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When we are to calculate the moment of inertial of an object, we basically sum all the infinitesimal moments of inertial of all small elements, so since $m_i /V_i = \sigma$ where $\sigma$ is the density of that small element, which will be constant for the time being.Therefore

$$dm r^2 = \sigma dVr^2$$.For an sphere, $dV = r^2 sin(\phi) dr d\phi d\theta$, and hence

$$I = \frac{\int_0^{2\pi} \int_0^\pi \int_0^R \sigma r^4 sin(\phi) dr d\phi d\theta}{\int_0^{2\pi} \int_0^\pi \int_0^R r^2 sin(\phi) dr d\phi d\theta}$$

But after calculation these integrals, I get

$$I = \frac{3MR^2}{5}$$.

I check many times that there is no problem in the calculation, but I can not find the problem in the construction since every source that I looked says that

$$I = \frac{2MR^2}{5}$$.

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  • $\begingroup$ Solid sphere or hollow sphere? $\endgroup$
    – SchrodingersCat
    Dec 11, 2016 at 12:04
  • $\begingroup$ @SchrodingersCat As you can see from the limits of the integral, it is a solid sphere. $\endgroup$
    – Our
    Dec 11, 2016 at 12:05
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    $\begingroup$ With this calculation you are solving the moment of inertia of the sphere with respect to its centre, not with respect to an axis. Since your distance $r$ in $dm\,r^2$ is measured with respect to the centre. Instead you have to write this differential as $dm(r\sin{\phi})^2$ $\endgroup$
    – HBR
    Dec 11, 2016 at 12:26
  • $\begingroup$ @HBR That is what I needed, thanks. $\endgroup$
    – Our
    Dec 11, 2016 at 12:31
  • $\begingroup$ your are welcome @Leth!! $\endgroup$
    – HBR
    Dec 11, 2016 at 12:32

1 Answer 1

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In order to find the moment of inertia of a solid sphere with respect to the $z$-axis you should integrate: $$I = \iiint_B (x^2+y^2) (\sigma dV)=M\frac{\iiint_B (x^2+y^2) dV}{\iiint_B dV}\\ =M\frac{\int_0^{2\pi} \int_0^\pi \int_0^R (r \sin(\phi))^2\cdot r^2\sin(\phi) dr d\phi d\theta}{\int_0^{2\pi} \int_0^\pi \int_0^R r^2 \sin(\phi) dr d\phi d\theta}=\frac{2MR^2}{5}.$$

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  • $\begingroup$ I could understand if you state what $z$ is and why we used $z^2$ ? $\endgroup$
    – Our
    Dec 11, 2016 at 12:21
  • $\begingroup$ But for a sphere, $x^2 + y^2 + z^2 = R^2$, so instead of $x^2 + y^2 $, shouldn't we use $R^2 - z^2$ ? $\endgroup$
    – Our
    Dec 11, 2016 at 12:27
  • $\begingroup$ by the way, it should be $\cos(\phi) $ not $sin(\phi)$ $\endgroup$
    – Our
    Dec 11, 2016 at 12:29
  • $\begingroup$ You have to integrate the square distance from the $z$-axis which is $x^2+y^2$. $\endgroup$
    – Robert Z
    Dec 11, 2016 at 12:29
  • $\begingroup$ Ok, I understood. $\endgroup$
    – Our
    Dec 11, 2016 at 12:31

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