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Moment of inertia for a cone

  1. Oct 19, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the moment of inertia of a right circular cone of radius r and height h and mass m


    2. Relevant equations

    I = ∫r2 dm
    V = 1/3*π*r2*h

    3. The attempt at a solution
    Assume density is p

    dm = p dv
    divide both sides by dr
    dm/dr = p dv/dr

    dm/dr = p (d/dr * 1/3*π * r2*h)
    so
    dm/dr = p(2/3)*πrh
    so:
    dm = (2/3)pπrh dr

    Sub that into the moment of inertia equation

    ∫(2/3)pπrh dr = I

    I = (1/3) pπr2h
    p = m/v
    I = (1/3)(m/v)πr2h
    I = v*(m/v)
    I = m

    What am I doing wrong?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 19, 2011 #2
    Hint: Set the problem up as a double integral problem. Lay the problem out with the cone's longitudinal axis being the x-axis. Use a ring for your dV.

    If you need more hints, let me know.
     
  4. Oct 19, 2011 #3

    vela

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    You forgot the factor of r2 multiplying dm. What you found was [itex]\int dm[/itex], which unsurprisingly turns out to equal the mass of the cone.

    About what axis are you supposed to be calculating the moment of inertia?
     
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