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About this Lesson
- Type: Video Tutorial
- Length: 11:41
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- Posted: 07/01/2009
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This lesson was selected from a broader, comprehensive course, Physics I. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/physics. The full course covers kinematics, dynamics, energy, momentum, the physics of extended objects, gravity, fluids, relativity, oscillatory motion, waves, and more. The course features two renowned professors: Steven Pollock, an associate professor of Physics at he University of Colorado at Boulder and Ephraim Fischbach, a professor of physics at Purdue University.
Steven Pollock earned a Bachelor of Science in physics from the Massachusetts Institute of Technology and a Ph.D. from Stanford University. Prof. Pollock wears two research hats: he studies theoretical nuclear physics, and does physics education research. Currently, his research activities focus on questions of replication and sustainability of reformed teaching techniques in (very) large introductory courses. He received an Alfred P. Sloan Research Fellowship in 1994 and a Boulder Faculty Assembly (CU campus-wide) Teaching Excellence Award in 1998. He is the author of two Teaching Company video courses: “Particle Physics for Non-Physicists: a Tour of the Microcosmos” and “The Great Ideas of Classical Physics”. Prof. Pollock regularly gives public presentations in which he brings physics alive at conferences, seminars, colloquia, and for community audiences.
Ephraim Fischbach earned a B.A. in physics from Columbia University and a Ph.D. from the University of Pennsylvania. In Thinkwell Physics I, he delivers the "Physics in Action" video lectures and demonstrates numerous laboratory techniques and real-world applications. As part of his mission to encourage an interest in physics wherever he goes, Prof. Fischbach coordinates Physics on the Road, an Outreach/Funfest program. He is the author or coauthor of more than 180 publications including a recent book, “The Search for Non-Newtonian Gravity”, and was made a Fellow of the American Physical Society in 2001. He also serves as a referee for a number of journals including “Physical Review” and “Physical Review Letters”.
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We've talked about the center of mass, of real-life objects, and we've seen a formula that allows us to calculate the center of mass of some system that's built-up out of a bunch of chunks of mass, like a barbell that's got a mass here and a mass here. But what do you do when you've got a real-life object, something with a complicated shape, mass is distributed all over it?
This is a rigid body - that means it's solid and it's got a definite shape that won't change with time. How can we find the center of mass of this thing? Well, we've got a formula and we can use that. The mathematics is obviously going to be a little bit tougher. But remember center of mass is also physics, and we can use a physical method as well. So let's start with that.
The center of mass is the point on the object where, in many ways, this complicated thing behaves as though its mass was at that spot. So think about what happens when you take a point mass and you hang it. It might bob around for a while at first, but when you let it settle down, the mass is going to be right below the support point. If it's off to the side, there will be unbalanced forces, gravity is down, and tension is at some funny angle. The net force will bring it towards the bottom and then, with a little friction, it will come to a halt and finally you get the stable situation.
So when mass is concentrated at a point, that point is always hanging directly below the support. So if this object has a center of mass somewhere, then, when I hang it, the center of mass should lie right below the support point, so that's the physical trick to locating the center of mass. I'm just going to hang it from some arbitrary spot - I can pick any spot I like, and it acts like a pendulum. It's kind of acting like a point mass x when you hang it, it swings around for a while, settles down, and now the center of mass of this object must be somewhere directly below the support. I could hang this, call this a plumb bob - that just means something that hangs straight down, and if you'll look at this line, I claim that the center of mass of the solid object must lie somewhere along this vertical line. So I've just drawn this line here. So somewhere along this line is the center of mass, but I don't know exactly where. It could be up here, it could be down there; all I know is that the thing settles down into this position.
So let me hang it from another point. I'll just pick a random spot and, once again, it acts like a pendulum, so the center of mass must be somewhere down below. Once again, I let it settle down and I draw the straight vertical line from the support. So the center of mass has to lie on this line, but I already argued that it has to lie on this line. So there's only one place for it to be, it's at the intersection of those two points. So here's the center of mass.
What happens if I hang it from the center of mass? Well, that's an interesting question. If the complicated object is behaving as though all of its mass was located right there, it's completely stable. There's no net forces or twists on it. If I hang it in another configuration, it's stable. It's not going to bob around like a pendulum, because it's acting as though its mass is right there. This is true in three dimensions as well. If I put my finger underneath the center of mass, it balances because it's acting as though all of its mass was right there, right on top of my finger.
So you can see why the center of mass is a useful quantity. Let's try to calculate this mathematically, because, after all, maybe you have a big object or an inconvenient object that you can't hang like this and you might still want to know where its center of mass is.
So we go back to our formula. The position of the center mass is defined to be - you have to break the object up into a bunch of little chunks and you label them, chunk number one, chunk number two, and you have mr plus mr. If you have discrete chunks, this is a discrete sum. But what happens when you have this funky shape? You can pick your origin to be wherever you want. What we're saying is break it up into little chunks, little, teeny imaginary chunks of differential size, dm[i]. So this is a tiny chunk with position vector r[i] and what I want to do when I'm summing this up is sum up an infinite number of infinitesimal little chunks, and that should tell me the center of mass.
So the infinite sum of a bunch of little infinitesimal terms, that's an integral. This is the integral of r times dm, that's the numerator, and the denominator, of course, is just the total mass. So this is the same formula, which is useful when you have a continuous distribution of mass. It looks a little scary with these vector signs in there, but remember what vector signs really mean - it means that the x component of the left side equals the x component of the right side. In other words, the x center of mass of an object is 1 over the total mass times the integral of xdm. And, similarly, for y center of mass, similarly for z, if you've got a three-dimensional object.
Now, you stare at this integral and you say, "This is a weird-looking integral. It's not like one I've ever seen before. x is not a function of M. How do you do that integral? Can you just pull x out?" No, because position and the mass, they may be connected to one another. I can't really give you any better formula than this. This formula is kind of a guideline and, for every given problem where you need to figure out the center of mass, you've just got to kind of look at the problem and think about what this formula means. You want to add up the little chunks of mass at position x and sum them. So let's do an example and see how this works.
Let me pick some funny continuous distribution of mass. How about a delta wing? So you're looking down on this wing. You can imagine a big flat wing. Let me assume it's uniform, so the mass is distributed all over this green-shaded area. This is a funny triangle, it's got width L, that's here, and it's got height 2L. So this upper triangle is an L by L right triangle and so is the lower one. This is a very special case and we want to find the center of mass. If you're designing an airplane with a delta plane, you certainly might want to know where the center of mass is, and it maybe too big and bulky to hang and measure the way we did physically.
So we need to find x center of mass and y center of mass. y center of mass is easy, just look at the symmetry of the problem; it's the same on the top as it is on the bottom, so I don't have to do any fancy integrals. y center of mass is surely going to be zero, it's going to be right here on this center line. So all we've got to do is find x center of mass. I've got this big X here. I'm labeling the x axis, which is starting at the tip and running horizontally off to the side. And remember the idea; I want to find the mass of little chunks that have a well-defined x. If you want to do this integral, you want to find chunks, dm, that have a particular x. So it doesn't have to be a little square, it can be this big vertical band, because every point on that band has the exact same x. So if I could just find the little infinitesimal mass of this infinitely thin strip, if I could find the mass dm of this dark green strip, that would be dm. I'd multiply it by x, and then I could sum that up over all the different strips that make this thing up. So the width of this band is dx.
So I want to find the mass of that band. How do you find the mass of a band if you know that mass is spread out uniformly? If I knew the mass per area and I could figure out the area of that band, then I'd be all set. Mass of the little band would be mass per area times the little area of the band. So that's really what I need, is the area of this little band, and I look at it and I see it's pretty close to a rectangle, and the area of a rectangle is just height times width. And the width is just dx. So a little differential area is height times width and width here is dx. And what's the height of this band? Well, now it's a little geometry puzzle. Let me stare at this for a second. Let's just look at the top half. I see an L by L right triangle and the little triangle here is similar to that one. So L is to L as this height is to x, which is how far across my little band is. So this height is x and this height is double that, 2x. So area is 2x times dx. That's the little d area, and so the dm is just mass per area for the whole thing, and that's called the mass density in two dimension, times the little differential area, 2xdx. And what's the total area of this triangle? Well, it's base times height. The base is 2L, so it's of 2L is L times L, that's just L^2 is the area of this particular triangle. So this becomes M over L^2 times 2xdx. That's my dm and what I want for x center of mass, I need to integrate x times dm. So I have to multiply this by x. Everything else is constants and they come out, but I have to integrate x^2dx.
So I have the integral of 2M over L^2 x^2dx, and what are my limits of integration? I started x equal to zero and I worked my way over to x equals L, so it's zero to L, and I get 2M over L^2. The integral of x^2 is x^3 over 3. And you evaluate that at L and you subtract it at zero and you get L^3 over 3. In other words - sorry, this is just the integral. Remember that the center of mass is this integral divided by mass. So x[cm] is the integral of xdm divided by M. The M's cancel and I get 2/3 L. And I look back at the picture and I say, "Yeah, I believe that."
Vertically the center of mass is right in the center, but horizontally there's a lot more mass over on the right side than there is over on the tip, so I would expect the center of mass to be shifted somewhat off to the right. So getting 2/3 L makes some intuitive sense to me.
The center of mass is an important point on physical objects. You need to know it and sometimes when the objects are simple you can calculate it directly just by that little finite sum. When they're continuous objects, it's nice if you can just hang them and find it physically, but if you have to, the integral formula, r[cm] equals 1 over M, integral of rdm, will give it to you. You just have to draw a picture carefully and think about how to set up this integral.
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