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Center of mass of solid cone

  1. Oct 9, 2012 #1
    I am using the textbook called Classical Mechanics by John R. Taylor.

    Z = 1/M ∫ ρ z dV = ρ/M ∫ z dx dy dz

    On page 89, example 3.2, it says:

    "For any given z, the integral over x and y runs over a circle of radius r = Rz / h, giving a factor of πr2 = πR2z2 / h2."

    I wish the book would show the steps. Can someone please help me understand this? I want to know what the limits of integration for dx and dy are.
     
  2. jcsd
  3. Oct 9, 2012 #2
    I think I need to relearn the whole concept of integration again... let me give this a try.

    At height h the radius of the cone is Rz / h. That is the radius of each "disk" stacked on top of another to form the cone.

    The area of that disk with height dz is πR2z2 / h2 .... the z2 stays inside the integral and we get ∫ z3 dz...

    But I want to know what the limits for dx and dy were.
     
  4. Oct 10, 2012 #3
    Okay I figured it out

    dx dy => r dr dθ

    Limits for dr is 0 up to R..... dθ is 0 to 2π
     
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