Hi, this is module 33 of Two-Dimensional Dynamics, and so for today we're gonna go ahead and solve an engineering problem or engineering problems using the equations of motion that we developed last time for a body in two-dimensional planar rigid body motion. And so these are the equations that we came up with last time. And so, again, I'm going to use this graphical tool, on the left hand side I'm gonna do my free body diagrams to show my forces and my moments, and on the right hand side I'm gonna use these motion vectors, or effective vectors mx double dot, my double dot, and I alpha. And so here's a body, and so this body, again, would be restricted to only rotating about the z-axis. That's the only motion that I'm looking at right now. On the left hand sid,e I put on my forces acting on the body at any effective moments, or excuse me, any moments that are applied to the body. And then on my kinetic diagram, I have my two orthogonal ma vectors. In this case I've chose mx double dot and my double dot. And I also have my rotational vector, I alpha. And so you'll also recall from a knowledge of equivalent force systems, that I can sum my moments about any point on the body, and I'll take into effect any applied moments, and the moments due to the forces acting on the body. And then on the right hand side, if I sum moments about P instead of C, I can still do that. I get the effective moment, I alpha, plus these effective forces with their moments arm. So from r to P to C cross with these effective forces, which are mx double dot and my double dot. And so that'll allow us not only to sum moments about C but about any point on the body. Okay, so let's go ahead and look at an example. This is a cylinder, and it's got a mass m, a uniform cylinder, mass m, radius r, it's released from rest. You have a coefficient of friction on the surface that's mu, and we want to determine the motion for the mass center for this body, and we want to assume that the coefficient of friction is large enough to prevent slipping of this cylinder or wheel. And so let's go ahead and look at a demo. And, here's my wheel, and the first thing I want to ask you is, which direction is the wheel gonna go, up or down the plane? And so, go ahead and answer that, and if I let go. Whoops it went uphill. That's kind of weird. That's my magic trick for today. You can think about that, why did that happen? You saw it went uphill. And also, it even started to slide, so that's not good, cuz we said we would have no slipping. So here's a better wheel. When I let it go, it's gonna roll down the plane, no slipping. And I wanna find the motion of the mass center, and predict that. Okay, how are we going to approach this problem? That's not trivial. It's always difficult to get started. And so what you should say is I wanna apply these equations to this problem, and so I'm gonna use my graphical tool of FBD = KD and I'd like you to begin by drawing a free body diagram, and this is the result you should have come up with. You have your weight. You have your friction force that's going to oppose impending motion to keep it from slipping. And we have our normal force. And then on the right hand side we're gonna have our kinetic diagram that's going to display these effective forces and effective motions, or these motion vectors, if you will. I've chosen x down the plane and y into the plane as my orthogonal coordinates. So I'm gonna have an mx sub c double dot down the plane. And my sub c double dot into the plane. And, by the right-hand rule, I'm gonna have an I sub zz c alpha counter-clockwise, but since I'm only going to have planar motion, planar rotation, I'm just gonna abbreviate that as I sub c alpha. Now, what you should realize is that I can have an acceleration of the mass center down the plane, but that mass center is not going to have any acceleration into the plane, cuz it's going to go parallel to the slope, and so this kinetic vector will zero out. And now what do we do? And so what you should say is we're gonna go ahead and apply these equations, and I'll start with summing forces in the x direction on my free body diagram and set it equal to the forces on my kinetic diagram in the x direction, so I will have the mg force. The x component is the sine component. So I've got mg sin beta. And then I've got my friction force. It's gonna be negative in accordance with my sign convention. The normal force is only in the y direction, and over here the only effective force I have is the mx sub c double dot down the plane, so it's in the positive direction. And we're gonna call that equation 1. Now, the next step is what? And what you should say is okay, apply the next equation. So we're going to sum forces in the y direction. I'd like you to go ahead and do that on your own, and come on back, and so this is the result you would come up with. Notice that I chose down and to the right positive, so this would've been -N + mg cos beta = 0, because there's no acceleration in the y direction, but that's the same as N- mg cos beta =0, so it ends up the same. We'll call that equation 2, and now we have our body. We have one more equation that we're going to work with, and that's the moment equation. So I'm going to go ahead and sum moments about point C on the free body diagram, set it equal to the sum of the moments about C on the kinetic diagram. I'll call counterclockwise positive. And so I get, if I sum moments about C, the line of action of the weight force and the normal force goes through point C, so I just get F times its moment arm which is r, It's gonna tend to cause a counterclockwise rotation that's positive, equals my I alpha. Notice the effective force of mx double dot ghost. It's line of action goes through C so it's not fgonna cause a moment on the right hand side. And then I substitute in I sub c around this zz axis, that's the mass moment inertia. I would look it up in textbook reference or online or I could do actually the triple integral again. But this is the result. The I, the mass moment of inertia of the cylinder about the mass centered through the z axis for a uniform cylinder is one-half mr squared, and that's theta double dot sub c, we'll call that equation 3. One thing I wanna note is, we're not limited to just summing moments about C. If I go back here. With my generic example. You can see that I not only can sum moments about C, I can sum moments about any point. And so I'm gonna continue with summing moments about C. But I would like you, as an exercise on your own, to go ahead and sum moments about another point, see if you get the same result. So you might choose P as the point of contact, and sum moments about that. See if you can get the same result. But again, I'm gonna leave that as an exercise on your own. We're gonna continue on. I now have my x equation, my y equation, and my moment equation. And so I can sub 3 into 1. The result I get is shown here. You can see now that the masses cancel out. I got a mass in every term. I would know what sine beta is, I know what beta is, it would be a given. I know what the acceleration due to gravity is. And so I end up with two unknowns, theta sub c double dot and x sub c double dot. So I have one equation and two unknowns, so I'm gonna need some more information. My question to you is, what other information do I know about this problem to finish it up? And what you should say is that you're going to have to use kinematics to relate theta double dot sub c and x double dot sub c, which are our two unknowns. So the kinematics we can use are for a no-slip wheel. If we go back to the module where I did rolling wheels, we can use those kinematic relationships. And for a no-slip wheel, we found that x sub c double dot, the linear acceleration of the mass center, is equal to r times theta sub c double dot, r times the angular acceleration of the mass center. So I can now substitute, we'll call this equation 4, if you will. This is equation 4, but we'll substitute this into equation 4, and I get. Mg sin beta minus one half mr theta double dot sub C squared and mr theeta sub C double dot on the right hand side, since I substituted this in, and so I can cancel the m's again and the result I have is shown here. Okay, so continuing on now, I'll solve for theta double dot C. I've got g sin beta. If I carry this term to the right-hand, side I get = three-halves r theta sub C double dot, then I can solve for theta sub C double dot. That equals 2g sin beta over 3r. But in the problem I wanna find the motion of the mass center, so I've gotta put this in term of x for the motion of the mass center. I can use my kinematic relationship again. I've got theta sub C double dot = x sub C double dot over r. If I substitute that in, I get x sub C double dot = 2g sine beta over r, or excuse me, over 3. Now, before I continue on to integrate this and find the motion of the mass center, let's go ahead and look at the friction. This was equation 3, if you recall, I can now cancel the rs. And I end up with f equals one half mr theta sub C double dot. Let's go ahead and substitute in for theta sub C double dot. And I get f = one half mr times 2g sin beta over 3r. Once again, the r's cancel and I get f = mg sin beta over 3. So I now have an expression for x, our x double dot, and an expression for the friction. Here's the expression for x. If I wanna find the motion of the mass center, I'm going to have to integrate this. So I've got the integral would be x sub C dot, taking acceleration to velocity equals 2g is a constant, sine of beta is a constant, over 3, so that's gonna be multiplied by time. Plus when I integrate I get a constant of integration. However, since this body is released form rest, that goes to 0. I can integrate again to get displacement. So I've got x sub C, let's go ahead and do that in black, x sub C = now I've got 2g sin beta over 3, when I integrate t I get t squared over 2. And I'm gonna get another constant of integration, C2 but again, since I'm going to call my datum where I released it 0, this C2 o goes to 0. And so I end up with the answer with the answer to my problem, which is what is the motion of the mass center, x sub C is going to be equal to g times t squared over 3 sin beta, again because these twos cancelled out. And so that's one of our answers, or actually the answer. But before I finish up the problem, let's also look at the friction, and so here's the expression for the friction that I came up with. Here's the normal force, given that came from the equation for the sum of the forces in the y direction. How much friction am I going to need for the no-slip condition? Well, let's look at that. The maximum friction I will be able to get is mu times N for the Coulomb friction, so I know that the friction has to be less than or equal to that max friction I'm able to get. Okay, so that's gonna be less than or equal to mu sub N. And so I'll substitute in for F, that's mg sin beta over 3 is less than or equal to mu sub N. I'll substitute in for N as mg cos beta. And that leads me to mu has to be greater than or equal to tan beta over 3. And that makes physical sense, because as beta gets larger, that means I'm gonna have to have a higher coefficient of friction to prevent slipping. Okay, so these are rather involved problems. But you get better and better by practicing. And so I've got a worksheet for you to do. Here's a crank arm, a mechanical engineering type mechanism. So go ahead and practice this on your own. I'd also ask you to look for other references for problems like this. The more you practice these problems, the better you'll get, and so I'll see you next time.
