Here's a concept question about basic physics.

I=∫(r^2)dm follows from lim(∑mr^2)

Why does it naturally follow that r^2 is being integrated with respect to m rather than the other way around?

Question

Here's a concept question about basic physics.

I=∫(r^2)dm follows from lim(∑mr^2)

Why does it naturally follow that r^2 is being integrated with respect to m rather than the other way around?

anonymous · Answer

I by the way is rotational inertia. Should have specified.

jamesj · Answer

Because we take a little bit of mass, $ \Delta m_i $ and we find its radius r.  Hence the moment of inertia is the sum over all those little bits of mass
$$ I = \sum_i m_i r^2(m_i) $$ and it makes sense to think of r as being a function of which piece of mass we're looking at, not the other way around.

anonymous · Answer

My main issue is why it isn't correct to imagine a little bit of radius squared, to find its mass.

I think I've always thought of the issue as you put it, I'm just wondering why the alternative is definitely incorrect. Since it shouldn't (?) evaluate out to the same thing.

jamesj · Answer

It's intuitively more obvious to build it up this way.

But if you've calculated any moments of inertia lately, you'll know that always end up figuring how to write dm as f(r) dr so you can actually integrate.

anonymous · Answer

I guess I haven't calculated any lately; just a bit of self-confusion while reviewing my old textbooks.