Free falling rotational motion

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SUMMARY

The discussion centers on the dynamics of a post undergoing free falling rotational motion when cut from its lowest point. The user derived the velocity of the upper part of the post upon impact with the ground using energy conservation principles, arriving at the equation v = √(3gL). The method involves equating potential energy loss to kinetic energy gain, specifically applying the conservation of mechanical energy to a rotating body. The key takeaway is the application of rotational dynamics and energy conservation to solve for the velocity of a falling object.

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vabamyyr
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There is a post with length 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 don't know how to prove it? Can anyone help me?
 
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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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