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I'm trying to calculate the inertia tensor for a circular arc, as shown in this image:

arc_diagram

It starts at the x-axis and lies entirely in the XY plane. It has a mass per unit length of $\rho$ and spans an angle of $\theta$ (which could be anything up to $\pi$).

I've found many resources which give the inertia tensor for a full hoop, or for a segment, but not this case. I think the off-diagonal terms are zero, and that $I_{zz}$ is just a fraction of the polar inertia for a full hoop, so: $$I_{zz} = \frac{\theta}{2\pi}\left(\rho 2 \pi r\right) r^2 = \rho \theta r^3$$ ... but I can't work out $I_{xx}$ or $I_{yy}$, or even if they're the same as each other.

G Cab
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DoktorDemento
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2 Answers2

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Hint:

You need the following integrals

$$M=\int_{\theta=0}^\Theta R\,d\theta=R\Theta$$ $$I_{xx}=\int_{\theta=0}^\Theta R^3\sin^2\theta\,d\theta$$ $$I_{yy}=\int_{\theta=0}^\Theta R^3\cos^2\theta\,d\theta$$

which are fairly easy.

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The reference system is not centered at the barycenter of the arc, and it is not oriented along the symmetry axes.

Therefore, you did right for $I_{zz}$, but for the rest you do have off diagonal elements, and you cannot do else than calcutating all the nine (well, six) components.

G Cab
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