University:
California State University, Northridge
Course:
ME 584 | Modeling and Simulation of Dynamic Systems
Academic year:

2010
Views:

38

Pages:

17
Author:

beddgell23k

1 Lagrange’s Equation chp3 2 Lagrange’s Method • Newton’s method of developing equations of motion requires taking elements apart • When forces at interconnections are not of primary interest, more advantageous to derive equations of motion by considering energies in the system • Lagrange’s equations: – Indirect approach that can be applied for other types of systems (other than mechanical) – Based on calculus of variations – finding extremums of quantifies expressible as integrals chp3 3 Lagrange’s Equations d  T  T R U     Qi   dt  qi  qi qi qi where : qi : independent coordinates necessary to describe system's motion at any instant Q i : corresponding loading in each coordinate U  f1 (qi ) : potential energy in terms of coordinates T  f 2 (qi2 ) : kinetic energy in terms of systemmasses, mass inertias, linear/ang ular veloc ities R  f 3 (qi2 ) : energy dissipation due to viscous friction chp3 4 Deriving Equations of Motion via Lagrange’s Method 1. Select a complete and independent set of coordinates qi’s 2. Identify loading Qi in each coordinate 3. Derive T, U, R 4. Substitute the results from 1,2, and 3 into the Lagrange’s equation. chp3 5 Example 11: Spring-Mass-Damper System Independent coordinate: q = x Substitute into Lagrange’s equation: chp3 6 Example 12: Pair-Share: Restrained Plane Pendulum • A plane pendulum (length l and mass m), restrained by a linear spring of spring constant k and a linear dashpot of dashpot constant c, is shown on the right. The upper end of the rigid massless link is supported by a frictionless joint. Derive the equation of motion. chp3 7 Example 12: Pair-Share: Restrained Plane Pendulum q   , Q  0, 1 R  c(a) 2 2 1 1 T  mv 2  m(l) 2 2 2 1 U  mgl  mgl cos   k (b ) 2 2 d T T R U (  )   Q dt     ml 2  ca 2  mgl sin   kb2  0 Linearize sin    for small  ml 2  ca 2  mgl  kb2  0 chp3 8 Example 13: Bead on a Spinning Wire Hoop • • • • One degree of freedom, q1=angle  as an independent coordinate Velocity of bead: R . ; Velocity of hoop:   R sin   1 2 1 2 Kinetic energy: 2 T  mv  m R   R sin   2 2  Potential energy relative to its position at the bottom of the hoop (when the hoop is not rotating and  = 0), is U  mgR  mgR cos  • •  R = 0, Q = 0 Substitute into Lagrange’s equation: Solving for the angular acceleration: R  d T T R U (  )   Q dt     mR2  mR2 2 sin  cos   mgR sin   0 •  g      2 cos    sin  . R  chp3 9 Example 14: Pair-Share: Copying machine • Use Lagrange’s equation to derive the equations of motion for the copying machine example, assuming potential energy due to gravity is negligible. q1=y, q2= θ Q1 = F, Q2 = 0 θ y chp3 Example 14: Pair-Share: Copying machine 10 chp3 11 Example 15: Mass Spring Dashpot Subsystem in Falling Container • A mass spring dashpot subsystem in a falling container of mass m1 is shown. The system is subject to constraints (not shown) that confine its motion to the vertical direction only. The mass m2, linear spring of undeformed length l0 and spring constant k, and the linear dashpot of dashpot constant c of the internal subsystem are also shown. • Derive equation(s) of motion for the system using – x1 and x2 as independent coordinates – y1 and y2 as independent coordinates chp3 12 Example 15: Mass Spring Dashpot Subsystem in Falling Container Let q1  x1 , q2  x2 be positions of m1 and m 2 with respect to horizontal ground . Q1, 2  0 1 1 m1 x12  m2 x 22 2 2 1 U  k ( x1  x2 ) 2  m1 gx1  m2 gx2 2 1 R  c ( x1  x 2 ) 2 2 Substitute into Lagrange' s Equation : T  d T T R U ( )    Q1 dt x1 x1 x1 x1 m1 x1  c ( x1  x 2 )  k ( x1  x2 )  m1 g  0 d T T R U ( )    Q2 dt x 2 x2 x 2 x2 m1 x2  c ( x 2  x1 )  k ( x2  x1 )  m2 g  0 chp3 13 Example 15: Mass Spring Dashpot Subsystem in Falling Container Let q1  y1 be inertial position of m1 with respect to horizontal ground q2  y2 be position of m2 relative to m1 and from unstretche d length of spring Q1, 2  0 1 1 m1 y12  m2 ( y1  y 2 ) 2 2 2 1 2 U  ky2  m1 g y1  m2 g ( y1  y2 ) 2 1 2 R  cy 2 2 Substitute into Lagrange' s Equation : T d T T R U ( )    Q1 dt y1 y1 y1 y1 (m1  m2 ) y1  m2 y2  m1 g  m2 g  0 d T T R U ( )    Q2 dt y 2 y2 y 2 y2 m2 y1  m2 y2  cy 2  ky2  m2 g  0 chp3 14 Example 16: A Block Sliding on a Wedge • • • Block mass m sliding down a wedge mass M Independent coordinates, q1 and q2, are shown, q1 is along the plane and is measured relative to the (moving) wedge. q1 Velocity of the wedge is q2 , but velocity of the block has components from both q2 α a q1 and q2: 2 2 2 vblock  vx2  v y2  q2  q1 cos a   q1 sin a    • Total kinetic energy is • The potential energy of the wedge can be taken as zero, and the block is 1 1 2 2 2  T  Mq2  m q2  q1 cos a   q1 sin a  2 2 U  mgq1 sin a • Substitute into Lagrange’s equations and differentiate wrt to q1 and q2 q1  g sin a . m 1 cos 2 a M m q2   m q1 cos a . M m chp3 Example 17: Pair-Share: Mass Pendulum Dynamic System • A simple plane pendulum of mass m0 and length l is suspended from a cart of mass m as sketched in the figure. The motion of the cart is restrained by a spring of spring constant k and a dashpot constant c; and the angle of the pendulum is restrained by a torsional spring of spring constant k, and a torsional dashpot of dashpot constant ct. Note that the constitutive relations for the torsional spring and torsional . dashpot are linear expressions τ=ktθ and τ=ctθ, respectively, where τ is the torque and θ is the change in the angle across the element from its undeformed configuration. • The torsional spring is undeformed when the pendulum is in its downwardhanging equilibrium position. Also, m0 is acted upon by a known dynamic force F0sinωt, which remains horizontal, regardless of the angle θ and the motion of the cart. The system is constrained to remain in the plane of the sketch and the cart remains on the bed throughout its motion. Derive the equation(s) of motion for the system. 15 chp3 Example 16: Pair-Share: Mass Pendulum Dynamic System Two independent coordinates : q1  x, q2   Q1  F0 sin t and Q2  ( F0 sin t )l cos  Rx  1 2 1 cx and R  ct 2 2 2 1 2 1 kx  kt 2  m0 gl (1  cos  ) 2 2 T  Tcart  T pendulum U 1 2 mx 2 1 1 T pendulum  m0 ( x  l cos  ) 2  (l sin  ) 2  m0 x 2  (l) 2  2 xl cos  2 2 Substitute into Lagrange' s Equations : (m  m ) x  cx  kx  m l ( cos    2 sin  )  F sin t Tcart   0  0  0 m0l  ct  kt  m0 gl sin   m0 xl cos   ( F0 sin t )l cos   16 chp3 17 Comparison of Newton’s and Lagrange’s Methods Newton’s (Direct Approach) Lagrange’s (Indirect Approach) Accelerations required Velocities required Generally vectors required Generally scalars required Free-body Diagrams useful Free-body diagrams not useful All forces considered Workless forces (constraints) forces not considered All forces handled via same expression Conservative and non-conservative forces handled separately Intermediate forces more readily available Intermediate forces less readily available

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