Hi, this is Module 6 of 2D dynamics. Let's look at today's learning outcomes. We're gonna look at the definition of kinetics. We're gonna state Newton's laws for particles. We're gonna derive Euler's first law for the motion of the mass center of bodies, and then finally we're gonna calculate the location of the mass center for a composite body. So, kinetics of particles and mass centers of particles. So far we've looked at the kinematics of particles and systems of particles. And that was the geometrical aspects of the motion, relating things like position, velocity, acceleration and time. Now, kinetics relates to forces acting on a particle or a body to the changes in the motion of that particle or body. And so, here's, as a review from my earlier courses, here's Newton's first law. Remember now, Newton's laws apply to particles, and so the first law says if I have a balance of forces on a particle, then the particle is at rest or remains in constant velocity, okay? And so what I want you to do now is to write out in your own words, what's Newton's second law for particles? What you should have come up with is that Newton's second law says that the time rate of change of the linear momentum or the momentum of a particle is equal to the sum of forces acting on the particle. So no longer do we have a balance of forces, we don't have the sum of the forces equals zero. But we have some change in the momentum, and we're going to be talking about in this course classical dynamics, where there are no relativistic effects and so we have constant mass. And I can take the mass, which is constant, outside of the derivative, and I get mass times acceleration, so this is your famous F=ma for particles. And then finally, write down in your own words what Newton's third law is. Okay, so Newton's law says if I have two particles, the forces between those particles are equal in magnitude, opposite in direction, and along the same line of action, or collinear. So here's my notation. This is the force on particle 1 due to particle 2 is equal to negative, because it's in the opposite direction, the force on particle 2 by particle 1. And so equal and opposite forces. So now let's extend these laws for particles to Euler's laws for bodies now. We want to look at bodies as a whole or systems of particles and then on to bodies. And so here I have a body composed of capital N particles, where capital N is a large number. I've only drawn a few of these particles, but I could have millions of these little particles in this body. And so that's a system of particles. There's a bunch of external forces acting on this system of particles or body. So I'm saying that F sub i is the net external force acting on particle i. And here's particle i right here. I've labelled it m sub i because it has a mass m sub i. Here's particle j, so I would say there are i through N particles in the body, so a total of N particles. Now in addition to the external forces felt by particle i, the ith particle also feels these internal forces from all the other particles in the body. And so now you can write the Newton's second law for the ith particle, and so for the ith, try to do it on your own and then come back and see how you did. And so what we have here is the external forces acting on particle i plus the internal force is acting on particle i, or the total forces acting on particle i equals the mass times the acceleration of particle i, by Newtons second law. And then we can for the entire system of particles or body, we can sum up all of those equations. So we're gonna sum N equations over N particles. And now I put the summation sign on here, in addition to all of these individual particles, we sum them up from i = 1 to N for the total number of particles in our body. So my next question to you is let's look at this second term, and can you simplify this second term for me? And after you think about that for a while, come on back. So what you should have realized is that by Newton's third law, this term is equal to zero, because if I go through my indices, from i=1 and j=1, I have f12, f21, remember they cancel out. f13, f31, they cancel out. f23 and f32 cancel out, and so if I sum up over the entire body, because each of those internal forces are equal and opposite, they end up zeroing out. And so all I'm left with is the total summation of external forces on our system of particles is equal to the total of these particles' mass times acceleration. And so here again is that same equation, and now I've drawn my system of particles or body here again, and I've got, I'm looking at it from an inertial reference frame where I'll call that point O and the origin in which I'm looking at it from, the body from. And I've got a position vector R to each of the particles. This is a position vector from O to the ith particle, and then I also have written the position vector from O to the mass center of the body. And so since I've shown the position vector from O to the ith particle, I can sum up over the entire body again and pull the double integral out, because we know that for the acceleration it's the second derivative of the position. And so these are equivalent. Once I've done that, let's look at the definition of center of mass. If I sum the mass of each of the particles times their positions vector, so I go to this particle and this particle and this particle, sum them all up and then I divide by the total mass of the body, that by definition of the mass center will give me a vector from O to C, the mass center of the systems of particle or the body. Okay, and I can rearrange that. And so I have m times rOC. Instead of a summation sign, if I change this to a continuous body so I have an infinite number of particles, then it becomes the integral of the position vectors over each little piece of mass or differential piece of mass in the body. Okay, we can now use this definition and put in for the summation of mR for a continuous body is equal to m rOC, or the integral of R dm. And so this is two forms for a continuous body, I can write, and then I come up with what's called Euler's first law. If I take the second derivative, remember mass is constant, we're talking about classical dynamics. And so the second derivative of the vector from O to the mass center is just the acceleration of the mass center. So what this says is, for a system of particles or body, the sum of the external forces acting on that system of particles or body is equal to the mass times a very particular point, mass times the acceleration of the mass center. And so you can see Euler's first law for bodies is completely analogous to Newton's second law for particles, for a particle we don't have to have the subscript of C here, but for the body we have to specify the particular point that we're talking about the acceleration. And, so this is the equation that describes the motion of the mass center of bodies. So since we're gonna have to find the mass center of bodies in solving these problems, I've got a review problem here for you from my previous courses, Introduction and Applications in Engineering Mechanics. What I want you to do is consider this body, which is a composite of a uniform sphere and a uniform cylinder, and each has a density of rho. And I want you to find the mass center of that composite body. So find where the mass center is located from the origin of this xy-axis. And you can do that as an exercise on your own. I've got the solution in the module handouts, and I will see you next time.