Consider the given system which shows a chain $AB$ of length $l$ and the end $A$ is held at rest.

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Suppose we release the chain at time $t=0$. How do we find the velocity of the chain when the end $A$ is leaving the tube?

My approach was to use the work energy theorem. Let velocity of chain at time $t$ when it is leaving the tube be $v$. Also assume the chain has uniform mass distribution with linear mass density $\lambda$. $$W_{\textrm{gravity}}+W_{\textrm{normal}}=\Delta K=\frac{1}{2}(\lambda l)v^{2}$$ Work done by normal force can be taken as 0 as the point of application of force undergoes no displacement. $$W_{\textrm{gravity}}=\lambda(l-h)g\frac{h}{2}+\lambda(h)g\frac{h}{2}=\frac{\lambda ghl}{2}$$ This gives $v=\sqrt{gh}$ which is a contradictory result.

What is my mistake in the given argument and is there any discrepancy by writing the work-energy theorem (work done by net force as change in kinetic energy) here?

  • $\begingroup$ The collision of the chain with the ground is inelastic. If it were elastic the chain would have to bounce upwards from the ground like a bouncing ball. This means energy is lost as heat so we cannot apply conservation of energy to the problem. $\endgroup$ Sep 22 '21 at 6:07
  • $\begingroup$ There is a detailed discussion of this in the question Missing force in the system. $\endgroup$ Sep 22 '21 at 6:13
  • $\begingroup$ @JohnRennie Is the chain really colliding with the ground? I thought the links of the chain undergo elastic collision. Keep in mind, the end B of the chain is already touching the ground at $t=0$. $\endgroup$
    – Pravimish
    Sep 22 '21 at 8:44

According to me your answer is mostly correct, however, I would have considered that the chain undergoes circular motion in the very last bit of its motion. Hence,

$$ \frac {mv^2}{r} = mg\cos\theta $$ (considering v as the final velocity, and $r$ as the radius)

Hence, $$v=\sqrt {rg}\cos\theta$$

This is however my approach and I welcome any counter-opinions!


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