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Lagrange equations of motion for hoop rolling down moving ramp.

  1. Oct 12, 2009 #1
    1. The problem statement, all variables and given/known data
    A hoop of mass m and radius R rolls without slipping down an inclined plane of mass M, which makes an angle [tex]\alpha[/tex] with the horizontal. Find the Lagrange equations and the integrals of the motion if the plane can slide without friction along a horizontal surface.

    2. Relevant equations
    L = T - U

    T = kinetic energy
    U = potential energy

    Inertia tensor of hoop = mr[tex]^{2}[/tex]

    3. The attempt at a solution
    Here's what I believe to be the situation.
    homework3-1.jpg

    I know that I have to break up the kinetic energy into its separate components, but I'm not entirely sure how. I have 2 examples to work with, one being a block sliding down a moving ramp and the other being a hoop rolling down a stationary ramp.

    Here's the block.

    homework3-1b.jpg

    Here's the hoop.
    homework3-1a.jpg

    With the block, we break it down into velocity along the x-axis and the y-axis for both the block and the ramp (with the ramp's y-velocity being 0). But with the hoop, we break up the velocity into an x'-axis (axis at the same angle as the ramp) and a [tex]\theta[/tex]. Both of the examples make sense to me, though combining them in this way leaves me a little confused.

    If I were to just solve with what I knew now, I would combine the rotational velocity, the velocity along the x-axis, and the velocity along the y-axis into the kinetic energy.
     
  2. jcsd
  3. Oct 12, 2009 #2

    kuruman

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    When you do Lagrangian problems, a good starting point is to decide on your generalized coordinates and then write any constraints relating them. So what are your generalized coordinates and what are your constraints?
     
  4. Oct 12, 2009 #3
    I'm going to use the hoop on the non-moving ramp as my reference for this.

    It has the equation of constraint being what describes the rolling without slipping, so
    f = dx' - r d[tex]\theta[/tex] = 0
    f = [tex]\dot{x}[/tex]' - r [tex]\dot{\theta}[/tex] = 0

    And y-ramp = 0.

    And the generalized coordinates are x' and [tex]\theta[/tex].

    That makes sense to me, but since the ramp is moving I'll have to describe the x-axis movement in terms of the x'-axis movement or describe the x'-axis movement in terms of the x-axis and y-axis, which is where I'm getting stuck.

    And I'm not sure why but whenever I preview the post it's not what I typed (in terms of the greek letters).
     
  5. Oct 12, 2009 #4

    kuruman

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    It is better to use x for the horizontal position of the ramp (more conventional) and s for the coordinate down the plane, relative to the plane. Keep angle θ as you have it. Can you write expressions for the kinetic energy T and the potential energy U in terms of these coordinates?
     
    Last edited: Oct 12, 2009
  6. Oct 12, 2009 #5
    Wow, that really helps. I forget what example it was when I was taking notes in class but I forgot with this approach you can kind of just use whichever axes you wish and it will still work.

    That being said, this is what I have come up with.

    T = 1/2 M [tex]\dot{x}[/tex][tex]^{2}[/tex] + 1/2 m [tex]\dot{s}[/tex][tex]^{2}[/tex] + 1/2 m r[tex]^{2}[/tex] [tex]\dot{\theta}[/tex][tex]^{2}[/tex]
    U = m g s sin[tex]\alpha[/tex]
     
    Last edited: Oct 12, 2009
  7. Oct 12, 2009 #6
    After much working it through, looking at sample problems, and asking for help, I finished it! Thank you for your help kuruman!
     
  8. Oct 12, 2009 #7

    kuruman

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    I hope you wrote the translational kinetic energy of the center of mass of the rolling hoop correctly. It is not simply

    [tex]\frac{1}{2}m\dot{s}^2[/tex]

    It is

    [tex]\frac{1}{2}m(\dot{s}cos\alpha+\dot{x})^2+\frac{1}{2}m(\dot{s}sin\alpha)^2[/tex]

    :wink:
     
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