Chain falling out of a horizontal tube onto a table

[NTesla]

Homework Statement

The question is from Irodov Q 1.184. See the Screenshot.

Relevant Equations

I'm trying to solve it using energy conservation. My thinking is that since all the surfaces are frictionless and only conservative gravitational force is acting, therefore, ME must be conserved.

My attempt:
Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE.
PE of hanging part = $\frac{1}{2} \frac{m}{l}gh^{2}$.
PE of part in the tube = $\frac{m}{l}(l - h)gh$.

Final ME = $\frac{1}{2}\frac{m}{l}gh^{2}$ + $\frac{1}{2}\frac{m}{l}hv^{2}$.

Since Initial ME = Final ME. Therefore, $\frac{1}{2}\frac{m}{l}hv^{2}$ = $\frac{m}{l}(l-h)gh$. Solving this gives: $v = \sqrt{2g(l-h)}$. But the answer in the book is: $v = \sqrt{2ghln(\frac{l}{h})}$.

 

[Steve4Physics]

Homework Statement: The question is from Irodov Q 1.184. See the Screenshot.

@NTesla, the last part of the question on the screenshot is missing. So here’s what (I believe) it should say:

"... At a certain moment, end-A of the chain is released. What is the speed of end-A when it reaches the right-hand end of the tube?"

Since Initial ME = Final ME

Are you sure about that? Suppose the whole chain, not just part of the chain, had reached the floor and come to rest. What would the final ME be? If it is different to the initial ME, how do you account for this difference?

 

[NTesla]

the last part of the question on the screenshot is missing

Yes, thanks. Edited. The missing screenshot has been added in the original post.

 

[NTesla]

Are you sure about that? Suppose the whole chain, not just part of the chain, had reached the floor and come to rest. What would the final ME be? If it is different to the initial ME, how do you account for this difference?

Yes, that's a valid point.

 

[jbriggs444]

There are a number of conundrums that deal with the unexpected behavior of falling ropes and chains. It seems clear that you are not intended to worry about the fine details involved with those.

The intent is that you pretend that the portion of the chain that reaches the table is quietly removed from existence as it gets there.

Spoiler

If a falling link strikes the table at an angle (perhaps nudged to an angle by the portion of the chain that had already fallen) then the strike will impart spin. That spin will pull down on the remainder of the falling chain. Almost paradoxically, the upward force from the table below has the net effect of a downward force on the falling chain.

https://en.wikipedia.org/wiki/Chain_fountain

 

[haruspex]

When writing "ln(x)” in LaTeX, you can make it clearer with "\ln": $\ln(x)$ v. $ln(x)$. This avoids confusion when $l$ or $n$ is also a variable.