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?


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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, John Rennie, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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

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, $\frac{2}{5} MR^2$.

  • 1
    $\begingroup$ yea.. I integrated it and got the answer.. thank you so much! $\endgroup$ – Sriram V Aug 1 '15 at 19:31

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.

  • $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.