Why does a coin take 2 full rotations around another coin?

[OmCheeto]

Fun maths problem:

I decided to let my spreadsheet graph the distance traveled by a circle of varying radius traveling along a flattened(unrolled?) circle of 1/(2π) radius and it said the answer was: distance = 2πr + 1

I thought; "There it is! The 'Plus 1' !!!!"

Then I plotted the same thing, only I swapped which variable was the variable and my spreadsheet then said the answer was: distance = 2πr + 2π.

I almost cried.

 

[Vanadium 50]

Devin-M, I think all that really needs to be said regarding this problem was expressed in the following post by PeroK:

"Everything that needs to be said has been said, but not everyone has said it yet." - Morris Udall.

 

[DaveC426913]

It just makes more intuitive sense to my mind ...

If the correct answer made intuitive sense, this PF thread would not exist.
Nor would the video that triggered it.
Nor would the article that detailed the plight of the boys who found the error.


What I'm not sure about is what your current issue is.
Are you disputing the offered answer?
Are you simply musing out loud about how difficult it is to intuit the offered answer?

 

[jbriggs444]

I recently stumbled across this video from Veritasium where a similar problem is discussed. It analyzes the case of a 1r coin rolling around a 3r coin.

That case arises from a multiple choice question on on a past SAT test did not offer either correct answer to the ambiguously worded question.

 

[PeroK]

I recently stumbled across this video from Veritasium where a similar problem is discussed. It analyzes the case of a 1r coin rolling around a 3r coin.

That case arises from a multiple choice question on on a past SAT test did not offer either correct answer to the ambiguously worded question.

That video is part of the discussion here:

https://www.physicsforums.com/threads/deceptively-simple-geometry-question-on-sat-test.1057821/

 

[Devin-M]

If the correct answer made intuitive sense, this PF thread would not exist.
Nor would the video that triggered it.
Nor would the article that detailed the plight of the boys who found the error.


What I'm not sure about is what your current issue is.
Are you disputing the offered answer?
Are you simply musing out loud about how difficult it is to intuit the offered answer?

I'm not disputing any of the offered answers. Yes, I'm simply musing out loud how difficult it is to intuit the offered answer.

That said I believe I found what appears to me to be an *intuitive* way to think of it:

-Suppose we have a radius 1 coin and a radius 10 coin.

-We "unwrap" the circumference of the radius 1 coin into a straight line.

-We roll the radius 10 coin along the straight line and find it completes 1/10th of a rotation traversing the straight line

-Now we "re-roll" the radius 1 coin back into a circle and roll the radius 10 coin along the radius 1 coin without slipping and find it completes 1 and 1/10th rotations. That is 11 times more rotations than rolling the radius 10 coin rolling along the straight line.

Now the intuitive part:

-In the straight line case if we measure the area "swept" by the tangent going through the radius 10 coin's contact patch and extending to the opposite side of the coin, this area is 125.6

-In the circular case, if we measure the area "swept" by the tangent going through the radius 10 coin's contact patch and extending to the opposite side of the coin, the area is 1382.3

-1382.3 area (circular path) is exactly 11 times greater than 125.6 area (straight path), and the radius 10 coin rotates exactly 11 times more rotations on the circular path than the straight path.

-In other words, on the circular path, the coin covers 11 times more area, and to accomplish this has 11 times more rotations.

 

[sdkfz]

This made my head hurt, so I just did an "experiment" with two Lego gears - with pins inserted into one hole each so I would not lose track of rotation.

What really made it sink in was realising:
At the start, the right side of gear A was touching the left side of gear B.
Keeping gear A still and moving gear B around it ...
Half way around, the left side of gear A was touching the right side of gear B.

Maybe that's blindingly obvious but it made clear to me the added effect of moving one of the object around the other. (Secondly, letting both gears move equally, they each (as same radius) of course did just one rotation each.)

 

[Ibix]

Maybe that's blindingly obvious but it made clear to me the added effect of moving one of the object around the other. (Secondly, letting both gears move equally, they each (as same radius) of course did just one rotation each.)

That's the point I was trying to make in #30. The Lego gears is a good demo idea.

 

[OmCheeto]

That's the point I was trying to make in #30. The Lego gears is a good demo idea.

But that was my argument. The lego/spirograph gear shows that the outer ring (in the 1:1 size scenario) only went around the inner circle once.

source: https://nathanfriend.io/inspiral-web/

Given the choices of answers presented in the SAT, I would have chosen 3.
I mean seriously, how many times have we been told:

e = mc² is WRONG!
pe = mgh is WRONG!

This was just another example of holier than thou adolts trying to mess with teenaged minds.

ps. I have hated word problems since I was 5, when my 4rd grade level sister was teaching me maths and asked me how many pies she would have if she started with 3 and gave me 4. I laughed and accused maths of being stupid.

 

[Ibix]

asked me how many pies she would have if she started with 3 and gave me 4.

$e^{i\pi}$, amusingly.

 

[Devin-M]

The lego/spirograph gear shows that the outer ring (in the 1:1 size scenario) only went around the inner circle once.

When the “spirograph pen” reaches the point circled in red, the moving gear has completed 1 full rotation, and is only half way around the other gear…

 

[OmCheeto]

When the “spirograph pen” reaches the point circled in red, the moving gear has completed 1 full rotation, and is only half way around the other gear…

View attachment 336796

Yes, I'm quite aware of that fact.

 

[A.T.]

It has nothing to do with travelling "further".

Have you watched the video?

Have you watched it? At 9:45 it explains that the distance the circle center travels determines the number of rotations the circle makes:

 

[DaveC426913]

Have you watched it? At 9:45 it explains that the distance the circle center travels determines the number of rotations the circle makes:

Circle centre is another matter.

My demonstration (in post 6) - and my dialogue - is not addressing circle centre, but perimeter*.

There are several ways to "solve" the apparent paradox. My demonstration disambiguates the issue of trying to tracking distance (of perimeter), which seemed to be confusing Devin-M.

So, while you are not technically wrong in your reference to 9:45 of the video, it is out of context of my solution, and therefore I'd say it constitutes a red herring.

* _{British/Canadian: perimetre?}

 

[A.T.]

There are several ways to "solve" the apparent paradox. My demonstration disambiguates the issue of trying to tracking distance (of perimeter), which seemed to be confusing Devin-M.

Yes, the image below that @Devin-M posted tracks the wrong point. To directly get the right quantitative answer, it should track the center of the large circle. But the general idea, that the circle rotates more, because it is travelling further, is on the right track.

 

[Devin-M]

But the general idea, that the circle rotates more, because it is travelling further, is on the right track.

Another intuitive way to think about it, in the lego gears case, for me, is suppose you “unwind” the outer geared surface of a lego gear so that it becomes a straight path— watch the tips of the neighboring gear teeth get closer together.

 

[DaveC426913]

Another intuitive way to think about it, in the lego gears case, for me, is suppose you “unwind” the outer geared surface of a lego gear so that it becomes a straight path— watch the tips of the neighboring gear teeth get closer together.

Hm. That's only the case for teeth of inordinate length. There is no reason why the height of the teeth can't approach zero, which means the amount by which they get closer together likewise approaches zero.

In fact, the scenario works by friction alone; teeth are superfluous.

Since we can eliminate the effect without altering the scenario, it's got to be a spurious factor.

 

[Devin-M]

But if the teeth are lengthened, then the tips of the neighboring teeth can be considerably closer when the path is unwound & straightened.

 

[DaveC426913]

But if the teeth are lengthened, then the tips of the neighboring teeth can be considerably closer when the path is unwound & straightened.

Teeth are not relevant to the problem or solution; they are a distraction (one that you added late in the game). And, after all, coins don't have interlocking teeth.

Witness the fact that the teeth can (and should) be removed completely and absolutely nothing about the problem will change.

 

[Devin-M]

Teeth are not relevant to the problem or solution; they are a distraction (one that you added late in the game). And, after all, coins don't have interlocking teeth.

Witness the fact that the teeth can (and should) be removed completely and absolutely nothing about the problem will change.

Here’s why I believe they are relevant and illustrative of the problem.

In the case of a radius 1 stationary coin and the radius 10 moving coin, the factor of change of # of rotations between the straight line case versus the circular case is directly proportional to the change in area swept by the tangent that extends through the radius 10 coin in straight vs circular cases.

In other words the area swept by the tangent is 11 times greater in the circular case & the coin rotates more times by a factor of 11 times greater.

The line forming center of base to tip of a gear tooth is also a tangent to the inner coin. In fact if the gear teeth on the radius 1 coin were length 20, if we rotate the radius 1 coin 1 rotation, then a length 20 tooth will sweep the same area as the tangent of the radius 10 coin.

So essentially the changing distance between the tips of the teeth in the straight line vs circular case can visually illustrate the changing area swept by the tangent of the radius 10 coin around the radius 1 coin… a change in area in which the factor of change is directly proportional to the factor of change in number of rotations.

 

[Lnewqban]

... I'm simply musing out loud how difficult it is to intuit the offered answer.

-Suppose we have a radius 1 coin and a radius 10 coin.

 

[A.T.]

In the case of a radius 1 stationary coin and the radius 10 moving coin, the factor of change of # of rotations between the straight line case versus the circular case is directly proportional to the change in area swept by the tangent that extends through the radius 10 coin in straight vs circular cases.

In other words the area swept by the tangent is 11 times greater in the circular case & the coin rotates more times by a factor of 11 times greater.

The line forming center of base to tip of a gear tooth is also a tangent to the inner coin. In fact if the gear teeth on the radius 1 coin were length 20, if we rotate the radius 1 coin 1 rotation, then a length 20 tooth will sweep the same area as the tangent of the radius 10 coin.

So essentially the changing distance between the tips of the teeth in the straight line vs circular case can visually illustrate the changing area swept by the tangent of the radius 10 coin around the radius 1 coin… a change in area in which the factor of change is directly proportional to the factor of change in number of rotations.

I don't see how your construction that involves areas is more intuitive, even if it was correct. It seems that this is just a unnecessarily complicated attempt to get to the distance traveled by the center of the coin, which is indeed directly related to the number of rations, as explained in the Veritasium video at 9:45.

What the "teeth" can offer, is a visualization of the local normal vectors, which can be used to define a local frame of reference moving along with the contact point. Relative to the local normal vector, the rolling coin has the same number of rotations as for straight rolling. But if it rolls around a coin, the normal vector itself rotates along the path in the same direction, so it adds one rotation. For rolling inside a circle the normal vector rotates opposite to the rolling rotation, so it subtracts one rotation.

 

[Averagesupernova]

We start with the above.

Rolling it around so the dull coin is on the bottom gives us an inverted dull coin relative to the first pic.

This last image is of course wrong. The part of the coin with Washington's nose would have to continually contact the shiny coin on its trip around. This action is not rolling.
-
When intuition gets you in trouble as it often does, work out a path to the answer by showing how it cannot work as your intuition says it would.

 

[davenn]

Have you watched the video?

yeah, saw it about a week ago
I love Destin's videos, they are incredibly informative :)

 

[Mister T]

on the circular path than the circumference “unwound” into a straight path.

The difference is that following the circular path causes it to make one additional rotation.

Note that the moon makes one rotation when it makes one revolution around Earth, because the same side of the moon always faces Earth.

 

[A.T.]

Note that the moon makes one rotation when it makes one revolution around Earth, because the same side of the moon always faces Earth.

Exactly. When you take pure sliding along a straight line (no rotation), and merely roll that entire scenario so the line becomes a circle (preserving the contact point of the coin with the line), the you already have 1 rotation, like the tidally locked Moon has.

 

[sophiecentaur]

Yes, I'm simply musing out loud how difficult it is to intuit the offered answer.

I was out of the loop when this thread ran, originally. This quoted statement only goes to show how useful Maths is. Intuition is a false friend but (the appropriate) Maths will not let you down.

 

[painter]

This one baffles me, I still can’t get my head around it (no pun intended).

Ahh. May I ask you that what does"Why does a coin take 2 full rotations around another coin? "mean?
I cannot get the point .
Does coins really take 2 full rotations around another coin then it will be stationary??

 

[jbriggs444]

Ahh. May I ask you that what does"Why does a coin take 2 full rotations around another coin? "mean?
I cannot get the point .
Does coins really take 2 full rotations around another coin then it will be stationary??

Take two coins. Place them flat on a table, one above the other:

Roll the top coin around the bottom coin without slipping so that it completes a complete circle. Watch how many times the picture on the moving coin rotates [relative to the fixed orientation of the table].

 

[Averagesupernova]

Ahh. May I ask you that what does"Why does a coin take 2 full rotations around another coin? "mean?
I cannot get the point .
Does coins really take 2 full rotations around another coin then it will be stationary??

Have you actually read this entire thread?

 

[Mister T]

Does coins really take 2 full rotations around another coin then it will be stationary??

You have to apply a force to make the coin move. Anytime you stop applying the force the coin stops moving (due to friction).

When you move the coin you have to rotate it yourself in such a way that it doesn't slip against the coin in the center.

 

[sophiecentaur]

in such a way that it doesn't slip against the coin in the center.

Use gears instead; no significant friction involved.

 

[Mister T]

Use gears instead; no significant friction involved.

Some coins have ridges that act like little gear teeth. But I meant friction between the coins and the surface they rest on.

 

[DaveC426913]

You have to apply a force to make the coin move. Anytime you stop applying the force the coin stops moving (due to friction).

Some coins have ridges that act like little gear teeth. But I meant friction between the coins and the surface they rest on.

This is over-thinking the issue. The coins do not slip as they move against each other. How this happens is irrelevant and a distraction.

 

[Vanadium 50]

Before duplicating the entire thread, maybe we should give @painter an opportunity to read it.