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Homework Help: Moment of inertia of a hollow sphere

  1. Dec 17, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the moment of inertia of a hollow sphere with mass m and radius R and uniform density

    2. Relevant equations

    Since the hollow sphere is an area, the density is mass divided by area, so:

    [tex]I = \int r^2 dm = \frac{m}{A}\int r^2 dA[/tex]

    3. The attempt at a solution

    . The total area is 4pi r^2, so here is what I got

    [tex]dA = 2\pi \sqrt{R^2-r^2}dr[/tex]

    [tex]I = \frac{m}{4\pi R^2} \int_{-R}^{R} r^2(2\pi\sqrt{R^2-r^2})dr[/tex]

    [tex]I = \frac{m}{R^2} \int_{0}^{R} r^2\sqrt{R^2-r^2}dr[/tex]

    From here I made the substitution [tex]r = R\sin{\theta}[/tex] and got

    [tex]I = mR^2 \int_{0}^{\frac{\pi}{2}} \sin^2\theta\cos^2\theta d\theta[/tex]

    And that evaluated to pi/16, which brings me to my problem

    the correct answer is supposed to be [tex]I = \frac{2mR^2}{5}[/tex]
    Last edited: Dec 17, 2007
  2. jcsd
  3. Dec 17, 2007 #2
    Is it a hollow sphere(a spherical shell) or a solid sphere? Because the answer that you say is supposed to be correct is for a solid sphere, unless I've made some error. They're both pretty similar and I think you're going to get into trouble if you don't use curvilinear coordinates, at least it was much easier for me to do it that way. My response is to a spherical shell which will not give the answer you say is correct. However, a solid sphere can be done in a similar way and does give the answer you say is correct.

    First, something to notice is that in spherical coordinates a volume element is [tex] R^2 \sin (\theta) d\theta d\phi dR[/tex]. For a spherical shell R is constant and the volume element becomes an area element [tex] dA= R^2 \sin (\theta) d\theta d\phi [/tex]. Can you now do the integral keeping in mind that in your initial given equation r is the perpendicular distance from an axis passing through the center of mass? (I assumed it was moment of inertia wrt the center of mass)
  4. Dec 17, 2007 #3
    You are right. I am sorry. I looked up the wrong answer in the chart.

    It should be [tex]\frac{2mR^2}{3}[/tex]
    I will look into the spherical coordinate method


    Yes, it worked. The error was my r was not actually representing the perpendicular distance.

    Here is what I did:

    [tex]\frac{m}{A}\int r^2 dA[/tex]

    [tex]A = 4\pi R^2[/tex]
    [tex]r = R\sin\phi[/tex] (simple spherical geometry)
    [tex]dA = 2\pi r dz = 2\pi * R\sin\phi * R d\phi[/tex] (definition of radian and the statement above)
    [tex]dA = 2\pi R^2 \sin\phi d\phi[/tex]

    plugging in

    [tex]\frac{m}{4\pi R^2} \int_{0}^{\pi} R^2\sin^2\phi * 2\pi R^2 \sin\phi d\phi[/tex]

    [tex]\frac{mR^2}{2} \int_{0}^{\pi}\sin^3\phi d\phi[/tex]

    That integral evaluates to 4/3, and bam, it works
    Thanks a lot
    Last edited: Dec 17, 2007
  5. Nov 22, 2009 #4
    How would you get 4/3 for the intergral, i used substitution rule and always got 2/3 for the answer.
    ∫_0^π▒〖sin〗^3 ϕ dϕ=∫_0^π▒〖(1-〖cos〗^2 ϕ)〗sinϕ dϕ
    u= cosϕ du=-sinϕ dϕ
    so ∫_1^0▒〖-(1-u^2 )du〗=∫_0^1▒〖1-u^2 du〗 = 2/3
    what is my problem?
  6. Nov 22, 2009 #5


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    Homework Helper

    You didn't convert the bounds of integration correctly. cos(pi)=-1, not 0.
  7. Nov 22, 2009 #6
    Thanks alot....why I am so stupid....-_-llll
  8. Jan 3, 2011 #7
    HI, I understand everything, except that in the solution, dz is written as R d\phi. why is that? shouldnt it be Rsin(phi)d(phi) or even, if we take x = R*cos(phi) (from that same simple geometry) than dz would be -Rsin(phi)d(phi)
    Last edited: Jan 3, 2011
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