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Torsional Pendulum

  1. Dec 2, 2009 #1
    1. The problem statement, all variables and given/known data

    https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/fall/homework/Ch-15-SHM/torsion-pendulum/4.gif [Broken]


    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?

    L = 1m
    L(final) = 0.76m
    T = 5s

    2. Relevant equations

    Moment of Inertia: I = (1/12)*M*L^2
    period: T = 2pi*sqrt(I/K) Where K is the torsional constant

    3. The attempt at a solution

    I first found the torsional constant:

    K = [(1/12)*M*(1^2)]/[(T/(2pi))^2] = 0.13459M

    So no I have K = 0.13459M

    T(final) = 2pi*[sqrt((1/12)*M*(0.76^2)/0.13159M)] = 3.8s

    The answer looks correct but I'm not sure, any ideas? Thanks!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 2, 2009 #2
    instead of going for so much calculation...
    u could have done it like this....

    becauz..

    T=2*pi*(I/K)^0.5

    and u have I=(M*L^2)/12

    so jus take the ratio

    T1/T2=L1/L2 :smile:

    where T2 is the time period when the length is 0.76
    directly arrives at your answer ...
     
  4. Dec 2, 2009 #3
    Oh yeah it does.....cool.
     
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