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Moment of Inertia of a non-uniform density paraboloid

  1. Dec 20, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the moment of inertia of a paraboloid f(x,y)=x^2+y^2 whose density function is ρ(r)=cr=dm/dv. use mass M and height H to express your answer

    2. The attempt at a solution

    I took the double integral ∫∫r^2 ρ r dr dθ to find the I of a single disk as a function of r

    Beyond that I need some help.
     
  2. jcsd
  3. Dec 21, 2011 #2

    ehild

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    Is it a three dimensional or a two-dimensional object?
    Explain the notations. What is f(x,y)? What are r and theta?

    ehild
     
  4. Dec 21, 2011 #3

    gneill

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    It appears to be a standard notation. If f(x,y) is plotted on the z-axis then the object is the surface of revolution of a parabola (a cross section of the object in any plane containing the z-axis is a parabola).

    I'd suggest switching to cylindrical coordinates and expressing r in terms of z (or h since the problem wants the results in terms of Height and Mass). Integrate over h. A mass element is then a ring of radius [itex] \sqrt{h} [/itex] with the appropriate density for that radius.
     
  5. Dec 21, 2011 #4

    ehild

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    Gneill, I know it! I asked the OP and wanted him to find it out.

    ehild
     
  6. Dec 21, 2011 #5

    gneill

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    My apologies. Sorry for putting my muddy footprints all over your nice clean thread :blushing:
     
  7. Dec 21, 2011 #6
    Yes, it is standard notation. Sorry if that wasn't clear. I integrated with z=r^2, dz=2rdr

    I found c to be (5M)/(2πr^5). Final result was ()MH, can anyone verify this for me?

    ∫∫r^2 cr r dr dθ =
    c∫∫r^4 dr dθ =
    (c2πr^5)/5
    ∫(c2πr^5)/5 2rdr =
    (c4πr^7)/35

    M=∫ρ dv
    V=∫πr^2 dz = ∫πr^2 2r dr = (1/2)πr^4
    dv=2πr^3dr
    M = ∫2πcr^4dr = (2πcr^5)/5
    c = 5M/(2πr^5)

    Plugging in c and setting r^2 = H
    (2/7)MH
     
  8. Dec 21, 2011 #7

    ehild

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    You get both I and M by volume integral, integrating with respect to θ, r and z between appropriate bounds. I do not see the bounds of integration. Make them clear, please. I do not think your result for M is correct.


    ehild
     
  9. Dec 21, 2011 #8
    Idisk = [itex]
    \int_0^{2\pi}\int_0^r r^2\rho\,r\,\mathrm{d}r\,\mathrm{d}\theta
    [/itex]

    M = [itex]\int_0^V\rho\,\mathrm{d}v[/itex]

    Change of variable to r gives different bound in the next part

    M = [itex]2\pi\int_0^r\rho\,r^3\,\mathrm{d}r[/itex]

    which I think holds true with the relationship z=r2
     
  10. Dec 22, 2011 #9

    ehild

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    No, it is not true. z=r2 is true only for one of the boundary surfaces of the paraboloid, not for the whole volume.

    You need the mass and the moment of inertia of a paraboloid - a three dimensional object, bounded by surfaces f(x,y)=x^2+y^2 and g(x,y)=H.
    If you integrate with respect to r , the bound of the definite integral can not be r. It has to be a specified value.
    If you use cylindrical system of coordinates, the volume element is dV=rdrdθdz and
    [tex]M=\int_0 ^{r_{max}}\int_{z_1(r)}^{z_2(r)} \int_0 ^{2 \pi}{\rho r d\theta dz dr }[/tex]

    Find the proper bounds z1, z2 and rmax. As the integrand depends only on r,

    [tex]M=\int_0 ^{r_{max}}{\rho r \left( \int_{z_1(r)}^{z_2(r)}{dz}\right)dr} \int_0 ^{2 \pi}{d\theta }=2\pi \int_0 ^{r_{max}}{ \rho(r) r \left(z_1(r)-z_2(r)\right) dr}[/tex]

    ehild
     
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