Calculating Time for Sliding Chain through Hole

[derravaragh]

Homework Statement

A flexible chain of mass M and length L lies on a frictionless table, with a very short portion of its length L0 hanging through a hole. Initially the chain is at rest. Find a general equation for y(t), the length of chain through the hole, as a function of time. (Hint: Use conservation of energy. The answer has the form y(t) = Ae^(γt) + Be^(-γt) where γ is a constant.) Calculate the time when 1/2 of the chain has gone through the hole. Data: M = 1.4 kg; L = 2.6 m; L0 = 0.3 m.

Homework Equations

The Attempt at a Solution

This problem has been asked before, and I've worked through it, but I can't seem to get it right. I've solved the position function to be:
x(t) = (x0/2)e^(√(g/l)t) + (x0/2)e^(-√(g/l)t)

which can then be solved for t to get:
t = √(L/g)*arcosh((L-x0)/x0)

Since I want the time when half the chain has gone though, I solve using L = 1.3 m, but I can't seem to get it. Any help on where I've gone wrong would be appreciated.

 

[TSny]

x(t) = (x0/2)e^(√(g/l)t) + (x0/2)e^(-√(g/l)t)

which can then be solved for t to get:
t = √(L/g)*arcosh((L-x0)/x0)

Shouldn't the variable x appear on the right hand side of your equation for t? You might double check how you got the expression for t.

Since I want the time when half the chain has gone though, I solve using L = 1.3 m, but I can't seem to get it. Any help on where I've gone wrong would be appreciated.

Doesn't L denote the total length of the chain? So, you aren't free to let it be 1.3 m.

 

[derravaragh]

I realized where I went wrong, I had solved for the time t when the entire chain had fallen over the side, so I went back and solved for when x(t) = L/2. Thank you for your help.