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Moment of thin Rod at pivot when acted upon by a Force

  1. Apr 2, 2016 #1
    This is of my own interest/ practice.

    1. The problem statement, all variables and given/known data
    A thin rod (of width zero, but not uniform) is pivoted freely at one end about the horizontal z axis , being free to swing in the xy plane (x horizontal, y vertically down). Its mass is m and its CM is a distance a from the pivot.
    The rod is struck with a horizontal force F which delivers an impulse F Δt = ξ a distance b below the pivot.
    Find the impulse η delivered to the pivot.


    2. Relevant equations

    F Δt = ξ @b
    L = T × Δt = r⋅J
    p = mvCM
    T = r × F
    v = ωr

    3. The attempt at a solution
    So first I found the angular momentum in terms of impulse. L = Iω = b ξ (j×i = -k).
    Using the relation, I solve for angular velocity: ω = bξ/I and find the linear momentum.

    Now here's my trouble, my friend is telling me I should add impulse and momentum for the total impulse delivered to the pivot. But I don't understand why that would be true.
     
  2. jcsd
  3. Apr 2, 2016 #2

    TSny

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    Homework Helper
    Gold Member

    I don't think that's quite right.

    Consider any system and suppose it has zero total linear momentum intitially. Then suppose a net external impulse Jnet acts on the system. Let Pf be the final linear momentum of the system just after the impulse acts. What is the relation between Jnet and Pf?

    For your problem, how would you write Jnet in terms of the impulses ξ and η? Remember, η is the impulse delivered to the pivot.
     
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