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Kinetic Energy of Springs with Mass

  1. Oct 14, 2005 #1
    Just to start with this equation:
    [tex]\frac{1}{6} m_s v^2[/tex]
    Incase my latex is broken, its (1/6)(m_s)(v^2). Where m_s is the mass of the spring, and V is the velocity of the spring at it's tip.
    I'm looking for a derivation for this equation, because it was just handed to me and I don't like using equations without knowing where they came from. Can anyone point me in the right direction?
    Thanks in advance :shy:
  2. jcsd
  3. Oct 15, 2005 #2

    Doc Al

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    Staff: Mentor

    Think of the spring (length L) in segments of mass dm. One end of the spring is fixed, the other moves at speed [itex]v_s[/itex]. So a mass element at position x from the fixed end has speed [itex](x/L) v_s[/itex], and thus a KE of [itex]1/2 (dm) (x/L)^2 v_s^2[/itex]. Express dm in terms of dx: [itex]dm = (m_s/L) dx[/itex] and integrate the previous expression (from 0 to L) to find the total KE.
  4. Oct 15, 2005 #3

    So, essentially we are just summing up all the tiny dKE's. That makes a whole lot of sense now. Thanks Doc.

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