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Problem set up

I'm trying to work out the angle subtended by the stretching of a conical spring and I have a solution but I'm not 100% sure about it so would appreciate some feedback.

An Archimedes spiral spring is shown above with radii at various points. What I would like to come up with is an equation for the angle that each part of the spring gets stretched by as a function of the radius and the distance from rest $x$. I know that for a normal spiral spring of constant radius $R_{ss}$ if it is stretched by a small amount $dx$ then it will subtend an angle $d\theta$ therefore $$ dx=R_{ss}d\theta \\ \int_0^{x}dx=R_{ss}\int_0^\theta d\theta \\ \therefore \theta=\frac{x}{R_{ss}} $$

For a spiral conical spring with radius $R_{cs}$ that starts at $R_{cs0}$ and grows to $R_{cs1}$ what I did was integrate the radius: $$ dx=dR_{cs}d\theta \\ \int_0^{x}dx=\int_{R_{cs0}}^{R_{cs1}} \int_0^\theta d\theta dR_{cs} \\ \therefore \theta=\frac{x}{R_{cs1}-R_{cs0}} $$ It seems too simple though hence why I would like to check here what the thought is on the solution.

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No, Its not right. If the cone becomes more, steep at the limit when $$ R_{cs1} \Longrightarrow R_{cs0} \quad \theta \Longrightarrow \infty $$

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  • $\begingroup$ My apologies but I'm not sure I understand your answer. Since it's an Archimedean spiral the cone doesn't become more steep because the slope of the cone is constant. $\endgroup$ – enea19 Oct 16 '19 at 7:42
  • $\begingroup$ @enea19 ok fine, but lets assume we started with a more steep cone, it is going to have smaller theta, but by your calcs it will have a bigger one. $\endgroup$ – kamran Oct 16 '19 at 19:53
  • $\begingroup$ Aah, I see what you mean. But isn't that ok? I mean, a conical spiral spring that has less than a single turn is not very useful. Mathematically I understand the problem, but in practice it would never happen. $\endgroup$ – enea19 Oct 22 '19 at 16:41

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