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A little while ago, I was fidgeting with a ball and some elastic bands, and I noticed that it seems like the elastic bands have to lie along a great circle (or near one) in order to not snap off. Is this generally true? If so, why? And if not, under what conditions does an elastic band snap off a ball?

Here are my thoughts so far:

If I have a curve and I put a small frictionless ball on some point of it where the slope (i.e. the derivative of the curve) is non-zero then I push down on the ball (the derivative should be zero with respect to 'down'), it'll slide down the surface:

enter image description here

On the other hand, if I put the ball down on a point where the slope is zero and push down, it won't move:

enter image description here

Furthermore, I then have a solid of revolution made of some curve and I put such a ball on a point on it then

  1. if the slope of every section of the surface that passes through that point is zero, then the ball should also stay put
  2. if even one such section has a non-zero slope, the ball will slide down

enter image description here

Now, let's pretend an elastic band is made up of a bunch of little balls which some force presses down on:

enter image description here

Then when the rubber band lies on a great circle condition (1) is fulfilled for every 'ball' and the rubber band will stay put.

On the other hand, if it's not on a great circle, a 'cross-section view' of the whole setup would look like this:

enter image description here

so the 'balls' slide down and the band snaps off.

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    $\begingroup$ Your last picture is all that is needed. As long as it is not a great circle, or close enough to a great circle such that the asymmetry-induced-pushing-off-force is tiny, then this last picture of yours shows that there will be a significant pushing-off force, and that then is sufficient to guarantee that the non-great-circle band will slide off. There is no need to invoke "a bunch of little balls". The only other thing that you might want to include, is friction between the band and the ball that will help mitigate sliding off when slightly deviating from a great circle. $\endgroup$ Commented Oct 28 at 20:31
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    $\begingroup$ There's a nice physics problem lurking here. Suppose an elastic band has tension $k(s-s_0)$ when its length (i.e. the circumference) is $s > s_0$ (and the natural length $s_0$ is constant). Suppose the coefficient of friction between band and surface of sphere is $\mu$. Find the limiting case where the band just stays on a sphere of radius $R$ with $2\pi R > s_0$. $\endgroup$ Commented Oct 29 at 16:41
  • $\begingroup$ This is not my area of expertise but I wonder if there's a fun solution out there based in part on the Hairy ball theorem. $\endgroup$ Commented 2 days ago

2 Answers 2

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The force of the rubber band points radially inward to the center of the circle made by the band. As it interacts with the surface of the ball, this can be decomposed into two forces - one that acts perpendicular to the ball's surface, and one that acts parallel to it. The perpendicular force describes how hard the band is pushed into the ball, and the resulting normal force creates friction between the band and the ball. If the limit of this friction force is greater than or equal to the parallel component force pushing the band across the surface of the ball, the band will not go anywhere.

If the band is put over a diameter of the ball, there is only a perpendicular component and no parallel component, so the band won't go anywhere (even on a frictionless ball). As the band is placed further off-center, the parallel component increases and the perpendicular friction component decreases, and eventually the band slips.

Assuming the band is smaller than the ball, how far off-center the band can be placed will be determined by the coefficient of friction between the band and ball.

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    $\begingroup$ A further complication is that in addition to the possibility of sliding, there is also the possibility of the band starting to "roll". $\endgroup$ Commented Oct 29 at 7:00
  • $\begingroup$ @PeterGreen I don't expect that should be a problem for a typical flat rubber band - if the inner section of the band would slip and "roll" over the outer section, the outer section would already be slipping anyway. I agree it could be a problem for a band with a circular cross-section - in fact, the band would roll off anywhere except the diameter, as the elastic force creates a torque around where the band touches the ball. $\endgroup$ Commented Oct 29 at 14:16
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    $\begingroup$ "there is only a perpendicular component and no parallel component" -- that is not wrong, but it would be clearer to express it as "the parallel component is zero". $\endgroup$ Commented Oct 29 at 14:45
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    $\begingroup$ (even on a frictionless ball) - I think that the discussion actually gets more complicated in a frictionless environment. The exact shape of the rubber band becomes important, whether the rubber band happens to be twisted, how much "exactly" is the rubber band exactly on the equator... $\endgroup$ Commented yesterday
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To add to Nuclear Hoagie's excellent answer: If you have a convex surface that is not a perfect ball, the stable loops around which a rubber band would remain stationary without any friction (and hence the loops the rubber band must be "close" to in the presence of some friction) must exactly be closed geodesics on surface. This is because the elasticity of the rubber band seeks to reduce the loop's length, and geodesics are locally length-minimizing-- there are no shorter loops "nearby" for the band to monotonically deform into.

On a sphere, the closed geodesics are precisely the great circles.

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    $\begingroup$ Since in spheres the closed geofesics are unstable critical points of the length functional, these would not generically be visible if you don’t have friction. $\endgroup$ Commented 2 days ago
  • $\begingroup$ @s.harp Indeed, thanks for this important comment. I was too hasty in saying "there are no shorter loops 'nearby'". This is always true if you fix endpoints (or any two points in the loop), but it's obviously not true for free loops on a sphere-- closed geodesics need only be at a critical point, not necessarily a local minimum. $\endgroup$ Commented yesterday

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