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Finding moment of inertia of cone

  1. Feb 17, 2017 #1
    1. The problem statement, all variables and given/known data
    Why is it that when you integrate to find the moment of inertia of a cone standing on its vertex (like a spinning top) with height h mass M and radius R do you integrate the R limits as 0 to (R/h)z in the triple integral (cylindrical coordinates) below?
    2. Relevant equations
    I = moment of inertia
    D = density (M/πR2h)
    ρ = R distance from rotation axis (limits from 0 to (R\h)z)
    φ = 2π the angle swept (limits 0 to 2π)
    z = h the height of the cone (limits 0 to h)


    3. The attempt at a solution
    I = ∫ρ2dm = D∫∫∫ρ3dρdφdz
     
  2. jcsd
  3. Feb 17, 2017 #2

    haruspex

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    I assume you mean the r limits, not R limits.
    It depends on the order of integration. If the integration wrt r is the last step then the range is 0 to R. If it is an earlier step then the maximum value of r is constrained by the current value of z in the outer integral.
     
  4. Feb 17, 2017 #3
    Yeah this is the case but I don't understand where it came from. Why doesn't it involve sines? How do you see it is (R/h)z from the picture? For the problem is was given as capital R for radius
     
  5. Feb 17, 2017 #4

    haruspex

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    R is the maximum radius, i.e. the radius at height h. For the integral, you need a variable for the radius at height z. r seems a reasonable choice.
    If the angle of the cone (slope to vertical) is θ then tan(θ) = R/h = r/z.
     
  6. Feb 18, 2017 #5
    Okay thanks
     
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