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Centre of mass of a solid hemisphere.

  1. Jan 3, 2009 #1
    Hi there, I can't get my head round how to do the math for this problem. I'm sure it's not as hard as I think...
    1. The problem statement, all variables and given/known data
    Show that the CoM of a uniform solid hemisphere of radius r lies at a distance (3/8)r from the centre of the flat face.
    You may find it convenient to regard the hemisphere as consisting of a very large number of extremely thin discs of varying radii stacked on top of each other.
    2. Relevant equations
    I'm thinking I'll need the equation for CoM, involving an integral: R=(1/M)[tex]\int[/tex]rdm. This I think can be modified to: R=(1/M)[tex]\int[/tex]r[tex]\rho[/tex]dV, where rho is the density at radius r.
    3. The attempt at a solution
    N/A

    Thanks very much for any help!
    ~Graham :wink:
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 3, 2009 #2

    Doc Al

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    Staff: Mentor

    So far, so good. To avoid confusion, I recommend you use a different variable of integration than r--let's say y. (Imagine the hemisphere axis to be along the y axis from y = 0 to y = r.) What's the volume of a thin disk located at position y?
     
  4. Jan 3, 2009 #3
    Ok, since the discs are extremely thin, I'd approximate the volume of the disc to be its area. So: A(y) = Pi.(r(y))2.
     
  5. Jan 3, 2009 #4

    tiny-tim

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    Homework Helper

    Hi Graham! :smile:
    You're ignoring the hint …

    if you use discs stacked on top of each other (with height z, say), you can go straight to integrating over dz, instead of dV. :wink:
     
  6. Jan 3, 2009 #5

    Doc Al

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    Staff: Mentor

    The disks are thin, but not zero thickness! :bugeye: Hint: dV = Area dy.
     
  7. Jan 3, 2009 #6
    Ah I think I see...so if dV = Ady, then I can put that into my original eqn. for CoM? R = (1/M)[tex]\int[/tex][tex]\rho[/tex]ydV. Then R = (Pi.[tex]\rho[/tex]/M)[tex]\int[/tex]y3dy...I think :redface:
     
  8. Jan 3, 2009 #7

    Doc Al

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    Staff: Mentor

    You're on the right track, but not quite there yet. What's the radius of a disk at position y? (Draw yourself a diagram.)
     
  9. Jan 3, 2009 #8
    yep! It worked! Thanks a lot guys, I'd have spent all afternoon trying to pick my way through :)
     
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