# Uniform chain falls off table Diff EQ

I really need some assistance setting up this problem. any assistance would be a Godsend:

a uniform heavy chain of length a initially has length b hanging off of a table. The remaining part of chain a - b, is coiled on the table. Show that if the chain is released, the velocity of the chain when the last link leaves the table is $\sqrt{2g\frac{a^3 - b^3}{3a^2}}$

Okay, so this is a variable mass problem, so momentum is constantly changing:

$$F_{ext}=m(t)g=\frac{dP}{dt}= \frac{d(m(t)v(t))}{dt}=ma +v\dot{m}$$

$$mg = ma + v\dot{m}$$

Gravity acts on the mass hanging off of the table, mass can be written as a function of length, as can velocity (where $\lambda$ is the linear mass density)

$m(t) = \lambda l(t)$
$\dot{m(t)} = \lambda v(t)=\lambda \dot {l(t)}$
$v(t) = \dot {l(t)}$
$a(t) = \ddot{l(t)}$

$$\lambda l(t)g=\lambda l(t) \ddot{l(t)} + \dot {l(t)} \lambda \dot {l(t)}$$

Assuming this all to be correct $\implies$ $0 =l(t)(g -\ddot{l(t)}) +\dot{l(t)}^2$

I've tried solved this DE, but I don't know many methods for non linear DEs.

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Quick follow up question from a physics student - when I saw this problem, my first instinct was to use conservation of energy (compare initial and final potential energies of the chain in terms of the length hanging over). The result by doing the problem this way, however, is incorrect. Why does conservation of energy not apply? – Shivam Sarodia Dec 18 '12 at 23:57
Energy would not be easy to deal with because both kinetic and potential energy depend upon the mass, which is changing with time. Potential energy also depends upon the center of mass which is also changing in time. In theory, energy would work (lol) but it would not be practical. – Cactus BAMF Dec 19 '12 at 0:04
Are you sure there is no missing information? As Draksis points out, this is really very easy to set up in terms of energy, but the velocity then comes out to be $v=\sqrt{g(a^2-b^2) / a}$. – Jaime Dec 19 '12 at 1:38
I just threw together some quick Python code to solve Cactus BAMF's DE for a = 30 and b = 10, and the final answer (13.757104646790195) agreed well with Cactus BAMF's provided solution. The energy argument seems to therefore be flawed, but I don't see why it would be. – Shivam Sarodia Dec 19 '12 at 2:54
I gave this a shot with energy and got the same thing as Jaime... haven't checked the momentum method yet, but I'm suspecting there's something wrong there... – Kyle Oman Dec 19 '12 at 4:24

With $\ell(t)$ as the length of the chain hanging off the table, the differential equation $$\ell(g-\ddot \ell)=\dot \ell^2$$ from the question can be rewritten as $$y\dot y=g\ell^2 \dot\ell,$$ where $y=\ell \dot\ell$. Then, integrating over the appropriate time interval will yield a final velocity of$$v_f=\sqrt{\frac{2g}{3}\frac{a^3-b^3}{a^2}}.$$

However, the differential equation above is incorrect, as it fails to take into account the tension in the chain. The correct equation should be $$\ddot\ell=\frac{g}{a}\ell,$$ giving a final velocity of $$v_f=\sqrt{\frac{g}{a}(a^2-b^2)},$$ which is in agreement with the conservation of energy.

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Beat me to it :) Glad to see my intuition was correct. – Kyle Oman Dec 19 '12 at 16:18

I'll add the conservation of energy based solution just for completeness. I wrote the change in gravitational potential energy based on the picture that the piece of chain that begins on the table ends hanging vertically from a point $b$ below the table (and the rest of the chain is unchanged).

$\Delta U_g = -g\lambda\int_0^{a-b}(b+x)dx$

$\Delta K = \frac{1}{2}\lambda av^2$

Where $\lambda$ is mass per unit length of the chain.

$\frac{1}{2}av^2 = g[bx + \frac{1}{2}x^2]_0^{a-b}$

$v^2 = \frac{2g}{a}(ba-b^2+\frac{1}{2}a^2-ab+\frac{1}{2}b^2)$

$v = \sqrt{\frac{g}{a}(a^2-b^2)}$

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The problem was imperfectly posed, which leaves me trying to answer multiple issues; I'll try to answer all, but the order of answering is completely arbitrary. An ideal chain has inelastic links, but this would not affect a stretched coiled chain. Nevertheless, a stretched coiled chain would not give the answer you get from simple energy conservation because centripetal acceleration would have the chain whipping all over the place, so the vertical drop would be smaller than you have calculated. In addition, the varying velocities of different sections of the chain would result in the mean speed being substantially smaller than 2.v^2/KE. Even a chain stretched perpendicular to the table edge would end up not falling vertically, because the tension in the horizontal section on the table would in due course be insufficient to stop the links moving horizontally as they reach the edge of the table. The energy "solution" would actually require that the chain be contained in an inverted L-shaped tube.

CactusBAMF started one version of the expected answer. leongz showed one soution method. An alternative is the substitution ℓ = pt^2 + qt + b, and equate the square-law, linear, and constant multipliers.

Returning to the posing of the original problem, there are constraints on the construction of the chain and also on its disposition. The construction must not allow any form of kick-back between the elements of the chain, or we will see increased initial velocity and failure to follow vertically at the edge of the table. The appropriate dimensional limit of zero-friction zero-inertia balls separated by flexible threads would do the job, and the "coil" would require zero total length along the chain (all links need to be collapsed, or we get a finite length moving on the table; this is more of a scrunch than a coil...

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