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Problem finding the center of mass of remaining portion ,when a part of it is removed

  1. Aug 21, 2012 #1
    Suppose there's a hemisphere of radius R (say) and a right cone of same radius R but ht. R/2 is scooped out of it then i have to find the center of mass of the remaining part.

    Here's how i approached...

    clearly by symmetry, Xcm = 0

    Now, Let M be the mass of the hemisphere so,

    Density per unit volume, ρ = M/(2/3.π .r3) x 1/3.π.r2.(r/2) = M/4

    Now, Ycm of remaining portion = {M(3R/8) - M/4(R/6)}/{M-M/4} = 4R/9

    Thus, C.M of the remaining portion = (0,4R/9)

    But the result given by the source is 11R/24 from base...!!

    Now where am i wrong????
     
  2. jcsd
  3. Aug 22, 2012 #2

    mfb

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    Staff: Mentor

    Re: problem finding the center of mass of remaining portion ,when a part of it is rem

    The center of mass of the cone is at R/8.
    R*(3/8 - 1/4*1/8)/(1-1/4)=11/24 R
     
  4. Aug 22, 2012 #3
    Re: problem finding the center of mass of remaining portion ,when a part of it is rem

    Thanks a lot....
    my fault was actually i assumed the cone to be hollow but it is solid...!!
    so h/4 not h/3 is the location of its center of mass from base... got it now..!!
     
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