[MUSIC] Hello in this video, we are going to study gyroscopic motion one of the 3D rigid body kinetics. Okay in the previous chapter we learned how the moment equation for the 3D has been expressed in terms of instead of simple alpha form but we have been which we learned in 2D, 2D motion. 3D motion has many other terms for the product of inertia or other omegas right? So, your moment equation is not going to be simple as i alpha instead you have a angular acceleration term as well as some coupling of omegas, okay. But to make your life simple, the most of the previous example what we have covered is a case where you only have a one omega, okay? So if you just suppose that you only have omega z left, then we'll leave out all the other terms. And what you can have is what's going to be the effect for the i for a moment of inertia and productive inertia term. So, to generate the product of inertia term, you should have a structure being asymmetry right? Otherwise the symmetric case those terms will be zero. So, since you only have one omega, those other omega related term will cancel out, and then to make your life a little bit more simpler mostly the constant omega case was given. So, what you can have is only moment component in the i directions and j direction will generate out the angular momentum change based on ijk term. In this example, now let's switch what if you have i, is really really simple case. So if you have i is kind of of symmetric case, you have product of inertia term goes out and if you thinking about the this case then you only have ixx term. So that's going to be pretty similar to the 2D case, angular momentum is going to be simply i omega. But in this case we are going to introduce the your moment is not only the i alpha, involving the other terms, saying like this one. So if you can eliminate all the Hy and Hz term, you still have this term left, okay? So this is term like your angular momentum in x direction and multiply by the omega in the other way, like a z or the y. So this is what we call the gyroscopic motion and lets take a look with this example. Okay, for about simple rod rigid body motion, I suppose there is a tilt with the theta and there is a gravity exist, and then those gravity will exert the moment in the clockwise direction, so directed through the screen. And then your motion is going to be clockwise motion as well. So the cause and the result are in the same direction, right? It's kind of pretty intuitive. However, suppose that you have a 3D structure rod, kind of cylindrical shape and it is a spinning initially. And if your gravity takes the role of rotating the whole system clockwise what is going to happen? Instead of just rotating clockwise, it will be rotating through the screen, okay? So this 3D structure so it is a tough to describe but there is going to be an other results, coming, compared to the 2D motion, okay? Same for the other example like if you have a disk through the shaft and then if I'm rotating with counter clockwise and that means I'm applying the moment that way the cause. And usually it's going to be a rotate counterclockwise rotation emotion with the omega and the alpha. However, it has initial spinning, if I apply the moment this way, the result is going to be the other way, this whole structure will turn around like through the vertical axis. So your results is going to be somewhat different from the case where the two dimensional. So for gyroscopic motion is the change in the orientation of the rotational axis of the rotating body. So like this one, this is a rotating body and then it's a spinning axis is along its shaft and the shaft is rotating with respect to the vertical axis. So this is what we call the gyroscopic motion or precession. How this movement behavior happen, okay? So for an instant there is a initial rotation in motion through this shaft and then there is, due to its tilt posture, there is a gravity applied and then that will actually apply the moment with respect to this contact point okay? The direction that's coming out of the screen. And this is actually a perpendicular to its original angular velocity. And suppose that this initial angular momentum has been changed due to small amount of the momentum applied. So it turned out to be this angular momentum at a time later. It's going to be another moment applied and another change for the angular momentum. So that ended up having circular motion of the angular momentum so the resultant precession occurs, so this is how it works. This is a video clip for describing gyroscopic precession and either you can find it from the YouTube or if you Google gyroscopic precession this video will pop up. This is a pretty well made one, so I strongly encourage you to take a moment to watch it. Okay, so to understand this precession, let's remind what happened actually your force direction is perpendicular to your motion for a two-dimensional motion when you think about the particle initially moving with the velocity, V. And if your force is applied perpendicular to the motion, what will happen? There is a rotational motion occurs, right? Since your time multiplication to the force is going to be the derivative of the linear momentum mv, the initial mv with the applied Fdt, the small amount of the linear momentum change were resultant to the linear momentum changes from G to G plus. And say this angle between these two linear momentum is d theta. Then this for this very small theta, this change could be approximated as initial mv d theta which is going to be equal to the Fdt. If you divide both term with the dt what you can have is with the direction for the theta, vertical the z-axis, you will have force is going to be d theta dt mv with the cross product. If you look at the vector direction for the theta and the v you will ended up having the direction for the F. So, you have omega cross product mv which is when the it has a uniform circular motion. Centripetal acceleration. So to interpret this, this is a cause. Applied force will generate the motion which is initially in this status has been rotating as a resultant motion, okay? Similarly to this, when you're thinking about the three dimensional structures which is the angular momentum this way. And then if you applied a moment coming out of the screen, okay? Then your change of the angular momentum is going to be from here to here with the amount of d phi, okay? So for a small d phi, this length has been approximated i omega d phi. And that, sorry, d psi, and it's going to be same as M dt. And then if you divided both sides with the time, dt that are you are going to have a moment is going to be d psi d t. Which is defined here capital mega cross product of I omega. So you will have omega cross product I omega. So sometimes, since these are two omega term, you can switch it as a P, okay? So again, this moment is a cause for your initial spinning, it's going to be rotating x axis. So it's just a resultant rotation, precession. You can also interpret this and different ways. Say, when you have a certain physical quantity, if you want to take a derivative of this with the axis. Small x y z is rotating with capital omega, primary omega. Then you will have omega cross self and the relative derivative of the self value. So suppose that you have angular momentum Ip, which is defined that the coordinate rotating capital omega. Then your derivative will be capital omega cross product Ip plus Ip dot at the relative coordinate. And then whenever you have a constant angular velocity of your structure. Then what you will have as moment is going to be capital omega cross Ip which is equivalent to this. Okay, let's do the example. You have a spinning wheel through the shaft. And then if I apply the force clockwise directing this way, what will happen with the spinning structure. Instead of just rotating this way, it will precess through the vertical axis. When you have a spinning disk and try to apply that moment clockwise, what will happen you will have a precession this way. If you have a spinning wheel and apply the, I guess suppose that you are riding a bike. And if you're turning the handle right in side, so that it's rotating this way, instead of actually rotating that way. You will have a precession like a tilting to the left hand side. So, if a stationary object, your moment will cause the change for the angular momentum, which is parallel to, wait, where's my cursor? Parallel to our moment, right? However, whenever you're initially spinning, your moment will have a resultant rotational motion. Which is perpendicular to the moment applied. That's the precession. Okay, so is the precession faster when spins are faster? What do you think? Intuitively, yes or no? Okay, from the equation, due to the same amount of moment. Your precession, amount of the precession is going to be inversely proportional to your speeding speed. And with this does make sense because you have a large in a rotation or inertia, which is Ip. It is less likely you're doing the omega large precession. Okay, let's solve the example for the gyroscopic motion. Here, you have a disk spinning with small omega. Which is due to the rotational motion of the whole assembly through the disk attached to the shaft. Shaft has to being attached to the U-shaped clamp, which had a hinged joint at O, through the main shaft. Main shaft is rotating capital omega. And then to that, this disk is spinning. And then at its surface, it's rolling without friction. So in that case, what's going to be the normal force from the floor? What's going to be the external force apply to this system? This is what you're supposed to solve. The very first step, let's set the coordinate, okay, for the primary rotation. And I'm going to set the coordinate at O. And for the secondary rotation was for the disk, I'm going to set the coordinate at the center of the disk. Now let's draw the free body diagram and apply the moment equals. Instead of I alpha you will have angular momentum vector and take a derivative of this. So we're just like this. So you have I xx, omega x, I yy omega y, I zz omega z, and all the other cross term, okay? Now, so this is the case where your handling the very simple I. So your disk is metric so to calculate the inertia you only have left the I xx, okay? At the center of mass and deleted out all the others, right? And then that will actually make the moment equation simpler. Because you have Hy and Hz turns out to be zero. So what you can have is I omega as an angular momentum. And the moment is going to be the I alpha with the same y direction of components. Plus, you have a perpendicular motion, which is going to be multiplied by initial angular momentum. Not necessarily of the initial, angular momentum, multiplied by the precession motion here, capital omega. This could be also interpreted as when you have a single I omega term, then your time derivative will be omega cross I omega plus I omega dot. Which is equal the i component here and which is going to be j component here. Okay, so since this is the equations of motion in three dimensional rigid body here, and then let's draw the free body diagram. So for the disk and the shaft there's going to be a two context, one with the pivots and one with the ground. So there are three forces here, three forces there. And the gravity and for the Newton's equation you have x and y and z components for the motion of the center mass of the disk, okay? Now, for the moment equation, this is a simplified form. So your x component for the moment will generate the I alpha, which is this one, H x dot here. And your y component, the moment it's going to be generated omega cross I omega, which is H x omega z term here. And x directions for that direction, the only force that applies that contributes to the moment for the x direction is the fy. With a distance of d, so this is fy d. And for the moment in the y directions, in that direction, let's say all these forces here will be crossing the y axis, so maybe this force And this force will contribute to that moment for the y direction. Now, since you're supposed to find the normal force which is related to the Rz. You still have to figure out what's going to be the omega, okay? So here, we are going to use the no slip condition here. So we will remember in chapter six where there were no slip condition. Your linear acceleration has a relationship with the angular acceleration, like this. And since you have acceleration relationship you also can extend it as a velocity relationship. So your v in y direction is minus d omega, which is minus omega here for the positive x direction this way. And since the center of the disc is also a part of the whole assembly, which is rotating by the primary of capital omega. So this is going to be also the omega cross R, that did l, displacement l here. So you will ended up having omega l for the vy. And this is going to be equal to each other. So we'll ended up getting small omega. The wheel spinning speed in terms of capital omega and the distance of the disk from the main axis and the radius. Now since you know what's going to be the omega here. You have all known variables to the right hand side so that you can calculate the Rz. Okay, since this is the constant value now I know that the alpha turns out to be zero. And assuming it's a smooth surface you don't have a friction here, okay? So what you have is all the known variable to the right hand side will be determining the Rz. And thus Rz is related to the normal force, what you are supposed to find. And ended up getting normal force is equal to the mg plus this would, excuse me, rotational term. So if this is as a stationary here, normal force is going to be equal to the mg. But due to the rotation, the precession motion, you have increased normal force. So this is the case where you have two omegas and then what's going to be the resultant? Normal force, actually the moments apply to to make this happen. Let's little bit change the problem. Instead you're given capital omega precession motion and the spinning. We are going to solve what's going to happen when you're having initial spinning and the external moment, okay? This is the same case where we have seen except that this is on the air so there's no contact with the ground. And the other one has a little bit different joints here, this is a hinge joint. So you only have a reaction force exist, and this is the fixed joint which will have a reaction force and torque. Okay, so what's going to be the motion for each assembly? And then what's going to be the bending moment at B? Okay, for the free body diagram, for the left hand side a very same and similar to the previous example. Except there is no contact force in the ground. And we are going to just similarly use the equations of motion. In the y direction I just copy the equations of motion, what we have obtained in the previous slides. Instead of having a contribution for the normal force, you only have a contribution for the Rz force. Which is equivalent to the mgl here, right? There's no normal force term here, we used to be have a -NL term here. And then there is no normal force because it's on the air. So you have a y direction on moment apply, which will make the precession exist, okay? For the right hand side, okay, so what is actually happening is you have initial Ip here. Your initial Ip here, and your moment due to the gravity's applying perpendicular to your initial angular momentum. So you are going to have ended up having perpendicular precession motion. Okay, for the right inside, you have a same free body diagram, a similar free body diagram. Except that there is a moment at this fixed joint is applied, okay? So your equations of motion, y directional moment equations, they're going to be the same. Except that you have added moment term here, okay? Now, to figure that out, what's going to be the moment applied at B joint. We should focus on the shaft, okay, the massless shaft. So due to the free-body diagram for this massless shaft, we could find it, okay, the Rz is going to be equal to mg. And moment Mb is going to be the same as mgl, okay? And then if you plug that into here, well, what'll happen is, the spin actually cancel the left-hand side of the equation. So that your capital omega turns out to be zero, means no precession. Which does make sense, because when you have a disk rotating with a fixed axis there's no precession. Why you have a hinge a joint is to free to rotate, then there's going to be a precession motion here. Okay, so in this chapter, we briefly go through how you can describe, how you can understand the gyroscopic motion. Basically, you have a moment equation. And mostly you have, in this example you have a simple system. We have a simple I due to the symmetry. Your angular momentum is only single components I omega, same as a two dimensional. However, whenever you have a moment applied, there could be some precession motion which is perpendicular to the I vector could occur, okay? Or if you have a spinning initial angular momentum nonzero, and if there is a rotation perpendicular to that. What that means is you should have a moment applied to make that happen. Thank you for listening.
