1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Finding time taken to fall off table

  1. Apr 30, 2009 #1
    1. The problem statement, all variables and given/known data
    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)

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

    3. 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.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 1, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper

    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:)
    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)
  4. May 2, 2009 #3
    Using conservation of energy I get:

    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

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

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

    Again, I am stuck here
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook