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A chain of Length L is fixed at one end to a point and the rest of the chain is hanging such that the other end of the chain just touches the ground. Find the potential energy of the chain given its mass is uniformly distributed and equal to M. (assume g is constant).

Now this question is pretty easily solvable by find the centre of mass of the chain (which is at h = L/2) and using formula for U = mgh. (PE = MgL/2).

But I tried using calculus (for practice) to find PE of chain, however I was not able to do so, please help me.

This is what I tried:

let dl be be a small length of the chain. mass of dl part of chain = M*dl/L

assume it is at height H above the ground. so its infinitesimal PE: dU = M*dl/L * g * h

now to integrate, I am not able to determine the limits for integration, also I am not able to eliminate "h" from this equation.
I think I have made some mistake, please tell me how can I proceed from here

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  • $\begingroup$ Is $h$ the same thing as $l$? What's the PE of a chain of zero height (length)? $\endgroup$
    – PM 2Ring
    Commented Oct 14, 2021 at 11:12
  • $\begingroup$ @PM2Ring wow thanks youtu.be/xkMZfIsoQ5I $\endgroup$
    – Zombie Killer
    Commented Oct 14, 2021 at 17:12

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now to integrate, I am not able to determine the limits for integration, also I am not able to eliminate "h" from this equation.

There is no need to eliminate $h$ from the equation. Actually if you take the infinitesimal length as $dl$ it won't work as there is no relation between $dl$ and $h$. Since an infinitesimal quantity is very small thus $dl = dh$ and if you take $dh$ as the infinitesimal length then you don't need to eliminate h from the equation.
Now, the potential energy of the system will be sum of potential energy $dU$. And by integration: $$P.E. = \int_0^{L} \frac{Mdh}{L}gh $$ $$P.E. = \frac{Mg}{L}\int_0^{L} hdh $$ $$P.E. = \frac{Mg}{L}\left[\frac{h^2}{2}\right]_0^L$$ $$P.E. = \frac{Mg}{L}\left(\frac{L^2}{2}\right)$$ $$P.E. = \frac{MgL}{2}$$ And the question which arises after solving your question is why should we take the infinitesimal quantity of the variable (as we took $dh$ instead of $dl$):
Well I also don't know the answer but as I said they are very small quantities but this is not a very solid reason but atleast it's satisfactory.
If you need any more help please let me know.
Hope it helps!! :)

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  • $\begingroup$ thanks a lot, now everything is clear. I think I also found out why dl = dh, assume a point on chain located at height h from ground. there is a very small segment of chain with size dl .The infinitesimally small segment's one end is at height h and its length is dl , we can equivalently say that the height of other end is h+dl. So instead of saying length of segment is between l and l + dl we can say the segment is located between h and h + dh . l = h as full chain is vertical with lowest point at h=0. $\endgroup$
    – Zombie Killer
    Commented Oct 14, 2021 at 16:17

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