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Pendulum in Simple Harmonic Motion

  1. Dec 7, 2008 #1
    1. The problem statement, all variables and given/known data

    A 20 cm long (L) wrench swings on its hook with a period of 0.90s. When the wrench hangs from a spring of spring constant 360 N/m (k), it stretches the spring 3.0 cm (x). What is the wrench's moment of inertia about the hook.
    A diagram also shows that the centre of mass for the wrench is 14cm (d) from the hook (i.e pivot point).

    2. Relevant equations

    Fsp = -ks
    Fg = mg
    I = Icm + md^2

    3. The attempt at a solution

    Fsp = -10.8 N
    Fsp = Fg gives m = 1.1 kg
    Icm for a rod with a pivot in one of the edges is 1/3mL^2
    Therefore, I = 1/3mL^2 + md^2 = 0.036 kg m^2

    However, that answer is not correct, the correct answer according to the textbook is 0.031 kg m^2. Any help is appreciated. Thank you!
  2. jcsd
  3. Dec 7, 2008 #2


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    Homework Helper

    I think you have assumed a moment of inertia initially as being a rod and this is not what they are suggesting.

    What they give you is information about a physical pendulum.

    Such a pendulum can be determined to have as a period of oscillation

    T = 2π(I/m*g*L)½

    To find I then:

    I = m*g*L*T2/2π
  4. Dec 7, 2008 #3


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  5. Dec 7, 2008 #4
    Oh now I understand! Thank you so much! :)
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