Vibrations Fundamentals and Practice Appb Maintaining the outstanding features and practical approach that led the bestselling first edition to become a standard textbook in engineering classrooms worldwide, Clarence de Silva''s Vibration: Fundamentals and Practice, Second Edition remains a solid instructional tool for modeling, analyzing, simulating, measuring, monitoring, testing, controlling, and designing for vibration in engineering systems. It condenses the author''s distinguished and extensive experience into an easy-to-use, highly practical text that prepares students for real problems in a variety of engineering fields.
de Silva, Clarence W “Appendix B” Vibration: Fundamentals and Practice Clarence W de Silva Boca Raton: CRC Press LLC, 2000 Appendix B Newtonian and Lagrangian Mechanics A vibrating system can be interpreted as a collection of mass particles In the case of distributed systems, the number of particles is infinite The flexibility and damping effects can be introduced as forces acting on these particles It follows that Newton’s second law for a mass particle forms the basis of describing vibratory motions System equations can be obtained directly by applying Newton’s second law to each particle It is convenient, however, to use Lagrange’s equations for this purpose, particularly when the system is relatively complex A variational principle known as Hamilton’s principle, which can be established from Newton’s second law, is the starting point in the derivation of Lagrange’s equations This appendix outlines some useful results of dynamics, in both Newtonian and Lagrangian approaches The Newtonian approach uses forces, torques, and motions, which are vectors Hence, it is important to deal with vector mechanics in the Newtonian approach The Lagrangian approach is based on energy, which is a scalar quantity Scalar energies can be expressed in terms of vectorial positions and velocities The subject of dynamics deals with forces (torques) and motions The study of motion alone belongs to the subject of kinematics This appendix starts with vectorial kinematics, and then addresses Newtonian mechanics (dynamics) and finally Lagrangian dynamics B.1 VECTOR KINEMATICS B.1.1 EULER’S THEOREM (SEE FIGURE B.1) Every displacement of a rigid body can be represented by a single rotation θ about some axis (of unit vector υ) Important Corollary Rotations cannot be represented by vectors unless they are infinitesimally small Proof Give a small rotation δθ = δθυ about the υ-axis, the corresponding displacement (of point P) is δr = δθ × r The new position (of P) is r1 = r + δθ × r Give another small rotation δφ = δφµ about the µ-axis; the new position (of P) is r2 = r1 + δφ × r1 The combined displacement ∆r = r2 − r = r1 + δφ × r1 − r = r + δθ × r + δφ × [ r + δθ × r ] − r = (δθ + δφ) × r + δφ × (δθ × r ) 14 4244 ( ) δ2 ©2000 CRC Press FIGURE B.1 Single rotation of a rigid body about an axis The second term can be neglected for small rotations Hence, ∆r = (δθ + δφ) × r B.1.2 ANGULAR VELOCITY (SEE FIGURE B.2) AND VELOCITY AT A Q.E.D POINT OF A (B.1) RIGID BODY From Figure B.1, it is clear that ω= Angular velocity lim δθ δt → δt (B.2) Note: Since this definition uses small rotations δθ, it follows that ω is always a vector The velocity of P relative to O (see Figure B.1) is given by vP O = lim δr lim δθ × r = =ω×r δt → δt δt → δt (B.3) Theorem For a rigid body, ω is unique (does not vary from point to point on the body) Proof Suppose that, on the contrary, angular velocities ω1 and ω2 are associated with points C1 and C2 (Figure B.2) Then, v P O = vC1 O + v P C1 = vC1 O + ω1 × r Also, v P O = vC1 O + vC2 C1 + v P C2 = vC1 O + ω1 × ( r − s) + ω × s Hence, ω × r = ω × ( r − s) + ω × s ©2000 CRC Press FIGURE B.2 Uniqueness of the angular velocity of a rigid body It follows that (ω − ω ) × s = Since (ω1 – ω2) is not parallel to s in general, because P is an arbitrary point, one has ω1 = ω B.1.3 RATES OF UNIT VECTORS ALONG AXES (SEE FIGURE B.3) Q.E.D OF ROTATING FRAMES General Result Suppose that i1, i2, i3 are unit vectors along the three orthogonal axes of a frame rotating at ω The rates are the velocities of these vectors about the origin of the frame Hence, from equation (B.3), one obtains di1 = ω × i1 , dt di2 = ω × i2 , dt di3 = ω × i3 dt (B.4) Some special cases (natural frames of reference) are considered below Cartesian Coordinates Suppose that the frame is free to move independently in the x, y, and z directions only This corresponds to a translatory motion (no rotation) The rates of the unit vectors in this moving frame are 0; that is, ω = di x di y diz = = =0 dt dt dt ©2000 CRC Press (Cartesian) (B.5) FIGURE B.3 Some natural coordinate frames: (a) Cartesian coordinates; (b) polar coordinates (2-D); (c) spherical polar coordinates; and (d) tangential-normal coordinates (2-D) Polar Coordinates (2-D) Suppose that the frame is free to move independently in the r and θ directions, but an increment in the r direction (i.e., δr) causes no rotation Hence, ˙ ω = θi z From equation (B.4), ©2000 CRC Press dir ˙ = θiθ dt (2 - D polar) (B.6) diθ = −θ˙ ir dt Spherical Polar Coordinates The coordinates are (r, θ, φ) as in Figure B.3 To find natural ω for the frame: Give δr to frame ⇒ No rotation Give δθ to frame ⇒ Rotation δθiφ Give δφ to frame ⇒ Rotation δφiz Hence, ω = θ˙ iφ + φ˙ iz But iz = cos θir − sin θiθ Hence, ω = φ˙ cos θir − φ˙ sin θiθ + θ˙ iφ (B.7) From equation (B.4), dir ˙ = θiθ + φ˙ sin θiφ dt diθ = −θ˙ ir + φ˙ cos θiφ dt diφ dt (Spherical polar) (B.8) = − φ˙ sin θir − φ˙ cos θiθ Tangential-Normal (Intrinsive) Coordinates (2-D) The coordinates are (s, ψ) as in Figure B.3 Note that s is the curvilinear distance along the path of the particle (from a reference point Po), and ψ is the angle of slope of the path To find ω of the natural frame, Give δs ⇒ No rotation Give δψ ⇒ Rotation δψiz Hence, ˙ z ω˙ = ψi ©2000 CRC Press The derivatives of the tangential and normal unit vectors it and in are given by dit = ψ˙ in dt (Tangential - normal) (B.9) din = − ψ˙ it dt Note: The vector cross-product of the two vectors a = a x i + a y j + az k b = bx i + by j + bz k can be obtained by expanding the determinant B.1.4 ACCELERATION EXPRESSED IN i j k a × b = ax ay az bx by bz (B.10) ROTATING FRAMES Spherical Polar Coordinates Position: r = r ir Velocity: v= di dr = r˙ ir + r r dt dt Substitute equation (B.8), and obtain ( ) ( ) v = r˙ir + rθ˙ iθ + r sin θφ˙ iφ Acceleration a = dv Hence, by differentiating equation (B.11) and using (B.8), one obtains dt ar ˙˙ r − rθ˙ − r sin θφ˙ d 2˙2 r θ − r sin θ cos θφ˙ a = aθ = r dt a d r sin θφ˙ φ r sin θ dt ( ) ( Tangential-Normal Coordinates (2-D) (See Figure B.4) The velocity is always tangential to the path Hence, ©2000 CRC Press (B.11) ) (B.12) FIGURE B.4 Velocity representation in tangential-normal coordinates v= dr ds = i = vit dt dt t (B.13) By differentiating equation (B.13) and using (B.9), one obtains dv at v˙ v ds a= = = a vψ˙ v n ρ See Figure B.4 Note : dv dv ds vdv = = dt ds dt ds dψ dψ ds v = = dt ds dt ρ ρ= ©2000 CRC Press ds = radius of curvature dψ (B.14) FIGURE B.5 Representation of a vector with respect to a rotating frame B.2 NEWTONIAN (VECTOR) MECHANICS B.2.1 FRAMES OF REFERENCE ROTATING (SEE FIGURE B.5) AT ANGULAR VELOCITY ω Newton’s second law holds with respect to (w.r.t.) an inertial frame of reference (normally a frame fixed on the earth’s surface or moving at constant velocity) The rate of change of vector B w.r.t an inertial frame is related to the rate of change w.r.t a frame rotating at ω by dB ∂B = +ω×B dt ∂t rel (B.15) From equation (B.15), the following results can be obtained for velocity v and acceleration a of point B (see Figure B.5) v= dRo + ω + v rel dt24 (B.16) vfrm 6centripetal 4748 d Ro ˙ a= + ω × r + ω × ω × v rel + arel (ω × r ) + 214 24 dt4444 4244444 afrm (B.17) Coriolis Note: vfrm and afrm are velocity and acceleration, respectively, of a point just beneath P and fixed to rotating frame ©2000 CRC Press FIGURE B.6 Vector motion of a particle in space B.2.2 NEWTON’S SECOND LAW FOR A PARTICLE f = OF MASS m (SEE FIGURE B.6) dp dt (B.18) = ma (Linear momentum principle) dR dt Cross multiply equation (B.18) by r The torque about B is Note: Linear momentum p = m τB = r × f = r × dp d dr ×p = ( r × p) − dt dt dt Now, r × p = hB = angular momentum about B Also, dr dR dR dR ×p= − vB × m = − mv B × = −v B × p dt dt dt dt This cross-product vanishes if either B is fixed (vB = 0) or B moves parallel to the velocity of m Hence, the angular momentum principle: B = â2000 CRC Press dhB + vB ì p dt in general (B.19a) FIGURE B.7 (a) Dynamics of a system of particles, and (b) dynamics with respect to the centroid τB = dhB dt B fixed, or B moves parallel to the velocity of m if B.2.3 SECOND LAW FOR A SYSTEM OF PARTICLES — RIGIDLY FLEXIBLY CONNECTED (SEE FIGURE B.7) F= (B.19b) OR dP dt (B.20) = MaC if constant mass (Linear momentum principle) ∑ f = resultant external force P = ∑ m v = Mv = total linear momentum Note: F = i i i C C = centroid of the particles Now, using the procedure outlined before for a single particle, and summing the results, one obtains r × f = ∑ (r × p ) − ∑ ×p ∑ dt dt 1424 123 dri d i τB But ©2000 CRC Press i i dH B dt i i ∑ dt × p = ∑ dt dri dR − v B × mi i = −v B × dt dRi i ∑m i dRi = −v B × P dt v B = 0, or if B moves parallel to C (special case; B ≡ C) = − Mv B × vC = Hence, we have the angular momentum principle τB = τB = d H + vB × P dt B d H dt B in general (B.21a) B fixed, or if B moves parallel to C (special case; B ≡ C) (B.21b) Note: (See Figure B.7(b)) ∑ r × m v =∑ (r + c ) × m v = r × ∑m v + ∑c × m v 12 4 14 4244 HB = i c i i c i i i P i i i i i Hc Hence, H B = Hc + rc × P (B.22) B.2.4 RIGID BODY DYNAMICS (SEE FIGURE B.8) — INERTIA MATRIX ANGULAR MOMENTUM AND Note: Equation (B.20) for a system of particles is also the convenient form of the linear momentum principle for rigid bodies But a more convenient form of equation (B.21) is possible using ω — the angular velocity of the rigid body dr Angular momentum about O is a given by Ho = ri × mi i = mi ri × (ω × ri ) dt Now, using the cross-product relation ∑ a × (b × c) = b( a ⋅ c) − c( a ⋅ b) one obtains ©2000 CRC Press ∑ FIGURE B.8 Motion of a rigid body with respect to a fixed frame Ho = [ I ]o ω (B.23a) (Angular momentum about fixed point O) where the inertia matrix is [ I ]o I xx = Symmetric = = I xz I yz I zz I xy I yy [ ∑ m (r [1] − r r ) T i ∫ i i i ( dm r [1] − rr T ) Discrete body (B.24) Continuous body ] Note: In Cartesian coordinates, riT = xi , yi , zi , and ri = ri = xi2 + yi2 + zi2 I xx = ∑ m (r I xy = − i i ) − xi2 = ∫ dm(r − x2 ) ∑ m x y = −∫ dmxy i i i etc Angular momentum about the centroid: Hc = = ∑c × m i i dri dt dr dc mc) ∑ m c × dt + dt = (1∑ 424 c i i i i i =0 by definition of centroid Now ©2000 CRC Press × drc + dt ∑ m c × dt dci i i ∑ m c × dt = ∑ m c × (ω × c ) dci i i i i for rigid body i = [I]c ω as for equation (B.23) Hence, Hc = [ I ]c ω (B.23b) (Angular momentum about centroid) Equations (B.23a) and (B.23b) can be written B fixed, or B is the centroid H B = [ I ]B ω B.2.5 MANIPULATION OF INERTIA (B.23) MATRIX (SEE FIGURE B.9) Parallel Axis Theorem — Translational Transformation of [I] a rc = b = position of centroid w.r.t a frame at B c If the axes of the two frames are parallel, [ I ]B = [ I ]C ( ) b2 + c2 + M Symmetric (− ab) (c +a ) ( (− ac) (− bc) ) a + b 2 (B.25) Rotational Transformation of [I] If a nonsingular square matrix [C] satisfies [C][C]T = [1] (B.26) then it is an orthogonal matrix, and the transformation of coordinates from r to r′ through r ′ = [C]r (B.27) is called an orthogonal transformation It can be verified using equation (B.24) that [ I ′] = [C][ I ][C]T (B.28) Principal Directions (Eigenvalue Problem) Principal directions ≡ Directions in which angular momentum is parallel to the angular velocity Then, ©2000 CRC Press FIGURE B.9 (a) Planar coordinate transformation (b) Mohr’s circle for moments of inertia H = λω ω Substitute equation (B.23): [ I ]ω = λω If the corresponding direction vector is u (i.e., ω = ωu), then [ I ]u = λu (B.29) Equation (B.29) represents an eigenvalue problem The nontrivial solutions for u are eigenvectors and represent principal directions Since [I] is symmetric, three independent, real solutions (u1, u2, u3) exist for u The corresponding values of λ are I1, I2, and I3 These are termed principal moments of inertia The matrix of normalized eigenvectors u1T [C] = u2T u3T (B.30) is an orthogonal matrix The orthogonal transformation of coordinates rp = [C]r (B.31) rotates the frame into the principal directions and hence diagonalizes the inertia matrix [C][ I ][C] T ©2000 CRC Press I1 = I2 0 I3 (B.32) Mohr’s Circle In the two-dimensional (2-D) case of rigid bodies, the principal directions and principal moments of inertia can be determined conveniently using Mohr’s circle In Figure B.9(b), the direct inertias Ixx and Iyy are read on the horizontal axis, and the cross inertia Ixy on the vertical axis If the inertia matrix is given in the (x, y, z) frame, two diametrically opposite points of the circle are known These determine the Mohr’s circle The inertia matrix in any other frame (x′, y′, z′), rotated by angle θ about the common z-axis in the positive direction, is obtained by moving through 2θ counter closewise on the circle This procedure also determines the principal moments I1 and I2, and the principal direction B.2.6 EULER’S EQUATIONS (FOR A RIGID BODY ROTATING AT ω) First consider a general body (rigid or not) for which angular momentum about a point B is expressed in terms of directions of a frame that rotates at ω Then, from equation (B.15), one obtains d ∂H B H = + ω × HB dt B ∂t rel (B.33) Now consider a rigid body rotating at ω If the body frame is oriented in the principal directions, then from equation (B.23), one obtains I1 HB = 0 ω1 I1ω1 ω = I2 ω I3 ω I3ω I2 B fixed, or if B moving parallel to C (special case B = C ) From equation (B.19b), the Euler’s equations (B.34) are obtained: I1ω˙ − ( I2 − I3 )ω ω τ B = I2 ω˙ − ( I3 − I1 )ω 3ω1 I3ω˙ − ( I1 − I2 )ω1ω Note: • For principal directions only • B fixed or moving parallel to centroid ∂H B = I1ω˙ , I2 ω˙ , I3ω˙ ∂t rel [ (B.34) ] T (because I1 , I2 , I3 are constant relative to the body frame) B.2.7 EULER’S ANGLES Consider two Cartesian frames F and F′ in different orientations One can rotate F to coincide with F′ in three steps, as follows: ©2000 CRC Press Step 1: Rotate by angle ψ about the z-axis The orthogonal transformation matrix for this rotation is cos ψ [ψ ] = − sin ψ sin ψ cos ψ 0 (B.35) Step 2: Rotate by angle θ about the new y-axis The orthogonal transformation matrix for this rotation is cos θ [θ] = sin θ − sin θ cos θ (B.36) Step 3: Rotate by angle φ about the new x-axis The orthogonal transformation matrix for this rotation is: 1 [φ] = 0 0 cos φ − sin φ sin φ cos φ (B.37) The same point P can be expressed as vector r in F, or vector r′ in F′ It follows that the two are related through r ′ = [φ][θ][ψ ]r (B.38) In the Euler angle representation, the angular velocity can be considered to consist of the following components: ψ˙ about the z-axis of the original frame θ˙ about the y-axis of the intermediate frame φ˙ about the x-axis of the final frame It follows that the angular velocity expressed in F′ is φ˙ 0 0 ˙ ω ′ = [φ][θ][ψ ] + [φ][θ]θ + [φ]0 0 ψ˙ 0 0 0 φ˙ = [φ][θ] + [φ]θ˙ + 0 ψ˙ 0 0 ©2000 CRC Press (B.39a) Also, angular velocity expressed in F is φ˙ 0 0 T ˙ T T ω = + [ψ ] θ + [ψ ] [θ] 0 0 ψ˙ 0 (B.40a) φ˙ − ψ˙ sin θ ˙ ω ′ = θ cos φ + ψ˙ sin φ cos θ ψ˙ cos φ cos θ − θ˙ sin φ (B.39b) φ˙ cos θ cos ψ − θ˙ sin ψ ω = φ˙ cos θ sin ψ + θ˙ cos ψ ψ˙ − φ˙ sin θ (B.40b) The resulting expressions are: Once ω is expressed in this manner, v and a can be expressed in terms of Euler’s angles (ψ, θ, φ) and their time derivatives This is the basis of using Euler angles to write equations of motion Note: The set of Euler angles described here is known as the (3, 2, 1) set or Type I Euler angles Other combinations are possible For example, if the first rotation is about the x-axis, the second rotation about the new y-axis, and the final rotation about the new x-axis, then one has the (1, 2, 1) set B.3 LAGRANGIAN MECHANICS B.3.1 KINETIC ENERGY AND KINETIC COENERGY Consider a single particle Kinetic energy T= ∫ v ⋅ dp (B.41) Kinetic coenergy T* = ∫ p ⋅ dv (B.42) Note: T + T* = v · p In classical mechanics, the constitutive relation between velocity v and linear momentum p is linear: p = mv Hence, ©2000 CRC Press (B.43) T = T* = 1 mv ⋅ v = mv 2 (for a particle) (B.44) Note: For this reason, it is not necessary to distinguish between T and T* But in Lagrangian mechanics, traditionally T* is retained For a system of particles, T* = ∑m v ⋅v i i (B.45) i For a rigid body rotating at ω (see Figure B.8), T* = ∑ m (ω × r ) ⋅ (ω × r ) i i i Now, using ( a × b) ⋅ c = (b × c) ⋅ a = (c × a) ⋅ b T* = ∑ [ ] mi ri × (ω × ri ) ⋅ ω 144 42444 Ho See equation (B.23a) One can write, T* = T ω Ho = ω T [ I ]o ω 2 (B.46) (For a rigid body rotating about fixed O) In Figure B.7(b), suppose that B is fixed Premultiply equation (B.22) by 1/2 ωT (or take the dot product with 1/2 ω) and obtain vc 67 4 m}vc T 1 T T = ω Hc + ω ⋅ ( rc × P ) = ω [ I ]c ω + (ω × rc ) ⋅ P 2 4243 24 4244 * From eq (B.23) Using vector identity Hence, with vc denoting the magnitude ΈvcΈ, T* = 1 Mvc2 + ω T [ I ]c ω 2 (B.47) (For a rigid body with centroid C ) B.3.2 WORK AND POTENTIAL ENERGY When the points of application ri of a set of forces fi move by increments δri, the incremental work done is given by ©2000 CRC Press δW = ∑ f ⋅ δr i (B.48) i Forces can be divided into conservative forces and nonconservative forces The work done by conservative forces can be given by the decrease in potential energy (this is one definition for potential energy) Hence, δV = −(δW )conserv (B.49) Note that conservative forces are nondissipative forces (e.g., spring force, gravity) Examples For masses mi located at elevations yi in gravity (acceleration g), V= ∑ m gy i (B.50) i For springs of stiffness ki stretched through xi, the potential energy (elastic) is V= ∑k x i i (B.51) B.3.3 HOLONOMIC SYSTEMS, GENERALIZED COORDINATES, AND DEGREES OF FREEDOM Holonomic constraints can be represented entirely by algebraic relations of the motion variables Dynamic systems having holonomic constraints only are termed holonomic systems For any system (holonomic or non-holonomic), the number of degrees of freedom (n) equals the minimum number of incremental (variational) generalized coordinates (δq1, δq2, …, δqn) required to completely describe any general small motion (without violating the constraints) The number of (nonincremental) generalized coordinates required to describe large motions may be greater than n in general But, for holonomic systems: Number of dof = Number of independent generalized coordinates B.3.4 HAMILTON’S PRINCIPLE For a holonomic system, the Lagrangian L is given by L = T* − V (B.52) Consider the variation integral tf δH = ∫ t0 δL + n ∑ Q δq dt j j (B.53) j =1 in which Qj are the nonconservative generalized forces corresponding to the generalized coordinates qj For a motion trajectory, t0 and tf are the initial and the final times, respectively Hamilton’s principle ©2000 CRC Press states that this trajectory corresponds to a natural motion of the system if and only if δH = for arbitrary δqj about the trajectory B.3.5 LAGRANGE’S EQUATIONS Note that L is a function of qj and q˙ j in general, because V is a function of qj and T* is a function of qj and q˙ j Hence, L = L(q1 , q2 , K, qn , q˙1 , q˙ , K, q˙ n ) (B.54) Then, it follows from Hamilton’s principle [equation (B.53) with δH = 0], that: d ∂L ∂L − = Qj dt ∂q˙ j ∂q j j = 1, 2, K, n (B.55) These are termed Lagrange’s equations, and represent a complete set of equations of motion Note: Newton’s equations of motion are equivalent to the Lagrange’s equations To determine Qj, give an incremental motion δqj to the system with the other coordinates fixed, and determine the work done δWj Then, δWj = Q j δq j (B.56) This gives Qj EXAMPLE Figure B.10 shows a simplified model that can be used to study the mechanical vibrations that are excited by the control-loop disturbances in a single-link robot arm The length of the arm is l, the mass is M, and the moment of inertia about the joint is I The gripper hand (end effector) is modeled as a mass m connected to the arm through a spring of stiffness k The joint has an effective viscous damping constant c for rotary motions The motor torque (applied at the joint) is τ(t) Generalized Coordinates This is a two-degree-of-freedom holonomic system The angle of rotation θ of the arm and spring deflection x from the unstretched position are chosen as the generalized coordinates Generalized Nonconservative Forces Keeping x fixed, increment θ by δθ The corresponding incremental work due to nonconservative ˙ forces is δWθ = τ(t )δθ − cθδθ Note that the damping torque c θ˙ acts opposite to the increment δθ Thus, the generalized force is given by Fθ = τ(t ) − cθ˙ Keeping θ fixed, increment x by δx The corresponding incremental work due to nonconservative forces is δWx = Hence, Fx = ©2000 CRC Press FIGURE B.10 Lagrangian The total kinetic coenergy (= kinetic energy in these Newtonian systems) is T* = ( ) ( ˙2 Iθ + m lθ˙ + x˙ 2 ) 2 Note: lθ˙ + x˙ is not exact (there is a nonlinear term that is neglected) The potential energy (due to gravity and spring) is given by: ( ) 1 V = Mg − l cos θ + mg −l cos θ + ( x + x ) sin θ + kx Note that the centroid of the arm is assumed to be halfway along the link It follows that the Lagrangian is given by: L= ( ˙2 Iθ + m lθ˙ + x˙ 2 ) M + + m gl cos θ − mg( x + x ) sin θ − kx Lagrange’s Equations From equation (B.55) one obtains [ )] ( For θ : d ˙ M Iθ + m lθ˙ + x˙ l − − + m gl sin θ − mg( x + x ) cos θ = τ(1) − cθ˙ dt For x : d m lθ˙ + x˙ − [mg sin θ − kx ] = dt ©2000 CRC Press [( )] If the steady-state configuration of the link is assumed to be vertical, for small departures from this position (e.g., due to control loop disturbances), the equations of motion are obtained by linearizing for small θ (i.e., sinθ = θ, cosθ = 1) The corresponding equations are I + ml ml ml ˙˙ θ c + 0 m ˙˙ x θ˙ M + m gl + x˙ mg mg θ τ(t ) − mgx = k x Note: For equilibrium at θ = 0, one needs τ = mgx0 In other words, this term represents the static torque needed at the motor joint in order to maintain equilibrium at the (θ = 0, x = 0) position If the system was symmetric, τstatic = Note also that this set of equations of motion is of the form My˙˙ + Cy˙ + Ky = f (t ) and the matrices M, C, and K are symmetric The mass matrix M is not diagonal in the present formulation It is possible, however, to make it diagonal simply by eliminating the x˙˙ term in the θ term in the second equation of motion through straightforward first equation of motion and the ˙˙ algebraic manipulation Natural frequencies and mode shapes of the undamped system can be determined in the usual manner (see Chapter 5) If the feedback gain of the control loop is such that at least a pair of eigenvalues is complex and has an imaginary part that is approximately equal to a natural frequency of the structural system, then that mode can be excited in an undesirable manner even by a slight disturbance in the control force Such situations can be avoided by modifying the control system (e.g., by changing the gains) or the structural system (for example, by adding damping, and by changing stiffness and mass) so that the structural natural frequencies would not be near the resonances of the feedback control system Related issues of design and control for vibration are addressed in Chapter 12 ©2000 CRC Press ... and velocities The subject of dynamics deals with forces (torques) and motions The study of motion alone belongs to the subject of kinematics This appendix starts with vectorial kinematics, and. .. + 214 24 dt4444 4244444 afrm (B.17) Coriolis Note: vfrm and afrm are velocity and acceleration, respectively, of a point just beneath P and fixed to rotating frame ©2000 CRC Press FIGURE B.6 Vector... principal directions and principal moments of inertia can be determined conveniently using Mohr’s circle In Figure B.9(b), the direct inertias Ixx and Iyy are read on the horizontal axis, and the cross