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Angular Momentum and Torque

  1. May 4, 2012 #1
    Considering a Rigid Body/Angular Momentum/Torque

    We know that Torque(ext) = dL/dt

    Now with respect to stationary point S:
    L(s, cm) = Ʃ(ρi x mivi)
    and that dL(cm)/dt = Ʃτ(ext, CM)

    Now with respect to ANY point, P, that is accelerating:
    L(s,p) = L(cm) + ρ(cm) x Mv(cm)
    And hence,
    Ʃτ(ext, p) = dL(rel_p)/dt + ρ(cm) x Ma(p)
    Ʃτ(ext, p) = dL(rel_cm)/dt + ρ(cm) x Ma(cm)

    Can someone explain to me why this happens?:
    Why this happens? : Ʃτ(ext, p) = dL(rel_p)/dt + ρ(cm) x Ma(p)

  2. jcsd
  3. May 4, 2012 #2


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    If you would take the trouble to describe all your variables you've a better chance of a response.
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