Hi. In this video, we are going to study how we can formulate Moment Equations of three rigid body in 3D. Moment equations in 3D is not going to be explained by a simple, IE alpha form because angular momentum is somewhat complicated form. So even though F equals ma at the acceleration of the center of mass can be applied for the 3D, but for the moment equation can be reformulated and also note that when we formulate the moment equation, we always first set the reference point to define the momentum, moment and angular momentum. So in the previous section, we learned how we can formulate angular momentum. Here, it's going to be somewhat multiplication for the moment of inertia tensor in the omega vector, right? So when we take a derivative since i is the geometric determined via geometry of the structure, so it's a constant value. So can we just take the time derivative of the omega's to obtain the H dot? If the system is under the rotation of the axis, so the axis is defined as a relative coordinate which is rotating, so the i j k vector is not going to be time derivative, is not zero anymore then you have to also consider the time derivative of the i, j, k vectors. So this doesn't hold. Therefore, in this case, you have to consider the omega cross H term if your angular momentum is defined at a coordinate which is rotating. So for omega cross H term, you can do the cross-product multiplication with the addition of time derivative of H relative. So finally, you ended up moment equation as a form of time derivative and omega cross H. Then if we have assuming that those rotation axis is at the part of rigid body and then also we assume those capital omega is the same as small omega, that just body rotation, then we can also simplify their form as a capital omega to the small omega. Let's solve the example. A uniform slender bar, slender bar means you don't have to worry about the moment of inertia of this bar. Don't think of it as a cylinder, just the mass and the length rotates with a constant angular velocity and it's been supported by the bearing A and B, find the force at varying B due to rotating imbalance. Most of the problems, equations of motion problems in Chapter 7 is about the rotating imbalance problem. Because that's a simple way that you can think of the how moment is applying resulted in momentum, angular momentum change and omega terms involved. So let's solve the problem, put the axis on the point o, and there are steps like defined a coordinate, draw the free-body diagram and obtain the equations of motion. Here, instead of moment of cos i alpha, you have to bring up the relationship with the angular momentum, which is I_xx will make x, I_y omega y, I_zz omega z and all the other term and the cross term, and also a time derivative of angular momentum and if this is a rotating coordinate, you have to also consider omega cross H term here. Now let's figure that out, where are the forces? There are two contexts at the two bearings, there are three force component per each contact and the gravity and those are generating translational motion and rotating motion of this mass. Graphical relationship has been transferred as a equations of motion in x and y and z translational motion of the center of mass and to obtain the moment because, angular momentum change relationship, note that center or masses just keep rotating here like this is been rotating this way. So we'd better have it switch the reference point which is something like fixed one, like a point o. So let's obtain the equations of motion, the moment equation with respect to the o. So to do so, you should figure that out, what's going to be the age of o and to do so, first step is figure that out, what's going to be the omega relationship? There's only z-components of the omega, so omega x and omega y is a zero, so that will make this equation a little bit simpler and you should plug in all those variables from their H_x dot H_x is currently here. So this Is are the constant value based on the geometry, so you only have omega z dot term here and then you can multiply the H_y and omega z. H_y contains omega z part, so if you multiply by the omega z, you will have omega z square term such as, you know, obtained all the parameters here from the equations, basic angular momentum and moment equation and since the omega is been rotating with constant speeds, omega z dot term goes out. So what you can then have ended up is double I_yz and z square term. For the two-dimensional, you have a moment equals I alpha term, but here you have a moment equals I and omega square term. Now to figure that out, what are the forces that will generate the moment in x-direction? Think about that. There's a y directional forces will generate the moment, so the mg and the f_By will generate the angular momentum in the x directions and same for the y direction. There is a moment by the gravity, but the reaction force at the pairing is a lot larger because this reaction force due to the motion, the angular motion of the mass, the unbalanced mass, right? So this term is relatively large, so usually we ignore the moment due to the gravity and this is how we can get the bearing forces at the B because we all know I components and the omega component. So the omega of the pairing force will be the vector sum of this x and y component. Now, how we can obtain the I_yz and I_xz? By definition, I_yz is a yz multiplication of that particular mass and the integrated over the body, right? Since we are only looking at this part, the rest of the part is just stays at the pivoted as a bearing. So in yz projection, I have this amount so it's going be cosine theta, l cosine theta. So z has a constant value b, but the y is varying from zero to L cosine theta. So you take the integral of the mass as density over length of row and dy and then you can do the math, you can ended up having I_yz in terms of theta, same for I_xz. If you look at this bar in the xz plane, or you can have is this component, this length. So you will have L sine theta components with the distance from the x axis that with the amount of p. So where you can have is z components as constant value b, but x could vary from zero to l sine theta. So you can actually do the math and will ended up getting this value as a function of theta of the product of inertia and then if you plot that in back into the original equation you will be able to get the benefit of the bearing force at joined p. Okay. Let's solve another similar problem. In this case, there is a disk of mass m, but the center of mass is located with the shift from the center like with the moment of d and then now is rotating with a constant angular velocity omega and what's going to be the bearing force at A and B due to the rotating imbalance? First, we should set the coordinate, may be at the center of the disc and draw the free body diagram and obtain the equations of motion. F equals MA, moment equals instead of I Alpha, you will have an angular momentum and take a derivative depths over time. Angular momentum has this relationship. So that your moment equation looks like a time derivative, relative angular momentum derivative, and if your angular momentum description is based on the rotational coordinate, you should have Omega cross H term here. The free body diagram, since there are two varying, you have a three forces per each and the gravity term and they will generate the rotational motion and the translation motion. So first, you have a three X, Y, Z is all about the center of mass equations of motion and to obtain the moment equation, instead of getting it with respect to the G, let's calculated with respect to the O. So to do so, you have to first calculate the what kind of the angular momentum to take a derivative of this to obtain the moment. So first, you should check what are the Omegas here. Here is Omega has only Z component, so you have some parameters is about the Omega X and Omega Y has been disappeared and time derivative with them is just a time derivative angular momentum and Omega cross H term since this axis is rotating with the disk. If you just plug in all those parameters, and take a derivative, time derivative of this term and H_Y component and multiply by the Omega, you'll ended up this equation, and since this disk is rotating with constant angular velocity, Omega Z dot term goes out to be zero. So what you can have is similar to the previous example. It's going to be I_yz Omega Z square and then I_xz Omega Z square. Also, you can calculate the moments due to the bearing forces. So for X directional force that all X directional moment is due to the force in the Y component, perpendicular to the X-axis with the amount of displacement. T1 will be maybe the H, T2 is going to be H plus L. We just generally just put it as a, T1 and T2. In Y direction, there are forces in the vertical way. So it will be the vertical or whatever the perpendicular to the Y-axis. So the force in X component in bearing A bearing B and the gravity will contribute the moment to the Y-direction. Then they are all be equal to the I_ and Omega Z square term. Now again, compared to the bearing force, the reaction force to due to the rotating imbalance is a lot larger than the gravity. So we can ignore the contribution of the gravity to the moment. Since Omega Z is given and I_yz, I_xz that you are able to calculate, we might be able to figure that out the appearing forces. However, let's first calculate the I first. I_yz, since we set the coordinate on the surface of the disk with where the center of mass lies, I_yz, the center of mass Z coordinate is zero. So I_yz and I_xz are all zero. So we have equations about the bearing forces in the Y direction and X direction equals to be zero. Since there are four unknowns and only two equations, we need more information. So think about it. In the previous case, the coordinate has been set to one of the pairing joints so that you can eliminate those unknowns when you obtain the moment equations. So let's do it for this case as well. So instead of setting the coordinate at the center of the disk, maybe we could move it to the one of the bearing. So that all the moment contributed by the bearing for set A can be disappeared. Then those moment equation will be all the moment contributed by the bearing force f_B. X direction and Y direction will generate the angle of dynamic imbalance. But here, it is I is defined at point A instead of O. So let's calculate I_yz and if you look at this gravity center of mass point with XZ plane projection, it looks like it's just a single particle with the distance D from the Z-axis and H from the X-axis, right? So your I_yz is going to be mdh, but your I_yz, since this is setting the coordinates aligned with their center of mass, at Y direction our coordinate is 0. So I_yz turns out to be zero. So if you plug that in, you can figure that out. The Y-direction on bearing force f_B turns out to be zero, while your X components will have this certain amount. Once you've figured that out IBF force in the bearing B separately, you can also use the information to the previous equations about joint bearing force A and bearing force B and calculate what's going to be the bearing force at A. So in this video, we have gone through how we can formulate the moment equation. The moment equation simply a time derivative angular momentum and if the angular momentum is defined by the coordinate which is rotating with Omega, you will have to also consider this term. Rest of them are just calculation for the Omega and moment of inertia for the angular momentum and its time derivative. Thank you for listening.