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I'm quite accustomed with integration and all those calculus involved in finding moment of inertia as well as center of mass. But a random thought is wiggling in my mind. Why do we take $dm$ instead of $dr$ in finding moment of inertia and center of masses of continuous bodies?

DanielSank
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Roger Michealson
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1 Answers1

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I take it that your $dm$ means $$ \int_V r^2 dm = \int_V r^2 \rho(\vec{r}) dV $$

And you might be thinking why it wouldn't look something like $$ \int_0^r m(r) d(r^2) = \int_0^r m(r) 2r dr $$

It's quite clear they are not the same, one is a 3 dimensional integral while the other is one. I suppose one could define a "moment of inertial within distance r" to make the second equation work out to the same answer but that would miss the point.

The point is, as with many physical quantities defined as an integral over a continuous space, to understand the quantity measured on an infinitesimal element first. For a very small block of mass $m$ far away at a distance r, the moment of inertial is defined as $$ I=m r^2 $$ Now, we all know that an integral is nothing but a sum of many infinitesimal elements. But what is it that is divided into many small elements? Is it r? No, the block is far away and r is a large number! Is it m? Yes, because the block is very small! To turn the above equation into an infinitesimal element ready to be integrated, only one version make sense. $$ dI=r^2 dm $$ While the other version $$ dI=m d(r^2) $$ doesn't mean what you want to calculate. It's inconceivable as a small element and you don't know what it means to add it up.

Liuke LYU
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