# Find drag force on link of rotating chain

Given a closed chain with a total length of 1.2m rotating at 1'800 rpm and a total mass of 0.4kg, what is the drag force pulling on one chain link?

I originally thought that since no link size was given I need to assume the link sizes to be infinitely small. Thanks to the answers below I now know this won't work. Yet I am still baffled as to how I could calculate the drag force without it, sure I could give a function of the drag force that is dependent on the link size, but looking at the parameters given I think I should be able to calculate the actual force.

Here is a video of a very similar experiment as we conducted and are asked to describe now: http://www.univie.ac.at/elearnphysik/video/PhysikI/rotKette_648x480.flv

I am glad for any hints and explanations.

Edit: Rewrote question to match exactly the problem description

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If the link is infinitely small, then the force will also be infinitely small. I think what you're looking for is force density – yohBS Mar 8 '12 at 19:56
I was not able to view the video, and I assumed the chain was fastened in the center and was being swung around. I just noticed you said a closed chain. Is the chain in the shape of a circle, spinning like the rim of a wheel? If so, I need to change my answer. – Mike Dunlavey May 11 '12 at 1:04
What does "drag force" mean? Usually it means a friction force. In this case, it is impossible to calculate regardless of chain size--- there is no drag necessary. What is the source of the drag? Is it the screwdriver? Does the question ask for the tension in the chain? – Ron Maimon May 11 '12 at 5:18
I love this experiment! Indeed, the force does not depend on the size of the link. Let me see if someone has already answered correctly... – Carl Brannen Jul 10 '12 at 10:34

I will assume in this answer that "drag" means tension. You are asked to find the tension in the chain as it is rotating. This is independent of the link size, so long as the links are not a significant fraction of the circumference.

If you have a hoop of mass density per unit length $\rho$ and circumference C (so that $\rho C = M$ where M is the total mass), rotating with rotational velocity $\omega$, the centripetal force on a segment of length l is the mass times the rotational velocity squared times the radius, or

$$F_c = \rho l w^2 {C\over 2\pi}$$

If the chain is at tension T, the two endpoints of the segment pull in with a total force of

$${Tl\over C}$$

Setting the two forces equal, the l drops out (as it must) and gives the tension:

$$T = (\rho C) \omega^2 {C\over 2\pi} = M \omega^2 {C\over 2\pi}$$

or $\omega= 30 {1\over s}$, $M=.4 \mathrm{kg}$, $C = 1.2 m$, this is about 68N.

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Think about this: in the limit as $m\to 0$, the net force also goes to zero, since $F_\text{net}=ma$ and the acceleration is clearly finite. So it's not very useful to calculate the force on an infinitesimal piece of the chain, since you just get zero.

If you're really being asked to find the centripetal force on a single link of the chain, you will have to involve the size of a link in some manner. You could use the length of a link, or the mass, or anything else you can work with, but there will have to be some sort of extensive property involved.

Alternatively, as yohBS pointed out in a comment, you can calculate the force density, which would be force per unit length, or per unit mass, or whatever. This is similar to assuming a unit-length chain link.

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Thanks David, updated the question with a 1:1 problem description as I have it on my paper. Agree though that infinitely small chain links sounds like a bad idea – Robin Mar 11 '12 at 14:26
@Robin if you see this: unfortunately I don't have anything else to add, though I apologize for not saying so earlier. The problem as written does seem kind of strange. – David Z Apr 10 '12 at 14:57

It seems to me as if the whole chain is just experiencing a centripetal force. For a given radius, speed and mass the force would be $$F = mv^2/r$$ This is independent of the size or number of individual links, as long as we can assume that the mass distribution is concentrated on a circle with radius $r$. Now from a steady state situation you cannot have any acceleration along the direction of rotation, as this would speed up or slow down the chain. If you could disconnect all links instantaneously the total force necessary to keep them on a circle is $F$, each individual element requires $F/n$, with $n$ being the number of link elements. So there is no way around some kind of normalization.

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Well, let's see. Centripetal acceleration is $r\omega^2$, and if the density of the chain is $D$ kg per meter of radius, then the centripetal force holding a small piece of chain of length $dr$ in a circular path is $f = Dr\omega^2dr$. If the length of the chain is $l$ then the stress at any radius $r$ is the integral of $fdr$ from $r$ to $l$ which looks something like $D\omega^2 (l^2-r^2)/2$. It shouldn't matter if it's a chain or a cable.

$D = 0.4kg/1.2m$, $l = 1.2m$, and $\omega = 2 \pi 1800/60$ radians/sec.

You take it from there.

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I didn't immediately see an answer that is easy to understand so I'll write this one up. This is a type of engineering problem that you'll see over and over in engineering school and it's best thought of in engineering terms.

Suppose your chain has two weak links, 180 degrees apart. We want to know how fast we can turn the wheel without having the chain break at those links.

We treat the two halves of the chain as if they were solid objects. The question now becomes: what is half the centripetal force between the two halves of the chain? Of course these two forces are opposite and equal. So we compute half the centripetal force of 180 degrees worth of chain.

Force is a vector. The chain at position $(r,\theta)$ where $r$ is the radius of the wheel has a centripetal force with a magnitude of $$(M\;d\theta/2\pi)\;r\;\omega^2$$ in the direction of the radius (or against it whatever if you care, go ahead and edit this), where $M$ is the mass of the chain and $\omega$ the rate of rotation in radians. Writing it as a vector we have: $$F_\theta = \frac{Mr\omega^2\;d\theta}{2\pi}(\cos(\theta),\sin(\theta)).$$ Integrating this from $\theta=0$ to $\theta=\pi$ gives (ignoring signs which I hate) a total force of: $$F_{tot} = \frac{Mr\omega^2}{2\pi} 2 = Mr\omega^2/\pi,$$ and so the tension in the chain is: $$F_{tension} = \frac{Mr\omega^2}{2\pi} 2 = Mr\omega^2/(2\pi),$$

This is the same as Ron Maimon's answer and I see that his derivation is correct and simpler. I'll leave this up as I think it is more intuitive.

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