[BLANK_AUDIO] Hi, this is Module 17 of Two Dimensional Dynamics. Our learning outcome for today is to use the instantaneous center of zero velocity, which we discussed last module, to find of velocity of bodies in planar motion, two dimensional motion. And so, here was the information that we, we came up with last time for the IC. We said that if we can locate the IC, any point on a body is going to be equal to a mega of that body crossed with a position vector from the IC to Q. And so, let's go back and look at the crank arm piston cylinder problem that we looked at previously using the relative velocity equation. This just, the IC method just give us another tool for solving these equations, which goes from knowns to unknowns. We know the angular velocity of body B1, we know that point O is a fixed point. So that's the IC for body B1. And, so I can then write an equation using the IC. I can say that the velocity of A is equal to, theta dot body B1 in the K direction, crossed with R from the IC of body B1 to point A. Or the velocity of A is equal to again theta dot B1 is 10 clockwise which is going to be in the minus K direction, so I've got minus 10 K crossed with I. I go minus three in the I direction, plus four in the J direction. And, I get the velocity of A then, is equal to 40 I plus 30 J inches per second. Okay, so we keep going towards unknowns again. To go from A to B, I'm going to use the IC method so I need to locate the IC for body B2 or the bar from A to B. We know that the velocity of A is up and to the right. 40 I plus 30 J. I can drop. We also know that the velocity of B then is going to have to be to the right because it's a slider mechanism or a piston. So this is VB. I can now drop perpendiculars from both of those velocity vectors. Look where they intersect and that will be my IC. So if I drop a perpendicular from this velocity vector and a perpendicular from this velocity vector. This is perpendicular. This is perpendicular. Where they cross, will be the IC for body B2. And we know that this is on a four on three slope, okay. From our dimensions here we also know that this distance is 12.4. Lets do this in a different color here. So this is 12.4 inches from B over to A. And I, therefore know that this distance from O to B is 9.4 inches, because that's minus 3. And I want to locate the location of the I for, body B2. I need to know this distance or this height, which I'll call H, by similar triangles I know that H is to 4 as 9.4 is to 3 and so H ends up being 12.53 inches, and that's the distance from here to here. So I've got the location of the IC for body B2. I can now use it to finish up the problem. So here again, here's the location of the IC, it's 12.53 inches below B for body, bar A B. I, I have already found the velocity of A. It's 40 I plus 30 J or 50 inches per second total, on, at this direction. So I've got the velocity of A, by my instantaneous center equation is equal to theta dot B2. Body B2 in the K direction. That's omega. Crossed with R from, the I for body B2, to point A. Okay. And when I do that I get velocity of A was found to be 40 I plus 30 J. The angular velocity of bar B2 is unknown, so that's theta dot B2 K. Crossed with and going from I to A, the position vector is minus 12.4 in the I direction, and plus 16.53 in the J direction. So this is going to be minus 12.4 I plus 16.53 J. And I can now do my cross products so I get 40 I plus 30 J equals let's see, K cross I is minus J so that, excuse me, K cross I is plus J. But I have a minus 12.4 times theta dot.B2 so that's going to be minus 12.4 theta dot B2 in the J direction and then I've got K crossed J is a minus I 16.53 theta dot B2 in the I direction. Now I can match components solve for the unknown which is theta dot B2. I'll match I components. [BLANK_AUDIO] When I do that, I've got 40 Equals minus 16.53 theta dot B2. Or theta dot B2 [COUGH] equals minus 2.42. Just as we found with the relative velocity equation. So we should get the result, if we don't we know we'd done something wrong. It's interesting to note that if you match J components, you'll get the same result for theta dot B2 which you should. Finally we, we found the velocity of A. We found the angular velocity of bar B2. Now using that angular velocity in the IC I can find the velocity for any point on bar AB. So, the one I want to find is the velocity, the piston pin B. And so I'm going to write my relative velocity equation again. And I'll get the velocity of B is equal to theta dot B2 in the K direction crossed with a position vector from the IC for body B2 to point B. And we know that theta dot B2 is minus 2.42. We solved for that R from I of B2 to B is just 12.53 in the J direction. And if I do that cross products, I've got K cross J is minus I. But I've got a minus 2.42 times 12.53 so minus a minus is a plus. I get 30.3 and it's in the I direction inches per second for the velocity of bar B. And so we get the same answer as we did when we used the relative velocity equation. Okay, that's solving a, a relative velocity velocities on a planar rigid two dimensional body and I want to give you a chance to do it again using the IC method. So here's another worksheet for you, very similar but it's great practice for you and once you finish this, I have the solution in the, in the module handouts, and I'll see you next time. [BLANK_AUDIO]