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Rotational Energy Conservation

  1. Nov 6, 2007 #1
    A thin, unirform rod (I=1/3ML^2) of length L and mass M is pivoted about one end. A small metal ball of mass m=2M is attached to the road a distance d from the pivot. The rod and ball are realeased from rest in a horizontal position and allowed to swing downward without friction or air resistance.

    a) show that when the rod reaches the vertical position, the speed of its tip is:
    v = [tex]\sqrt{3gL}\sqrt{(1 + 4(d/L))/(1 + 6(d/L)^{2})}[/tex]

    b) At what finite value of d/L is the speed of the rod the same as it is for d=0? (This value of d/L is the center of percussion, or "sweet spot" of the rod.)

    For part a i know its Ei = Ef, but im not sure what you should be using to calculate it..
    The rod has Ei = MgL, and [tex]Ef=1/2I\omega^{2} + Mg(L/2)[/tex] But i don't know what i should calculate for the ball, potential at the top and rotational kinetic at the bottom?

    i haven't even looked at part b yet.
     
  2. jcsd
  3. Nov 7, 2007 #2
    bump?
     
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