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Find the rotational inertia of a sphere suspended from the ceiling

  1. Feb 21, 2008 #1
    A solid uniform sphere of radius R and mass M has a rotational inertia about a diameter that is given by (2/5)MR2. A light string of length 2R is attached to the surface and used to suspend the sphere from the ceiling. Its rotational inertia about the point of attachment at the ceiling is
    - 47/5mR^2 (not sure how to find it)


    Three identical objects, each of mass M, are fastened to a massless rod of length L as shown. The rotational inertia about one end of the rod of this array is
    -5/4mL^2
    - again no clue how to get this
     
  2. jcsd
  3. Feb 21, 2008 #2

    Doc Al

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    Staff: Mentor

    Consider the parallel axis theorem.
     
  4. Feb 21, 2008 #3
    I looked at that stuff but I dont see how to use to solve these problems.

    For the second one I kinda guessed and got it but I dont think its the right way.

    I used 1/12ml^2 for the center mass and 1/3ml^2 for the far right mass and added it together. then since there are 3 masses I multiplied by 3 and for 5/4mL^2 but I dont think thats how i"m supposed to find it.
     
  5. Feb 21, 2008 #4

    Doc Al

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    Look at it again. For the first problem, you have the rotational inertia about the center of mass but you need it about the point of attachment. Sounds like a perfect opportunity to use the parallel axis theorem!

    For the second problem, please attach a diagram.
     
  6. Feb 21, 2008 #5
    Second problem

    [​IMG]
     
  7. Feb 21, 2008 #6

    Doc Al

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    Another perfect opportunity to apply the parallel axis theorem. Alternatively: What's the rotational inertia of a point mass M at a distance D from an axis?
     
  8. Feb 21, 2008 #7
    well for the first one I tried

    2/5mr^2 + mL^2
    2/5mr^2 + 2mL^2 and got 12/5mr^2 which is wrong...honestly I got no clue how this parallel axis thereom stuff works, if u got any tips I'd appreciate it.
     
  9. Feb 21, 2008 #8

    Doc Al

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    OK, what did you use for "L"? That should be the distance between the center of mass to the point of attachment (in terms of R, of course). (Draw yourself a diagram, unless one is provided--in which case, attach it.)
     
  10. Feb 21, 2008 #9
    I used 2R for L since thats how long it says the string is, and no diagram with the problem :(
     
  11. Feb 21, 2008 #10

    Doc Al

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    Careful: The problem says "A light string of length 2R is attached to the surface"

    Draw your own diagram! (Just for yourself.)
     
  12. Feb 21, 2008 #11
    I just dont see it...I drew the diagram, but I dont see what I'm trying to find...
     
  13. Feb 21, 2008 #12

    Doc Al

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    Your diagram should show a sphere hanging from a string attached to the ceiling. The string is length 2R and attaches to the surface of the sphere. Where's the center of the sphere?
     
  14. Feb 21, 2008 #13
    the center would be another distance R. so

    2/5mr^2 + 3mL^2 = 17/5mr^2 which is still wrong :(
     
  15. Feb 21, 2008 #14
    ah, since the length is 3R it would be M(3R)^2 so it would be 9MR^2 so..

    2/5mr^2 + 9mr^2 = 47/5mr^2 would that be the proper way to do it?
     
    Last edited: Feb 21, 2008
  16. Feb 21, 2008 #15
    For the second one...since it is rotating on the far left mass, I can disregard it since the inertia would be of the masses rotating around it right? So...

    the middle mass mr^2 = m(L/2)^2 = 1/4mL^2
    the far right mass mr^2 where r would be the length to the point of rotation so mL^2

    1/4mL^2 + mL^2 = 5/4mL^2



    I think I did the two problems the correct way, if there is anything you see shady about my method I'd love to know. THX for your patience and help lol. I think i'm starting to understand this some. Thx again! Peace.
     
  17. Feb 21, 2008 #16

    Doc Al

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    Your latest solutions to both problems are perfectly correct. Well done!
     
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