# calculation of tension in curved chain for chain fountain (Mould effect)

https://royalsocietypublishing.org/doi/10.1098/rspa.2013.0689

A chain of some sort goes upwards, turns in a half circle (according to the drawn diagram), and goes back down.

Focusing on the half circle, the equation given by the paper to account for its centripetal force is

"In our initial analysis, we assume the curved region at the apex of the fountain is small enough, and has tight enough curvature, that the centripetal acceleration is much larger than the gravitational acceleration (v2/r≫g). The centripetal acceleration is then provided by the tension in the chain. If the (local) radius of curvature is r and the tension in the curved region is Tc" Also the lambda is mass per unit length

Why is Tc divided by r in the left half of the equation? If it was force per unit length then I would expect Tc to be divided by (pi)r instead of just r.

Consider a small element of chain with length $$dl$$ at the top of the arc.

Its mass is $$\lambda\,dl$$.

It is curved through an angle $$dl/2r$$ on each side of the vertical.

So the downward force on it is $$2T_c\sin(dl/2r) \approx T_c\,dl/r$$ (the first 2 is because there are two free ends of the element)

The acceleration downwards is $$v^2/r$$ (ignoring gravity).

Newton's second law gives the equation in the paper.

• May I ask where the dl goes in the end? – never took courses but why Feb 4 '20 at 18:19
• It just cancels out from both sides of "F = ma." The mass and force are both proportional to $dl$. – alephzero Feb 4 '20 at 20:02