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My Derivation

I have derived moment of inertia of solid sphere along diameter but my textbook says that moment of inertia is:

$$\frac{2MR^2}{5}$$ What is the mistake in my derivation?

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closed as off-topic by ACuriousMind, John Rennie, Qmechanic Aug 1 '15 at 19:31

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  • $\begingroup$ ive been trying this for a long tym and still I am not getting any other answer $\endgroup$ – Sriram V Aug 1 '15 at 18:48
  • $\begingroup$ Your work is hard to read. Could you either take a higher-res picture, or tex it up? $\endgroup$ – Jahan Claes Aug 1 '15 at 19:05
  • $\begingroup$ yea sure.. ill upload a high res pic.. plz wait for a few mins! $\endgroup$ – Sriram V Aug 1 '15 at 19:07
  • $\begingroup$ I've made an effort to solve the problem... but why have you put it on hold? $\endgroup$ – Sriram V Aug 1 '15 at 19:34
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The mistake is in the second line, in the calculation of the differential mass element. The differential mass element in this case is a disc, of radius $r$ where $r = R \cos\theta$ as you have correctly used.

However, the thickness of this differential disc is NOT $ R d\theta$ but $Rd\theta cos\theta$. Try to wrap your head around this. $Rd\theta$ is the length of a tiny, tiny arc of radius $R$ and angle $d\theta$ and in the infinitesimal limit it can be approximated to a straight line, that is, a chord but notice that this chord is still not along the z axis (id est, the vertical axis). So, the shape that you have described is not a disc at all. It is a conical frustum instead. (You can verify this by trying to calculate the volume of the sphere using your formula. You'd see that it doesn't come to $\frac{4}{3}\pi R^3 $.) And the moment of inertia for a frustrum is not $\frac{mr^2}{2}$.

What you need to do is take the vertical projection of this chord, and that is where the $cos\theta$ comes in. Try the integration again with this new differential element and you will have landed at correct moment of inertia, $\frac{2}{5} MR^2$.

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    $\begingroup$ yea.. I integrated it and got the answer.. thank you so much! $\endgroup$ – Sriram V Aug 1 '15 at 19:31
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I can't really follow your work, but here's one way to do it.

Take the axis to be the $z$ axis. The distance of a point $\left(r,\theta,\phi\right)$ in the sphere from the $z$ axis is $r \sin \theta$, so

$$ \begin{eqnarray} I &=& \int dV \rho \left(r \sin \theta\right)^2 \\ &=& \rho \int_0^{2\pi} d\phi \int_0^\pi d\theta \int_0^R dr \ r^2 \sin \theta \ \left(r \sin \theta\right)^2 \\ &=& 2 \pi \rho \int_0^\pi d\theta \ \sin^3 \theta \int_0^R dr \ r^4 \end{eqnarray} $$

You can take it from here.

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  • $\begingroup$ I don't think the OP knows multiple integration, so it's best to steer clear of derivations from definition. $\endgroup$ – Aritra Das Aug 1 '15 at 19:32
  • $\begingroup$ I know what you mean @AritraDas, though OP is able to avoid the volume integral because he uses the moment of inertia of a cylinder. If you want to calculate the moment of inertia of a cylinder, you have to do a double integral. $\endgroup$ – Eric Angle Aug 1 '15 at 21:40
  • $\begingroup$ No, he could avoid the double integration again by using the moment of inertia of a ring and just one integration. However, a formal derivation should be as you have done it, using 3 integrations. $\endgroup$ – Aritra Das Aug 2 '15 at 5:00
  • $\begingroup$ That's true @AritraDas. $\endgroup$ – Eric Angle Aug 3 '15 at 0:02

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