Hello, in this video, we are going to study 3D rigid body kinetics, Newton's law, equations of motion. This has three-parts, angular momentum, moment equation, and work and energy. And first, we are going to study how we can formulate the angular momentum in 3D rigid body. To make the long story short, this is a quick summary for how the kinetic relationship will be explained in 3D. So for Newton's law, f equals ma, actually the f equals the change for the linear momentum, and moment is going to be change for the angular momentum. Was expressed by ma center of mass and I center mass and the alpha. But this does not hold for 3D, specially for the moment equation because angular momentum is not going to be just as simple as I omega in 3D. It is expressed like this way, okay? So it has Ixx Omega x. Ixy Omega y. I xz Omega z. And I vector and j and K as well. This angular momentum looks a little bit complicated. There moment equations looks like the moment is going to be the change for the angular rate. And this is the time derivative, the angular momentum. Okay, we are going to learn about this later. And for the kinetic energy, for the particle, you have one-half mv square. And for the rigid body, what has been added? Yeah, rotational term. I Omega term. I Omega square term has been added. For the 3D Rigid body, this going to be changed a little bit like you have a v center of mass square. Same for this one. But for this one, since now a, angular momentum, is not simply I Omega, you will have omega dot and angular momentum with respect to the g. Okay, let's derive how we can formulate the angular momentum. Angular momentum is defined by the r cross linear momentum, mv. And whenever you have a particle is moving with the Omega, your v could be expressed by Omega cross r. And now, we are going to check is that going to be I Omega in 3D? Suppose that you have a point A in 3D, and it's rotating Omega. And you want to describe it from a reference point, a fixed point like x from O. So it's going to be x, y, z with the velocity like Omega cross r. Then, your angular momentum by definition is going to be H of o is going to be r cross mv. And then, integral over the mass, okay? Now, suppose that you are putting the reference point at the center of mass. Then, what you can have is the r is going to be changed as a rho. Rho is the distance from the particle with respect to the center of mass, okay? So the same format. And so anyway, either rho or the r, depending on the people point that you're defining your angular momentum, is going to be, suppose it expressed by x comma y comma z. And let's say this is x. Then, derivative of the angular momentum, dH, is going to be the term that you want to integrate, which is x cross product v, which is Omega cross x and dm. And those are calculated by the cross product of formulation, like this. And then, if you also do the cross product of the x, you will ended up having this form, okay? Now, let's rewrite this in terms of Omega and the others. What you can have is y-squared, z-squared, Omega x minus xy Omega y, minus xz Omega ztm I vector, etc., okay? Now, since you have dH, to obtain the the angular momentum, you have to do the integral, okay? So what is the integral of the y-square, z-square, and the dm? That's the definition of moment of inertia, right? So Ixx is the distance of that particle over the integral of the mass, right? And Ixy defined by xy dm integral, that's what we call the product of inertia. So with that it definition, okay, and then I here could be defined either with respect to the reference point O or G. Depending on that, you may have I-bar here when it is defined from the center of mass. Now, anyway, so by plugging in those definition of the moment of inertia, what you can have finally is going to be Ixx Omega x, Iyy Omega y, Izz Omega z for the diagonal term. And for the cross term, it's going to be minus Ixy Omega y minus xz Omega z, okay? And then, those are those are matching for the product of inertia term, okay? This is how you could derive the angular momentum with respect to the O or with respect to the G. So here, we can express it as this is a multiplication of the I inertia tensor, 3 by 3 one, multiply by the Omega vector. Okay, so this is what you are supposed to memorize. Now, what is the relationship between the H of o and H of G? So a H of o defined by r cross Omega cross r. And r could be split out as r-bar, the distance from the origin to the center of mass and the rho. Rho is the distance from the center of mass to the particle. And rho plus r4 plus rho here, again. And if you expand the equation like this, and then if you distribute the cross product term, what you can have is this one. And due to the cross product relationship, a cross, b cross, c, this is been canceled, okay? Now, what you can have is this one, which is r cross v, right, at the center of mass. So this is the angular momentum of a center or mass, assuming that the whole rigid body is a single particle. And this one, by definition it's H of G, right? So you can obtain the relationship between H of o and H of G such as H of o is going to be H of G plus angular momentum of the center of mass. Sometimes it'll be easier for you to calculate H of G instead of H of o. In that case, you can actually calculate H of G and angular momentum center of mass to find out H of o, okay? Let's solve the problem, then. We have three small spheres, small sphere means you can consider it as a particle, of mass m are welded to the horizontal bar, here. And which is rotating at the z axis. And you're supposed to find the angular momentum, H, with respect to the o. So this is a three dimensional rigid body, so let's bring up the formula or the equations of how angular momentum in 3D looks like. Do you remember? Okay, H of o has a three diagonal term, which is Ixx Omega x. Iyy omega y, Izz omega z, right? And then the cross term is I minus Ixy, omega y for the second and then omega z for the third, right? Okay, so to calculate this, because these are somewhat complicated. So first, you should specify what's the omega in your condition. What's going to be the I. You have to specify the I and then you can calculate the age. Maybe we can change the problem by having reached body system like this. So instead of have a particle the small sphere, you might have a certain body which has its own inertia like like this. Okay, so let's solve the problem first. You should specify the omega here. You only have omega z, right? So you have omega equals omega z and omega x and then omega y turn is zero. So you can delete all the terms related to omega x and then omega y. Good news, so your angular momentum is going to be simpler like are you have only three terms of regarding to the omega z. Now Ixz, what was the definition of Ixz? Yes, it's integral xz dm and yz is integral yz over dm. So this is all defined from the original, because we are calculating H of O. Now, let's first put the coordinate for each mass. R1 is going to be here. It's 0, 1, 2. And r2 is 1, 0, k. And r3, 0, 0, -k. To calculate Ixz, you have to calculate Ixz of each mass and add them together. Okay, to calculate the Ixz, let's view this as x, z-coordinate, then it's going to be a particle has a distance distance from the z-axis and y-axis with the amount of b, okay? If you build this with the y, z-coordinate. It's going to be located at y equals 0 part, right? So if you calculate this, this is going to be mb square. This one is zero and Izz is going to be mb square. In the same manner, you can calculate here. Ixz is zero. Iyz is zero and Izz is also zero. How about this particle this particle? Ixz, zero. Yz, you have a distance to be here. One be there. So 2mb square and Izz, you have an mb square. So once you calculate or Ixz, Iyz, Izz for each mass, you can just add them together to finally or calculating the whole moment of inertia. And since you know the omega z and you know all the I components, you can just pluck that in. So you will end up having, obtaining angular momentum. Okay, since I said, this one could be treated as a single particle based on the problem. You might think, well, this is what we have covered in chapter four systems of particle. So can we just calculate the angular momentum of each mass and add them together? Yes, you can do that. So you can do a sigma for I equals 1,2, 3 here and then ri multiply cross product mivi and the vi here is going to be omega cross r. So you can do one for each like r1, r1 omega cross r1 or 2 omega cross r2. R3 omega cross r3. And if you calculate them together, you will have omega 0 as mb square omega. I, 2j minus 2k which is equivalent to the one that you obtained using this 3D formula. Let's solve another example. Do you have a cylinder of mass m radius d and length l and it's rotated omega and the whole shaft assembly is also rotating with another capital primary omega here, okay? In the previous chapter for the kinematics, we are interested in calculating the acceleration of the a or angular acceleration of the drum is a and so on. But in this case, since we are working on the kinetics the very first step find the angular momentum H of O over cylinder with respect to point here. Okay, so can you bring up the formula or the equations how you can obtain the age of O. Yes, okay, so first user set the accord specify the omega. Omega is a primary here, secondary there and then your age angular momentum relationship is a Ixx omega x, Iyy omega y, Ixd omega z minus xyxz omega y and omega z and there are cross matching for the moment of inertia. And whenever you have a H of O as a relationship with H of G and angular momentum of the center of mass. Now first, specify the omega, specify the I and calculate the H. Okay, so since we define the primary and secondary omega, let's formulate it. Primary omega has omega capital I and secondary has omega small j. And at this moment, at this instant, this small xijk is same as a large capital Ijk. So your primary omega is this omega I, small I. And when you take the derivative Ijk vector, you have to multiply quick capital omega cross product. Now, so they are your total omega is going to be capital omega I and small omega j. So to formulate it omega x, omega y, omega z component. You have a omega x is capital omega. Omega y is small omega and there is 0 omega d term is zero. So if you delete omega z, you only have six terms left and this is what we have done for the specified omega. Now, let's specify the I. Okay, since you're working on, you have to find the cylinder. You have to identify what kind of the Ixx, xy, etc., for the cylinder. We respect to the O here, right, but it will be a lot easier. If you could calculate the age, we respected the G, the center of mass. So even though you're supposed to find the angular momentum with respect to O, maybe you may use this relationship finding H of G and calculate the angular momentum of a center of mass treating this as a particle, then this might be easier. Okay, so what is I bar xx and zz? So if you look up the table, you would say, the cylinder, one of the inertia of the cylinder with the to the central axis is a one-half. r square. And whatever's along the height of the the cylinder, that's this. So if you plug that in, those are corresponding to Ix and Izz, which is one-fourth md-square, one-twelfth ml-square where Iyy is one-hald md-square. And since it's symmetric with respect to center of mass, our product of inertia terms turns out to be 0. So by plugging into them, those terms out to be 0, so you only have to term Ixx Omega x, Iyy Omega y. So you can just plug that in. And how about this term? Since after you're calculating the H of G, you have to calculate this one. v-bar is speed, velocity of the center of mass, so you have to define the r-bar vector. Here is Hk and One-half j minus in direction l over 2j in the Omega. I mean, your speed for the center of mass is going to be Omega cross r, right? So finally, you ended up having H of o as H of G. This is H of G. And r cross product mv. v here is capital Omega cross product r-Bar. Now, let's solve another problem similar to the one in the previous. We have a disc, point A. Previously we were interested in the kinematics chapter interested in calculating the acceleration of the a or angular acceleration of the disk. In this case, we are supposed to find the angular momentum of the disk with respect to point O, okay? The one different thing of this example from the previous one is this one has three Omegas, okay? So your disc is rotating the Omega, and the shaft is rotating with Omega 2, and the whole structure is vertically rotating with capital Omega. So the primary Omega is capital Omega, and the secondary is Omega 2, and the third one is a small Omega. And then, I can put the axes, primary and secondary here at o, and the third one here at B, okay? So how we can formulate the angular momentum? Same, Ixx Omega x, Iyy Omega y in the diagonal term. And the cross term, minus xy Omega y. This one matches with this, and so on. Okay, the very first step, specify the Omega, specify the I, and just do the calculator, just the math. Okay, Omega, capital Omega is capital Omega j. Capital Omega 2 is Omega 2 capital I. At this instant, that capital I is going to be same as minus small i, okay? And if you take the derivative of this ij vector for the Omega 2, you have to consider capital Omega cross term here. And at that instant, your capital J is the same as small j. For the third Omega, you have Omega small k. And if you take the derivative of the small k, this is under the influence of capital Omega and capital Omega 2. So those two terms are doing the cross product to obtain the k dot. Now, we are done with specifying the Omega. So your Omega is adding them together in terms of small ijk vector. So the first one is Omega x, and second one Omega y, and third one is Omega z. Nothing is a 0, so nothing can be deleted yet. Now, since you we are supposed to find the angular momentum of the disc, we have to calculate the I of this disc with respect to the o. That's tough. So let's assume center of mass located at the center of the disk. Then it'll be a lot easier to calculate H of G instead of H of o. So we are going to use this relationship again. So H of G plus angular momentum of the center of mass, which is r-bar cross product mv-bar. Now, what is Ixx-bar yy and zz and the product of inertia? Look at the table, and then you can find the relationship for the disc. So that axis through the center of the disk is Izz. So Izz is one-half md-square. But for the y and x axis, which is along this height, you only have this term. Because for the disc, we assuming l is a 0. Or those product of inertia terms start with 0 due to the symmetry. So those has been canceled. What you have is only the diagonal term here. So next step, that's how you can get the H of G. And then, now you can calculate the center of mass angular momentum. So what's going to be the v-bar? v-bar is going to be Omega cross r-bar, right? And here is under the influence of capital Omega. And then, capital Omega 2, it's going to be Omega plus Omega 2 cross product r. But here, these two terms are in parallel. So what you'll have left is just capital Omega cross r term. So finally, you ended up having H of G here, or the diagonal term. And the r cross product Omega cross r. Okay, so this is the end for part one, equations of motion in 3D. F equals ma is simple, but moment equals I-alpha is ot as simple as a two-dimensional rigid body because angular momentum has a form of moment of inertia tensor matrix multiplied by the Omega vector. So as a first step, we learn how we can formulate the angular momentum. The next step, you're going to learn how we can make it as moment equation form. Moment equals angular momentum change form. Thank you for listening.
