- #1

etotheipi

- 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!

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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