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So I know the polar moment of inertia of a solid cylinder is: $$ I= \frac{1}{2} mr^2 $$

My question arises with the polar moment of inertia uses for solid cylinders in my mechanics of materials book, which is: $$ J=\frac{\pi}{2}r^4 $$ Don't these describe the same thing? Or am I mistaken in this, or simply overlooking something?

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  • $\begingroup$ it seems that what you have is the "geometric" moment of inertia for a homogeneous cylinder $\endgroup$
    – Phoenix87
    Jan 16, 2015 at 21:18
  • $\begingroup$ One is a mass moment of inertia and the other is polar moment of area. Having said that I always thought that $J_{z} = I_{xx} + I_{yy}$. $\endgroup$ Jan 17, 2015 at 0:23

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Obviously $J$ is something different from $I$ because the units would be different. OTOH, they are related. Consider this idea and see if you can see what their relationship is.

The mass of the cylinder is given by $m = \rho V$, and the volume of a cylinder is $V=LA$, where $L$ is the height (or length) of the cylinder and $A=\pi r^2$, the cross sectional area.

I think you can take it from there to find the connection.

EDIT: J is the polar moment of inertia of the area, not the mass. $$ J=\int r^2 dA $$

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  • $\begingroup$ I could already see that relationship. What I'm after is why they are both called polar moments of inertia a cylinder when they are obviously not equal. $\endgroup$
    – John
    Jan 16, 2015 at 22:16
  • $\begingroup$ Engineering texts often talk about the moment of inertia of the area. In that case they integrate a coordinate over the area of the object, and no mass appears. Your I in the question is a mass moment of inertia. Your J is an area polar moment. (J specifically refers to a polar moment of area.) Physicists typically use the mass moment of inertia, and engineers just try to confuse everything ;) $\endgroup$
    – Bill N
    Jan 19, 2015 at 17:54
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For a cylinder with density $1/\ell$, the mass is $\pi r^2$ and the moment of inertia $\frac{\pi}{2}r^4$. This makes a bit more sense for a disk than for a cylinder (you don't have to fret over the $1/\ell$ term). I hope you can now see the relationship between the two formulas.

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  • $\begingroup$ @bobie - you have a very good answer showing the c.o.m. transformation already. I would have written a similar answer, but I see no need in this case. $\endgroup$
    – Floris
    Jan 18, 2015 at 23:00
  • $\begingroup$ Oops... responded to the wrong comment. $\endgroup$
    – Bill N
    Jan 19, 2015 at 17:53

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