# 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).

## 3 Answers

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.
• Thank you. I suppose then that there is very little friction in this system? Also, I am having a difficult time determining a method for calculating the amount of force being exerted on the floor. – dotancohen Jul 8 '13 at 18:29
• Hey, I added the answer in the edit. – udiboy1209 Jul 9 '13 at 13:49
• Wow, thanks, that is a great edit! It is so elegant once I see it! I assume that v is the initial velocity imparted by the hand? Thank you! – dotancohen Jul 9 '13 at 13:59
• Just misconception here, can the normal do work? – Mauricio Nov 26 '17 at 11:34

We made a precise experiment on the Newton Beads. We varied the height of the container and measured the lifting height and the velocity of the chain for four cases: $H=1.46, 1.96, 5.22, 9.15\,\text{m}$. The measured velocities are $4.3, 5.4, 7.6, 10.1\,\text{m/s}$, respectively.

This is simply explained from the following consideration. The equation of the energy conservation is $$\frac{mv^2}{2} + mgz - T + \text{loss}= \text{constant},$$ where $m$ is the linear mass density, $v$ velocity of the chain, $g$ the gravitational acceleration, $z$ the height from the ground and $T$ the tension.

The losses are

1. Entrance loss, which is the loss of energy due to the starting the motion, which is assumed to be $amv^2/2$, in which $a$ is a constant to be decided from the experiment.

2. Exit loss, which is the loss of kinetic energy by hitting the ground and is $mv^2/2$.

In the container, the energy is expressed as $$0+mgH+0+0.$$

Just before hitting the ground, it is $$\frac{mv^2}{2}+0+0+ \frac{amv^2}{2}.$$ After hitting the ground, it is $$0+0+0+(1+a)\frac{mv^2}{2}.$$ Equating these, we get $$v=\sqrt{2gH/(1+a)}.$$ From the experimental result stated above, we have $a=0.73$, and so $v=\sqrt{2gH/1.73}$. This formula can explain the result beautifully. Please try it.

The curve of the chain is proved to be the inverse catenary by considering the force balance. The equations are a little lengthy to write down here.

This phenomenon has a close relationship with the siphon, which is a pipe flow. In the pipe flow, there are three kind of loses, the entrance loss, the frictional loss, and the exit loss. In our Newton Beads, the friction loss due to air is proved to be unimportant.

Takuya Matsuda, Emeritus Professor of Kobe University, Japan

Hideo Konami, Professor of Kyoto Women University, Japan

http://jein.jp/jifs/scientific-topics/975-topic53.html

• I'm going to have to work on my Japanese, that looks like just what I'm looking for! Welcome to physics.SE! – dotancohen Dec 25 '13 at 11:47
• Welcome to Physics.SE. This site supports Latex commands (based on MathJax I believe). You should improve your answer, by using them. – Ali Dec 25 '13 at 12:27

Dr. Matsuda, There is a more complete analysis of the chain fountain by Dr. John Biggins that you might be interested in that also predicts the catenary shape (height and width) of the fountain at http://arxiv.org/pdf/1401.5810v1.pdf.

He and Dr. Wagner also produced a video that explains their view of the pushing effect at http://www.youtube.com/watch?v=-eEi7fO0_O0. Gordon Judd

• Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/1401.5810 – Qmechanic Jan 27 '14 at 15:57