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I'm not expecting a definitive answer. But I would like someone to explain which are the main forces that interact in this situation:

An ideal cylindrical column that is at first vertical is pushed just enough so it loses its balance and falls on its side. At every moment there's a point in the circumference of the base of the column that is in contact with the ground. The column is made of some material that we know it will break, and the floor is much harder, and won't break.

In how many pieces will the column break? At which points will the column break? Which are the constants that are meaningful in this scenario?

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I am certainly not going to give you a definitive answer. The following is more an overly simplified starting point.

A very simple model that mimics a collapsing tower with zero tensile strength (basically a tower of loose blocks) consists of a vertical array of point masses with the point mass at height $h$ undergoing parabolic trajectories as follows: $$x =\sqrt{gh} \ t$$ $$z = h - \frac{1}{2}g \ t^2$$

For fully inelastic collisions with the ground ($z=0$), the following motion results:

enter image description here

Note that the parabolic profile for lateral velocities ($v_x = \sqrt{g h}$) is chosen such that the line of blocks doesn't get compressed ($x^2+z^2 \ge h^2$ for all $t \ge 0$).

The end result of the collapse is a shattered tower positioned on the ground and elongated (strained) by a factor $\sqrt2$.

In real life the tower will have finite tensile and finite shear strength. As a result the above model breaks down. As a qualitative starting point for further modeling, it might provide useful though.

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Consider, the strongest forces will be at the end that was falling, with compressive force in the vertical direction in the center (from the impact) and tensile horizontal forces from the momentum of the sides that didn't hit the ground, pulling at the middle.

So, assuming you're landing on an ideally flat, level surface, and assuming a column with isotropic mechanical properties, weak enough to break under these forces, the column would break in both vertical and horizontal directions (i.e. into many pieces). I would expect the resulting scatter to be wedge-shaped due to the horizontal tensile force.

R\-Jason

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It depends on how many cracks the column has in it and what material it is made out of. That is the main criteria for when a material breaks. The break will depend where the cracks (internal or external) of the cylinder are. See for example http://en.wikipedia.org/wiki/Fracture_mechanics

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    $\begingroup$ I think the question asks for more detail than just a link to Wikipedia. $\endgroup$ – udiboy1209 Jul 21 '13 at 8:47
  • $\begingroup$ I understand that in a real situation this is relevant, but I was refering to a ideal column, meaning that it has no cracks whatsoever. The link is interesting though. $\endgroup$ – luisl Jul 21 '13 at 13:51

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