Tension in a rotating ring under gravity

[ricles]

Homework Statement

A metal chain is put on rotation over a horizontal surface with enough velocity not to collapse. The chain's mass density $\rho$ is known, as well as its radius $R$ and the velocity of its center of mass, $v$. Find the tension in the chain.

Relevant Equations

The linear and rotating motion equations

I know the solution for the problem of the tension on a rotating ring without gravity (tha is, $\frac{mR\omega^2}{2\pi}$) - that I find simple enough. But I'm at a loss how can I change it to do with gravity :/
Any help is appreciated! (and apologies for the bad drawing)

 

[wrobel]

Homework Statement:: A metal chain is put on rotation over a horizontal surface with enough velocity not to collapse. The chain's mass density \rho is known, as well as its radius R and the velocity of its center of mass, v. Find the tension in the chain.
Relevant Equations:: The linear and rotating motion equations

But I'm at a loss how can I change it to do with gravity :/

why do you think that such a motion exists under the gravity?

 

[ricles]

why do you think that such a motion exists under the gravity?

it doesn't spontaneously, but yet it is thusly put in motion - you can picture yourself rotating a short light chain with your fingers and letting it go on the floor, for instance, and in principle you should be able to rotate it fast enough that it doesn't collapse when you let it go. but I'm having trouble figuring out how gravity - once you let it go on the floor - affects the tension on the chain (which, I suppose, it does)

 

[Steve4Physics]

Homework Statement:: A metal chain is put on rotation over a horizontal surface with enough velocity not to collapse. The chain's mass density \rho is known, as well as its radius R and the velocity of its center of mass, v. Find the tension in the chain.

It is not obvious that the rolling shape would be circular (as opposed to elliptical for example) or that the tension would be constant throughout. How do you know that each of these (implicit assumptions) is correct?

My gut feel is that the situation is much more complicated than you think. You may find this article about rolling rubber bands (not dissimilar to chains in some ways) interesting.
https://physicsworld.com/a/rolling-rubber-bands-stretch-students/

 

[ricles]

Thanks a lot for the reference!
Indeed, the problem seems more complicated in general.
But, in principle, shouldn't you be able to rotate the chain (or ribbon, or what have you) quickly enough so that it doesn't deform/collapse after you let it go, for at least some time? Maybe it will deform and then collapse after some time, but then I guess there should be a $\Delta t$ s.t. for $t < Δt$ the deformations are negligible enough... maybe you can read this as a transient-state problem

 

[Steve4Physics]

Thanks a lot for the reference!
Indeed, the problem seems more complicated in general.
But, in principle, shouldn't you be able to rotate the chain (or ribbon, or what have you) quickly enough so that it doesn't deform/collapse after you let it go, for at least some time? Maybe it will deform and then collapse after some time, but then I guess there should be a $\Delta t$ s.t. for $t < Δt$ the deformations are negligible enough... maybe you can read this as a transient-state problem

It it were rotating fast enough to keep a circular shape, I think this would require that the centripetal acceleration be much bigger than the acceleration due to gravity $(R\omega^2\text{ }\gg\text{ } g)$. This would mean that tension would still be (to a good approximation) $\frac{mR\omega^2}{2\pi}$ . And, of course, the centre of the circle would have speed $R\omega$.

For smaller rotational speeds, you could probably get stability. In the absence of frictional losses the rolling would continue indefinitely. But I think the shape and tension-distribution would be much harder to calculate,

 

[Steve4Physics]

why do you think that after taking into account the gravity it still would be a circle?

As noted, that would only apply at sufficiently high rotational speeds.

The simple answer is that it is a requirement of symmetry. A more detailed explanation is this:

For a chain rolling along the ground, the forces on a link are:
F₁ - the tension pulling on one side of the link.
F₂ - the tension pulling on the other side of the link.
F₃ - the link's weight.
F₄ - (for the link currently in contact with the ground) the reaction force from the ground. (I think this equals the chain's total weight.)

As ω gets bigger, F₁ and F₂ get bigger but F₃ and F₄ remain constant. If we make ω big enough then F₃ and F₄ become negligible compared to F₁ and F₂. So we can ignore F₃ and F₄. That means (in terms of forces) we are effectively in the same situtation as if g=0. There is no preferred direction for the forces on the various links so we have circular symmetry. A circular shape is the simplest shape satisfying this.

Interestingly other possible shapes could exist. These are standing waves around the chain a bit like these:
https://spark.iop.org/sites/default/files/image/setup-standing-waves-on-a-closed-loop.gif
A standing wave pattern would be rotating around the centre. But standing waves are not relevant here.

 

[haruspex]

If it were rotating fast enough to keep a circular shape

Not possible. It would be like inflating a tyre so much that it remains perfectly circular under load.

 

[Steve4Physics]

Not possible. It would be like inflating a tyre so much that it remains perfectly circular under load.

Yes indeed. I should have said "approximate circular shape". The bigger ω/g gets, the closer the shape approximates to a circle.. But ω can't be infinite.

 

[ricles]

But then, being impossible doesn't stop people from considering things like perfect circles, that rotate while having only one point in contact with the floor - and that could still provide useful results.
I agree, considering the situation presented, $\frac{mR\omega^2}{2\pi}$ is a good enough approximation - gravity doesn't quite play a role here, but if/when it did, the geometry would indeed be a tad more complicated.
Thanks a lot for the clarifying answers! :)
Cheers

 

[ricles]

Interestingly other possible shapes could exist. These are standing waves around the chain a bit like these:
https://spark.iop.org/sites/default/files/image/setup-standing-waves-on-a-closed-loop.gif
A standing wave pattern would be rotating around the centre. But standing waves are not relevant here.

(maybe) Interestingly, my next question (that I wouldn't post here, but since you alluded to it) would be:
now, if there was a pump on the road, and the chain went rotating over it

the pump would generate a pulse that propagates through the chain. What would be the velocity of the propagating pulse? But now I guess it would be just like finding the propagating velocity of a pulse on a tensioned string - not too interesting

 

[Steve4Physics]

Agreed. You would possibly get $v = f(R)\sqrt{\frac{T}{\mu}}$ where f(R) allows for the propagation in a curved path rather than in a straight line.

 

[haruspex]

Yes indeed. I should have said "approximate circular shape". The bigger ω/g gets, the closer the shape approximates to a circle.. But ω can't be infinite.

Well, I'm not convinced it works at all with a chain. As each link hits the ground, flat, it loses all its KE. Unless there's something subtle going on where the links rise again (as in the chain fountain), seems to me the whole thing will collapse pretty quickly, no matter what the rotation rate.
For elastic bands it is different; they have some elastic resistance to being curved.

 

[hutchphd]

My gut feel is that the situation is much more complicated than you think.

the pump would generate a pulse that propagates through the chain. What would be the velocity of the propagating pulse? But now I guess it would be just like finding the propagating velocity of a pulse on a tensioned string - not too interesting

No it is very interesting. The relationship you calculated for the tension in a ring without gravity can be written as $$T=\rho v^2$$ Notice this relationship is independent of R and recapitulates that for a propagating pulse on a string. So as long as the top is locally circular the chain does the tension adjustment for you.

 

[ricles]

No it is very interesting. The relationship you calculated for the tension in a ring without gravity can be written as $$T=\rho v^2$$ Notice this relationship is independent of R and recapitulates that for a propagating pulse on a string. So as long as the top is locally circular the chain does the tension adjustment for you.

I'm not sure I follow you. Do you refer to the tension calculated as $\frac{mR\omega^2}{2\pi}$? Then it doesn't quite seem to be independent of R (we have $\frac{mR\omega^2}{2\pi} = \frac{mR\left(\frac{v}{R}\right)^2}{2\pi} = \frac{mv^2}{2\pi R}$, which does seem to depend on R, right? - more precisely, it is inversely proportional to the circunference).

Or do you mean it doesn't depend on "$R$", read as a vector that represents each point's position in the circle? And, in that sense, the tension doesn't depend on $R$ (but it most likely would, if the shape assumed wasn't circular)

 

[ricles]

Well, I'm not convinced it works at all with a chain. As each link hits the ground, flat, it loses all its KE. Unless there's something subtle going on where the links rise again (as in the chain fountain), seems to me the whole thing will collapse pretty quickly, no matter what the rotation rate.
For elastic bands it is different; they have some elastic resistance to being curved.

Oh, but wouldn't you have a chain to be close enough to an elastic band if the chain's links were small enough? Or, even if they weren't, wouldn't the links not hit the ground flat and lose their KE if they were rotating fast enough (due to inertia)?

 

[Steve4Physics]

I found someone who's done it!
https://phys420.phas.ubc.ca/p420_08/Anthony Toljanich/intro_theory_demonstration.pdf
EDIT: Maybe not done it! It seems to be a student's report planning the experiment. Sorry.

 

[Steve4Physics]

A rolling circular hoop has instantaneous velocities: 2v at top, v at centre , 0 at bottom. To a good approximation the speed of a link coming into contact with the ground will be zero so there is relatively little kinetic energy to lose on 'impact'. For a real-world rolling chain, the main energy losses probably would be due to friction in the joints between links.

 

[hutchphd]

I'm not sure I follow you. Do you refer to the tension calculated as mRω22π? Then it doesn't quite seem to be independent of R (we have mRω22π=mR(vR)22π=mv22πR, which does seem to depend on R, right? - more precisely, it is inversely proportional to the circunference).

Yes but $$m/2 \pi R $$ is called $\rho$ the linear mass density of the chain. Poof ! there goes R

 

[ricles]

Yes but $$m/2 \pi R $$ is called $\rho$ the linear mass density of the chain. Poof ! there goes R

feels like I was a blind for not seeing it
thanks!

I found someone who's done it!
https://phys420.phas.ubc.ca/p420_08/Anthony Toljanich/intro_theory_demonstration.pdf

also, thanks Steve for the resources ^^