Here is a circular coin with diameter D

Figure A

From its starting position the small coin goes completely around a bigger circular body of diameter 4D without slipping, always in contact with the bigger circle, rotating around its own center point and returns to its original position. Both bodies are in a horizontal plane.

Figure B

In this case the coin does the same thing EXCEPT the small coin is in a vertical plane (standing up as shown, perpandicular to the big circle). The big circle is lying in a horizontal plane. The small coin goes around the bigger circle without slipping, keeping contact with the bigger circle and rotating. It returns to its original position.

Figure C

In this case the small coin goes around a SQUARE shape with each side = 4D. Again it does not lose contact with square, does not slip and is rotating around its own center. It returns to its original position

Figure D

The small coin is placed INSIDE a circular shape with diameter 5D. Again it goes around (inside) without slipping, keeping contact with the bigger circle all the time and rotating around its own center point. It returns to its original position

In which case the small coin makes more rotations?

In other words in which case the rotational distance travelled by any point on the circumference of thee small coin is greater than the other cases or are they all the same?

enter image description here

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As far as I know one of these cases may have been addressed before. So please no computers.

  • $\begingroup$ For part B, what is the thickness of the small coin? $\endgroup$ – Jaap Scherphuis Jul 18 '19 at 12:42
  • $\begingroup$ You can assume it is very small, non consequential $\endgroup$ – DrD Jul 18 '19 at 12:44
  • $\begingroup$ Just a quibble. In B both of the circles must have zero thickness for there to be no slipping, either against the "ground" or against the larger circle. $\endgroup$ – Weather Vane Jul 18 '19 at 12:46
  • $\begingroup$ Hypothetical geometric case I guess $\endgroup$ – DrD Jul 18 '19 at 12:47
  • $\begingroup$ In B, is it the central point of the coin the bit in contact with the circle? $\endgroup$ – Gamora Jul 18 '19 at 12:56

My answer is

diagram C


In A the circumference of the small coin is 4 times that of the large circle, so the coin makes $4$ revolutions. But there will be +1 revolution because of the whole coin rotating once around the large disk, making $5$ revolutions.

In B the same answer, it doesn't matter that the coin is on edge and it also makes $4$ revolutions, but what does matter is that the additional revolution as in A, happens around a different axis. So here, the answer is $4$ revolutions.

In C the circumference of the coin is $ \pi \times D$. The perimeter of the square is $16 \times D$. So the coin makes $\frac{16 \times D}{\pi \times D} = 5.093$ revolutions. But at each corner the coin must rotate through 90° (in the same direction) without making any progress along the square's circumference. This is a clearer example of what happens continuously in A, but here in four steps. The coin makes $6.093$ revolutions.

In D it doesn't matter whether the coin is inside or outside, the coin makes $5$ revolutions.
But what does matter, is that if it were on the outside, it would gain 1 revolution. On the inside it is rotating in the opposite direction and loses 1 revolution. So here, there are $4$ revolutions.

Of those, the greatest number of rotations is C.

Note: I modified my answer to take account of the +1 or -1 rotation, but before then my answer was still C, which is the largest number of revolutions whether or not the compensation is made.

Finally, there are two questions asked, and I am unsure whether the second question is the same as the first, despite the "in other words" remark.


Diagram D is perhaps the most controversial with rotation in different rotations.
The question has the tags geometry and no-computers so I decided to be practical.

I raided the recycle bin (not the computer's!) and made a rough-and-ready experiment.
The circle is five times the diameter of the coin.
To prevent slippage, I nipped and curved a piece of sandpaper tucked under the edge.

enter image description here

A quarter circuit was enough to prove that the coin rotated exactly one revolution.

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This is the smart alec answer:

I am assuming that in case B the point of contact is in the middle of the small coin at first, so the coin flips over as it goes around the large coin.
B uses the most rotations, because the coin (probably) never returns to its original position. The sideways perimeter of the small coin is $2D+2W$ (where $W$ is the width), and this probably does not divide exactly into the circumference of the large circle, which is $4\pi D$. Once the coin has travelled around once, it will likely not be in the same position, as will not be lying flat. It will therefore have to go around several times at least, and unless $W$ is specifically chosen to make it work, the coin will never be in the correct orientation at the top of the large circle again.

The more normal answers:

A: the large circumference is $4$ times the small coin circumference, so the small coin rotates 4 times relative to the point of contact, plus one because the point of contact goes around the small coin once, for a total of $5$ revolutions.
B: First assume as above that the small coin touches the large coin in its centre. Now the small coin perimeter is about $2D$ (ignoring the width), so it rotates about $1+\frac{4{\pi}D}{2D} = 1+2\pi \approx 7.28$ times. If on the other hand the point of contact is on the rim of the small coin, then it is not possible to really provide a single number for its rotations. It rotates 4 times around an axis through its centre, but that axis in turn rotates once as it goes around the large coin. Those are two different rotations in 3D space that do not add up to a single number.
C: The large coin perimeter is $16D$, so the small coin rotates about $1+\frac{16D}{\pi D} = 1+\frac{16}{\pi} \approx 6.09$.
D: This time the small coin rotates $5$ times relative to the point of contact (similar to case A but with a larger coin), but the point of contact goes in the opposite direction, for a net rotation of $4$ times.

So given the assumption for how case B works, the right answer is still B. If that assumption for B is wrong then case B is not directly comparable to the others, but whichever way you look at those rotations, there are no more than $5$, which means that C would be the right answer.

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  • $\begingroup$ Did you actually try it? Might surprise you $\endgroup$ – DrD Jul 18 '19 at 13:36
  • $\begingroup$ @DEEM Did you actually read my answer? I have clarified it now with the assumption I was making. $\endgroup$ – Jaap Scherphuis Jul 18 '19 at 13:55
  • $\begingroup$ Surely the small coin perimeter in case B is $\pi$D instead of 2D, you are rolling the coin on its edge, not continuously flipping it over against the big coin. $\endgroup$ – Bass Jul 18 '19 at 18:10
  • $\begingroup$ @Bass I did both options. $\endgroup$ – Jaap Scherphuis Jul 18 '19 at 18:15
  • $\begingroup$ Ah yes, so you did. Reading the question again, I agree that "rotational distance travelled by any point on the circumference of the small coin" should also specify the axis in order to get a more reasonable result. $\endgroup$ – Bass Jul 18 '19 at 18:20

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