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I have seen an interesting problem as follows:

Both ends of a rope with the mass of $\rho$ kg/m and length of $L$ are attached to a horizontal surface (see photo attached). At $t=0$, one end of the rope detaches from the surface and starts falling vertically. When this end falls the height $X$, what is the tension on the other end of the rope?

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I also have seen two different solutions from a professor which result in two different answers!

Approach 1 is based on falling part of the rope is freely falling (acceleration $= g$).

Approach 2 is based on energy conservation $KE + PE = $ Constant (no loss).

Now I would like to ask

1- Is the falling part of the rope really freely falling?

2- Is energy conserved when the rope is falling (for $X \lt L$)?

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    $\begingroup$ Note to those voting to close: the post isn't asking how to solve the problem. $\endgroup$ Jul 6 at 21:21
  • $\begingroup$ Chain Drop Answer 2 $\endgroup$
    – mmesser314
    Jul 7 at 2:57
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The rope is not freely falling, and energy is conserved:C. W. Wong, K. Yasui, Falling chains, American Journal of Physics 74 (2006) 490, and the earlier M. G. Calkin, R. H. March, The dynamics of a falling chain I, American Journal of Physics 57 (1989) 154.

A point not fully explained in the cited references is that when the free end of the rope is falling at speed $v$, the tension in curved bit of the rope is $T={\mu v^2}/{4}$ on both sides of the fold. Consequently the falling bit of the rope has a force of $T={\mu v^2}/{4}$ pulling it down in addition to gravity. To see that this is so recall that in the absence of gravity a chain moving with speed $u$ can maintain an arbitrary planar shape because the centripetal acceleration of its links is automatically provided by the tension according to $$ \frac{\mu u^2}{r}= \frac{T}{r}, \quad (\star\star) $$ so the tension takes the value $T= \mu u^2$ independent of the radius of curvature $r$. For our falling rope, and in the reference frame that is descending with the fold at $u=v/2$, the rope/chain is moving through the fold at $u=v/2$. If we can ignore the effect of gravity and the non-inertial reference frame, the tension throughout the fold must be $T= \mu (v/2)^2=\mu v^2/4$. Ignoring these effects is a safe approximation for our sharp fold because when $r$ is small the forces in $(\star\star)$ completely dominate all other forces.

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