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Lagrangian of a Pendulum on a rotating circle

  1. Nov 6, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the Lagrangian of a simple pendulum of mass m whose point of support moves uniformly on a vertical circle with constant angular velocity.

    (So basically there is a circle around the origin that spins with a constant angular velocity and the pendulum is attached to the end of the circle.)

    2. Relevant equations
    [tex]\frac{d}{dt}\frac{∂L}{∂ \dot{q}_k}=\frac{∂L}{∂q_k}[/tex]

    3. The attempt at a solution
    I am stuck right in the beginning in trying to find the kinetic energy T. I know that the kinetic energy of a pendulum is..
    [tex]T=\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)[/tex]
    But since r remains constant on the pendulum because it is rigid, the equation reduces to..
    [tex]T=\frac{1}{2}mr^2\dot{\theta}^2[/tex]

    I am unsure how to add the rotating circle in which the pendulum is a attached to. If the problem stated that there was a mass where the pendulum was attached to, it would be easier I think. Should I view this circle as mass-less? How would I get the kinetic energy then? Can anyone provide some help?? :(??
     
  2. jcsd
  3. Nov 6, 2011 #2

    diazona

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    That's the expression for KE of a pendulum attached to a fixed support, expressed in polar coordinates. It's convenient to use polar coordinates because the origin of the coordinate system can be set at the support of the pendulum, and the fixed length of the pendulum corresponds to one of the coordinates being fixed.

    But when you have a pendulum whose support is not fixed, the reason for using polar coordinates disappears. There's no reason to expect that the fixed length of the pendulum will correspond to any one coordinate being fixed. And you can't have a coordinate system with a moving origin. (Well, you can, but you have to introduce fictitious forces and it gets really complicated...)

    When in doubt, it's usually easiest to start from scratch with Cartesian coordinates. So try writing down the position of the pendulum bob as a function of time in Cartesian coordinates.
    Your expression will still involve [itex]r[/itex] and [itex]\theta[/itex], but they will only be physical parameters of the state of the pendulum, not coordinates.
     
  4. Nov 6, 2011 #3
    Okay so using your advise, this is my work.
    (I just scanned my work because I didn't feel like typing it up. If it is too sloppy let me know!)

    http://img256.imageshack.us/img256/2551/imglvk.jpg [Broken]

    EDIT: Nah this isn't right either, It still isn't in Cartesian coordinates AND I assumed both angles would be the same at all times, when in fact they are not. :\
     
    Last edited by a moderator: May 5, 2017
  5. Nov 6, 2011 #4
    Okay so if I used what I got in the picture, but this time make [tex]θ_1 ≠ θ_2[/tex], I get this as the Lagrangian..

    [tex]L=\frac{1}{2}m[a^2+l^2-2alcos(θ_1-θ_2)]-acosθ_1-lcosθ_2[/tex]

    From which I get the two equations...
    [tex](mal)sin(θ_1-θ_2)-asinθ_1=0[/tex]
    [tex]-(mal)sin(θ_1-θ_2)-lsinθ_1=0[/tex]

    Am I close?? :\??

    EDIT: *Sigh* I forgot to do the chain rule and include [itex]\dot{θ_1}[/itex] and [itex]\dot{θ_2}[/itex] :\
     
    Last edited: Nov 6, 2011
  6. Nov 8, 2011 #5

    diazona

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    Sorry about the delay... anyway, your Lagrangian seems almost right, just check the sign of your definition of [itex]y[/itex].

    Before you move on to computing the equations of motion, what do you know about [itex]\theta_1[/itex], specifically about how it changes in time?
     
  7. Nov 8, 2011 #6
    Thats okay! :)
    Theta 1 dot would be constant no? Would that change the lagrangian in anyway? :eek:?
     
  8. Nov 8, 2011 #7

    diazona

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    Yes it would. Well, not so much the Lagrangian, but rather the equations of motion. Think about this: if [itex]\dot\theta_1[/itex] is a constant, then what is [itex]\partial\dot\theta_1[/itex] in the E-L equation?
    [tex]\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial\dot\theta_1} = \frac{\partial L}{\partial\theta_1}[/tex]

    Also don't forget: if [itex]\dot\theta_1[/itex] is constant, what can you say about [itex]\theta_1[/itex]?
     
  9. Nov 8, 2011 #8
    Ah! So [itex]\partial\dot\theta_1[/itex] would be equal to zero yes? Meaning that whole first term would be zero? And if [itex]\dot{θ_1}[/itex] is constant then [itex]θ_1[/itex] is linear??? (Only thing I can think of!) Ill crunch through the Lagrangian later today to see what I can get.
     
  10. Nov 8, 2011 #9

    diazona

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    Hint: [itex]\partial\dot\theta_1 = 0[/itex], so when you see [itex]\frac{\partial L}{0}[/itex] what should you be thinking?

    It is indeed true that if [itex]\dot x[/itex] is constant, [itex]x[/itex] is a linear function of time.
     
  11. Nov 8, 2011 #10
    Wouldn't that mean it is undefined? I an not entirely sure what that would mean. :confused:
     
  12. Nov 8, 2011 #11
    I did this exact same problem about a month ago in my theoretical mechanics class, I'll try to remember this thread and if you get stuck give you some tips on what I did to solve it.
     
  13. Nov 8, 2011 #12

    diazona

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    Yeah, exactly. The point is that when you find yourself dividing by zero in the Euler-Lagrange equation, it means you're writing an equation for something that isn't actually a variable of motion - in other words, you shouldn't be writing that equation at all. And in fact you don't need to, because you already know how [itex]\theta_1[/itex] behaves without having to solve for it. So go back and replace [itex]\theta_1[/itex] with an explicit function of time in the Lagrangian, then see where you need to go from there. (I suggest using a different letter such as [itex]\omega[/itex] for [itex]\dot\theta_1[/itex], so you remember that it's a known parameter)
     
  14. Nov 10, 2011 #13
    r(x,y) = (Rcosωt + lsinβ, Rsin(ωt) -lcosβ)

    take the time derivative to find velocity

    it falls out pretty nicely from r(x,y)
     
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