nicksauce

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**1. Homework Statement**

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. Homework 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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