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Physical pendulum hole

  1. Apr 6, 2009 #1
    1. The problem statement, all variables and given/known data

    You are given a meter stick and asked to drill a hole in it so that when pivoted about the hole the period of the pendulum will be a minimum. Where should you drill the hole


    2. Relevant equations

    T=2*pi*sqrt(I/mgd)

    3. The attempt at a solution

    So I use parallel axis theorem for I and get I=I(com)+mh^2=(1/12)mL^2+mh^2 = m*((1/12)+h^2) because L=1 as in a meter stick.

    Plug in the formula I have T= 2*pi * sqrt( (1/12)+h^2 / gh) because m cancels out and d=h because h is the distance from center of mass.

    How should I move one to get the final answer ???
     
  2. jcsd
  3. Apr 6, 2009 #2

    rl.bhat

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    Take differentiation of T with respect to h and equate to zero.
     
  4. Apr 6, 2009 #3
    Ok, thanks, I got like 0.238, don't know if it sounds reasonable. Can you check that for me please ??
     
  5. Apr 6, 2009 #4

    rl.bhat

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    Check it again. I am getting different answer.
     
  6. Apr 6, 2009 #5
    I punched the whole expression in my calculator and set it equal 0 and use the solver. That's what it gave me. What did you get ?
     
  7. Apr 6, 2009 #6

    rl.bhat

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    0.288675 m
     
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