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Homework Help: Did I do this right or get lucky? Deriving spherical shell moment of inertia

  1. Jan 25, 2010 #1
    1. The problem statement, all variables and given/known data
    Basically just to find moment of inertia of a spherical shell


    2. Relevant equations
    Moment of Inertia = Int(r^2 dm)


    3. The attempt at a solution

    I am in 12th grade and dont know much about integrals? Did I get lucky or am I right?

    I = Int(r^2dM)

    dM= (phi)(dA)

    I = Int(r^2 phi dA)

    Pull out phi, which equals M/4piR^2

    I started by looking at half a sphere so I made that M/2piR^2. To get dA I figured you think of a bunch of hoops all going up to one point with a circumferance of 2piR, as you get higher up the radius changes however. A triangle says that R sin theta will get you the new radiuses, so for dA I wrote the integral of 2 pi R sin theta, dTheta. I put that in as dA, this R is a constant so now I have (M 2 pi R)/(2 pi R ^2) on the outside which cancels to M/R

    Now I have

    (M/R) Int(r^2 sin Theta, dTheta, dR) the R goes from 0 to R the theta from 0 to pi/2. The integral of sin Theta over that integral is 1, so it ends up

    (MR^3)/(3R)

    R's cancel

    MR^2/3

    That's half a sphere so multiply by 2

    2/3 M R^2

    Was that luck or does that work?
     
  2. jcsd
  3. Jan 25, 2010 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    I would use Cartesian co-ordinates. Put the centre of the sphere at the origin. Divide the sphere into thin rings of area [itex]dA = 2\pi y dz[/itex] and express y as a function of z.

    See this calculation for a solid sphere: http://hyperphysics.phy-astr.gsu.edu/hbase/isph.html#sph

    AM
     
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