussed earlier with the pole hose. Or b1 is access of least inertia which at least linearly without energy loss is also stable. So you see in here also mathematical linearized proof of the stability of the min and max inertia case. The intermediate one gets positive and negative, definitely not stable. And the non-0 one in the non-dimensionalized terms. These are the conditions we need.