Feynman Physics Lectures V1 Ch19 1962-01-09 Center of Mass Moment of Inertia

Feynman Physics Lectures V1 Ch19 1962-01-09 Center of Mass Moment of Inertia

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Unformatted text preview: 19 Center of Mass; Moment of Inertia 19—1 Properties of the center of mass In the previous chapter we found that if a great many forces are acting on a complicated mass of particles, whether the particles comprise a rigid or a nonrigid body, or a cloud of stars, or anything else, and we find the sum of all the forces (that is, of course, the external forces, because the internal forces balance out), then if we consider the body as a whole, and say it has a total mass M, there is a certain point “inside” the body, called the center of mass, such that the net resulting external force produces an acceleration of this point, just as though the whole mass were concentrated there. Let us now discuss the center of mass in a little more detail. The location of the center of mass (abbreviated CM) is given by the equation (19.1) This is, of course, a vector equation which is really three equations, one for each of the three directions. We shall consider only the x—direction, because if we can understand that one, we can understand the other two. What does XCM = Emmi/2m, mean? Suppose for a moment that the object is divided into little pieces, all of which have the same mass m; then the total mass is simply the number N of pieces times the mass of one piece, say one gram, or any unit. Then this equation simply says that we add all the x’s, and then divide by the number of things that we have added: XCM = min/mN = in/N. In other words, XCM is the average of all the x’s, if the masses are equal. But suppose one of them were twice as heavy as the others. Then in the sum, that x would come in twice. This is easy to understand, for we can think of this double mass as being split into two equal ones, just like the others; then in taking the average, of course, we have to count that x twice because there are two masses there. Thus X is the average position, in the x-direction, of all the masses, every mass being counted a number of times proportional to the mass, as though it were divided into “little grams.” From this it is easy to prove that X must be somewhere between the largest and the smallest x, and, therefore lies inside the envelope including the entire body. It does not have to be in the material of the body, for the body could be a circle, like a hoop, and the center of mass is in the center of the hoop, not in the hoop itself. Of course, if an object is symmetrical in some way, for instance, a rectangle, so that it has a plane of symmetry, the center of mass lies somewhere on the plane of symmetry. In the case of a rectangle there are two planes, and that locates it uniquely. But if it is just any symmetrical object, then the center of gravity lies somewhere on the axis of symmetry, because in those circumstances there are as many positive as negative x’s. Another interesting proposition is the following very curious one. Suppose that we imagine an object to be made of two pieces, A and B (Fig. 19—1). Then the center of mass of the whole object can be calculated as follows. First, find the center of mass of piece A, and then of piece B. Also, find the total mass of each piece, M A and M 3. Then consider a new problem, in which a point mass M A is at the center of mass of object A, and another point mass M B is at the center of mass of object B. The center of mass of these two point masses is then the center of mass of the whole object. In other words, if the centers of mass of various parts 19—1 19—1 Properties of the center of mass 19—2 Locating the center of mass 19—3 Finding the moment of inertia 19—4 Rotational kinetic energy Fig. 19—1. The CM of a compound body lies on the line joining the CM’s of the two composite parts. of an object have been worked out, we do not have to start all over again to find the center of mass of the whole object; we just have to put the pieces together, treating each one as a point mass situated at the center of mass of that piece. Let us see why that is. Suppose that we wanted to calculate the center of mass of a complete object, some of whose particles are considered to be members of object A and some members of object B. The total sum Zmixi can then be split into two pieces—the sum ZAmix, for the A object only, and the sum 23mm- for object B only. Now if we were computing the center of mass of object A alone, we would have exactly the first of these sums, and we know that this by itself is M A X A, the total mass of all the particles in A times the position of the center of mass of A, because that is the theorem of the center of mass, applied to object A. In the same manner, just by looking at object B, we get M BX B, and of course, adding the two yields MX: z mixi + Z mixi A B MXCM = MAXA + MBXB. (19.2) Now since M is evidently the sum of M A and M B, we see that Eq. (19.2) can be interpreted as a special example of the center of mass formula for two point objects, one of mass M A located at X A and the other of mass M 3 located at X B. The theorem concerning the motion of the center of mass is very interesting, and has played an important part in the development of our understanding of physics. Suppose we assume that Newton’s law is right for the small component parts of a much larger object. Then this theorem shows that Newton’s law is also correct for the larger object, even if we do not study the details of the object, but only the total force acting on it and its mass. In other words, Newton’s law has thepeculiar property that if it is right on a certain small scale, then it will be right on a larger scale. If we do not consider a baseball as a tremendously complex thing, made of myriads of interacting particles, but study only the motion of the center of mass and the external forces on the ball, we find F= ma, where F is the external force on the baseball, m is its mass, and a is the acceleration of its center of mass. So F = ma is a law which reproduces itself on a larger scale. (There ought to be a good word, out of the Greek, perhaps, to describe a law which reproduces the same law on a larger scale.) Of course, one might suspect that the first laws that would be discovered by human beings would be those that would reproduce themselves on a larger scale. Why? Because the actual scale of the fundamental gears and wheels of the universe are of atomic dimensions, which are so much finer than our observations that we are nowhere near that scale in our ordinary observations. So the first things that we would discover must be true for objects of no special size relative to an atomic scale. If the laws for small particles did not reproduce themselves on a larger scale, we would not discover those laws very easily. What about the reverse problem? Must the laws on a small scale be the same as those on a larger scale? Of course it is not necessarily so in nature, that at an atomic level the laws have to be the same as on a large scale. Suppose that the true laws of motion of atoms were given by some strange equation which does not have the property that when we go to a larger scale we reproduce the same law, but instead has the property that if we go to a larger scale, we can approximate it by a certain expression such that, if we extend that expression up and up, it keeps reproducing itself on a larger and larger scale. That is possible, and in fact that is the way it works. Newton’s laws are the “tail end” of the atomic laws, extrapolated to a very large size. The actual laws of motion of particles on a fine scale are very peculiar, but if we take large numbers of them and compound them, they approximate, but only approximate, Newton’s laws. Newton’s laws then permit us to go on to a higher and higher scale, and it still seems to be the same law. In fact, it becomes more and more accurate as the scale gets larger and larger. This self-reproducing factor of Newton’s laws is thus really not a fundamental feature of nature, but is an important historical feature. We would never discover the fundamental laws of the atomic particles at first observation because the first observations are much too crude. In fact, it turns 1 9—2 out that the fundamental atomic laws, which we call quantum mechanics, are quite different from Newton’s laws, and are difficult to understand because all our direct experiences are with large-scale objects and the small-scale atoms behave like nothing we see on a large scale. So we cannot say, “An atom is just like a planet going around the sun,” or anything like that. It is like nothing we are familiar with because there is nothing like it. As we apply quantum mechanics to larger and larger things, the laws about the behavior of many atoms together do not reproduce themselves, but produce new laws, which are Newton’s laws, which then continue to reproduce themselves from, say, micro-microgram size, which still is billions and billions of atoms, on up to the size of the earth, and above. Let us now return to the center of mass. The center of mass is sometimes called the center of gravity, for the reason that, in many cases, gravity may be considered uniform. Let us suppose that we have small enough dimensions that the gravitational force is not only proportional to the mass, but is everywhere parallel to some fixed line. Then consider an object in which there are gravitational forces on each of its constituent masses. Let m,- be the mass of one part. Then the gravitational force on that part is m,- times g. Now the question is, where can we apply a single force to balance the gravitational force on the whole thing, so that the entire object, if it is a rigid body, will not turn? The answer is that this force must go through the center of mass, and we show this in the following way. In order that the body will not turn, the torque produced by all the forces must add up to zero, because if there is a torque, there is a change of angular momentum, and thus a rotation. So we must calculate the total of all the torques on all the particles, and see how much torque there is about any given axis; it should be zero if this axis is at the center of mass. Now, measuring x horizontally and y vertically, we know that the torques are the forces in the y-direction, times the lever arm x (that is to say, the force times the lever arm around which we want to measure the torque). Now the total torque is the sum 7' = 2 migxi = 82 mixis (193) so if the total torque is to be zero, the sum 2mm- must be zero. But 2mm,- = M X, the total mass times the distance of the center of mass from the axis. Thus the x—distance of the center of mass from the axis is zero. Of course, we have checked the result only for the x-distance, but if we use the true center of mass the object will balance in any position, because if we turned it 90 degrees, we would have y’s instead of x’s. In other words, when an object is supported at its center of mass, there is no torque on it because of a parallel gravitational field. In case the object is so large that the nonparallelism of the gravitational forces is significant, then the center where one must apply the balanc- ing force is not simple to describe, and it departs slightly from the center of mass. That is why one must distinguish between the center of mass and the center of gravity. The fact that an object supported exactly at the center of mass will balance in all positions has another interesting consequence. If, instead of gravitation, we have a pseudoforce due to acceleration, we may use exactly the same mathe- matical procedure to find the position to support it so that there are no torques produced by the inertial force of acceleration. Suppose that the object is held in some manner inside a box, and that the box, and everything contained in it, is accelerating. We know that, from the point of View of someone at rest relative to this accelerating box, there will be an effective force due to inertia. That is, to make the object go along with the box, we have to push on it to accelerate it, and this force is “balanced” by the “force of inertia,” which is a pseudoforce equal to the mass times the acceleration of the box. To the man in the box, this is the same situation as if the object were in a uniform gravitational field whose “g” value is equal to the acceleration a. Thus the inertial force due to accelerating an object has no torque about the center of mass. This fact has a very interesting consequence. In an inertial frame that is not accelerating, the torque is always equal to the rate of change of the angular mo- mentum. However, about an axis through the center of mass of an object which 19—3 Fig. l9—2. A right triangle and a right circular cone generated by ro- tating the triangle. is accelerating, it is still true that the torque is equal to the rate of change of the angular momentum. Even if the center of mass is accelerating, we may still choose one special axis, namely, one passing through the center of mass, such that it will still be true that the torque is equal to the rate of change of angular momentum around that axis. Thus the theorem that torque equals the rate of change of angular momentum is true in two general cases: (1) a fixed axis in inertial space, (2) an axis through the center of mass, even though the object may be accelerating. 19—2 Locating the center of mass The mathematical techniques for the calculation of centers of mass are in the province of a mathematics course, and such problems provide good exercise in integral calculus. After one has learned calculus, however, and wants to know how to locate centers of mass, it is nice to know certain tricks which can be used to do so. One such trick makes use of what is called the theorem of Pappus. It works like this: if we take any closed area in a plane and generate a solid by moving it through space such that each point is always moved perpendicular to the plane of the area, the resulting solid has a total volume equal to the area of the cross section times the distance that the center of mass moved! Certainly this is true if we move the area in a straight line perpendicular to itself, but if we move it in a circle or in some other curve, then it generates a rather peculiar volume. For a curved path, the outside goes around farther, and the inside goes around less, and these effects balance out. So if we want to locate the center of mass of a plane sheet of uniform density, we can remember that the volume generated by spinning it about an axis is the distance that the center of mass goes around, times the area of the sheet. For example, if we wish to find the center of mass of a right triangle of base D and height H (Fig. 19—2), we might solve the problem in the following way. Imagine an axis along H, and rotate the triangle about that axis through a full 360 degrees. This generates a cone. The distance that the x-coordinate of the center of mass has moved is 21rx. The area which is being moved is the area of the triangle, %HD. So the x-distance of the center of mass times the area of the triangle is the volume swept out, which is of course 7rD2H / 3. Thus (21rx)(%HD) = l/37rD2H, or x = D/ 3. In a similar manner, by rotating about the other axis, or by symmetry, we find y = H /3. In fact, the center of mass of any uniform triangular area is where the three medians, the lines from the vertices through the centers of the opposite sides, all meet. That point is 1/3 of the way along each median. Clue: Slice the triangle up into a lot of little pieces, each parallel to a base. Note that the median line bisects every piece, and therefore the center of mass must lie on this line. Now let us try a more complicated figure. Suppose that it is desired to find the position of the center of mass of a uniform semicircular disc—a disc sliced in half. Where is the center of mass? For a full disc, it is at the center, of course, but a half-disc is more difficult. Let r be the radius and x be the distance of the center of mass from the straight edge of the. disc. Spin it around this edge as axis to generate a sphere. Then the center of mass has gone around 27rx, the area is 1rr2/2 (because it is only half a circle). The volume generated is, of course, 47rr3/3, from which we find that (27rx)(%1rr2) = 47rr3/3, or x = 4r/37r. There is another theorem of Pappus which is a special case of the above one, and therefore equally true. Suppose that, instead of the solid semicircular disc, we have a semicircular piece of wire with uniform mass density along the wire, and we want to find its center of mass. In this case there is no mass in the interior, only on the wire. Then it turns out that the area which is swept by a plane curved line, when it moves as before, is the distance that the center of mass moves times the length of the line. (The line can be thought of as a very narrow area, and the previous theorem can be applied to it.) 19—4 19—3 Finding the moment of inertia Now let us discuss the problem of finding the moments of inertia of various objects. The formula for the moment of inertia about the z-axis of an object is I = 2 mix? + y?) 01' I = [(x2 + yz) dm = [(x2 + y2)p dv. (19.4) That is, we must sum the masses, each one multiplied by the square of its distance (x,2 + y?) from the axis. Note that it is not the three-dimensional distance, only the two-dimensional distance squared, even for a three-dimensional object. For the most part, we shall restrict ourselves to two-dimensional objects, but the formula for rotation about the z-axis is just the same in three dimensions. As a simple example, consider a rod rotating about a perpendicular axis through one end (Fig. 19—3). Now we must sum all the masses times the x-distances squared (the y’s being all zero in this case). What we mean by “the sum,” of course, is the integral of x2 times the little elements of mass. If we divide the rod into small elements of length dx, the corresponding elements of mass are propor- tional to dx, and if dx were the length of the whole rod the mass would be M. Therefore dm = de/L and so L Md M L ML2 _ 2 x — _ 2 = - I—Ax L —L/0xdx 3 (19.5) The dimensions of moment of inertia are always mass times length squared, so all we really had to work out was the factor 1/3. Now what is I if the rotation axis is at the center of the rod? We could just do the integral over again, letting x range from —%L to +%L. But let us notice a few things about the moment of inertia. We can imagine the rod as two rods, each of mass M/2 and length L/2; the moments of inertia of the two small rods are equal, and are both given by the formula (19.5). Therefore the moment of inertia is _ 2(M/2)(L/2)2 _ ML2 I 3 _12 (19.6) Thus it is much easier to turn a rod about its center, than to swing it around an end. Of course, we could go on to compute the moments of inertia of various other bodies of interest. However, while such computations provide a certain amount of important exercise in the calculus, they are not basically of interest to us as such. There is, however, an interesting theorem which is very useful. Suppose we have an object, and we want to find its moment of inertia around some axis. That means we want the inertia needed to carry it by rotation about that axis. Now if we support the object on pivots at the center of mass, so that the object does not turn as it rotates about the axis (because there is no torque on it from inertial effects, and therefore it will not turn when we start moving it), then the forces needed to swing it around are the same as though all the mass were concen- trated at the center of mass, and the moment of inertia would be simply 11 = MREM, where RCM is the distance from the axis to the center of mass. But of course that is not the right formula for the moment of inertia of an object which is really being rotated as it revolves, because not only is the center of it moving in a circle, which would contribute an amount 11 to the moment of inertia, but also we must turn it about its center of mass. So it is not unreasonable that we mu...
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