[usamamuneeb]

The image in the attachment shows a ball X roling over 8 central balls without slipping, from position a to position b (a and b are not to be considered as balls). How many complete revolutions will the ball X make, until it reaches position b?

[Agent-X]

I think four. I could be wrong though.
Maybe two.
Quarter spin per ball new ball.
looks that way...

Anyway, not the forum for this.

You could think of these things in terms of sin and cosecant.

[kveldulf980]

Actually it makes a third of a revolution every time and there are 9 so you would be one turn from completing three revolutions.

[cayalee]

lol, i think the real question here is "who cares?" :¬D

[usamamuneeb]

Actually, when I solved this, I thought I had made a new law of Geometry - Usama's theorem like Pythagora's Theorem!!!!!

Of course, that sounds ridiculous, but I am quoting the answer in the next reply, so anyone please tell me if I have made a great breakthrough!!

BTW, all of the answers above are wrong!

[usamamuneeb]

Consider the following example which I claim to be MY law (laugh your head off):

See Example.png. It shows a circle rolling over another from position A to B. As we can see, it comes into contact with 60*/360* = 1/6 of the circumference of circle.



Therefore, my law = when a circle rolls over others in a line of equal circumference, radius and specs, the part of the circles on the bottom which come in contact with the moving circle = 1/6 of their circumference.



Suppose the circumference of one circle = 360 cm.
If we calculate the distance moved by the circle, it is (1/6) x 10 x 360 cm

= 600 cm. (I multiplied 1/6 by 10 as 2/6 of the circumference of first and last circle are in contact with the rolling one!)

Divide the distance by its circumference: 600 / 360
= wait calculator is opening . . . = 1 whole 2/3 revolutions!

[usamamuneeb]

lol, i think the real question here is "who cares?" :¬D

but maybe he is right in some sense!

[kveldulf980]

Well, your "law" is right in that the circles create those 60* angles, that's clear enough, but simply performing the experiment in your head wll show you that the math after it is wrong. The circle both "spans" 60* and then rolls 60* each move, so each move, from "valley" to "valley", covers 120*, or 1/3 of the circle. Since there are 9 valleys, or 8 rolls, until it touches circle "b" (or 10 valleys and 9 rolls until it rolls over it), there will be 8 * 120 = 960, so 960 / 360 = 2 and 2/3 (or 9 * 120 = 1080, so 1080 / 360 = 3 if you want to get over "b").

An easier way to do it is to think of the circle on top as a triangle with corners given the values 1, 2, and 3 as such:

1
2 3

(they should be an equilateral triangle)

Then just simply realize that the 8 moves are 2 and 2/3 rotations of the triangle, and there you have it. Now what you have done is given a visual representation of the rotational subgroup of the dihedral group D6, or, for those of you who know your abstract algebra, the group A3 of even permutations, which is a much more interesting result. So now for a more fun problem, what if you reflected the circle after every roll, which permutation of the triangle will we end up with?

[kveldulf980]

Incidentally, I meant a reflection about a verticle bisector, for anyone who would care.

[minthaka]

It would turn 480 degrees, because it needs 60 degrees to turn from 1 gap to another. The reason: when a ball is over and between 2 others, their centers make an equilateral triangle, so the angular distance between the two tangent point is round 60 degrees. Since we have 8 gaps to go, that means 8X60 = 480 degrees or 1 and 1/3 spins.
The task-giver missed the fact that there's only 8 gaps to go instead of 10, but the basic principle is O.K.
Q.E.D.