Hi, this is Module 31 of Three Dimensional Dynamics. And today's learning outcome is to solve another problem for rigid body three dimensional rotational motion. And I'll let you know up front, this is a rather long module, but this is a really neat problem to solve. And I think it's worth the time. Also there's a lot of math involved, but you can stop the tape and go through the math step by step. I'm gonna go rather quickly, but the beauty of the module is again that you can stop it, make sure you understand it before you go on. And so, here's a summary of the equations of motion we came up with for different situations. We looked at a body fixed frame with a origin at the mass center, or the pivot point. Or we have an intermediate framed with the origin of the animate frame at the mass center, or a pivot point. Now we've done these two problems with intermediate frames in the last couple of modules. We haven't done a problem with a body fixed frame yet. And so we're gonna do a problem this module with a body fixed frame. And so we're gonna look at spin stabilization for a satellite. And so here's a picture of a typical satellite. I'm going to use a different satellite in here in the, for the module, I'm gonna use this book as my satellite. And so you gotta imagine yourself out in space. And so there wouldn't be any gravity, but this book is gonna spin out here in space and that's gonna be my satellite. And so we'll go ahead and analyze that. So here's my situation. I've got my inertial frame, and I've got my book here. And for this problem, I'm going to go ahead and weld my frame, my body-fixed frame into the body itself. And so this is the equations that we came up with, Euler's equations for a body fixed frame with the origin at the mass center. Okay, so here we are again. Now what happens out in space if, again, we have no moments, so it's torque free? What happens to my equations? And so what happens is the left hand side the moments go to zero and these are the equations I end up with for the equations of equilibrium for my satellite. Okay, we're also gonna assume steady motion. Constant spin stabilization around the z-axis. And so what that means is omega about the z-axis is gonna be a constant value, I'll call it a capital omega z. Then what I'm gonna do is, I want to investigate what happens to the system if it's perturbed from this steady motion. So what happens if a space rock or some piece hits the satellite and starts perterming it from its constant spin stabilization. And, so I'm gonna let omega x equal some perturbation, small perturbation, delta omega x, omega y, the y component of the angular velocity have some small perturbation omega Delta omega y. And then omega z has its constant spin stabilization plus some perturbation delta omega sub z. So here's my equations of motion for torque free motion in my satellite. Here's my angular velocity components where they've been perturbed. And I can sub now 4, 5 and 6 into 1, 2 and 3 and I come out with the following three equations. And I let you do that math on your own. Now we're gonna neglect the higher order terms because we have this constant rotation of my satellite spin stabilization. And I'm perturbing it just slightly. And those pertubations are gonna be very small. And so I take a very small value and I multiply it by itself, that's gonna be a much smaller order of magnitude than the other terms. And so in this problem here wherever I have two small terms multiplied together I'm gonna neglect those, and so I can neglect this term. And here's another two small pertubations together. I'm gonna neglect that term. And down here, this term is gonna fall out. And so that's what I'm left with. Okay. So if I take equation nine now and I've eliminated this very small term here, neglected this higher order term, I end up with delta omega z dot is equal to zero. So what does that tell me about the perturbation of the angular velocity about the z-axis? And what you should say is that means that it has to be constant, since its derivative is equal to zero. And so, remember I had a constant spin stabilization, which I called capital Omega sub z. Then I perturbed it and added Delta Omega sub z, but Delta Omega sub z is also constant. So let's just call that overall spin stabilization a combination of Omega sub z plus the perturbation. And I'm gonna remove the z subscript and just call it capital Omega for the spin stabilization. So again, here we are back at equation eight. Again, I'm gonna go through a lot of math here, but I'm gonna rephrase this equation so that I solve for delta, omega y dot. Again, I've put in capital omega for my spin stabilization, instead of omega sub z, because now it includes that constant perturbation around the z-axis. Let's take the time derivative now of equation seven, and I get, remember from my body fixed frame, my moments of inertia stay constant. So only my perturbations are going to change with respect to time so this is the equation that I come up with here. Again, I can solve for a mega Y dot for this equation. So I have two expressions for a mega Y dot,. So I'm now going to sub this up into here. And when I do that this is the result I get. So I have an equation now for the perturbation around the x-axis. So here's that equation again. Rather complicated, but you see that all of these terms here in front of delta omega x are constant. You have the constant spin stabilization, all of the mass moments of inertia are constant. So let's just call that term p squared, a constant value squared. And if I substitute that in now, then I get this equation, a lot easier to work with mathematically. Not so, doesn't even look so complex. So here it is and we see, if you remember from your math, now. This is a little bit more advanced math than I've done before but we have a second order ordinary differential equation. Where P squared is this value here. And this is, if you work with second order ordinary differential equations, maybe a more familiar form would for you be something like x double dot + ax=0. Where x double dot is the same as Delta, omega, x double dot. P squared is the same as a and x is the same as delta, omega, x. And if you've solved a second order differential equation like this, ordinary differential equation like this, you can solve this one. And so I'd like you to go ahead and follow along here. The way we do that is we assume a solution to solve. One way we do it. There's several ways to [LAUGH] solve different equations. The easiest way for this problem is just to assume a solution. And since we're trying to solve for delta, omega x, and the equation involves delta mega x and its second derivative. We wanna assume a form of the function that comes back the same no matter how many derivatives you take. And so a nice assume solution that does that is an exponential type solution. So I've got delta omega sub x = a e to the lambda t. And I substitute now, if this assumed solution is going to work. I have to substitute in my equations and equate them all to zero. I'll just take the second derivative here for this term. When I do that, I get lambda squared comes out when I take two derivatives times a e to the lambda t, and then p squared times a e to the lambda t. And I can solve for what I call my roots now, or the lambdas. And if you solve for the lambdas, you find that lambda could be. When you set lambda squared plus P squared equal to 0. For P squared greater than 0, I get plus or minus I sub P, where i is an imaginary number. Or plus or minus P for P squared. [LAUGH] Excuse me, for P squared less than 0. Okay. So, here's my ordinary differential equation. Here's what P is equal to. I assume this solution. I found out that these with the roots I got depending on whether P squared was greater than 0 or P squared was less than 0. And so let's look at the solutions now. For P squared greater than zero, my lambdas are plus or minus i sub P, where i is an imaginary number, and so I put those lambdas in. And I get delta omega x =, I get two terms here because I have two roots. And I'll call the first term, a1 times e to the ipt, and then + a2 times e to the ipt. And you can show that a complex exponentials are the same as harmonic motion, sines and cosines. And I'll actually do this math to show this equality here in a future course. And I end up with b1 cosine Pt + b2 sine Pt. And if you've worked with differential equations before, I'm sure in your math class you also showed how complex exponentials are harmonic motion. Harmonic bounded motion, that's just an oscillation back and forth. And it's gonna be stable. By the way a1 and a2 are not the same as b1 and b2. The other solution I get is with P squared less than zero where I get plus or minus P where P is just a real value. If I substitute that in, I get two terms again, a1 times e to the Pt, and a2 times e to the minus Pt. And so my question to you is what happens to this term as time increases? And what you should say is that goes to zero because that's a decaying exponential. And what happens to this term as time goes to zero? And so, that's going to be an increasing exponential. And so an increasing exponential means delta omega sub z is going to go off to infinity and therefore it's gonna be unstable. So this goes to infinity as time goes to infinity, so it's unstable. So depending upon what these values end up with, whether P squared is greater than 0, or less than 0. For my particular situation I can end up with a stable solution, or an unstable solution for my motion or my perturbations that take place. Okay. So, let's do a stability table here and look at what happens. And so here's my satellite. First of all, for the principle moments of inertia, which one is going to be the largest given the orientation of my coordinate axis? Which one is going to be the next largest? And then, excuse me, which one is gonna be the largest? Which is gonna be the next largest? And then which one is gonna be the smallest? Around the x, around the y or around the zx? Put those in order. And for this orientation, remember mass molar inertia is the amount of the mass how far from the axis about which you're rotating? So in this case around the X-axis I have the least amount of mass, around the Y-axis I have more mass, and the Z-axis I have even greater amount of mass. And so I sub zz is going to be greater than I sub xx. And I sub zz is gonna be greater than I sub yy in this orientation. So if I sub zz is greater than I sub yy, that means that this term is negative. This term is positive. But I have a negative out here, and so that means my value for p squared is gonna be greater than 0. Well the system should be stable. And so if I spin around the z-axis, my satellite should be stable. Let's look, okay. So my z-axis would be coming in this direction and if I spin this, you can see that that's a stable motion. So my satellite would continue like that forever. Okay, let's look at another orientation now. Let's say that Izz is less the Ixx and less then Iyy. And so what that says is, this will be my z-axis now. This will be my x-axis, and this will be my y-axis. And so now I sub zz is the smallest mass moment of inertia. And so we're spinning around the smallest mass moment inertia. I zz is less than I yy, so this is a positive value. I cz is less than I xx, and so this is a negative value. A negative times a negative gives me p squared greater than 0. So again, that should be stable motion. And so let's look. Now I'm spinning about this axis, the axis where the mass moment inertia is the smallest. And you can see again that that's stable motion. Okay. Now let's look at the other case, which is where I sub zz is the middle mass moment of inertia. And so in this case, here's my z-axis. This is my x-axis, and this is my y-axis. So, we've already spun it around this axis. We saw it was stable. We’ve spun around the smallest spatial inertia axis. We saw it was stable. Now we’re gonna spin it around the middle axis. And in this case, since I cz is in the middle, I've got I sub yy minus I sub zz. Well I sub yy is bigger than I sub zz. And so this is going to be a positive number, and I sub zz is less than. Let's see, let's do this again. I sub yy is greater than I sub zz. So this is a positive number. And then I sub xx is less than I sub zz. So this is a positive number, but there's a negative number out here so p squared is going to be less than 0. And so this is an unstable situation. And that occurs also if I sub y is greater than I sub zz, or excuse me, less than I sub zz and that's also less than I sub xx. Because in this case, I've got now I sub z yy is less than I sub zz, so that's a negative value. I sub zz is greater than I sub xx. So that's a negative value. So I get negative, negative, negative. Overall, P squared is gonna be less than zero or unstable. So for both of these situations, where Z is the intermediate mass moment of inertia. The value in between. I should get unstable motion. And so, let's go ahead and look at my demonstration one more time. So I rotated about the z-axis, it was stable. I rotated about the z-axis, when the mass moment of inertia was the smallest and it was stable. And now I'm gonna rotate it around the intermediate axis. And you're gonna see now that my book is gonna wobble. It's gonna go unstable, and you can see it, hopefully through the camera here. And I'm not doing [LAUGH] I'm not doing this, this is not a trick, I'm not doing it myself. But you can see how it doesn't stay constant spin stabilization. And you should try that on your own Just wrap a rubber band around a book like this, and you can see that unstable motion. All right, let's look at one more example. Let's look at a football, and here's a picture of a football spinning. Our video of a football spinning. So here's my situation. I've got my body fixed coordinate in the football. And in this case I sub cz is going to be my smallest moment of inertia. I sub cz the mass moment of inertia but the z axis is gonna be smallest. You can see because of symmetry I sub x and I sub y are going to be equal. And so if I put that in my equation, I get I sub c is smaller than I sub c yy, so that's a positive value, i sub zz is less than i sub xx, so that's a negative value. A negative times a negative is positive. So p squared is greater than zero. So if I spin around the z-axis, I should get stable motion. And that's why a football spins, that's why we throw it and it's stable around that z-axis. The other thing we saw in the video clip was that the person that threw the football, and it wasn't me, I would of thrown a perfect spiral. But the person that threw this spiral, it is wobbling a little bit but it's like a harmonic motion. Remember, that's a bounded harmonic motion, but it's stable. And so you can actually see that dynamics. Okay, and so one last example here that I wanted you to do on your own is a race car. Here's a picture of a couple race cars on a curved banked turn. And here's a sketch of that. We have the race car has a mass m. M is travelling a constant speed v sub c around a curved bank. The curve of radius for that curve bank is capital R and the bank angle is theta. And so what we're asked to find here is the normal force acting at the upper and lower set of wheels. And so you can do that on your own then come on back and, excuse me, look at the module handouts. And you can see the solution and we'll continue next time.
