The Coin Problem

[JustBecause]

Think about it from the perspective of the center of the small coin.

Yes, its circumference is 1. If you roll it on a flat surface for three inches, the center travels 3 inches, in a straight line

But think about how far the center is traveling when it goes around the big circle. The center is now traveling in a circle, and that circle's circumference has changed.

Instead of moving in just a straight line, equal to the circumference of itself 2*pi*r = 1, the center is now moving in a circle with a circumference of 2*pi*(R+1) = whatever (more than 1). That is where the extra movement is coming from.

No it’s not. It comes from the fact that it’s getting an extra rotation because the line it’s traveling on is a circle.

 

[dafarmer]

No it’s not. It comes from the fact that it’s getting an extra rotation because the line it’s traveling on is a circle.

An old saying, you need to be hit up along side your head, a 3 inch circumference coin is 3 .inches long and a 1 in circumference is 1 inch long

 

[CloneJD]

An old saying, you need to be hit up along side your head, a 3 inch circumference coin is 3 .inches long and a 1 in circumference is 1 inch long

3 revolutions about the coin’s own axis/center plus 1 revolution about the other coin’s axis/center = 4 revolutions.

 

[besserheimerphat]

The starting point on the dime, at the bottom of FDRs neck, only touches the big circle 3 times - once at the top, once after 120° and once after 240°. However, we're not talking about how often FDRs neck touches the big circle. We're talking about how often FDRs neck is at the bottom of the dime. At 90° along the big circle, FDR has only rolled 3/4 of a full revolution. But it's ALSO rolled along 1/4 of the big circle's circumference. So FDRs 3/4 rev + 1/4 rev from the big circle means FDR has traveled 360° total and is now right-side up even though it hasn't yet made contact along the dime's full circumference.

At the bottom of the big circle, FDR has rotated 1.5 revs + .5 revs from the big circle. That's two total revs so now the top of FDRs head is against the big circle but he's right-side-up from our perspective.

At 270° along the big circle, FDR has traveled 2.25 revs + .75 revs from the big circle. The back of FDRs head is against the big circle but he's right-side-up from our perspective.

At 360° along the big circle, FDR has traveled 3 revs + 1 rev from the big circle.

 

[Rogue52]

I *think* I’m right about this which gives the answer for any size radius for either circle. The video earlier in this thread assumed the rotating circle has fractional radius of the stationary circle.

“Rotating Circle” has radius a
“Stationary Circle” has radius b

Number of Rotations = (Distance traveled of a single point on Rotating Circle) / (Circumference of Rotating Circle)

(Circumference of Rotating Circle) = 2*pi*a

For the distance traveled of a single point on the rotating circle, the easiest point to consider is the circle’s center which travels the circumference of a circle with radius = a + b.

So…

(Distance traveled of a single point on Rotating Circle) = 2*pi*(a+b)

Final Solution:

Number of Rotations = (Distance traveled of a single point on Rotating Circle) / (Circumference of Rotating Circle)
Number of Rotations = (2*pi*(a+b))/(2*pi*a)
=(a+b)/a
= 1 + b/a

For the example in this thread, a = 1, b = 3, so the answer is 1 + 3/1 = 4.

 

[ISUTex]

View attachment 107390

4. Duh.

 

[FDWxMan]

No it’s not. It comes from the fact that it’s getting an extra rotation because the line it’s traveling on is a circle.

Well, yes. That's what I'm saying. Just a way to try to help people visualize that their circular path is adding to the problem