Stretching of a rotating spring

[L0r3n20]

TL;DR Summary: Are the k and the w linked?

Yesterday I came across this problem:

A mass is attached to a spring and the system rotates (one of the spring end is fixed) in an horizontal plane. Given the mass m, the value of k, the length of the spring l_0 and the angular velocity w, compute the stretching.

I worked out the formula, which turns out to be

$x = \frac{ m \omega^2 \ell_0}{k - m\omega^2}$

(Sorry I don't how to implement latex code)
Now the question: why can't I choose ANY value for w? In principle, the faster the rotation, the longer the stretching... In this case it seems there's a limit for w (which is suspiciously equal to the value of the pulsation for mass-spring). Can someone explain why are these quantities linked?

 

[berkeman]

Can you upload a diagram of the problem? It's a little confusing how a mass in a horizontal plane will stretch a rotating/torsion spring. Use the "Attach files" link below the Edit Window.

Also, I'll send you some hints on how to use LaTeX via PM now. (see the "LaTeX Guide" link below the Edit window)

 

[gleem]

Did you use F = - kX?

 

[L0r3n20]

Can you upload a diagram of the problem? It's a little confusing how a mass in a horizontal plane will stretch a rotating/torsion spring. Use the "Attach files" link below the Edit Window.

Also, I'll send you some hints on how to use LaTeX via PM now. (see the "LaTeX Guide" link below the Edit window)

Sure, here it is! And thank you for your pm! :)

 

[L0r3n20]

Did you use F = - kX?

Yes I did. I set the elastic force equal to the centripetal force.

 

[Delta2]

I agree with your solution as long as $l_0$ is the natural length of the spring.

What happens is that the spring cant provide the necessary force after that breakpoint because the centripetal force grows by $m\omega^2 x$ while the spring force grows only by $kx$ so if $\omega$ becomes too big such that $m\omega^2>k$ there can never be $m\omega^2x=kx$ (it will be $m\omega^2x>kx$, for any x and of course certainly not the even worst condition $m\omega^2(l_0+x)=kx$.

 

[gleem]

Yes I did. I set the elastic force equal to the centripetal force.

I mean $\: \: m(l_{0}+x)\omega ^{2}=(-1)kx$

There is no difference in the application of Hooke's Law between this situation and a vertical spring with a weight attached.

 

[Delta2]

I mean $\: \: m(l_{0}+x)\omega ^{2}=(-1)kx$

This is not correct, for x positive, because we know that the centripetal force is in the same direction with the spring force, this equation implies they have opposite directions.

For x negative I think we cannot allow negative x for this problem because then the spring force becomes with outward direction while the centripetal force is always inward.

 

[gleem]

This is not correct, for x positive, because we know that the centripetal force is in the same direction with the spring force, this equation implies they have opposite directions.

Duh, I should have also noted the problem with the behavior of x with ω