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Problem in moment of inertia

  1. Dec 20, 2007 #1
    1. The problem statement, all variables and given/known data
    a table teennis racket has a handle of mass m and length l and a flat circular disc of radius r and mass m attached to the handle . the moment of inertia of the racket about an axis perpendicular to its plane and passing through its center of mass of is .....?

    2. Relevant equations
    i think probably the eqs of MI of a thin rod of length l , and of a circular disc.

    3. The attempt at a solution
    i thought that by taking out the moment of inertia of rod and circular disc separately i will b able to solve it by later adding the answers . but i ended up with this instead [tex]\frac{ML^2}{3}[/tex] + [tex]\frac{3MR^2}{2}[/tex]

    whereas the answer is :
    [tex]\frac{ML^2}{12}[/tex] +[tex]\frac{MR^2}{2}[/tex] + [tex]\frac{M(L+2R)^2}{8}[/tex]
     
    Last edited: Dec 20, 2007
  2. jcsd
  3. Dec 20, 2007 #2

    Shooting Star

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    Use parallel axis theorem. Isn't anything mentioned about the shape of the handle? Then you have to take it as a thin rod.
     
  4. Dec 20, 2007 #3

    Dick

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    I don't know how you got your first answer but in the second one the first and second terms are the contributions from the rod and the disk and the last comes from the parallel axis theorem. Find the distance of the center of mass of each from the common center of mass. Hi, Shooting star. Nice medal!!
     
    Last edited: Dec 20, 2007
  5. Dec 20, 2007 #4
    did everything yet the answer does not come!
     
  6. Dec 20, 2007 #5
    yeah but can u plz enlighten me on how 2 to get the common centre of mass . and plz state ur technique which i can use for other ques like this as well . thanks for the same . :)
     
  7. Dec 20, 2007 #6

    Dick

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    Pick a coordinate system. Find COM (center of mass) of each. Then the common COM is (COM1*M1+COM2*M2)/(M1+M2).
     
    Last edited: Dec 20, 2007
  8. Dec 20, 2007 #7

    Shooting Star

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    Suppose the CM lies on the handle at dist x from where the handle meets the disk. Then,

    m(r+x) = (m/L)(L-x)[(L-x)/2].

    (EDIT.)
     
    Last edited: Dec 20, 2007
  9. Dec 20, 2007 #8

    Shooting Star

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    Thank you for noticing.
     
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