Dynamics 7

One of the most interesting of all problems in dynamics is that of gyroscopic motion. This motion occurs whenever the axis about which a body is spinning is itself rotating about another axis. Although the complete description of this motion involves considerable complexity, the most common and useful examples of gyroscopic motion occur when the axis of a rotor spinning at constant speed turns (precesses) about another axis at a steady rate. Our discussion in this article will focus on this special case

The force equation for a mass system, rigid or nonrigid, Eq. 4/1 or 4/6, is the generalization of Newton's second law for the motion of a particle and should require no further explanation. The moment equation for three-dimensional motion, however, is not nearly as simple as the third of Eqs. 6/1 for plane motion since the change of angular momentum has a number of additional components which are absent in plane motion.

The kinematic analysis of a rigid body which has general threedimensional motion is best accomplished with the aid of our principles of relative motion. We have applied these principles to problems in plane motion and now extend them to space motion. We will make use of both translating axes and rotating reference axes.

The momentum properties of a rigid body may be represented by the resultant linear-momentum vector G through the mass center and the resultant angular-momentum vector HG about the mass center, as shown in Fig. 7/13. Although HG has the properties of a free vector, we represent it through G for convenience

The relative-motion terms represent the effect of the rotation about B and are identical to the velocity and acceleration expressions discussed in the previous article for rotation of a rigid body about a fixed point. Therefore, the relative-velocity and relative-acceleration equations may be written (7/4) where and are the instantaneous angular velocity and angular acceleration of the body, respectively

We observe now that for the two cases of Figs. 7/12a and 7/12b, the position vectors i and ri are given by the same expression xi yj zk. Thus, Eqs. 7/8 and 7/9 are identical in form, and the symbol H will be used here for either case. We now carry out the expansion of the integrand in the two expressions for angular momentum, recognizing that the components of are invariant with respect to the integrals over the body and thus become constant multipliers of the integrals. The cross-product HO [r ( r)] dm HG [ ( )] dm m v HG v Σmii Σ[i mi ( i )] v SECTION B KINETICS x G Y X Z y v - mi z i ρ ω x y mi z ω ri Y X O Z (a) (b) Figure 7/12 c07.qxd 2/10/12 2:29 PM Page 539 expansion applied to the triple vector product gives, upon collection of terms, Now let (7/10) The quantities Ixx, Iyy, Izz are called the moments of inertia of the body about the respective axes, and Ixy, Ixz, Iyz are the products of inertia with respect to the coordinate axes. These quantities describe the manner in which the mass of a rigid body is distributed with respect to the chosen axes. The calculation of moments and products of inertia is explained fully in Appendix B. The double subscripts for the moments and products of inertia preserve a symmetry of notation which has special meaning in their description by tensor notation.* Observe that Ixy Iyx, Ixz Izx, and Iyz Izy. With the substitutions of Eqs. 7/10, the expression for H becomes (7/11) and the components of H are clearly (7/12) Equation 7/11 is the general expression for the angular momentum either about the mass center G or about a fixed point O for a rigid body rotating with an instantaneous angular velocity . Remember that in each of the two cases represented, the reference axes x-y-z are attached to the rigid body. This attachment makes the Hz Izxx Izyy Izzz Hy Iyxx Iyyy Iyzz Hx Ixxx Ixyy Ixzz (Izxx Izyy Izzz)k (Iyxx Iyyy Iyzz)j H ( Ixxx Ixyy Ixzz)i Izz (x2 y2) dm Iyz yz dm Iyy (z2 x2) dm Ixz xz dm Ixx (y2 z2) dm Ixy xy dm k[ zxx zyy (x2 y2)z] dm j[ yxx (z2 x2)y yzz] dm dH i[(y2 z2)x xyy xzz] dm 540 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies *See, for example, the first author's Dynamics, 2nd Edition, SI Version, 1975, John Wiley & Sons, Art. 41. c07.qxd 2/10/12 2:29 PM Page 540 Article 7/7 Angular Momentum 541 moment-of-inertia integrals and the product-of-inertia integrals of Eqs. 7/10 invariant with time. If the x-y-z axes were to rotate with respect to an irregular body, then these inertia integrals would be functions of the time, which would introduce an undesirable complexity into the angularmomentum relations. An important exception occurs when a rigid body is spinning about an axis of symmetry, in which case, the inertia integrals are not affected by the angular position of the body about its spin axis. Thus, for a body rotating about an axis of symmetry, it is frequently convenient to choose one axis of the reference system to coincide with the axis of rotation and allow the other two axes not to turn with the body. In addition to the momentum components due to the angular velocity of the reference axes, then, an added angular-momentum component along the spin axis due to the relative spin about the axis would have to be accounted for.

We will first describe gyroscopic action with a simple physical approach which relies on our previous experience with the vector changes encountered in particle kinetics. This approach will help us gain a direct physical insight into gyroscopic action. Next, we will make use of the general momentum relation, Eq. 7/19, for a more complete description.

When a body rotates about a fixed point, the angular-velocity vector no longer remains fixed in direction, and this change calls for a more general concept of rotation.

A more general formulation of the motion of a rigid body in space calls for the use of reference axes which rotate as well as translate. The description of Fig. 7/10 is modified in Fig. 7/11 to show reference axes whose origin is attached to the reference point B as before, but which rotate with an absolute angular velocity which may be different from the absolute angular velocity of the body

A good background in the dynamics of plane motion is extremely useful in the study of three-dimensional dynamics, where the approach to problems and many of the terms are the same as or analogous to those in two dimensions. If the study of three-dimensional dynamics is undertaken without the benefit of prior study of plane-motion dynamics, more 7/1 Introduction Section A Kinematics 7/2 Translation 7/3 Fixed-Axis Rotation 7/4 Parallel-Plane Motion 7/5 Rotation about a Fixed Point 7/6 General Motion Section B Kinetics 7/7 Angular Momentum 7/8 Kinetic Energy 7/9 Momentum and Energy Equations of Motion 7/10 Parallel-Plane Motion 7/11 Gyroscopic Motion: Steady Precession 7/12 Chapter Review CHAPTER OUTLINE 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies c07.qxd 2/10/12 2:28 PM Page 513 time will be required to master the principles and to become familiar with the approach to problems.

Although a large percentage of dynamics problems in engineering can be solved by the principles of plane motion, modern developments have focused increasing attention on problems which call for the analysis of motion in three dimensions. Inclusion of the third dimension adds considerable complexity to the kinematic and kinetic relationships. Not only does the added dimension introduce a third component to vectors which represent force, linear velocity, linear acceleration, and linear momentum, but the introduction of the third dimension also adds the possibility of two additional components for vectors representing angular quantities including moments of forces, angular velocity, angular acceleration, and angular momentum. It is in three-dimensional motion that the full power of vector analysis is utilized.

Although the development in this article is for the case of rotation about a fixed point, we observe that rotation is a function solely of angular change, so that the expressions for and do not depend on the fixity of the point around which rotation occurs. Thus, rotation may take place independently of the linear motion of the rotation point. This conclusion is the three-dimensional counterpart of the concept of rotation of a rigid body in plane motion described in Art. 5/2 and used throughout Chapters 5 and 6.

Before applying Eq. 7/19, we introduce a set of coordinates which provide a natural description for our problem. These coordinates are gr k2p G x y z z y x (a) (b) (c) O θ z y x G Figure 7/19 c07.qxd 2/10/12 2:29 PM Page 561 shown in Fig. 7/20 for the example of rotation about a fixed point O. The axes X-Y-Z are fixed in space, and plane A contains the X-Y axes and the fixed point O on the rotor axis. Plane B contains point O and is always normal to the rotor axis. Angle measures the inclination of the rotor axis from the vertical Z-axis and is also a measure of the angle between planes A and B. The intersection of the two planes is the x-axis, which is located by the angle from the X-axis. The y-axis lies in plane B, and the z-axis coincides with the rotor axis. The angles and completely specify the position of the rotor axis. The angular displacement of the rotor with respect to axes x-y-z is specified by the angle measured from the x-axis to the x-axis, which is attached to the rotor. The spin velocity becomes p ˙.

But for the rigid body, vi i, where i is the relative velocity of mi with respect to G as seen from nonrotating axes. Thus, we may write where we have factored out from the first summation terms by reversing the order of the cross product and changing the sign. With the origin at the mass center G, the first term in HG is zero since Σmii 0. The second term with the substitution of dm for mi and for i gives (7/8) Before expanding the integrand of Eq. 7/8, we consider also the case of a rigid body rotating about a fixed point O, Fig. 7/12b. The x-y-z axes are attached to the body, and both body and axes have an angular velocity . The angular momentum about O was expressed in Art. 4/4 and is HO Σ(ri mivi), where, for the rigid body, vi ri. Thus, with the substitution of dm for mi and r for ri, the angular momentum is

Consider a body with axial symmetry, Fig. 7/19a, rotating about a fixed point O on its axis, which is taken to be the z-direction. With O as origin, the x- and y-axes automatically become principal axes of inertia along with the z-axis. This same description may be used for the rotation of a similar symmetrical body about its center of mass G, which is taken as the origin of coordinates as shown with the gimbaled gyroscope rotor of Fig. 7/19b. Again, the x- and y-axes are principal axes of inertia for point G. The same description may also be used to represent the rotation about the mass center of an axially symmetric body in space, such as the spacecraft in Fig. 7/19c. In each case, we note that, regardless of the rotation of the axes or of the body relative to the axes (spin about the z-axis), the moments of inertia about the x- and y-axes remain constant with time. The principal moments of inertia are again designated Izz I and Ixx Iyy I0. The products of inertia are, of course, zero.

Consider now the motion of a symmetrical rotor with no external moment about its mass center. Such motion is encountered with spacecraft and projectiles which both spin and precess during flight. Figure 7/21 represents such a body. Here the Z-axis, which has a fixed direction in space, is chosen to coincide with the direction of the angular momentum HG, which is constant since ΣMG 0. The x-y-z axes are attached in the manner described in Fig. 7/20. From Fig. 7/21 the three components of momentum are 0, HG sin , HG cos . From the defining relations, Eqs. 7/12, with the notation of this article, these components are also given by I0x, I0y, Iz. Thus, x x 0 so that is constant. This result means that the motion is one of steady precession about the constant HG vector. With no x-component, the angular velocity of the rotor lies in the y-z plane along with the Z-axis and makes an angle with the z-axis. The relationship between and is obtained from tan I0y/(Iz), which is (7/29) Thus, the angular velocity makes a constant angle with the spin axis. The rate of precession is easily obtained from Eq. 7/27 with M 0, which gives (7/30) It is clear from this relation that the direction of the precession depends on the relative magnitudes of the two moments of inertia. ˙ Ip (I0 I) cos tan I0 I tan HGy /HGz HGz HGy HGx HGz HGy HGx gr k2p ˙ I ˙p. ˙ mgr/(Ip) ˙ mgr I ˙p (I0 I) ˙2 cos mgr 564 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies β ω θ z ω ω y φ ⋅ HGz HGy HG Z y z x G Figure 7/21 c07.qxd 2/10/12 2:29 PM Page 564 Article 7/11 Gyroscopic Motion: Steady Precession 565 If I0 I, then , as indicated in Fig. 7/22a, and the precession is said to be direct. Here the body cone rolls on the outside of the space cone. If I I0, then , as indicated in Fig. 7/22b, and the precession is said to be retrograde. In this instance, the space cone is internal to the body cone, and and p have opposite signs. If I I0, then from Eq. 7/29, and Fig. 7/22 shows that both angles must be zero to be equal. For this case, the body has no precession and merely rotates with an angular velocity p. This condition occurs for a body with point symmetry, such as with a homogeneous sphere.

Consider now the rotation of a rigid body about a fixed axis n-n in space with an angular velocity , as shown in Fig. 7/2. The angular velocity is a vector in the direction of the rotation axis with a sense established by the familiar right-hand rule. For fixed-axis rotation, does not change its direction since it lies along the axis. We choose the origin O of the fixed coordinate system on the rotation axis for convenience. Any point such as A which is not on the axis moves in a circular arc in a plane normal to the axis and has a velocity (7/1) which may be seen by replacing r by h b and noting that h 0. The acceleration of A is given by the time derivative of Eq. 7/1. Thus, a ˙ r ( r) (7/2) v r rA rB rA/B vA vB aA aB z y x B A rA rB rA/B vA vB Figure 7/1 c07.qxd 2/10/12 2:28 PM Page 514 where has been replaced by its equal, v r. The normal and tangential components of a for the circular motion have the familiar magnitudes an ( r) b2 and at b, where Inasmuch as both v and a are perpendicular to and it follows that 0, 0, 0, and 0 for fixed-axis rotation.

Consider now the steady precession of a symmetrical top, Fig. 7/18, spinning about its axis with a high angular velocity p and supported at its point O. Here the spin axis makes an angle with the vertical Z-axis around which precession occurs. Again, we will neglect the small angular-momentum component due to the precession and consider H equal to Ip, the angular momentum about the axis of the top associated with the spin only. The moment about O is due to the weight and is sin , where is the distance from O to the mass center G. From the diagram, we see that the angular-momentum vector HO has a change dHO MO dt in the direction of MO during time dt and that is unchanged. The increment in precessional angle around the Z-axis is Substituting the values MO sin and d /dt gives mgr sin Ip sin or mgr Ip mgr d MO dt Ip sin mgr r 560 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies Ω Ω Ω p M Figure 7/16 Ω Ω M Ω p Ω p M Ω Ω y dH H Hy H′ Hz M dt = dH z x dψ Ω = ψ · j Figure 7/17 dHO = MO dt HO ≈ Ip θ θ z Z x X MO mg O G y Y dψ r _ Figure 7/18 c07.qxd 2/10/12 2:29 PM Page 560 Article 7/11 Gyroscopic Motion: Steady Precession 561 which is independent of . Introducing the radius of gyration so that I mk2 and solving for the precessional velocity give (7/25) Unlike Eq. 7/24, which is an exact description for the rotor of Fig. 7/17 with precession confined to the x-z plane, Eq. 7/25 is an approximation based on the assumption that the angular momentum associated with is negligible compared with that associated with p. We will see the amount of the error associated with this approximation when we reconsider steady-state precession later in this article. On the basis of our analysis, the top will have a steady precession at the constant angle only if it is set in motion with a value of which satisfies Eq. 7/25. When these conditions are not met, the precession becomes unsteady, and may oscillate with an amplitude which increases as the spin velocity decreases. The corresponding rise and fall of the rotation axis is called nutation.

Figure 7/1 shows a rigid body translating in three-dimensional space. Any two points in the body, such as A and B, will move along parallel straight lines if the motion is one of rectilinear translation or will move along congruent curves if the motion is one of curvilinear translation. In either case, every line in the body, such as AB, remains parallel to its original position

Figure 7/10 shows a rigid body which has an angular velocity . We may choose any convenient point B as the origin of a translating reference system x-y-z. The velocity v and acceleration a of any other point A in the body are given by the relative-velocity and relative-acceleration expressions [5/4] [5/7] which were developed in Arts. 5/4 and 5/6 for the plane motion of rigid bodies. These expressions also hold in three dimensions, where the three vectors for each of the equations are also coplanar

Figure 7/14 shows a symmetrical rotor spinning about the z-axis with a large angular velocity p, known as the spin velocity. If we apply two forces F to the rotor axle to form a couple M whose vector is directed along the x-axis, we will find that the rotor shaft rotates in the xz plane about the y-axis in the sense indicated, with a relatively slow angular velocity known as the precession velocity. Thus, we identify the spin axis (p), the torque axis (M), and the precession axis (), where the usual right-hand rule identifies the sense of the rotation vectors. The rotor shaft does not turn about the x-axis in the sense of M, as it would if the rotor were not spinning. To aid understanding of this phenomenon, a direct analogy may be made between the rotation vectors and the familiar vectors which describe the curvilinear motion of a particle.

Figure 7/15a shows a particle of mass m moving in the x-z plane with constant speed v v. The application of a force F normal to its linear momentum G mv causes a change dG d(mv) in its momentum. We see that dG, and thus dv, is a vector in the direction of the normal force F according to Newton's second law F G˙, which may be written as ˙ Spin axis Precession axis Torque axis y F F M p z x Ω ψ ψ = · ⏐ ⏐ Figure 7/14 558 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies c07.qxd 2/10/12 2:29 PM Page 558 Article 7/11 Gyroscopic Motion: Steady Precession 559 F dt dG. From Fig. 7/15b we see that, in the limit, tan d d F dt/mv or F In vector notation with the force becomes which is the vector equivalent of our familiar scalar relation Fn man for the normal force on the particle, as treated extensively in Chapter 3. With these relations in mind, we now turn to our problem of rotation. Recall now the analogous equation M which we developed for any prescribed mass system, rigid or nonrigid, referred to its mass center (Eq. 4/9) or to a fixed point O (Eq. 4/7). We now apply this relation to our symmetrical rotor, as shown in Fig. 7/15c. For a high rate of spin p and a low precession rate about the y-axis, the angular momentum is represented by the vector H Ip, where I Izz is the moment of inertia of the rotor about the spin axis.

Figure 7/17 shows our same rotor again. Because it has a moment of inertia about the y-axis and an angular velocity of precession about this axis, there will be an additional component of angular momentum about the y-axis. Thus, we have the two components Hz Ip and Hy I0, where I0 stands for Iyy and, again, I stands for Izz. The total angular momentum is H as shown. The change in H remains dH M dt as previously, and the precession during time dt is the angle d M dt/Hz M dt/(Ip) as before. Thus, Eq. 7/24 is still valid and for steady precession is an exact description of the motion as long as the spin axis is perpendicular to the axis around which precession occurs

For a rigid body, the relative term becomes the kinetic energy due to rotation about the mass center. Because is the velocity of the representative particle with respect to the mass center, then for the rigid body we may write it as i, where is the angular velocity of the body. With this substitution, the relative term in the kinetic energy expression becomes Σ1 2mi ˙i 2 Σ1 2mi ( i )( i ) ˙i ˙i ˙ r v 1 2mv 2 1 2m˙ r ˙ r 1 2v G v T 1 2mv 2 Σ1 2mi ˙i 2 HP HG r G mv 542 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies P G HG G = mv r Figure 7/13 c07.qxd 2/10/12 2:29 PM Page 542 Article 7/8 Kinetic Energy 543 If we use the fact that the dot and the cross may be interchanged in the triple scalar product, that is, P R, we may write Because is the same factor in all terms of the summation, it may be factored out to give where HG is the same as the integral expressed by Eq. 7/8. Thus, the general expression for the kinetic energy of a rigid body moving with mass-center velocity and angular velocity is (7/15) Expansion of this vector equation by substitution of the expression for HG written from Eq. 7/11 yields (7/16) If the axes coincide with the principal axes of inertia, the kinetic energy is merely (7/17) When a rigid body is pivoted about a fixed point O or when there is a point O in the body which momentarily has zero velocity, the kinetic energy is T This expression reduces to (7/18) where HO is the angular momentum about O, as may be seen by replacing i in the previous derivation by ri, the position vector from O. Equations 7/15 and 7/18 are the three-dimensional counterparts of Eqs. 6/9 and 6/8 for plane motion.

If a succession of photographs were taken, we would observe in each photograph that the rotation axis would be defined by a new series of sharply-defined dots and that the axis would change position both in space and relative to the body. For rotation of a rigid body about a fixed point, then, it is seen that the rotation axis is, in general, not a line fixed in the body.

If points A and B in Fig. 7/10 represent the ends of a rigid control link in a spatial mechanism where the end connections act as ball-andsocket joints (as in Sample Problem 7/3), it is necessary to impose certain kinematic requirements. Clearly, any rotation of the link about its own axis AB does not affect the action of the link. Thus, the angular velocity n whose vector is normal to the link describes its action. It is necessary, therefore, that n and rA/B be at right angles, and this condition is satisfied if 0

If the coordinate axes coincide with the principal axes of inertia, Eq. 7/11 for the angular momentum about the mass center or about a fixed point becomes (7/13) It is always possible to locate the principal axes of inertia for a general three-dimensional rigid body. Thus, we can express its angular momentum by Eq. 7/13, although it may not always be convenient to do so H Ixxxi Iyyy j Izzzk *See, for example, the first author's Dynamics, 2nd Edition, SI Version, 1975, John Wiley & Sons, Art. 41. c07.qxd 2/10/12 2:29 PM Page 541 for geometric reasons. Except when the body rotates about one of the principal axes of inertia or when Ixx Iyy Izz, the vectors H and have different directions.

If we use Fig. 7/2 to represent a rigid body rotating about a fixed point O with the instantaneous axis of rotation n-n, we see that the ˙. Article 7/5 Rotation about a Fixed Point 517 ω = r = × r r = constant v ω α = = × = constant Ω Space cone Rigid cone O A r O ω ω Ω ω ω · · ⏐ ⏐ ⏐ ⏐ Figure 7/9 O Body p cone Space cone ω α = ω ⋅ Figure 7/8 A n ω O Space cone Body cone Figure 7/7 c07.qxd 2/10/12 2:28 PM Page 517 velocity v and acceleration a of any point A in the body are given by the same expressions as apply to the case in which the axis is fixed, namely, [7/1] [7/2] The one difference between the case of rotation about a fixed axis and rotation about a fixed point lies in the fact that for rotation about a fixed point, the angular acceleration will have a component normal to due to the change in direction of , as well as a component in the direction of to reflect any change in the magnitude of . Although any point on the rotation axis n-n momentarily will have zero velocity, it will not have zero acceleration as long as is changing its direction. On the other hand, for rotation about a fixed axis, has only the one component along the fixed axis to reflect the change in the magnitude of . Furthermore, points which lie on the fixed rotation axis clearly have no velocity or acceleration.

In Art. 4/3 on the dynamics of systems of particles, we developed the expression for the kinetic energy T of any general system of mass, rigid or nonrigid, and obtained the result [4/4] where is the velocity of the mass center and i is the position vector of a representative element of mass mi with respect to the mass center. We identified the first term as the kinetic energy due to the translation of the system and the second term as the kinetic energy associated with the motion relative to the mass center. The translational term may be written alternatively as where is the velocity of the mass center and G is the linear momentum of the body

In Art. 5/7 we also developed the relationship (Eq. 5/13) between the time derivative of a vector V as measured in the fixed X-Y system and the time derivative of V as measured relative to the rotating x-y system. For our three-dimensional case, this relation becomes (7/7) When we apply this transformation to the relative-position vector rA/B rA rB for our rigid body of Fig. 7/11, we obtain or which gives us the first of Eqs. 7/6. Equations 7/6 are particularly useful when the reference axes are attached to a moving body within which relative motion occurs. Equation 7/7 may be recast as the vector operator (7/7a) where [ ] stands for any vector V expressible both in X-Y-Z and in x-yz. If we apply the operator to itself, we obtain the second time derivative, which becomes (7/7b) This exercise is left to the student. Note that the form of Eq. 7/7b is the same as that of the second of Eqs. 7/6 expressed for aA/B aA aB.

In Art. 7/7 it was mentioned that, in general, for any origin fixed to a rigid body, there are three principal axes of inertia with respect to which the products of inertia vanish. If the reference axes coincide with the principal axes of inertia with origin at the mass center G or at a point O fixed to the body and fixed in space, the factors Ixy, Iyz, Ixz will be zero, and Eqs. 7/20 become (7/21) These relations, known as Euler's equations,* are extremely useful in the study of rigid-body motion.

In Fig. 7/15d we see that during time dt the angular-momentum vector Ip has swung through the angle d , so that in the limit with tan d d , we have Substituting d /dt for the magnitude of the precession velocity gives us (7/24) We note that M, , and p as vectors are mutually perpendicular, and that their vector relationship may be represented by writing the equation in the cross-product form (7/24a) which is completely analogous to the foregoing relation F m v for the curvilinear motion of a particle as developed from Figs. 7/15a and b. M I p M Ip d M dt Ip or M I d dt p H˙, H˙ F m v mv˙. ˙j, F (F = G) x y z dG = d(mv) G = mv (a) = j x y z mv F dt = d(mv) (b) M dθ ω x y z M dt = d(Ip) Ip (d) (c) dψ (M = H) y H = Ip dH = d(Ip) z x Ω = ψ · j · · θ · Figure 7/15 c07.qxd 2/10/12 2:29 PM Page 559 Equations 7/24 and 7/24a apply to moments taken about the mass center or about a fixed point on the axis of rotation.

In the foregoing discussion of gyroscopic motion, it was assumed that the spin was large and the precession was small. Although we can see from Eq. 7/24 that for given values of I and M, the precession must be small if p is large, let us now examine the influence of on the momentum relations. Again, we restrict our attention to steady precession, where has a constant magnitude

Infinitesimal rotations, however, do obey the parallelogram law of vector addition. This fact is shown in Fig. 7/5, which represents the combined effect of two infinitesimal rotations d1 and d2 of a rigid body about the respective axes through the fixed point O. As a result of d1, point A has a displacement d1 r, and likewise d2 causes a displacement d2 r of point A. Either order of addition of these infinitesimal displacements clearly produces the same resultant displacement, which is d1 r d2 r (d1 d2) r. Thus, the two rotations are equivalent to the single rotation d d1 d2. It follows that the angular velocities 1 1 and 2 2 may be added vectorially to give 1 2. We conclude, therefore, that at any instant of time a body with one fixed point is rotating instantaneously about a particular axis passing through the fixed point.

Initially, we neglect the small component of angular momentum about the y-axis which accompanies the slow precession. The application of the couple M normal to H causes a change dH d(Ip) in the angular momentum. We see that dH, and thus dp, is a vector in the direction of the couple M since M which may also be written M dt dH. Just as the change in the linear-momentum vector of the particle is in the direction of the applied force, so is the change in the angular-momentum vector of the gyro in the direction of the couple. Thus, we see that the vectors M, H, and dH are analogous to the vectors F, G, and dG. With this insight, it is no longer strange to see the rotation vector undergo a change which is in the direction of M, thereby causing the axis of the rotor to precess about the y-axis.

Relative to the plastic cylinder of Fig. 7/6, the instantaneous axis of rotation O-A-n generates a right-circular cone about the cylinder axis called the body cone. As the two rotations continue and the cylinder swings around the vertical axis, the instantaneous axis of rotation also generates a right-circular cone about the vertical axis called the space cone. These cones are shown in Fig. 7/7 for this particular example.

Similarly, it is only the component n* of the angular acceleration of the link normal to AB which affects its action, so that 0 must also hold.

Substitution of the angular-velocity and angular-momentum components into Eq. 7/19 yields (7/26) Equations 7/26 are the general equations of rotation of a symmetrical body about either a fixed point O or the mass center G. In a given problem, the solution to the equations will depend on the moment sums applied to the body about the three coordinate axes. We will confine our use of these equations to two particular cases of rotation about a point which are described in the following sections.

The angular acceleration of a rigid body in three-dimensional motion is the time derivative of its angular velocity, In contrast to the case of rotation in a single plane where the scalar measures only the change in magnitude of the angular velocity, in three-dimensional motion the vector reflects the change in direction of as well as its change in magnitude. Thus in Fig. 7/8 where the tip of the angular velocity vector follows the space curve p and changes in both magnitude and direction, the angular acceleration becomes a vector tangent to this curve in the direction of the change in

The array of moments and products of inertia which appear in Eq. 7/12 is called the inertia matrix or inertia tensor. As we change the orientation of the axes relative to the body, the moments and products of inertia will also change in value. It can be shown* that there is one unique orientation of axes x-y-z for a given origin for which the products of inertia vanish and the moments of inertia Ixx, Iyy, Izz take on stationary values. For this orientation, the inertia matrix takes the form and is said to be diagonalized. The axes x-y-z for which the products of inertia vanish are called the principal axes of inertia, and Ixx, Iyy, and Izz are called the principal moments of inertia. The principal moments of inertia for a given origin represent the maximum, the minimum, and an intermediate value of the moments of inertia.

The correct spatial relationship among the three vectors may be remembered from the fact that dH, and thus dp, is in the direction of M, which establishes the correct sense for the precession . Therefore, the spin vector p always tends to rotate toward the torque vector M. Figure 7/16 represents three orientations of the three vectors which are consistent with their correct order. Unless we establish this order correctly in a given problem, we are likely to arrive at a conclusion directly opposite to the correct one. Remember that Eq. 7/24, like F ma and M I, is an equation of motion, so that the couple M represents the couple due to all forces acting on the rotor, as disclosed by a correct free-body diagram of the rotor. Also note that, when a rotor is forced to precess, as occurs with the turbine in a ship which is executing a turn, the motion will generate a gyroscopic couple M which obeys Eq. 7/24a in both magnitude and sense

The gyroscope has important engineering applications. With a mounting in gimbal rings (see Fig. 7/19b), the gyro is free from external moments, and its axis will retain a fixed direction in space regardless of the rotation of the structure to which it is attached. In this way, the gyro is used for inertial guidance systems and other directional control devices. With the addition of a pendulous mass to the inner gimbal ring, the earth's rotation causes the gyro to precess so that the spin axis will always point north, and this action forms the basis of the gyro compass. The gyroscope has also found important use as a stabilizing device. The controlled precession of a large gyro mounted in a ship is used to produce a gyroscopic moment to counteract the rolling of a ship at sea. The gyroscopic effect is also an extremely important consideration in the design of bearings for the shafts of rotors which are subjected to forced precessions

The position vectors and their first and second time derivatives are where rA/B remains constant, and therefore its time derivative is zero. Thus, all points in the body have the same velocity and the same acceleration. The kinematics of translation presents no special difficulty, and further elaboration is unnecessary.

The resultant of all external forces acting on a rigid body may be replaced by the resultant force ΣF acting through the mass center and a resultant couple ΣMG acting about the mass center. Work is done by the resultant force and the resultant couple at the respective rates and where is the linear velocity of the mass center and is the angular velocity of the body. Integration over the time from condition 1 to condition 2 gives the total work done during the time interval. Equating the works done to the respective changes in kinetic energy as expressed in Eq. 7/15 gives (7/22) These equations express the change in translational kinetic energy and the change in rotational kinetic energy, respectively, for the interval during which ΣF or ΣMG acts, and the sum of the two expressions equals T.

The selection of the reference point B is quite arbitrary in theory. In practice, point B is chosen for convenience as some point in the body whose motion is known in whole or in part. If point A is chosen as the reference point, the relative-motion equations become where rB/A rA/B. It should be clear that and, thus, are the same vectors for either formulation since the absolute angular motion of the body is independent of the choice of reference point. When we come to ˙ aB aA ˙ rB/A ( rB/A) vB vA rB/A ˙ aA aB ˙ rA/B ( rA/B) vA vB rA/B AB aA aB aA/B vA vB vA/B Z ω rA rB rA/B z Y A B O y x X Figure 7/10 This time-lapse photo of a VTOL aircraft shows a three-dimensional combination of translation and rotation. © Jim Sugar/CORBIS c07.qxd 2/10/12 2:29 PM Page 527 528 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies Z rA rB rA/B Y A B O X ω Ω (Axes) (Body) z y x Figure 7/11 the kinetic equations for general motion, we will see that the mass center of a body is frequently the most convenient reference point to choose.

The treatment presented in Chapter 7 is not intended as a complete development of the three-dimensional motion of rigid bodies but merely as a basic introduction to the subject. This introduction should, however, be sufficient to solve many of the more common problems in threedimensional motion and also to lay the foundation for more advanced study. We will proceed as we did for particle motion and for rigid-body plane motion by first examining the necessary kinematics and then proceeding to the kinetics

The work-energy relationship, developed in Chapter 4 for a general system of particles and given by [4/3] was used in Chapter 6 for rigid bodies in plane motion. The equation is equally applicable to rigid-body motion in three dimensions. As we have seen previously, the work-energy approach is of great advantage when we analyze the initial and final end-point conditions of motion. Here the work done during the interval by all active forces external to the body or system is equated to the sum of the corresponding changes in kinetic energy T and potential energy V. The potentialenergy change is determined in the usual way, as described previously in Art. 3/7. We will limit our application of the equations developed in this article to two problems of special interest, parallel-plane motion and gyroscopic motion, discussed in the next two articles.

The x-y-z axes here are taken as fixed in space and do not rotate with the body. In part a of the figure, two successive 90 rotations of the sphere about, first, the x-axis and, second, the y-axis result in the motion of a point which is initially on the y-axis in position 1, to positions 2 v v ˙ a a˙ ˙, ˙ r ˙. ˙ r Article 7/5 Rotation about a Fixed Point 515 ω n n O A n n A b h r v= ω × r ω a t = ω ⋅ × r a z y x Fixed axis an = ω × ( ω × r) ω ⋅ = α Figure 7/2 P A G A′ x z y x′ y′ Figure 7/3 c07.qxd 2/10/12 2:28 PM Page 515 and 3, successively. On the other hand, if the order of the rotations is reversed, the point undergoes no motion during the y-rotation but moves to point 3 during the 90 rotation about the x-axis. Thus, the two cases do not produce the same final position, and it is evident from this one special example that finite rotations do not generally obey the parallelogram law of vector addition and are not commutative. Thus, finite rotations may not be treated as proper vectors.

These vectors have properties analogous to those of a force and a couple. Thus, the angular momentum about any point P equals the free vector HG plus the moment of the linear-momentum vector G about P. Therefore, we may write (7/14) This relation, which was derived previously in Chapter 4 as Eq. 4/10, also applies to a fixed point O on the body or body extended, where O merely replaces P. Equation 7/14 constitutes a transfer theorem for angular momentum.

To aid in visualizing the concept of the instantaneous axis of rotation, we will cite a specific example. Figure 7/6 represents a solid cylindrical rotor made of clear plastic containing many black particles embedded in the plastic. The rotor is spinning about its shaft axis at the steady rate 1, and its shaft, in turn, is rotating about the fixed vertical axis at the steady rate 2, with rotations in the directions indicated. If the rotor is photographed at a certain instant during its motion, the resulting picture would show one line of black dots sharply defined, indicating that, momentarily, their velocity was zero. This line of points with no velocity establishes the instantaneous position of the axis of rotation O-n. Any dot on this line, such as A, would have equal and opposite velocity components, v1 due to 1 and v2 due to 2. All other dots, ˙ ˙ ˙ 516 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies 3 1 1,2 3 x y z x y z 2 O O x then y θ θ y then x θ θ (a) (b) Figure 7/4 ω = ⋅ θ θ d 1 θ d 1 θ d 2 θ d 2 θ d θ O A r d θ d 1 × θ r d θ × r d 2 × θ r Figure 7/5 v1 v2 A n P O ω 1 ω 2 Figure 7/6 c07.qxd 2/10/12 2:28 PM Page 516 such as the one at P, would appear blurred, and their movements would show as short streaks in the form of small circular arcs in planes normal to the axis O-n. Thus, all particles of the body, except those on line O-n, are momentarily rotating in circular arcs about the instantaneous axis of rotation

We must first examine the conditions under which rotation vectors obey the parallelogram law of addition and may, therefore, be treated as proper vectors. Consider a solid sphere, Fig. 7/4, which is cut from a rigid body confined to rotate about the fixed point O.

We now apply Eq. 7/19 to a rigid body where the coordinate axes are attached to the body. Under these conditions, when expressed in the x-y-z coordinates, the moments and products of inertia are invariant with time, (H˙z Hxy Hyx)k (H˙y Hzx Hxz)j ΣM (H˙x Hyz Hzy)i H˙ (H˙xi H˙y j H˙zk) H ΣM dH dt xyz H H˙, ΣM H˙ ΣF G˙ c07.qxd 2/10/12 6:36 PM Page 550 Article 7/9 Momentum and Energy Equations of Motion 551 and . Thus, for axes attached to the body, the three scalar components of Eq. 7/19 become (7/20) Equations 7/20 are the general moment equations for rigid-body motion with axes attached to the body. They hold with respect to axes through a fixed point O or through the mass center G.

We now consider a rigid body moving with any general motion in space, Fig. 7/12a. Axes x-y-z are attached to the body with origin at the mass center G. Thus, the angular velocity of the body becomes the angular velocity of the x-y-z axes as observed from the fixed reference axes X-Y-Z. The absolute angular momentum HG of the body about its mass center G is the sum of the moments about G of the linear momenta of all elements of the body and was expressed in Art. 4/4 as HG Σ(i mivi), where vi is the absolute velocity of the mass element m

We now examine the conditions under which the rotor precesses at a steady rate at a constant angle and with constant spin velocity p. Thus, and Eqs. 7/26 become (7/27) From these results, we see that the required moment acting on the rotor about O (or about G) must be in the x-direction since the y- and z-components are zero. Furthermore, with the constant values of , and p, the moment is constant in magnitude. It is also important to note that the moment axis is perpendicular to the plane defined by the precession axis (Z-axis) and the spin axis (z-axis). We may also obtain Eqs. 7/27 by recognizing that the components of H remain constant as observed in x-y-z so that 0. Because in general ΣM H, we have for the case of steady precession (7/28) which reduces to Eqs. 7/27 upon substitution of the values of and H. By far the most common engineering examples of gyroscopic motion occur when precession takes place about an axis which is normal to the rotor axis, as in Fig. 7/14. Thus with the substitution /2, z p, , and ΣMx M, we have from Eqs. 7/27 M Ip [7/24] ˙ ΣM H (H˙)xyz (H˙)xyz ˙, ΣMz 0 ΣMy 0 ΣMx ˙ sin [I( ˙ cos p) I0 ˙ cos ] p constant, ˙ p 0 constant, ˙ ¨ 0 ˙ constant, ¨ 0 ˙ ΣMz I d dt ( ˙ cos p) ΣMy I0( ¨ sin 2 ˙˙ cos ) I ˙( ˙ cos p) ΣMx I0( ¨ ˙2 sin cos ) I ˙( ˙ cos p) sin c07.qxd 2/10/12 2:29 PM Page 563 which we derived initially in this article from a direct analysis of this special case. Now let us examine the steady precession of the rotor (symmetrical top) of Fig. 7/20 for any constant value of other than /2. The moment ΣMx about the x-axis is due to the weight of the rotor and is sin . Substitution into Eqs. 7/27 and rearrangement of terms give us We see that is small when p is large, so that the second term on the right-hand side of the equation becomes very small compared with If we neglect this smaller term, we have which, upon use of the previous substitution and mk2 I, becomes [7/25] We derived this same relation earlier by assuming that the angular momentum was entirely along the spin axis.

We now make direct use of Eq. 7/19, which is the general angularmomentum equation for a rigid body, by applying it to a body spinning about its axis of rotational symmetry. This equation is valid for rotation about a fixed point or for rotation about the mass center. A spinning top, the rotor of a gyroscope, and a spacecraft are examples of bodies whose motions can be described by the equations for rotation about a point. The general moment equations for this class of problems are fairly complex, and their complete solutions involve the use of elliptic integrals and somewhat lengthy computations. However, a large fraction of engineering problems where the motion is one of rotation about a point involves the steady precession of bodies of revolution which are spinning about their axes of symmetry. These conditions greatly simplify the equations and thus facilitate their solution

We now make use of Eqs. 5/11, 5/12, 5/13, and 5/14 developed in Art. 5/7 for describing the plane motion of a rigid body with the use of rotating axes. The extension of these relations from two to three dimensions is easily accomplished by merely including the z-component of the vectors, and this step is left to the student to carry out. Replacing in these equations by the angular velocity of our rotating x-y-z axes gives us (7/5) for the time derivatives of the rotating unit vectors attached to x-y-z. The expression for the velocity and acceleration of point A become (7/6) where vrel and arel are, respectively, the velocity and acceleration of point A measured relative to x-y-z by an observer attached to x-y-z. We again note that is the angular velocity of the axes and may be different from the angular velocity of the body. Also we note that rA/B remains constant in magnitude for points A and B fixed to a rigid body, but it will change direction with respect to x-y-z when the angular velocity of the axes is different from the angular velocity of the body. We observe ˙xi ˙yj ˙zk ¨xi ¨yj ¨zk aA aB ˙ rA/B ( rA/B) 2 vrel arel vA vB rA/B vrel ˙i i ˙j j k˙ k n rA/B n rA/B *It may be shown that n if the angular velocity of the link about its own axis is not changing. See the first author's Dynamics, 2nd Edition, SI Version, 1975, John Wiley & Sons, Art. 37. ˙n c07.qxd 2/10/12 2:29 PM Page 528 Article 7/6 General Motion 529 further that, if x-y-z are rigidly attached to the body, and vrel and arel are both zero, which makes the equations identical to Eqs. 7/4.

We see that the body cone rolls on the space cone and that the angular velocity of the body is a vector which lies along the common element of the two cones. For a more general case where the rotations are not steady, the space and body cones are not right-circular cones, Fig. 7/8, but the body cone still rolls on the space cone.

When all particles of a rigid body move in planes which are parallel to a fixed plane, the body has a general form of plane motion, as described in Art. 7/4 and pictured in Fig. 7/3. Every line in such a body which is normal to the fixed plane remains parallel to itself at all times. We take the mass center G as the origin of coordinates x-y-z which are attached to the body, with the x-y plane coinciding with the plane of motion P. The components of the angular velocity of both the body and the attached axes become x y 0, z 0. For this case, the angularmomentum components from Eq. 7/12 become and the moment relations of Eqs. 7/20 reduce to (7/23) We see that the third moment equation is equivalent to the second of Eqs. 6/1, where the z-axis passes through the mass center, or to Eq. 6/4 if the z-axis passes through a fixed point O. Equations 7/23 hold for an origin of coordinates at the mass center, as shown in Fig. 7/3, or for any origin on a fixed axis of rotation. The three independent force equations of motion which also apply to parallelplane motion are clearly Equations 7/23 find special use in describing the effect of dynamic imbalance in rotating machinery and in rolling bodies.

When all points in a rigid body move in planes which are parallel to a fixed plane P, Fig. 7/3, we have a general form of plane motion. The reference plane is customarily taken through the mass center G and is called the plane of motion. Because each point in the body, such as A, has a motion identical with the motion of the corresponding point (A) in plane P, it follows that the kinematics of plane motion covered in Chapter 5 provides a complete description of the motion when applied to the reference plane

When the magnitude of remains constant, the angular acceleration is normal to . For this case, if we let stand for the angular velocity with which the vector itself rotates (precesses) as it forms the space cone, the angular acceleration may be written (7/3) This relation is easily seen from Fig. 7/9. The upper part of the figure relates the velocity of a point A on a rigid body to its position vector from O and the angular velocity of the body. The vectors , , and in the lower figure bear exactly the same relationship to each other as do the vectors v, r, and in the upper figure

With the description of angular momentum, inertial properties, and kinetic energy of a rigid body established in the previous two articles, we are ready to apply the general momentum and energy equations of motion. Momentum Equations In Art. 4/4 of Chapter 4, we established the general linear- and angular-momentum equations for a system of constant mass. These equations are [4/6] [4/7] or [4/9] The general moment relation, Eq. 4/7 or 4/9, is expressed here by the single equation ΣM where the terms are taken either about a fixed point O or about the mass center G. In the derivation of the moment principle, the derivative of H was taken with respect to an absolute coordinate system. When H is expressed in terms of components measured relative to a moving coordinate system x-y-z which has an angular velocity , then by Eq. 7/7 the moment relation becomes The terms in parentheses represent that part of due to the change in magnitude of the components of H, and the cross-product term represents that part due to the changes in direction of the components of H. Expansion of the cross product and rearrangement of terms give (7/19) Equation 7/19 is the most general form of the moment equation about a fixed point O or about the mass center G. The 's are the angular velocity components of rotation of the reference axes, and the Hcomponents in the case of a rigid body are as defined in Eq. 7/12, where the 's are the components of the angular velocity of the body

ree vectors for each of the equations are also coplanar. In applying these relations to rigid-body motion in space, we note from Fig. 7/10 that the distance remains constant. Thus, from an observer's position on x-y-z, the body appears to rotate about the point B and point A appears to lie on a spherical surface with B as the center. Consequently, we may view the general motion as a translation of the body with the motion of B plus a rotation of the body about B.

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