Re: dropped slinky

[Bob LaMontagne <rlamont@POSTOFFICE.PROVIDENCE.EDU>]

Bob Sciamanda wrote:

> Excellent, John!
>
> Applying my simplified model:
> the system consists of the two end masses connected by a spring of seven
> masses. thus my predicted acceleration is 2g + g(7/1) = 9g !
>
> A much simpler model applies:
> The upper mass (turn) is initially in equilibrium under your force F and
> the total spring weight M.  When your F vanishes, the mass m accelerates
> under influence of the total string weight Mg, so that  a = Mg/m = 9g in
> your setup.

This seems to also support the original poster (Carl Mungan). If the slinky is
subdivided into 100 segments, each of mass m=M/100, then the acceleration will be
100g, just as he claimed!

Bob at PC

This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.