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Nickel Landed on Edge

Date: 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/ 
Associated Topics:
High School Probability

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