newtonian mechanics - Beads in 'perpetual motion'?

This page has an interesting video of beads 'syphoning' out of a glass beaker:

https://www.youtube.com/watch?v=6ukMId5fIi0

The host has a few explanations for the effect, but none of them sound plausible to me. The beads are in 'perpetual motion'? 'Shock wave'? What could cause an entire bead chain to syphon out of a beaker like water does out of a syphon? It certainly does not look to me that there was enough initial investment in energy to move the chain as is done. On the other hand, considering that only a small portion of the chain is moving at any one moment in time, the whole system could be considered to be moving very slowly thus meeting the initial energy investment requirement (very low velocity).

Answer

If you observe closely, the chain does not accelerate much once it starts to come out of the beaker, and so we can say that energy is not continuously being provided to the chain. If you think the part of the chain getting lifted up from the beaker requires energy, it indirectly comes from the work done by normal force from the ground in stopping the equivalent part of the chain which fell on the ground. You can think of it as the chain passing on energy to the next part, taking it from the part which came to rest.

As to where it gets the initial velocity from in the first place(it is obviously not being given be the person holding the beaker), I believe that it comes from the work done by gravity, because the center of mass of the chain has lowered significantly due to the heap of chain formed on the ground.

But the center of mass still keeps moving down as the heap on the ground grows bigger and bigger. So gravity is still doing work. I cannot say for sure where this work goes, but I think this is what makes the loop over the beaker grow bigger and bigger. At some point this work might be compensating for all the loss in energy, and maybe that's why the loop stops growing.

Edit: The Force exerted on the ground

1. There will be the weight of the chain already on the floor.


2. There will also be the impulse of the part of the chain currently falling which comes to rest. Impulse is given by $Ndt$ where $N$ is the normal reaction by the ground, and $dt$ is the infinitesimal time in which it is acting. We can say that $$Ndt = \delta P$$ $$Ndt = dm v$$ where $dm$ is the mass of the chain which falls on the ground in time $dt$. Assuming constant linear mass density $\lambda$ we get $$dm = \lambda dx$$ Thus $$\frac {dm}{dt} = \lambda \frac{dx}{dt} = \lambda v$$ Thus we get $N$ as $$N = \frac {dm}{dt} v = \lambda v^2$$ This $N$ is the extra force the ground needs to apply to stop the moving chain.