tronvillain

I have yet to read any of the replies, so there might be a convincing explanation of why I am wrong, but I am going to have to say "Yes, the rolling ball can come back rotated relative to its original orientation." Well, it depends on exactly what was meant by "rotated" - the rolling ball will end up with exactly the same point resting on the fixed ball's north poll, but the orientation the rest of the ball has will depend entirely on how far it was rolled along the equator.

Jayjay

Shouldn't this be:

----- ..2..
|---| 6-4-1
----- ..5..

instead? Followed by


..5..
1-4-6
..2..

-----
|---|
-----

i.e. the same as the start...

Jesse

OK, longer version of my proof--if you want to follow it, you pretty much have to draw pictures for each of the steps, it's a little too complicated to see it all in your head, at least at first.

A term I need to define first: a "great circle" is a circle on the sphere that has the center of the sphere as its own center, like the equator or the lines of longitude on a globe. When the spheres move, they always roll along great circles.

Start out with a picture like fando's initial diagram earlier, with a red ball on top of a black ball and a stick pointing out the right side of the red ball. To keep track of angles, define three axes--a z-axis which points up (going through the centers of both balls when they're in their starting position), with +z in the direction of the red ball and -z in the direction of the black ball; an x-axis with -x pointing to the left of the picture and +x pointing to the right (so the stick is coming out of the red ball pointing in the +x direction); and finally, a y-axis with -y pointing out of the picture towards the viewer, and +y pointing in away from the viewer.

Now roll the red ball down to the equator of the black ball. The great circle that the red ball is rolling on always rotates by twice the angle it moves along the black ball, so the moving 90 degrees along the black ball will cause the stick to rotate 180 degrees, so it's now pointing in the -x direction. The point where the stick comes out of will now be the point of contact between the two balls. Meanwhile, the line from the center of the red ball to the center of the black ball is parallel to the x-axis.

Now roll the red ball n degrees along the black ball's equator. The red ball will rotate 2n degrees around its z-axis, so the stick will still be in the x-y plane, but it will be pointing 2n degrees clockwise from the -x direction. At the same time, the line from the center of the red ball to the center of the black ball will be rotated n degrees around the z-axis, so it will also be in the x-y plane, but offset by n degrees from the x-axis.

Now, to make the visualization easier, rotate the whole system by n degrees counterclockwise around the z-axis, so that the line from the center of the red ball to the center of the black ball is once again parallel to the x-axis. This will mean that the stick is now pointing only n degrees clockwise from the -x direction. At the end we'll rotate the whole system back n degrees clockwise to cancel this out.

Call the point where the stick is coming out "A", and call the current point of contact between the red ball and the black ball "B". Now look at the great circle which contains both A and B, which currently will look like the "equator" of the red ball. What happens to this circle as the red ball rolls back up to the pole of the black ball? The answer is that it will rotate 180 degrees around an axis which passes through the center of the circle and which is parallel to the y-axis. So where the stick was formerly n degrees clockwise from the -x direction, it will now be n degrees counterclockwise from the +x direction, which is equivalent to an angle of 180-n degrees clockwise from the -x direction.

Now, since we earlier rotated the whole system n degrees counterclockwise, we now have to rotate it n degrees clockwise to cancel that out. This will leave the stick pointing 180 degrees clockwise from the -x direction, which means its pointing straight out in the +x direction just like when we started. Thus, the stick will not have changed positions. And it's pretty obvious that the original point of contact between the red and black balls, at the "south pole" of the red ball, will be the point of contact at the end too. So if neither of these points have changed position, there cannot have been any net rotation of the red ball around any axis.

tronvillain

Well, I read the posts and for a moment I was still convinced that the answer was "Yes." Then I did a little more work on my mental model and realized that the answer was "No." If you draw longitudinal lines on both balls, there will be a corresponding line on each. Obviously, if you roll down a line to the equator and then roll along the same line nothing is going to happen, but if you roll the rolling ball along the equator you are just matching up corresponding lines, which means it will also have no effect. Thanks Friar, that was an interesting little excercise.

Silent Acorns

Arggh! I made a mistake in the "equatorial" part of my trips. The sphere will walways return to its original position if it moves in 90 degree increments (as Jayjay shows above - you posted before I could retract - double argh!).

However, for non-90 degree rotations it will NOT return to the same poistion. This is a lot more difficult to show but I'll try:

START
looking from the TOP:


..3..
1-5-6
..4..

Step 1: bring down to front

TOP view:

-----
|---|
-----
..4..
1-2-6
..3..

Step 2: move 45 degrees along equator
TOP view:

-----
|---|
-----
.......6..
.....4-2-3
.......1..

Step 3: return to top
TOP view

. /\
1/ .\3
/ 5 .\
\. . /
4\ ./6
. \/

The red cube is now viewed "edge on" between the 4 and the 6 when viwed from the front. Definitely not the original orientation.

Jesse

Neat, that's a very elegant way to prove it--thanks tronvillain.

Jayjay

Hmm, I find tronvillain's mental model much more comprehensible than Jesse's or Principia's. It also seems obvious now that it works for arbitrary "rolls" around the fixed globe, does it not?

Principia

OK, I see it Jesse. I did forget that the orientation of the moving ball also changes with its position on the fixed ball, in addition to the no-slip rotation.

Silent Acorns

OK, I screwed up again. I was using an eraser instead of a cube and it screwed up the last step. When I tried it again with cubes the final position was the same as the original.

I now believe that the answer is "no"

Silent Acorns

I just read Tonvillan's proof. Much better than what I was doing (far too easy to screw up).

:notworthy