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Moment of Inertia of a Hemispherical Shell

  1. Dec 31, 2011 #1
    Prompt: Find the moment of inertia about the z-axis of the hemispherical shell of problem II-6.

    Additional Info.:

    -Problem II-6 states: the distribution of mass on the hemispherical shell z=(R^2-x^2-y^2)^1/2 is given by o(x,y,z) = (o/R^2)(x^2+y^2) where σ0 is a constant. Find an expression in terms of o and R for the total mass of the shell.

    -My solution to II-6 (verified by the text): 4(pi)R2σ0/3.

    Solution Attempts:


    1. Looked up a formula for the moment of inertia of a hemisphere, 2mr^(2)/5, where m is mass and r is radius. This gives us 8(pi)R^(4)o/15, which is off by a factor of 2.

    2. Found another formula ∫ r^2 dm. Simplified ∫ dm into σ0R2∫∫sin3ϕ dϕ dθ. Took r to mean distance from the z-axis and not radius (it was not specified where I found this formula). Let r=psinϕ, or in this case r=Rsinϕ. Determined ∫ r^2 dm = σ0R4∫∫sin5ϕ dϕ dθ. Tried to integrate this and got:
    -5 Cos[x]...5 Cos[3 x]...Cos[5 x]
    ---------- + ----------- - ----------
    ....8...............48..............80
    over the integral [O,pi/2], which equals zero.

    Real Solution: 16(pi)R^(4)o/15
     
  2. jcsd
  3. Dec 31, 2011 #2

    vela

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    It looks like you just integrated incorrectly. Mathematica says
    $$\int_0^{2\pi}\int_0^{\pi/2} \sin^5\phi \,d\phi\,d\theta = \frac{16\pi}{15}$$which is what you want.

    Actually, it looks like you integrated correctly but mistakenly thought cos 0 is 0, but it's equal to 1.
     
  4. Dec 31, 2011 #3
    cos(0) is equal to 1? No way! Haha, thanks very much vela. A teacher of mine once said a mathematician spends most of his time searching for that missing factor of 2.
     
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