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Center of mass of cone using cylindrical coordinates

  1. Nov 9, 2011 #1
    1. The problem statement, all variables and given/known dataSet up intergral expression for center of mass of a cone using cylindrical coordinates with a given height H and radius R



    2. Relevant equations
    rdrddθdz is part of the inter grand. M/V=D volume of cone is 1/3π(r^2)H



    3. The attempt at a solution

    dm=Kdv dv=drdθdx K is just a constant because it is uniform density and mass. ∫∫∫zdrdθdz
    z=z. Equation for cone in cylindrical coordinates is Z=(H/R)r. The top is bounded by Z=H. The angle is from o to 2π and the radius is square root of H^2+R^2. Once I take this integral I then divide by the same integral but instead there is no extra z in the inter grand. After all is set in done I get the wrong answer. I believe my initial set up might be wrong. Can anyone verify that or give a helpful suggestion thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 9, 2011 #2

    LCKurtz

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    Hard to follow but what did you use for your lower limit on the inside dz integral?

    [Edit: Woops I misread your order of integration... wait a sec...Ok, if you integrate dr first, r goes from zero to the cone. What did you use for r on the cone?
     
  4. Nov 9, 2011 #3
    for r I used the square root of H^2+R^2
     
  5. Nov 9, 2011 #4

    LCKurtz

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    But that is a constant. The radius varies as z moves up the cone. Use the equation of the cone to get r in terms of z for your lower limit.
     
  6. Nov 9, 2011 #5
    I used that for the dr for the dz I used H/R(r) as the lower limit.
     
  7. Nov 9, 2011 #6

    LCKurtz

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    Your integrals are showing r dr dθ dz in that order. In that order you would integrate z last and its limits must be constant. Order matters so what order are you using?

    [tex]\int_{lower}^{upper}\int_{lower}^{upper}\int_{lower}^{upper} (...)rd?d?d?[/tex]

    Quote this in your reply and fill in between the brackets and the question marks with what you did. Then it will be easy to see what you did wrong.
     
  8. Nov 9, 2011 #7
    [tex]\int_{0}^{2∏}\int_{0}^{R}\int_{(H/R)r}^{H} (...) zrdzdrdθ[/tex]
     
  9. Nov 9, 2011 #8

    LCKurtz

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    OK. That integral is correct for your numerator when calculating [itex]\bar z[/itex]. When you divide it by the volume you should get [itex]\bar z[/itex] providing you don't make any algebra mistakes. Good luck, I have to go now.
     
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