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Hi, and welcome to module 19, of Three Dimensional Dynamics.

Today's learning outcome will be to take the the review that

we did in the last couple of modules about angular momentum,

and we looked at it from my two d course, and

now we're going to extend this to rigid bodies in 3D motion.

And we're also going to review the moments of inertia and products of

inertia, which we saw in the review for my 2D dynamics course as well.

And so again, moment of momentum is also called

angular momentum, if we look at a system of particles

here, so I have a bunch of particles, each one

of those particles has a linear momentum, mass times velocity.

And then if I cross the R vector with each of these

linear momentums, we call that the angular momentum, and it's about point P.

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So we can now extend this for a rigid body, rigid continuous body undergoing.

We're not going to restrict it to two-dimensional motion now.

We're going to let it go under three-dimensional motion.

And so, Angular Momentum, or Moment of Momentum of body B,

this is body B, about point P, this is a point P

an arbitrary point P, is defined as now instead of a summation, it's the

integral of R, these little position vectors, crossed with vdm.

Which vdm, again is the linear momentum of each differential piece of mass.

Now we have to be careful, because this velocity has to be, for,

for the definition of Linear Momentum, it has to be an Absolute Velocity.

So it has to be defined with an Inertial

Reference Frame, which I'm calling this frame F down here.

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Okay, so here, here it is again.

And we need to find this velocity v, the Absolute Velocity of this point v.

And so what I'd like you to do, is try to do that on your own and then come on back.

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And what you should say is to find this Absolute Velocity

v we're going to use two points on the same body B.

We're going to use point p, and we're going

to find the Absolute Velocity of point p from the

Inertial Reference Frame, and then that's going to be plus

omega of the body crossed with this little r here.

That will give me the velocity of the point out here.

So that's the relative velocity equation, which we used

all the way back to my 2D Dynamics course.

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Now we can substitute that in.

So I'm going to substitute that in here, and this is the result I get.

And I've broken it into two parts.

I've got the VP part down here, and omega cross R part here.

And you'll notice that I drop the omega B

with respect to F, and I'm just calling it omega.

This is the angular velocity of the body B with respect to the frame F.

And so here now I've moved the dm over to the integral

of r, and since vp is not going to be integrated,

that's the velocity of a point p, I've left this term alone.

What I want you to do next, is tell me what this integral of

r dm, give me in another expression for that, or another way of writing that.

And we've done this before in my earlier course courses.

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And so what you should say here is, okay, I'm taking the integral.

Now the, this r is the position vector to each piece

of mass in the body, integrated over the entire body.

And so that's the same as the total mass of the body,

with a position vector directed from r to from P to C.

Okay?

So that's a position vector from p to, let's say that c is here.

Okay?

And so, that's a position vector from r, from p to c.

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And that's probably a little bit off.

Let me, let me rewrite that.

I'll put my, put my, my my Position Vector

now, I'll put it more towards the center, position c.

And so that's a vector r from P to C.

So I'm taking, again I'm going r crossed with

each little piece of dm, I'm integrating it over

the body, that's the same as the total mass

times a position vector from r p to c.

All right.

Okay so, this is where we've left off.

Now r little r, as we look around the body, is going to be as we go

to each point along the body, that's going to to be, this is that r right here.

That's going to be, have some component in the x direction, have

some component in the y direction, and some component in the z direction.

Now as far as omega is concerned, that's the Angular Velocity of the body,

and so it's going to have an x component, a y component, and a z component.

And again, I can substitute those in, so I'm

going to put this in here, and this in here,

also this goes in here, and I'm going to

use that vector identity which I've used several times before.

And if you do the math, what you'll find is,

you now get H of p equals this first term.

M from [COUGH] times r, the position vector, from

P to C, crossed with the absolute velocity of P.

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And then, as I put these in, my other terms look like this.

And so my next question, and we just reviewed this in the

previous two modules for my 2D course, is what is this term here?

What is the integral of x squared plus y squared dm.

And what you should say is, that's defined as what we call I sub zzp,

which is the Mass Moment of Inertia about the z axis through point p.

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And my next question is, what is this integral here?

Minus the integral of xzdm.

And what you should say is, that's a product of inertia.

That's I sub xz about point p, or the product of

inertia with respect to the x and z axis, through point p.

And similarly now, we've got this term

here, integral, minus the integral of yzdm, that's

also a product of inertia, with respect to the y and z axis through point p.

And so now you see, instead of having these

three terms which we did have in 2-D dimensional dynamics.

We now we end up with another.

This is a Mass Moment of Inertia.

This is another product of Inertia.

This is another product of Inertia.

Another mass of moment of Inertia.

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And so, I'm going to write this in a little bit of shorthand.

I've got H sub p now, is equal to M times r from P to

C crossed with Vp, and then I've got plus now.

I've got, this is Ixx, through point p.

And that's times omega sub x.

And then I have plus I, x, y through point p.

Omega sub y plus I x z through point p omega

z, and all of those are in the I direction.

And then I've got plus, this is a product

of Inertia x y, so that's I sub xy through

point p omega x, plus I sub.

This is actually xy or yx.

And it's the same.

This is also Iyy through point P omega y.

And then we have plus Iyz through point p omega z,

and all of those terms are in the j direction.

And then I have my final line here.

I've got I, this is Ixz.

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This is yz through point P omega y.

Plus finally this is a product of inertia again,

Izz through point P omega z in the k direction.

And so that's just a little short hand using the symbols for

my products and Mass Moments of Inertia instead of the full integrals.

And so, if I write that out again it looks like this.

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And so, it gets a little long, and so if you remember back to your

linear algebra days, and matrix vector algebra.

You can put this in matrix vector form, and so I have this first term plus now.

I've written my Inertia properties, my products of inertia

and Mass Moments of Inertia and my matrix, and then

I have my Angular Velocity components in a vector,

and you can see that these are the same now.

I've got this term plus Ixx P omega x.

Okay?

Plus I, yz omega y.

Plus I xz omega z, but when I see I've missed an omega z here, a couple of times.

So this should be omega z, and this should omega z, and

this should be omega z, so we don't want to forget those.

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And then i've got plus I xy omega x, plus I yy

omega y, plus I yz omega z, and they're in the j direction.

And then finally the last one, I xz omega x, plus I yz omega y, plus I zz omega z.

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Now this matrix is called I sub p, and it's the Inertial Property Matrix.

It's, or sometimes it's referred to as the Inertia Matrix.

You'll notice that it's symmetric.

And, also, you'll notice that it includes in the integrals, information

about the mass and also, the shape and geometry of the body.

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'Kay, so here it is again.

Now we can consider a couple of specific points.

If point p actually is the Mass Center, so

let's say that point p coincides with the Mass Center.

Okay?

So this is C and that's, this p is gone, and so C equals p.

Well, then, if C equals p, R from p to c goes to 0.

And so, all I end up with is H about point c is equal to just I, the

Inertia Matrix for point c, times the Angular, Velocity.

And there's one more special point.

We can also look at a point p where it's fixed in an Inertial Reference Frame.

And so V sub p.

Let's say that we have a hinge here, so V sub p is fixed

in an Inertial Reference Frame, and we'll call that the same as point O.

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Different from this O down here.

Since now V of p has zero velocity since it's fixed, we

just end up with H about point O is equal to I omega.

So that's, a couple of simple forms of this formula.

Very specific cases that we will use a lot.

And so this portion of the, of the term

goes away, if we're talking about two spacial points.

If p is this mass center, or if p is a fixed point in a Inertial Reference Frame.

And that's where we'll leave off this module, come back next time and finish up.