Hi, you all!
A little more about the continuous version of the worm on the
rubber band (continuation of my message of yesterday). For completeness
I repeat the variables.
The rubber band is on the x-axis. To simplity the equations, I take
x=0 at the TRACTOR END of the rubber band (so, x=0 is moving with the
tractor, a galilean coordinate). Initial position of the worm is
x(0)>0. The tength of the rubber band is L=L_0(1+kt). The worm is
crawling with speed -v (on the stretched rubber band).
Then, as I showed in the message of yesterday,
x(t) = (1+kt)(x(0) - (v/k) ln(1+kt)).
Assuming x(0), v and k all positive, the main question is to find the
time t* when the worm reaches the tractor, i.e. x(t*)=0. This will
obviiously happen when x(0) = (v/k) ln(1+kt*) and we obtain
t* = (exp(kx(0)/v) - 1)/k = finite!
OBSERVATION: This result is somewhat counterintuitive (at least to
me). The reason is probably that we think of the stretching as an
exponential growth, but here the stretching is NOT exponential but
LINEAR in time. Hence the percentage stretching is decreasing
(dL/L = (k/(1+kt)) dt. On the other hand, the speed of the worm
is constant, hence faster than the stretching, after a certain time.
Good luck wisualizing it! Emilio