Welcome to module 15 of two dimensional dynamics. Today's learning outcome is to solve for the velocities of a planar rigid body in motion using the relative velocity equation that we developed in the previous modules. So, here's a picture of what we developed before we have a generic body, two points on the body, and the relative velocity equation relates the linear velocity of those two points on the body to the angular motion of the body and it's geometry. So, here's a worksheet of an actual problem. This is the piston cylinder mechanism that I used as a demo before, and this problem is just a mirror image of this piston cylinder. I have a crank arm here, which is body B1, which just is a portion of my wheel, but it's the same body. It's got a fixed point at O, so it's rotating about a fixed axis. I've got a connector arm here, B2, connected to a piston that's undergoing Linear translation. I know in the problem that the crank arm has an angular velocity of 10 radians per second clockwise, but in my model here, since it's a mirror image, it's counterclockwise. We want to find, at this given instance, what the linear velocity is of the piston itself. Okay, so let's go ahead and talk about a strategy on how to solve this problem. Generally, I would think the best approach is to pick a body, one of the bodies in the problem, and go from knowns to unknowns. So, in this case, the body I know something about is body B1, the crank arm. I know that it has a ten radiant per second clockwise angular velocity, and I also know that it's a velocity at point O, since its fixed pin, is zero. So let's go between O and A, and write our relative velocity equation. So, I have velocity at A, which is unknown, is equal to the velocity at O plus theta dot from O to A in the K direction, crossed with R from O to A. So, in this case I know that the velocity of O is zero. I'm given that theta dot, or the angular velocity of the crank arm, b1, is 10 radians per second, clockwise, and if I go 10 radians per second, clockwise, by the right hand rule, that is into the board, or in the minus k direction. So I've got plus minus 10 K crossed with R from O to A. The position vector in going from O to A is minus three in the I direction, and plus four in the J direction. So that's gonna be minus three I plus four J. So, the velocity of A then Is equal to K cross J is minus I, but I've got minus 40 times minus I, so that's plus 40 I, and then K cross I is J minus three times minus 10 is plus 30. So I get plus 30 J inches per second. Okay, so I know now the velocity of this point of body B1, the angular velocity of body B1, and the linear velocity of point A. I want to keep going toward the unknown that I want to solve for, which is the velocity of pin B, and so let's continue on. So here's my result so far. Now I'm going to go from what is now known as the velocity of A to velocity B, using the body B2. So, now I have the velocity of B is equal to the velocity of A plus theta dot AB in the K direction crossed with R from A to B. That's just an application of the relative velocity equation. We know what A is, we don't know what VB is. We don't know what theta dot AB is. We can find RAB from the geometry. Do we know anything? We want to reduce as much as possible, the number of unknowns as we can. So, do we know anything special about the velocity of point b that would allow us to reduce some of the unknowns? What you should have thought about and arrived at was the velocity of B is in the I direction only. So, the only unknown for the velocity of B is the magnitude, not the direction. So, I can then write the velocity of B In the I direction is equal to the velocity of A, which we found to be 40 I plus 30 J, plus theta dot AB in the K direction, the angular velocity of body B2, crossed with R from A to B. We're gonna have to use a little geometry here, and so I know by Pythagoras' theorem rise is four, the hypotenuse is 13, so this distance is 12, and so in going from A to B, I'm gonna 12.4 in the I direction and down four. So, we're going to have crossed with 12.4 I minus four J. Okay? So, let's continue on again. Here's the result that I have. Let's do the cross products out here. We've got VB I equals 40 I plus 30 J, and then I've got K cross I is J, so I'm gonna get plus 12.4 theta dot AB J, and then K cross J is minus I, but I've got minus I times minus four times theta dot AB or plus four theta dot ABI. Now I'd like you to think about, what do I do next? How can I finish up the problem? I know the direction of the velocity of B, I need to find the magnitude of it. What you should arrive at is that you should use match components, like we have in past courses and past modules. So, I'm gonna match the J components to start. On both sides of the equation. On the left-hand side of the equation, I don't have any J components, so that's zero equals, on the right-hand side I've got 30 plus 12.4 times theta dot AB. What that allows us to arrive at is that theta dot AB, if I solve for it, equals minus 2.42, or theta dot AB, the angular velocity of bar B2 as a vector, is minus 2.42 K radiants per second. So, what that means is this bar B2 here is turning with an angular velocity of 2.42 K radiants per second. Since, it's negative K, it's a clockwise rotation. Finally, we need to solve for the velocity of B. How do I do that, the magnitude of velocity B? What you should say is match the I components. When I do that, on the left hand side I have VB, on the right hand side I've got forty plus JJI component four times theta dot AB, and I know that theta dot AB, I just solve for it, is minus 2.42. So, I arrive at VB equal to 30.3. That's the magnitude of VB. I know its direction. So, my final answer is VB as a vector is equal to 30.3 inches per second in the I direction. I've solved my problem. I wanna go ahead and give you a chance to solve a similar problem, so here's a worksheet for you to do on your own. I've included the solution in the module handouts, and I'll see you next time.
