UA - ME 431: Lagrange's Equations, Example

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  • Опубліковано 1 лют 2025

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  • @fcvgarcia
    @fcvgarcia Рік тому

    Thanks for sharing! Great explanation!

  • @eulerthegreatestofall147
    @eulerthegreatestofall147 9 місяців тому

    Good example, although, I was thinking that the potential energy associated to the second spring must be 1/2*k*(x2-x1)^2 not 1/2*k*(x2-2*x1)^2 as you indicated. I do really appreciate you elaborate more on that. Thanks!

    • @quinndd
      @quinndd  9 місяців тому

      Ha. The displacement of the top of disk 1 is twice its center, so y1 = 2*x1. With z = x2 - y1, then z = x2 - 2*x1. The coordinate relations are described on slide 3.

    • @eulerthegreatestofall147
      @eulerthegreatestofall147 9 місяців тому

      @@quinndd Many thanks!

  • @fayslmatar2088
    @fayslmatar2088 10 днів тому

    Hi Dane,
    Thank you for the great content you share-it's very helpful!
    I’m currently working on a system involving a two-wheel tractor and need to derive its mathematical model using the Lagrange method. Despite multiple attempts and spending significant time on this, I’ve ended up with equations that don’t seem to make sense.
    Could you please assist me with this? Any guidance would be greatly appreciated.

    • @quinndd
      @quinndd  9 днів тому

      Hi. Please feel free to reach out to me at quinn@uakron.edu

  •  Рік тому

    I know that this video is kind of old, but I just found your channel and I love your content.
    One big doubt: Why for the kinetic energy you do not take into account the displacement of the discs? I mean, 0.5·m·v² plus the rotational energy.
    Thanks!

    •  Рік тому +1

      Nevermind. I just compared the kinetic energy when calculated with respect of O and G. I answered my self.
      Thanks anyway for this awesome and free content!

    • @quinndd
      @quinndd  Рік тому

      @ I'm glad that you figured it out! If you have any other questions please let me know.

  • @Alexander-qi7pc
    @Alexander-qi7pc Рік тому

    Hey , why have we ignored the Translation KE term // 0.5mx'^2 while solving for total kinetic energy

    • @quinndd
      @quinndd  Рік тому

      Hi! The inertia is taken with respect to the point fixed in the ground, instead of the mass center. When taken with respect to the ground, the KE incorporates both the translational KE and the rotational KE with respect to the mass center.