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If a coin rolls without slipping around another coin of the same or different size, how many times will it rotate while making one revolution?

The proof given is like this:

Cut the curve open at some point and uncoil it into a straight line segment. Rolling the circle along this segment it will rotate (length curve)/(length circle) times.

Keeping the circle attached to one end of the segment, we then recoil the segment back into thecurve, which contributes the final rotation.

I can understand the first part of the proof, but I couldn't understand the second part, any ideas? Also, I am also interested in different approaches for proving this one.

Problem/Solution source.

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Both ways are the same thing,only difference is that in one of them the line segment is stationary and in other one the circle is stationary while the segment is wrapped around it.(by relative motion funda :) – Tomarinator May 24 '12 at 14:24
@Willie Wong:Hm, assuming anticlockwise movement and from what I can understand from that .gif file in wike page shouldn't the winding number in this case is 1? – Quixotic May 24 '12 at 14:25
Winding number explains the "extra" twist. In general if you "roll a coin" where the coin has circumference $C$ along a closed curve $\gamma$, the total number of turns the coin goes through is the winding number of $\gamma$ plus $|\gamma| / C$. The latter is the number of turns in the "co-moving" coordinate system, and the former is the number of turns the coordinate system itself made. – Willie Wong May 24 '12 at 14:29
up vote 6 down vote accepted

I suggest that you first regard the case with "total slipping".

Let a coin slide around another coin while having always the same point attached to the central coin. Note that the outer coin makes exactly one rotation in this scenario.

Now, roll out the inner circle and let the outer coin slide along. Clearly, it does not rotate at all.

So, there is one rotation of the outer coin that comes just from going around the inner coin. The rest of the rotation can then be found by looking at the rotation along the line and combining the two.

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You do it in three steps (where I just recall the first two steps, which you understood).

  1. The two circles touch at one point. You cut one of the circles open at that point and unravel it to make it into a line. Let us assume we do this such that the line is horizontal and the circle sits at the left end.

  2. You roll the second circle along that line to the right end.

  3. You bend the line back to a circle.

For step three note that the circle is "attached" to the right end. You fix the left end of the line take the right end of the line and start to move it. The circle - which is still attached to that end - will make another whole turn.

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Sorry, but I still don't understand why are we joining the right end after the the second step. – Quixotic May 24 '12 at 14:30

It is maybe easiest if you imagine an "axis" joining the two centers of the circles all the time (so one end of the axis doesn't advance at all an the other end turns a full circe), and a camera mounted on this axis filming the circles.

Since the camera makes a full tour in one direction (call it positive), it will see the fixed circle make a complete tour in the negative direction. If that fixed circle has radius $R$ and the other radius $r$, then at the point of contact this negative rotation produces a displacement of $2\pi R$ in the backwards direction, which will cause the other circle to make $R/r$ tours in the positive direction (so that it also produces a backward displacement of $(R/r)2\pi r=2\pi R$ at the point of contact, corresponding to "no slipping"). But since the camera also made a complete tour in the positive direction while filming, it is clear that the mobile circle made $(R/r)+1$ tours in reality (i.e., with repect to a fixed observer).

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