Kinetic energy, let's start here. Now we did this for blobs in space. And for blobs in space we were able to show kinetic energy was nothing but the kinetic energy off the center of mass plus kinetic energy about the center of mass. But this second term contains both deformational energy, if panels are flexing and doing or fuel is sloshing. And the rotational part, right? Now let's use the same assumption. r dot for a rigid body, it doesn't go to zero, but the r dot just became omega crossed r, the body relative derivative of zero. So we're also just going to focus on the rotational side. This is basically your translational energy which comes out of your orbits class. We're not going to focus on orbits in here, just the rotational dynamics, so we're not going to cover that. And this is where I plugged in r dot is omega cross r. So you get a bunch of cross products, dot product. And you know right away there must be some magic trick identity that we can rearrange this. That's kind of what I'm doing here, almost actually. First, and I'm using trig here, this omega is a constant as seen of the integration. If it's rigid, every point has to be rotating at the same rate. So this omega I can just take outside. There's a dot product. Okay I did use a vector identity already to rewrite this into this form, and now I could take your omega outside to the dot product. Here I'm still left with omega in the middle but we have the trick that a cross b is minus b cross a. That puts it on the right hand side. And you can take that one out as well. In fact if you go back and look at Hc, this whole definition is nothing but Hc. In fact this term becomes nothing but your inertia tense times omega. And the inertia tense times omega was nothing but angular momentum about that point c. So it's nice, we now have, rotational energy can simply be written as one half omega dotted with angular momentum. Those are two vectors dotted together. When you dot them, mathematically make sure they're taken with respect to the same frame. But here is written in a frame agnostic way. So angle rotation, so that's the same thing, H is I omega, you might remember as rotational energy being inertia over 2 times angular rate squared for fixed axis rotation. This is the 3D version of that. You would have one-half omega transpose I omega. And that's basically inertia omega squared over 2, essentially that's what it is. Power. We had earlier, if the time derivative from a blob, the first part translation was just force times the inertial velocity, again, that's the translational sign. What we do care about here is T dot. So if you differentiate this term, you will end up with this. Now to get this form, you need to have the rotational equations of motion, which we haven't derived yet. But I will let you, this is something you should make a note in exam, definitely go back and make sure you can go from here to here, but it simplifies. But the beauty of this is, for translation, we have force started with translational velocity. With rotation, you have torque dotted with angular velocity. It's really the same type of term that we're going to have. And that's what makes, this is what gives you your power rate. To the system. When is T-dot going to be zero? Let's ignore translation, we'll just focus on this. When is my power equation going to be zero? Sheila, what do you think? Without reading more, I want to hear you think. >> The angular velocity is zero or- >> You could, if you're not spinning. >> Center of mass is acceleration is zero, the Rc. >> Well we're going in translation right now. >> Okay. >> Just focus on this one, the rotational side. What makes this second term go to zero? Omega could be zero, yes. It's kind of a trivial dynamical system then, yes, if the energy is zero then the energy rate is going to be zero. But that works, yep. What else could you do? >> [INAUDIBLE] Make Lc zero. >> So what is Lc here? [INAUDIBLE]. >> No? What is LC here? >> Torque. >> It's a torque. So this is a torque about point c. And in fact when we get this torque, this torque is due to external forces when we derived all this stuff. So for this dynamical system, if your external torque on that rigid body is zero, you won't dissipate energy. This makes a great integration check. So as you guys are writing numerical code, you've got something that's spinning and there's no external torque on this system. I know energy has to be preserved. If we had H dot equal to L, that was the same thing. The angle momentum as seen by the inertial observer, that's what that dot meant right, inertial derivative. As seen by inertial observer, H has to be a constant vector if no external torque is acting on it. These things become fantastic integration checks. This is for a single rigid body. But even if you do multi-body, complex panels flapping, reaction wheels, gyros spinning, twisting, this stuff still has to hold, which is really nice. And now if there is a torque acting on that system. I can actually predict very quickly how my kinetic energy needs to change. And this becomes a nice integration check, because I can compute kinetic energy from the simulation numerically and then differentiate it to get T dot. And compare, I know I applied this gravity gradient torque, this is my spin rate at the time. Energy should have changed by this many joules per second, right? So these tools becomes very, very useful validation tools, but also for analysis you will see. So torque-free motion is when momentum is inertially preserved and kinetic energy is preserved, because this is zero. The third option actually could exist, in that the dot product goes to zero. Maybe you have very particular torques being applied, such that no matter what your spin rate is, right now the torques are orthogonal to that. In that case you have a non-working force acting on it. That's what they call it because it doesn't, not doing work means it doesn't change my energy state. So there's three ways, either omega zero, Lc zero, or those two things are orthogonal. Then T dot goes to zero. And again over, if you have to work down between points and time, you can do this with a time integral or you could also do it in the spatial integral.