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Center of mass of a cone

  1. Oct 13, 2011 #1
    1. The problem statement, all variables and given/known data

    Calculate the X_com of the cone of mass M in terms of quantities given in the picture. The density of the cone is uniform.

    See the attachments for the picture.

    2. Relevant equations



    3. The attempt at a solution

    When I did it I got

    X_com = (3L)/2

    and I am unsure if this is correct or not. The infinitely small sections are circular disks of area pi R^2 were r is the radius of the infinitely small disk and a volume just an infinitely small width dx times the area. therefore the density

    rho = dm/dV = dm/(pi R^2 dx)

    I sort of got confused when I did because the radius changes with respect to x and during the middle of the test I sort of rushed this problem and thought that it just canceled out in the end but believe I may be wrong
     

    Attached Files:

  2. jcsd
  3. Oct 13, 2011 #2

    SteamKing

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    You have shown that the center of mass of the cone in the x-direction lies outside of the cone. Your answer is not correct.

    Yes, the radius changes with respect to x, but it does so in a predictable manner.

    Care to reformulate your solution to the c.o.m.? Concentrate on writing the moment equation using the origin as the reference point.
     
  4. Oct 13, 2011 #3
    what is the moment equation?
     
  5. Oct 14, 2011 #4

    SteamKing

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    It's an essential component in your c.o.m. calculation.

    d(moment)/dx = x * dm
     
  6. Dec 12, 2011 #5
    Im not exactly sure what is meant by moment in your equation. Can someone tell me. I recalculated it except this time I got 3/4 R. I believe the answer is 3/4 L though. I'm not exactly sure what I did wrong. Thanks for any help.

    sorry that a four in the denominator magically disappeared from my work towards the end and magically reappeared. I just realized this. Everything else is correct though I believe, except some reason I got 3/4 R instead of 3/4 L
     

    Attached Files:

    Last edited: Dec 12, 2011
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