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College Physics

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Calculate the rotational inertia of a meter stick, with mass 0.58 kg, about an axis perpendicular to the stick and located at the 43 cm mark. (Treat the stick as a thin rod.)

I know I = (1/12)(M)(L^2) and then I need to use the parallel-axis theorem so I need to add Mh^2 but that is not working. I get 0.156 doing this method but it is not correct.

  • College Physics -

    Check your units (are they meters?) and your decimal point.

    The center of the stick is at the 0.50 m mark and the moment of inertia about that point is (1/12)ML^2 = 0.0483 kg m^2.
    (L = 1.0 m and M = 0.4833 kg)

    Using the parallel axis theorem, the moment of inertia about an axis h = 0.07 m away is higher by
    M*h^2 = 0.0028 kg/m^2, for a total of 0.4861 kg/m^2

  • College Physics -

    Ah, I see what I may have done wrong. I was not using .07m for h. Unfortunately, 0.4861 for the overall answer still seems to be wrong.

  • College Physics -

    Oops I found the problem, it seems you were off by a decimal point. I did the math and got .051 for a final answer, thank you for the help.

  • College Physics -

    I made a typo error saying that I used a mass of M = 0.4833 kg. I actually used 0.58 kg

    I may have made a computational error somewhere but I don't see where. .

  • College Physics -

    (1/12)(.58)(1^2) = .0483
    +
    (.58)(.07^2) = .0028

    So final answer about .0511 if that helps finding out your error.

  • College Physics -

    You are right. I moved a decimal point and added a wrong (1/2)ML^2 value. Thanks for catching that.

  • College Physics -

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  • College Physics -

    Your Open QuestionShow me another ยป
    A uniform meter stick (a ruler 100 cm long) is supported by a fulcrum at its 20 cm mark balances when a 200 g?mass is suspended at 0cm.Mass of meter of stick is at the middle 50cm mark what is the mass of the meter stick in grams.
    Just need to know how to set this up can some one walk show me so I know for another similiar problem I have.

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