Free falling rotational motion

  • Thread starter vabamyyr
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Main Question or Discussion Point

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?
 
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Answers and Replies

  • #2
Doc Al
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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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