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I have found equation for moment of inertia $(J)$. I'm calculating $J$ for hemisphere, with rotational axis $Z$.

$$ J = \iiint\limits_V r^2 \cdot \rho \cdot dV $$

But if $\rho$ is constant (homogenous), I can do: $$ J = \rho \cdot \iiint\limits_V r^2 \cdot dV $$

Which is: $$ J = \rho \cdot V $$ $$ J = m $$

Am I right?

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Tip: It is always a good idea to check if your equations are dimensionally consistent. In plain words, check what is the SI-units for moment of inertia and mass, respectively? –  Qmechanic Apr 15 '13 at 12:10

3 Answers 3

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No. the integral $\int r^2 dV$. It is not $V$.

$\int dV$=$V$.

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No. Be careful about the integral $\int r^2 dV$. It is not $V$.

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As others have said $$\iiint _V r^2 dV \neq V$$

The volume element $dV$ in cylindrical coordinates (which is a convenient coordinate system for moment of inertia calculations) is: $$ dV = r \, dr \, d\theta \, dz$$

There are 3 integrands ($r,\theta,z$) so you will need to integrate each with appropriate boundaries that describe a hemisphere (be careful as the $r$ above is not the radius of the hemisphere, but it is the $r$ in your equation!)

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Hi ejrb. Welcome to Physics.SE. Indeed, it's good to help other users through answering. But while dealing with homework questions, you can always have a look at our policy on answering such questions. Please make sure that you don't give away the answer (or solve it for the author) ;-) –  Waffle's Crazy Peanut Apr 15 '13 at 13:24
    
Thanks for the pointer. I'll crop this answer back to meet the guidelines –  ejrb Apr 15 '13 at 13:49

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