1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    R's cancel


    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

    User Avatar
    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

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook