1. Not finding help here? Sign up for a free 30min 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 help needed

  1. Nov 16, 2008 #1

    fluidistic

    User Avatar
    Gold Member

    I've been trying to calculate the moment of inertia (with respect to the center of mass) of several rigid bodies (including 2 dimensional ones) but I never reached any good answer.
    For example a cube whose edges are equal to [tex]a[/tex].
    My work : From wikipedia and at least 2 physics books, [tex]I_{CM}=\int r^2 dm[/tex]. From my notes it's equal to [tex]\int _{\Omega} \rho \zeta ^2 dV[/tex].
    My alone work : [tex]dV=dxdydz[/tex] and as [tex]\rho[/tex] (the density) is constant I can write [tex]I_{CM}=\rho \int_{-\frac{a}{2}}^{\frac{a}{2}} x^2 dx \cdot \int_{-\frac{a}{2}}^{\frac{a}{2}} y^2 dy \cdot \int_{-\frac{a}{2}}^{\frac{a}{2}}z^2dz[/tex]. Calculating this I get that [tex]I_{CM}=\rho \left( \frac{a^3}{12^3} \right)=\frac{M}{V}\cdot \frac{V}{12^3}=\frac{M}{12^3}[/tex].
    I never took calculus III yet so I never dealt with triple integrals and even double ones. I'm guessing that I'm multiplying the 3 integrals and that instead I should be adding them or something like that, but I don't know why at all. I'd be glad if you could help me.
    P.S. : I considered the origin of the system as being the center of mass of the cube.
     
  2. jcsd
  3. Nov 16, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    moment of inertia must be measured about an axis

    Hi fluidistic! :smile:

    (have an integral: ∫ and a rho: ρ and a squared: ² :wink:)
    i] r² = x² + y² + z², not times, so your integral would be

    ∫∫∫x²dxdydz + ∫∫∫y²dxdydz + ∫∫∫z²dxdydz

    (but that's still wrong, because:)

    ii] moment of inertia must be measured about an axis

    I expect you've been misled by 2D questions which ask for the moment of inertia "about a point" … but they really mean about the axis coming straight out of the page through that point …

    so the integral for the z-axis, say, would be ∫∫∫x²dxdydz + ∫∫∫y²dxdydz :smile:
     
  4. Nov 16, 2008 #3

    fluidistic

    User Avatar
    Gold Member

    Thank you very much tiny-tim.
    I haven't been mislead by anything but myself.
    I'll try to do it and do a lot of more complicated cases. I really like this part of Physics.
    EDIT : I'm not sure I got it right. Can you confirm if the answer is [tex]\frac{Ma^2}{6}[/tex]. I had not to forget about [tex]\rho[/tex].
     
    Last edited: Nov 16, 2008
  5. Nov 16, 2008 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

  6. Nov 16, 2008 #5

    fluidistic

    User Avatar
    Gold Member

    Thank you very much once again.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?