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Moment of Inertia of a thin uniform wire

  1. Nov 11, 2009 #1
    1. The problem statement, all variables and given/known data
    A thin uniform wire is bent into a rectangle. The short, vertical sides are of length a, and the long, horizontal sides are of length b. If the total mass is 31.00 grams, a = 25.00 cm and b = 44.30 cm, what is the moment of rotational inertia about an axis through one of the vertical wires?
    prob29new.gif


    2. Relevant equations
    I = 1/3mL^2


    3. The attempt at a solution
    I tried using the above equation, which is the moment of inertia for a solid rectangle. I knew it wasn't going to be right but I don't have an equation for a hollow rectangle. I tried modifying it by using 1/4 or 1/6 because I think the moment of inertia would be smaller for a hollow rectangle than a solid one.
     
  2. jcsd
  3. Nov 11, 2009 #2

    Doc Al

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    Staff: Mentor

    Think of the object as composed of four thin rods.
     
  4. Nov 11, 2009 #3
    So the top and bottom would be I= 1/3 mL2 but it's only a portion of the mass, right? The right side would be I = mh2 and the h would be the length b. The only thing I can think of for the left side is the formula for a solid cylinder, but for that you need a radius.
     
  5. Nov 11, 2009 #4

    ideasrule

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    Homework Helper

    Because the wires are assumed to be very thin, the left side has nearly no moment of inertia. You can assume I=0.
     
  6. Nov 11, 2009 #5

    Doc Al

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    Staff: Mentor

    Right.
    Right.
    Think of it as being very thin.
     
  7. Nov 11, 2009 #6
    Yes, I got it! Thanks a lot. I'm home sick with the flu so this means a lot that someone would help me.
     
  8. Nov 10, 2010 #7
    I'm having trouble with this problem, too. I don't understand how to get the masses of the individual parts of the rectangle.
     
  9. Nov 10, 2010 #8

    Doc Al

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    Staff: Mentor

    You have the total mass. Figure out the mass of each side, realizing that the mass is proportional to the length.
     
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