Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Uniform Solid Sphere

  1. Dec 2, 2011 #1
    Find the moment of inertia of a uniform solid sphere of mass,m and radius,a about an axis through its centre.
    I have tried to solve it but I get the different answer, I don't know where I have done mistake. Please! check and correct my solution below:-
    Consider a volume element, dv of the sphere; this has mass,mdv/(4/3)∏a^3.
    I = ∫(mdv/(4/3)∏a^3)r^2
    Where r is the distance of the volume element, dv from the axis, and
    I is the moment of inertia
    I = m/(4/3)∏a^3∫r^2dv
    We know, the volume of the sphere,v is:
    v = (4/3)∏r^3
    dv = 4∏r^2dr
    I = 3m/(4/3)∏a^3∫r^2(4∏r^2)dr
    = 3m/a^3∫r^4dr under limit [0,a]
    = 3m/a^3[r^5/5] under [o,a]
    I = (3/5)ma^2 --------------WRONG!!!

    The correct answer is: I = (2/5)mr^2
  2. jcsd
  3. Dec 2, 2011 #2


    User Avatar
    Science Advisor

    For starters you seem to mixing r and a.
  4. Dec 2, 2011 #3

    Doc Al

    User Avatar

    Staff: Mentor

    Here r is the distance to the axis.
    Here r is the distance to the center.
  5. Dec 3, 2011 #4
    My intention was to find the volume element, dv of the sphere about the axis. But, since the axis was through its centre that's why I said
    v = (4/3)∏r^3
    Then, after differentiating w.r.t variable, I got the volume element, dv
    dv = 4∏r^2dr
    So, if I am wrong here,what should I do, so as to get the volume element,dv?
  6. Dec 3, 2011 #5


    User Avatar
    Science Advisor
    Gold Member

    I'd use cylinder coordinates [itex]\rho,\varphi,z[/itex]. The sphere is given by [itex]\rho \in [0,R][/itex], [itex]\varphi \in [0,2 \pi[[/itex], and [itex]z \in [-\sqrt{R^2-\rho^2},+\sqrt{R^2-\rho^2}[/itex]. Let [itex]n=\text{\const}[/itex] be the mass density. Then the moment of inertia is

    [tex]\Theta=n \int_0^{R} \mathrm{d} \rho \int_0^{2 \pi} \mathrm{d} \varphi \int_{-\sqrt{R^2-\rho^2}}^{\sqrt{R^2-\rho^2}} \mathrm{d} z \; \rho^3.

    There one factor of [itex]\rho[/itex] comes from the volume element [itex]\mathrm{d}^3 x=\rho \mathrm{d} \rho \mathrm{d} \varphi \mathrm{d} z[/itex].

    The integrals over [itex]z[/itex] and [itex]\varphi[/itex] are trivial. You find

    [tex]\Theta=4 \pi n \int_0^R \mathrm{d} \rho \; \rho^3 \sqrt{R-\rho^2}.[/tex]

    This remaining integral becomes

    [tex]\Theta=\frac{8 \Pi}{15} n R^5 = \frac{8 \pi}{15} \frac{3m}{4 \pi R^3} R^5=\frac{2}{5} m R^2.[/tex]
  7. Dec 3, 2011 #6
    Thanks Vanhee71 for your solution although it is the new idea to me.
  8. Dec 3, 2011 #7

    Doc Al

    User Avatar

    Staff: Mentor

    Unfortunately, vanhees71 just did the problem for you which deprives you of figuring it out on your own.

    The key point is to choose a volume element for which you can easily write down the moment of inertia. You chose spherical shells. (And in the process, mixed up the radius of those shells with the distance of a mass element to the axis of rotation.) You could do the problem that way, but you'd have to know the moment of inertia of a spherical shell. The better way is use cylindrical shells, which are easy. What's the moment of inertia of a cylindrical shell?
  9. Dec 3, 2011 #8
    The moment of inertia of the Cylindrical shell is:
    I = MR^2
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook