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Homework Help: Torsion Pendulum Oscillation (SHM)

  1. Jun 17, 2010 #1
    1. The problem statement, all variables and given/known data

    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?

    2. Relevant equations

    T=2pi(sqrt(I/K))
    I=(1/12)ML^2

    3. The attempt at a solution

    I tried solving for m in terms of k and then plugging that into a new equation solving for T with the length being .76, but that didn't work. Any suggestions?

    Thanks!
     
  2. jcsd
  3. Jun 17, 2010 #2

    tiny-tim

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    Hi SuperCass! :smile:

    (have a pi: π and a square-root: √ and try using the X2 tag just above the Reply box :wink:)

    Hint: by how much is I reduced? :wink:
     
  4. Jun 18, 2010 #3
    (1-.76)² ?
    So since the formula is T=2π√(I/k), would I multiply the original T by (1-.76)?


    (Thanks for your signature!)
     
  5. Jun 18, 2010 #4

    tiny-tim

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    No, I is proportional to ML2, and both M and L are reduced. :wink:
     
  6. Jun 18, 2010 #5
    Remember that you've been told that the rod is uniform, that means that its mass is proportional to its length! If we define a mass per unit length, [tex]\mu\equiv \frac{M_0}{L_0}[/tex] then the mass of a rod of length [tex]\ell[/tex] made of the same material would be..?
     
  7. Jun 18, 2010 #6
    .76 of the original mass?
     
  8. Jun 19, 2010 #7
    Exactly! That makes for [tex]m \propto \ell[/tex] which implies [tex]I \propto \ell^3[/tex]
     
  9. Jun 20, 2010 #8
    Oh!
    I got it now!
    So I just do the original period times the square root of .76^3!

    Thank you all very very much!!
     
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