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I Lagrangian at a rolling sphere

  1. Oct 23, 2016 #1
    A sphere is rolling inclined wall (θ radian).
    and the momentum of that sphere is

    L = 1/2 mv^2 + 1/5 mv^2 + mgx sin(θ) = 7/10 m v^2 + mgx sin(θ)
    ∂L/∂v = p
    7/5 m v = p

    but I cant understand why the factor of mv is 7/5.

    p is the linear momentum of sphere.

    which means the factor of mv must be always 1 (my thought)

    p = mv

    ↑that was an unchanging truth of my physical world.

    why rolling effect to the momentum? angular momentum of the sphere has no relation with p direction!!

    Attached Files:

  2. jcsd
  3. Oct 23, 2016 #2


    Staff: Mentor

    Well, p isn't really the momentum from Newtonian analysis. It is the conjugate momentum from Lagrangian analysis. They have similar names, but they are only equal if ##KE=0.5 m v^2##
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