MITOCW | 10. Equations of Motion, Torque, Angular Momentum of Rigid Bodies The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: OK. I've been giving out the money cards for a few of the lectures, and two or three questions came up in those that I haven't addressed so far. I'm calling them loose ends. And I'm going to pick up a couple of those today, I think they'll help you consolidate the knowledge around the quiz. So I'm going to tie up a little loose ends there. And then the lecture topic I started last time, which is making this transition from thinking about angular momentum of particles to using the full angular momentum equations for rigid bodies, where we talk about mass moments of inertia and products of inertia. And that's where we'll pick up, there again today. Because that's where we're going for the next few lectures. OK. Let's pick up with a first example here. This is on the topic of basically finding equations of motion. And there's been a little confusion with people, who have asked me what do you mean find? Do you mean solve, et cetera. And so I'm going to go through, and just a real quick example, skipping some of the steps because my purpose is emphasizing the steps, not working out all the details. So finding equations of motion. Where does it begin? One of the really important steps is this, determine the number of independent coordinates you need. Because when you've done that, that tells you, basically-- it really starts finding the number of degrees of freedom. Should have put this in a different order. Degrees of freedom tells you the number of independent coordinates you need. This is 1, 2, and then 3, that leads you to the number of equations of motion that you need. So this is really an important step. Secondly, draw a free body diagram. And third, apply summation of forces. External vector equations gives you mass times acceleration, and summation of torques gives you DHDT plus this V cross P term. So this is just kind of the step by step. So 1 let's apply it briefly. We've talked a lot about things on hills, so here's a cart. It's got wheels attached by a cord to a second mass that's sliding. m1, m2, doesn't stretch the cord in between. Let's think of these things as rigid bodies. So how many degrees of freedom? How many possible degrees of freedom? For the maximum possible, you have how many rigid bodies? 2. How many degrees of freedom for rigid bodies possible? 6 each. So we're at 6 times m plus 3 times n minus the number of constraints is the number of independent degrees of freedom. This is the number of rigid bodies, number of particles, so we have 6 times 2, 3 times 0 minus constraints, so this one comes out 12 minus the constraints. You have to figure out the constraints quickly. We're not going to allow rotation in any of the three directions on either. They're on carts, they're big, they're sliding, they're not rolling or any of that. So no rotation for 3, no rotation for 3 more. That's minus 6. So c equals minus 6, or c equals 6 for rotation. And then what else can we say? There's no-- I'll designate this the y direction so we can talk about directions here. And I'll designate this x in general. No acceleration at all in the y direction, right. Can't move in the y. So that gives us 1, 2 for each mass, plus 2 more, that's 8 constraints that we've come up with. Now a 9th constraint is the fact that these two are tied together. And so if you had just temporarily assigned a coordinate here, x1, and another one here, x2, we know for a fact that x1 has got to be equal to x2, and that gives you yet one more constraint. So that's 9. So we might just stop there. We say OK, that's a total of 9, so the number of degrees of freedom, 12 minus 9 is 3, and that implies that you need three equations of motion. Some confusion comes, you know, if something's not-- let me rephrase that. We haven't talked anything about the z direction. I haven't described any constraints in the z direction. If this is me in a car and I'm dragging a sled down a hill, or I'm in a vehicle and I'm dragging a sled down a hill. I don't know if you've ever been in a vehicle with a trailer on an icy road in the winter time, that's a dicey maneuver, going down a hill trying to put brakes on. So this thing 2 could conceivably move in this direction. And you can either constrain it to be that way to make the problem simple, or you can just say it's possible, so we have three equations of motion, of which two, for now, the summation of the forces, external in the-- since I've got x and y, z must be this way-- in the y direction. And we'll make it in z direction, this is coordinate system 1, so this would be z1 or z1. The summation of forces, z1, is m1 z1 double dot. But I'm going to set that equal to 0, I just know there's no forces. So this becomes a trivial equation of motion. And I add another one, summation of forces on the second mass. This is on m2 in the z2 direction, is m2 z2 double dot, and we set that equal to 0 also. So what it boils down to, I had 3 degrees of freedom, 2 trivial equations of motion, leaving me with just 1 equation of motion that's going to be meaningful. Yeah. AUDIENCE: If x1 is equal to x2, would that mean that the rod has to be entirely along the x-axis? So that would mean that-- PROFESSOR: So he asked if x1 equals x2, does that mean they both have to be along the x-axis? I'm assuming that. So I really am assuming this thing's going down the hill. I'm making a point about this z direction thing because it's just a subtlety that you have to decide on when you're figuring out how to actually analyze the situation. If you really were thinking about what happens when a vehicle is going down a steep, icy hill towing a trailer, maybe put the brakes on, maybe not. You could probably say well, unless I really have a disaster, we're not going to get rollovers and things like that. But you could imagine that it can get out of the x direction, right. It could start sliding into z. It's probably not going to go anywhere in the y, that's still a good assumption. So this is about modeling, and how complicated do you make the modeling. You actually have to make quite a few modeling decisions when you go to do this, and we tend, in class and examples, tend to really oversimplify problems so that we can do them. So I've boiled this down to where I'm going to end up with one significant equation of motion. And that one's going to say that the summation of the forces in 3 the x direction-- and I'm just going to write mx double dot here, not putting down m1s or m2s because we're going to have to go to free body diagrams to figure out how to apply this. Free body diagrams. First mass. So this is m1 here. m1g, a normal force, no doubt a friction force, I'll call that f1, and a tension in the cord. And I'm assuming that tension's going to always be there. So if, again, in a situation where the trailer starts overtaking the car and the rope goes slack, we're not going to consider that one today. So there's your free body diagram for the first one and we're going to need to break mg into a couple of components. We're going to need the slope here, and that translates into an angle here. So there's our first free body diagram. And our second free body diagram, second mass, tension, normal force, another m2g. And this one's on wheels, and we're going to consider this one frictionless. So we don't have any friction force holding it back. So the reason I've kind of-- this seemed like a ridiculous simple problem, but the point I want to make is not the solution, not the particular problem, but an issue that crops up. How many unknowns are there in this problem. AUDIENCE: 2. PROFESSOR: Well, you know n1 immediately, or n2, or the friction force, or the tension, or, for that matter, the x double dot we're looking for. There's actually-- you start off this problem with five unknowns.