Hi, this is module 33 of Two Dimensional Dynamics and so for today we're going to 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 this, these are the equations that I we came up with last time. And so again I'm going to use this graphical tool on the left hand side. I'm going to do my free body diagrams to show my, my forces and my moments and on the right hand side I'm going to use these motion vectors or effective vectors m x double dot, m y double dot and i alpha. And so here's a, a body and so this body again would be restricted to only rotating about the z axis. That's, that's the only motion that I'm looking at right now. On the left hand side I put on my forces acting on the body and any effective moments are, excuse me, any moments that are applied to the body. And then on my kinetic diagram I have my, two orthogonal M A vectors. In this case I've chosen M X double dot and M Y 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 arms, so R from p to c crossed with these effective forces which are m x double dot, and m y double dot. And so that will 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 of the mass center for this body, and we want to assume that the coefficient of friction is large enough to prevent slipping of this, this cylinder or wheel. And so let's go ahead and look at a demo. And here's my, my wheel, and the first thing I want to ask you is which direction is the wheel going to 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 the day. So let's now take. You can think about that. Why did that happen? You saw it it went uphill and also it even started to slide, so that's not good, because we said we would have no, no no slipping. So here's here's a better wheel. When I let it go, it's going to roll down the plane, no slipping. And I want to find the motion of the mass center, and predict that. Okay, how are we, how are we going to approach this problem? That's not trivial. It's, it's always difficult to get started. And so what you should say is, I want to apply these equations to this problem. And so I'm going to use my graphical tool of FBD equals KD. And I'd like you to begin by drawing the free body diagram. And this is the result you should've come with. You have your, 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 going to have our kinetic diagram that's going to display these effective forces and effective motion or effective moments, 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 going to have an m x sub c double dot down the plane and m y sub c double dot into the plane. And by the right hand rule, I'm going to have an I sub zzc alpha counter-clockwise. But since I'm only going to have planar motion, planar rotation, I'm just going to 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,' because 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 going to go ahead and apply these eq, equations and I'll start with summing forces in the x direction on my free body diagram so, 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 sine of beta, and then I've got my friction force. It's going to 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 going to call that equation one. 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 have been minus n plus mg cosine beta equals zero, because there's no acceleration in the y direction, but that's the same as n minus m g cosine beta equals zero. So it ends up the same. We'll call that equation two and now we have our body. We have one more equation that we're going to work with, and that's the, 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, I'll call counter clockwise, 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 going to tend to cause a counter-clockwise rotation, so that's positive, equals my I alpha. Notice the effective force of m x double dot goes, it's line of action goes through c so it's not going to cause a moment on the right hand side. And then I substitute in i sub c around this zz axes. That's the mass moment inertia. I would look it up in the textbook reference, or online, or I could do actually the triple integer again. But this is the result. The i, the mass moment inertia of a cylinder about the mass center through the z axis for a uniform cylinder is one half m r squared. And that's theta sub double dot, theta double dot sub c. We'll call that equation three. One thing I want to note is we're not limited to just summing moments about c. If, 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 going to continue with summing moments about c, but I would like you on a, as an exercise on your own to go ahead and sum moments about another point and see if you can get the same results. 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 going to leave that as an exercise on your own. We're going to continue on. I now have my x equation, my y equation, and my moment equation and so I can sub three into one. The result I get is, is is shown here. You can see now that the masses cancel out. I have 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, and, and two unknowns. So I'm going to need some more information. And 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, are two unknowns. So, the kinematics we can use are for a no-slip wheel. If you go back to the module where I did rolling wheels, we can use those, 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 four if you will. And sub this is equation four, but we'll substitute this into equation four. And I get m sub g sine beta minus one half m r theta double dot sub c squared and m r theta sub c double dot on the right hand side since I, I substituted this in. And so I can cancel the Ms again, and the result I have is shown here. Okay, so continuing on now I'll solve for beta double dot c. I've got g sine of beta. If I carry this term to the right hand side, I get equals three halves or three halves r, theta sub c double dot. And I can solve for theta sub d, c double dot. That equals 2g sine beta over 3r. But in the problem, I want to find the motion of the mass center, so I've got to put this in terms of x for the motion of the mass center. I can use my kinematic relationship again. I've got theta sub c double dot equals x sub c double dot over r. If I substitute that in, I get x sub c double dot equals 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, the friction. This was equation three if you'll recall. I can now, 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 equals one half m r times 2g sine beta over 3r. Once again the Rs cancel. And I get f equals m g sine beta over 3. So I now have an expression for x, so double the, or x double dot and a fric expression for the friction. Here's the expression for x. If I want to find the mass, 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 going to be multiplied by time. Plus when I integrate, I get a constant of integration. However, since this, a body is released from rest, that goes to zero. [BLANK_AUDIO] 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, equals. Now I've got 2 g sine beta over 3. When I integrate t, I get t squared over 2. And I'm going to get another constant of integration, c2. But, for the, it's, again, since I'm going to call my datum when I release it, 0, this C sub 2 goes to 0. And, so I end up 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 sine of 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 [COUGH] that came from the equation for 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 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 going to be less than or equal to mu sub n. And so, I'll substitute in for f. That's mg sine of beta over 3 is less than or equal to mu sub n. I'll substitute in for n as mg cosine beta, and that leads me to mu has to be greater than or equal to tangent beta over 3. And that makes physical sense, because as beta gets larger, that means I'm going to 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, a crank arm, a mechanically 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.
