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Center of mass of a solid hemisphere (where's the error)?

  1. Aug 21, 2012 #1
    On finding the center of mass of a solid hemisphere i came up with some different result.

    Here's what i did...

    consider a small ring at a distance r from the center of the hemisphere and one more ring at a distance
    of r+dr from center of the ring.

    let, mass of the small element formed by the two rings be dm
    so,

    dm = http://data.artofproblemsolving.com/images/latex/1/8/9/189c6482a7f0b088a7d84607b3df5837533486eb.gif [Broken]

    Clearly by symmetry w.r.t origin taken Xcm = 0

    now, Ycm = 1/M http://data.artofproblemsolving.com/images/latex/f/4/e/f4e81964f7df0126c6c05827f0d5f0b972fc5f45.gif [Broken] = 1/M http://data.artofproblemsolving.com/images/latex/9/a/a/9aabc0767384d5b00970c757ff71f80cd37d11eb.gif [Broken]

    but we all know this is maybe absurd... its 3R/8 actually....

    don't give me an alternate way..... i know it gives the right result...
    but where am i wrong in my soln?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 21, 2012 #2

    Jano L.

    User Avatar
    Gold Member

    You made a little error: the thickness of the disk is not

    $$
    r d\theta
    $$

    but

    $$
    r\sin \theta d\theta
    $$
    - the disk has its height in y direction.
     
  4. Aug 21, 2012 #3
    ya... got it.. but the thickness of the small element is rcosθdθ and not rsinθdθ maybe. I am quite sure as it gave me the correct result :) !!
     
  5. Aug 22, 2012 #4

    Jano L.

    User Avatar
    Gold Member

    Ah, you're right, it's cos theta.
     
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