Falling Tower Problem: Watch How to Solve It in This Video

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The discussion revolves around the Falling Tower Problem, where participants express confusion about the mechanics involved. Inertia is highlighted as a key concept, with a reference to a Wikipedia article for further understanding. The complexity of the problem is noted, particularly regarding friction and the tipping of the tower as the bottom block moves. The conversation suggests that the tipping may not be the primary concern intended by the problem's creators. Overall, the focus is on understanding the dynamics of the blocks and the consequences of the tower's movement.
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Homework Statement
Identical discs are stacked on on top of another to form a freestanding tower. The bottom disc can be removed by applying a sudden horizontal force such that the rest of the tower remains standing. Investigate the phenomenon and determine the conditions that allow the tower to remain standing.
Relevant Equations
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I actually have no clue. Found a video that demonstrates the problem:

Can someone help pls?
 
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To find the exact conditions will be exceedingly complex.
There's friction to think about, but you should cope with that.
But once the bottom block is halfway out from the block above, that block and the tower above it will start to tip. As the bottom block continues, the torque on the tower above it increases. How far will it have tipped when the bottom block is finally clear? What will be the consequence when that leaning tower hits the ground?
You should at least assume the blocks are completely inelastic.

I suggest they do not intend you to worry about the tipping at all.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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