Gyroscope Precession and Nutation: Rigid Body Rotation

    This is a work in progress...please do not read until I am finished. Thanks.

    [The CM-Gyroscope_RigidBodyRotation.pdf and CM-Gyroscope_SymmetricTop.pdf files provided below do contain the full and complete derivations of the formulae describing the rotations of rigid bodies.]

    Gyroscope Precession and Nutation: Rigid Body Rotation

    This page discusses rigid bodies, what they are, and the derivations of Euler Angles, Euler's Equations, and the free precession of a rigid body. Included is the topological reason why a body will only rotate stably (precess) around two of its three principal axes while around the third axis it tumbles.

    • Rigid Body Rotation

    • Rigid Body
    To begin with, just what is a ``rigid body''? Now everyone has some intuitive idea what is meant by a rigid body, why it is just a body that doesn't change its shape [Theoretically, a `rigid body' cannot exist for any number of reasons. But one, in particular, concerns us here, and that is Einstein's Special Theory of Relativity. If a rigid body were to experience an impulse on one side, then in order for the body not to flex or deform, the opposite side from the impulse would have to instantaneously begin accelerating. But according to Special Relativity, the opposite side would have to wait at least for light to cross the width of the body before it could start its acceleration. Thus Special Relativity rules out the existence of a `rigid body'. Nevertheless, the concept is quite useful as long as we realize that it cannot be attained perfectly, even in a theoretical sense.]. Let's make this intuition a tiny bit more rigorous, shall we?

    In a plane, the distances of point i from points 1 and 2 specify two circles of radius r_{1i} and r_{2i}. Two circles having distinct centers, that is, 1\ne2, can intersect in two points. Thus the distances from two points does not fix point i. The distance from a third point, r_{3i}, specifies a third circle that obviously passes through point i. This third circle also intersects the circles about points 1 and 2 in two additional points, but these points will be distinct as long as point 3 is not collinear with 1 and 2. Thus three distances from three non-collinear points uniquely determines the position of point i. The same argument applies in 3-space but now the equal distance circles becomes spheres and two distinct spheres intersect in a circle instead of two points. The spheres about points 1 and 2 intersect in a circle, and the spheres about points 2 and 3 intersect in a different circle which intersects the first circle in at most two points. As long as the three initial points are not collinear, then the intersection of the spheres about points 1 and 3 results in a third distinct circle which can intersect the first two in only a single point, as shown in the above Figure.
    Any particle in a rigid body has three space coordinates associated with it, say particle i has coordinates r_{i}=(r_{i_1},r_{i_2},r_{i_3}). A rigid body is defined as an n particle system with special constraints on the coordinates of the n particles. Now n particles will have a total of 3n coordinates. But this does not mean that an n particle rigid body has 3n degrees of freedom. If r_{ij} denotes the distance from particle i to particle j, that is r_{ij}=\Vert{r}_j-{r}_i\Vert = \sqrt{({r}_{j}-{r}_{i})\cdot({r}_{j}-{r}_{i})}, then the rigid body constraints can be described as n(n-1)/2 equations of the form r_{ij}=k_{ij}, for all pairs i,j with i>j and the k_{ij} are constants.
    These constraints just express the fact that in a rigid body the interparticle distances are constant, otherwise the body would not be ``rigid''. The reason that there are only n(n-1)/2 of these constraints instead of the n^2 total number of (i,j) pairs is because we don't have to specify both r_{ij} and r_{ji} constraints since the distance from particle i to particle j is the same as the distance from j to i: r_{ij}=r_{ji}. Also, the distance from particle i to itself is automatically zero, thus we only need constraints for pairs of i,j where i>j, and there are only n(n-1)/2 of these. Also notice that for all n>=8, then n(n-1)/2>3n, thus we cannot simply subtract the number of constraints from 3n to get the number of degrees of freedom for a rigid body. This is because the rigid body constraints are not independent of one another.
    In fact, once we have the coordinates of three non-collinear particles, say r_1, r_2, and r_3, then the distances of any other particle, say at r_i for i>=4, to these three particles serve to constrain particle i to constant distances from all other particles j within the system. This results from just simple Euclidean geometry that says any two distinct circles can only intersect in at most two points. Thus the circle of radius r_{1i} centered on 1 and the circle of radius r_{2i} centered on point 2 intersect not only at point i but also at one other point. A third non-collinear particle 3 is required so that a circle centered on 3 of radius r_{3i} intersects the other two circles at i but also intersects the circle at 1 in a different location than the circle centered at 2. Thus the three distances from three non-collinear points are enough to uniquely specify the position of particle i. In other words, instead of each new particle i requiring distance constraints to all of the other particles of the body, r_{ij} for all j<i, it only requires three distance constraints in order to be uniquely positioned within the rigid body.
    So each new particle, r_i for i>=4, adds three new coordinates, but it also adds three new distance constraints to the initial three particles r_1, r_2, and r_3. The three distance constraints cancel the three new degrees of freedom for the added particle. So all of the particles from number 4 to n add no new degrees of freedom, each of their three coordinates are canceled by three distance constraints. Thus it appears that we have only nine degrees of freedom, the three coordinates for each of the initial three noncolinear particles, r_1, r_2, and r_3. But even these nine coordinates are not independent since they must satisfy the interdistance requirements amongst themselves, of which there are three interparticle distances to constrain them: r_{12}, r_{23}, and r_{13}. Therefore, in total, there are only 6 coordinates remaining that are free of constraints, that is, there are six independent degrees of freedom for a rigid body. Let's see how these six degrees of freedom arise in practice, shall we?
    In 3-space, a point is denoted by a 3-vector. Thus we require three fixed and perpendicular axes defining three Euclidean space coordinates, represented by the unit vectors \hat{i}=(1,0,0), \hat{j}=(0,1,0), and \hat{k}=(0,0,1). A rigid body similarly requires three perpendicular axes affixed to it defining three body coordinates, represented by the unit vectors \hat{i'}, \hat{j'}, and \hat{k'}. The origin of the body coordinates, (0',0',0'), must be defined in the space coordinates, that is, (0',0',0')=(x_0,y_0,z_0).
    This takes three independent variables. We also must specify the directions of the body axes relative to the space axes. Direction cosines are perhaps the simplest way of doing this. These are the cosines of the angles between every pairing of a space axis with a body axis. Thus each body axis makes an angle with each of the three space axes. Thus the projections along each of the space axes is determined by the dot products (see the ``Geometric Interpretation of the Dot Product'' Example),

      • Euler Angles

    Until I have time to finish the discussion of Rigid Body Rotations on this wiki page, the mathematical details are provided in the following PDF file.
    Here is the PDF file discussing Rigid Body Rotations: Download file "CM-Gyroscope_RigidBodyRotation_pw107.pdf"

    • Euler's Equations

    Until I have time to add the discussion of Spinning Symmetric Tops to this wiki page, the mathematical details are provided in the following PDF file.
    Here is the PDF file discussing Spinning Symmetric Tops: Download file "CM-Gyroscope_SymmetricTop_pw107.pdf"

    For a freely rotating rigid body, there are no external torques and thus Euler's Equations reduce to

    The solutions to these differential equations indicate that stable rotations are possible around two of the three principal axes of a body, but around the third principal axis, the one having the intermediate moment of inertia, rotation is not possible, rather the rigid body tumbles. You can see this yourself by attempting to rotate a paperback novel around its three principal axes. Around the intermediate moment of inertia axis the book tumbles instead of rotating smoothly. See the PDF file, CM-Topology_Rotating_Book.pdf, for a full discussion of the topological cause of this: Download file "CM-Topology_Rotating_Book_pw107.pdf"

    Below the paperback book is held so as to spin it around its intermediate moment of inertia axis...
    ...but when the book is tossed into the air while attempting to spin it, it does not rotate smoothly but rather it tumbles...

    ...try this experiment for yourself...you will be able to smoothly spin the book around its longest principal axis as well as its shortest principal axis, but around the intermediate length principal axis the book will tumble when you toss it into the air, as shown above.

    The topological analysis of the phase portrait is a very powerful tool for determining the behaviors of dynamical systems. What is a phase portrait? Crudely, it is just a graph in the coordinate space of the body of the possible trajectories available to the system. Thus the phase portrait diagrams the possible dynamical behaviors for the body. For a rigid body having three distinct principal moments of inertia, the angular portion of the phase portrait may be drawn in a subspace of 3 dimensions, as shown below. The trajectories drawn in this Figure shows the possible free rotations for the rigid body. Now there is a relatively clean way of understanding the genesis of this phase portrait.

    I know that we are getting ahead of ourselves by showing this phase portrait, but I just want to point out that closed trajectories, which represent precessional motions, occur around the 2- and 3-axes, but around the 1-axis no closed trajectories are possible (no precession). This phase portrait is found by recognizing that Conservation of Energy means that the energy associated with the rotary motions is a constant. This rotational kinetic energy, E, is given below:
    Since the right-hand side of the above equation is just the equation of an ellipsoid, we see that the energy level surfaces are thus ellipsoids having semiaxes of \sqrt{2E\lambda_{\kappa}} for \kappa={i,j,k}. Every phase space trajectory of the rotating book having a given constant energy E must therefore reside on this ellipsoid. I have attempted to draw a representation of this energy constant ellipsoid as a black outline in the above phase portrait.
    At the same time, Conservation of Angular Momentum means that
    must also be a constant since the angular momentum L is conserved. This equation is represented by a sphere of radius L=\sqrt{\sum_{\kappa={i,j,k}}L_{\kappa}^2}. Because angular momentum is also conserved, any phase space trajectory must also reside on this sphere.
    In other words, all trajectories of the system must reside on both the constant energy ellipsoid and the constant angular momentum sphere. Hence, combining these two criteria means that any phase space trajectory of the system must be given by the intersection of this E energy level ellipsoid and a L angular momentum sphere. These intersections, for a given E with various L values, are drawn on the ellipsoid in the above phase portrait as red, blue, and green trajectories. Notice that two axes have closed elliptical orbits (red and blue) around them while the third axis cannot be surrounded by a closed trajectory (the green trajectories leave the immediate neighborhood of the 1-axis and thus do not form closed loops around it). This means that there are no stable rotations possible around the 1-axis; in words, the rigid body tumbles around its axis having an intermediate moment of inertia. In addition, if the E ellipsoid and L sphere do not intersect, then no actual trajectory for those particular values of E and L exists. In other words, at the energy E every trajectory would have an angular momentum value different from the L value.

    In order to treat the case where the applied torques are nonzero, we must consider in detail the motions of the spinning body and thus we revisit the definition of the Euler angles.

    Any arbitrary rotation of a body may be obtained by a sequence of three rotations. By convention, we first perform a counterclockwise rotation about the k, or z, axis by an angle of \alpha. This is shown in Part A of the above Figure. Next a counterclockwise rotation by \beta around the new j' axis is performed leaving j''=j' (the new j''-axis is known as the line of nodes). Part B illustrates this rotation which takes the original k=k' axis to a new k''-axis that makes an angle of \beta with the old k-axis. Part C shows the final rotation through an angle of \gamma about the new k''-axis that thus rotates the i'' and j'' axes around to new locations i''' and j'''. These three Euler angles, (\alpha,\beta,\gamma), are thus capable of specifying any arbitrary rotation of the body from its original orientation.

    Copyright (c) 2010-2016 Craig G. Shaefer, all rights reserved.


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