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Moment of Inertia of a Physical Pendulum?

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data
    I have a uniform ruler, 80cm long weighing 50g. I'm attaching a weight at one end, 1cm from the bottom. I'm assuming that its a point mass to make things easier. This weighs 10g. The ruler has an axis 5 cm from the top. What's the moment of inertia?

    2. Relevant equations


    3. The attempt at a solution
    None yet :S I'n pretty clueless about how to do this
     

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    Last edited: Nov 23, 2008
  2. jcsd
  3. Nov 23, 2008 #2

    fluidistic

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    Gold Member

    Find out the center of mass of the system. After you've done that, apply Steiner's theorem to determine the moment of inertia of the system. Consider the pendulum as being a point situated in the center of mass.
     
  4. Nov 23, 2008 #3
    I can work out the centre of mass (it's 33.5cm from the bottom) but I don't know how to work out the moment of inertia from there.
    Also, the pendulum can't be thought of as a point mass otherwise it turns into a simple pendulum (although I wish it could, this would be a hell of a lot easier :frown: ).
     
  5. Nov 23, 2008 #4

    fluidistic

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    Sorry from there you don't have to use Steiner's theorem. It's even more simple.
    The moment of inertia with respect to an axis passing by the upper point of the pendulum is worth [tex]\int r^2 dm[/tex]. You have a point so it is simply worth [tex]d^2M[/tex] where M is the mass of the system and d is the distance between the upper point and the center of mass.
    I hope I'm not wrong...
     
  6. Nov 23, 2008 #5

    Doc Al

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    Staff: Mentor

    Can you find the rotational inertia of the ruler by itself about the given axis? (You'll need the parallel axis theorem, also known as Steiner's theorem as fluidistic said.) The rotational inertia of the mass by itself?

    Find the separate rotational inertias and add them up for the total.
     
  7. Nov 23, 2008 #6
    Well using the standard result for the moment of inertia at the centre of mass I get this:
    I = 1/12 * ML[tex]^{2}[/tex] + Mr[tex]^{2}[/tex]
    I = 1/12 * 0.05 * 0.8[tex]^{2}[/tex] + 0.05 * 0.415[tex]^{2}[/tex]
    I = 0.01128

    Then for moment of inertia of the weight I get this:
    I = m * x[tex]^{2}[/tex]
    I = 0.01 * 0.74[tex]^{2}[/tex]
    I = 0.005476

    So Total Inertia = 0.0168
    Does this seem right?
    I've tried subbing it into the Time Period equation (T = 2[tex]\pi[/tex][tex]\sqrt{\frac{I}{Mgr}}[/tex] )
    but I get different values to my experiment results. It's out by around 0.3 seconds, I don't think my results are that bad :S Have I done the calculation properly?
     
  8. Nov 23, 2008 #7

    Doc Al

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    Staff: Mentor

    Good.
    What's the distance between the center of the ruler and the axis?

    Looks good.

    You'll need to recalculate that.
    What did you use for the distance between center of mass and axis of rotation?
     
  9. Nov 23, 2008 #8
    Ah right so the r in my second line of working should be 0.35m. However, it's still a little off :(
    I'm using 0.415 as the distance between CoM and the axis.
    I'm starting to think that my experimental values are off...
     
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