1
$\begingroup$

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 $1/2$.

$\endgroup$
  • $\begingroup$ 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 '14 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 '14 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. $\endgroup$ – John Alexiou Jul 31 '14 at 0:14
  • $\begingroup$ Related: physics.stackexchange.com/a/111081/392 $\endgroup$ – John Alexiou Jul 31 '14 at 16:33
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I want to comment on two things : $\endgroup$ – Fadi Jul 31 '14 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 '14 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 '14 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 $\endgroup$ – John Alexiou Jul 31 '14 at 16:33
  • 1
    $\begingroup$ 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 ${\rm d}I = -{\rm d}m [\vec{r}\times][\vec{r}\times]$. $\endgroup$ – John Alexiou Jul 31 '14 at 16:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.