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I've tried to find the moment of inertia of a cylinder rotating about an axis parallel to its base (i.e about the 'End diameter') as one can see here . But when I checked my results with different references ,I've found that it's incorrect!.I need a help to figure out where I did it wrong.

since $$I=\int\limits x^2.dm$$

$$dm = \rho.dv$$ where

$$dv= r.d\theta . dr .dh$$ &

$$\rho=\frac{M}{\pi R^2 H}$$

$M$:cylinder total mass. $H$:total height of cylinder. $R$:the radius of the cylinder.

The distance of each infinitesimal element from the axis of rotation would be :

$$\sqrt{r^2+h^2}$$

Therfore,

$$I= \frac{2M}{ R^2 H}\int\limits_{0}^{R} \int\limits_{0}^{H}\, r(r^2+h^2)dh\,dr$$

and this gives a result of : $$\frac{M~H^2}{3}+\frac{M~R^2}{2}$$

Which is obviously wrong since the second term should be multiplied by a factor of $1/2$.

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  • $\begingroup$ How do you get the distance to be sqrt(r^2 + h^2) in all cases? It seems to me that is incorrect except for points on a plane down the center of the cylinder, perpendicular to the axis. Otherwise it would depend on the angle that the radius to the element makes with the rotation axis $\endgroup$
    – BowlOfRed
    Jul 30, 2014 at 22:43
  • $\begingroup$ In the website your linking, they use the parallel axis theorem to find the moment of inertia. The parallel axis theorem says $ dI_z=dI_x + dI_y$ where x,y and z are any orthogonal axis $\endgroup$
    – aPhysicist
    Jul 30, 2014 at 23:16
  • $\begingroup$ Check the factor in front of the double integral. It should be equal to $\rho = \frac{M}{\pi R^2 H}$ since $I=\int \rho d^2 \,{\rm d}v$ where $d$ is the perpendicular distance from the axis of rotation. $\endgroup$ Jul 31, 2014 at 0:14
  • $\begingroup$ Related: physics.stackexchange.com/a/111081/392 $\endgroup$ Jul 31, 2014 at 16:33

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I think I understand what you want. Let us call $z$ the axis along the cylinder and $x$, $y$ the other two directions. If the center of the rod is at point C and the end at A you want $I_{xx}^A = \int \rho ({y^2+z^2}) {\rm d} V$

With $m=\int \rho \,{\rm d}V =\rho \pi R^2 H $ and ${\rm d}V = r\,{\rm d}\theta{\rm d}r{\rm d}z$ and $(x,y,z)=(r \cos\theta, r \sin\theta,z)$

$$ \begin{align} I_{xx}^A &= \frac{m}{\pi R^2 H} \int_0^H \int_0^R \int_{-\pi}^{\pi}\, ({r^2\sin^2\theta+z^2})\,r\,{\rm d}\theta\,{\rm d}r\,{\rm d}z \\ & = \frac{m}{\pi R^2 H} \int_0^H \int_0^R \,\pi r (2 z^2+r^2)\,{\rm d}r\,{\rm d}z \\ & = \frac{m}{ R^2 H} \int_0^H \, \frac{ R^2 (4 z^2+R^2)}{4}\,{\rm d}z \\ & = \frac{m}{ R^2 H} \left( \frac{ H^3 R^2}{3} + \frac{ H R^4}{4} \right) \\ & = m \frac{H^2}{3} + m \frac{R^2}{4} \end{align}$$

which matches the hyperphysics answer.

NOTE: Any separation along the $x$ axis does not play into the integral because this is axis of rotation and it just represents a parallel displacement. BTW the parallel axis theorem is a misnomer as it should be the perpendicular axis theorem.

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  • $\begingroup$ I want to comment on two things : $\endgroup$
    – Fadi
    Jul 31, 2014 at 15:39
  • $\begingroup$ 1- This have nothing to do even with Perpendicular Axis theorem,since this theorem applies on rigid bodies that are 'entirely laying within a plane' (e.g, desks). $\endgroup$
    – Fadi
    Jul 31, 2014 at 15:45
  • $\begingroup$ 2- This integral from $-\pi$ to $\pi$ , is confusing me.since there is $d\theta$ in the $dv$ term , the integral should sweep all across the cylinder to cover all the volume.i.e, from $0$ to $2\pi$ ! . Notice that they both give the same results, but it's a concept issue for me. $\endgroup$
    – Fadi
    Jul 31, 2014 at 15:50
  • $\begingroup$ @Fadi the parallel axis theorem has to do with calculated mass moment of inertia about an axis away from the center of mass. Here the end axis is $\frac{L}{2}$ away from the COM. I mentioned this because of the definition of the mass moment tensor arises from the parallel axis theorem. See physics.stackexchange.com/a/111081/392 $\endgroup$ Jul 31, 2014 at 16:33
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    $\begingroup$ To define the inertia tensor you need $$I= \int \begin{pmatrix} y^2+z^2 & - x y & - x z \\ - x y & x^2+z^2 & -y z \\ -x z & -y z & x^2+y^2 \end{pmatrix} \,{\rm dm}$$ most of the terms fall out for a symmetric shape. This is a result of the vector form of the parallel axis theorem ${\rm d}I = -{\rm d}m [\vec{r}\times][\vec{r}\times]$. $\endgroup$ Jul 31, 2014 at 16:51

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