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Timeline for Moment of Inertia of Torus

Current License: CC BY-SA 4.0

11 events
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Nov 30 at 23:50 comment added John Alexiou I used a free version of the 3D CAD software called SpaceClaim. The free version is called DesignSpark Mechanical
S Nov 23 at 13:39 history suggested Sahaj CC BY-SA 4.0
fix some formatting
Nov 23 at 12:33 review Suggested edits
S Nov 23 at 13:39
Nov 23 at 12:29 comment added Sahaj Hello John. What software did you use for the image?
Oct 20 at 15:45 comment added John Alexiou @HugdeRoda - if it didn't, I would have made a mistake somewhere.
Oct 20 at 15:41 comment added Hug de Roda It's nice to see that the moment of inertia coincides with that of a thin loop when $R\to 0$.
Apr 20, 2017 at 14:18 comment added Luan Nico thanks for your reply! :)) The wikipedia result matches yours, since we are talking about the z axis: $\frac{m}{4}(3R^2+4\rho^2)$ (it just put the $\frac{1}{4}$ inside). About the volume element, I used $dV = A \cdot dr'$, where $dr'$ is a linear element from the circunference; if I use $\rho d\theta$, I get the same result; I integrate $r'$ from $0$ to $2 \pi \rho$ (circ.), and I integrate $\theta$ from $0$ to $2 \pi$.
Apr 19, 2017 at 23:30 comment added John Alexiou Honestly I couldn't follow exactly. The logic does not flow smoothly. Also I think the volume of the red circle is ${\rm d}V = {\rm d}A \rho {\rm d}\theta = \pi R^2 \rho {\rm d}\theta$ and this is not what you have. Also the wikipedia value you mention is not what the correct result is.
Apr 19, 2017 at 22:08 comment added Luan Nico I understood your answer, but after reevaluating mine, I still can't figure out what's wrong... Could you please comment on that?
Apr 19, 2017 at 21:17 history edited John Alexiou CC BY-SA 3.0
added 296 characters in body
Apr 19, 2017 at 21:08 history answered John Alexiou CC BY-SA 3.0