Hi everyone. In this video, we are going to now integrate the equations of motion over time so that we can obtain the impulse and momentum equation relationship for the rigid body. To briefly review how those equations of motion and work energy relationship has been changed for the rigid body structure, all the forces are going to be equivalent to the total mass and the center of mass acceleration. The total moment with respect to the G is going to be the I with respect to the G and the Alpha. If you integrate the equations of motion over the displacement, you can obtain the work and energy relationship especially with the moment and d Theta integration and I Alpha and d Theta integration. As I said in the previous supplementary video, you can actually integrate the displacement, Newton's equations over the displacement, mathematically it hold. However, this is not the definition of work of the rigid body. For the rigid body work, you have to integrate over a single particle and make a summation over the rigid body. So by doing so, you can obtain the kinetic energy of a single particle and it's sum, and which will ended up giving you the kinetic energy as a form of one-half mv square plus I Omega square. So again, for the work energy relationship, you could use the moment equation integral over the Theta, and then the kinetic energy you can get is the transnationals kinetic energy of the center of mass and the rotational part written by I Omega square. Now, how this impulse-momentum relationship will change would be written in rigid body case. To make the long story short, the answer is you can just consider the center of mass position. So all the linear impulse to the rigid body is going to be the linear inputs at the center of mass later and the initial, and angular impulse is also same as whatever the angular momentum at the later status minus the initial state. Note that the moment and the angular momentum sub reference points should match for left-hand side and the right-hand side. Now, let's briefly go over how those relationship, what I just mentioned, is hold. So for a single particle, the linear momentum is defined by a single mass and the single velocity. Therefore, the whole rigid body you can integrate the single-particle's linear momentum, which will be m_i v_i, and then we'll see if that's going to later simplify total mass and the center of mass velocity. To describe the v, we have split it out of the v as a vector from the fixed point O to the center of mass, and the distance of the particle with respect to the center of mass. Now, when you take the derivative of this row item, since this is a rigid body, the distance doesn't change, so just as length change would be zero. However, if the rigid body is rotating with angular velocity Omega, then Rho, the time derivative of this one, includes a term Omega cross r. So I have Omega cross r term included so that the final form of linear momentum will be total mass, and the velocity of the center of mass, and Omega cross m_i Rho i. By definition of center of mass this turns out to be zero, so finally, you will have a linear momentum of the rigid body is going to be same as total mass multiply the center of mass velocity. When you take the derivative of both sides, you will end up having all the forces will be the change of linear momentum of the rigid body at the center of mass. Now, with this equation, you can do a time integral over left and right hand-side. So the linear impulse, FTT, is going to be the change for the linear momentum at the center of mass later minus initial. We're going to go over how this angular momentum relationship will be varied over a different reference point. First, moment equation with respect to center of mass is going to be M_G equals I_G Alpha, and then if you time integrate them to obtain the angular impulse-momentum relationship, whatever you have an initial angular momentum I_G of Omega and added angular impulse will keep you the angular momentum later. Suppose that you are defining the moment equation reduces to any fixed point O, you have M_O is going to be I_O Alpha, those two subscription match. Then if you do the time integral, you have initial angular momentum with respect to the O and added angular impulse with respect to the O, will give you the change of angular momentum with respect to the O. Now, let's say any arbitrary point P, what will happen? Basically, it's going to be the same, moment with respect to the P is going to be same as I_P Alpha. However, for the general case where a_P has any some arbitrary acceleration term, then when you set the coordinated P, there will be inertia force applied for center of mass with the Omega and ma_P. So when you calculate this moment with respect to P, you should also consider moment of to the fictitiously inertia force at the center of mass. So if you include all those moment P with respect to the P, including the inertia forces moment, then it's going to be equal to the I_P Alpha. So subscript matches here and there, and then just to the time integrals. So your initial angular momentum with respect to the P is going to be same as angular momentum with respect to P later with the summation of all the external moment applied to the P during those time integral. So this is pretty much the brief overview of how linear and angular impulse momentum relationship has been changed for the rigid body system. Now, let's work on the example. There is a disk of radius R and mass m and then with the center there was a hub which has a cords wrapped around it, and then through the core there is the external force applied here. Initially the disk is rolling down to the incline. After the T is applied, what would happen? Those downward rotational velocity will be getting smaller and smaller due to the force to the right hand side, the upward. So at a certain point, it'll stop. So during those behavior, this disk is kept rolling without slipping. So you are supposed to find the angular velocity at time t_Naught and how long it or take the wheel to come to a stop. This is the problem. The first step to set the coordinate. As I said before, the coordinate should be located somewhere at the contact point, absolute coordinate. Before we draw the free body diagram, let's briefly review what are the conditions for rolling condition without slipping and with slipping. Without slipping, there is a kinematic relationship between the acceleration and the Alpha angular acceleration. While we don't know the friction information until we actually solve the problem, when the slip occurs, those acceleration and the angular exertions are two independent variable, whereas the friction force is specified as a dynamic friction. Here there are two contact, one with the string and one with the ground. So they are a three-body diagram shown like this, and then it'll generate a translational and rotational motion of the disk. So the equations of motion 3x directional forces like tension, friction, and gravitational components. In the y direction, there is only gravitational [inaudible] force. When you obtain the moment equation [inaudible] the center of mass, there's only torque generated by the tension and the friction. There are two terms here like this. Now, we're supposed to find the acceleration to actually solve the velocity. Since this is the no slip and then the friction is unknown, since this is a no slip condition, let's use the kinematic relationship between a and the Alpha. If you plug that in, we are going to obtain the equations about the unknown friction and the Alpha, and another unknown friction and the Alpha. So if you eliminate those unknowns, you will be able to obtain the equations about Alpha, and everything else is known so that you can do the time integral to obtain the Omega, which is a time integral of the Alpha at time t_Naught, like this. So all the constant value multiplied by the t_Naught is equations about Omega that we want to know and it's initial Omega which could be determined by initial velocity. Now, we can actually do the same thing with the impulse momentum equation. So initial mv and then linear impulse is going to be final mv initial I Omega, and then angular impulse is going to be final I Omega. So where should we set the point P here? Well, if we set the point P here at the ground contact point 0, there is no velocity, so we don't have to worry about the inertia force. So let's set the contact point 0 as our reference point to define the angular momentum. So for the linear impulse-momentum relationship, initial mv, which is mv_Naught and the initial impulse due to the attention and the friction and the gravitational components is going to be generating the linear momentum changes from minus mv_Naught to mv_2. For the angular impulse-momentum relationship, initial I Omega added by angular impulse is going to be final I Omega. Then here we don't know the frictional force here if you'd like to apply the linear impulse-momentum relationship, but by taking the moment with respect to point 0, the frictional force does not contribute to the moment with respect to the 0. So what we could get for the angular impulse are all known variable like, you know the geometry, you know the tension, and you know the other inclined components. So this is a solvable. So if you equate this angular impulse momentum, you can get this. It's all like initial angular impulse and initial angular moment, and angular impulse will generate later angular momentum. So this is how you solve the Omega at time T_Naught, and this is the equivalent to the solution you obtained by integrating the equations of motion. Let's solve another problem. Similarly, a uniform disc is located, and here, instead of putting in the inclined surface it's on the treadmill, treadmill which has a velocity on it. Then after awhile like initially rest disc is placed on the treadmill, and then later this disc will keep rolling. This disc will rotate and then reach to the steady state so that the velocity of the disc is same as the velocity of the belt. So in that case, what's going to be the time, t, required for the disc to reach its steady state status? The very first step is, just set the coordinate here. Well, are you going to set the coordinate here or at the center? It doesn't matter because the coordinate doesn't move. I am setting the coordinate not on the belt, but the place where it's a stationary. This case, two contact; one with the ground, one with the pivot point, so there are five forces. If you can obtain the equations of motion for two x directional forces and three y rational forces, and your moment equation related to t is only a moment generated by the friction. Now, and you are supposed to find the Alpha. To find the Alpha you have to know the friction. To know the friction, let's see, I have to know the reaction force to calculate the halves, but there those unknowns as well. So there are too many unknown. So to solve this, you need more information. I may use the no slip condition like a kinematic relationship, but this is the case where the top part is rotating moving this way and bottom part is moving the other way, and those actually matches with the speed for the treadmill. So this is the case that slip occurs. So instead of not being able to use these kinematic relationship, what we can have is I can specify the friction as a kinetic friction. So f is going to be Muk and normal force. So I can use that information. But still to obtain the normal force, I have to know what's going to be the y direction of reaction force, so let's do more work. Since this is a two body, one is with the disc and the other is a linkage, the links connects the disc to the pivot. So let us draw the free body diagram of this one. Since this one is fixed system, so there is no movement, so the zero acceleration; so what you can have is, if you could calculate the moment with respect to point A, this pivot point, what you can have is torque by the reaction force R, and torque by the gravity will be balanced out, because it doesn't rotate. Therefore, from there you can obtain the value of R of y, which you can plot back into original equations of motion to obtain the normal force. Once you obtain the normal force, you can obtain the friction force because it's going to be kinetic friction. Therefore, what you can end up finally is all known variable equals to the I Alpha. So to obtain the Omega, you can just do the time integral of the both sides of the equation, and can be able to obtain the Omega in terms of time. Note that at steady state, the contact point velocity of the disc is same as the velocity of the belt. We can do the same thing using the impulse-momentum relationship. Initial mv and the linear impulse is going to be final mv. Initial I Omega and the angular impulse is going to be final I Omega. So where should we put point P here? Where should we put the reference point? Reference point will be either this contact point or the central point. But the central point that we don't have to worry about the moment due to the unknown reaction forces. So I'm going to set the P here at the center point. Since it doesn't move, you don't have to consider the inertia force. So only the I Omega initial and all the moment with respect to the center is going to be generating the change of the angular momentum 2, I Omega 2. For linear impulse-momentum relationship, there is no speed here, no speed later, so you're linear impulse actually turns out to be zero. For the angular momentum, initially it has I_0, Omega 1, and then it set initially at rest, so it's going to be zero. The angular impulse is going to be due to the friction, and that will generate the change of the angular impulse from zero from start to it's final value. So if you equate these parameters, you can have those angular impulse will generate the change with the final angular momentum. At steady state, you could use the speed information about the contact point of the disc and the treadmill. Note that this answer is equivalent to the solution you obtained by integrating the equations of motion shown in the previous slide. With that, that ends the impulse and momentum equation part 1, and we're going to do more examples in part 2 later. Thank you for listening.