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BruceW
#16
Dec26-12, 03:54 PM
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hmm. Tricky one. Well, I'm pretty sure that the whole point of saying the angular momentum "about a point" is equivalent to calculating the angular momentum, given that the origin is the point about which we want to find the angular momentum.

So I think the angular momentum about the COM is simply angular momentum, given that our origin is the COM. And in our original reference frame, the rod was rotating around the end. So in a reference frame where the COM stays at the origin, the angular momentum will simply be
[tex] \omega \frac{mL^2}{12} [/tex]
(in other words, same as what the angular momentum would be for a system where the rod is rotating around it's COM.)