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Moment of inertia of a spherical shell

  1. Nov 27, 2013 #1
    Hey. There's one thing I've always been wondering about when it comes to deriving the expression for the moment of inertia of a spherical shell.

    Namely, why is the length of the infinitesimal cylinder used in the derivations (like here ) equal to ##R d \theta##, instead of ##R d \theta \cdot \sin(\theta)##? Afterall, ##R d\theta## isn't the actual height of the cylinder. The vertical component (the component in the direction of the axis of rotation) of ##R d\theta## is the length, which equals to ##R d \theta \cdot \sin(\theta)##.
     
  2. jcsd
  3. Nov 27, 2013 #2

    D H

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    That's not the best of derivations.

    They are accounting for the distance between the band and the axis of revolution. The problem with the derivation is that the authors are switching back and forth between cartesian (dx) and spherical (dθ). Choose one and stick with it!

    Using dx, that little band is a spherical segment. Regardless of location on a sphere, the area of a spherical segment of height h on a sphere of radius R is ##A = 2\pi R h##, or ##dA = 2\pi R dx## in the case of an infinitesimal segment. The mass of that infinitesimal segment is ##dm = \rho dA = \frac{M}{4 \pi R^2}\,2\pi R\,dx = \frac M {2R} dx## . The distance to the axis is given by ##r^2=R^2-x^2##, and thus the moment of inertial is ##\int_{-R}^R (R^2-x^2) \frac M {2R} dx = \frac 2 3 MR^2## .

    Using dθ, that little band is the surface of an open spherical sector. The area of that infinitesimal sector is ##dA = 2\pi R\sin\theta\, R\,d\theta##, so the mass is thus ##dm = \rho dA = \frac{M}{4 \pi R^2}\, 2\pi R\sin\theta\, R \, d\theta = \frac M 2 \sin\theta\,d\theta##. The distance to the axis is ##r = R\sin\theta## and thus the moment of inertial is ##\int_0^{\pi} (R^2\sin^2\theta)\,\frac M 2 \sin\theta\,d\theta = \frac 2 3 MR^2## .
     
  4. Nov 27, 2013 #3
    oh I'm such an idiot. of course! thanks, I see it now :) It's because the infinitesimal cylinder rolled up must have a length ##R d \theta## for it to cover the surface area of the circle sector.

    thanks :)
     
  5. Sep 8, 2016 #4
    Really Thnx for your explication, but I can't understand why dA is equal to 2πRsinθRdθ and why are you using cilindrical coordenates instead of spherical.

    I've tried to solv for dA = sinϕdϕdθ and what i've got was ρ4 π r²
     
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