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F = MA 2010 Exam # 24 (MoI after shift in CoG)

  1. Jan 28, 2013 #1
    1. The problem statement, all variables and given/known data
    http://www.aapt.org/physicsteam/2010/upload/2010_FmaSolutions.pdf
    See Number 24


    2. Relevant equations
    CoM = M1x1 + M2x2 / (M1+M2)
    MOI_disk: 1/2mr^2

    3. The attempt at a solution
    So what I was thinking was find the new CoM then use the parallel-axis theorem.
    To find the new CoM:
    MsXs + MbXb / (Ms + Mb) = 0, where s is the small unshaded part, and b is the big, shaded portion.
    MsXs = - MbXb
    Xb = -Ms/MbXs
    Xb = (-R)(ρ∏R^2) / ρ(∏(2R)^2 - ∏R^2)
    Xb = R/3

    Parallel Axis Theorem will smaller mass:
    1/2MR^2 + 1/9MR^2
    11/18MR^2

    Yeah I'm pretty lost. I feel like there is a much better solution.
     
  2. jcsd
  3. Jan 28, 2013 #2
    I like your idea of using the parallel axis theorem. Naively I might consider the moment of inertia for each circle about each circle's center. But we know that the circle cut out, it never rotated about its center. Before being cut out, it rotated about the larger circle's center. So I would start by finding the moment of inertia of the small circle rotating about the large circles axis. Does that make sense? If you get that you will have the moment of inertia for the large disk about its center and one for the small disk about the large disk's center. Of course you want neither of those, you want the moment of inertia for a large disk with a small disk cut away.
     
  4. Jan 28, 2013 #3
    Ok going along with what you said:
    The big disk is simply 1/2Mr^2
    Little disk about big disk's center:
    1/2mr^2 + m(r/2)^2
    1/2mr^2 + mr^2/4
    3/4mr^2 is MoI about the big disk center for little guy.

    But we have two different masses and if we bust out ρ it won't end up cancelling.
     
  5. Jan 28, 2013 #4
    Should be uniform density right? So 'm', the little disk's mass should be proportional to 'M' in the same way the area's of each are proportional. One square meter of the stuff always has the same mass. Figure out what the ratio of masses is and then you can eliminate 'm'.
     
  6. Jan 28, 2013 #5
    because of the area ratio: m should be 1/4M
    This makes it:
    3/32MR^2

    Subtracting 1/2MR^2 (MOI big disk) from 3/32MR^2 (MOI little disk about big disk's center) gives 13/32MR^2. Aha!

    Thanks for your help.
    By the way, in general can you find the moment of inertia this way; that is, if you have a hole and a shape, you can take the MoI of the original shape about its center and subtract the MoI of the hole around the shape?
     
  7. Jan 28, 2013 #6
    Yes, I think so. Make sure that you stick to the same axis by using the theorem. Some things that might cause a problem would be irregular shapes or non-constant mass density. Each of those would need some calculus. But if you have "nice" shapes and "nice" mass density (constant or maybe linear) then this scheme should work.

    Makes sense right? Envision spinning the disk around on a stick. Then think about spinning the little disk mounted on it's edge. Each of those has a resistance to being spun. This comes from the mass needing force and energy to get up to speed. Each little piece has its own need of force and energy, each little piece has its own contribution to the resistance to being spun, each little piece has its own moment of inertia. If you take away that piece, you take away that resistance and you literally subtract away the moment of inertia for that piece.
     
  8. Jan 28, 2013 #7

    tms

    User Avatar

    Since the moment of inertia is defined as
    [tex]I = \int r^2\,dm,[/tex]
    and since integrals are a fancy form of addition, moments of inertia are additive.
     
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