[BLANK_AUDIO] Hi, this is module 41 of two dimensional dynamics. We're continuing to solve this work energy problem that we've started for the last, or we've worked on for the last two modules. So far, we found the total work done on the system. Now, we need to find the change in kinetic energy. Let's begin by finding the kinetic energy when the position, when the unbalanced wheel is in the initial position. And so, what I want you to do is write an expression for the entire kinetic energy. All the things that will have kinetic energy in our problem in position one. So write that, those terms out. And what you're going to say is, okay, the wheel, the mass center of the wheel, has a linear velocity, plus there's a rotation of the unbalanced wheel. And so the wheel has both translational kinetic energy and rotational kinetic energy. Likewise, if we're looking at this from the perspective of the mass center of the arm. The mass center of the arm has a linear velocity and the arm is also rotating. So it also has a translational kinetic energy and a rotational kinetic energy. Looked at from the center of the wheel or the mass center of, excuse me, the mass center of the arm. Okay, so, let's now work on this wheel. Well, before we even do that it looks like we're going to need some of these mass moments of inertia, one for the wheel and one for the arm. So go ahead and do the mass moment of inertia for the wheel given the information that we have in the problem. And so we're given a radius of gyration which is 0.07 meters with respect to the axis through the mass center of the unbalanced wheel. And so I can just use our, our equation that we've talked about before. The mass moment of inertia for the wheel then will be m times the radius of curve radius of gyration squared or 80 kilograms times 0.07 squared which ends up being 0.392 kilogram meters squared. Now also find the mass moment of inertia for the arm. And the arm is a uniform slendered bar. If you look in any reference you'll find that the mass moment of inertia for a slender bar about its mass center is one 12th ml squared. The mass is 35. The length of the arm is or the bar is 0.6 meters and so that comes out to be 1.05 kilograms meters squared. Okay, so we have our expression for our initial kinetic energy. All the terms we now have our, our mass moments of inertia for both the arm and the wheel. And we need to start finding the velocities. Well, we're told that the initial angular velocity of body A, which is the wheel, is two point radians per second clockwise. And so we know this magnitude of the velocity is the omega of the wheel is 2.5 radians per second. What I want you to do now is define vc1 of the wheel given that angular velocity. So, we've got now this term. We found this term. We just were given this term. And now I want to find vc1 of the wheel. And you can do that on your own. Come on back. So we can use the the instantaneous center of zero velocity. This is the geometric center, but we can see on our problem the mass center here. Where we want to find a, the the magnitude of the velocity is over here. And so, we know that at any instant in time and at this snapshot in time the velocity is going to be r, or omega cross r, where omega is the angular velocity of the wheel, and r is the distance from the instantaneous center out to that mass center. So this is going to be the direction of the velocity. The radius here is going to be, here we go here again, here is the ic, there's point c. The radius is going to be by Pythagoras' theorem, the square root of the radius of the wheel, which is 0.1 squared plus the distance from here to here which is 0.05 meters squared. And so, here is our expression. The velocity at c1, position one, is equal to omega which is 2.5 times r from i to c. Which is the square root of 0.1 squared plus 0.05 squared. And it's going to be tangent to the path. But all we're really interested in is the magnitude here. And so that comes out to be 0.2795 meters per second. So now we have that term. All right? Let's continue on. I've written those results here. Now let's go back. We need to, we've taken care of all these terms for the wheel, now we need to go back. We have the ic of the arm, but we need to find omega of the arm and vc of the arm. So, we have an expression. We know omega of the wheel. We have an expression that relates it to omega of the arm. Remember, the omega of the, of the wheel was theta dot, and the omega of the arm was phi dot. And so, we had a relationship between theta dot and phi dot, recall that from a previous module. And here is that relationship, so phi dot corresponds to omega of the arm, theta dot corresponds to omega of the wheel. We know that omega of the wheel in the initial position is 2.5, so I can solve for omega of the arm. That's going to be r over l, or the radius of the, the unbalanced wheel is 0.1, divided by the length of the arm, which is 0.6, times the 2.5. And omega of the arm then, is 0.4167 radians per second. So we now know what this omega of the arm is. Given omega of the arm, we can then find the velocity of the center of the arm, which is going to be right here. So go ahead and do that on your own. Come on back when you've got it. And you should say, okay, the magnitude of the velocity of the center of the arm is omega cross r, or omega r, so that's going to be omega is what we just found, r is from the instantaneous center for the arm out to the mass center, which is a distance of 0.3 for a uniform bar, or one half the length. And so we get the vc of the arm is 0.125. We've got all of our terms now. And we can collect everything. And find our total initial kinetic energy. Here I've substituted all the values in, and I get that the initial kinetic energy of the system is equal to 4.71 newton meters. And that's a good stopping point. We'll come back next time. We'll find the final kinetic energy. Then I'll have known all the work terms. I'll know the initial kinetic energy. I'll know the final kinetic energy. And we should be able to solve the problem.