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Chain in a tube

  • Thread starter nicksauce
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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.
 
Last edited:

Answers and Replies

tiny-tim
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… negative: trigonometric … positive: exponential

Hi nicksauce! :smile:

For negative parameter, it's trigonometric, for positive parameter it's exponential …

which btw agrees with the T + V = 0 method (I take it you're not allowed to use that in this question?). :smile:
 
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Hi! I got the same equation for L. Solving for the equation of motion I get

[tex] f(t)'' - f(t)*c^2 = 0 [/tex]

which has as a solution

[tex] f(t) = c_1 * e^{c*t} + c_2 * e^{-c*t} [/tex]

where c is a constant. This doesn't seem right, since I have to solve f(t)=1, but I don't know c1 or c2. Help plz? :s
 
Last edited:
tiny-tim
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hi mathman44! :smile:

(try using the X2 and X2 icons just above the Reply box :wink:)

you can find c1 and c2 from the initial conditions :wink:

(usually they're given as f(0) = f'(0) = 0)
 
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hi mathman44! :smile:

(try using the X2 and X2 icons just above the Reply box :wink:)

you can find c1 and c2 from the initial conditions :wink:

(usually they're given as f(0) = f'(0) = 0)
Thanks for the response tiny_tim!

I get f(0) = f_initial = c1 + c2. Also from f'(0)=0 I get that c1 = c2. Hence c1 = f_ini/2 = c2. But that doesn't work, I'm still unable to solve for "t" in the equation f(t)=1, I'm getting complex solutions. Help please! (could my solution to the ODE be incorrect? Doesn't the solution have to be an exponential since f is positive at all times?).
 
Last edited:
tiny-tim
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hi mathman44! :smile:

in this case the initial condition is f(0) = f0 :wink:

(or is it 1 - f0 ? :confused:)
 
206
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Maybe this helps :s

eiplog.jpg


"f" is definitely the proportion of the chain hanging on the incline!
 

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