Welcome to Module 30 of Three Dimensional Dynamics. Today, we're gonna solve for another motion of a three dimensional body using the equations that we developed earlier. As a summary of those equations again, we had Euler's equations for a body fixed frame, which we haven't done yet, but I"m gonna do in a later module, one of those examples. We could have the body fixed frame with its origin at the mass center, or about a pivot point. Last time, I looked at a problem with an intermediate frame attached at the mass center, and this time I'm going to look at the intermediate frame attached to a pivot point. So let's go ahead and look at an example here. Here is my example. I've got my wheel. I'm going to spin the wheel and I'm going to have it rotate about this pivot point where I'm holding my hand, and we're going to call that point O. Again, here's my pivot point. And so I'm not much of an artist. [LAUGH] But here is a diagram of the situation. I've got my wheel spinning with omega s, and I've got that procession as you saw. I'm going to use my intermediate frame here at point O, the pivot point. And this is a picture of me with my little smiley face here in the inertial frame. Okay. For this problem, we're going to go ahead and we're going to say the radius of the wheel is again big R, but we're going to neglect the width. And we're going to let omega x, this procession angular velocity and omega s, the spin angular velocity be a constant. And so I'm going to use this intermediate frame. Again not body fixed to the wheel spinning but out here at point O, the pivot point. And again I can do that. The key here is that the inertia properties, the mass moment to inertia and products of inertia, if there were any. In this case there's not, but they're gonna be the same for this F frame or a frame that would be attached to the body itself because of the symmetry of the wheel. And what we are gonna wanna do, what we're solving for is, to find the rate of procession omega x in terms of the other properties or parameters in the problem. L and R and things of that nature, and the mass. Okay. So, here's my equation of motion. I need to start off again by finding the angular momentum, this time about point O. That means I'm going to have to have the angular velocity of the wheel with respect to the I frame, and so do that on your own, come on back, and see how you did. And so here's omega B with respect to I. Omega s around the j frame again and omega x around the i frame. The I matrix now is written here. I have to use the parallel axis theorem because I've neglected the thickness, but IXX around maximum inertia, around the x axis is equal mR squared over 4 through point c, there at the mass center. But then I have to bring it back to my origin of my frame O, and so, or my frame F which has it's origin at O. So I got to take the mass times the distance squared using the parallel axis theorem or plus mL squared. The mass moment inertia about the y-axis is the same for the B frame [COUGH] or for the frame out here at the mass center or back here at O so I don't need to use the parallel axis theorem. But again, I need to use it for taking the mass moment inertia about the z-axis back to point O. And for my omega matrix, my angular velocity of the wheel, again, it only has i and j components. I can multiply that out. And this is the result I get. And so, here is my expression. I have my angular momentum and I'm ready to finish up the problem here. And so, I've got, first of all I need to find the derivative of the angular momentum in the F frame. And so with respect to the F frame, again, these mass moments in inertia are constant, and I also said that omega x and omega s is gonna be constant. So what that says, is that this term is actually gonna be equal to 0. Next, I want you to write the angular velocity of the intermediate frame F, with respect to the inertial frame. And that's just going to be omega x in the I direction, and so I can now do the math, and I get the sum of the moments about point O. In fact, let's just do this rather quickly here. For some of the moments about point O, we only have our weight here. So what's the some of the moments about point O going to be equal to? And what you should say is, you should be real good at this by now, given my earlier classes and the earlier work you've done, you've got MG the force times the distance L, and it's gonna be around the K axis. So it's just gonna be mgL around the k-axis equals we've got omega xi for omega F with respect to I crossed with now my angular momentums that's mR squared over 4 plus mL squared times omega xi plus mR squared over 2 omega sj. And if I multiply that out, I get mgL times k is equal to i cross i is zero, i cross j is k, so I get omega x omega s, mR squared over 2. And that's also in the k direction. And so what we see is here the ms cancel and so I get my procession. Omega x is equal to, I got 2, bring it over to the other side, gL over omega s R squared. And so that is the rate of procession, with respect to my other parameters. And so, let's see if that makes physical sense. Let's, first of all, say what happens if L increases. So if L increases what happens to the precession angular velocity. And so what you should say is it increases. And let's look at the demonstration. As I spin this wheel, for a very short L, it spins rather slowly but for a larger L, it spins faster. So, if I'm real close here with my pivot point, it spins very slowly but as I go out with a further distance L, the procession angular velocity speeds up. So in this case, we've got L increasing omega x increases. Now, what happens if we increase R? Well in that case, and I can't increase R, my wheel itself, so I can't do that demo. But since R is in the denominator, that would say that omega x would decrease. And then finally, what happens if I increase the spin angular velocity? And again, if I increase the spin angular velocity, it's in the denominator so that's going to decrease my procession angular velocity. And so, here again, demonstration. So if I increase my spin, as that spin goes up, omega x goes down and vice versa. As the spin becomes less, the procession becomes faster. And so as this thing slows down, you can see that it starts to process around a whole lot more quickly. And you can actually do that on your own, using a top. Okay, spin a top. And as it processes, you know, as it gets to the, the spin gets less and less and less, all of a sudden it starts processing faster and faster and faster, until it falls down. And so you can do that as an experiment on your own. One other qualitative thing, we know if you wanted to just figure out which way is gonna process, is it gonna process to the right, or is gonna process to the left, or clockwise, or counter clockwise, you can do it very simply qualitatively. I have my point O out here, I have my weight down. And so my moments, remember, we have the summation of the moments is equal to the time derivative of the angular momentum. And so my moment by the right hand rule forces down, moment arms this direction so I've got R cross F, so my moment is going to be in this direction. Back toward me. The H factor starts out in this direction and since the sum of the moments is equal to the time rated change of the angular momentum, the angular momentum vector is gonna have to go in this direction. And that's why it'll start out in this direction. After a period of time, it's gonna go this, this, and that's why it processes in the direction it does. And you can do that for any spinning object sort of intuitively or qualitatively. And so that's a good demonstration, another neat example, and we'll come back and do another one next time.