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Free falling rotational motion

  1. Dec 8, 2004 #1
    There is a post with lenght L and it is being cut down from its lowest point. So the post starts a free fall with circular trajector. The question is that what is the v of upper part of post when it hits the ground?

    I used energetic method and mvL= 0,5mv^2 plus or minus something.....
    i have a feeling that it might be [tex] mgL= \frac 1 2 mv^2 - \frac 1 2 mgL \Longrightarrow v=\sqrt{3gL} [/tex]

    but i dont know how to prove it? Can anyone help me?
    Last edited: Dec 8, 2004
  2. jcsd
  3. Dec 8, 2004 #2

    Doc Al

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    Staff: Mentor

    It's just conservation of mechanical energy, treating the falling post as being in pure rotation about the lowest point:
    [tex]\Delta {PE} = - \Delta {KE}[/tex]
    [tex]mg\Delta h_{cm} = - 1/2 I \omega^2[/tex]
    ... etc...
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