# A chain falling off of a table

Homework Statement:
A chain of mass ##m## and length ##l## rests on a frictionless table with a fraction ##\alpha## of its length initially hanging vertically over the edge of the table. The chain is released, and we're asked to determine the length of chain beneath the surface of the table at some time ##t##.
Relevant Equations:
N/A
This question came up in a lecture, and I wasn't really satisfied with how it was solved. Specifically, they assumed that the hanging part of the chain has zero horizontal velocity. What they did was essentially write down the equation ##m\ddot{x} = \frac{mg}{l} x##, which under the above assumption can be obtained either through conservation of energy, or by "resolving along the chain", or by recasting the problem as a completely vertical chain falling in a region where ##\vec{g} \neq \vec{0}## below the height of the table, but ##\vec{g} = \vec{0}## above the height of the table. But, whatever way you set up the equation, this approach clearly wrong since horizontal momentum isn't conserved under this assumption.

So when I was playing around with it, I added a new coordinate ##y## which is the horizontal distance from the edge of the table. The potential energy of the chain is then something like$$U(t) = -\int_{0}^{l} \rho x g \sqrt{ \left(\frac{\partial x}{\partial \lambda} \right)^2 + \left(\frac{ \partial y}{ \partial \lambda} \right)^2} d\lambda$$where ##(x(\lambda, t), y(\lambda, t))## is a parameterisation of the chain, with material coordinate ##\lambda \in [0, l]## and time ##t##. The kinetic energy of the whole chain is$$T(t) = \frac{1}{2} \int_{0}^{l} \rho (\dot{x}^2 + \dot{y}^2) \sqrt{ \left(\frac{ \partial x}{ \partial \lambda} \right)^2 + \left(\frac{\partial y}{ \partial \lambda} \right)^2} d\lambda$$Then maybe$$\mathcal{L} = \int_{0}^{l} \rho \left[ \frac{1}{2} (\dot{x}^2 + \dot{y}^2) + x g \right] \sqrt{ \left(\frac{\partial x}{\partial \lambda} \right)^2 + \left(\frac{ \partial y}{ \partial \lambda} \right)^2} d\lambda$$But now I get stuck, I don't know how to get an equation of motion out of that mess. I wondered if anyone could give any pointers as to how to solve for ##x(\lambda, t)## and ##y(\lambda, t)##. Thanks!

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This is a frequently misposed problem. As you observe, it ought to specify something like a smooth L-shaped tube leading down through a hole in the table, so that the horizontal momentum is converted to vertical.
The misposed version is very unlikely to be tractable.

• etotheipi
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It seems you have seen Lagrangians, so why not set it up the usual way. Take generalized coordinates ##x## = length of chain on table, ##y## = overhanging length and define linear mass density ##\lambda=m/l##. Then $$T=\frac{1}{2}\lambda x ~\dot x^2+\frac{1}{2}\lambda y ~\dot y^2~;~U=-\lambda y ~g~\frac{y}{2}.$$Apply the constraint ##x+y=l## and simplify to get the Lagrangian and hence the equation of motion,$$\mathcal{L}=\frac{1}{2}\lambda l \dot y^2+\frac{1}{2}\lambda gy^2 ~\Rightarrow~\ddot y-\frac{g}{l}y=0.$$

• etotheipi
It seems you have seen Lagrangians, so why not set it up the usual way. Take generalized coordinates ##x## = length of chain on table, ##y## = overhanging length and define linear mass density ##\lambda=m/l##. Then $$T=\frac{1}{2}\lambda x ~\dot x^2+\frac{1}{2}\lambda y ~\dot y^2~;~U=-\lambda y ~g~\frac{y}{2}.$$Apply the constraint ##x+y=l## and simplify to get the Lagrangian and hence the equation of motion,$$\mathcal{L}=\frac{1}{2}\lambda l \dot y^2+\frac{1}{2}\lambda gy^2 ~\Rightarrow~\ddot y-\frac{g}{l}y=0.$$

That's fine but this is assuming the hanging part of the chain is vertical, which can't satisfy conservation of momentum. I think the chain will have some arbitrary curved shape below the table, which leads to a more complicated Lagrangian. So far I have been unsuccessful in extracting any solutions from that Lagrangian

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Yes, the assumption is that some feature exists that smoothly converts horizontal momentum to vertical as @haruspex noted. A normal force of variable direction from vertical to horizontal is needed at the point of contact in order to continuously provide the necessary impulse for the conversion of the momentum. If the direction changes discontinuously, momentum is lost.

My own bias is that, with problems such as this, a smooth transition with energy conservation is implicitly assumed. A "supple chain" can be imagined as going over an "ideal pulley" having a radius that approaches zero but never gets there.

• etotheipi
But how do we solve it when there is no such assumption, i.e. where we just have a chain sliding off of a table? I did think that it might be useful to guess a functional form of ##(x(\lambda, t), y(\lambda, t))## and substitute this into the Lagrangian, but I'm struggling to come up with a suitable ansatz

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Would the centrifugal effect suffice?

• Hamiltonian