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Homework Help: Torsion Pendulum

  1. Jan 26, 2004 #1
    A uniform meter stick is hung at its center from a thin wire. It is twisted and oscillates with a period of 5 s. The meter stick is then sawed off to a length of 0.76 m, rebalanced at its center, and set into oscillation.

    With what period does it now oscillate?

    Ok, I know that the period for a torsion pendulum T=2pi*sqrt(I/k)

    Where I is moment of inertia
    and K is the torsion constant

    Two things I am stuck on.
    To calculate the moment of inertia => 1/12 ML^2
    I need to konw the mass of the rod...unless this mass drops out of the calculations, I don't know what to do

    Also, the torsion constant. I do not konw what that is either, unless that too also drops out of my calculations. A little guiadance here greatly appreciated.

    Attached Files:

  2. jcsd
  3. Jan 26, 2004 #2


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    The mass does drop out of the calculations. Use the data point you are given to find k.

    You know that

    [tex]T = 2 \pi \sqrt{\frac{(1/12) ML^2}{k}}[/tex]

    is valid when T = 5 and L = 1. You can solve this expression for k. Leave k in terms of M.

    Now use the equation to solve for the situation when L = 0.76. M cancels. I get 3.8 seconds.

    Check this answer for sanity: 3.8 / 5 is the same ratio as 0.76 / 1. In other words, the period shrinks linearly with a decreasing length of rod. Does this make sense?

    1) The period decreases at the rod is made smaller. This is as expected.

    2) The period is linear with respect to the length. An inspection of the formula reveals that L is first squared, then its square root is taken. Yes, the period should be linear in L.

    The answer makes sense.

    - Warren
  4. Dec 6, 2009 #3
    sorry but, I have the same problem with the same numbers and 3.8 seconds is not the answer according to the computer program that grades the problem.

    Also when I solve for K as you stated, K= (T/2pi)^2*(1/12)ML^2
    When you insert that back into the original equation T=2pi*sqrt((1/12)ML^2/k)
    then the (1/12)ML^2 cancels.....which leaves the new period, T, equal to 2pi*sqrt(1/(t^2/4pi^2))
    This yields an answer of 7.895, which is also wrong. I don't see what is wrong here, but 3.8 seconds was also not correct. Any additional advice? Thanks!
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