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Homework Help: Moment of Inertia for three spherical masses

  1. Apr 9, 2010 #1
    1. The problem statement, all variables and given/known data

    2) Three spherical masses are located in a plane at the positions shown in the figure below. A
    has mass 39.3 kg, B has mass 35.9 kg, and C has mass 17.6 kg.
    a5.jpg
    Calculate the moment of inertia (of the three masses) with respect to the z-axis perpendicular to the xy plane and passing through the origin. Assume the masses are point particles; e.g., neglect the contribution due to moments of inertia about their center
    of mass. Answer in units of kg X m.

    2. Relevant equations
    I = sum of mr2
    I = (2/5)mr2

    3. The attempt at a solution
    1. The problem statement, all variables and given/known data
    (39.3)(0.52) = 9.825 kg[tex]\dot{}.[/tex]m2
    (35.9)(0.52) = 8.975 kg[tex]\dot{}.[/tex]m2
    (17.6)(0.52) = 4.4 kg[tex]\dot{}.[/tex]m2
    (9.825+8.975+4.4) = 23.2


    I = (2/5)mr2
    I = (2/5)9.8252 = 3.93
    I = (2/5)8.9752=3.59
    I = (2/5)4.42=1.76
    Add them all up, I got 9.28 kg[tex]\dot{}.[/tex]m2

    AND I got it WRONG. ;(
    PLEASE HELP!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 9, 2010 #2

    Doc Al

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    Staff: Mentor

    What's the distance of each mass to the axis? You have them all set to 0.5. Why?


    You were told to "neglect the contribution due to moments of inertia about their center
    of mass". So forget about this and just treat them as point masses.
     
  4. Apr 9, 2010 #3
    What do you mean treat them as point masses? How do I do that? Thanks.
     
  5. Apr 9, 2010 #4

    Doc Al

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    Staff: Mentor

    Use this:
    Not this:
     
  6. Apr 9, 2010 #5
    so my answer would just be 23.2? What's wrong with my radius? Isn't it 1/2 = 0.5? Thanks.
     
  7. Apr 9, 2010 #6

    Doc Al

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    Staff: Mentor

    No.
    r is the distance from the axis for each mass. Where do you get 0.5 from?
     
  8. Apr 9, 2010 #7
    Would you use the Distance formula to find the radius between the Z Axis and the Point?
     
  9. Apr 9, 2010 #8
    don't know if this is right but have you thought of finding the center of mass then from that (x,y) coordinate use the parallel axis theorem Inew= Icm+h^2*(total mass) where h is the distance between the new axis of rotation and the center of mass.

    but then again it did say, "neglect the contribution due to moments of inertia about their center of mass"
    and the problem says to treat the particles as point masses so no need for the sphere thing,

    so then basically, sumI= mr^2+m2r2^2+....
     
  10. Apr 9, 2010 #9
    I don't think you need to worry about using the parallel axis theorem, since there wasn't an actual change of axis. I think you second part of your answer is correct, because you are just finding the moment of inertia of the system.
     
  11. Apr 9, 2010 #10
    agreed :) just make sure for the r to use the sqrt(x^2+y^2) for the radius since the question would be to rotate everything around the origin, so they will be following a constant radius.
     
  12. Apr 10, 2010 #11
    I did d or r = sqrt of x2+y2 for A, B, and C
    A = sqrt of -22+2.52 = 3.202
    B = sqrt of 12+.52 = 1.118
    C = sqrt of 4.52+22 = 4.924
    then I used
    I = sum of mv2 = m1r1+ m2r2 =m3r3
    so I added them all up and got 874.50
    Is this reasonable? Am I doing it right?
    Thanks guys. ;)
     
  13. Apr 10, 2010 #12

    Doc Al

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    Staff: Mentor

    Good!
    Too many typos here. I assume you did I = Σmr².
    Looks like you did the right thing.
     
  14. Apr 10, 2010 #13
    Re: [SOLVED]Moment of Inertia



    opps, sorry for the typos. YAY, I got it right.
     
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