1.7 Lagrange's equations 23 m20-ml202 sin 0 cos0+mglsin=0 For small oscillations near 0.the ea atio can be simplified using the approximations sin and cos=1:this leads to 8+90-220=0 Introducing=号，the pendulum frequency i+(号-22)0=0 One sees that the centrifugal force introduces a negative stiffness.Figure 9 e品 1.9(c)shows the 1.7.5 Example 3:Rotating spring mass system A spring mass system is rotating in the horizontal plane at a constant velocity Fig.1.10(a).The system has a single d.o.f.,described by the coordinate u measuring the extension of the spring.The absolute velocity of the point mass m can be conven projected in the moving frame (,)the components are (,u).It follows that T=2m2+(u2) Once It is not necessary to include the kinetic coenergy of the rotating mecha nism,because it is constant,and will disappear when writing the Lagrange equation.Since the system is not subjected to gravity,the potential en- ergy is associated with the extension of the spring:assuming that the spring force is equal to zero when u=0, V= The Lagrangian reads L=r-v=mi2-k-m
1.7 Lagrange’s equations 23 ml2 ¨θ − ml2Ω2 sin θ cos θ + mglsin θ = 0 For small oscillations near θ = 0, the equation can be simplified using the approximations sin θ ≃ θ and cos θ = 1; this leads to ¨θ + g l θ − Ω2θ = 0 Introducing ω2 0 = g l , the pendulum frequency ¨θ + ³ ω2 0 − Ω2 ´ θ = 0 One sees that the centrifugal force introduces a negative stiffness. Figure 1.9(c) shows the evolution of the frequency of the small oscillations of the pendulum with Ω; the system is unstable beyond Ω = ω0. 1.7.5 Example 3: Rotating spring mass system A spring mass system is rotating in the horizontal plane at a constant velocity Ω, Fig.1.10(a). The system has a single d.o.f., described by the coordinate u measuring the extension of the spring. The absolute velocity of the point mass m can be conveniently projected in the moving frame (x, y); the components are ( ˙u, uΩ). It follows that T∗ = 1 2 m h u˙ 2 + (uΩ) 2 i Once again, there is a quadratic contribution, T2 ∗, and a contribution independent of the generalized velocity, T0 ∗ (potential of centrifugal force). It is not necessary to include the kinetic coenergy of the rotating mechanism, because it is constant, and will disappear when writing the Lagrange equation. Since the system is not subjected to gravity, the potential energy is associated with the extension of the spring; assuming that the spring force is equal to zero when u = 0, V = 1 2 ku2 The Lagrangian reads L = T∗ − V = 1 2 mu˙ 2 − 1 2 (k − mΩ2)u2
24 1 Lagrangian dynamics of mechanical systems leading to the Lagrange equation mi+(k-m22)u=0 i+(2-22)u=0 form of the】 evious (a) () k1 Fig.1.10.Rotating spring-mass systems (a)Single axis(b)Twaxis 1.7.6 Example 4:Gyroscopic effects Next,consider the system of Fig.1.10(b),where the constraint alongy=0 has been removed and replaced by another spring orthogonal to the pre vious one.This system has 2 d.o.f.;it is fully described by the generalized dg axes s rot g.1.10(b obal s to the kinetic coenergy of the point mass T=ml在-g++ expanding T,one gets
24 1 Lagrangian dynamics of mechanical systems leading to the Lagrange equation mu¨ + (k − mΩ2)u = 0 or u¨ + (ω2 n − Ω2)u = 0 after introducing ω2 n = k/m. This equation is identical to the linearized form of the previous example; the system becomes unstable for Ω > ωn. u k m x y Ò Ò y x c1 m k1 k2 (a) (b) Fig. 1.10. Rotating spring-mass systems (a) Single axis (b) Two-axis. 1.7.6 Example 4: Gyroscopic effects Next, consider the system of Fig.1.10(b), where the constraint along y = 0 has been removed and replaced by another spring orthogonal to the previous one. This system has 2 d.o.f.; it is fully described by the generalized coordinates x and y, the displacements along the moving axes rotating at constant speed Ω. We assume small displacements and, in Fig.1.10(b), the stiffness k1 and k2 represent the global stiffness along x and y, respectively. We assume viscous damping along x, with damping coefficient c1. The absolute velocity in the rotating frame is ( ˙x − Ωy, y˙ + Ωx), leading to the kinetic coenergy of the point mass T∗ = 1 2 m h ( ˙x − Ωy) 2 + ( ˙y + Ωx) 2 i As in the previous example, we disregard the constant term associated with the rotation at constant speed of the supporting mechanism. Upon expanding T∗, one gets
1.7 Lagrange's equations 25 T=攻+T+6 with Tg=5m(2+的) Ti=m(xy-iy) T0=5m22(x2+2) 一 appears i th can be The potential y is a the springs:with the assumption of small displacements, v=h2+ D=12 The Lagrange equations read mi-2m2g+q立+x-m22x=0 mj+2m.2t+2x-m22y=0 or,in matrix form,with =(,y)T, M+(C+G)g+(K-22M)g=0 (1.55) where w=[ac=88=g are respectively the mass,damping and stiffness matrices,and [0-2m2 G=2m2 0 (1.56) is the anti-symmetric matrix of gy oscopic forces,which couples th tion in the ts magn itude is proportional to the inertia(m)
1.7 Lagrange’s equations 25 T∗ = T∗ 2 + T∗ 1 + T∗ 0 with T∗ 2 = 1 2 m( ˙x2 + ˙y2) T∗ 1 = mΩ(xy˙ − xy˙ ) T∗ 0 = 1 2 mΩ2(x2 + y2) Note that a contribution T∗ 1 of the first order in the generalized velocities appears for the first time; it will be responsible for gyroscopic forces [the system of Fig.1.10(b) is actually the simplest, where gyroscopic forces can be illustrated]. The potential V is associated with the extension of the springs; with the assumption of small displacements, V = 1 2 k1x2 + 1 2 k2y2 The damping force can be handled either by the virtual work, δWnc = −c1xδx ˙ or with dissipation function (1.53). In this case, D = 1 2 c1x˙ 2 The Lagrange equations read mx¨ − 2mΩy˙ + c1x˙ + k1x − mΩ2x = 0 my¨ + 2mΩx˙ + k2x − mΩ2y = 0 or, in matrix form, with q = (x, y)T , Mq¨+ (C + G) ˙q + (K − Ω2M)q = 0 (1.55) where M = " m 0 0 m # C = " c1 0 0 0 # K = " k1 0 0 k2 # are respectively the mass, damping and stiffness matrices, and G = " 0 −2mΩ 2mΩ 0 # (1.56) is the anti-symmetric matrix of gyroscopic forces, which couples the motion in the two directions; its magnitude is proportional to the inertia (m)
26 1 Lagrangian dynamics of mechanical systems and to the rotating spee e again,the V.G K d C e matrice can be written the various energy terms appearing m MQ =Gq -2Ma (1.57) V-3Kq D-C Note that the modified potential V+=V-T6=59'(K-22M）g (1.58) islonger positive definite ifmm -22g+(-22)x=0 前+22e+(品-22）y=0 To analyze the stability of the system,let us assume a solution of the form z=Xept,y=Yept;the corresponding eigenvalue problem is [p2+w-n2 22p Nontrivial solutions of this homogenous system of equations require that the determinant be zero,leading to the characteristic equation p+2p2(2+22)+(-22)2=0 The roots of this equation are
26 1 Lagrangian dynamics of mechanical systems and to the rotating speed Ω. The contribution −Ω2M is, once again, the centrifugal force. Note that, with the previous definitions of the matrices M, G, K and C, the various energy terms appearing in the Lagrangian can be written T∗ 2 = 1 2 q˙ TMq˙ T∗ 1 = 1 2 q˙ TGq T∗ 0 = Ω2 2 qTMq (1.57) V = 1 2 qT Kq D = 1 2 q˙ T Cq˙ Note that the modified potential V + = V − T∗ 0 = 1 2 qT ³ K − Ω2M ´ q (1.58) is no longer positive definite if Ω2 > k1/m or k2/m. Let us examine this system a little further, in the particular case where k1 = k2 = k and c1 = 0. If ω2 n = k/m, the equations of motion become x¨ − 2Ωy˙ + ³ ω2 n − Ω2 ´ x = 0 y¨ + 2Ωx˙ + ³ ω2 n − Ω2 ´ y = 0 To analyze the stability of the system, let us assume a solution of the form x = Xept, y = Y ept; the corresponding eigenvalue problem is " p2 + ω2 n − Ω2 −2Ωp 2Ωp p2 + ω2 n − Ω2 # (X Y ) = 0 Nontrivial solutions of this homogenous system of equations require that the determinant be zero, leading to the characteristic equation p4 + 2p2(ω2 n + Ω2) + (ω2 n − Ω2) 2 = 0 The roots of this equation are
1.8 Lagrange's equations with constraints 27 =-(n-2)2,p=-un+22 、"a2c of th 32( is plot is ofter n do ot be oscopic force w 2 1.8 Lagrange's equations with constraints Consider the case where the n generalized coordinates are not indepen- dent.In this case,the virtual changes of configurationmust satisfya set of m constraint equations of the form of Equ.(1.18): ∑=0 l=1,.m (1.59 The number of degrees of freedom of the system is n-m.In Hamilton's principle (1.45),the variations 6i are no longer arbitrary,because of Equ.(1.59),and the step leading from Equ.(1.45)to (1.46)is impossible. This difficulty can be solved by using Lagrange multipliers.The technique consists of adding to the variational indicator a linear combination of the constraint equations 三*宫u)-宫（医a0 (1.60)
1.8 Lagrange’s equations with constraints 27 p2 1 = −(ωn − Ω) 2, p2 2 = −(ωn + Ω) 2 Thus, the eigenvalues are all imaginary, for all values of Ω. Figure 1.11 shows the evolution of the natural frequencies with Ω (this plot is often called Campbell diagram). We note that, in contrast with the previous example, the system does not become unstable beyond Ω = ωn; it is stabilized by the gyroscopic forces. ! Ò !n !n Fig. 1.11. Campbell diagram of the system of Fig.1.10(b), in the particular case k1 = k2 and c1 = 0. 1.8 Lagrange’s equations with constraints Consider the case where the n generalized coordinates are not independent. In this case, the virtual changes of configuration δqk must satisfy a set of m constraint equations of the form of Equ.(1.18): X k alkδqk = 0 l = 1, .m (1.59) The number of degrees of freedom of the system is n − m. In Hamilton’s principle (1.45), the variations δqi are no longer arbitrary, because of Equ.(1.59), and the step leading from Equ.(1.45) to (1.46) is impossible. This difficulty can be solved by using Lagrange multipliers. The technique consists of adding to the variational indicator a linear combination of the constraint equations Xm l=1 λl ÃXn k=1 alkδqk ! = Xn k=1 δqk ÃXm l=1 λlalk! = 0 (1.60)