Physics Asked by Kuke on January 6, 2021
I just wanted to ask if this equation can be used when finding rotational inertia for the thin spherical shell. I am not sure if it is done the right way.
Since we know that rotational Inertia for three dimensional objects can be written as $I=iiint _{E} left( x^{2}+y^{2}right)cdotrholeft( x, y, zright)cdot dV$. It is true that this is rotational Inertia for the z-axis, but as this is a sphere it does not matter which axis we choose. We assume that $rholeft( x, y, zright)=rho$, so density is constant.
Then we know that surface can be computed as $A=iint _{D} left| r_{alpha}times r_{beta}right| dA$. Combining these expressions our integral looks like this: $I=rhocdotiint _{A}left(x^{2}+y^{2} right)cdot left| r_{alpha}times r_{beta}right| dA$. as $rleft( alpha,betaright)=leftlangle xleft( alpha,betaright), yleft( alpha,betaright), zleft( alpha,betaright) rightrangle$ and $r_{alpha}=leftlangledfrac{partial x}{partial alpha}, dfrac{partial y}{partial alpha}, dfrac{partial z}{partial alpha} rightrangle,$ and $r_{beta}=leftlangledfrac{partial x}{partial beta}, dfrac{partial y}{partial beta}, dfrac{partial z}{partial beta} rightrangle$.
If we use spherical coordinates, as $x=rcdot sinleft(alpha right)cdot cosleft( betaright), y=rcdot sinleft(alpha right)cdot sinleft( betaright), z=rcdotcosleft(alpha right)$.
Then $left| r_{alpha}times r_{beta}right|= r^{2}cdotsinleft( alpharight)$, and $left(x^{2} + y^{2}right)=r^{2}cdotsinleft(alpha right)^{2}$.
Our integral then transforms to $I=rhocdotint _{0}^{2pi}int_{0}^{pi} r^{2}cdotsinleft( alpharight)cdot r^{2}cdotsin^{2}left( alpharight) dalpha dbeta=rhocdotint _{0}^{2pi}int_{0}^{pi} r^{4}cdotsin^{3}left( alpharight)dalpha dbeta=dfrac{8}{3}cdotpicdot r^{4}cdotrho.$
As we know the surface of a sphere is $4pi r^{2}$, and that $m=rhocdot A$, so
$I=dfrac{2}{3}r^{2}m$.
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