Nickel Landed on EdgeDate: 10/11/2002 at 00:36:13 From: Daniel J Hyman Subject: Nickle landed on edge Dear Dr. Math, Here's the story: I was sweeping at work when I swept a nickle out from under the shelves on the ground; as I hit the nickle, it took flight. As I watched it hit the ground I thought to myself that it would land heads. But when it came to a complete halt, it was standing on its edge. I was amazed because I have never seen this happen in real life. My question is: what are the odds of such a thing occurring? Date: 10/11/2002 at 10:18:47 From: Doctor Mitteldorf Subject: Re: Nickle landed on edge When I was your age, there was a weekly TV show called Twilight Zone, and I remember one episode in which a man was selling newspapers on the street, with a cigar box full of coins. One morning, a man took a paper and threw a dime into the box. It landed on its edge, and the paperseller started noticing that he could read people's thoughts. All day, he heard voices telling him exactly what others were thinking. Of course, it drove him crazy; he wasn't used to it, he misinterpreted the thoughts as if they were words that were actually expressed, he got angry and afraid... Then the same man came back after work to buy his evening paper and threw in another dime, and the dime that had been standing on edge all day was knocked over by the same customer, and the paper-seller lost his ability to read minds. Thank goodness. Well, there's no one mathematical answer to your question, but there are ways to think about it that may give us a reasonable estimate. If you stand a coin on edge, how far would you have to tip it for it to fall over? Suppose the coin has a thickness t and a radius r. The center of gravity is a distance r above the ground, and if it tilts through an angle greater than t/(2r), the center of gravity will be out past the edge, so it will fall over. If the coin is spinning as it falls, and it stops spinning when friction finally slows it to a stop, then it has a probability of t/(pi*r) of being in the right range to remain standing. The pi comes from pi radians in half a circle, and the 2 in 2r went away because the coin could be tipped up to t/(2r) either to the right or to the left. I think in reality this number is much too generous. The point is that while a coin is close to being "flat" on the ground, it can bounce and twist and roll in ways that dissipate its spin via friction; whereas the friction while it's just tipping from its edge is much lower. In order for a coin to land on its edge, I'd think it would have to have very little angular momentum, that is to say, spinning imperceptibly. A flipped coin would never do this. If you throw the coin instead of flipping it, it has a chance that's closer to the calculated formula above. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ |
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