How Good Is Your Intuition, 2? (Another Mind-Teaser)
 

I am told this question was on a recent SAT test.

Two coins lay flat on a table top .

The small one has radius 1, the large one has radius 3 .

The large coin is fixed, no motion (translation or rotation).

The small coin is placed so that it's edge touches the edge of the large one. The small coin is then rolled
clockwise around the edge of the large coin. The coins' edges do not slip where they touch (think of gears).

How many times will the small coin rotate around its own center, before the center of the small coin returns to its original location?

For all you weasels:
- the coins are round and they cannot be deformed, Schumi
- there is no effing thermal expansion, Randy
- rotation and translation are measured relative to the tabletop, so no good telling me about how the earth rotates or the milky way rotates
-

3 times.

what he said.

It is not quite that simple . . .

Doh! 1.5 times maybe.

ok ~3 times

damn

Schfifty-five....

< disregard >What am I doing wrong then?

C = 2πr or πd

C of big coin = 6π

C of little coin = 2π

One rotation takes the point on the edge of the small coin so it's 2π or 1/3 of the way around the large coin's circumference, the second takes you around to 2/3, and the third rotation takes you around so that the starting point on the outside of the small coin has traveled to the same point on the outside of the large coin. If the two outside points have again met2, the center should be in the same location as it started.
< /disregard, crap, it's all wrong >

C = 2 pi r, where r= 3

= 6 pi total distance around

4 times

Yeah, that's what I got, but it's not really relevant in this case. The question is how many rotations (360° or 2π) will the small coin need to get back to the point where it started. Well, the circumference of the small coin will wrap around the large coin 3 times, but that's not 3 rotations, thats 4 full rotations.

Suppose the edge of the large coin were cut off, and straightened into a line segment. It has length 2 * pi * 3. The circumference of the small coin is 2 * pi * 1. So the small coin needs to roll 3 times to get from one end of the line segment to the other. Exactly as you said.

But - the edge of the large coin is not a line segment. It is a circular arc that makes a full 360 degree loop. Thus the small coin makes 1 full rotation just to follow this arc. Even if the small coin were slid around the large coin's edge (the same point on the small coin's edge is always touching the large coin), it would make 1 full rotation by the time it returned to its original location.

3 + 1 = 4. The small coin makes 4 full rotations around its own center.

Test this. Take 2 quarters. Place them so that the 6 o'clock of the rotating quarter touches the 12 o'clock of the fixed (bottom) quarter. Orient the rotating quarter so that the top of Washington's head points at 12 o'clock. Now roll the rotating quarter around the fixed quarter until it is back at its starting point. How many times does the top of Washington's head point at 12 o'clock? Once, or twice?

Now, imagine the small coin were rotated around the "inside" of the large coin's edge. What's the answer then?

I'm told the professor who wrote this SAT question got the wrong answer himself.

I think the thing here is, our intuition is familiar with round things rolling on straight things (wheels on a road). We are not accustomed to thinking of round things rolling on other round things. Unless you're an engineer designing planetary gears.

Or, you could get out your paper and pens and a vernier caliper, cut two paper circles that are very close to being 3:1, and then work it out after you realized that your first answer is wrong, but you can't mentally figure out/visualize the right answer. :rolleyes:

:D

I don't even want to know how you did the earth-wire problem . . .

..9

When I imagine it, on the first rotation (relative to the fixed) the small coin has rotated only 2/3... so 2.

edit: make that 4/3. so 4 total ...the small would have gone thru 4/3 to get to that first 2/3 position.

should have read Steve's answer/ thinking, so I wouldn't have been so quick to answer. ...but what fun would that be?

2

Les

Yes, 3 - 1 so 2.

Isn't the center of the coin always in the same position, if it is rotating around the center? That's like saying how many revolutions must the Earth make before the North Pole is back in its original location - the North Pole is always in the same location.

I mean, make a dot on the table under the center of the small coin. Then start rolling the small coin around the large coin. The center of the small coin will move away from the dot, travel in a circle, and eventually return to the dot.

Arggghhh!! You baxtard!

I think the answer is two times.

Sorry for the poor quality sketch. The contact point between the coins "travels" a distance of 6pi. The smaller coin has a circumference of 2pi. Therefore the smaller coin "rolls" three times during the trip. But point B only circumnavigates point A twice. It would circumnavigate three times if the trip had been in a straight line. But one doesn't happen because the distance traveled was a circle.http://forums.pelicanparts.com/uploa...1295448084.jpg

Here is a practical application of that problem.

Say you want to test your inertial measurement unit while it is standing still by seeing how fast the earth is going around. You want to figure the rotation speed out and don't have a WGS sitting around. The earth goes around the sun 1 time per year, and rotates relative to the sun once per day, 365.25 times a year. It keeps coming out wrong.

The earth makes about 366.25 rotations per year. The once around the sun is another rotation.

Good diagram, but it shows 4 rotations. The point on the small coin marked "B" starts at 3 o'clock (relative to the center of the small coin), it passes through 3 o'clock at 1/4 of the way around the large circle, again at 1/2 way, again at 3/4 way, and finally when it completes the trip - that is 4 times.

Maybe think about it this way. Suppose the large coin had radius zero, it is just a point. Rotate the small coin around that point - it makes 1 rotation even though the circumference of the large coin is zero. So you have to add 1 rotation to the 3 rotations that the small coin has to do just to travel around the circumference of the large coin.

Me and Zarniwoop have a artificial universes in our offices. I synthesized a spare earth to do the experiment. Getting the wire the right length and fused without a visible joint was the hard part. The quarter/penny experiment was much easier.

Nice try, but as has been said, your drawing is missing a couple of important points. I made two circles in a 3:1 diameter ratio and worked this out. Your rotation is off. If you start at the top of the big circle, then the spot where the two circles touch will be back at the bottom of the small circle when it is at the 3 o'clock position on the big circle. Each full rotation is at 12, 3, 6, and 9 o'clock positions.

Reply to jyl: Thanks for the correction - Four times.

More visualization...

note the pelican would be on its head again before the first third. ;)

. . .starting on its head, and going clockwise.. :cool:

http://forums.pelicanparts.com/uploa...1295455171.jpg

No visualization needed.

Three circumferences of small coin = one circimference of large coin.

Rolling small coin for three circumferences (three revolutions) brings it to the end of the circumference of the the large coin.

Rolling it one more circumference (revolution) puts it back in its original position, relative to the large coin.

Four revolutions is the answer.

what coins?

updated. visualization...


http://forums.pelicanparts.com/uploa...1295457317.jpg

Such strong (post edit) confidence you have. What happened to your 'oops' post? :cool:

The earth does go around 366 1/4 times each year.

Suppose the small coin is rolled along a path whose length is N-times the circumference of the small coin, but that path is not the edge of the large coin - it is some other squiggly line. What's the rule that lets us figure out how many rotations the small coin makes?

http://forums.pelicanparts.com/uploa...1295459519.jpg

Immediately after posting it I noticed a typo and at the same time I heard my doorbell. It was the termite inspector arriving for his 9:00 appt.

I had to show him where my concern was and rather than leave the faulty post up I changed it to "oops", let the termite guy in, showed him where the droppings were and he told me it wasn't termites, it was ants. Didn't take very long. he left, and I reposted.

When I read the OP I typed out my answer as concisely and quickly (thus the typo) as I could. That took, oh, two minutes?

How much time did you spend playing with your graphic? :cool:

Rotations = length of line / circumference of coin.

Then you have to consider whether the net direction change of the line is in the direction of rotation of the coin or opposite it.
If in the same direction, add one turn for every 360 degree net change in orientation of the line or subtract one turn for every 360 degree net change on the opposite direction.

So for a straight line: exactly what Darisc said.
For a line such as the one above, subtract about one rotation.

Les

Who can argue with that? :)