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Rotational Inertia (Moment of Inertia) of a Rod

  1. Feb 8, 2008 #1
    1. The problem statement, all variables and given/known data
    A very light (meaning don't consider mass of the rod) rod is placed along the x axis. It has a mass m1=2.0kg at x=0, a mass m2=1.50kg at x=50cm, and a mass m3=3.0kg at x=100cm.

    Find the moment of inertia of the system about a pivot point at x=0.


    2. Relevant equations

    I=(1/3)M(L^2) -- Parallel Axis Theorem

    I= integral of[(r^2)dm]?

    3. The attempt at a solution

    Because the mass of the rod does not matter, I'm thinking of finding the center of mass between the three masses and treating that as one whole mass at a certain point x. Does that idea make any sense? Any help would be greatly appreciated!
     
  2. jcsd
  3. Feb 8, 2008 #2

    Doc Al

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    Staff: Mentor

    That would be a bad idea. The rotational properties depend on the distribution of mass, not just the center of mass. Hint: What's the rotational inertia of a point mass about some axis?
     
  4. Feb 8, 2008 #3
    It is I=MR^2. If I find the sum of the point masses of inertia I get the moment of inertia? Does it matter that it's on a rod?
     
  5. Feb 8, 2008 #4

    Doc Al

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    Right.
    Absolutely.
    Nope. (You need something to connect the masses as a rigid structure.)
     
  6. Feb 8, 2008 #5
    Got it! Thanks Doc Al! Now I don't understand why there is a Parallel Axis Theorem for a rod with the equation I=(1/3)M(L^2). What would that be used for?
     
  7. Feb 8, 2008 #6

    Doc Al

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    Staff: Mentor

    Not sure I understand your question. The parallel axis theorem applies to any object, not just a rod.

    Starting with the rotational inertia of a rod about its center of mass (what's the formula for that?), the parallel axis theorem will allow you to find the rotational inertia of the rod about any other parallel axis--including about one end, which is what (1/3)M(L^2) is for. (Try it--it's easy.)
     
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