This is Module 29 of Three Dimensional Dynamics. Today we're gonna actually use that theory we developed in the last couple of modules and solve for the motion of a rigid body undergoing 3D rotation of motion. So as a summary, we had a number of equations to deal with these problems. One was if we used the body fixed frame, we can fix that frame at the mass center, the origin at the mass center, or about some pivot point O. And then we said there was two other situations when we could use an intermediate frame that's not body fixed, and we can put the intermediate frame's origin at the mass center C again. Or at a pivot point, O. So let's begin by going back and looking at our retracting landing gear example. So here's the body of my aircraft. And I'm gonna apply a moment to rotate the body up into the aircraft. And my wheel is spinning. And so let's look at a demo. So again, here's my landing gear. It's rotating, the wheel itself. And then I'm gonna apply a moment to bring it up into the body of the aircraft. So, [COUGH] previous to this, we only looked the kinematics of this problem. We looked at the angular velocities and the angular accelerations but now, we're going to look at the forces and moments acting on the body. In fact, this is a moment about the x axis that's being applied, and how that's gonna change the motion of the body. Now this is actually a situation where I'm going to use an intermediate frame. And I'm fixing my intermediate frame to this arm here that's attached to the wheel. So here's my frame. It does not rotate with the wheel, so it's not body-fixed. It's attached to the arm itself. And what we wanna do is, we're gonna find, as a result of this moment being applied, what is the rate of change of the spin for the wheel and what are the moment reactions at C. Because I have a bearing here and I have to design for that bearing. So I wanna make sure I know what kind of forces and moments are acting there. So here's our situation again and we're using this expression for the equations of motion with an intermediate frame. Here's my diagram. You'll notice that I've added some dimensions here. I've given the radius of the wheel Distance R. And the thickness or the width of it is going to be capital H. And so the first thing we're gonna do is find the angular momentum of the wheel itself. And you gotta be careful here because when we find the angular momentum of the wheel, it's the angular momentum with respect to the inertial frame. And angular momentum is the inertia matrix times the angular velocity, so we have to find the angular velocity of the wheel. And so I want you to do that on your own, and come on back. And so the angular velocity of the wheel is the angular velocity of B with respect to I. And we can use the addition formula to find that. We'll say it's equal to the angular velocity of B with respect to the intermediate frame. And then the intermediate frame F with respect to the inertial frame. And we see that the angular momentum of B with respect to F is omega S or omega spin in the J direction. And omega F with respect to I is omega X around the I axis. And so now we have to find the mass moment of inertia matrix. And this is the step which is a key to using an intermediate frame. Because we see that for this frame attached to the I matrix, or excuse me, attached to the arm here, the F frame, in this orientation which does not spin with the wheel, the moments of inertia and products of inertia. In fact, this is a principle axis, so you only have moments of inertia. They do not change with respect to the wheel itself. They're expressed, both of these come out to be the same inertia matrix, so I about point C for the B matrix is the same as I about point C for the F matrix, and that's because of the symmetry of the wheel itself, and so we can use this intermediate frame. And so, you can go to a reference, find the moments of inertia for a wheel or a disk. This is the equation you'll come up with. I'm just, you got I about the X axis and I about the Z axis are going to be the same. You can see by symmetry this is the value. I'm just gonna call that a value I, just some number. And then I, about the Y axis, is gonna be mR squared over 2, and we're just gonna call that J, again, some number. And so if I, I also have my products of inertia equal to 0 because of symmetry, in the problem, when you look at this disk. And so here is my I matrix, it has just the mass moment inertia on the diagonal. Okay, so we now have our expression for the angular momentum. We found the angular velocity of the wheel with respect to the inertial frame. We found the I matrix for the wheel. We can substitute those in and we get this expression. The inertial matrix times the angular velocity and if I multiply that out I get I times omega X in the i direction. Capital I times omega X in the i direction. And then plus capital J times omega S in the little j direction and so here's where we're at. We have our expression for the equations of motion. I found the angular momentum for my body and now I can go ahead and substitute in and so I've got sum of the moments about c is equal to. The first thing we wanna do is take the derivative of this angular momentum now with respect to the F frame. With the respect to the F frame remember, the inertia matrix does not change, it's constant. However the angular velocity vector can change with respect to time. So I'm gonna have I, which is a constant times omega x dot plus J times omega s dot. And then I have omega F with respect to I. Now this is the omega of my intermediate frame. With respect to the inertia matrix, and so write out what that is on your own. And what you should say is, okay, for the F frame rotating with respect to the inertia frame, all I'm going to get is omega x about the I axis. So this is just omega xI. And so I end up with now omega xi crossed with H sub C itself, which is I omega xi plus J omega s j. And then I can do that math, and I get the sum of the moments about c is equal to I omega x dot plus J omega s dot plus I cross I is 0, I cross J is K, so I get plus omega x omega s J around the K axis. So, here we are. We've got my expression for the equations of motion for this system. I wanna find the rate of change of omega s, and I wanna find the reactions at point C. I know what the right-hand side is. How am I gonna find out what the left-hand side is? And what you should say to find the moments acting on the wheel, we're gonna have to draw a free body diagram. And so what I want you to do now is to draw a free body diagram of the wheel on your own for the moments about point C. And here is my expression. Let's look at a demo again. At the bearing, at point C here, I have no acceleration in the x, y, or z direction. X is the axis pointed towards you Y, is this axis, Z is up. So I'm gonna have a force, reaction force in the x, y and z direction. Now, I cannot rotate from this bearing. I cannot rotate around the x axis. I cannot rotate around the z axis. So I'm gonna have moment reactions about the x and the z axis, however, I can rotate about the y axis and so there's gonna be no moment reaction about the y axis. So here it is expressed on my free body diagram here. I have C sub x, C sub y, C sub z for the force reactions. I've got a moment reaction about the Z axis, a moment reaction about the X axis, but there's no moment reaction about the Y axis because it's free to spin. And the other thing I need to include on my free body diagram is the weight, mg. So, that's a good free body diagram. That's going to go in the right hand side of my equation, and so what I find is that I've got the sum of the moments about C is equal to M sub Cx around the i axis plus M sub Cy around the k axis, yeah k axis. And so there is written again. And so, here is my system, here is the equation, the equation of motion form I'm going to use with an intermediate frame. Here's the result I came up with. For the right hand side I just used my equation of, excuse me, my free body diagram to find the moments on the left hand side. And if I sub those in now, here's the expression I arrive at. And so, what I wanna do now is I wanna equate components to solve my answers, and so let's first equate the I axis components. So I've got moment about C around the x axis is equal to the i component, and the right hand side is I omega x dot. And so what that says is the more moment I have putting on the landing gear to pull it up into the body. The faster the rotation will be. And that makes physical sense if you look at my picture. The next equation we have, on the left hand side we have no J components, so 0. On the right hand side I've got J omega S dot. And so omega S dot has to equal 0. What does that tell me about omega S? And what you should say is omega S has to be a constant. That's my rate of spin, angular velocity or spin and so that answers the rate of change of omega S for this problem and finally I wanna find the moment reactions at C. And so we got one of them. We want to find the other one around the z axis. So I've got the moment through point C about the z axis is equal to J omega x omega s k. And this is a real interesting result. Because although I only have angular velocities around the x axis and around the y axis, I actually feel a moment about the K axis, or I develop a moment about the K axis, which is unexpected, not intuitive. And as I mentioned earlier in the course, this is called the gyroscopic moment, and so it's kind of a twisting moment. And that's what it's equal to. Let's look at a demo again, and so as I spin this thing and bring it up to the body what I do is, I feel this gyroscopic motion about the z axis. And I could feel it, but you can't feel it. And so what I'm gonna follow this up with is a demonstration that you can do on your own to experience this gyroscopic motion. So this is a demonstration that you can actually do at home if you have a bicycle. This is my bicycle. I actually commute to work at Georgia Tech, and so I've got my, this is my outfitted bicycle. I've got a horn on here. This is actually an 18 wheeler truck horn that's powered by a 12 volt battery so that I can make sure I get attention for the traffic. I've got a radio here, I've got lights, I've got my pannier bags, so I can ride in style. But what I want you to do with the bike, to feel this gyroscopic moment, is I'm gonna put it up on the end. And then I'm gonna ask Ray to come up and go ahead and spin that front wheel. And if you spin that front wheel and you turn the bike like this, you can actually slightly feel this gyroscopic or twisting moment. One thing I thought of just recently is that actually the heavier your wheel the more mass that you have further from the axis about which you're rotating the more gyroscopic moment you'll feel. Also, the faster you spin it the more gyroscopic motion you'll feel. And so if you don't feel it right away what you may wanna do is try to add more mass to the outside of your wheel. Maybe wrap some towels or something, wet towels around the outside, or something that you could add more mass out there. And go ahead, you can actually feel that gyroscopic moment. So that's a great demonstration for you to do on your own.
