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Moment of inertia of hemispherical shell

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data
    The distribution of mass on the hemispherical shell:

    z=(R2-x2-y2)1/2

    is given by

    σ(x,y,z)=(σ0/R2)(x2+y2)

    where σ0 is a constant. Find the moment of inertia about the z-axis of the hemispherical shell.
    2. Relevant equations
    I=∫r2dm


    3. The attempt at a solution
    r2 is just R2. ∫dm in spherical coordinates is:

    σ0R2∫∫sin3θ dθ d(phi)

    with boundaries 0 to pi/2 for θ, and 0 to 2pi for phi. The completed definite integral representing the total mass is then:

    (4pi/3) σ0R2.

    I feel pretty confident about that, since it is given in the back of the book as the first step in the problem. What I don't understand is why the moment of inertia is anything more than just this value times another R2 to give:

    (4pi/3) σ0R4.

    The answer is supposed to be:

    (16pi/15) σ0R4.

    So, I'm guessing there's something more to the calculation of moment of inertia. I've been looking through the different derivations of moments but can't see how to get the right factor. I could use a hint.
     
  2. jcsd
  3. Oct 20, 2011 #2
    Oh, I understand why I can't do it that way. r in the moment of inertia formula is measured from the center of mass and is not equal to R in this case.
     
  4. Oct 20, 2011 #3
    What are the units of sigma0?
     
  5. Oct 20, 2011 #4
    None given, but the sigma function is a mass per area so I guess it represents an average mass density, or "baseline" mass density that then varies according to the more general function.
     
  6. Oct 20, 2011 #5
    Yay, got it.

    ∫∫r2σdA

    r in the formula for moment of inertia can be put into spherical coordinates and integrated over the whole hemisphere (also in spherical).

    σ0R4∫∫sin5dθ d(phi)

    The first integral makes the 8/15 factor needed and the second of course is 2 pi. I guess this was a fairly simple problem after all.
     
  7. Dec 28, 2011 #6
    I follow this, but do you mean to say rotational axis, not center of mass? The center of mass would be a point. If r is distance from the rotational axis then we can let r=p*sin[itex]\phi[/itex], or in this case r=R*sin[itex]\phi[/itex].

    I follow this and it makes sense, but when I integrated sin^5[itex]\phi[/itex], I got

    -5 Cos[x] 5 Cos[3 x] Cos[5 x]
    --------- + ----------- - ----------
    8.................48.................80

    which, when integrated from [o,∏/2], equals zero.
     
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