I am dreaming of three balls on a hoop.

Three solid balls of radii a, b, and c are placed on a hoop of radius d. The hoop is a thin, rigid wire. The balls have been drilled so they fit perfectly on the wire hoop and slide without friction. The hoop passes through the center of gravity of each ball. The hoop is oriented vertically so the balls slide to the bottom.
The following information is known:

1) a, b, and c are all < d,


2) d is large enough so that each ball is entirely below the center of the hoop,

3) the central ball is large enough so that it touches the other two balls,

4) the balls are made of the same material so that their weights are proportional to their volumes,

5) the forces that the balls exert on each other and the hoop are directed along the lines determined by their centers.

After the balls come to rest, what is the angle between the vertical and a line from the center of the hoop to the center of each ball?

Inspired by Three Balls in a Bowl.

Create an origin at the center of the hoop, Y= vertically down, 
X= to the right. Assume that the balls are in the order A, B, C, 
on the hoop (from left to right).  Further name the angles from the 
center of the each ball to the Y-axis (vertical) as Ta, Tb, Tc, 
respectively.  The Y coordinate of the c.g. of the 3-ball system then is:
 
Ycg = d*(a^3 cos(Ta) + b^3 cos(Tb) + c^3 cos(Tc)) / (a^3 + b^3 + c^3).
The density constant, k, can be ignored as would appear as a linear 
term in both the numerator and denominator.
 
As Bractals points out, the idea is to maximize Ycg.  This is equivalent 
to saying that gravity will pull the 3-ball system as low as possible on 
the hoop.  The 3 unknowns are Ta, Tb, and Tc.  However they are 
related.  Note that the radius of each ball forms a chord on the hoop. 
Draw an isosceles triangle using the radius of the ball as the base, 
each of the other sides is length d.  The angle opposite the base can
 then be shown to be arcos (1-(r^2)/2/d^2) where r is the 
radius of the ball in question.  Therefore for any position of the three 
balls, as placed in order, above:
 
Ta = Tb – arcos(1-(a^2)/2/(d^2)) – arcos(1-(b^2/2/(d^2))    &
Tc = Tb + arcos(1-(b^2)/2/(d^2)) + arcos(1-(c^2)/2/(d^2))
(angles are measured  counter clockwise from the 6 o'clock position 
(on the Y-axis) on the hoop)
 
Substituting these into the equation for Ycg, the only variable is Tb. 
 I'll leave the calculus to someone else, but I did create an Excel 
spreadsheet with the  above formulae, and using Solver to maximize
 Ycg it seems to work for some easily understood cases:
 
a=b=c=X   à  Ta=-Tc, Tb=0, Ycg is smaller for increasing X
a=c=X, b=Y --> Ta=-Tc, Tb=0
a=0, b=c à Ta=Tb=-Tc
etc. 

Edited on November 27, 2006, 9:02 pm

Edited on November 27, 2006, 9:13 pm

Edited on November 30, 2006, 7:02 pm

Posted by Kenny M on 2006-11-27 20:57:52