Centrifugal force and elastic deformation

[ClassicalMechanist]

Homework Statement

Consider a spring of natural length L_0 with constant k which rests on a horizontal frictionless surface. The spring is attached at one end to a fixed post and at the other end to a mass m. Suppose the spring is rotating around the post in a circle with angular velocity w. What is the new length of the spring?

The above problem is simple enough, and I have found a solution to it. Now consider the following related problem: you have a circular elastic band of natural radius R_0, mass m, and spring constant k. It is rotating on frictionless surface about its center with angular velocity w. What is the new radius of the rubber band?

I am trying to use my ideas from the first problem to solve the second problem.

Homework Equations

F=-Δx
F_c=mw^2R

The Attempt at a Solution

First I will show my solution to the first problem. It is as simple as setting the centripetal force equal to the restoring force of the spring: F_s=kΔx=F_c=mw^2L, where Δx=L-L_0. Solving I find,

L=(kL_0)/(k-mw^2)

As for the second problem, the main difficulty is how to model an elastic band. One idea is to think about it as the sum of many small springs attached to each other at the center of the circle, and extending to the boundary of the circle, with a small mass element on the other end (in other words, each mass element dm of the band can be thought of as attached to it's own spring).

I am not really sure how to proceed this way, so another idea is to use Young's modulus and think about the stress/strain of the rubber band as it is deformed by the centrifugal force.

Another idea is to use energy. The rotational energy of the band is 1/2Iw^2, where I=mR^2 (the band can be thought of as a hoop). Not sure about the elastic potential energy of the band.

 

[haruspex]

Consider an element subtending a small angle $d\theta$ at the centre. If the tension in the band is T, what is the radial force on the element?

 

[ClassicalMechanist]

Consider an element subtending a small angle $d\theta$ at the centre. If the tension in the band is T, what is the radial force on the element?

Drawing a circle and doing using some geometry we find the radial force is F_r=2Tsin(dθ/2), which is also equal to the centripetal force on the element: F_c=dm*w^2*R. Thus it suffices to find the tension T, presumably in terms of the spring constant k. I'm not sure how to do this...

 

[haruspex]

Drawing a circle and doing using some geometry we find the radial force is F_r=2Tsin(dθ/2), which is also equal to the centripetal force on the element: F_c=dm*w*R^2. Thus it suffices to find the tension T, presumably in terms of the spring constant k. I'm not sure how to do this...

Use a small angle approximation for sine. In the limit it will be exact.

You need to relate the overall extension to the tension etc.

 

[ClassicalMechanist]

Clearly the tension T is a function of the radius R, the natural radius R_0, and k. Naively I guess that T=k(R-R_0) and doing the math I get

R=2*pi*kR_0/(2*pi*k-mw^2)

Interestingly this is of a similar form to the answer I gave to the first problem.

Now my only assumption is T=k(R-R_0), which seems reasonable, because it is the analogous to Hooke's law for a spring, however it could be off by a constant factor.

 

[haruspex]

R=2*pi*kR_0/(2*pi*k-mw^2)

If you allow for a rapid spin, and hence a large increase in radius, I get that r is the solution of a quadratic.
For modest spins, the linear approximation is $\Delta r=\frac{r_0\rho\omega^2}{2\pi k}$.

T=k(R-R_0)

Yes, it must be so.

 

[ClassicalMechanist]

If you allow for a rapid spin, and hence a large increase in radius, I get that r is the solution of a quadratic.
For modest spins, the linear approximation is $\Delta r=\frac{r_0\rho\omega^2}{2\pi k}$.

Yes, it must be so.

I don't believe that r is the solution of a quadratic. I actually made a mistake in post #3 which I later corrected. I originally wrote F_c=dm*w*R^2 and corrected it to F_c=dm*w^2*R. When I used the first formula I got a solution with a quadratic as well, which leads me to believe you did not catch my mistake.

I believe the correct derivation is as follows: as you said, we can use the small angle approximation, so the radial force on the small mass element dm is F_r=T*dθ, which equals the centripetal force F_c=dm*w^2*R. So T*dθ=dm*w^2*R, and summing over all mass elements we get T*2pi=mw^2R. Substitute T=k(R-R_0) and the result follows immediately.

By the way, what is the justification for T=k(R-R_0)?

 

[haruspex]

By the way, what is the justification for T=k(R-R_0)?

No, sorry, that was wrong. It should be $T=2\pi k(r-r_0)$. The increase in length is $2\pi (r-r_0)$.

I don't believe that r is the solution of a quadratic.

No, you're right. (Not having a good day! Need to be more careful with my scribblings.)
Putting that together I get the rather surprising $r=\frac{4\pi^2kr_0}{4\pi^2k-m\omega^2}$.
Note the consequence as omega increases.

 

[ClassicalMechanist]

Right that was my other question, as w increases, r becomes negative. What is the interpretation of this?

 

[haruspex]

Right that was my other question, as w increases, r becomes negative. What is the interpretation of this?

It's not so much that it becomes negative... on the way it would become infinite. That is, for any given k and m there is a critical rate at which the radius will just keep increasing. Not surprising, when you think about it.

 

[ClassicalMechanist]

It's not so much that it becomes negative... on the way it would become infinite. That is, for any given k and m there is a critical rate at which the radius will just keep increasing. Not surprising, when you think about it.

It's sort of surprising to me...I mean what happens when you go past that critical rate?

 

[haruspex]

It's sort of surprising to me...I mean what happens when you go past that critical rate?

Consider r >> r₀. Required tension for the centripetal acceleration is proportional to r; actual tension is also almost proportional to r. So there comes a point at which getting wider to increase the tension is self-defeating.

 

[Don Sant]

Hi guys, Good discussion. It helped me too. I have a quick question.
@haruspex From your comment 11: we have k = [m*w^2*r] / [4*pi*pi*(r-r0)].

1) If I am not mistaken, this stiffness is in the radial direction. Since we have considered a Rubberband rotating at 'w', don't you think the stiffness will be tangential to the rotation?
2) If we consider the tangential direction, don't you think the cross-section of the rubber band has something to do with stiffness? (width, thickness,; if width = thickness==> square cross section)

Please correct me if i am wrong!

 

[haruspex]

If I am not mistaken, this stiffness is in the radial direction.

Not sure what you mean. The constant k was defined treating the rubber band as elastic in the usual way, i.e. In respect of a stretched perimeter.

don't you think the cross-section of the rubber band has something to do with stiffness?

For the purposes of the question, I treated the rubber band as an ideal spring. In reality rubber behaves quite differently.

 

[Chestermiller]

Hi guys, Good discussion. It helped me too. I have a quick question.
@haruspex From your comment 11: we have k = [m*w^2*r] / [4*pi*pi*(r-r0)].

1) If I am not mistaken, this stiffness is in the radial direction. Since we have considered a Rubberband rotating at 'w', don't you think the stiffness will be tangential to the rotation?
2) If we consider the tangential direction, don't you think the cross-section of the rubber band has something to do with stiffness? (width, thickness,; if width = thickness==> square cross section)

Please correct me if i am wrong!

A more accurate description: If the rubber band is regarded as a non-linear purely elastic material, then the tension can be represented by:
$$T=A_0\sigma_E(r/r_0)$$where $\sigma_E$ is the "engineering stress" function of the hoop stretch $r/r_0$ and $A_0$ is the initial cross sectional area of the rubber band. From a force balance in the radial direction: $$T=\frac{m}{2\pi}\omega^2 r$$where m is the mass of the rubber band: $$m=\rho (2\pi r_0)A_0$$where \rho is the density of the rubber. Therefore, combining these equations, we have:$$\frac{\sigma_E(r/r_0)}{(r/r_0)}=\sigma^*(r/r_0)=\rho \omega^2r_0^2$$. To use this analysis to predict the expansion of the rubber band, one would have to measure the function $\sigma (\lambda)$ as a function of the stretch ratio $\lambda$ from uniaxial tensile tests on the rubber.

 

[Don Sant]

Yes, I think this is what I was looking for, Thank you. Is there any TextBook that you would recommend, there should be graphs in those that I can refer to.

 

[Don Sant]

Not sure what you mean. The constant k was defined treating the rubber band as elastic in the usual way, i.e. In respect of a stretched perimeter.

For the purposes of the question, I treated the rubber band as an ideal spring. In reality, rubber behaves quite differently.

Yes, I understood that sir. In the above formula, I think you considered a spring-mass system attacher to a pole, rotating at an angular velocity. (or in other words-- a small mass "dm" connected to a spring of stiffness k spinning at "w", Integrated from "0-2pi"). I believe this still won't give us an understanding how rubber expands with respect to "w", but that equation will show how length (or radius) of the spring changes due to "w".

I needed to relate the rate at which cross section of the rubber band changes with respect to "w"

Please correct me if I am wrong.

 

[Chestermiller]

Yes, I think this is what I was looking for, Thank you. Is there any TextBook that you would recommend, there should be graphs in those that I can refer to.

The key to all this is related to the fundamental "rheological behavior" of the material (in this case, rubber). You can Google rubber elasticity, or get a textbook on rubber elasticity or polymer material behavior.