With what speed does the tip of a 25 meter high tree hit the ground?

[spsch]

Homework Statement

With what speed does the tip of a 25-meter high tree hit the ground?
Assume the tree has a moment of inertia of a long thin rod.

Relevant Equations

$mg* \frac {l}{2} = \frac {1}{6} *m*l^2 * (\frac {v^2} {l^2})$

Hi, I am still working on the same exams problems, the second last one is giving me a bit of a headache after I found this https://physics.stackexchange.com/q...r-an-upright-rigid-body-to-fall-to-the-ground which contradicts my solution.
Can I assume that when the tree hit's the ground all of it's potential energy is converted to rotational energy?
I was thinking this is why they mention the moment of inertia being that of a long thin rod.

So I went with:
$mg* \frac {l}{2} = \frac {1}{6} *m*l^2 * (\frac {v^2} {l^2})$

and got: v= $\sqrt {3*g*l}$

 

[PeroK]

Problem Statement: With what speed does the tip of a 25-meter high tree hit the ground?
Assume the tree has a moment of inertia of a long thin rod.
Relevant Equations: $mg* \frac {l}{2} = \frac {1}{6} *m*l^2 * (\frac {v^2} {l^2})$

Hi, I am still working on the same exams problems, the second last one is giving me a bit of a headache after I found this https://physics.stackexchange.com/q...r-an-upright-rigid-body-to-fall-to-the-ground which contradicts my solution.
Can I assume that when the tree hit's the ground all of it's potential energy is converted to rotational energy?
I was thinking this is why they mention the moment of inertia being that of a long thin rod.

So I went with:
$mg* \frac {l}{2} = \frac {1}{6} *m*l^2 * (\frac {v^2} {l^2})$

and got: v= $\sqrt {3*g*l}$

That looks correct. I'm not sure where the contradition is.

 

[spsch]

That looks correct. I'm not sure where the contradition is.

Hi, Thanks. I used the formula from that link and got a different speed. Around 19, whereas my solution gets about 27 meters per second.
I probably made a mistake in plugging in the numbers.

Thank you so much for checking!