Addendum:
If the spring has mass M, then I get an acceleration of the upper mass (m)
of:
2g + gM/m .
I think this is a worth model of the slinky with the two masses (m)
representing the top and bottom turns.
Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net http://www.velocity.net/~trebor
----- Original Message -----
From: "Bob Sciamanda" <trebor@VELOCITY.NET>
To: <PHYS-L@lists.nau.edu>
Sent: Friday, November 22, 2002 2:10 PM
Subject: Re: dropped slinky
| A simplified first model:
| Consider two equal masses joined by a spring of negligible weight. Let
| the system hang from your hand. The upper mass is in equilibrium under
| the influences of your upward force (F), the downward pull of the upper
| weight (mg), and the downward pull of the stretched spring (kx). The
| lower mass is in equilibrium under the downward pull of its weight (mg)
| and the upward spring force (kx) - so that kx = mg. If your supporting
| force suddenly vanishes, the acceleration of the upward mass is downward
| and equals (kx + mg)/m = 2mg/m =2g.
This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.