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Why is angular momentum for rotation about the center of mass independent of origin of the coordinate system?

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  • $\begingroup$ Because you are calculating it according to a definite origin, i.e. in the center on mass frame! $\endgroup$ – Ali Jul 21 '13 at 14:28
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$$\vec L = \sum_i \vec {r_i} \times \vec{p_i} $$

Now changing coordinates to CM, i.e. $\vec{r_i} \rightarrow \vec{R_{CM}}+\vec{\mathfrak{r}_i}$, where $\vec{\mathfrak{r}_i}$ is the new coordinates of the particle according to the CM frame.

$$\vec{L}=\sum_i \left(\vec{R_{CM}}+\vec{\mathfrak{r}_i} \right)\times \vec{p_i}=\vec{R_{CM}} \times \sum_i \vec{p_i}+\sum_i \vec{\mathfrak{r}_i}\times\vec{p_i}$$

Now the first term is zero, because $\sum_i \vec{p_i}$ is zero in the CM frame; and the latter is just the angular momentum in the CM. Ergo:

$$\vec{L}=\vec{L_{CM}}$$

I think this is what OP meant in the question.

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  • $\begingroup$ +1, it would be more readable if you used $r'_{i}$ for the new coordinates. $\endgroup$ – Physiks lover Jul 21 '13 at 17:27
  • $\begingroup$ @Physikslover I changed my notation, how does it look now? $\endgroup$ – Ali Jul 22 '13 at 14:29
  • $\begingroup$ Looks nice - very arty ;) $\endgroup$ – Physiks lover Jul 22 '13 at 14:52

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