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I'd like to understand the physics of a chain fountain (https://www.youtube.com/watch?v=_dQJBBklpQQ) and in order to do so, I am reading this paper credited for solving the mystery: http://rspa.royalsocietypublishing.org/content/470/2163/20130689

Yet, I have trouble with the first equation, specifically $\frac{T_C}{r}=\frac{ \lambda v^2 }{r}$. Where does it come from? The right part seems like a representation for the centripetal force, but that's everything I am able to decipher.

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  • $\begingroup$ The article indeed states that the centripetal acceleration is then provided by the tension in the chain, so what is it about that statement/derivation that is confusing to you? $\endgroup$
    – Kyle Kanos
    Commented Feb 9, 2015 at 18:26
  • $\begingroup$ I can't see it in a mathematical way. I don't know where the $force/radius$ term comes from, I've never seen something like this learning mechanics. $\endgroup$
    – neverneve
    Commented Feb 9, 2015 at 18:57
  • $\begingroup$ [The reason of anti gravity of bead chain fountain and the reason of lift ](youtu.be/m7uic8nNJ60) Yes, you can choose English subtitles $\endgroup$
    – enbin
    Commented Nov 23, 2017 at 11:13
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/70345/2451 , physics.stackexchange.com/q/138270/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Feb 3, 2021 at 22:25

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You're right about $\frac{\lambda v^2}{r}$, it is the centripetal force per unit length. The $\frac{T_C}{r}$ term is the component of tension force per unit length, acting perpendicular to the chain.

Let's show this. Consider a small, curved segment of the chain with mass per unit length $\lambda$. The curve can be described as subtending an angle $\theta$ of a circle with radius $r$.

The directions of the tensions $T_1$ and $T_2$ (both equal to $T_C$) at the two ends of the chain segment differ by the angle $\theta$. Aligning our axes opposite $T_1$, we have:

$$F_\| = -T_1\cos0 + T_2\cos\theta \approx 0$$ $$F_\perp = -T_1\sin0 + T_2\sin\theta = T_2\sin \theta \approx T_2\theta$$

Where I've used the small angle approximations $\cos\theta \approx 1$, $\sin\theta \approx \theta$. Dividing by the $\text{length} = r\theta$ of the chain segment:

$$ \frac{F_\perp}{\text{length}} = \frac{T_2\theta}{r\theta} = \frac{T_C}{r}$$

And that's it.

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