Welcome to module 16 of two-dimensional dynamics. Today's learning outcomes are to define the instantaneous center of zero velocity, or something we will call the IC as an abbreviation. And to locate the instantaneous center of zero velocity or the IC on bodies undergoing planer motion. So the instantaneous center of zero velocity is a point about which a body seems to be rotating at any given instant or instantaneous, like a snapshot in time. It has zero velocity, and there is only one instantaneous center per body per instant of time. The location of the IC can actually be on or off the body, and we call that the extended body. In locating the ICs, I'll use this symbol I with a circle around it. And we'll have three techniques that I'm going to teach you on how to locate it. One is to look for a point of zero velocity on a body. Another is to look at the perpendiculars to the velocity vectors on a body or to use geometry and similar triangles to find the IC. So let's start with our relative velocity equation between two points. In this case now we're going to go between the IC. And point Q. Looking at the rotational angular velocity of the body and a position vector from the IC to Q. But by definition, the IC has 0 velocity, and so this term goes away. And we end up with the velocity of any point on the body is just equal to the angular velocity of the body, crossed with a position vector from the instantaneous center of the body to that point q. So, let's go ahead and do some practice on locating IC's. First let's start with a no slip wheel. So I have a wheel here. If it's undergoing, it's no slipping at the surface, contact surface here. And as we roll along, the point right where the wheel touches the surface at this point in time is the point of zero velocity. And so I've labeled it here I. Now, some of you may not see that all so well. So I've got a little bit of a demonstration here to show you why that's the case. I have a little track dozer here. And on this track dozer, my track surface. If I put my finger right on the track where it touches the surface, the no slip surface, you can see that my finger does not move. So my finger has to have zero velocity. And if I go back here now and if I morph that track to where less and less of the surface is touching the ground. You'll see that right at the point where it touches the ground at that instant in time, you actually do have zero velocity. And so, at that period of time every point on the wheel appears to be rotating about that instantaneous center. Okay, so that's the tool for locating the IC using a point of zero velocity. The next tool is to look at the perpendicular, the velocity vectors on the body. And so I want to find the IC for bar BC here And so I know that the velocity of B because AB is a RAFA bar, rotation about a fixed axis, the velocity b can only go up, or down. It's restricted to those directions at that instant in time. And the velocity of C, since CD is also a bar, C can only go left or right. So velocity of C is restricted to go left or right. And so now, to find the IC for body BC, Given those restrictions on those velocities, what I can do is, I can drop a perpendicular to the velocity vector. So, I draw a perpendicular from the velocity vectors. And, where that crosses is my. IC. That's my instantaneous center. So you can see if bar AB was rotating up, this would go up. Point C would go to the right. And the bar BC would be seemingly rotating at this instant about the IC. Similarly in the opposite direction, if BB was going down, BC would go to the left. And bar BC would again rotate about the instantaneous center at this snapshot in time. So that's the second way for locating ICs. The last way is to use geometry. In this case, I'm going to drop a line along the velocity vectors. I know that the IC, this should be a straight line. I know that the IC should be perpendicular to these velocity vectors. And what I find by similar triangles is that the IC is going to be 8, and I'm not drawing this very well to scale, but it's going to be 8 meters below point B. Because by similar triangles now. I'm going to have 0 velocity down here. As I go up I have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So 8 is to 5 as 16 is to 10. And this is the location, as I said, of the IC. Okay, let's have you do some on your own. So I want you to locate the IC for this pulley and come on back and we'll look at it together. Let me explain what's going on here first. We've got a cable or a cord wrapped around the inner drum of this pulley, and then on the outer drum we have another cable that's attached to this block b. And so this thing is unwinding and rolling down, and I want you to find the IC for the pulley itself. And what you should have come up with is that the IC is located right here, because that's a point of zero velocity on the pulley because it's attached to this cord. The cord has zero velocity. Every other point on the pulley is rotating about this point. Now sometimes It's difficult for folks to visualize that and so what I like to do is turn the problem 90 degrees and here is my cord, which is fixed in space, so it's not moving. Since it's not moving and it's touching this point on the pulley, you can see it's very similar to a wheel rolling on a no slip surface. That's the point of zero velocity. Okay, so that's one example. So here's another example I'd like you to do. I'd like you to locate the IC for bar AB. The way to do this one is to look at the perpendicular to the velocity vectors. We know that a, the piston, can only move in the i direction, so the velocity of a can either be right or left. The velocity of b also can only be right or left because this is a RAFA bar. And now if I drop perpendiculars from those two velocity directions, what I see is that the perpendiculars never intersect, which means that there is no IC for this bar A B and think about what that means. And so what you should have deduced is that what that means is that at this instant in time, at this snapshot in time, this bar A B is in rotation for this particular rotation. Now an instant later, an instant before it would be turning, but at this snapshot it is only translating and there is no IC. Okay one more problem, I want you to find the ic for this wheel and take some time on your own. Again, we are going to drop a perpendicular from the velocity vectors. I'll also draw a line between the tips of the velocities and this is going to be the ic. To do that I'm going to use similar triangles. I want to find this distance x, where this cross is. And so I know by similar triangles that x is to the total distance which is 20 for the diameter of the wheel as four is to the total change in velocity over those two points, which is twelve. So x equals four twelfths times twenty or 6.67 meters, and so this x is 6.67 meters, which means this distance here is 20 minus that, or 13.33 meters. So this is a case where we found the instantaneous center using geometry, or similar triangles. Okay, that's it for this lesson and we'll see you next time.