Hanging a bait
A 10m long fishing line is held steady close to the surface of the water. A bait and a small lead weight is hanging at the end of the fishing line.
The water moves with a steady velocity of 0.1m/s. The drag force on the fishing line causes the bait to move in the direction of the flow by 1m relative to its position when the water was not moving. The drag force on the bait is negligible.
What happens to the displacement of the bait if the fishing line's diameter is doubled?
Hints and numbers: Calculate the Reynolds number, and estimate the magnitude of the change of the drag force on the fishing line. One may assume that the displacement is proportional to the drag force. The diameter of the fishing line is 0.2mm and its specific density is approximately the same as the water's. The viscosity of water is 1.05 Nm/s .
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1 solution
The Reynolds number is R e = ρ v d / η = 1 9 , where ρ ad η are the density and the viscosity of the water, v is the speed of the flow and d is the diameter of the fishing line. At this very low Reynolds number (up to about R e = 1 0 0 ) the flow is approximately laminar (with two stationary eddies on the back side of the object) and the force per unit length, acting on the fishing line, is approximately independent of the diameter.
While the laminar (Stokes) flow around a sphere is widely discussed in elementary textbooks, the flow around an infinite cylinder is barely mentioned. Historically, this question was quite a bit of puzzle. Stokes himself could not solve the problem. The simple dimensional analysis that yields F ∝ η v r in the case of a sphere of radius r would lead to F / L ∝ η v for the cylinder of length L , with no room for d , the diameter. The exact solution of the problem has a weak (logarithmic) dependence on the diameter, and it is still true that the force per unit length is approximately independent of the diameter.