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I was thinking that what could possibly be the angular momentum about centre of mass of a rod of mass M and length L hinged about it's one end and rotating with angular velocity W....unable to think help!

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    $\begingroup$ "angular momentum about centre of mass " while "hinged about it's one end"? $\endgroup$ – Steeven Oct 17 '16 at 18:50
  • $\begingroup$ As Steeven points out, your question is contradictory. Have you done an internet search for moment of inertia J of rod? Angular momentum is $L=J\omega$. $\endgroup$ – sammy gerbil Oct 17 '16 at 20:36
  • $\begingroup$ but it would have angular momentum that may or may not be 0 about any point in space so how is it contradictory ..? $\endgroup$ – Yash Awasthi Oct 18 '16 at 13:01
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A rod rotating with $\vec{\omega}$ is hinged on one side at point A. The center of mass is at point C.

  1. The velocity of the center of mass is $\vec{v}_C = \vec{\omega} \times \vec{r}_{AC}$
  2. The linear momentum of the rod is $\vec{p} = m \vec{v}_C$
  3. The angular momentum about the com is $\vec{L}_C = {\rm I}_C \vec{\omega}$
  4. The angular momentum about the pin is $\vec{L}_A = \vec{L}_C + \vec{r}_{AC} \times \vec{p}$
  5. The MMOI about the pin is found from the angular momentum about the pin $$\vec{L}_A = {\rm I}_C \vec{\omega} + m \vec{r}_{AC} \times \vec{\omega} \times \vec{r}_{AC} = {\rm I}_A \vec{\omega}$$

Does this help you think?

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  • $\begingroup$ but since it is performing pure rotation about pin it's angular momentum = (I(about axis through pin ) x W) $\endgroup$ – Yash Awasthi Oct 18 '16 at 13:03
  • $\begingroup$ The MMOI through the pin is found by 4) above. $\endgroup$ – ja72 Oct 18 '16 at 13:16

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