0:05

Let us do some more fun.

Â Kinetic energies for particles, you guys know that.

Â That's mass over two times velocity squared essentially,

Â that's what you've had.

Â This is the vectorial form of that.

Â I have dm, so instead of mass I just have dm over two.

Â That's essentially what that does.

Â Times my inertial speed squared.

Â I'm doing this in a vectorial way, so R dot is my, you

Â know if R is my inertial position vector, I need to have my inertial velocity.

Â That's the first inertial time derivative of R, that's an R dot.

Â I'm dotting it with itself,

Â that's going to give me my inertial velocity magnitude squared.

Â It's just a vectorial way to write it, but it's a very general way to write this.

Â So that's it.

Â Now that's for one dm, I need it for the whole blob.

Â You simply integrate again over the body.

Â That's where the integral comes in.

Â It's just summing them all up.

Â 0:59

So good, now we have kinetic energy.

Â That's not the most convenient form.

Â We don't typically write it this way because it's kind of hard.

Â We would like to break it up in terms of rc and kinetic energy off the center of

Â mass and then we're looking at kinetic energy about the center of mass and

Â that will help us separate rotational stuff, deformational stuff and

Â translational stuff.

Â That's what you'll see popping out of this.

Â So let's do this.

Â 1:23

R dot.

Â R was defined as Rc + r, so R dot, you just put dots everywhere.

Â Again, vectorial form, dead simple, right.

Â The 3.6 comes back.

Â Don't do part b any more.

Â Do part a where we only take time derivatives and vectorial parts.

Â You plug this in and you carry out the various dot products.

Â You guys know how to do basic dot products in calculus.

Â I'm not going to do that again.

Â So we end up with these terms.

Â And, as Kevin was saying earlier, in these body intervals, anything that's RC and

Â RC dot, those are all going to be things that are fixed over the body interval.

Â Because every point in that blob has the same center of mass.

Â Therefore, it has the same center of mass velocity.

Â So we can always treat these quantities as constants over the integration.

Â So when you carry, you know, this is a step I'm not doing here,

Â I'm purposely not, because I want you guys to, you should do this yourself.

Â [COUGH] Run through the notes, that's how it's going to stick.

Â If we do that, these can be pulled out.

Â Same thing gets pulled out here.

Â And at the end there's a factor of two with the half that cancels, and

Â you have also this other term.

Â 3:03

integral of the radius times the- >> That's the vector.

Â It's not radius.

Â Let me correct quickly.

Â That's the position vector.

Â >> Position vector, yeah, that's [INAUDIBLE].

Â >> Radius would be a scalar.

Â >> [CROSSTALK] And so that's equal to zero.

Â >> Yes. >> Take

Â the derivative of that that should still be zero.

Â >> Exactly without the dot this is nothing but the center of mass definition.

Â We rewrote that instead of the mass average location also saying that if

Â all locations are relative to the center of mass those summing ups have to give

Â you zero.

Â So integral of little r times dn is zero.

Â So therefore, the derivative of zero is still going to be zero.

Â So this one, the center of mass location.

Â This is, basically, using the center of mass definition.

Â That one will go to zero leaving us with this term and this term.

Â Now this term is nothing but total mass again.

Â So I have a big M.

Â So mass over 2 velocity squared, so you can see for

Â the blob, the center of mass acts like a particle again, which is really nice.

Â And if you're doing an orbit problem,

Â this would be your orbital energy of your spacecraft basically.

Â And then the second part is this integral, body integral of r dot, so

Â that's the position vector, of each DM relative to the center of mass.

Â Get its inertial derivative though.

Â That's important.

Â This is a dot.

Â This is not a body frame derivative.

Â That's the inertial derivative and we sum them up.

Â So this is basically your kinetic energy about the center of mass.

Â And in this form it contains two parts.

Â One part of kinetic energy comes from rotation.

Â If all the stuff is rotating then these things could be spinning, and

Â that gives you all that kind of kinetic energy.

Â The other part is because this is blobs, you might have deformational energy,

Â where things are changing shape as well and you're redistributing mass.

Â I'm not gaining or losing mass in this system, but I'm just redistributing.

Â So if you had fuel slosh bouncing around, that's one example you know.

Â So you might have deformational energy happening here.

Â So that's what that is, but

Â this is the kinetic energy of a jello, written in the very form, basic form.

Â Now, we very often want to have power,

Â that means the time derivative of kinetic energy.

Â We have that expression, broken up into the translational part of the cm and

Â energy about the cm.

Â If you differentiate this, you come up with these here.

Â Let me show you that one trick.

Â If you don't see that, how to go from there.

Â 5:35

I will simply have 2 times x times x dot, right?

Â Very simple.

Â Here, if you have this vectorial quantities I mean these expressions

Â we see a lot of vectors dotted with vectors again, especially themselves.

Â If you have this and you take a time derivative and here, I'm going to

Â take an inertial time derivative, so I can just put dots over my vectors.

Â With the chain rule, you would have x dot,

Â dotted with x, plus x dotted with x dot, right?

Â 6:20

>> [INAUDIBLE] Reverse the order.

Â >> Right.

Â The answer here, the dot part is actually a scalar.

Â So if you do this in vector form a dotted with b is the same thing as b

Â dotted with a.

Â If you're doing it in matrix form this would have been written as a trans x

Â transpose x.

Â Right?

Â It's equivalent.

Â And in that case too, the answer is a scalar.

Â So any scalar you can transpose and it's the same scalar.

Â So it allows you to reverse the matrix order with the right transposes on it.

Â This is nothing but the vector form of it.

Â So here, here we just end up with 2 times x times x dot, or

Â you could've rewritten this by putting the dot on the first one.

Â Whatever's more convenient in your analytical development.

Â 7:25

So we can plug in this becomes nothing but f.

Â Over here little r dot we have to do a little more math.

Â We knew that big R is Rc plus little r, so you just put two dots over everything.

Â Now we have the inertial derivatives of all these quantities, and

Â I can plug that one in.

Â So that will break up this integral as being Rc double dot, dotted with R dot.

Â And, sorry, there's an R double dot and a Rc double dot that plugs in.

Â You break up the math, this is the form that you're going to have.

Â So this part is actually pretty much done.

Â So the power equation gives you force dotted with velocity.

Â The shift makes sense, hopefully with high school physics,

Â if you look at power stuff.

Â That's typically always the force times a velocity measure.

Â That's the instantaneous power being produced mechanically.

Â 8:22

>> How about the one at the end?

Â >> Yep, this one again, right?

Â Due to the center of mass property.

Â These things will pop up everywhere and we love that, because zero is great.

Â It really simplifies the stuff.

Â So we're going to see that one go to zero and this is nothing but

Â mass times the initial acceleration of an infinitesimal element that was nothing but

Â dF, that little force.

Â So on the body, this is your general power expression that you would have to do and

Â you can write, if you have to energy change between two points,

Â you would integrate these equations over time.

Â And this is one form or you can do a substitution of variables.

Â This is a classic thing you've probably seen where you can also do it force times

Â distance gives you the energy take put into this system.

Â So, you can replace it.

Â There's different forms you can write this.

Â This is the general one.

Â What we'll use in class will be the rigid body specific one, but

Â I just wanted to highlight first the general formulation.

Â We're going to then write it later on.

Â because this one doesn't make much sense practically for you right now.

Â Once you make it rigid, we'll have a turn that would be very very clear.

Â This is how we compute this.

Â