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Moment Of Inertia Integration

  1. Nov 14, 2009 #1
    1. The problem statement, all variables and given/known data
    Use integration to determine the moment of inertia of a right circular homogeneous solid cone of height H, base radius R, and mass density [tex]\rho[/tex] about its symmetry axis

    2. Relevant equations
    Volume of cone = [tex]1/3*pi*r^2*h[/tex]
    I = [tex]\int r^2 dm[/tex]
    [tex]\rho = m/v[/tex]

    3. The attempt at a solution
    [tex]m=\rho v[/tex]
    [tex]m=1/3 \rho \pi r^2 h[/tex]
    [tex]dm=2/3 \rho \pi r h dr[/tex]

    [tex]\rho = m/v[/tex]
    [tex]\rho = \frac{m}{(1/3 \pi r^2 h)}[/tex]

    [tex]I=\int r^2 dm[/tex]
    [tex]I=\int \rho \frac{2}{3} \pi r^3 h dr[/tex]
    [tex]I = \frac{1}{6} \rho \pi r^4 h dr[/tex]
    (now substituting rho out)
    [tex]I = \frac{m \pi r^4 h}{2 \pi r^2 h}[/tex]
    MY ANSWER: [tex]I = \frac{mr^2}{2}[/tex]
    CORRECT ANSWER: [tex]\frac{3mr^2}{10}[/tex]

    I can't figure out where my error is
     
  2. jcsd
  3. Nov 14, 2009 #2
    Not completely sure, and i didn't look too intensely, but shouldn't the [tex]\frac{2}{3}[/tex] in [tex] I=\int\rho\frac{2}{3}\pi r^{3}hdr [/tex] be [tex] \frac{1}{3}[/tex]? Once again, I could be wrong.
     
  4. Nov 14, 2009 #3

    rl.bhat

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    Homework Helper

    Consider a disc ofmass m, radius r,thickness dh at a distance h from the vertex of the cone.
    Its moment of inertia about the axis of symmetry is 1/2*m*r^2.----(1)
    Volume of the disc = π*r^2*dh
    Density of the cone ρ= M/(1/3*πR^2*H) = 3M/π*R^2*H
    There mass of disc m = (3M/π*R^2*H)* π*r^2*dh -------(2)
    From the simple geometry it can be shown that
    h/H = r/R or dh/H = dr/R or dh = (H/R)*dr
    Substitute the value of dh in eq(2) and substitute the value of m in eq.(1).
    Now find the integration between the limits r = o to r = R.
     
  5. Nov 15, 2009 #4
    Thanks! That explains it
     
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