Consider a round jar with no lid. The radius of the opening in the jar is 10. A coin (with a radius R < 10) can be placed flat on the edge of the jar without falling. The coin can be placed such that its center is on the rim of the jar, or it can be moved toward the center of the jar.
What is R, when the coin is as close to the center of the jar as possible without falling in, and the inside edge of the coin is at the center of the jar's opening?
I am assuming that the coin will tip-in if the center of gravity is on the "inside" of the two intersection points of the coin's edge and the jar's edge. Thus the closest balance occurs when the coin's center is on the chord between the two intersection points.
A quick calculation on the resulting diagram gives a coin radius of 52. Yes?
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Posted by owl
on 2004-10-24 15:58:03 |
Maybe Solution
