for a cylinder, let's say with constant density with radius $3$ and height $10$ so

$$ \rho(r, \theta, y)=1 $$

so $$ dm = \rho\left(r, \theta, y\right)r \, dr \, d\theta \, dy = r \, dr \, d\theta \, dy $$

in that case the inertia at the end of the rod is (from here)

$$ I_{xx} = \int_m \left( y^2 + z^2 \right) \, dm=10060.91 $$

where $z = rsin(\theta)$

but (from here)

$$ \frac{1}{3}mL^2 = 9424.77 $$


$$ I_{xx} = \int_m \left( y^2 + z \right) \, dm=9424.77 $$

if $z=sin(\theta)$

why is this?

  • $\begingroup$ In cylindrical coordinates, $z$ should be a direct variable, and the only angle should be azimuthal, so that $x=\rho \cos(\alpha)$ and $y=\rho \sin(\alpha)$. $\endgroup$ – FGSUZ Jul 31 at 19:33
  • 1
    $\begingroup$ A typesetting details here is that you want the differentials to be identifiable as distinct entities. So you set them off with a thin space (dr\, d\theta to get $dr\, d\theta$). I'm one of the people who also like to typeset the "d" upright (so \mathrm{d}r to get $\mathrm{d}r$), but that is much less universal. $\endgroup$ – dmckee Jul 31 at 19:33
  • $\begingroup$ @FGSUZ in my case $y$ is the direct variable and $x, z$ are not. As in this post math.stackexchange.com/questions/3259247/…. $\endgroup$ – fullnitrous Jul 31 at 19:43
  • $\begingroup$ Oh, could you then draw the axes and the limits of integration? Thanks. By the way, I think there's a missing squaring in your last formula $\endgroup$ – FGSUZ Jul 31 at 19:49
  • $\begingroup$ @FGSUZ on purpose because it works $\endgroup$ – fullnitrous Jul 31 at 20:13

So it looks like you are doing the integral $$\int_0^{10}\int_0^{2\pi}\int_0^3\left(y^2+r^2\sin^2\theta\right)r\,\text dr\,\text d\theta\,\text d y$$

So you are assuming your cylinder is oriented along the y-axis, and $\theta=0$ lines up with the positive x-axis. The value $y^2+z^2$ is the distance a point is from the x-axis. Therefore, based on your limits of integration your cylinder's base is in the x-z plane, and you are calculating the moment of inertia about the x-axis.

Looking at your link then, you want to use the equation in the lower left corner along with the parallel axis theorem, not $I=1/3ML^2$ for the thin rod. More explicitly: $$I_{xx}=\frac14MR^2+\frac1{12}ML^2+M\left(\frac12L\right)^2=\frac14MR^2+\frac13ML^2$$

This is the equation you want for your moment of inertia. Plug in your numbers and it all works out.

Why does your integral with the $z$ instead of $z^2$ work out to be $1/3ML^2$ like the rod on one end? Well if you don't square $z$ in the integrand, then this term integrates to $0$ due to the $\sin\theta$ term from $0$ to $2\pi$. Therefore the integral is equivalent to just integrating over $y^2$. This would be the same as if you were looking at a rod for $R=0$. This is why you get $1/3ML^2$ for your incorrect integral

  • $\begingroup$ no you don't understand, the problem is that the integration gives another value that the equation for a special case. i need the integration to give me the same answer. i do not care about the special case equation, i am just using it for verification. also it is not the equation in the lower left for the origin is not in the middle. $\endgroup$ – fullnitrous Jul 31 at 22:17
  • $\begingroup$ here is how the cylinder looks like imgur.com/V70wTa9 $\endgroup$ – fullnitrous Jul 31 at 22:22
  • $\begingroup$ @fullnitrous Yes I deduced that figure from the integral. My answer is what you want. Please read it more carefully. $\endgroup$ – Aaron Stevens Jul 31 at 22:24
  • $\begingroup$ but the equation in the bottom left has the rotation axis in the center which i do not. $\endgroup$ – fullnitrous Jul 31 at 22:25
  • $\begingroup$ @fullnitrous Which is why you use that with the parallel axis theorem. As stated in my answer. $\endgroup$ – Aaron Stevens Jul 31 at 22:29

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