482 Dynamics of Mechanical Systems Equation (14.2.15) shows that the small disturbance gets larger and larger. Thus, θ = π is an unstable equilibrium position. In the next several sections, we will use the foregoing technique to explore the stability of several other mechanical systems. 14.3 A Particle Moving in a Vertical Rotating Tube Consider the system consisting of a particle free to move in the smooth interior of a vertical rotating tube as depicted in Figure 14.3.1 (we considered the kinematics and dynamics of this system in Section 8.5). If the angular speed Ω of the tube is prescribed, the system has one degree of freedom represented by the angle θ . From Eq. (8.5.15) we see that the equation of motion is then: (14.3.1) where r is the tube radius. If the particle P has reached an equilibrium position, will be zero. The equilibrium angle will then satisfy the equation: (14.3.2) It is readily seen that the solutions to this equation are: (14.3.3) Thus, there are three equilibrium positions. In the following paragraphs, we consider the stability of each of these. Case 1: θθ θθ = θθ θθ 1 = 0 Consider first the equilibrium position θ = 0. By introducing a small disturbance θ * about θ = 0, we have: (14.3.4) FIGURE 14.3.1 A vertical rotating tube with a smooth interior surface and containing a particle P . ˙˙ sin cos sinθθθ θ−+ () =Ω 2 0gr ˙˙ θ −+ () =Ω 2 0sin cos sinθθ θgr θθ θθ π θθ== == == () − 12 3 12 0, , cosand grΩ θθ=+0 * n O r n n R n θ P(m) 3 2 r 1 n n Ω θ 0593_C14_fm Page 482 Tuesday, May 7, 2002 6:56 AM Stability 483 Then, to the first order in θ * , sin θ and cos θ may be approximated as: and (14.3.5) By substituting from Eqs. (14.3.4) and (14.3.5) into (14.3.1), we obtain: (14.3.6) By referring to the solutions of Eqs. (14.2.5) and (14.2.12), we see that if [( g / r ) – Ω 2 ] is positive, the solution of Eq. (14.3.6) may be expressed in terms of trigonometric functions and thus will be bounded and stable. Alternatively, if [ g / r – Ω 2 ] is negative, the solution of Eq. (14.3.6) will be expressed in terms of exponential or hyperbolic functions and thus will be unbounded and unstable. Hence, the equilibrium position θ = 0 is stable if: (14.3.7) Case 2: θθ θθ = θθ θθ 2 = 0 Consider next the equilibrium position θ = π. A small disturbance about θ = π may be expressed as: (14.3.8) Then, sinθ and cosθ may be approximated to the first order in θ * as: (14.3.9) By substituting from Eqs. (14.3.8) and (14.3.9) into (14.3.1) we obtain: (14.3.10) In this case, the coefficient of θ * is negative for all values of Ω. Therefore, the solution of Eq. (14.3.10) will involve exponential or hyperbolic functions; thus, the equilibrium posi- tion is unstable. Case 3: θθ θθ = θθ θθ 3 = cos –1 (g/rΩΩ ΩΩ 2 ) Finally, consider the equilibrium position θ = cos –1 (g/rΩ 2 ). Observe that this equilibrium position will not exist unless g/rΩ 2 is smaller than 1. That is, Ω 2 must be greater than g/r for equilibrium at θ = cos –1 (g/rΩ 2 ). sin sin ** θθθ=+ () =0 cos cos * θθ=+ () =01 ˙˙ ** θθ+ () = [] =gr Ω 2 0 gr gr><ΩΩ 22 or θπθ=+ * sin sin cos cos ** * θπθθ θπθ =+ () =− =+ () =−1 ˙˙ ** θθ−+ () [] =Ω 2 0gr 0593_C14_fm Page 483 Tuesday, May 7, 2002 6:56 AM 484 Dynamics of Mechanical Systems If Ω 2 > g/r, a small disturbance about the equilibrium position may be expressed as: (14.3.11) Then, to the first order in θ * , sinθ and cosθ and may be approximated as: and (14.3.12) By substituting from Eqs. (14.3.11) and (14.3.12) into (14.3.1) we obtain: (14.3.13) Because the right side of Eq. (14.3.13) is a constant, the stability (or instability) of the equilibrium position is determined by the sign of the coefficient of θ * . Specifically, the equilibrium position is stable if the term [Ω 2 (sin 2 θ 3 – cos 2 θ 3 ) + (g/r)cosθ 3 ] is positive. If the term is negative, the equilibrium position is unstable. From the definition of θ 3 in Eq. (14.3.3), we see that: and (14.3.14) Hence, the coefficient of θ * in Eq. (14.3.13) becomes: (14.3.15) Therefore, the equilibrium position is stable if [1 – (g/rΩ 2 ) 2 ] is positive or if: (14.3.16) Comparing the inequality of Eq. (14.3.16) with that of Eq. (14.3.7) we see that they are opposite. Also, as noted earlier, we see that Eq. (14.3.16) is a necessary condition for the existence of the equilibrium position θ = θ 3 = cos –1 (g/rΩ 2 ). That is, for slow tube rotation θθ θ=+ 3 * sin sin sin cos ** θθθ θθθ=+ () =+ 333 cos cos cos sin ** θθθ θθθ=+ () =− 333 ˙˙ sin cos cos sin cos sin ** θθθ θθ θθ θ +− () + () [] =− () Ω Ω 22 3 2 33 2 33 3 gr gr cos , cosθθ 3 22 3 224 ==gr g rΩΩ sin cos 2 3 2 3 224 11θθ=− =− () grΩ ΩΩ Ω Ω ΩΩ Ω ΩΩ 22 3 22 33 2224 2224 222 22 2 1 1 sin cos cosθθ θ−+ () =− () [] − () + () =− () [] gr g r gr gr gr Ω 2 > gr 0593_C14_fm Page 484 Tuesday, May 7, 2002 6:56 AM Stability 485 (specifically, Ω 2 < g/r), there are only two equilibrium positions: θ = θ 1 = 0 and θ = θ 2 = π, with θ = θ 1 = 0 being stable (see Eq. (14.3.7)) and θ = θ 2 = π being unstable (see Eq. (14.3.10)). For fast tube rotation (specifically, Ω 2 > g/r), there are three equilibrium positions: θ = θ 1 = 0, θ = θ 2 = π, and θ = θ 3 = cos –1 (g/rΩ 2 ), with θ = θ 1 = 0 and θ = θ 2 = π being unstable (see Eqs. (14.3.7) and (14.3.10)) and θ = θ 3 = cos –1 (g/rΩ 2 ) being stable (see Eq. (14.3.16)). That is, the third equilibrium position does not exist unless the tube rotation is such that Ω 2 > g/r, but if it does exist, it is stable. 14.4 A Freely Rotating Body Consider next an arbitrarily shaped body B that is thrown into the air, rotating about one of its central principal axes of inertia. Our objective is to explore the stability of that motion; that is, will the body continue to rotate about the principal inertia axis or will it be unstable, wobbling away from the axis? To answer this question, consider a free-body diagram of B as in Figure 14.4.1, where G is the mass center of B; m is the mass of B; k is a vertical unit vector; F * and T * are the inertia force and couple torque, respectively, of a force system equivalent to the inertia forces on B; –mgk is equivalent to the gravitational forces on B, with g being the gravity constant; and R is an inertial reference frame in which B moves. In the free-body diagram, we have neglected air resistance; thus, the gravitational (or weight) force –mgk is the only applied (or active) force on B. From Eqs. (7.12.1) and (7.12.8), we recall that the inertia force F * and couple torque T * may be expressed as: (14.4.1) where a is the acceleration of G in R; ωω ωω and αα αα are the angular velocity and angular acceleration, respectively, of B in R; and I is the central inertia dyadic of B (see Sections 7.4 to 7.9). From the free-body diagram we then have: (14.4.2) and (14.4.3) FIGURE 14.4.1 Free-body diagram of freely rotating body. Fa TI I ** =− =− ⋅ − × ⋅ () m and ααωωωω −+=mgkF * 0 T * = 0 B G F k T -mgk * * 0593_C14_fm Page 485 Tuesday, May 7, 2002 6:56 AM 486 Dynamics of Mechanical Systems By inspection of Eqs. (14.4.1) and (14.4.2), we have: (14.4.4) Thus, G moves as a projectile particle having a parabolic path (see Section 8.7). Note further that points of B not lying on the central principal inertia axis of rotation will not have a parabolic path. That is, as B rotates it rotates about the central inertia axis. From an inspection of Eqs. (14.4.1) and (14.4.3) we have: (14.4.5) Let n 1 , n 2 , and n 3 be mutually perpendicular unit vectors fixed in B and parallel to the central principal inertia axes of B. Let ω and α be expressed in terms of n 1 , n 2 , and n 3 as: (14.4.6) Because the n i (i = 1, 2, 3) are fixed in B the α i are derivatives of the ω i (see Eq. (4.4.6)). That is, (14.4.7) By substituting from Eqs. (14.4.6) and (14.4.7) into (14.4.5), we obtain: (14.4.8) (14.4.9) (14.4.10) where I 11 , I 22 , and I 33 are the central principal moments of inertia. Equations (14.4.8), (14.4.9), and (14.4.10) form a set of three coupled nonlinear ordinary differential equations for the three ω i (i = 1, 2, 3). To use these equations to determine the stability of rotation of the body, let B be thrown into the air such that B is initially rotating about the central principal inertia axis parallel to n 1 . That is, let B be thrown into the air such that its initial angular velocity components ω i are: (14.4.11) (14.4.12) (14.4.13) ak=−g II⋅+×⋅ () =ααωωωω 0 ωω αα =++= =++= ωωω ω ααα α 11 22 33 11 22 33 nnnn nnnn ii ii αω ii = ˙ −+ − () =III 11 1 2 3 22 33 0 ˙ ωωω −+ − () =III 22 2 3 1 33 11 0 ˙ ωωω −+ − () =III 33 3 1 2 11 22 0 ˙ ωωω ω 1 =Ω ω 3 0= ω 2 0= 0593_C14_fm Page 486 Tuesday, May 7, 2002 6:56 AM Stability 487 By inspection, we see that the ω i (i = 1, 2, 3) of Eqs. (14.4.11), (14.4.12), and (14.4.13) are solutions of Eqs. (14.4.8), (14.4.9), and (14.4.10). To test for the stability of this solution, let small disturbances to the motion occur such that: (14.4.14) (14.4.15) (14.4.16) where as before the ( * ) quantities are small. Then, by substituting these expressions into Eqs. (14.4.8), (14.4.9), and (14.4.10), we have: (14.4.17) (14.4.18) (14.4.19) By neglecting products of small quantities, these equations take the form: (14.4.20) (14.4.21) (14.4.22) Equation (14.4.20) has a solution: (14.4.23) Equations (14.4.21) and (14.4.22) may be solved by eliminating one of the variables (say, ) between the equations. Specifically, from Eq. (14.4.21), we have: (14.4.24) Then, by substituting into Eq. (14.4.22), we obtain: ωω 11 =+Ω * ωω 22 0=+ * ωω 33 0=+ * −+ − () =III 11 1 2 3 22 33 0 ˙ *** ωωω −+ + () − () =III 22 2 3 1 33 11 0 ˙ ** * ωω ωΩ −++ () − () =III 33 3 1 2 11 22 0 ˙ *** ωωωΩ ˙ * ω 1 0= −+ − () =III 22 2 3 33 11 0 ˙ ** ωωΩ −+ − () =III 33 3 2 11 22 0 ˙ ** ωωΩ ωω 110 ** ,= a constant ω 3 * ωω 3 22 33 11 2 ** ˙ = − () I IIΩ − − () +− () = II II II 33 22 33 11 211222 0 Ω Ω ˙˙ ** ωω 0593_C14_fm Page 487 Tuesday, May 7, 2002 6:56 AM 488 Dynamics of Mechanical Systems or (14.4.25) Inspection of Eq. (14.4.25) shows that the disturbance will remain small and the motion of B will be stable if the coefficient of is positive, and that this will occur if I 11 is either a maximum or minimum movement of inertia. If I 11 is an intermediate valued moment of inertia (that is, if I 33 < I 11 < I 22 or I 22 < I 11 < I 33 ), the motion will be unstable. 14.5 The Rolling/Pivoting Circular Disk Consider again the rolling circular disk (or “rolling coin”) as discussed earlier in Sections 4.12 and 8.13 and as shown in Figure 14.5.1. As before, D is the disk, with radius r, mass m, mass center G, contact point C, and orientation angles θ, φ, and ψ. Recall that the condition of rolling requires that the contact point C of D has zero velocity relative to the rolling surface. Recall further that by setting the moments of forces on D about C equal to zero, we obtained the governing equations: (14.5.1) (14.5.2) (14.5.3) In the following paragraphs, we consider the stability of the motions represented by elementary solutions of Eqs. (14.5.1), (14.5.2), and (14.5.3): straight-line rolling, rolling in a circle, and pivoting. FIGURE 14.5.1 Rolling circular disk. ˙˙ ** ωω 2 2 11 22 11 33 22 33 2 0+ − () − () =Ω IIII II ω 2 * ω 2 * 4565 0 2 gr () −+ + =sin ˙˙ ˙ ˙ cos ˙ sin cosθθψφ θφ θ θ 33 5 0 ˙˙ ˙˙ sin ˙ ˙ cosψφθφθθ++ = ˙˙ cos ˙ ˙ φθψθ+=20 D Y N Z N n G n L N X ψ θ φ n 3 2 1 1 2 3 C 0593_C14_fm Page 488 Tuesday, May 7, 2002 6:56 AM Stability 489 Case 1: Straight-Line Rolling Recall from Eq. (8.13.20) that if D is rolling in a straight line with constant speed the angles θ, φ, and ψ are (see Figure 14.5.1): (14.5.4) By inspection we readily see that Eq. (14.5.4) forms a solution of Eqs. (14.5.1), (14.5.2), and (14.5.3). Suppose that the disk D encounters a small disturbance such that θ, φ, and have the forms: (14.5.5) where as before the quantities with a ( * ) are small. By substituting from Eq. (14.5.5) into Eqs. (14.5.1), (14.5.2), and (14.5.3), we obtain: (14.5.6) (14.5.7) (14.5.8) Equations (14.5.7) and (14.5.8) may be integrated, leading to: (14.5.9) (14.5.10) where and are small constants. By substituting from Eq. (14.5.10) into Eqs. (14.5.5), eliminating φ * , we obtain: (14.5.11) or (14.5.12) where is the small constant . As before, the motion is stable if the coefficient of θ * is positive. That is, the motion is stable if: (14.5.13) θφφ ψψ==0 0 ,, ˙˙ ,a constant, = a constant 0 ˙ ψ θθφφφψψψ=+ = + = +0 00 *, *, ˙˙ ˙ * 4560 0 gr () −+ =θθψφ ** * ˙˙ ˙ ˙ ˙˙ * ψ=0 ˙˙ ˙ ˙ * * φψθ+=20 0 ˙ * * ψ=c 1 ˙ ˙ ** * φψθ+=2 02 c c 1 * c 2 * 45620 02 0 gr c () −+ − () =θθψ ψθ ** * * ˙˙ ˙˙ ˙˙ ˙ ** * θψ θ+ () − () [] =12 5 4 5 0 2 3 gr c c 3 * 65 0 2 2 ˙ * ψ c ˙ ψ 0 3> gr 0593_C14_fm Page 489 Tuesday, May 7, 2002 6:56 AM 490 Dynamics of Mechanical Systems Thus, if the angular speed of D exceeds , D will remain erect and continue to roll in a straight line. If the angular speed of D is less than , the motion is unstable. D will wobble and eventually fall. Case 2: Rolling in a Circle Next, suppose D is rolling in a circle with uniform speed such that θ, , and are (see Figure 14.5.1): (14.5.14) By inspection of the governing equations, we see that Eqs. (14.5.2) and (14.5.3) are then identically satisfied and that Eq. (14.5.1) becomes: (14.5.15) Equation (14.5.15) provides a relationship between θ 0 , 0 , and 0 . By inspection of Figure 14.5.1 we see that if 0 and 0 are positive, then Eq. (14.5.15) requires that θ 0 be negative. That is, the disk will lean toward the interior of the circle on which it rolls. To test the stability of this motion, let the disk encounter a small disturbance such that θ, , and have the forms: (14.5.16) where, as before, the quantities with the ( * ) are small. Then, sinθ and cosθ are: (14.5.17) and (14.5.18) By substituting from Eqs. (14.5.16), (14.5.17), and (14.5.18) into Eqs. (14.5.1), (14.5.2), and (14.5.3) and by neglecting quadratic (second) and higher order terms in the ( * ) terms we obtain: (14.5.19) (14.5.20) gr3 gr3 ˙ φ ˙ ψ θθ φφ ψψ== = 00 0 , ˙˙ , ˙˙ 46 5 0 00000 2 00 gr () ++ =sin ˙ ˙ cos sin cosθψφ θφθ θ ˙ φ ˙ ψ ˙ φ ˙ ψ ˙ φ ˙ ψ θθ θ φφ φ ψψ ψ=+ =+ = + 00 0 ** * , ˙˙ ˙ , ˙˙ ˙ sin sin sin cos ** θθθ θθθ=+ () =+ 000 cos cos cos sin ** θθθ θθθ=+ () =− 000 46 5 4 56 6 6 10 5 5 00000 2 00 0 0000000 0000 22 00 2 gr gr () ++ + () −+ + − ++− sin ˙ ˙ cos ˙ sin cos cos ˙˙ ˙ ˙ cos ˙ ˙ cos ˙ ˙ sin ˙˙ sin cos ˙ cos ˙ * ** * * ** θψφ θφθ θ θθ θψφ θψφ θψφθθ φφ θ θ φθ θ φθ ** sin 2 0 0θ= 33 5 0 00 0 ˙˙ ˙˙ sin ˙ ˙ cos ** * ψφθφθθ++ = 0593_C14_fm Page 490 Tuesday, May 7, 2002 6:56 AM Stability 491 and (14.5.21) In view of Eq. (14.5.15), the sum of the first three terms of Eq. (14.5.19) is zero, thus they may be neglected. Also, Eqs. (14.5.20) and (14.5.21) may be integrated as: (14.5.22) and (14.5.23) where and are constants. Solving Eq. (14.5.23) for we have: (14.5.24) Then, by substituting into Eq. (14.5.22), we have: (14.5.25) Finally, by solving for and by substituting for and into Eq. (14.5.19) (without the first three terms) we have: (14.5.26) where λ and κ * are defined as: (14.5.27) and (14.5.28) We recall from our previous analyses that stability will occur if the coefficient λ of θ * in Eq. (14.5.26) is positive; as a corollary, instability will occur if λ is negative. (λ = 0 represents a neutral condition, bordering between stability and instability.) Recall also from Eq. (14.5.15) and by inspection of Figure 14.5.1, that if 0 and 0 are positive, then θ 0 must be negative. Hence, from Eq. (14.5.27) we see that λ is positive; thus, stability occurs, if: (14.5.29) ˙˙ cos ˙ ˙ ** φθψθ 00 20+= 33 5 00 01 ˙ ˙ sin ˙ cos ** * * ψφθφθθ++ =c ˙ cos ˙ ** * φθψθ 002 2+=c c 1 * c 2 * ˙ * φ ˙ ˙ cos ** * φψθ θ=− + () 2 02 0 c 363 5 0200 01 ˙˙ tan ˙ cos ** * * * ψψθ θφθθ+− + () +=cc ˙ * ψ ˙ * φ ˙ * ψ ˙˙ ** θλθ κ+= λψ φψθφ θ = () + () + − () D 12 5 14 5 45 0 2 00 0 0 2 0 ˙ ˙ ˙ sin cosgr κψ φθφθ* ˙ ˙ cos ˙ sin ** * = () + () + () D 65 25 45 02 01 0 02 0 cc c ˙ ψ ˙ φ 12 5 4 5 14 5 0 2 0 2 0000 () +> () − () ˙ ˙ cos ˙ ˙ sinψφ θ φψ θgr 0593_C14_fm Page 491 Tuesday, May 7, 2002 6:56 AM [...]... Control Systems, Schaum’s Outline Series, McGraw-Hill, New York, 1967, 114 ff 14.3 Davis, S A., Feedback and Control Systems, Simon & Shuster, New York, 1974, 262 ff 14.4 Skelton, R E., Dynamic Systems Control: Linear Systems Analysis and Synthesis, Wiley, New York, 1988, chap 7 0593_C14_fm Page 510 Tuesday, May 7, 2002 6:56 AM 510 Dynamics of Mechanical Systems 14.5 Meirovitch, L., Elements of Vibration... orientation of D be defined by the angles θ, φ, and ψ as in Figure 14.6.1 The presence of the mass at Q makes it necessary to know the kinematics of Q The kinematics of Q, however, are directly dependent upon the kinematics of D Therefore, in developing the kinematics of Q it is helpful to review and summarize the kinematics of D as previously developed in Section 4.12 In developing the kinematics and dynamics. .. with that of Problem P14.5.3 Ω FIGURE P14.5.8 A pivoting ring P14.5.9: Repeat Problem P14.5.8 for a ring with radius 30 cm Compare the result with that of Problem P14.5.4 0593_C15_fm Page 513 Tuesday, May 7, 2002 7:05 AM 15 Balancing 15.1 Introduction As with stability, balancing is a subject of great interest to engineers and to designers of mechanical systems — particularly in the design of rotating... each of the ci (i = 1,…, 8) is positive (see Eq (14.6.45)), the inequality of Eq (14.6.60) ˙ will be satisfied if the spin rate φ 0 is sufficiently large and the inequality of Eq (14.6.59) ˙ is satisfied That is, the first inequality of Eq (14.6.58) will be satisfied if φ 0 is sufficiently large and if: c2c6 − c4 c5 − c1c8 > 0 (14.6.61) 0593_C14_fm Page 502 Tuesday, May 7, 2002 6:56 AM 502 Dynamics of Mechanical. .. Page 506 Tuesday, May 7, 2002 6:56 AM 506 Dynamics of Mechanical Systems Then, we obtain: a1b1λ4 + ( a3b1 + a1b3 − a2b2 )λ2 + a3b3 = 0 (14.7.8) Equation (14.7.8) is identical to Eq (14.6.52), which was obtained by the simultaneous solution of Eqs (14.6.40) and (14.6.41) It happens, in infinitesimal stability procedures in the analysis of small disturbances of a system from equilibrium, that we can generally... concepts of balancing Readers interested in more detail or in more technical aspects should consult references at the end of the chapter or in an engineering library In classical dynamical analyses of machines and mechanical systems, the machines or systems are generally modeled as ideal bodies with perfect geometry and uniform mass distribution — and, specifically, with no imbalance In reality, of course,... small distance δ away from A–A Figure 15.2.2 depicts an end view of the rotor showing G and a point O on the axis A–A 513 0593_C15_fm Page 514 Tuesday, May 7, 2002 7:05 AM 514 Dynamics of Mechanical Systems δ D S A O G A FIGURE 15.2.1 A cylindrical shaft with a mounted circular disk FIGURE 15.2.2 End view of rotor showing mass center offset To “balance” the system we need to add or remove weight to... forces on a body, or a mechanical system, by a single force F* passing through the mass center G together with a couple with torque T*, then F* and T* are (Equation (7.12.1) and (7.12.8)): ˆ F* = −MaG (15.3.3) T * = − I ⋅ α − ω × (Ι ⋅ ω ) (15.3.4) and 0593_C15_fm Page 516 Tuesday, May 7, 2002 7:05 AM 516 Dynamics of Mechanical Systems P1 (m) FIGURE 15.3.3 Dynamic balancing of the shaft of Figure 15.3.2 Q... 0593_C14_fm Page 508 Tuesday, May 7, 2002 6:56 AM 508 Dynamics of Mechanical Systems c B k m FIGURE 14.7.2 Damped linear mass–spring oscillator n x x* away from equilibrium (x = 0) simply produces a bounded oscillation of the form (see Eq (13.3.4)): x * = A* cos k m t + B* sin k m t (14.7.18) where A* and B* are small constants But, because the amplitude of the disturbance oscillation does not diminish... differentiate Eq (14.6.5) to obtain the angular acceleration of D in R as: R ˙˙ ˙ ˙ ˙˙ ˙˙ ˙ ˙ α D = (θ − ψφ cos θ)n1 + ( ψ + θφ cos θ + φ sin θ)n 2 (14.6.8) ˙˙ ˙ ˙ ˙˙ + ( ψθ + φ cos θ − φθ sin θ)n 3 Recall that because D is pivoting (a special case of rolling; see Section 4 .11) on a surface in the X–Y plane, the velocity of the center G of D in R is (see Eq (4 .11. 5)): R ˙ ˙ ˙ V G = Rω D × rn 3 = r( ψ + φ sin θ)n1 . stability of several other mechanical systems. 14.3 A Particle Moving in a Vertical Rotating Tube Consider the system consisting of a particle free to move in the smooth interior of a vertical rotating. 22 0 ˙ ** ωωΩ ωω 110 ** ,= a constant ω 3 * ωω 3 22 33 11 2 ** ˙ = − () I IIΩ − − () +− () = II II II 33 22 33 11 2112 22 0 Ω Ω ˙˙ ** ωω 0593_C14_fm Page 487 Tuesday, May 7, 2002 6:56 AM 488 Dynamics of Mechanical. will occur if I 11 is either a maximum or minimum movement of inertia. If I 11 is an intermediate valued moment of inertia (that is, if I 33 < I 11 < I 22 or I 22 < I 11 < I 33