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I have two coins - one has $N$ times the diameter of the other. I roll the smaller coin one full lap around the circumference of the larger coin like a gear, keeping contact it in with the edge of the larger coin and keeping the larger coin still.

How many rotations does the smaller coin make around its own center in the rest frame? Equivalently, if you drew an arrow on the small coin which was initially pointing East, how many times during the process would the arrow point North?

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    $\begingroup$ This is more of a puzzle than appears at first glance because the obvious answer is wrong. $\endgroup$ – xnor Jul 10 '15 at 0:52
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    $\begingroup$ @xnor: The solution technically is ambiguous, depending on how we interpret "revolutions", or equivalently, whether point $A$ is on the larger coin or on the smaller one. $\endgroup$ – COTO Jul 10 '15 at 3:01
  • $\begingroup$ @COTO Good observation, that was ambiguous. I tried to clean up the wording to clarify. $\endgroup$ – xnor Jul 10 '15 at 3:09
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    $\begingroup$ @xnor: Upon closer inspection, the obvious answer still looks right. What's your interpretation of "revolution"? I can't find one that does anything unintuitive or interesting. $\endgroup$ – Chris Burt-Brown Jul 10 '15 at 12:55
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    $\begingroup$ @ChrisBurt-Brown See this animation which shows that when $N=1$, the moving coin actually rotates $2=N+1$ times. $\endgroup$ – Mike Earnest Jul 10 '15 at 17:27
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Answer :

$N+1$

Explanation :

Circumference of bigger coin $= 2\pi N r$
Circumference of smaller coin $= 2\pi r$
Number of revolutions $= \frac{2\pi Nr}{2\pi r}=N$
But the smaller coin has also revolved about the center of the whole configuration, which increases the number of revolutions by 1
So total $= N+1$

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    $\begingroup$ As far as I can see, mathematically speaking, the $N+1$ is meaningless since the N and the 1 were counting two different kinds of revolution. (N = revolutions around self, 1 = revolutions around other coin). I suppose they are both revolutions though, so perhaps this is the intended answer in a "How many triangles are in the picture" sort of way. But I don't see any reason why adding the N and the 1 is actually correct. $\endgroup$ – Chris Burt-Brown Jul 10 '15 at 12:39
  • $\begingroup$ @Mike , thanks for the formatting !! Chris , yes , the wording of the question may be improved. I am sure OP wanted "N+1" [[ A question like "what is 2+2 ?" will have the obvious answer 4 , but if the question had a twist , and the answer was 5, then it becomes a puzzle !! ]] $\endgroup$ – Prem Jul 10 '15 at 14:18
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    $\begingroup$ @ChrisBurt-Brown: en.m.wikipedia.org/wiki/Coin_rotation_paradox $\endgroup$ – Deusovi Jul 12 '15 at 8:28
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Edit:

The number of revolutions would actually be increased by one (see this link provided by Mike), so the answer could be either $N+1$ or $1/N + 1$ if N is greater or less than 1 respectively.


The answer is actually:

Either $N$ or $1/N$ depending on if $N$ is greater or less than 1. (And $N$ is from solving $(2\pi*r*N)/(2\pi*r)$ using the formula for circumference).

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  • $\begingroup$ It is not wordplay: the question is asking how many revolutions the small coin makes around its own center. So, if you drew a red dot at its center, and a black dot on its edge, how many times does the black dot rotate around the red. $\endgroup$ – Mike Earnest Jul 10 '15 at 4:14
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    $\begingroup$ @MikeEarnest Aside from the frame of reference not being specified, I don't see how that interpretation would result in N+1 either. If you assume N to be 1 and draw the path of the red dot and the black dot, the black dot will be below, to the right, above, and to the left of the red dot in that order during its motion. And if you weren't making a case for N+1 (wasn't sure), it could be considered N+1 if you count all types of revolution, not just from a rotational perspective but an objective perspective (the small coin revolved around the big coin once). $\endgroup$ – Quark Jul 10 '15 at 4:56
  • $\begingroup$ I like this because it considers the case if $N < 1$ $\endgroup$ – justhalf Jul 10 '15 at 9:23
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    $\begingroup$ @Quark See this animation, for the $N=1$ case. After the moving coin has gone half way around the stationary one, the moving coin has made a full rotation around its own center. So, when it goes all the way around, it makes 2 rotations (i.e. $N+1$). $\endgroup$ – Mike Earnest Jul 10 '15 at 15:39
  • $\begingroup$ @MikeEarnest I finally see now and you are completely correct. I think you are underestimating your solution; I don't think anyone else has the same thing in mind that the animation shows. Everyone else is thinking of adding two different interpretations of revolution, while that animation shows the actual paradox at play. I think everyone here would benefit from an actual answer with that animation included. If you don't want to make one I can edit it into my answer. $\endgroup$ – Quark Jul 10 '15 at 16:52
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The answer

N+2 rotation around large coins center doesn't count
N+1

Why?

Every time the small coin rolls a distance of π it actually makes 1+1/N revolutions around its own center. The angle from the center of the small coin to the point at which the coins touch will be 2π/N more than the angle at the beginning.
enter image description here
So by the time it rolls Nπ then it has made N*(1+1/N)=N+1 revolutions around its own center. If you then add to this the one revolution it has made around the large coins center then you get N+2

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  • $\begingroup$ @CodeNewbie Technically using definitions from astronomy there is only 1 revolution around the center of the large coin. The small coin spinning while it revolves would be called rotations. However since the two words are used interchangeably in common everyday language I am counting both as revolutions. $\endgroup$ – Dean Brown Jul 10 '15 at 16:51
  • $\begingroup$ updated my answer now that the question was clarified to ask for rotations not revolutions $\endgroup$ – Dean Brown Jul 12 '15 at 8:25
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For the smaller coin the distance traveled in one revolution would be $\pi*Diameter$. Since we know the bigger coin has a diameter of $N*Diameter$ of smaller coin we know the total distance that needs to be traveled by the smaller coin is $N*Diameter*\pi$. That leaves us with the formula of $\frac{N*\pi*Diameter}{\pi*Diameter}$ which leaves us with $N$. So the answer is $N$ revolutions.

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Is the answer

one revolution

because

point A is on the small coin. So after one revolution of the small coin you are back at A.

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  • $\begingroup$ As the small coin rotates around, it is spinning, so the answer is more than one $\endgroup$ – Mike Earnest Jul 10 '15 at 2:54
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The answer is

1 revolution and $N$ rotations.
Since a revolution is one path around the bigger coin, and a rotation would be about its own axis. Based on the circumferential calculations in the other answers, the number of rotations is $N$ whereas the smaller coin only completes one lap around the bigger coin, so it makes just one revolution.

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  • $\begingroup$ The question is asking how many revolutions the small coin makes around its own center. So, if you drew a red dot at its center, and a black dot on its edge, how many times does the black dot rotate around the red. $\endgroup$ – Mike Earnest Jul 10 '15 at 4:13
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    $\begingroup$ @MikeEarnest, the question doesn't specify the frame of reference. $\endgroup$ – CodeNewbie Jul 10 '15 at 4:31

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