Challenge: chain on a table

Problem

Challenge: chain on a table. A chain of length $L = 2\;\text{m}$ and linear mass density $\lambda = 0{,}3\;\text{kg/m}$ rests on a table, with a small part ($\ell_0 = 0{,}2\;\text{m}$) already hanging over the edge. Calculate the velocity at which the chain completely leaves the table, neglecting friction. Use conservation of energy with the barycentre.

Solution

Setup. Total mass $M = \lambda L = 0{,}3\cdot 2 = 0{,}6\;\text{kg}$. The chain is inextensible: all segments have the same velocity $v$. We use conservation of energy, calculating the potential energy with the barycentre. Take the edge of the table as height $0$ (the hanging segment sits at negative heights).

With a hanging segment of length $x$, the potential energy is $E_\text{pot}(x) = (\lambda x)\,g\,(-x/2) = -\tfrac{1}{2}\lambda g\,x^2$ (the segment on the table is at height $0$).

Initial and final states: $E_\text{pot,i} = -\tfrac{1}{2}\lambda g\,\ell_0^2 = -\tfrac{1}{2}\cdot 0{,}3\cdot 9{,}81\cdot 0{,}2^2 \approx -0{,}059\;\text{J}$ $E_\text{pot,f} = -\tfrac{1}{2}\lambda g\,L^2 = -\tfrac{1}{2}\cdot 0{,}3\cdot 9{,}81\cdot 2^2 \approx -5{,}89\;\text{J}$

Conservation of energy ($\tfrac{1}{2}M v^2 = E_\text{pot,i} - E_\text{pot,f}$): $\tfrac{1}{2}\cdot 0{,}6\cdot v^2 = -0{,}059 - (-5{,}89) = 5{,}83\;\text{J}$ $v = \sqrt{\frac{2\cdot 5{,}83}{0{,}6}} \approx \sqrt{19{,}4}$ $\ev{v \approx 4{,}4\;\text{m/s}}$

Note on the result $v \approx 2{,}5\;\text{m/s}$ as the answer, but the correct energy balance with the barycentre (CM drop of about $0{,}99\;\text{m}$) gives $v \approx 4{,}4\;\text{m/s}$. The textbook's value of $2{,}5$ appears to be a typo.

The textbook gives

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Topics: Centro di massa Concepts: Centro di massa · Energia potenziale gravitazionale · Conservazione dell’energia meccanica Skills: Conservazione dell’energia