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Two points are on a disk that rotates about an axis perpendicular to the plane

  1. Dec 2, 2011 #1
    Two unequal masses m and 2m are attached to a thin bar of negligible mass that rotates about an axis perpendicular to the bar. When m is a distance 2d from the axis and 2m is a distance d from the axis, the moment of inertia of this combination is I. If the masses are now interchanged, the moment of inertia will be?

    This is the work I tried:

    The mases are m and 2m
    The distance of the mass m from the axis is 2d
    The distance of the mass 2m from the axis is d

    The moment of inertia of the masse before exchanged is

    I = m(2d)^2 + 2m(d)^2

    I = 6md^2

    The moment of inertia of the masse After exchanged is

    The distance of the mass m from the axis is d
    The distance of the mass 2m from the axis is 2d

    I = m(d)^2 + 2m(2d)^2

    I = 8md^2

    I thought that after the masses were interchanged, the I would be 2I, the difference between before and after. The answer was wrong. I think the answer might be (2/3)I but I'm not sure. help is much appreciated
     
    Last edited: Dec 2, 2011
  2. jcsd
  3. Dec 2, 2011 #2
    There's an error in your last calculation... I'=m(d)^2+2m(2d)^2=m(d)^2+8m(d)^2=9m(d)^2
    Meaning that the ratio between I' (after the masse were interchanged) and I is 9/6=3/2.
    But I'm not sure I've understood your problem...
     
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