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  1. Lecture 18: Review and Examples PS4 problem 2 The double pendulum A four degree of freedom problem

  2. 2. Consider the two degree of freedom system shown below. Suppose m1 = 1 kg = m2, and k1= 1000 N/m = k2. Find the motion of y1if the right hand mass is struck impulsively from the right so that the initial motion of the right hand mass is to the left. Suppose the blow to come from a 2 kg mass moving at 6 m/s and let the collision be perfectly elastic. k1 k2 m1 m2 y1 y2

  3. This is an impulse problem, so we need to work an initial value problem There is no particular solution for the problem as posed. k1 k2 m1 m2 For a change of pace, let’s do this using our new skills y1 y2 y1 and y2 are fine as generalized coordinates

  4. Steps 5-8 for y1 Steps 5-8 for y2 The vector equations of motion

  5. Seek exponential solutions Let y1 and y2 go like exp[st]

  6. The characteristic polynomial with the numbers is The frequencies are w1 = 19.54 and w2 = 51.17 and all we need to do is find the modal vectors, and we know how to do that for these simple 2 x 2 systems

  7. Look for the modal vectors Modal vectors are then

  8. And the homogeneous solution is and its derivative is

  9. The initial conditions are that k1 k2 m1 m2 where P denotes the amount of momentum transferred, here twice the input momentum because the collision is perfectly elastic y1 y2

  10. can be rewritten which we can solve to get

  11. Recall the problem statement 2. Consider the two degree of freedom system shown below. Suppose m1 = 1 kg = m2, and k1= 1000 N/m = k2. Find the motion of y1if the right hand mass is struck impulsively from the right so that the initial motion of the right hand mass is to the left. Suppose the blow to come from a 2 kg mass moving at 6 m/s and let the collision be perfectly elastic. The input momentum is 12, so we have P = 24 We can plug in all the numbers and plot the displacements of the two masses

  12. Red denotes the mass that was struck, blue the other mass

  13. So what did I do? Found the equations of motion using the Euler-Lagrange process No damping No external forces because we address impulse problems using initial conditions The equations are linear, so there’s no need to linearize Homogeneous odes with constant coefficients, so I sought an exponential solution

  14. Let y1 and y2 go like exp[st] Write the differential equations in matrix form and use the homogeneous matrix equations to find the modal frequencies Find the modal vectors Write the general homogeneous solution in terms of the model frequencies and the modal vectors This solution contains four constants to be determined Apply the initial conditions to determine the coefficients in the general solution

  15. QUESTIONS?

  16. THE DOUBLE PENDULUM z y The two angles are the obvious generalized coordinates q1 m1 (y1, z1) Let’s see how we get there q2 m2 (y2, z2)

  17. z y constraints q1 m1 (y1, z1) their derivatives q2 m2 (y2, z2) combine for the first mass

  18. The second mass is messier Finally so which simplifies to

  19. z And we can combine all of this to get our energies and the Lagrangian y q1 m1 We can identify the angles with the qs without necessarily relabeling them (y1, z1) q2 m2 (y2, z2)

  20. For q1

  21. we have a cancellation We can linearize this

  22. For q2 We’ll have a similar cancellation when we combine (6) and (7) as we did for the first Euler-Lagrange equation

  23. which we can also linearize

  24. This is the basis for the forced problem and for the damped problem for both of these we want to have the free modes and the matrix The development here is for zero damping Some of the following may be a bit telegraphic; I’ll put up the Mathematica when we’ve got through it.

  25. Free modes: Natural frequencies and eigenvectors Because there is no damping, I can assume that the linear solutions will be harmonic We convert to matrix form as before — remember how? We’ll need this matrix for the particular solution as well for now all I care about is the homogeneous solution

  26. Find the natural frequencies by setting the determinant equal to zero Finding these symbolically leads to very large expressions choose m1 = 1, m2 = 2, l1 = 1/2, l2 = 1, all in SI units The roots are (in floating point)

  27. The matrix equation becomes The symbolic modal vectors are We get the actual eigenvectors by substituting for the frequency

  28. And we have the unnormalized eigenvectors in floating point The low frequency has both pendulums moving in the same direction The high frequency has them moving in opposition This is another illustration of the general rule The higher frequencies correspond to most complicated motion

  29. Here is the motion of both pendulums in the low frequency mode

  30. Here is the motion of both pendulums in the high frequency mode

  31. What happens when we force the system?

  32. z Add a torque to the upper pivot y This will require a particular solution and a set of initial conditions q1 m1 (y1, z1) The torque does work when the first pendulum moves It does not do work when the second pendulum moves The generalized force appears in the q1 equation q2 m2 (y2, z2) The differential equations are

  33. We can get a particular solution by choosing the torque to be There’s no dissipation, so both angles will be in phase with the forcing

  34. The particular solution is then of the form

  35. We find the coefficients by inverting the old matrix with w replaced by the forcing frequency wf the matrix its inverse and the coefficients

  36. the particular solution where

  37. The initial value problem requires initial values for the angles and their rates of change Let’s start from rest

  38. We can write the general homogeneous solution in terms of the modal frequencies and modal vectors as The initial conditions determine the coefficients the forcing frequency determines the nature of the response

  39. We have angles and their derivatives to match This goes just like tonight’s first example The contributions from the particular solution

  40. The contributions from the homogeneous solution

  41. The initial angles are zero, so we have and we have b1 = 0 = b2

  42. The initial rotations are also zero, and so we have These can be solved fairly easily, but clumsily for a1 and a2 I’ll spare you the details and simply take a look at the results

  43. Low frequency forcing

  44. High frequency forcing

  45. Near the low frequency resonance

  46. Near the high frequency resonance

  47. Here’s a new example

  48. Here’s a four degree of freedom problem with damping and external forcing What’s the best way to choose position coordinates? y y+ a + y2 y+ 2a + y3 k k c M M M f y2 and y3 represent relative motion q (y4,z4) The damper is symbolic representing viscous resistance to the motion of the train it does not limit translation

  49. Let’s look at my choice of coordinates for a moment Spring forces will be zero when y2 and y3 are zero so the force equations must be These are the difference between the position of the right end and the position of the left end minus the rest length, a, and they reduce to the simple expressions

  50. Energies Constraints Generalized forces When the third mass moves, f does work; when the first mass moves, the damper does work We’ll get our forces from them, and assign them when we have our coordinates