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I had a lab report to do based around moments of inertia and thought I'd remind myself of the parallel axis theorem, so I looked up the derivation. It goes something like:

$$ I= \int [(x+D)^2 + y^2]dm , $$

$$I = \int (x^2+y^2) dm + D^2\int dm + 2D\int xdm ,$$

where the last term is equal to zero. I was confused by this so I decided to look around a bit and none of the explanations have made sense to me. They explain that this term is equal to zero because the $x$ component of the center of mass is zero, if we have our origin as the center of mass. I can understand that, however, wouldn't that indicate that the x squared should also be equal to zero? I suppose the problem here is me not understanding exactly what these terms are saying.

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$\int xdm=0$ by definition of center of mass, you are right. However, $\int x^2 dm\neq 0$ since it is a summation of positive (or rather non-negative) terms, $x^2\geq 0$.

Example: imagine two points with mass $m$ located at $x=\pm1$. Their center of mass is at $x=0$. Then $\int xdm=-m+m=0$, and $\int x^2dm=m+m=2m$.

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  • $\begingroup$ Alright, so it's because all of these are actually definite integrals, correct?. And therefore, as you said, the x^2 integral is positive. Thank you! This makes much more sense than the explanations I read. It's not really that the x-component is zero, rather the sum of all the mass-components to the right is equal to the sum of all the mass-components to the left. $\endgroup$ – agaminon Apr 2 at 16:48
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The integral is similar to a sum, as $dx$ approaches zero.

$\int x dm$

does add up to zero at the centre of mass. If we take each $dm$ element as $1$, then it's similar to

$1+2+3+(-1)+(-2)+(-3) = 0$

(depending on what the $x$ and $dm$ weightings are, but the point is they cancel out).

$\int x^2 dm$

is different, it's simlar to adding like this $1^2+2^2+3^2+(-1)^2+(-2)^2+(-3)^2 = 28$

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