Sorry, but your analysis of the top of the slinky is wrong.
The bottom accelerates at 0 m/s^2 as you say
The center must accelerate at g. (Consider the net force on the net mass)
The top accelerates at 2g.
Timothy Folkerts
Department of Physics
Fort Hays State University
Hays, KS 67601
785-628-4501
"The only thing necessary for the triumph of evil is for good men to do
nothing." - Edmund Burke
"Carl E. Mungan"
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Gene Mosca mentioned the following interesting tidbit to me.
Hold a slinky from the top so that it stretches down vertically
(while standing on top of a table so the bottom end is free). It's
pretty well known that if you now release it, the initial
acceleration of the bottom end is zero. (The slinky collapses from
the top down before the bottom end begins to fall substantially. Try
this yourself if you haven't seen it before.)
The interesting new question is: What is the initial acceleration of
the top turn? Suppose the slinky has 100 turns.
The answer, which is as surprising as the zero for the bottom end, is
about 100g! (The top turn has 1/100 of the mass but has to support
the weight of the entire mass initially. Or consider a mass-weighted
average acceleration where only the top turn initially is out of
equilibrium.)
One might consider this a "slowed down" model for compressional waves
traveling through a solid rod when something sudden happens at one
end (hits the ground, dropped from a suspension point, etc).
--
Carl E. Mungan, Asst. Prof. of Physics 410-293-6680 (O) -3729 (F)
U.S. Naval Academy, Stop 9C, Annapolis, MD 21402-5026
mungan@usna.edu http://physics.usna.edu/physics/faculty/mungan/
This posting is the position of the writer, not that of SUNY-BSC, NAU or
the AAPT.
This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.