20 Rotation in space 20-1 Torques in three dimensions In this chapter we shall discuss one of the most remarkable and amusing 20-1 Torques in three dimensions consequences of mechanics, the behavior of a rotating wheel. In order to do this 20-2 The rotation equations using we must first extend the mathematical formulation of rotational motion the principles of angular momentum, torque, and so on, to three-dimensional space cross We shall not use these equations in all their generality and study all their conse- 20-3 The gyroscope quences, because this would take many years, and we must soon turn to other subjects. In an introductory course we can present only the fundamental laws and 20-4 Angular momentum of a solid apply them to a very few situations of special interest First, we notice that if we have a rotation in three dimensions whether of a igid body or any other system, what we deduced for two dimensions is still right. That is, it is still true that xFy- yFr is the torque"in the xy-plane, "or the torque "around the z-axis. "It also turns out that this torque is still equal to the rate of change of xpv -ypa, for if we go back over the derivation of Eq (18. 15)from Newtons laws we see that we did not have to assume that the motion was in a plane;when we differentiate xpy-ypr, we get xFy-yFr, so this the oren still right. The quantity xP, -DPz, then, we call the angular momentum belonging to the xy-plane, or the angular momentum about the z-axis. This being true, we can use any other pair of axes and get another equation For instance we can use the yz-plane, and it is clear from symmetry that if we just substitute y for x and z for y, we would find yF, - zF, for the torque and yp, -zp, would be the angular momentum associated with the yz-plane. Of course we could have another plane, the zx-plane, and for this we would find zFx -xF:= d/dt(zp2 -xp) That these three equations can be deduced for the motion of a single particle quite clear. Furthermore, if we added such things as xp,- yp, together for many particles and called it the total angular momentum, we would have three kinds for the three planes xy, yz, and zx, and if we did the same with the forces, we would talk about the torque in the planes xy, yz, and zx also. Thus we would have laws that the external torque associated with any plane is equal to the rate of change of the angular momentum associated with that plane. This is just a generalization of what we wrote in two dimensions But now one may say, " Ah, but there are more planes; after all take some other plane at some angle, and calculate the torque on that plane from the forces? Since we would have to write another set of equations for every such plane, we would have a lot of equations! "Interestingly enough, it turns out that if we were to work out the combination x'Fy-y'F2 for another plane, measuring the x, Fy, etc, in that plane, the result can be written as some combination of the three expressions for the xy, yz-and zx-planes. There is nothing new. In other words, if we know what the three torques in the xy, yz-, and zx-planes are, then he torque in any other plane, and correspondingly the angular momentum also can be written as some combination of these: six percent of one and ninety-two percent of another, and so on. This property we shall now analyze Suppose that in the xyz-axes, Joe has worked out all his torques and his angu- lar momenta in his planes. But Moe has axes x', y, z in some other direction. To make it a little easier, we shall suppose that only the x and y-axes have been turne Moe's xand y are new, but his z happens to be the same. That is, he has new planes, let us say, for yz and zx. He therefore has new torques and angular momenta which he would work out. For example, his torque in the x'y' plane would be equal to x'Fy-y' Fr and so forth. What we must now do is to find the relation ship between the new torques and the old torques, so we will be able to make a
connection from one set of axes to the other. Someone may say, " That looks just like what we did with vectors. And indeed, that is exactly what we are intending to do, Then he may say, " Well, isnt torque just a vector? It does turn out to be a vector, but we do not know that right away without making an analysis. So in the following steps we shall make the analysis, We shall not discuss every step in detail, since we only want to illustrate how it works. The torques calculated by Joe ry=xFy- yF (20.1) We digress at this point to note that in such cases as this one may get the wrong sign for some quantity if the coordinates are not handled in the right way. Why not write Tuz= zFy-yFz? The problem arises from the fact that a coordinate system may be either "right-handed"or"left-handed. Having chosen(arbitrarily)a sign for, say Try, then the correct expressions for the other two quantities may always be found by interchanging the letters xyz in either order 'y'sx'Fy'-y'l Ty'r=yF2-2'F z'Fr'-x'F2 Now we suppose that one coordinate system is rotated by a fixed angle 0, such that the z-and z'-axes are the same. (This angle 0 has nothing to do with rotating objects or what is going on inside the coordinate system. It is merely the relation ship between the axes used by one man and the axes used by the other, and is supposedly constant. ) Thus the coordinates of the two systems are related b y'= cos 0-x sin 8 (20.3) Likewise, because force is a vector it transforms into the new system in the same yay as do x, y, and z, since a thing is a vector if and only if the various components transform in the same wa Fr cos 0+ F sin Fy'= Fy cos 0- Fx sin e x', y, and zthe expressions(20.3), and for Fr, Fy, Fx those given by(20.4), all into(20. 2). So, we have a rather long string of terms for Try and (rather sur ingly at first) it turns out that it comes right down to xFy-yFr, which we re nize to be the y 0)(Fy Fr sin 8) e)(Fr cos 0+ Fy sin 0) xF,(cos+sin2)-yF-(sing cos26 +xFxCsin 0 cos 0 sin g cos 0) +yF,(sin 0 cos e
That result is clear, for if we only turn our axes in the plane, the twist around z in that plane is no different than it was before, because it is the same plane! What will be more interesting is the expression for Tyy because that is a new plane We now do exactly the same thing with the y' z -plane, and it comes out as follows Ty'?'=(cos 8-x sin O)Fz z(F Fr sin 8) =(yF-zFy)cos 8+(aFx -xFi)sin 8 Ty cos 8 Tzr sin 8. Finally, we do it for zx: T2r=z(Fr cos 6+ Fy sin 8) (x cos 0+ y sin O)Fz (zFx- xFz)cos 8-, -, =T cos 6-Tuz sin 8 We wanted to get a rule for finding torques in new axes in terms of torques in old axes, and now we have the rule. How can we ever remember that rule If we look carefully at(20.5),(20.6), and (20.7), we see that there is a close relation ship between these equations and the equations for x, y, and z. If, somehow, we could call Try the z-component of something, let us call it the z-component of T, then it would be all right; we would understand (20.5)as a vector transformation since the z-component would be unchanged, as it should be. Likewise, if we associate with the yz-plane the x-component of our newly invected vector, and with the zx-plane, the y-component, then these transformation expressions would Tr cOS , Ty=Ty cos 8-T sin 8, hich is just the rule for vectors Therefore we have proved that we may identify the combination of xFy - yFa with what we ordinarily call the z-component of a certain artificially invented vector. Although a torque is a twist on a plane, and it has no a priori vector char the mathematically it does behave like a vector. This vector is at right angles to plane of the twist, and its length is proportional to the strength of the twist. The three components of such a quantity will transform like a real vector So we represent torques by vectors; with each plane on which the torque is supposed to be acting, we associate a line at right angles, by a rule. But"at right angles"leaves the sign unspecified. To get the sign right, we must adopt a rule which will tell us that if the torque were in a certain sense on the xy-plane, then the axis that we want to associate with it is in the"up"z-direction. That is, some- body has to define"right"and"left"for us. Supposing that the coordinate system is x, y, z in a right-hand system, then the rule will be the following: if we think of the twist as if we were turning a screw having a right-hand thread, then the direction of the vector that we will associate with that twist is in the direction that the screw ould advance Why is torque a vector? It is a miracle of good luck that we can associate a single axis with a plane and therefore that we can associate a vector with the torque; it is a special property of three-dimensional In two dimensions. the torque is an ordinary scalar, and there need be no direction associated with it. In three dimensions, it is a vector. If we had four dimensions, we would be in great difficulty, because(if we had time, for example, as the fourth dimension)we would not only have planes like xy, yz, and zx, we would also have tx-, ty, and tz-pla anes There would be six of them, and one cannot represent six quantities as one vector in four dime We will be living in three dimensions for a long time, so it is well to notice that the foregoing mathematical treatment did not depend the fact that x
was position and F was force; it only depended on the transformation laws for vectors. Therefore if, instead of x, we used the x-component of some other vector it is not going to make any difference. In other words, if we were to calculate a,by-abx, where a and b are vectors, and call it the z-component of some new quantity c, then these new quantities form a vector c. We need a mathematical notation for the relationship of the new vector, with its three components, to the vectors a and b. The notation that has been devised for this is c a x b. We have then, in addition to the ordinary scalar product in the theory of vector analysis, a new kind of product, called the vector product. Thus, if c=ax b this is the same as writing (20.9) If we reverse the order of a and b, calling a, b and b, a, we would have the sign of c reversed, because c would be bzay-byaz. Therefore the cross product is unlike ordinary multiplication, where ab ba; for the cross product, b x a X b. From this, we can prove at once that if a= b, the cross product is The cross product is very important for representing the features of rotation and it is important that we understand the geometrical relationship of the three vectors a,b, and c. Of course the relationship in components is given in Eq. 20.9 and from that one can determine what the relationship is in geometry. The answer is, first, that the vector c is perpendicular to both a and b. (Try to calculate ca and see if it does not reduce to zero. Second, the magnitude of c turns out to be the magnitude of a times the magnitude of b times the sine of the angle between the two. In which direction does c point? Imagine that we turn a into b through an angle less than 1800; a screw with a right-hand thread turning in this way will advance in the direction of c. The fact that we say a right-hand screw instead of a left-hand screw is a convention, and is a perpetual reminder that if a and b are honest"vectors in the ordinary sense, the new kind of"vector"which we have created by a x b is artificial, or slightly different in its character from a and b because it was made up with a special rule. If a and b are called ordinary vectors, ve have a special name for them, we call them polar vectors. Examples of such vectors are the coordinate r, force F, momentum p, velocity v, electric field E, etc these are ordinary polar vectors. Vectors which involve just one cross product in their definition are called axial vectors or pseudovectors. Examples of pseudovectors are, of course, torque r and the angular momentum L. It also turns out that the angular velocity w is a pseudovector, as is the magnetic field B In order to complete the mathematical properties of vectors, we should know all the rules for their multiplication, using dot and cross products. In our applica tions at the moment, we will need very little of this, but for the sake of completeness we shall write down all of the rules for vector multiplication so that we can use he results later. These are (a)a×(b+c)=a×b+a×c, (b) (c)a·(b×c)=(a×b)·c, (d)a×(b×c)=b(a·c)-c(a·b), (e) a=0, (f)a·(a×b)=0. 20-2 The rotation equations using cross products Now let us ask whether any equations in physics can be written using the cross product. The answer, of course, is that a grea eat many equations can be so written. For instance, we see immediately that the torque is equal to the position
vector cross the force T=r×F This is a vector summary of the three equations Tx= yFz-zFy, etc. By the same token, the angular momentum vector, if there is only one particle present, is the distance from the origin multiplied by the vector momentum L=r X For three-dimensional space rotation, the dynamical law analogous to the law F=dp/dt of Newton, is that the torque vector is the rate of change with time of the angular momentum vector If we sum(20. 13)over many particles, the external torque on a system is the rate of change of the total angular momentum lTot/dt Another theorem: If the total external torque is zero, then the total vector angular momentum of the system is a constant. This is called the law of conserva- tion of angular momentum. If there is no torque on a given system, its angular momentum cannot change What about angular velocity? Is it a vector? We have already discussed turning a solid object about a fixed axis, but for a moment suppose that we are turning it simultaneously about two axes. It might be turning about an axis inside a box, while the box is turning about some other axis. The net result of such combined motions is that the object simply turns about some new axis! The wonderful thing about this new axis is that it can be figured out this way. If the rate of turning in the xy-plane is written as a vector in the z-direction whose length is equal to the rate of rotation in the plane, and if another vector is drawn in the y-direction,say, which is the rate of rotation in the zx-plane, then if we add these together as a vector, the magnitude of the result tells us how fast the object is turning, and the direction tells us in what plane, by the rule of the parallelogram That is to say, simply, angular velocity is a vector, where we draw the magnitudes of the rotations in the three planes as projections at right angles to those planes. As a simple application of the use of the angular velocity vector, we may evalu ate the power being expended by the torque acting on a rigid body. The power, of course, is the rate of change of work with time; in three dimensions, the power turns out to be p=r·a All the formulas that we wrote for plane rotation can be generalized to three dimensions. For example, if a rigid body is turning about a certain axis with angular velocity a, we might ask, "What is the velocity of a point at a certain radial position r? We shall leave it as a problem for the student to show that the velocity of a particle in a rigid body is given by v= wX r, where w is the angular velocity and r is the position. Also, as another example of cross products, we had a formula for Coriolis force, which can also written using cross products Fe 2mv X w. That is, if a particle is moving with velocity v in a coordinate system which is, in fact, rotating with angular velocity w, and we want to think in terms of the rotating coordinate system, then we have to add the pseudoforce Fc BEFORE AFTER 20-3 The gyroscope Fig. 20-1. Before: axis izontal Let us now return to the law of conservation of angular momentum axis is vertical; momentum about vertical w may be demonstrated with a rapidly spinning wheel, or gyroscope, as fol is still zero; man and chair spin in (see Fig. 20-1). If we sit on a swivel chair and hold the spinning wheel wit direction opposite to spin of the wheel axis horizontal, the wheel has an angular momentum about the horizontal axis w That this is true can be derived by compounding the displacements of the particles of the body during an infinitesimal time At. It is not self-evident, and is left to those wh