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Three Balls in a Bowl (Posted on 2005-09-07) Difficulty: 5 of 5
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)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: How about Two Balls in a Bowl | Comment 14 of 18 |
(In reply to How about Two Balls in a Bowl by Leming)

Two balls in a bowl:

We represent the cross section of the "bowl" by a circle, with centre D, radius d.

We locate balls A and B in this bowl such that they touch and denote the perpendicular to the chord in D through A and B by OD.

(1) OD = h

(2) AO + OB = a + b




                         /   |   

                      /      |     

                   /         |         


             A              O           G               B

         /                   |

      /                      |

  /                          |

A'                           |


We next extend DA and DB to the circumference of D so that AA' = a, BB' =b and DAA' = DBB' = d. The physical system therefore satisfies:

(3) (d-a)^2 = OA^2 + h^2

(4) (d-b)^2 = OB^2 + h^2

Eliminating OA and OB we can obtain the following :

h = 2 Sqrt (ab d^2 - bd a^2 - ad b^2)/(a+b)^2)

for 0 < a < b < a + b < d < (ab + b^2)/(b-a).

We denote the centre of gravity G, then

(5) OG = ((OB) b^3 - (OA) a^3))/(a^3+b^3)

In equilibrium the system rotates about D so that G lies under D. By geometry we can show that the angle the chord makes with the horizontal is equal to ODG and if we let t = Tan (ODG) = OG / h then we obtain

t = -1/2 Sqrt( -a^8 + 2 a^4 b^4 - b^8 + 2 a^7 d - 4 a^6 b d +

4 a^5 b^2 d - 2 a^4 b^3 d - 2 a^3 b^4 d +

4 a^2 b^5 d - 4 a b^6 d + 2 b^7 d - a^6 d^2 +

4 a^5 b d^2 - 8 a^4 b^2 d^2 + 10 a^3 b^3 d^2 -

8 a^2 b^4 d^2 + 4 a b^5 d^2 -

b^6 d^2)/((a b((a^2 - a b + b^2))^2((a + b -

d)) d))

Pretty isn't it?

Three balls is no nicer.


Edits to fix typos.


Edited on November 17, 2005, 4:22 am

Edited on November 17, 2005, 12:07 pm
  Posted by goFish on 2005-11-17 04:20:11

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