(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A uniform chain of length L and mass M is constrained to move in a frictionless tube. Initially a fraction (1-f0) of the chain rests in a horizontal section of the tube. The remaining fraction f rests in a section of the tube that is inclined downward from the horizontal at an angle theta. At t = 0, the chain is released and begins to slide down the tube. Find the time for the remainder of the chain to pass through the horizontal section of the tube.

2. Relevant equations

Lagrange equation

3. The attempt at a solution

We want to characterize the system completely by the parameter f.

[tex] T = \frac{1}{2}M\dot{l}^2[/tex]

where l = f * L

[tex] T = \frac{1}{2}ML^2\dot{f}^2[/tex]

[tex]V = -\int{mg dh}[/tex]

where m = f * M

[tex]V = -\int{mfg\sin{\theta}dl}[/tex]

[tex]V = -\int{mfg\sin{\theta}Ldf}[/tex]

[tex]V = \frac{-Mgsin{\theta}Lf^2}{2}[/tex]

Note we take the horizontal part of the chain to be at a gravitational potential of 0, thus it does not contribute.

[tex]L = T - V = \frac{1}{2}M\dot{f}^2L^2 + \frac{1}{2}Mg\sin{\theta}Lf^2[/tex]

Then I can plug into the Lagrange equation and solve. Does this solution seem to be correct? If I am not mistaken this Lagrangian gives trigonometric functions as the solution, which seems somewhat odd for this problem.

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# Homework Help: Chain in a tube

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