Dropping a Slinky (calculation)

Let’s do a quick bit of math related to Dropping a Slinky. Last time, I estimated that it takes about 0.3 seconds for the slinky to collapse. To get a more precise answer, note that however the slinky falls, its center of mass must accelerate downwards at gravitational acceleration.

Where is the slinky’s center of mass? When it’s just hanging, the slinky is in equilibrium, so the derivative of the tension is proportional to the density. Also, if we assume an ideal spring with zero rest length, the tension is inversely proportional to the density (why?). Therefore, we write

\frac{\mathrm{d}T}{\mathrm{d}x} = g \rho


T = \frac{\alpha}{\rho}


This can be solved to show that the density follows


\rho \propto \frac{1}{\sqrt{x}}


Integrating, we find that the center of mass is one third the way up the slinky. The time for the slinky to collapse is the same as the time for the center of mass to fall to the bottom, or


t = \sqrt{\frac{2 (1/3 l)}{g}}


This is the same answer, but modified by a factor of 0.81. Notice that this only depends on the “slinkiness” – the zero rest length ideal spring. We expect thick and thin slinkies of different stiffnesses to act in essentially the same way.


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