[BLANK_AUDIO] Hi and welcome to Module 13 of Two Dimensional Dynamics. Today's learning outcomes are to define Angular Momentum, and to develop Euler's 2nd Law which is also called the Moment Equation. So, today's lesson is quite a little bit of theory, but it's important theory that we're going to use throughout the course. And so I I hope you can stay with me and, and follow each of the steps. You should be able to. And so, let's look at Euler's 2nd Law. First of all, let's recall Newton's 2nd Law for an Inertial Reference Frame. Newton's 2nd Law says that the, some of the forces acting on a particle is equal to the time derivative of the linear momentum of that particle, or the momentum of that particle. Since a particle can only have linear momentum. And if I take a system of particles, I have a bunch of particles here. I can then cross, I can look at the, the system of particles from some Inertial Frame F, maybe there, the earth or some fixed frame Inertial Reference Frame. I can look at it and I can draw a position vector from the origin of that fixed frame to each of the particles. I've done that pair, I've crossed that position vector on both sides of my equation for Newton's 2nd Law. And I know from way back in my earlier course introduction to engineering mechanics, that R cross F is defined as a moment. So I've got the moment about o is equal to the time derivative of this stuff on the right hand side. And this stuff on the right hand side, its the time derivative of what we define as the angular momentum, and we give it the symbol capital H. And so its, its like the moment of momentum. It's called the moment equation, so we have R cross m v, m v is the momentum, so we have the moment of momentum or angular momentum. That's for a fixed referenced frame, let's look for any arbitrary point P. For an arbitrary point P, the angular momentum about point P is the angular momentum about the mass center, plus a position vector from this arbitrary point P to the mass center, crossed with the linear momentum of the entire body. And so H sub p again, angular momentum about point P, H sub C angular momentum about mass center C. And this is the moment of linear momentum about point P. So you can actually see that development for an arbitrary point P in any standard dynamics textbook. So let's go back now, and look at the system of particles as referenced from a point fixed in an Inertial Reference Frame. And this is the, what we had come up with. So, I have the sum of the moments about o, and I can use the product rule here to take the derivative. That's times, the timed derivative of the position vector crossed with m v, plus the position vector crossed with the timed derivative of m v which is m a. But you'll note that the position, the derivative of the position vector Ri, is the same as Vi. And so, this gives us Vi crossed with Vi, and we know that Vi crossed with Vi or Ri dot crossed with Vi is then going to be equal to 0. And so what I'm left with is, Euler's 2nd law for a point fixed in an Inertial Reference Frame is the sum of the moments about point O is equal to the summation of R cross m a. Or some of the moments about point O is equal to the time derivative of the angular momentum of point O, for the definition of angular momentum. So that's one result for Euler's 2nd law or The Moment Equation for a point fixed in an Inertial Reference Frame. The sum of the moments equals the time derivative of the angular momentum about point O. Similarly and again, similarly again, you can look at this in in any standard reference text for the development. You can find that Euler's 2nd Law about the mass center, is equal to the time derivative of the angular momentum about the Mass Center. So here are the results, the moment equation about a point fixed in Inertial Reference Frame about the Mass Center. However in general, you need to note that this form of some of the moments equals the time derivative of the angular momentum, only works for a point fixed in a Inertial Reference Frame or the Mass Center. It does not work for arbitrary point P. And so in general, the sum of the moments about P does not necessarily equal the time derivative of the angular momentum about P. To show what that is, let's look at or recall our knowledge of equivalent force systems from way back in my my first course introduction to engineering mechanics. We saw by a balance of moments that the sum of the moments about P, balances with some of the moments about C plus R cross F. So that's just a balance of equivalent force systems shown from this picture. Now in this case, in dynamics, we've developed that the sum of the moments about C is equal to the time derivative of the angular momentum about C. We know that force, the sum of the forces is the time derivative of the linear momentum or d dt m vC. And so we get the relationship for the sum of the moments about P is equal to the time derivative of the angular momentum about C, plus rPC cross, m aC. And so it's not in this form as shown up here, but for an arbitrary point it's in this form. So here's the summary. Sum of the moments about O or C is equal to time derivative about, of the angular momentum about O or C respectively. That form is not true for an arbitrary point, for an arbitrary point, this is, this is true and we'll use this, these equations quite a bit in the remainder of the course. And so it's important to have a, a good understanding of those. And I did want to make one other point. If you recall, let's look at, we had the sum of the forces equals the time derivative of the linear momentum for a body. And so, what you'll see is, this is completely analogous. [SOUND]. And so, this'll maybe give you a little bit better feel. Sum of the forces is equal to timed derivative of the linear momentums. Sum of the moments is equal to the timed derivative of the angular momentum, and so you see the, the connection there. And that's where we'll pick up next time. [BLANK_AUDIO]
