Hi, and welcome to Module 39 of Two-Dimensional Dynamics. Today's learning outcome is to go ahead and solve an engineering mechanics problem using the work-energy principle theory that we developed in the last few modules. And so here's where we left off. The work-energy principles says that if we have work of external forces and moments due to couples on a system, then it's gonna change the kinetic energy of that system. And so, we looked at several ways that work can be applied to the system. One was work done by a constant force, work done by a variable force, work done by gravity, work done by a linear spring, and finally work, now that we can have rotation, work due to a moment due to a couple. So let's look at a problem here. I have a system with a fixed cylinder, fixed to the ground. And I have an unbalanced wheel rolling on the outside of the cylinder, there's a bar pinned from the center of the fixed cylinder, up to the center of the unbalanced wheel. And there's a spring attached to that arm. And we're told that the unstretched length of that spring is 0.3 meters. We're given that the initial angular velocity of body A, the unbalanced wheel, is 2.5 radians per second clockwise at the top. And we're also told that there's enough friction to prevent the slipping of body A, the unbalanced wheel, on body B, the fixed cylinder. And we want to determine the spring constant, which is unknown, which will cause this body A to stop just prior to the 90 degree position down here. So it's gonna be just slightly less than 90 degrees, and it's gonna roll down starting with 2.5 radians per second, and then stop once it gets to this position. So, here again is work done on the system is equal to the change in kinetic energy. What I want you to start with is think of all the different ways in this system that work can be done, and list them out on a piece of paper and then we'll come back and we'll do them one by one. And so one of the ways that work can be done is due to the gravity on the unbalanced wheel. The unbalanced wheel has a mass center at the top. It has a mass center at the bottom, and it's gonna have gone down some height. And we're gonna get work done by that. Now, we know the initial position is C. But as it rolls down, and C is offset from the geometric center of the unbalanced wheel by 0.05 meters, but as it rolls down, I do not know, yet I'll have to find out, where is the position C? Because where position C is down here, the mass center of the unbalanced wheel is going to change the height that the unbalanced wheel has gone through. And so we know the position of the center of mass initially, but we must find the final position of the center of mass. And so let's go ahead and do that. And so here I've drawn a picture. I'm gonna have to know how many rotations this wheel has gone through. I'm gonna have to know the angular displacement so as I roll down I'll know exactly where it stops. To find the angular displacement, I'm gonna use kinematics first and look at the angular velocities, and then once I get angular velocities, I will be able to integrate them to get angular displacement. So here is my unbalanced wheel rolling outside the fixed cylinder. And you'll recall I've got two bodies here that have instantaneous centers of zero velocity. One is the bar, which is connected from the pin down here in the center all the way out to point Q. And the other body that has an instantaneous center is the wheel rolling on this surface with no slip. So this unbalanced wheel has an instantaneous center of zero velocity, right here at the point of contact. And so we can use the instantaneous center of zero velocity equation to find two different equations for the velocity of point Q. And so here was our theory for instantaneous center of zero velocity. We said that the velocity of Q will be the velocity of the instantaneous center for each of these bodies, one at a time, plus theta dot k, the angular velocity of that body, crossed with r from the instantaneous center to Q. And we know that the instantaneous center's velocity is 0, so therefore, it's just omega cross r. And so let's first look at it for arm D. For arm D, this is arm D here, it goes through an angle phi as it rolls down. So here's phi. And so the velocity is going to be phi dot, the angle of velocity of arm BD. So v of Q is the angular velocity of bar D crossed with r, which is distance l from point O up to point Q. So this is gonna be L phi dot, and the direction of the velocity, as we know, for a curved path, is always gonna be tangent to the path. So that's the velocity of Q in terms of phi dot. I would like you now to do the same thing for the velocity of Q in terms of theta dot, using the instantaneous center for the unbalanced wheel. And here's the result you should have got. You should've said that v of Q is equal to v of IC, which is 0 for point A, plus omega, which is theta dot in this case, crossed with r from the instantaneous center up to Q, which is capital R. Again, the direction is in the e tangential direction, and so we now have two expressions for the velocity of Q, one in terms of phi dot, and one in terms of theta dot. And so I can equate the two, and this is the relationship we get. But we want to know the angular displacement of the unbalanced wheel, so we can find out how many times it turns, and where c ends up at the bottom, or at the 90 degree point. And so what I'm gonna do, or what should I do to find that angular displacement? And what you should say is, you're going to have to integrate, and we'll integrate. We'll call phi zero, and theta zero, our datum, where those angles are equal to zero, at the top. And so though we end up with a relationship of L,the length of the bar, times phi ,this angle equals R which is the radius of the unbalanced wheel, times theta, which is the the angle of rotation for the unbalanced wheel. And so we know that phi is going to be just slightly less than pi over 2, so I'm gonna go ahead and use pi over 2, close enough. We could say it's 89.9999 degrees or whatever, but it's close enough to pi over 2 so it's gonna come down and stop right there. I've get phi equals pi over 2. If I substitute that in, I find theta is equal to l over R times phi, where l is given as 0.6 meters, R is given as 0.1 meters, and we've gone through pi over 2 for phi, so theta has had to have gone through 3 pi. So at this time, I'd like for you to try on your own, if this unbalanced wheel goes a rotation of 3 pi, where is point C once it arrives down at this position here? Where on the circumference, or not on the circumference, but where in the position of the unbalanced wheel is point C gonna be? And then come on back and we'll do it together. Okay, we have work done by gravity. We wanna find where C starts, we know where C starts, and we wanna find where C ends. We know the initial position. We can now find the final position. We know that theta has gone through 3 pi rotations, and so here's my wheel. My wheel started off at this side, on the right hand side, and it's gone through 3 pi rotations. So here we go. We go pi, 2 pi, 3 pi, and C ends up here. It's initially her, and it ends up in this position. So C in the final position is right there. And so what that means is my height change is from, is actually the length of the bar, which is 0.6 meters so I can substitute that in. I know that the unbalanced wheel weighs 80 kilograms, excuse me, its mass is 80 kilograms. I have to multiply by the acceleration due to gravity to find the weight. And so we have mg delta h. We can multiply it out. And we get the work done by the wheel, the unbalanced wheel, due to gravity in going from position 1 at the top to position 2 at the bottom is going to be 471 Newton-meters. Now, we're gonna go ahead and do another one of the work terms and you should have listed all these work terms. And so the next one I'm going to do is the work do to the arm, okay, arm D. Do it on your own. Come on back and let's see how you did. And for the work done by arm D, its mass center is right in, it says it's a uniform slender bar, so its mass center is gonna be right in there, the center, I'll call this C' for a different than C. So if the bar is 0.6 meters, C' is at 0.3 meters. Down here, C ' is gonna be at the horizontal, and so the change in height of the mass center for arm D is 0.3 meters. Its weight is going to be 35 kilograms times the acceleration due to gravity. I multiply that out, and I find that the weight of the arm, excuse me, the work done by the arm in going from position 1 to position 2 is 103 Newton-meters. And so that's probably a good stopping point. We're gonna come back the next module, continue to work the work terms. Once we find all the work terms, we're gonna do the change in kinetic energy. And our only unknown will be the spring constant of the spring, which we're asked to solve for to finish the problem. So, I'll see you next time.