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Re: Loaded dice



Surely we're simply talking here about shifting the centre of mass of the
die towards one face so that the stable position with that face down would
be favoured in interactions with the table?

Mark

At 16:16 02/12/02 -0500, you wrote:
You read my consternation correctly, John. Since all masses experience
the same acceleration in a gravitational field, there must be some other
reason for the heavier side to out-run the others. With most objects I
would look for differences in air resistance, because of shape. Not
applicable here. I am left with a vague appeal to a tumbling phenomenon
and am trying to remember applicable dteails of past studies of "rigid
body motion". The "tennis racket theorem" (only the words) just came to
mind! Please elucidate!
-Bob

Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor
----- Original Message -----
From: "John Mallinckrodt" <ajm@CSUPOMONA.EDU>
To: <PHYS-L@lists.nau.edu>
Sent: Monday, December 02, 2002 3:56 PM
Subject: Re: Loaded dice


| >Is it true that a dice is "loaded" by adding an internal weight near a
| >given face. Is this supposed to favor the dice falling with this face
| >"down"? Someone please explain(???)
|
| This would seem to me to be the case and it probably doesn't hurt if
| the dice are of the "rounded edge" variety. See, for instance,
|
| http://www.halfpast.demon.co.uk/dd.html
|
| But why do I think you must have some sort of a theory to suggest
| that this shouldn't work? ;-)
|
| --
| John Mallinckrodt mailto:ajm@csupomona.edu
| Cal Poly Pomona http://www.csupomona.edu/~ajm

Mark Sylvester
UWCAd
Duino Trieste Italy