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Homework Help: Torsional Oscillations Question

  1. Sep 25, 2008 #1
    1. The problem statement, all variables and given/known data
    A thin steel beam 8.0 meters long is suspended from a crane and is undergoing torsional osciallations. Two 75-kg steelworkers leap onto opposite ends of the beam, as shown in the figure (no figure given, they just jump straight towards the center of the beam).

    If the frequency of the oscillations diminishes by 20%, what is the mass of the beam?

    2. Relevant equations
    Iw_1 = Iw_2 + 2 (m R^2 w_2)
    I = (1/12) ML^2
    w = 2 (pi) (freq.)

    3. The attempt at a solution
    I (w_1) = I (w_2) + 2 (m R^2 w_2)
    I (w_1) = (w_2) (I + 2 (m R^2))
    I (w_1) = ((80%) w_1) (I + 2 (m R^2))
    I = (80%) (I + 2 (m R^2))
    I = 80% I + 80% 2 (m R^2)
    (2/10) I = (8/10) 2 (m R^2)
    I = 8 (m R^2)
    I = 8 (75 R^2)
    I = 600 R^2

    But I = (1/12) ML^2, so:

    600 R^2 = (1/12) ML^2
    M = 7200 R^2 / L^2
    M = 7200 (4^2) / (8^2)
    M = 1800

    Mastering physics does not accept this answer. What is wrong with my solution?
  2. jcsd
  3. Sep 25, 2008 #2


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    Homework Helper

    Hi akan,

    I think you are confusing the [itex]\omega[/itex] for angular velocity with the [itex]\omega[/itex] for angular frequency.

    This is a torsional pendulum and your book probably has derived formulas for the period, frequency, and angular frequency for them.
  4. Sep 25, 2008 #3
    w = 2 pi f
    so if f diminishes by 20%
    so does w
    is that right?
  5. Sep 26, 2008 #4


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    Homework Helper

    That's right; now what is the formula for w? For a simple pendulum it is:

    \omega =\sqrt{\frac{g}{\ell}}
    for a spring it is


    What is it for this torsional pendulum? (Rather than w, you might find a formula for the regular frequency or the period instead.)
  6. Sep 26, 2008 #5

    \omega =\sqrt{\frac{k}{I}}

    However, I do not know what k is, and I also don't know the initial [tex]\omega[/tex] to calculate it. So what do I do?


    Nevermind, I see. Haha.

    [tex]\omega 1 =\sqrt{\frac{k}{I}}[/tex]

    [tex]\omega 2 = \frac{80}{100} \sqrt{\frac{k}{I}} = \sqrt{\frac{k}{I + 150 R^2}};

    etc., gives M = 800 kg, which is correct.
    Last edited: Sep 26, 2008
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