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 a sphere with different methods

  1. Sep 24, 2015 #1
    Hi everyone, I am trying to find out the Moment of Inertia of a sphere which is all known to be
    I calculate this in 2 ways.
    One by triple integration and one by disk method.
    From the text books, moment of inertia should be in the form,
    dI = (r^2) dm,
    However, the text book, University Physics, propose the method of disk by suming up,
    dI = 1/2 (r^2) dm, which is the I of a extremely thin disk. It sounds logical.
    However, I just think in a way that the equation,
    dI = (r^2) dm,
    must be suitable for any situations.
    Thus, in the 2 ways, I just start by this and found that the triple integration way gives the right answer. But the disk method way just provide a doubled result.
    Can anyone try me why this is happened???
    Why, in the disk method, we should think in the way that suming up the dI of all thin disks, but not just start from the original equation, then
    dm = (ro) dV = (ro) (pi) (r^2) dz
    Treating the dV as the volume of a thin disk and add them up???

    Attached Files:

  2. jcsd
  3. Sep 24, 2015 #2


    User Avatar
    Gold Member

    Its easy if you taking thin hemispherical shells inside the sphere instead of taking disks.
  4. Sep 24, 2015 #3


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    Think of a semicircle ##y=\sqrt{r^2-x^2}## in the ##xy## plane of a Cartesian coordinate system. Then you can think of the sphere as the body swept out by rotating this semicircle around the ##x## axis. From the result you've given I assume that (a) you want the moment of inertia around a diameter of the sphere (you must always tell the rotation axis for which you want to calculate the moment of inertia) and that (b) the mass distribution of the sphere is homogeneous.

    Now for ##\mathrm{d}m## you can use infinitesimal zylinders, given by some ##\mathrm{d} x## in the above "construction" of the sphere as a rotating body around each ##x \in (-r,r)##.

    To proceed further we need the moment of inertia for an infinitesimally thin homogeneous disk of radius ##\rho## around an axis perpendicular to it going through its center. This is easy, because for that disk your
    $$\mathrm{d} m = \frac{m}{\pi \rho^2} \rho \mathrm{d} \rho \mathrm{d} \varphi.$$
    So we have
    $$I_{\text{disk}}=\frac{m}{\pi \rho^2} \int_{0}^{\mathrm{\rho}} \mathrm{d} \rho' \int_0^{2 \pi} \mathrm{d} \varphi \rho'^3 = \frac{m}{\pi \rho^2} 2 \pi \frac{\rho^4}{4} = \frac{m}{2} \rho^2.$$
    Now back to the sphere. You just have to add up all the infinitesimal disks to get the total moment of inertia.

    The mass ##\mathrm{d}m## of the disk around ##x## is
    $$\mathrm{d} m=\frac{m}{4 \pi r^3/3} \mathrm{d} x \pi \rho^2=\frac{3m}{4 r^3} \mathrm{d} x (r^2-x^2),$$
    since the radius of this disk is given by ##\rho=\sqrt{r^2-x^2}##. So the moment of inertia of the sphere is
    $$I_{\text{sphere}}=\int_{x=-r}^{x=r} \frac{\mathrm{d} m}{2} (r^2-x^2)=\frac{3m}{8r^3} \int_{-r}^r \mathrm{d} x (r^2-x^2)^2=\frac{2mr^2}{5}.$$
  5. Sep 24, 2015 #4
    The result is double when you use disks because you used the wrong formula for the moment of inertia of these tiny disks.
    The moment of a disk is 1/2mr^2 and not mr^2.
    If you take a infinitely thin disk, of mass dm, the moment of inertia is still (dm r^2)/2 where r is the radius of the disk. This is so because not all the parts of the disk are at the distance r from the axis.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook