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Center of mass of cone using cylindrical coordinates

  • Thread starter xdrgnh
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  • #1
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Homework Statement

Set up intergral expression for center of mass of a cone using cylindrical coordinates with a given height H and radius R



Homework Equations


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



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.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #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?
 
  • #3
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for r I used the square root of H^2+R^2
 
  • #4
LCKurtz
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for r I used the square root of H^2+R^2
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.
 
  • #5
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I used that for the dr for the dz I used H/R(r) as the lower limit.
 
  • #6
LCKurtz
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I used that for the dr for the dz I used H/R(r) as the lower limit.
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.
 
  • #7
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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_{(H/R)r}^{H}\int_{0}^{2∏}\int_{0}^{R} (...)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.
[tex]\int_{0}^{2∏}\int_{0}^{R}\int_{(H/R)r}^{H} (...) zrdzdrdθ[/tex]
 
  • #8
LCKurtz
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[tex]\int_{0}^{2∏}\int_{0}^{R}\int_{(H/R)r}^{H} zrdzdrdθ[/tex]
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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