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Lagrange' s eequations of a suspended set of rods

  1. Apr 18, 2004 #1
    Lagrange' s equations of a suspended set of rods

    Three identical rods of length l and mass m are hinged together so as to form three sides of a square in a vertical plane. The two upper free ends are hinged to a rigid support. The system is free to move in its own plane. Use Lagrange' s equations to obtain the equations of motion of the system.

    I just need help in setting up T and V...I think that I can handle the rest, but any help with getting T and V would be greatly appreciated..I am really desparate here...this problem has been frustrating me a lot. I tried to use theta and phui but they led me nowhere
     
    Last edited: Apr 18, 2004
  2. jcsd
  3. Apr 18, 2004 #2
    I don' t think that you would have to use theta that' s for sure, but that' s just me.
     
  4. Apr 18, 2004 #3
    Surely someone can help me. Just hlpe me to set up T and V and I can take it from there. Please I am really desparate.
     
  5. Apr 18, 2004 #4
    Really just something...anything
     
  6. Apr 18, 2004 #5

    turin

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    Homework Helper

    student1938,
    You seem desperate, so I'm going to jump in here preemptively.

    First, notice that this is a 1-D problem, so your lagrangian should be 1-D (the configuration of the rods is restricted to a rombus, for which only the angle between adjacent sides causes a change in the configuration energy). The Lagrangian is a function of only 1 q, 1 q_dot, and time: L = f(q,q_dot,t).

    You can use whatever you want to represent this single generalized coordinate, so long as it can completely specify the configuration and state of motion. 2 suggestions that immediately come to mind are the angle that the rods attached to the ceiling make with some reference plane or the height of the horizontal rod.
     
  7. Apr 18, 2004 #6

    Doc Al

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    I'll just add my two cents to turin's response:

    Assuming I have the picture correct, this assembly can swing from side to side (in its plane), but that's it. The bottom rod is always horizontal. There is only one degree of freedom. (I'll use the angle with respect to vertical as my coordinate.)

    Start by writing the potential energy of each rod as a function of the angle of the side rods to vertical: PE of each side rod is -mg(L/2)cosθ; the PE of the bottom rod is -mgLcosθ. (I'm measuring PE from the top support--you can measure from any point you choose.)

    That should get you started... :smile: (Now you work out the KE.)
     
  8. Apr 18, 2004 #7
    Thanks for the PE ..it provided a lot of insight...I even think that I might be onto something for the KE:

    ok for the KE, I get (1/2)*I*theta^2...now I will be the same for the two side rods and different for the bottom rod. For a rod I get (1/3)*m*l^2 but for the bottom what do I use?
     
  9. Apr 19, 2004 #8
    Any suggestions?
     
  10. Apr 19, 2004 #9

    Doc Al

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    I assume you mean KE = 1/2*I*ω^2. The side rods can be treated as purely rotating. Here's a hint for the bottom rod: does it rotate?
     
  11. Apr 19, 2004 #10
    So then KE is just (1/2)*I*omega^2. The bottom rod does not rotate so there is no KE like you said. IS that all that is needed for the KE? how would I get omega? I think that I should just leave it the way it is
     
  12. Apr 19, 2004 #11
    No wait the rod moves so there must be some KE....(1/2)*m*(dl/dtheta)^2 if you take it from the tip?
     
  13. Apr 19, 2004 #12

    Doc Al

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    I never said that! :eek: It's just pure translation, not rotation. It's center of mass moves along with the bottom tip of the side rods.
    Omega is just your generalized velocity d(theta)/dt. Leave it.
     
  14. Apr 19, 2004 #13
    For some reason, I am just unable to figure it out...according to me, based on the last hint, it should be (1/2)*m*(l*cos(theta))^2...I am saying l cos theta because that is the distance to the CM from the tip of the side rods.....does that make sense? Any suggestions?
     
  15. Apr 19, 2004 #14
    Anything please!
     
  16. Apr 19, 2004 #15

    Doc Al

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    velocity of the bottom rod

    Ask yourself: When the tip of the side rod moves, how does the bottom rod move? I say it tracks it perfectly, just displaced horizontally. So the velocity of the bottom rod equals the velocity of the tip of the side rod. Make sense? If so, then what is the speed of the tip of the side rod?
     
  17. Apr 19, 2004 #16
    the speed should just be d(lcos(theta))/dt right? I think thta lcos(theta) is the horizontal distance and we are dividing by time here to get speed right?
     
  18. Apr 19, 2004 #17

    Doc Al

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    total speed

    Where does the cos(theta) term come in? I want the total speed of the bottom rod, not a particular component. What's the total speed of the bottom tip of a side rod?
     
  19. Apr 19, 2004 #18
    Then it should just be l/t but then wht happens to t?
     
  20. Apr 19, 2004 #19

    Doc Al

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    Not sure what you are talking about. The tip moves at a speed of lω.
     
  21. Apr 19, 2004 #20
    oh did you use v = omega X r where r = l and the angle is 90 degrees.....now that makes sense. So then T = (1/2)*m*(l*omega)^2 and I can now use Lagrange' s Eqs right?
     
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