The slinky, released from rest hanging under its own weight, falls in a peculiar manner. The bottom stays at rest until a wave hits it from above. Two cases -- one unphysical one where the slinky is able to pass through itself, and the other where the coils of the slinky collide creating a shock wave travelling down the slinky -- are analysed. In the former case, the bottom begins to move much later than in the latter.
Deep Dive into The falling slinky.
The slinky, released from rest hanging under its own weight, falls in a peculiar manner. The bottom stays at rest until a wave hits it from above. Two cases – one unphysical one where the slinky is able to pass through itself, and the other where the coils of the slinky collide creating a shock wave travelling down the slinky – are analysed. In the former case, the bottom begins to move much later than in the latter.
Hang a slinky up so that it is supported only on top, and let it go. The bottom of the slinky stays at rest (does not move) for a lengthy time. Let us analyse the behaviour in two cases, one where the slinky coils can interpenetrate each other (a physically unrealistic situation), and one where the coils inelastically collide. For the former case see the earlier work by Calkin [1] who examines the case of a spring whose equilibrium lenght is non-zero so there is no inversion of the spring.
Set up a labeling of the slinky with coordinate y uniformly along the slinky. The density in this coordinate is given by ρ which is a constant. Let x be a vertical real space coordinate. Then the location of the point y along the slinky is given by x(t, y). The stretching of the slinky will be given by ∂x ∂y and the force due to this stretching is k ∂x ∂y . We can write the Lagrangian by
where g is gravitational acceleration. This gives the equation of motion
Intially the slinky is supported at its top end and is stationary. The solution is
where y = 0 is taken to be the bottom of the slinky and y = L the top. k ρ is v 2 , the wave velocity of sound (compression) waves on the slinky.
After it is released, the boundary conditions at the two ends must be that ∂x ∂y = 0 at the two ends so that there are no forces due to the stretched spring at the ends. The solution to the equations of motion are of the form, such that the velocity of any point on the spring is 0 at t=0 is
Ie, we have the gravitational fall of the slinky plus waves travelling to the left and to right. In order to have the correct boundary condition dx dt = 0 everywhere at t = 0 we require
or, choosing the integration constant appropriately
In order that at all times at y=0 we have ∂x ∂y = 0 we require
Combing with f (z) = h(z) for 0 < z < L we have f (z) = f (-z) = h(z) for -L < z < L. Finally, demanding that
or
Thus f and h are periodic with period 2L. Since at t=0, we have
we have
for -L < y < L and, being periodic with period 2L, this determins the value at other values of y. Thus, let us consider the complete solution . We can determine it for0 < t < L/v by dividing the interval 0 < y < L into two parts, 0 < y < L -vt and L -vt < y < L. For the former, we have
Ie, for y < L -vt, the bottom of the slinky, the slinky remains static. For the top L > y > L -vt, the solution is given by
Ie, for any point y above the junction, that part of the slinky moves with constant velocity gL v after the junction moves by. The junction point, y = L -vt occurs at
If one were to continue the solution for times longer than L v , one would find the same behaviour, namely the slinky is divided into two, with each section travelling at constant velocity, but with a moving boundary (the boundary travelling at v in the slinky internal coodinates). Ie, while the bottom is stationary, the top moves with velocity gL v , then the bottom moves with velocty 2 gL v ,while the top continues with its former velocity, then the top moves at 3 gL v , etc. iI.e., the slinky falls in steps.
Figure 1 shows the X as a function y and t at set intervals of t, showing this motion for the first complete cycle. At the end of the cycle, the whole slinky would be falling with a uniform velocity, and the two ends, y = 0 and y = L have changed ends, as if the slinky were now supported at the y = 0 and moving with velocity gL v downards. FIG. 1: The location of the “yth” slinky coil at time t for the “transparent” slinky at specific times Of course this is all nonesense, because this solution, no matter how interesting, assumes that one part of the slinky can interpenetrate another part of the slinky which is not true of any slinky I know. In figure 1, we see that after t = 0 there are two solutions for y for some values of x(t, y). I.e., two values of “coil” coordinate, y have the same location x. Or, another way of phrasing it, at the point y = L -vt, the slope ∂x ∂y changes from positive to negative. For y less than L -vt the slope is gy/v 2 while for larger values of y, it is g v 2 (y -L) < 0. But a slope change from positive to negative means that the slinky has passed through itself. Before that happens the coils will crash together, and the above solution becomes invalid. I.e., the speed of propagation of the junction between the parts of the slinky is NOT equal to speed of sound along the slinky.
Let us assume that, when the coils come together the collision is a perfectly inelastic collision. Let us assume that above some point Y (t) the coils are all together, while below that point, the slinky, as above, remains motionless. Then we have a mass M = ρ(L -Y ) above that transition point, while below it, we assume as above that the slinky remains motionless. Note that if dY dt < v this will clearly not be a valid solution. The velocity in x-space of that point Y (t) is
Thus the Momentum equation for that mass above the transition point is
or
The solution is
If c = 0, then for
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