18 1 Lagrangian dynamics of mechanical systems V=V(a,n:t) (1.4) the L=T"-V=L(gl,.qn:41,.n:t) (1.42) Let us now examine the virtual work of the non-conservative forces:ex pressing the virtual displacement r in terms of one finds me=∑R函=∑∑h0m=∑Qm 1.43) where -∑0 (1.44 ergy).Introducing Equ.(1.43)in Hamilton's principle(1.26),one finds VL=1=6L(,m,m.mi0+∑Q.6qdt =∑g:+)+∑o,lt and,upon integrating by part to eliminate using “=()盖(0）) one finds =紫-广∑品()-张-Q1业=04同 s are arbitrary (are independent),one must ha i=1.n (1.46)
18 1 Lagrangian dynamics of mechanical systems V = V (q1, ., qn;t) (1.41) From these two equations, it can be assumed that the most general form of the Lagrangian is L = T∗ − V = L(q1, ., qn, q˙1, ., q˙n;t) (1.42) Let us now examine the virtual work of the non-conservative forces; expressing the virtual displacement δxi in terms of δqi, one finds δWnc = X i Fi.δxi = X i X k Fi ∂xi ∂qk δqk = X k Qkδqk (1.43) where Qk = X i Fi ∂xi ∂qk (1.44) Qk is the generalized force associated with the generalized variable qk; they are energetically conjugate (their product has the dimension of energy). Introducing Equ.(1.43) in Hamilton’s principle (1.26), one finds V.I. = δI = Z t2 t1 [δL(q1, ., qn, q˙1, ., q˙n;t) + XQiδqi]dt = Z t2 t1 [ X i ( ∂L ∂qi δqi + ∂L ∂q˙i δq˙i) + XQiδqi]dt and, upon integrating by part to eliminate δq˙i, using ∂L ∂q˙i δq˙i = d dt µ ∂L ∂q˙i δqi ¶ − d dt µ ∂L ∂q˙i ¶ δqi one finds δI = X i [ δL δq˙i δqi] t2 t1 − Z t2 t1 X i [ d dt µ ∂L ∂q˙i ¶ − ∂L ∂qi − Qi]δqidt = 0 (1.45) The first term vanishes because δqi(t1) = δqi(t2) = 0, and since the virtual variations are arbitrary (qi are independent), one must have d dt( ∂L ∂q˙i ) − ∂L ∂qi = Qi i = 1, .n (1.46)
1.7 Lagrange's equations 19 grange's equation s,their nun r is equal to the mber n (1.43).Once the of the loc in s of the eralized coordinates has been found.Equ.(16)allows us to write the differential equations governing the motion in a straightforward way. 1.7.1 Vibration of a linear,non-gyroscopic,discrete system neral form of the kinetic coenergy of a linear non-gyroscopic,dis- T-Mi (1.47) where is a set of generalized coordinates and M is the matrix A s the fact that anv velocity distribution must lead to a non- ive value of the kinetic co energy:M is strictly positive definite if all the coordinates have an inertia associated to them,so that it is impossible to find a velocity distribution such that T=0.Similarly,the general form of the strain energy is V= (1.48) where K is the stiffness matrix,also symmetric and semi-positive definite. A rigid body mode is a set of generalized coordinates with no strain energy in the system.K is strictly positive definite if the system does not have L-T-V-Mi-Kz (1.49) the virtul work of the te).one gets the equation of rces can be written Wne applying the Lagrange equa- Mi+Kx=f (1.50) 1.7.2 Dissipation function
1.7 Lagrange’s equations 19 These are Lagrange’s equations, their number is equal to the number n of independent coordinates. The generalized forces contain all the nonconservative forces; they are obtained from the principle of virtual work (1.43). Once the analytical expression of the Lagrangian in terms of the generalized coordinates has been found, Equ.(1.46) allows us to write the differential equations governing the motion in a straightforward way. 1.7.1 Vibration of a linear, non-gyroscopic, discrete system The general form of the kinetic coenergy of a linear non-gyroscopic, discrete mechanical system is T∗ = 1 2 x˙ TMx˙ (1.47) where x is a set of generalized coordinates, and M is the mass matrix. M is symmetric and semi-positive definite, which translates the fact that any velocity distribution must lead to a non-negative value of the kinetic coenergy; M is strictly positive definite if all the coordinates have an inertia associated to them, so that it is impossible to find a velocity distribution such that T∗ = 0. Similarly, the general form of the strain energy is V = 1 2 xT Kx (1.48) where K is the stiffness matrix, also symmetric and semi-positive definite. A rigid body mode is a set of generalized coordinates with no strain energy in the system. K is strictly positive definite if the system does not have rigid body modes. The Lagrangian of the system reads, L = T∗ − V = 1 2 x˙ TMx˙ − 1 2 xT Kx (1.49) If, in addition, one assumes that the virtual work of the non-conservative external forces can be written δWnc = fT δx, applying the Lagrange equations (1.46), one gets the equation of motion Mx¨ + Kx = f (1.50) 1.7.2 Dissipation function In the literature, it is customary to define the dissipation function D such that the dissipative forces are given by
20 1 Lagrangian dynamics of mechanical systems Q:=-0 (1.51) If this definition is used,Equ.(1.46)becomes (1.52) where includes all the non-conservative forces which are not already included in the dissipation function.Viscous damping can be represented by a quadratie dissipation function.If one assumes D=iTCi (1.53) in previous section,one gets the equation of motion Mi+Ci+Kx=f (1.54） and s trate ome of the the ethod. 0X- o (b) (c) m Disk of mass m %1 1.7.3 Example 1:Pendulum with a sliding mass Consider the pendulum of Fig.1.8(a)where a mass m slides without fric neld gi a ear spring of s to the du the pen tem has two d ion of the mass along the bar)an
20 1 Lagrangian dynamics of mechanical systems Qi = −∂D ∂q˙i (1.51) If this definition is used, Equ.(1.46) becomes d dt µ ∂L ∂q˙i ¶ + ∂D ∂q˙i − ∂L ∂qi = Qi (1.52) where Qi includes all the non-conservative forces which are not already included in the dissipation function. Viscous damping can be represented by a quadratic dissipation function. If one assumes D = 1 2 x˙ T Cx˙ (1.53) in previous section, one gets the equation of motion Mx¨ + Cx˙ + Kx = f (1.54) where C is the viscous damping matrix, also symmetric and semi-positive definite. We now examine a few examples of mechanical systems, to illustrate some of the features of the method. o q1 q : 2 q : 1 (b) q1 q2 o (c) Disk of mass and moment of inertia m I q1 q2 g o (a) m k k Fig. 1.8. Pendulum with a sliding mass attached with a spring. (a) and (b): Point mass. (c) Disk. 1.7.3 Example 1: Pendulum with a sliding mass Consider the pendulum of Fig.1.8(a) where a mass m slides without friction on a massless rod in a constant gravity field g; a linear spring of stiffness k connects the mass to the pivot O of the pendulum. This system has two d.o.f. and we select q1 (position of the mass along the bar) and
1.7 Lagrange's equations 21 (angle of the pendum)the t its velocity can be express sed in gy ortho ronal dir that T=。m(最+2g) The potential energy reads V=-mgm cos2+ The first contribution taken at the pivot O)and the s (assumed unstretched when=).The Lag e st e equations are ()+=0 mi -mgz-mg cos g2+ka =0 no longer point m a disk g1.8g d body is the sum of the kinetic coenergy of translation of the total mass lumped at the center of mass and the kinetic coenergy of rotation around the center of mass): T=与m(G+呢)+1弱 djas the ame a 6=e2=B=MrB (M=);the additional contribution to the kinetic coenergy is Note that,this includes the translational energy as well as the rotationa e moment of o refe ers to the nergy -n91C0sq2 /2(the cente s is a mid length)
1.7 Lagrange’s equations 21 q2 (angle of the pendulum) as the generalized coordinates. It is assumed that, when q1 = 0, the spring force vanishes. The kinetic coenergy is associated with the point mass m; its velocity can be expressed in two orthogonal directions as in Fig.1.8(b); it follows that T∗ = 1 2 m( ˙q2 1 + ˙q2 2q2 1) The potential energy reads V = −mg q1 cos q2 + 1 2 k q2 1 The first contribution comes from gravity (the reference altitude has been taken at the pivot O) and the second one is the strain energy in the spring (assumed unstretched when q1 = 0). The Lagrange equations are d dt(mq2 1 q˙2) + mg q1 sin q2 = 0 m q¨1 − mq1q˙ 2 2 − mg cos q2 + kq1 = 0 If one assumes that the mass m is no longer a point mass, but a disk of moment of inertia I sliding along the massless rod [Fig.1.8(c)], it contributes an extra term to the kinetic coenergy, representing the kinetic coenergy of rotation of the disk (the kinetic coenergy of a rigid body is the sum of the kinetic coenergy of translation of the total mass lumped at the center of mass and the kinetic coenergy of rotation around the center of mass): T∗ = 1 2 m( ˙q2 1 + ˙q2 2q2 1) + 1 2 Iq˙ 2 2 The disk has the same potential energy as the point mass. Furthermore, if the rod is uniform with a total mass M and a length l, its moment of inertia with respect to the pivot is Io = Z l 0 ̺x2dx = ̺l3/3 = Ml2/3 (M = ̺l); the additional contribution to the kinetic coenergy is Ioq˙ 2 2/2. Note that, this includes the translational energy as well as the rotational energy of the rod, because the moment of inertia I0 refers to the fixed point at the pivot. The bar has also an additional contribution to the potential energy: −mg l cos q2/2 (the center of mass is at mid length)
1 Lagrangian dynamics of me anical system (a) (e) 、Instability w0 10 Fig.1.9.Rotating pendulum 1.7.4 Example 2:Rotating pendulum Consider the rotating pendulum of Figure 1.9(a).The point mass m is connected by a massless rod to a pivot which rotates about a vertical axis constant v th ite the kin int ig.1.9(b).One vertical axis with fixed,while the other one is tangent to the trajectory of the mass in the plane of the pendulum when it does not rotate about the vertical axis;the projected components are respectively Isin and 16.Being orthogonal,it follows that T=g【oP+2sim9月 Note that the first term is quadratic in [T2'in(1.39)],while the second term is independent of and appears as a potential of centrifugal forces To'in (1.39)].Taking the reference altitude at the pivot,the gravity potential is V =-gml cos and L=T-v [(i)+(12 sim+gmi cos0 The corresponding Lagrange equation reads
22 1 Lagrangian dynamics of mechanical systems (a) ò Ò l m Pivot Ò ò lsin ò lsin ò Ò l ò : ! Ò !0 !0 Instability (b) (c) Fig. 1.9. Rotating pendulum. 1.7.4 Example 2: Rotating pendulum Consider the rotating pendulum of Figure 1.9(a). The point mass m is connected by a massless rod to a pivot which rotates about a vertical axis at constant velocity Ω; the system is in a vertical gravity field g. Because Ω is constant, the system has a single d.o.f., with coordinate θ. In order to write the kinetic coenergy, it is convenient to project the velocity of the point mass in the orthogonal frame shown in Fig.1.9(b). One axis is tangent to the circular trajectory when the pendulum rotates about the vertical axis with θ fixed, while the other one is tangent to the trajectory of the mass in the plane of the pendulum when it does not rotate about the vertical axis; the projected components are respectively lΩ sin θ and l ˙ θ. Being orthogonal, it follows that T∗ = m 2 h (l ˙ θ) 2 + (lΩ sin θ) 2 i Note that the first term is quadratic in ˙ θ [T2 ∗ in (1.39)], while the second term is independent of ˙ θ and appears as a potential of centrifugal forces [T0 ∗ in (1.39)]. Taking the reference altitude at the pivot, the gravity potential is V = −gml cos θ and L = T∗ − V = m 2 h (l ˙ θ) 2 + (lΩ sin θ) 2 i + gml cos θ The corresponding Lagrange equation reads