# Falling rope

LCSphysicist
Homework Statement:
All below
Relevant Equations:
All below
A chain with length L and mass density σ kg/m is held in the position
shown in Fig. 5.28, with one end attached to a support. Assume that
only a negligible length of the chain starts out below the support. The
chain is released. Find the force that the support applies to the chain, as
a function of time.

I am trying hard to see how the things work here. Try by conservation of energy is, to me, wrong.
I think the support's force need to stop the falling part and yet bear the weight of the fallen part.
To show my vision:
this elementar mass at y fall actually twice y to come to rest.
It falls y in gt²2

So N = W + F = σyg + σv². But this seems wrong.

An idea; in a time ##t##, the chain falls down a distance ##\frac{1}{2}gt^2##. The two loops at the bottom each have a height of ##\frac{1}{4}gt^2##.

If you can work out the centre of mass ##\bar{y}## of the whole chain in terms of ##t##, then you can set ##m\frac{d^2 \bar{y}}{dt^2}## equal to the net external force on the whole chain .

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LCSphysicist and haruspex