The moment of inertia of a sphere of mass $M$ and radius $R$ can be calculated exactly (meaning, with certainty) using integrals. The formula we get is $\frac{2}{3}MR^2$. However, there's an other approach: you take the moment of inertia of a solid ball of radius $R$ and substract from it the moment of inertia of a solid ball of radius $R-dr$, with the fact, that what remains (a sphere) has a mass equal to $M$, in mind. However, by using this method, you stumble accross something like $(R-dr)^5$. By convention, you expand it and neglect every component proportional to $dr^k, k>1$, because while $dr$ is really small, $dr^2$, $dr^3$ etc. are even smaller, nearly zero. By doing so you eventually get $I = \frac{2}{3}MR^2$. Now, what bugs me about it: why one method which doesn't use approximations (integration) gives us the same result as the one which neglects something? Sure, $dr^2$ is unimaginably small, but why is it perfectly reasonable to fully ignore it?

  • 4
    $\begingroup$ Are you sure that your "approximation" is not just mimicking the definition of the Riemann integral, and doing the limits in there explicitly? $\endgroup$ – ACuriousMind Apr 29 '15 at 20:01

Inertia of a solid sphere is $\frac{2}{5} m R^2$ and that of a thin shell sphere is $\frac{2}{3} m R^2$.

By subtracting one solid sphere from an another and taking $\mathrm{d}r \rightarrow 0$ you exactly calculate the integral for the shell. There is no approximation here.

A thick walled sphere with $\mathrm{d}r > 0$ has different mass moment of inertia from a thin walled sphere. This is reflected by a factor that comes out of $R^5 - (R-\mathrm{d}r)^5$.

If I assume $\mathrm{d}r > 0$ but small enough for $(\mathrm{d}r/R)^2\approx0$ (with a 1st order taylor series) I get

$$ I = \frac{2}{3} m R (R-\mathrm{d}r) $$

Now this represents the inertia of a thin walled sphere, but not inifinesimally thin. Only finite thin section.

So depending on the level of approximation you choose it represents a different thing as it relates to the relative thickness of the shell to the radius.

In real life any calculation not of thick wall is an approximation because you cannot have an a wall of zero thickness. So $\frac{2}{3} m R^2$ is never an exact value.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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