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Moment of Inertia of Compound Disk

  1. Apr 3, 2013 #1
    1. The problem statement, all variables and given/known data

    A compound disk of outside diameter 152 cm is made up of a uniform solid disk of radius 41.0 cm and area density 3.30 g/cm2 surrounded by a concentric ring of inner radius 41.0 cm , outer radius 76.0 cm , and area density 2.10 g/cm2 .

    Find the moment of inertia of this object about an axis perpendicular to the plane of the object and passing through its center (in kg*m2).


    2. Relevant equations

    Moment of inertia of solid cylinder (a thin cylinder is a disk) = I = .5mr2


    3. The attempt at a solution

    This object is basically one inner disk with mass mi surrounded by an outer disk with mass mo. Finding the moment of inertia of each of these and adding them together should give the solution.

    mi in kg = (area * density)/1000 = (pi*412 *3.3)/1000 = 17.427

    mo in kg = [(area - inner area) * density]/1000 = [(pi*712 - pi*412) * 2.1]/1000 = 22.1671

    Using the moment of inertia for solid cylinder and adding yields:
    .5mi*.412+.5mo*.762= 7.87 kg*m2

    The answer given is 11.5 kg*m2. What am I doing wrong?
     
    Last edited: Apr 3, 2013
  2. jcsd
  3. Apr 3, 2013 #2

    haruspex

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    I assume the 8.76 is a typo for .76. But you have effectively taken the outer disk's mass as being spread over a disk of radius .76. In fact, it is concentrated between .41 and .76 radius, increasing the MI.
    The simplest approach would be to treat it as a solid disk radius .76 and density 2.1, plus a solid disk radius .41 and density 3.3-2.1=1.2.
     
  4. Apr 3, 2013 #3
    Oh, thanks! And yup, fixed the typo.
     
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