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Masses on an inclined plane and their Lagrangian

  1. Mar 22, 2014 #1

    ShayanJ

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    Consider the configuration below shown in the attached picture!
    The wedge can slide on the inclined plane and the cube on the wedge.Their motion is described by [itex] x_1 [/itex] and [itex] x_2 [/itex] respectively. There is no friction and the inclined plane doesn't move.
    Here's the Lagrangian of the system:
    [itex]
    \mathcal L=\frac{1}{2} M \dot{x}_1^2+\frac{1}{2}m(\dot{x}_1^2+\dot{x}_2^2-2 \dot{x}_1\dot{x}_2 \cos\theta)+(M+m)gx_1 \sin\theta
    [/itex]
    And the Euler-Lagrange equations are:
    [itex]
    M\ddot{x}_1+m(\ddot{x}_1-\ddot{x}_2\cos\theta)=(M+m)g\sin\theta\\
    m(\dot{x}_2-\dot{x}_1\cos\theta)=const
    [/itex]
    I wanna know,what is the constant of motion appearing above?
    One may expect it to be the component of total linear momentum parallel to the ground but it can't be that because it doesn't involve M.So what it is?
    Thanks
     

    Attached Files:

  2. jcsd
  3. Mar 22, 2014 #2
    Don't know the answer to your question but if there is no friction won't both masses accelerate together?
     
  4. Mar 22, 2014 #3

    ShayanJ

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    Of course the motion of M may cause m to move, whether there is friction or not.
     
  5. Mar 22, 2014 #4
    It's the horizontal component of the momentum of the mass m (The top block), obviously...
     
  6. Mar 22, 2014 #5

    ShayanJ

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    No, the horizontal component of the momentum of the mass m is [itex] \dot{x}_2 + \dot{x}_1\cos\theta [/itex].
    Anyway, still another question remains. So where is the horizontal component of the momentum of the wedge?That should remain constant too.
     
  7. Mar 22, 2014 #6
    You must've made a sign mistake somewhere than because that's the one quntity that 's conserved in this problem
    No because the normal force between the wedge and the fixed incline pushes the wedge sideways.
     
  8. Mar 22, 2014 #7
    If there is no friction, the small block will have no horizontal motion. Unless it had some initial horizontal speed.

    And are you sure you have that minus sign in the kinetic energy?
     
  9. Mar 22, 2014 #8
    Yes, there is a sign mistake in your Lagrangian. the term with a cosine in it should be positive.
     
  10. Mar 22, 2014 #9

    ShayanJ

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    Not, obviously both speeds have a common positive direction which means they should be added to give the speed relative to the inclined plane.

    The normal force is equal and opposite to [itex] (M+m)g\cos\theta [/itex], and so is canceled out.Otherwise there could be an acceleration normal to the inclined plane which causes the wedge to fly away!!!
     
  11. Mar 22, 2014 #10
    The sign mistake is in the Lagrangian
    You're getting confused. We're talking about horizontal components. [itex] (M+m)g\cos\theta [/itex] isn't horizontal. Gravity has no horizontal component.
     
  12. Mar 22, 2014 #11

    ShayanJ

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    Why do you tend backward when the bus you're standing in accelerates forward?

    Yeah, Looks like that should be positive.
    But still the problem remains that where is the x component of the momentum of the wedge?
    Or why isn't that conserved?
     
  13. Mar 22, 2014 #12

    ShayanJ

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    Yeah... I got that.
    In fact we're confusing two different horizontals!
    Gravity has no component parallel to the ground. But it has a component parallel to the inclined plane. When you talk about the normal force from the inclined plane to the wedge, the weight of the wedge does make a contribution.
     
  14. Mar 22, 2014 #13
    I already answered that question. You may not have believed my answer but it is in fact correct.
    You need to hone your intuition a bit more.
     
  15. Mar 22, 2014 #14
    When I say horizontal I usually mean the actual horizontal.
     
  16. Mar 22, 2014 #15

    ShayanJ

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    If you mean this:
    I already answered that too. If it wasn't canceled, it would give the wedge an acceleration normal to the inclined plane!!! It is canceled by the component of the weight of the wedge normal to the inclined plane.
     
  17. Mar 22, 2014 #16
    That's ridiculous now. We're talking about the HORIZONTAL component of the normal which cannot be canceled by gravity. You're stuck thinking about perpendicular and parallel components to the incline. I'm talking about HORIZONTAL component.
     
  18. Mar 22, 2014 #17
    Another way to look at it is that the normal is cancelled by one component of gravity, but the other component of gravity accelerates the wedge down the ramp and that acceleration has a horizontal component. Either way there is a horizontal acceleration.
     
  19. Mar 22, 2014 #18

    ShayanJ

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    Ohh...dammit...yeah...sorry. This was a problem with the word. It was like my mind wasn't interpreting the word horizontal correctly. That's a problem when you're not a native speaker. Sorry.You're right.
     
  20. Mar 22, 2014 #19
    For the horizontal direction, the small block moves (accelerates) in respect to the wedge but not in respect to the ground or the inclined plane.
     
  21. Mar 22, 2014 #20

    ShayanJ

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    Yeah,That's the thing one of the equations is telling us. Its horizontal component of momentum is conserved. I was talking about motion w.r.t. the wedge.
     
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