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Homework Help: Mass Moment of Inertia issues with half circle

  1. Nov 13, 2009 #1
    1. The problem statement, all variables and given/known data
    specific volume of steel = 489 lb/ft^3

    2. Relevant equations
    I tried using Ixo=1/4mr2+1/12ml2+md2

    3. The attempt at a solution
    I tried using d=70mm because of the parallel axis therm, but that was wrong.

  2. jcsd
  3. Nov 13, 2009 #2


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    Your equation for moment of inerta of a half cylinder is incorrect as well as your distance used for the parallel axis theorem. Recall that it is from the axis of rotation to the objects centroid.

    Review your equation and find the half-cylinders centroid.
  4. Nov 13, 2009 #3
    Hmm I rechecked the equation and it seems to be the correct one to use, unless I'm missing somethings here or am looking at it wrong? Because the line Xo is under the center of the half circle.

    As for the distance that makes sense so I used (4r/3pi) for the centroid and added that to the 70mm.
  5. Nov 13, 2009 #4


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    Every reference I see lists the moment of inertia of a half-cylinder about its axis the same as that of a full circle, noting that it is about its centroid:
    I_{x_0} = \frac{1}{2}mr^2

    Try that, using the parallel axis theorem appropriately.
  6. Nov 13, 2009 #5
    oye vay! Of course wrong axis. fml lol I'll try that then.
  7. Nov 14, 2009 #6
    Okay so I went back and tried it again.

    For Ix'x'=[tex]\frac{mr^2}{2}[/tex]+md2

    m=(7830kg/m3)([tex]\frac{volume of cylinder}{2}[/tex])=5.95kg




    This ended up being wrong.
  8. Nov 15, 2009 #7
    Any help?
  9. Nov 16, 2009 #8


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    That's what I get to. If this is an online homework assignement check your significant figures.
  10. Nov 16, 2009 #9
    It's online. I'll ask my professor about it then. This online thing has issues sometimes.
  11. Nov 17, 2009 #10


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    They always do. The luxury of open-book, open-note, open-friend, open-PhysicsForums.com is not without slight penalty.
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