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MITOCW | 7. Degrees of Freedom, Free Body Diagrams, & Fictitious Forces
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: Last, kind of under the announcements category, is I want to talk about the muddy cards. So I've used those many times in the past. Last time was the first time I handed them out. And your comments were great. This was really a good lecture to have handed them out. We covered lots of interesting important concepts. And so I'm going to review a couple of things that came up in the muddy cards. A couple of the most positive comments is people really like the demos, and they really like the explanations, especially with examples. People particularly commented it was really helpful to compute angular momentum from two different points. And you get the revelation that you get two very different answers. So that's a really important point. And somebody then asked the question, said, well, I thought that vectors were independent of the coordinate system that you select. It's true. A velocity ought to be a velocity no matter whether it's r theta or x, y, z. But why? That seems to kind of violate that notion that vectors should be independent of coordinates. And yet we computed an angular momentum with respect to one place and respect to a different place, and we got different answers. How do you resolve that? Yeah. AUDIENCE: It's sort of like since it's moving, the coordinates shouldn't matter. Like if it was equilibrium, it wouldn't matter where you would put the [INAUDIBLE]. PROFESSOR: OK, well, you're getting close. Here was the problem I think we did. And we chose this point, which I'll call 2, and this point, which I'll call 1. And we computed h1. I'll
just call h1 with respect to point 1. We'll call this A. We computed with respect to 1. And we computed the angular momentum of A with respect to 2. But in this case, angular momentum of a particle with respect to some location, origin of a coordinate system, is defined as r of the partial with respect to the coordinate system crossed with the linear momentum of the particle that you're-- we'll call it B here, just the name of the particle. The definition-- these are both vectors. You arch changing the vector. It's a different vector. Because this changes in the two parts. So it's not a constant vector at all. By its definition, it is just something different when you move to a different place and this piece changes. This piece is invariant, but this piece is not. And that's the answer to that one. Lots of people were still not clear about Coriolis. We'll work on that as time goes by. And people were interested in how to pick reference frames and so forth. Somebody made the suggestion, try using some colored chalk. It would help. I don't own any colored chalk. My assistant just walked in with some. She found some at the last minute. So I'll try doing that. That's a good idea. And someone else says, take a break in an hour. And that's a pretty good idea, too. I'll try to remember to do that. So the muddy cards are great. Please today we'll do the same thing. So let's start with this topic. It's a subject which we constantly use throughout the course doing dynamics. You have to be able to figure out coordinate systems, degrees of freedom, drawing free body diagrams. So I'm going to do a few quick examples-- coordinates, fbd's. I picked some examples here just to emphasize a few different points. Here we have a slope. I've got a wheel. It's a rigid body. I pick a preliminary coordinate system. Sometimes you do that just to help you think about it. And now let's talk about degrees of freedom. What do we mean by degrees of freedom? I'm going to define it as the number of independent coordinates necessary to
to the mass times the acceleration of the body in the z acceleration. It's a vector, vector component. And then you just say, oh well, that's a trivial equation of motion. And so you could deal with it that way. But we'll just assume that we have no motion in the z. And that gives us another constraint. Now we can assume again that there's constraints in the problem, or it's well behaved, in that the thing won't fall over. It won't roll over. And it won't change direction running down the hill. So we'll assume no rotation about the x or y axes. So that implies two more constraints. And finally, the constraints can come in many flavors. Finally we know that in this problem that I need to think in terms of a rotation. So there's a positive rotation in that coordinate system. So now I have a rotational coordinate that I can think about. But this now says that if there's no slip, I can say that the distance it rolls down the hill is minus r theta. That's a constraint. x and theta are not independent of one another. We're looking for the number of independent coordinates required to completely describe the motion. x and r are not independent because of this no slip condition. And so that implies yet another. OK, so we've got one, two, three, four, five constraints. And we said that the number of degrees of freedom in this problem is equal to 6 minus 5, which is 1. So you take the single coordinate. You could have told me that long ago that that's what it's going to take. But this is the sort of thinking you have to go through to come up with all these constraints. So this is going to take a single coordinate. It could be x. It could be theta. But you don't actually need them both. You use them both for a while, because it's convenient. But in the final analysis, you'll be able to write an equation of motion just in terms of x or just in terms of theta.
OK, free body diagram of our wheel-- we said no slip. So here's your slope. You know you've got mg. You know there's going to be a normal force from the slope. There may also be some tangential force that makes it impossible for it not to slip, some f. So there's the free body diagram, in this case. What if-- I'll do e here, or do a case ii here, slip allowed. Then how many constraints do we have? Just four. Because now you can no longer say that this is true. They're independent of one another. They take on values not controlled by this formulation. So 6 minus 4 gives you 2. And you're going to end up having to have both x and theta probably as your chosen coordinates to do the problem. And you'll end up with two equations of motion. So I've got a hockey puck here. I've drawn kind of a 3D perspective of this. So it has a coordinate system out here. The z-axis is going like that. So here's my z. Here's my x. Here's my y in the plane of the ice that this thing is sliding on. So this is my z- coordinate. And I have string wrapped around it. And I've got a piece of string coming off like that. And I'm pulling on it with some tension. Because otherwise the only constraints are it's sitting on this icy surface, which I'm going to assume is-- well, I don't even have to assume it's frictionless. I could. So let's figure out how many equations of motion we're going to need here. So the number of degrees of freedom, 6 times-- this is a rigid body. It's not a particle. And there's only one of them. So it's 6 times 1 minus C and the number of constraints. So can it move into the table, into the surface? No, so that's a constraint in z. Are there any constraints in the x or y? It can rotate about z. But it can't rotate about the x or y-axes. All right, so constraint into z is one. Can't rotate about x or y-- two, three. Are there any others? Who thinks I may have missed one? OK, I'd say we've got 6 minus 3. We're going to need three equations of motion to be able to actually describe the motion of this thing. And we'd probably use-- this is
AUDIENCE: [INAUDIBLE].
PROFESSOR: Pardon? I hear a no. I hear some yeses. What about at B? Is it constrained at B? In the-- AUDIENCE: y. PROFESSOR: y direction, OK. I didn't bring my big foam disk. But earlier in the term, I said that the definition of translation is that all points on a rigid body do what if you're rectilinear or curvilinear translation as opposed to rotation? AUDIENCE: Move in parallel. PROFESSOR: All points move in parallel, exactly right. That means that if we use real strict definitions of translation and rotation, that if I constrain the motion of any point on that object, that object is now not allowed to translate in that direction by the definition of translation. So this point constrains it in the x direction. This constrains it in the y direction. And we're going to-- so it's constrained in translation. And that implies 2. And b, we'll assume that z motion is 0. We'll just assume there's nothing going on out of the plane. So that gives me another one. And I'll assume no rotation. I'm not going to allow any rotation in this problem. I'm not interested in rotation about the x or about the y, about these axes. And that implies two more. So we have two, three, four, five. And I better not have any more than that or the thing can't move. So in this case, this is 6 minus 5 equals
I need a single coordinate to describe the motion. And if you look at it, you say, well, that's kind of intuitive and obvious. If I specify the x position here, I could figure everything out. If I know the length and the x, I could figure out where it is. If I know the y position and the length, I could figure out where it is. If I know theta, I could
figure it out. I only need one. But that strict definition of translation is really helpful here. This thing, this object, is in pure rotation. And if it's in pure rotation, it must rotate about some point. Where? You know how to find that? What's the velocity here? It's got gravity acting on it, so it's probably down. But it's parallel to the wall. It has to be. What's the velocity here? It's got to be parallel to the wall. If I draw perpendicular to that-- if I'm saying, this is rotation. All points in the body rotate at the same rate. But their speed is determined by the distance away from the center of rotation. But if it's pure rotation, there must be a center of rotation somewhere. And it must be perpendicular to any velocity vector. So you draw the perpendicular. You draw the perpendicular. And here is the instantaneous center of rotation, the ICR. There's a little short section in the book on that. When this thing drops down to here, same kind of arguments hold. But the center of rotation has changed locations. Yeah. AUDIENCE: So is this not translating x and y? PROFESSOR: So now let's talk about the center of mass. So she asked-- excuse me, I should repeat the question. You guys aren't holding me to that very well. Raise your hands if I don't repeat an important question. She says, is it not translating? We've determined that-- let me ask you, does the center of mass move? Does Newton's second law apply to the motion of the center of mass? AUDIENCE: [INAUDIBLE]? PROFESSOR: Yeah, it's got to. So the center of mass translates-- no doubt about that. Newton's second law applies to it. So we're not saying that there isn't motion of the system in the x and y. We're just saying that that motion is caused by rotation. It's not caused
is now a rigid body with respect to the ICR, dt. OK, in this problem, what are the torques? Where do the torques come from? Is there any torque caused by Nx? No, because there's no moment on it, right? It's pointing right at the center. Same thing-- no torque caused by this. You want to find equations that get rid of unknowns. So neither unknown appear in this equation. Where does the torque come from? Gravity, right? And it's going to be some Mg times a moment arm. And the moment arm is going to be like that. So it's an L over 2 sine theta. And the sine you'll have to figure out from an r cross and f. So you have an i cross j, gives you a k. But it's in the minus direction. So I think it'll come out minus. But I could be wrong. I did that on the fly. So we're not going to go further with this. But the instantaneous centers of rotation could be really handy. OK, a final example in this stuff-- how are we doing on time? Couple of carts, so the floor constrains the motion. And I've got a spring and a dashpot and an M1 and an M2. And I want to figure out how many-- yeah? AUDIENCE: I was just curious. Why is torque negative in your earlier solution? PROFESSOR: Let's just figure out r cross f. So my r-- you're saying, why is the torque negative? The r is L/2 in the i hat. And the gravity crossed with the force, which is Mg in the-- but it's minus Mg in the j hat. So i cross j is positive k. But the minus sign comes from there. AUDIENCE: What is that over to the left [INAUDIBLE]? PROFESSOR: Well, I'm working on the center of mass here. That's my equation. And this thing is L long. So half of the length must be L/2. And I'm interested in this side of the right triangle. AUDIENCE: Why are you interested in that side rather than the side that connects it to the-- PROFESSOR: Because this side crossed with that gives me 0. There's no moment. So this is a moment equation. I'm trying to compute moments. Yeah?
AUDIENCE: Should the moment come from the ICR load? PROFESSOR: Oh, you're right. I don't know what I'm thinking. I could have messed this up. It's got to be about the ICR. So the force is down. Ooh, it's got to be this one. I messed up-- good catch. So that's a cosine, still L/2. You've got a theta. You have a-- this is also theta. And we're looking now for this side. Eh, it's still sine theta, right? Does the sign still work out the right way? AUDIENCE: [INAUDIBLE]. PROFESSOR: Good, OK, I've got to keep rolling. I've got something else really fun I want to talk about. So let's do this example quickly. This is mostly to get you to think about free body diagrams. This-- two rigid bodies. The degrees of freedom quickly here-- 6 times 2. There's no particles-- minus c, so 12 minus c. Now how many? What does your intuition tell you? How many independent coordinates is it going to take to solve this problem? Hold up your fingers. I see two, one, two, one. OK, one or two. Well, I think you can find 10 constraints in this problem. If you assume it doesn't roll over and you assume it doesn't move, you can find 10. You're going to need two coordinates. Because just because you've got a spring and a dashpot, they don't fix. They don't say there's any particular relation between the motion of this and the motion of that. You're going to need an independent coordinate to describe the-- whoops, 2 times 6. This is 12 minus 10. You don't need an independent coordinate to describe the motion of each of these masses. And I'd probably choose a coordinate that, let's say, goes from the center of mass of this one. I'll call it x1, center of mass of this one call it x2. And because I've been doing problems, vibration problems and stuff like this, for a long time, I'll tell you it's smart to start your coordinates at 0 when you have zero spring forces. Or from the static equilibrium position-- that's the good place to start. So then your answer, if you're at the static equilibrium position, then any non-zero
AUDIENCE: Push back. PROFESSOR: Push back kx. And I let the sign be indicated by the direction of the arrow. And I'll use that when I write out my equation of motion. OK, now I'm going to assume a positive bx1, so its velocity in that direction. What does the dashpot do? AUDIENCE: [INAUDIBLE]. PROFESSOR: Resist or [INAUDIBLE]? AUDIENCE: Resist. PROFESSOR: Pushes back, right? OK, so you have another force here, bx1 dot. Now, this is if you have two bodies. You have to have two free body diagrams, one for each. But I need to know what's the effect of x2 and x2 dot on this body. Well, let's just go do the same thing. Let's let x2 be positive. Now it's the only motion. I have a positive movement of x2 that stretches the spring. What is that spring force applied to this? What direction is it in? This body is moving in that way. AUDIENCE: Positive x direction. PROFESSOR: Which way is the spring going to pull on this thing? It's going to pull on it, right? So k, by amount kx2, and of positive velocity, to the right. It makes the dashpot open up. What direction is the force that dashpot puts on this? AUDIENCE: Positive x. PROFESSOR: bx2 dot. And now in this problem, except for gravity, I've got a normal force f and an Mg down. But all the action, all the motion, is in the horizontal direction. I can write out an equation of motion for the first rigid body here. And that's a sum of the forces in the x direction on body one-- I'll give this a 1 here-- in the x is equal to M1 x double dot. And now I can just write it out. It is k x2, because that one's positive, minus x1 plus b x2 dot minus x1 dot. And that
has the signs right. And the whole key is just one at a time assume positive motions and deduce what happens, and then use the arrows, the direction of the arrows, to set the signs. And now there's your equation of motion. Now we can do the same thing, sum of the forces on 2 in the x 2 direction, M2 x double dot. And now we would do exactly the same thing. So positive motion of x1, what does it do over here? It gives me a force through the spring in which direction, positive or negative? AUDIENCE: Positive. PROFESSOR: Right, so now you just end up with a kx1, bx1 dot. And you'll find that kx2 bx2 dot. And you sum it up. And you'll end up-- this should switch around, k x1 minus x2 plus b x1 dot minus x2 dot. I've got two equations of motion. And they're mixed. So each one has both coordinates in it. So this problem has two questions of motion, and they're coupled. They're not independent. You have to solve them together. OK, now we're going to move on to a subject which has come up in conversation. People have asked about this lots of time. And they say, what about the centrifugal force? And you sometimes use the term "fictitious force." How many of you use or heard the word used "fictitious force"? And how many of you heard us say that centripetal acceleration is not a force, it's an acceleration? And yet we love to talk about this concept of centrifugal force, which doesn't exist. But it trips us all up. Because it's handy to think about it. And so we're going to talk about fictitious forces now. They're handy. But they are dangerous. You really have to understand your fundamentals if you're going to use the concept of fictitious forces without getting yourself in trouble. OK, what is a fictitious force? Well, Newton's law, let's start there, Newton's second. Sum of the forces external on the body equals a mass times acceleration. It's a vector equation. So we can break
summation of the forces in the y direction-- and in this problem, we would say it's the mass times the acceleration of point A with respect to O. And the sum of those forces-- you've got an N minus mg. And that's a pretty simple equation. Now I'm going to specify. It's given that acceleration of A with respect to O is 1/4 of g. So that's what the cables in the elevator are doing. It's making this thing move, and it's going up at 1/4 of g, the N acceleration. So it's getting faster and faster. OK, so if I solve this for N, I will get-- let me stop there for a second. It's normally what I would just do. But since we're talking about fictitious forces, I need to go through that step for a second. So now I say that, well, the summation of the forces in the y direction here minus M acceleration of A with respect to O, that total is equal to 0. And that then is N minus Mg minus M times g over 4. And now I have taken this upward acceleration. And I've treated it like a force. I've just moved it to this other side, set the whole thing equal to 0. This is the fictitious force. I'm going to say, OK, that's all the forces in the system, solve for N. And of course you get N is M times g plus g/4. And that's 5/4 mg. And so you read 25% heavier. It's a really trivial example. But the notion is that you think of this as a force that's been applied. It's the mass times acceleration with a minus sign in front of it. It'll always turn out like that. It's minus the mass times the acceleration. Now, the acceleration can come from Coriolis acceleration. It can come from centripetal acceleration. So if we do this in a rotating thing, this fictitious force might be the centrifugal force, which is minus M times the centripetal acceleration. And we'll do an example like that. All right, stop-- there we go. OK, trivial example-- let's see if we can find something a little harder. How are we doing on time? We're doing pretty good. So let's do an example where the notion is really quite powerful. Now, I showed you this last time. And we talked a lot about--
we did the time derivative is the angular momentum. And we computed the torques that this thing exerts around different axes with respect to the point of attachment to this. Well, this is an example which, if you're comfortable with fictitious forces, you can figure out those torques really rather quickly. So I'm going to-- these two problems are identical. This problem, or with a shaft running like that, are really exactly identical. But I'm going to pretend that my thing is made this way. Because it makes it easier to see where these torques are coming from. So here's my z. And I've got my rotation about the z-axis. Here's this mass, my coordinates. Here's my r hat. Here's my z k hat, is this distance here. And I'm going to set conditions in the problem, my rotation rate. It's also theta dot about the k z- axis. And omega dot, theta double dot-- also in the k. It's not restricted. So I'm allowing this thing to accelerate. But r dot equals r double dot is 0. So it's not changing a position. It's just fixed. And I need to figure out the forces on this thing and talk about how we might consider some of them as fictitious forces. So let's think about in the r direction-- summation of the forces here in this r hat direction. So it's one vector component. I don't have to carry along all of the other baggage. It's equal to-- and I'll call this B. This is going to be the point about which I care about things. There's a fixed coordinate system here, Oxyz. But then a rotating coordinate system-- we'll call this A. And it's going to have my coordinates. I'll use polar coordinates, r theta z. But this one rotates. But it also has its origin right there coincident with O. So the sum in the r direction, then, we have the mass times the acceleration of B with respect to O. And the acceleration of B with respect to O is the acceleration of-- we'll just write out the whole formula using cylindrical coordinates. r double dot minus r theta dot squared, this is in the r hat, plus r theta double dot plus 2r dot theta dot in the theta hat direction. That's my full expression for--
So looking-- a side view. Here's your bead. I'm going to draw it as an unknown here. There is an unknown force in the r direction that comes from this bar holding it. It's applying-- there's got to be a force that makes this go in a circle. And that bar is what provides it. It's the only thing [INAUDIBLE]. I'll just call this unknown and r. And I'm just drawing it in the positive direction. The way you can do this, if you're not sure the direction, draw it positive. And the sign that falls out tells you what the right answer is. There'll be some force in the z provided by the rod. There'll be Mg. And that should be it. If we did a top view, then there'd be an unknown in the theta direction. You'd also see the N in the r direction. And anything else? No, the Mg is down where you can't see it. So these are your two free body diagrams. So now, in this direction, this is the acceleration. And the external forces are just that. So the sum of the forces is Nr. Now I want to treat it, bring in this concept of a fictitious force. I'm going to move the acceleration term to the other side. Nr-- unknown positive. Now I'm going to move that acceleration term over here minus-- ah, but now it's minus the acceleration. So this term looks like that, has a minus sign. If you move it over here, that actually becomes plus, equals 0. Now this is your fictitious force. Sometimes people call them inertial forces. And it acts like it's pulling out on the object. So the force that the mass appears to apply to the rod is this centrifugal force pulling out. And of course now we can solve for Nr. And we find it's minus Mr theta dot squared, which we knew all along. It's the force applied to the mass by the arm as it spins around. It has to pull in on it to make it go in that circle. Now, we could do the same thing in the theta direction. The theta direction will have an Mr theta double dot. And when we look at the theta free body diagram, it's plus N theta. You could solve that, and you immediately come up with a solution for the force in the theta direction. I'm just going to write that one down. When you solve for this one, you get a minus sign. This one ends up plus, Mr theta double dot.
I'm not going to go through the gyrations of getting to this. You can do that one. Now, I want to do one quick thing with this. Once you develop confidence in knowing when you can use a fictitious force and not get in trouble, this is the sort of thing you might do. Here's my system. It's rotating. And I want a quick estimate of, what's the torque? What's the bending moment about this point caused by the fact that it is the centripetal acceleration? Well, centripetal acceleration is equivalent to having this fictitious force outward on this of an amount Mr theta dot squared. And this is the moment arm z. what's the torque that that causes about this point? This is just levers now, forces and lever arms. AUDIENCE: 0. PROFESSOR: No, not 0. About this point here-- this is O. And I've got that centrifugal force pulling out on this fictitious force. Yeah? AUDIENCE: zMR of theta dot squared? PROFESSOR: Yeah, in the-- this direction, which is theta, right? So that's the moment the torque about this point caused by that-- torque at O is minus Mr theta dot squared times z, the moment arm. And it's not minus. It's in that direction. So it's plus. It's in the theta hat direction, the torque, that that force is applying about this point. We could have solved this problem the way we did in the last lecture, very carefully going through the dh dt's and following it all out. And we would have gotten that answer for the torque except for a minus sign. Now, why the difference? This is the torque that the centripetal force causes down here. When you do dh dt, you get the torque required to make what's happening happen. It's just the opposite. This is the torque. This is putting about that. There must be an equal and opposite torque that this system puts on this arm out here to keep it from flopping out.
