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Inertia of weighted rings

  1. Dec 8, 2008 #1
    1. The problem statement, all variables and given/known data
    A compound disk of outside diameter 138 cm is made up of a uniform solid disk of radius 39.0 cm and area density 5.40 g/cm^2 surrounded by a concentric ring of inner radius 39.0 cm, outer radius 69.0 cm, and area density 2.60 g/cm^2.

    Find the moment of inertia of this object about an axis perpendicular to the plane of the object and passing through its center.

    2. Relevant equations
    I=m r^2
    area = pi r^2

    3. The attempt at a solution


    pi x 39^2 = 4778 cm2
    4778x5.40 = 25801.2g =25.801kg

    pi x 69^2 (-4778) = 10179cm2
    10179x2.60=26465.4g =26.465kg

    i now dont understand how to work out the inertia
  2. jcsd
  3. Dec 8, 2008 #2

    Doc Al

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    Staff: Mentor

    Look up the formula for the moment of inertia of a disk and a ring.
  4. Dec 8, 2008 #3


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    Homework Helper

  5. Dec 8, 2008 #4
    I= 0.5 m r^2
    I = 0.5 x 25.8 x 39^2 = 1920.9

    And for the other:
    I= 0.5 m(r1^2 + r2^2)
    I=0.5 x 26.465 x (39^2 +69^2) = 83126.6

    do i add these together? giving
    102747.465 ?
  6. Dec 8, 2008 #5


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    Homework Helper

    Yes the moments about the same point add for a compound moment.

    Not sure about your math.
  7. Dec 9, 2008 #6
    10.27 kg m^2 is the answer :)
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