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Lagrangian mechanics (problem with generalized coordinates)

  1. Jan 4, 2005 #1
    Dear friends,

    Well, I’ve got a problem to solve but I’m not gonna ask you to do it for me. Instead, what I need is an explanation of what I am doing wrong.

    The problem is as follows: we have a rod of mass m and length l hanging of a rail (don’t know how to call it). It moves as the diagram shows,. I am asked (using Lagrangian mechanics) to write down the movement equations. And once done so, imagine theta is constant, what is the equation of just the translation motion?

    The generalized coordinates I take are q and theta (alpha is fixed) and therefore the positions of the centre of mass of the rod is:

    X=qcos(alpha)-L/2sin(theta)
    Y= qsin(alpha)-L/2cos(theta)

    Now I write the Lagrangian (taking into account that the kinetic energy of the rod is the T of the center of mass plus the kinetic energy around the center of mass).

    Well, after cancelling all theta’s derivatives (cause theta is constant) I reach to this conclusion:

    1.equation mq’’=-mgsin(alpha)
    2. equation mq’’=g sin(theta)/cos(alpha+theta)

    Equation 1 makes sense, but the second doesn’t. That’s why I think I’m doing something wrong. Please help!
     

    Attached Files:

  2. jcsd
  3. Jan 4, 2005 #2

    dextercioby

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    The problem's not that simple as it looks.
    The "rail" (the stick :tongue2: ) is it fixed at the angle 'alpha'??If so,that simplifies things a little bit.
    The Lagrange function [itex] L=T-V [/tex]
    T is the kinetic energy of the rod,V is the potential energy.By the looks of it,the suspension point slips down the "rail" freely in the gravitational field of the earth.Let "q" be the generalized coordinate of the suspension point.
    The potential energy of the rod is
    [tex] V(t)=q(t)\sin\alpha-mg\cos \theta(t) [/tex]
    ,where \theta is the generalized coordinate chosen for the angular movement of the rod.
    The kinetic energy will be:Itheta dot
    [tex] T=\frac{I}{2}\dot{\theta}^{2}(t)=\frac{ml^{2}}{2}\dot{\theta}^{2}(t) [/tex]
    The fact that the suspension point is slipping down the "rail"contributes only to the potential energy of the rod.
    U'll need to find q=q(t) knowing it is moving in the gravitational field.To find it,make abstraction of the oscillating system and consider a point mass "m" in the suspension point and find it eq.of movement.

    Daniel.
     
    Last edited: Jan 4, 2005
  4. Jan 4, 2005 #3
    I don't get where you got the kinetic energy from. Don't you have to add the kinetic energy of the center of mass, which is
    1/2m(q'+L^2/4(theta)'-q'L(theta)'cos(alpha+theta))

    Thanks for your help
     
  5. Jan 4, 2005 #4

    dextercioby

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    Yes,sorry,i've misinterpreted the problem.Yes,then both of my expressions need modification.The KE for the center of the mass of the rod and the potential energy of the same point will have the expressions u posted.

    Daniel.
     
  6. Jan 4, 2005 #5
    Well, thanks again. But how do you interpret the result I got after applying Lagrange's equation and making all theta's derivatives zero (that's what the problem asks me to do)

    1.equation mq’’=-mgsin(alpha)
    2. equation mq’’=g sin(theta)/cos(alpha+theta)

    To me the 2. doesn't make sense. Or is it that sin(alpha)=-sin(theta)/cos(alpha+theta)?
     
  7. Jan 4, 2005 #6
    You have one generalized coordinate (fixing theta). How come you have 2 equations? My solution coinside with your first one.


    [tex] \ddot{q}=-g\sin(\alpha) [/tex]

    G.
     
  8. Jan 4, 2005 #7
    yes you're right. I must be stupid.
    I just calculated the equations considering theta not fixed. And afterwards I considered it fixed.

    Isn't it supposed to work this way, too?
     
  9. Jan 4, 2005 #8
    Gamma, actually the problem is aimed at calculating what happens to this particular solution of the equations when it suffers a little perturbation from this fixed point.
    That's why the teacher asked us to develop the equations considering

    (theta,theta',theta'')=(theta0,0,0)

    What am I suposed to do?
     
  10. Jan 4, 2005 #9

    dextercioby

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    Small perturbations from the equilibrium positions are considered linearizing the sytem of equations/equation.That means:
    [tex] \cos\theta(t)\rightarrow 1-\frac{\theta^{2}(t)}{2} [/tex]
    and similar
    [tex] \sin\theta(t)\rightarrow \theta(t) [/tex]
    It's standard trick to analyze small oscillations.

    Daniel.
     
  11. Jan 4, 2005 #10
    I'm sorry but I'm getting nowhere.

    I understood how to analize small oscillations but I can't integrate the result I get. Intuitively, what is it that I have to get out of this? A physical pendulum that is moving?

    Is it of any help the principle of conservation of Energy?
     
  12. Jan 5, 2005 #11
    How would you calculate q(t)?
     
  13. Jan 7, 2005 #12
    I'm still working on this but don't know how to solve the system of differential equations I get after linearializing it. Is it of any help calculating the Hamiltonian? Or applying the energy conservation principle?
     
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