[BLANK_AUDIO] Hi, this is Module 27 of Two Dimensional Dynamics. Today's learning outcome will be to solve an engineering problem using the theory of the acceleration of the same point expressed relative to two different reference frame or bodies in planar motion. And we developed that theory last time. Here's the equation we came up with. We saw that, now in addition to the tangential and a radial acceleration term, when we have the acceleration the same point relative to two different reference frames, we can get a Coriolis acceleration term. So let's do a problem, we'll go back to that problem we did before, for velocities and extended now to accelerations. We've got a vertical bar b and a horizontal bar r. We found out that the vertical bar b ended up having the same angular velocity as the vertical bar r, which was .2 radians per seconds clockwise. And we also found out, in our last problem when we did this for velocities that the relative velocity of P with respect to the reference frame attached to the B bar was point one meters per second in the J direction. So now we're given that the angular acceleration of the bar r, is 0.25 radiance per second squared counterclockwise and we're asked to, find the angular acceleration of the vertical bar B and the relative acceleration of P with respect to that vertical bar. So here's my problem. I've got my my moving frame attached to the bar B. Okay, welded to the bar, B, here's my development of my acceleration equation, and so I have the acceleration of P is equal to the acceleration of O prime. What's the acceleration of O prime going to be? And you should say zero. It's at a pin point. And by the way, if you can continue on and finish this problem, go for it. If you get stuck come on back and watch the video with me and work through it but go as far as you can. This is good practice on your own. Then I have a rail again this is a difficult concept, but if I shrink myself down from the perspective of this bar B which is my moving frame now. I've got my moving frame welded to that and I look up at that point p, as I'm going along with that bar, point p is only coming in or going away from me. It's only in the j direction from my perspective, or my reference frame. Difficult concept. You should, by now be able to grasp that. So we've got a rail can only be in the j direction for the moving frame. So it's going to be a rail in the j direction plus the angular velocity, excuse me, angular acceleration of the moving frame. Crossed with the position vector in the moving frame. You should be able to find that relatively easy from geometry. You know this distance is .5. You know this angle is 30, so you'll find out that this is 0.866 meters and so that's going to be crossed with Zero point eight six six in the j direction. And then I've got minus a mega squared a mega of the moving frame b was found to be the magnitude was minus point two since it was Clockwise and in the negative K direction. So this is minus 0.2 but I square it so it's still going to be positive times R. R again is 0.866J. And then I've got my last term here which is this Correalus acceleration term, so I've got plus. 2 times omega as a vector. Omega is the vector for my rotating reference frame, is minus 0.2 in the k direction. We found that from the velocity problem that we solved when we worked this problem. And then that's crossed with v REL, we found out from the velocity solution for this problem that v REL was 0.1 In the J direction for the moving B frame. So that's 0.1 J. Okay. You can go ahead and do that math, I'll leave that as an exercise on your own. And you'll get the result that A of P is equal to 0.04 minus 0.866 alpha. All in the I direction, plus a REL minus 0.03464 in the j direction. And so let's go ahead and look at that result. Here's the result. And you can see we have one equations. We got one, two, three unknowns. So we're going to need another, at least another equation or so. So, let's go ahead and continue on. Think about how you might proceed. Is there another way I can find the acceleration of p rather than looking at it from the perspective of these two frames that I've, I've drawn or from the perspective of the moving frame b? And what you should say is, oh yeah, I've got this bar R and I can look acceleration of P with respect to bar R where O and P are two points on the same body and so I can go back and use the equation between two points on the same body Instead of from the perspective of two different frames. And so, here's the theory that we came up with for that. The relative acceleration equation for two points on the same body. In this case we're going to have this is point p to q generically but here we're going to have point p point o to p. So this would be point o and this would be point p. This would be P and this would be o. And we're going from o to p and o to p. Okay. So I can adapt this for this particular problem. So I got the acceleration now of p from perspective of the r bar Is equal to the acceleration of o and the acceleration of o, you should say is going to be equal to zero because that's another pin point. Plus, we've got the angular acceleration of bar r now. So this is going to be plus Theta double dot, I'll super, sub-script it with R. Okay. Crossed with r form o to p. And then, minus. Now this is theta dot for R. This is minus theta dot squared for R, r from o to p. Okay so that's another equation written from o to p, same body, two different points on that body. So I've got a of p, and we said that the acceleration of the origin now, or this point o down here now, is zero. And then I've got plus, I've got theta double dot R. Well we're given the angular acceleration of bar R is 2.5 radians per second counter clockwise. So counter clockwise will be in the positive K direction by the right hand rule. So that's going to be plus... 0.25k crossed with r from o to p. R from o to p will be -0.5 in the x direction and plus 0.866 in the j direction. So this is going to be -0.5i plus 0.866j. And that takes care of that term. I've taken care of that term. We just have the last term, the radial or normal acceleration. So, I've got minus the quantity of point, theta dot sub r squared. Theta dot sub r was Point 2 radians clockwise. So that's going to be -0.2 squared, R from moda P again is -0.5. I plus 0.866 J. And again, I'll leave that math as an exercise for you to do. The result that you should come up with is a of p is now equal to -0.1965i and -0.15964j. Okay. So, here are the two equations that I came up with. I came up with the acceleration of p with respect to this moving frame b. And the acceleration of p with respect to the bar r so it's the two points on the same body. And so, I have two expressions for the acceleration of p. I can now equate those to find the unknowns, alpha and acceleration of R or the acceleration relative which I'm trying to find. So lets equate the I components first and again you should go ahead and work ahead if you can and then come back If you get stumped. So if I equate the I components I get, zero point zero four minus zero point eight six six alpha equals minus zero point one nine 65. And that will lead you to the result of alpha equals positive 2.73. So alpha as a vector is 2.73 radians per second squared in the K direction. So it's going to be counterclockwise. And that's one of our quest-, that's one of our required questions. What is the angular acceleration of the vertical bar, alpha. And then let's go ahead and equate the J components. And if I do that, I get A REL minus zero point zero three four six four Equals minus 0.15964 and that leads me to A REL is equal to minus 0.125 And as a vector that means a rail is equal to minus zero point one two five j meters per second squared. So that's the relative acceleration of point p. It's in a minus direction, minus j direction, from the perspective reference frame, of the vertical bar b. These are difficult problems. No doubt and so it takes a while to get get competent at solving these and so I've got a couple of practice problems for you to do. Try them on your own. Go as far as you can. if you get stuck, come back and kind of work through it step by step, but these are great problems. If you can do these You should be in good shape, and so, this is the follow-on from the problem we did before, for velocity, is now working with acceleration, and another follow-on to a previous problem we've used with velocities and working with accelerations, and I've put the modules, solution, or the handouts for the solution in the module handouts, so. And I'll see you next time.