This is not actually a homework question, but rather a question posed by a friend, so don’t expect the result to be pretty ;)(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

A spring of unstretched lengthL, spring konstantkand massMis whirled around with constant angular velocity [tex] \omega [/tex]. Find its new, stretched length.

3. The attempt at a solution

I’ve been through a couple of possible solution methods. I think the most promising one was to try and find the length density function [tex] \mu(r) [/tex]. I think it would be possible to find a differential equation that this density function must satisfy.

Firstly I thought that the stretch of the spring at some point a distrancerfrom the origin could be described in terms of this density function. More precisely I thought that in a lengthdrthe infinitisimal stretch would be

[tex] \delta r=\left(\frac{\mu_0}{\mu(r)}-1 \right)dr [/tex]

where [tex] mu_0 [/tex] is the unstretched length density. Furthermore, the centrifugal force on a lengthdrcould then be written as [tex] dF_c=\omega^2\mu(r)rdr[/tex].

Then I thought that the total spring force in an interval [tex] r[/tex] to [tex] r+\epsilon [/tex] must be the sum of the forces trying to stretch the piece minus the sum of the forces trying to compress it. This could be written like

[tex] k\int_r^{r+\epsilon} \left(\frac{\mu_0}{\mu(r)}-1 \right)= k\int_0^r \left(\frac{\mu_0}{\mu(r)}-1 \right)dr +\omega^2 \int_{r+\epsilon}^R \mu(r)rdr -k\int_{r+\epsilon}^R =\left(\frac{\mu_0}{\mu(r)}-1 \right) dr -\omega^2 \int_0^r \mu(r)rdr[/tex]

WhereRis the total length of the spring. Now, I tried taking derivatives on both sides with respect torbut this didn’t seem to yeild anything useful. Any suggestions or corrections would be greatly appreciated.

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# Homework Help: Rotating spring

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