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 Three Balls in a Bowl (Posted on 2005-09-07)
Here is a problem I have been developing. Maybe somebody can tell me if it can be solved or if more information is needed.

Three solid balls of radii a, b, and c are placed in a bowl whose inner surface is a hemisphere of radius d. The following information is known:

1) a < b < c < d,

2) d is large enough so that each ball touches a point on the inner surface of the bowl,

3) a is large enough so that each ball 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 bowl are directed along the lines determined by their centers.

After the balls come to rest, what is the angle between the plane determined by the centers of the balls and the horizontal in terms of a, b, c, and d ?

 No Solution Yet Submitted by Bractals Rating: 3.6667 (9 votes)

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 Solution too long but possible | Comment 9 of 18 |

We denote the three balls by centre and radius (A,a), (B,b) and (C,c) with c ¡Ý b ¡Ý a in the hemisphere (D,d) with d ¡Ý c ¡Ý b ¡Ý a.<o:p></o:p>

<o:p> </o:p>

Since A, B and C touch D, the plane ABC is circumscribed by some circular section of D.  Knowing this allows us to determine the origin and radius of the circle in terms of a,b,c.<o:p></o:p>

<o:p> </o:p>

The (CG) centre of gravity of the ABC system is given (relative to C) in terms of the vectors CA and CB by  CG = ((CA)a^3+(CB)b^3)/(a^3+b^3+c^3).  This is clearly on the plane ABC but not at its centre.<o:p></o:p>

<o:p> </o:p>

So, if we put our circular section with this CG into D (of negligible weight) D will roll until the CG is at its lowest possible point, i.e CG is directly below D. From this we can calculate the angle of intersection with the original horizontal.  However  as a single  function, the expression for this in terms of a,b,c and d is extremely long  (pages).<o:p></o:p>

 Posted by goFish on 2005-09-20 22:39:06

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