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I hate pendulums

  1. Nov 25, 2004 #1
    A student wants to use a L = 1.0 m stick as a pendulum. She plans to drill a small hole through the meter stick and suspend it from a smooth pin attached to the wall (Fig. 14-35). Where in the meter stick should she drill the hole to obtain the shortest possible period? Answer in number of meters from the upper end

    How short an oscillation period can she obtain with a meter stick in this way?

    http://www.webassign.net/gianpse3/14-35alt.gif

    My guess is 0 meter from the upper end will give the shortest period. Agree?
     
  2. jcsd
  3. Nov 25, 2004 #2

    Doc Al

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    No. But why guess? Figure it out. What's the period of a physical pendulum?
     
  4. Nov 25, 2004 #3
    T=2*pi*sqrt(I/mg(0.5-x))
    T=2*pi*sqrt( (1/12mL^2+m(0.5-x)^2) / (12*m*g*(0.5-x) )

    Canceling the m
    T=2*pi*sqrt( (1/12*L^2+(0.5-x)^2) / (12*g*(0.5-x) )

    So if x=0, the period is the smallest...why is this wrong?
     
    Last edited: Nov 26, 2004
  5. Nov 26, 2004 #4

    Doc Al

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    OK, but better use (L/2 - x) instead of (0.5 - x). Also, it seems that there's an extraneous 12 in your denominator.
    What makes you think that the period is smallest when x=0?
     
  6. Nov 26, 2004 #5
    Because it gives me the smallest T possible when I plug it in...0.50 would give me 0/0
     
  7. Nov 26, 2004 #6

    Doc Al

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    How do you know that? Did you check T for every possible value of x?
     
  8. Nov 26, 2004 #7
    The problem is...i don't...and I don't know how to either. :confused:
     
  9. Nov 26, 2004 #8

    Doc Al

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    If you know some calculus, you can find the value of x that minimizes the period by taking a derivative and setting it equal to zero.
     
  10. Nov 26, 2004 #9
    OK, I did that and got x=1.5 or -0.5. So -0.5 is the answer? But that doesn't make much sense since they want distance from the upperend of the stick...
     
  11. Nov 27, 2004 #10

    Doc Al

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    No, those answers are not correct. Two suggestions:
    (1) Correct your expression as I advised in post #4. Your answer should be in terms of L.
    (2) Redo the derivative.
     
  12. Nov 27, 2004 #11
    I'm still getting the same answers I got before. Is there any other way to do this since the derivative of 2*pi*sqrt( (L^2+(L/2-x)^2) / (12*g*(L/2-x) ) with respect to x is really difficult for me.
     
  13. Nov 28, 2004 #12

    Doc Al

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    That expression should be:
    [tex]T = 2 \pi \sqrt{\frac{L^2/12 + (L/2 - x)^2}{g(L/2 - x)}}[/tex]
    This will be a minimum when the expression within the square root is a minimum. So the only thing you need to take the derivative of is this:
    [tex]\frac{L^2/12 + (L/2 - x)^2}{(L/2 - x)}[/tex]
    Use the quotient (or product) rule. It's not as bad as it looks. :smile:
     
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