Hi, this is Module 46 of Two Dimensional Dynamics. And today we're gonna solve a nice real-world engineering problem using the Principle of Impulse-Momentum to determine what's called the Center of Percussion of a body. And so, here is the principle of momentum that we developed a couple lessons ago. We had our body, we put the initial momentum on this diagram plus whatever impulse acting on the body and that's gonna end up equaling our final momentum. So lets go ahead and look at an example. And this example is a simple model of an individual hitting a ball with a stick, as shown. And the question is, where should the ball hit the stick to eliminate what we'll call the sting or the transverse reaction of y. We call that the sweet spot. So if you swing the bat and you hit it just right in the sweet spot, then the y reaction is gonna go away. And we call that the Center of Percussion or the sweet spot. Let's go ahead and look at this example. In my intro video, I did the video of me hitting a golf ball and I said this same sort of concept works for a golf ball or it works for any kind of racket or sports striking tool. So I have a simple bat here. And so it's just a stick, so I'm playing stickball. I've got my arms and I'm gonna hit a ball. And I wanna find out where should I hit the ball? Where is gonna be the sweet spot or the Center of Percussion? So I don't get any y reaction out here. So if I come through, all of it it's gonna go straight through with no y reaction. The Center of Percussion or sweet spot. And so we're gonna go ahead and figure out where that spot should be. So I want you to work through the problem and I'll help you as we go along. So here is our diagram of the bat or the stick hitting the ball. We have my hands are going to be modeled as a hinge here. And I want you to write on here, the Initial Momentum and the Impulse and the Final Momentum. And when you've completed that, come on back and we'll see how you did. So here are my diagrams. The Initial Momentum, I've got an m y dot in the y direction at the mass center, an m x dot in the x direction and an I omega or an angular momentum about point c. Those are the three potential momentum vectors I would have. And then for my Impulse, I'm just gonna draw a good free-body diagram integrated over some period of time. And so I've got the force on the ball striking the stick. I've got my Rx reaction and I have my Ry reaction. And to eliminate the sting, we wanna get rid of this Ry reaction. When I say sting, for those that don't understand that concept, if it's a real cold day, like it is today in Atlanta, it's in the 20s, which is cold for Atlanta. If you're out striking a ball with a stick, if you don't hit it just right in that Center of Percussion, you'll feel it in your hands. And if you hit it well and you hit the sweet spot, it'll be very smooth and you won't feel this y reaction. So that's what I mean by sting. And with that Impulse then you have a Final Momentum and so I've shown my three momentum vectors, here. And so at this point, I want you to say okay, what do I do next? And what you should say is, I need to start summing my Impulse Momentum vectors to come up with some equations. I'm gonna sum the Impulse Momentum vectors in the y direction. I'll call up as positive here or in the J direction positive. So I've got my my c dot initial, okay? And then I have my integral over some period of time of Ry minus the force of the ball. Rx is not in the y direction. And then for my Final Momentum, I have m y dot final. And so that's a good equation for the Impulse Momentum in the y direction. Now let's do the same thing for a rotation about o, or the angular momentum about point o. And the reason I choose point o is because I get rid of the m x dot, which I'm not interested in. I'm not interested at this point in RX, and so if I do it about o, and you should have done this on your own, I get my angular momentum, initial about point o. Plus I have this linear momentum at point c, times its moment arm, which is L over 2, minus the integral of the moment arm to the force of the ball, which I'll call a distance d to the force of the ball, equals, again, my final angular momentum and my linear momentum in the y direction times its moment arm, which is L over 2. And then what I've done here is I've, since I'm saying this is just a uniform slender bar for my stick, it's maximum inertia, if you look it up in any reference, would be one-twelfth mL squared. About the mass center again, okay, I'm looking at it from the perspective of the mass center. As I've said in earlier slides, you can look at this from the perspective of moving the vectors down to o, and you get the same result. But I'm working with my momentum vectors at the mass center. And so I've got one-twelfth mL squared omega initial, plus, and I wanna put everything in terms of omega. And so I'll use my kinematics, and I say that the velocity of the center in the y direction is equal to the radius times omega, or in this case, the radius is L over 2 times omega, and I also substitute that same relationship over here. And so this is what I arrive at. And I'm going to simplify that a little bit more. I'll leave my Impulse on the left-hand side. I've pulled all of my momentums over to the right-hand side. And this is just one last simplification. Okay? And now let's go from here and we have two equations. We have an equation for the impulse momentum in the y direction and angular momentum here. And so that's where I start off on this slide, and what I want you to do is continue on, as far as you can, to boil this down to try to solve for d, cuz d is going to tell us where this sweet spot is. And so it tells where the ball has to hit for the Center of Percussion. And so we just do some mathematical manipulation here. For no sting, I said that Ry has to go away. And so we're gonna get Ry to go to zero. That lets the first equation go to the integral of F, actually minus the integral of F, of the ball, dt, equals, and I'll carry the Initial Momentum in the y direction to the other side. And then I'll put in my kinematics relationship here, and then I'm gonna multiply this equation by d, and so I get, integral of d of f of BALL dt is equal to minus mLd over 2, times omega final minus omega initial. And so this has been transformed into this equation, right here. So we're gonna end up with this equation and this equation. And so that's where we start off on this slide. And now I can add these two together to get rid of the force of the ball, which I don't know. And I get now, zero on the left-hand side and on the right-hand side, I get one-third m L squared omega final minus omega initial and minus mLd over two, omega final minus omega initial. I can now solve for d and so I go through and I get d equals two-thirds of L. And so what that tells me is for Ry to be equal to zero, to find the Center of Percussion or the sweet spot, the ball has to hit the stick at two-thirds from my hands, two thirds of the way out on the stick. And that should make physical sense to you, if you ever played stickball or baseball or something of that nature. That sounds like about the right spot for the sweet spot. So that's a real cool real-world problem in applications of dynamics and I hope you enjoyed it.