# Fluid rope trick investigated

Original entry by Joerg Fritz, AP225 Fall 2009

## Contents

## Source

"Fluid rope trick investigated"

L. Mahadevan, W.S. Ryu, A.D.T. Samuel: *Nature*, 1998, 392, pp 140 to 141

## Keywords

Buckling, Viscosity, Surface tension, Instability

## Summary

This paper studies a very rich problem at the interface between fluid dynamics and classical mechanics that can be observed in thousands of household every morning. If honey is poured from a sufficient height, it approaches the morning toast as a thin filament which twists and whirls steadily even if the pouring hand is completely static. This can be explained by the theory of buckling with very simple scaling laws.

## The Math

What sets the speed with which the honey rotates? We could imagine up to six parameters that could have an influence on the rotation speed. On an intuitive basis it could depend on fluid density <math>\rho</math>, the viscosity <math>\mu</math>, the
flow rate *Q*, the gravity constant *g*, the filament radius *r*, and the height *h* from which the filament is falling. Unfortunately these are too many to reach the desired result directly by dimensional analysis. We have to make use of physical arguments to arrive at the solution.

Observations tell us that the filament of honey starts to rotate once it impacts the toast from a height that is big enough to create coils on the toast. We would thus assume that the rotation is due to an instability to buckling in the filament very close to where it impacts on the toast. It has been previously shown that the onset of a buckling instability of a falling jet is determined by the competition of two effects, gravity and viscosity. We can compare the two time scales associated with them, the ratio of which gives us something like a Reynolds number for this problem: <math>Re = \frac{g r^3 \rho^2}{\mu^2}</math>. Only when this parameter fall below a critical value will the filament start to rotate (an instability exist).

Once the fluid filament starts to rotate the dominant balance is on of torques where inertial effects are compensated for by bending torque due to viscous stresses. The viscous force scales like <math>f_v = \int \sigma r dA \approx \mu U r^4 / R^2</math> where R is a characteristic radius of curvature, which is approximately the radius of the coil structure on the toast. The torque due to this force is <math> \left f_v r^2 R^2</math>

## Conclusion

working on it