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 Tipping Tube (Posted on 2005-06-23)
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?

 See The Solution Submitted by Bractals Rating: 3.7500 (4 votes)

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 What I think I know | Comment 1 of 11

The volume of the cylinder = 18ð(.2r+.01)
Small sphere 8788ð/375
Large sphere 19652ð/375

Since the densities are all the same multiply all by 375/ð to make the numbers nicer

cylinder mass = 1350r + 67.5
small sphere = 8788
large sphere = 19652

consider the large-sphere-on-top case.

The cylinder pushes down with 1350r+67.5 units of force and the edge of the cylinder pushes back equally.  The displacement is r so the torque in the direction is 1350r^2 + 67.5r.

The spheres each push sideways on the walls of the sphere with 19652(r-6)/sqrt(12r-r^2) units of force [work omitted].  The displacement is sqrt(12r-r^2) so the torque in the opposite direction is 19652(r-6)

The system will be at the edge of stability when these torques are equal.  The resulting quadratic has no real solutions, so something is wrong.

Maybe tomorrow...

 Posted by Jer on 2005-06-23 18:47:55

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