Friar Bellows

This is a little mind bender I read on a usenet group:

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Suppose we have two balls of the same radius. I'll call them the "rolling ball" and the "fixed ball".

The fixed ball is not allowed to move.

The rolling ball must touch the fixed ball - and it's allowed to roll without slipping or twisting on the surface of the fixed ball.

Now:

1) Start with the rolling ball touching the north pole of the fixed ball.

2) Roll it down to the equator along a line of longitude.

3) Then roll it along the equator for an arbitrary distance.

4) Then roll it back up to the north pole along a line of longitude.

The question is: when we carry out this process, can the rolling ball come back *rotated* relative to its original orientation ... or not?

tribalbeeyatch

Unless I'm missing something, the answer is "no". As I didn't find it particularly difficult to arrive at that answer, and this is being called a mind bender, I would guess that I am in fact missing something

Unbeliever

Is this why it's called "fixed"?

Without any markings on the "rolling ball" there would be absolutely no way to tell if it were "rotated".

hyzer

My answer - no. Because the rolling ball has made 3 changes of the two axies of rotation. An even number of turns would be required to bring it back to its original disposistion.

nermal

The question asked was: "can it return to any position other than it's original position"?

The answer is no. It will always return to its original orientation.

Ed

Psycho Economist

Let's define the orientation of the rolling ball at the start to be "down" (i.e. their north poles are touching). When it's been rolled so the point of contact is in touch with 45deg latitute of the fixed ball, it's pointing "out", the north poles are perpendicular to one another. When it gets down to 0deg latitude it's pointing "up", the north poles are pointing in the same direction.

No matter how long we roll it along the equator, the rolling ball is still going to be north-pole-up w.r.t. the fixed ball so when we roll it 90o back up, it will always be "down" again.

HOWEVER, if the rolling ball hasn't gone around the equator of the fixed ball an whole number of times, when it comes back up it will be in a new yaw orientation, as if it had been spun along an axis going through the north poles of both.

yguy

Without having looked at anyone else's answers, I say the movable ball is rotated with respect to the fixed ball's N/S axis by the number of degrees it traveled along the equator, and that it ends up touching the fixed ball at the same point it did before it was moved - so that its orientation with respect to the N/S axis is unchanged.

fando

Let's see if this works... the red ball has a stick poking out of it. The black ball is fixed.

It rotates around the N/S axis but that's it.

Jesse

That can't be right. The way you drew it, the same point on the red ball would be touching the black ball at all times--that would be sliding, not rolling. If the stick started out pointing right, I think it would end up pointing left, into the black ball, once it was on the equator (the second step in your series).

fando

oops, you're right... there was sliding. I can't portray the corrected version in ASCII though... but there's this:

It's harder to see the result, but I think the answer is no rotation. Not sure though.