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Z Center of mass of an ice cream cone

  1. Apr 3, 2012 #1
    1. The problem statement, all variables and given/known data
    The z com of the entire ice cream cone, the base being an inverted right circular cone, the top a hemisphere. Equal density throughout. Must solve using integrals and density relationship.


    2. Relevant equations
    The z com of the top hemisphere (ice cream) is equal to 3/8 R
    The z com of the lower section (cone) is equal to h/4


    3. The attempt at a solution
    m = density * volume
    mt/mb = (2/3*PI*R^3) / (1/3*PI*R^2*h) = 2R/h
    zcom = 1/M summation m*z
    (h/2R) = summ (2/3*PI*R^3) * (3/8*R) + (1/3*PI*R^2*h) * (h/4)

    To be honest, I have no clue what I am doing. I know I need to some how come up with a way, using the density relationship, to set up an integral and solve the total z com of the two solid components. I'm not sure if what I did above is on the right track. Please Help!
     
  2. jcsd
  3. Apr 3, 2012 #2

    BruceW

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    Homework Helper

    I'm not sure what you were calculating on the final line, but the one before that is useful. You wrote "zcom = 1/M summation m*z" Try writing this out, don't worry that you are not given the density, write it in (just use a symbol like rho), and hopefully you'll find that it cancels out.
     
  4. Apr 3, 2012 #3
    Hi BruceW,
    My last line was my summation of the two z com components. I'm really not sure where to go with it. I am supposed to set up an integral, I just do not know what to put in the integral.
     
  5. Apr 3, 2012 #4

    gneill

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    Staff: Mentor

    Your proposed solution involved using the "known" centers of mass of two geometric solids. In order to satisfy the question, it would be sufficient to derive those two centers of mass. You can do them separately.
     
  6. Apr 3, 2012 #5
    Those two separate center of masses were derived, not given. I just don't know how to use those two in order to set up an integral to solve the center of mass of them combined.
     
  7. Apr 3, 2012 #6

    gneill

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    Staff: Mentor

    There's no integration required if you already have the masses and positions of the centers of masses of two objects. Treat then like point masses at their centers of mass.
     
  8. Apr 4, 2012 #7

    BruceW

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    Hey Trish, welcome to physicsforums. gneill has the right idea for this problem. (although really you don't have the masses, but you do have the volumes, and since the density is the same, the density will cancel out).
     
  9. Apr 4, 2012 #8
    Thanks. So I am on the right track with the summation. It just really confused me that he said to use integration. Maybe he just meant on the individual parts.

    Thanks both for your help :)
     
  10. Apr 4, 2012 #9

    BruceW

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    The summation is close, but not quite right. The z values need to take account for the relative positions of the objects (and the fact that the cone is inverted), so they should not just be the z com values which were given
     
  11. Apr 4, 2012 #10
    Ok, thanks.
     
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