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Moment of Inertia

  1. Sep 9, 2007 #1
    1. The problem statement, all variables and given/known data
    There is a rod of length 0.5m, to the ends of which two finite point masses of 20g each (0.02 kg) each are attached. Find the moment of inertia along the axis passing through one end of the rod, neglecting the moment of inertia of the rod itself.


    2. Relevant equations
    I=M*R^2


    3. The attempt at a solution

    Since the Axis passes through one end, the moment of inertia is,

    I = 0.02*(0.5)^2+0.02*(0)^2
    = 0.02* 0.25 = 0.005
    = 5*10^(-3)

    But the answer given is twice the answer I've found (i.e. 10^(-2) ). In many similar questions, the answer is always twice what I get. Am I wrong or the answer printed is wrong?

    Thanks,
    Sleek.
     
  2. jcsd
  3. Sep 9, 2007 #2

    learningphysics

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    Homework Helper

    Are you sure the rod is 0.5m in your question? Your answer seems right to me.
     
  4. Sep 9, 2007 #3
    Yes, the rod is 0.5m. The question has two sub-questions (the one I pointed above is the second), and the answer from the first one matches whats given in the book. So the datas are correct.

    I'm confident that there have been some printing mistakes in the book (there are many, but mistakes in ClassWork problems are always rectified while solving them, HomeWork can be a problem.

    Thanks for the reply!

    Regards,
    Sleek.
     
  5. Sep 9, 2007 #4

    learningphysics

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    Homework Helper

    no prob. talk to your professor about this if you can.
     
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