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Finding time taken to fall off table

  • Thread starter Arctic
  • Start date
  • #1
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Homework Statement


A uniform chain of length L lies on a smooth table with part of its length b initially hanging over the edge. Show that the time required for it to fall off the table when released from rest is given by

t = √(L/g) arccosh (L/b)


Homework Equations


http://img10.imageshack.us/img10/2203/integral.png [Broken]


The Attempt at a Solution


Not sure how to begin, since the acceleration of the chain keeps increasing as b increases. b is in turn related to the acceleration of the chain.
 
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Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Welcome to PF!

Hi Arctic! Welcome to PF! :smile:

(have a square-root: √ and try using the X2 tag just above the Reply box :wink:)
A uniform chain of length L lies on a smooth table with part of its length b initially hanging over the edge. Show that the time required for it to fall off the table when released from rest is given by

t = (L/g)^0.5 arccosh (L/b)
Either use good ol' Newton's second law, or (slightly quicker, since it gives you a first-order equation :wink:) simply use conservation of energy. :smile:

(But if you use conservation of energy, don't forget you have to integrate to find the PE)
 
  • #3
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Using conservation of energy I get:

eq.latex?Initial%20PE%20=%20mg[(\frac{L}{2})(\frac{L-b}{L})+(\frac{L-b}{2})(\frac{b}{L})].gif

This is by breaking the chain into 2 parts: the part of length L-b lying on the table and the part b over the edge. I also take PE = 0 at height L/2 below the table to simplify calculations later

When the chain has fallen off the table, PE = 0 (since its center of gravity is at L/2 below the table) and
eq.latex?PE%20=%200.5mv^{2}=%20mg[(\frac{L}{2})(\frac{L-b}{L})+(\frac{L-b}{2})(\frac{b}{L})].gif


eq.latex?\Leftrightarrow%20v^{2}=%20g[(L-b)+(\frac{b(L-b)}{L})]=g(L-b)(1+\frac{b}{L}).gif

Not sure how to continue from here, assuming this is right that is.

As for Newton's 2nd law..
F=ma
mgb/L = ma
a = gb/L
dv/dt = gb/L
d2b/dt2 = gb/L
dt2 = d2b L/gb

Again, I am stuck here
 

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