Hi, this is module 19 of Two Dimensional Dynamics. Previously we looked at the velocities for two-dimensional motion of bodies. We looked at the linear velocity between two points in a body as related to the angular velocity of the body. Now we're gonna extend that for accelerations of a body in planar 2D motion. And so here was the theory that we did before. And so we're gonna go ahead and apply this acceleration equation to this problem, which we did for velocities. And so we have omega 1 is 10 clockwise. Omega 2 is 2.42 radians per second clockwise. We found that from the velocity problem. And then we have the angular acceleration is 5 radians per second counterclockwise on body B1. And we wanna find again the linear translation of piston B. So I'm gonna go ahead and apply the acceleration equation we came up with, and so I've got the acceleration of the unknown point A, is equal to the acceleration of the known point O, which is 0 because it's a pin. Plus, we're working on body OA. We're going from knowns to unknowns here, and so I'm gonna have theta double dot B1, and that's in the k direction, crossed with the position vector from the known to the unknown, or from O to A, and then I've got minus theta dot B1 squared, r from O to A. And so let's go ahead and put those values in, so I've got the acceleration of A is equal to, well, we do know that the acceleration of O is 0 because it's a pin, and then I have plus theta double dot k for body B1 is given. It's alpha 1, it's 5 radians per second counterclockwise, it's going to be in the positive k direction, so this is going to be 5k, crossed with r from O to A, is, I go, -3 in the x direction, +4 in the y direction, so -3i + 4j minus, now theta dot B1 squared, the magnitude of the angular velocity of bar B1 is 10, so this is gonna be 10 squared times r from O to A, that's again, -3i + 4j. If you do that multiplication, I'll let you do that on your own, you'll get the acceleration of A, now, is equal to 280i- 415j inches per second squared. So let's continue on from there. Here is our acceleration of A, which we just found, so I've gone from this known to this unknown, which is now known, and now I wanna go down to the acceleration of B, which is unknown, and what we're trying to solve for in the problem. So, I've got the acceleration of B now, is going to be equal to the acceleration of A plus theta double dot B2 k, crossed with r, now, from our known to our unknown, or A to B, minus theta dot B2 squared times r from A to B. Okay. Do I know anything special about the acceleration of B, to reduce the unknowns? Again, B is moving in rectilinear translation. So it can only have one component of acceleration, and that's gonna be in the i direction. So I can reduce a as a vector to a sub B in only the i direction, equals a sub A, which I found, which is 280i- 415j plus, now, I don't know what theta double dot B2 is, so that's gonna be an unknown. So I've got theta double dot B2 in the k direction, crossed with now r from A to B. We're gonna have to use a little bit of geometry here. We know this distance is 4 inches, cuz this is 13 and this is a 4 on 3 triangle with a 5-inch hypotenuse, so that's 4 inches. And so if you do your Pythagorean theorem, you'll find out that this side is 12.4 inches. And so in going from A to B, I go 12.4 in the i direction and -4 in the j direction. So this is 12.4i- 4j. And that takes care of that term. I still have my radial or normal acceleration term, so that's gonna be minus the magnitude of B2's velocity is 2.42, so I'm gonna get 2.42 squared, and that's going to be times r again, so that's 12.4i- 4j. And again, I'm gonna let you do that mathematics, multiply that all out. This is the result you should come up with aB in the i direction equals 207 + 4 theta double dot B2 in the i direction = -392 + 12.4 theta double dot B2 in the j direction. And so, you should be able to do that mathematics. Okay. So, here's that result, and my question is, okay, we have one equation, a vector equation, and we have two unknowns, how can I solve for those two unknowns? And you should be used to this by now. We're gonna go ahead and equate components. Let's start by equating the j components. And for the j components I get theta double dot B2 = 392 over 12.4 = 31.6 radians per second squared. And then I'll do the i components. Why don't you go ahead and do the i components on your own and come on back and see if you got the answer correct? And for the i components I've got aB = 207 + 4 times data double dot B2. But I just found data double dot B2 was equal to 31.6, and so this equals 333, or as a vector, aB = 333, all in the i direction, inches per second squared. And we've solved our problem. As I've said all along, practice is going to make you better and better at these engineering problems. So I have an angular velocity and angular acceleration problem for you to complete now. You have to find at the instant shown the angular acceleration of B2. I always like to put these problems in terms of real world systems, and you recall back, well, no, you may not recall back. But hopefully, you've been to amusement park rides. And you may have seen this amusement park ride, which is like a pirate ship going back and forth. And you can see that that's somewhat related. It's not exact, but there's somewhat of a similar configuration here. If we think of this as being the passenger compartment for the pirate ship ride, this would be a fixed vertical pole, in the case that I have here it's not fixed. But you can see that it's a similar type configuration. And why is that important? Why would we even care about accelerations on amusement park rides? And so what you should say is that the human body can only sustain so many G's of acceleration before you black out, or have problems. And so if you were a designer or an engineer working on these amusement park rides, you'd want to make sure that you stay within realistic standards. And so these have real world applications, these problems, and I think that it's important to remember that. And we'll see you next time.