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!

Moment of inertia of solid sphere

  1. Nov 17, 2009 #1
    Hi, i am trying to find the moment of inertia of a uniform density solid sphere about z-axis

    I = integrate => x^2 dm

    x = perpendicular distance from z-axis to anywhere in sphere
    so by pythagorus theorem, r^2 - z^2 = x^2

    since dm = p dV
    and V = 4/3 (//pi)r^3
    dV = 4(//pi)r^2 dr

    so I = integrate=> r^2 - z^2 pdV

    but the problem is z is a variable.

    so how do i convert z?

    assuming i put z = rcos θ,

    then i will have a θ variable now.

    i tried integrating θ from 0 to (pi) but the answer is wrong, its not 2/5mr^2

    so what should i do?

  2. jcsd
  3. Nov 17, 2009 #2

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2017 Award

    Look at your volume element.
  4. Nov 17, 2009 #3
  5. Nov 17, 2009 #4
    i have seened that, but they considered slices of circular disk

    and so they sum up the moment of inertia of each disk for the whole sphere

    but i am sure the method that i am doing can work also.. just that i don't know how
  6. Nov 17, 2009 #5
    whats wrong with the volume element?
  7. Nov 17, 2009 #6
    If you are not summing slices, what are you summing in your integration? I suppose maybe you could take ever-widening, and ever-shortening cylinders centered about an axis. The first piece would be a line, and the last would be a circle. Is that what you are doing?
    Last edited: Nov 17, 2009
  8. Nov 18, 2009 #7
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook