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Question: Calculate the moment of inertia of the cylinder defined below when the cylinder is rotated around the $x$-axis. The cylinder’s axis lies along the z-axis and is defined by $x^2+y^2=1$, $z ≥ 0$ and $z ≤ 2$ and has constant mass density $ρ$. State your answer in terms of the mass of the cylinder, $M$. (End of question)

My issue is setting up the integral for this. I understand we will need to use cyclindrical polar co-ordinates. Also, the moment of inertia is calculated by squaring the axis which is perpendicular to the rotation axis (Is this correct?).

So, how would I go about setting up the integral? This is what I have so far:

$\int_{z=0}^2\int_{\theta=0}^{2\pi}\int_{r=0}^1$[unkown integrand]$r drd\theta dz$

I want to say that the integrand is $r^2$, but I think that give the moment of inertia about the rotation axis. I would appreciate an explanation of what the integrand should be and why. I will have no trouble computing the integral after this stage.

Many thanks in advance.

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The moment of inertia is calculated by using: $$ I = \int _V \rho(x,y,z)\bar{r}^2 dv, $$ where $\bar{r}$ is the distance from the rotation axis. When you change to cylindrical Coordinates you need to take into account that: $$ dv = rdrd\theta dz$$ And the distance from the rotation axis is: $$ \bar{r}=\sqrt{y^2+z^2}= \sqrt{r^2\sin^2(\theta)+z^2}$$ With $\rho(x,y,z)=\rho$ you get: $$ I = \int_{z=0}^2 \int_{\theta=0}^{2\pi} \int_{r=0}^1 \rho (r^2\sin^2(\theta)+z^2) rdrd\theta dz $$

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  • $\begingroup$ Wait, are you sure it's $r^3$? We're rotating about the $x$-axis, not the axis of symmetry. $\endgroup$ – Mr Croutini Jun 5 '14 at 10:57
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    $\begingroup$ No, sorry. Changed my answer. $\endgroup$ – oholmer Jun 5 '14 at 13:28
  • $\begingroup$ How did you get the distance from the rotation axis? Also, if we were rotating about the $y$-axis instead, would the distance from the rotation axis be $\sqrt{x^2+z^2}$? $\endgroup$ – Mr Croutini Jun 5 '14 at 14:37
  • $\begingroup$ Also, why have you put $\sin^2(\rho)$? should it not be $\sin^2(\theta)$? $\endgroup$ – Mr Croutini Jun 5 '14 at 14:53
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    $\begingroup$ It is the shortest distance from the axis to that point. So that is correct, and for z-axis it's $\sqrt{x^2+y^2}$. I think you have understood. It should of course be $\theta$. $\endgroup$ – oholmer Jun 5 '14 at 15:02

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