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how do I find the moment of inerta of cylinder along central diameter which is the one that is $(h^2 + 3r^2)M /12$ ???

I used triple integral and I get

Which I I'm not sure what I done wrong..Any help will be appreciated. Thanks

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By definition of moment of inertia $I_{xx}$,

$$ I_{xx} = \int (y^2+z^2) dm = \int (r^2 \sin^2 \theta +z^2) \frac{M}{V} dV = \frac{M}{V} \int_{-H/2}^{H/2} \int_0^{2\pi} \int_0^R (r^2 \sin^2 \theta +z^2) r dr d\theta dz $$ what you doing wrong is that the limit of integration that you choose for $dz$ and the factor $\frac{M}{V}$. The integration on $z$ must run from $-H/2$ to $H/2$ by symmetry. And by substitute $V = \pi R^2 H$, you'll have the desired result.

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  • $\begingroup$ hi @Sou thanks for helping $\endgroup$
    – kevin
    Commented Oct 22, 2017 at 7:43
  • $\begingroup$ but @Sou I'm a little bit confused I mean for the moment of inertia along central axis instead of central diameter, do I still use the same triple definite integral also? $\endgroup$
    – kevin
    Commented Oct 22, 2017 at 7:48
  • $\begingroup$ oh wait, I actually still calculating from centre of cylinder so it should be still -H/2 to H/2 if I'm not wrong?? $\endgroup$
    – kevin
    Commented Oct 22, 2017 at 7:50
  • $\begingroup$ @john : Yes. Whatever components of I that you want to calculate that involve z, it should be from -H/2 to H/2. $\endgroup$ Commented Oct 22, 2017 at 7:55

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