Newtonian Analysis of a Folded Chain Drop

Abstract

Consider a chain of length L that hangs in a U shape with end A fixed to a rigid support and free end E released from rest starting from the same initial height (call it y = 0) as A. Figure 1 sketches the chain after end E has fallen a distance y. Points O and A are assumed to be close enough to each other and the chain flexible enough that the radius of curvature r at the bottom point C can be taken to be negligibly small (compared to the length of the chain). The problem is to compare the speed of descent v(y) = dy/dt of the free end E of the chain to the speed vfree(y)=2gy of a free-falling point mass that has descended the same distance y. If v(y) > vfree (y) for all y > 0, then, in a race to fall any arbitrary distance Y (where 0 < Y < L), the chain end E will always beat a simultaneously released point mass, because the fall time t for E will be shorter than tfree for the point mass, t=∫0Ydyv(y)<tfree=∫0Ydyvfree(y).(1)Experimentally, this outcome of the race is observed in a “Veritasium” video.