This is module seven of two dimensional dynamics. Today's learning outcome will be to solve the motion of the mass center of bodies using particles and bodies, actually using Newton-Euler motion equations. And so, this is where we left off last time. We took Newton's second law for particles and we extended it to the first law for bodies, where the sum of the external forces acting on the body is equal to mass times the acceleration of the mass center of the body. And so going back to my statics courses, Introduction to Engineering Mechanics and Applications of Engineering Mechanics, we used a graphical tool called a free body diagram, or FBD, in representing the external forces Forces acting on the body. And we're going to continue to use that in this course. And so here's my body. I've got the external forces acting on the body. In this case, I've made sure that the line of action of all the forces are concurrent, so that they're going through a single point, since we're only talking about translational motion at this point. Later on, we'll be talking about rotational motion and we'll be able to look at non-concurrent force systems. But, in addition to the free body diagram, what I like to do what helps me in learning and, and visualizing is to have another graphical tool for the right-hand side of the equation and I call that my kinetic diagram. And so here's my body again and my kinetic vectors are the vectors associated with emotion or these MA vectors, which are essentially effective forces, if you will. And so, since this M A is a vector, I'm going to break it into its two scaler components in the X and Y two dimensional plane. So I have MA C in the X direction, MAC in the Y direction. So the, this picture, is just a graphical representation of this Euler's first law. And, and I just, as I say, I like that, it help me keep my sign straight and helps me when I'm solving these problems. So let's go ahead then and and do a problem. So let's look we have, in this case I'll look at a particle, there will be some worksheets later on doing bodies, but for the particle we're going to use the Newton's Second Law it moves along the outside of the curved surface as shown and so there is a force acting on the particles going along these sur, curved surface and we're going to say that there's no friction there. And what we're going to do is, we want to find the condition that will ensure that this particle P will stay in contact with the, the, the curved surface. And so I always like to try to think in my own mind how does this relate to the real world? Why am I doing these types of problems? And you know, the other day I was riding past an amusement park. It's it's in the middle of the winter up in upstate New York where my family is from. But I saw this amusement park and I said oh,you know, that's a good picture of of this type of problem. Maybe I can think of the, the the compartment or the the pa, passenger compartment as, as the particle riding along this track. And we want to find out what the condition is so that this particle, or passenger compartment doesn't fly off the track. now, obviously, we're fortunate because, in these rides we have constraints that make it sure it doesn't fly off the tracks. But, it, it, it, it, it, it's a, it's, you know, it, it, it, it makes it a more realistic problem. So, let's go ahead and do it then. What I'm going to do is I'm going to use my graphical tool of FBD equals KD. And I want to draw my body of interest, which is my particle in this case. I'm going to draw it for both the free body diagram and the kinetic diagram. Go ahead and try to on your own do the free body diagram, and then I'll do it with you together. So for the free body diagram I've got this applied force, F. I've got a normal force, N, acting up from the surface. And I've got the weight of the particle acting down which is mg, and then let's put in our angles here. So our dimensions, this angle is alpha, and this angle is alpha. And that's a good free body diagram. Now for the kinetic diagram, since this, particle is moving along a cur,uh, a curved surface what I'm going to do is, a convenience and a coordinates to use because it's moving along a curved surface, would be normal and tangential components as we've seen in the earlier modules. And so for this body, I gotta have two orthogonal mass times acceleration, vectors and so, I'm going to have mass times acceleration tangent, tangential. And I'm going to have, let's see. There we go. We have, that's the congressional component of acceleration and then I have the normal component of acceleration or radio component of acceleration directed radially inward times it's mass. That's mass times acceleration, normal. And, again mass times acceleration normal, if you go back and, review the earlier modules we saw that the acceleration normal was related to the velocity by the magnitude of the velocity squared over the radius of curvature row and so the last thing I want to do is put in my my angles again here, so this is the angle alpha. And this is the angle alpha. And so I now have a good free body diagram equals kinetic diagram. So what would you do next? And so what you should say is okay, let's go ahead and apply equations of equilibrium. Now, instead of just summing forces in the normal and tangential direction and setting them equal to zero, we have to set them equal to my effective force or my kinetic vectors that I've drawn on my kinetic diagram. So, let's go ahead and sum forces in the normal direction. So I'm going to say, I'll, I'll call up and to the right positive. And, I've got some of the forces in the normal direction on my free body diagram equals some of the forces in the end direction, normal direction on my kinetic diagram. I want you to go ahead and, and write that equation on your own. Come on back and we'll do it together. So you should have gotten, okay N is in the positive normal direction that I've chosen for my sign convention. F is in the tangential direction so none of it's component, contributes to this equation. I've got minus mg and it's the component is the cosine, alpha component, so this is minus mg cosign alpha and that's good for the left-hand side, equals in this case I've got; minus M times acceleration normal or minus M, magnitude of V squared over Rho. And so, that's the equation and so, the next thing I want to ask you is, think about it what will be the condition for the normal force if we want to ensure that P will stay in contact with the curved surface? Okay, the condition you should have come up with is the normal force is going to have to be greater than or equal to zero to have the particle maintain contact with the surface. If it becomes less than zero that means that the particle has flown off the surface. And so to maintain contact, we must have N greater than or equal to zero. And so in that case we have N equals mg cosine alpha, minus M magnitude of V squared, over Rho. And let's see, that has to be greater than or equal to zero. I'll cancel out the Ms, and so I find out that the velocity here must be less than or equal to rho g times cosine alpha, and we have to take the square root of that. And so that's our answer. That's the condition. If the velocity of my particle is less than the quantity of square root of Rho G, cosine alpha, and then the particle will be remaining on the surface. So it goes faster than that, it's going to fly off and so, that's a, a, a nice problem and we'll come back next time to do some more problems. So I'll see you then.