A man places a circular tube upright on a table. He then places a solid ball into the tube followed by a smaller solid ball. The tube stays upright. He removes the balls and places them again with the smaller ball being placed in first. The tube tips over. In both cases the man holds onto the tube until the balls come to rest and then lets go. The radii of the balls are 2.6 and 3.4 centimeters. The length of the tube is 18.0 centimeters and its thickness (external radius minus internal radius) is 0.1 centimeter. The balls and tube are made of the same material - so their weights are proportional to their volumes. Assume the points of contact between the balls, table, and tube are frictionless. What are the minimum and maximum values for the internal radius of the tube?

(In reply to re: What I think I know by Leming)

I think we were both wrong about the displacement of the centers of the spheres.

The horizontal displacement of the spheres is 2r-6, this is the diameter of the cylinder less the radii of the spheres (I used r-6 before)

The direct length between the centers is 6.  Using pythagoean thm for the vertical displacement gives sqrt(24r-4r^2)

To address Hugo's question:  each sphere pushes againt the opposite side of the cylinder with an equal force proportional to the mass of the larger sphere divided by the tangent of the angle the segment connecting the centers forms with the horizontal.  

Posted by Jer on 2005-06-24 12:58:39