Let us do some more fun. Kinetic energies for particles, you guys know that. That's mass over two times velocity squared essentially, that's what you've had. This is the vectorial form of that. I have dm, so instead of mass I just have dm over two. That's essentially what that does. Times my inertial speed squared. I'm doing this in a vectorial way, so R dot is my, you know if R is my inertial position vector, I need to have my inertial velocity. That's the first inertial time derivative of R, that's an R dot. I'm dotting it with itself, that's going to give me my inertial velocity magnitude squared. It's just a vectorial way to write it, but it's a very general way to write this. So that's it. Now that's for one dm, I need it for the whole blob. You simply integrate again over the body. That's where the integral comes in. It's just summing them all up. So good, now we have kinetic energy. That's not the most convenient form. We don't typically write it this way because it's kind of hard. We would like to break it up in terms of rc and kinetic energy off the center of mass and then we're looking at kinetic energy about the center of mass and that will help us separate rotational stuff, deformational stuff and translational stuff. That's what you'll see popping out of this. So let's do this. R dot. R was defined as Rc + r, so R dot, you just put dots everywhere. Again, vectorial form, dead simple, right. The 3.6 comes back. Don't do part b any more. Do part a where we only take time derivatives and vectorial parts. You plug this in and you carry out the various dot products. You guys know how to do basic dot products in calculus. I'm not going to do that again. So we end up with these terms. And, as Kevin was saying earlier, in these body intervals, anything that's RC and RC dot, those are all going to be things that are fixed over the body interval. Because every point in that blob has the same center of mass. Therefore, it has the same center of mass velocity. So we can always treat these quantities as constants over the integration. So when you carry, you know, this is a step I'm not doing here, I'm purposely not, because I want you guys to, you should do this yourself. [COUGH] Run through the notes, that's how it's going to stick. If we do that, these can be pulled out. Same thing gets pulled out here. And at the end there's a factor of two with the half that cancels, and you have also this other term. Now one of these will go to zero. Medal. Which one of these is going to be zero? >> Tom Tom. >> Why is this one going to be zero? That's always my second question. First part is 50/50. Might be lucky. Who recognizes this part? Trevor? >> Well, we just went over that the body integral of the radius times the- >> That's the vector. It's not radius. Let me correct quickly. That's the position vector. >> Position vector, yeah, that's [INAUDIBLE]. >> Radius would be a scalar. >> [CROSSTALK] And so that's equal to zero. >> Yes. >> Take the derivative of that that should still be zero. >> Exactly without the dot this is nothing but the center of mass definition. We rewrote that instead of the mass average location also saying that if all locations are relative to the center of mass those summing ups have to give you zero. So integral of little r times dn is zero. So therefore, the derivative of zero is still going to be zero. So this one, the center of mass location. This is, basically, using the center of mass definition. That one will go to zero leaving us with this term and this term. Now this term is nothing but total mass again. So I have a big M. So mass over 2 velocity squared, so you can see for the blob, the center of mass acts like a particle again, which is really nice. And if you're doing an orbit problem, this would be your orbital energy of your spacecraft basically. And then the second part is this integral, body integral of r dot, so that's the position vector, of each DM relative to the center of mass. Get its inertial derivative though. That's important. This is a dot. This is not a body frame derivative. That's the inertial derivative and we sum them up. So this is basically your kinetic energy about the center of mass. And in this form it contains two parts. One part of kinetic energy comes from rotation. If all the stuff is rotating then these things could be spinning, and that gives you all that kind of kinetic energy. The other part is because this is blobs, you might have deformational energy, where things are changing shape as well and you're redistributing mass. I'm not gaining or losing mass in this system, but I'm just redistributing. So if you had fuel slosh bouncing around, that's one example you know. So you might have deformational energy happening here. So that's what that is, but this is the kinetic energy of a jello, written in the very form, basic form. Now, we very often want to have power, that means the time derivative of kinetic energy. We have that expression, broken up into the translational part of the cm and energy about the cm. If you differentiate this, you come up with these here. Let me show you that one trick. If you don't see that, how to go from there. If you have a vector, x dotted with x. If I have a scalar x squared and then I take a time derivative of the scalar I will simply have 2 times x times x dot, right? Very simple. Here, if you have this vectorial quantities I mean these expressions we see a lot of vectors dotted with vectors again, especially themselves. If you have this and you take a time derivative and here, I'm going to take an inertial time derivative, so I can just put dots over my vectors. With the chain rule, you would have x dot, dotted with x, plus x dotted with x dot, right? Why is this the same as 2 times x dotted with x dot? Why can I simply add these two things up? >> [INAUDIBLE] Reverse the order. >> Right. The answer here, the dot part is actually a scalar. So if you do this in vector form a dotted with b is the same thing as b dotted with a. If you're doing it in matrix form this would have been written as a trans x transpose x. Right? It's equivalent. And in that case too, the answer is a scalar. So any scalar you can transpose and it's the same scalar. So it allows you to reverse the matrix order with the right transposes on it. This is nothing but the vector form of it. So here, here we just end up with 2 times x times x dot, or you could've rewritten this by putting the dot on the first one. Whatever's more convenient in your analytical development. So practice that if you haven't. That's how I go from R dot R dot. The one half cancels with the two. Same thing here. Little r dots there's a one half in front of it that cancels with the two that comes out of the differentiation. This is what you end up with. Now we're going to apply some stuff. We know from the super particle theorem that the mass time initial acceleration of the center of mass is equal to the net external force. So that's good. So we can plug in this becomes nothing but f. Over here little r dot we have to do a little more math. We knew that big R is Rc plus little r, so you just put two dots over everything. Now we have the inertial derivatives of all these quantities, and I can plug that one in. So that will break up this integral as being Rc double dot, dotted with R dot. And, sorry, there's an R double dot and a Rc double dot that plugs in. You break up the math, this is the form that you're going to have. So this part is actually pretty much done. So the power equation gives you force dotted with velocity. The shift makes sense, hopefully with high school physics, if you look at power stuff. That's typically always the force times a velocity measure. That's the instantaneous power being produced mechanically. These other parts, we're going to have to look at. Now here, anybody see a term that's going to go to zero? >> How about the one at the end? >> Yep, this one again, right? Due to the center of mass property. These things will pop up everywhere and we love that, because zero is great. It really simplifies the stuff. So we're going to see that one go to zero and this is nothing but mass times the initial acceleration of an infinitesimal element that was nothing but dF, that little force. So on the body, this is your general power expression that you would have to do and you can write, if you have to energy change between two points, you would integrate these equations over time. And this is one form or you can do a substitution of variables. This is a classic thing you've probably seen where you can also do it force times distance gives you the energy take put into this system. So, you can replace it. There's different forms you can write this. This is the general one. What we'll use in class will be the rigid body specific one, but I just wanted to highlight first the general formulation. We're going to then write it later on. because this one doesn't make much sense practically for you right now. Once you make it rigid, we'll have a turn that would be very very clear. This is how we compute this.