# Rotating spring

1. Sep 13, 2010

### Jolsa

This is not actually a homework question, but rather a question posed by a friend, so don’t expect the result to be pretty ;)

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

A spring of unstretched length L, spring konstant k and mass M is whirled around with constant angular velocity $$\omega$$. 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 $$\mu(r)$$. 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 distrance r from the origin could be described in terms of this density function. More precisely I thought that in a length dr the infinitisimal stretch would be

$$\delta r=\left(\frac{\mu_0}{\mu(r)}-1 \right)dr$$

where $$mu_0$$ is the unstretched length density. Furthermore, the centrifugal force on a length dr could then be written as $$dF_c=\omega^2\mu(r)rdr$$.

Then I thought that the total spring force in an interval $$r$$ to $$r+\epsilon$$ 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

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

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

Last edited: Sep 13, 2010
2. Sep 14, 2010

### kuruman

Why not use Hooke's Law F = -kx and Newton's Second Law? You know that the centripetal acceleration is a = ω2r, so ...

3. Sep 17, 2010

### Jolsa

Well, the problem is that the spring does not stretch uniformly. This means that the mass will not be uniformly distributed along the spring, and that the centrifugalforce is different at different points on the spring.

4. Sep 17, 2010

### zhermes

Use what Kuruman suggested, but apply it to a differential element of the spring length, and some distance r from the axis. You should be able to find a closed form solution for the total length.

5. Sep 17, 2010

### kjohnson

You can still use hookes law on an elemental basis, just remember to use the stiffness of each element not the total stiffness k. Then essentially guess the center of mass of the spring, which will allow you to calculate the centrepital force applied at the boundary. From that force find the internal force distribution and use the force distribution on an elemental basis to find how much each element elongates. Once you know the elongation you can find the center of mass. If the calculated center of mass matches the guessed center of mass it is solved. So it appears to be a differential equation..

6. Sep 17, 2010

### Jolsa

How would I calculate the stiffness of each element? $k=\frac{L}{dl}k_0$?

Last edited: Sep 17, 2010
7. Sep 17, 2010

### kjohnson

yeah i got it to be (delta k)=(k total)(length)/(delta x)

8. Sep 20, 2010

### kjohnson

In my earlier post I forgot to take into consideration that even on an element basis the deformation must also be taken into account. This is because to use a=r $$\varpi$$ ^2 for any element you must first identify the radius of that element. This radius does not simply equal its initial position x on the spring. Instead r=(x+$$\delta$$(x)) where $$\delta$$(x) is the function for the deformation we are trying to solve. This is because the element is initially at position x, but after the spring deforms it moves out an amount equal to all the deformation before it or $$\delta$$(x). This ends up leading to a second order differential equation that must be solved. I got a solution, but something is wrong (i think its a negative sign somewhere) because my $$\delta$$(x) starts at x=0 and then goes negative meaning the net deformation is compression...I will make another post when I get it all figured out.

9. Sep 21, 2010

### obafgkmrns

What a great problem! Here’s one approach:

Designate position measured from the point of rotation in the unstretched spring by “s” and let “s” be a variable attached to the spring’s material. Let the length of the unstretched spring be “L”. Let “x” be distance from point of rotation to a position in the stretched spring. Then

x(s) = s + e(s)

where e(s) describes the accumulated “stretch” at "s". The tension “T” in the spring at a point “s” is then

T(s) = k * de(s)/ds

where k is the spring constant. T(s) must equal the integral of the centrifugal forces of all mass elements “dm” from “s” to the end of the spring. If the unstretched spring has a mass density per unit length of “p”, then

dm = p * ds

so that

k * de(s)/ds = Integral s to L (w^2 * (s + e(s)) * p * ds)

where “w” is the angular velocity. Differentiate that equation once to get a second order ODE whose solutions are sines & cosines. Boundary conditions are

e(0) = 0 & de(L)/ds = 0

Which will eliminate the cosine solution & establish the sine's spatial frequency. Surprizingly, there are an infinite number of solutions!

10. Sep 22, 2010

### Jolsa

Great thinking! But I don't see how you get infinitly many solutions. The solution to the differential equation is

$e(s)=A\sin(\eta s)+B\cos(\eta s)-s$

where

$\eta^2=\frac{\omega^2 \mu}{k}$

Applying the first boundary conditions yeilds B=0, but the second boundary condition seems to yield

$A \eta \cos(\eta L)-1=0$

And thus

$A=\frac{1}{\eta \cos(\eta L)}$

Now, this doesn't seem to be correct as it suggests that if you increase the length density of the string, the spring stretches less?

Last edited: Sep 22, 2010
11. Mar 23, 2011

### Jolsa

For anyone interested, I have attached the problem (with additional sub-problems that I made up). I have also included the (presumably) correct solution to the questions.

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