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enter image description here
I have several questions about angular momentum.

1.I know there are 2 parts of angular momentum - translational and rotational. For a body with respect to a fixed point, translational angular momentum is $MV_{cm}R_{\perp}$. The rotational angular momentum is defined as $I\omega$. Here, what is $I$ and is $\omega$ the $\omega$ of the rigid body or with respect to the fixed point? I believe $I$ is of CM and $\omega$ is of entire body. Why is this so? Also does this not mean that the angular momentum of a body will be same with respect to every point at equal perpendicular (to linear momentum) distance? (Refer to the picture)

2.Let us say a body is pure rolling. For a point on the body, let's say $x$ distance below CM, what will be the angular momentum of the body?

3.For generalizing the above question, if I want to calculate angular momentum about a point, do I have to first make the point's velocity zero by using relative frame? For example, if a square is translating. What will be the angular momentum of the body about a vertex?

Thank you!

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  • $\begingroup$ clarify your question a bit more. And where is the picture? $\endgroup$ – manshu Nov 16 '15 at 13:43
  • $\begingroup$ First part of my question I can rephrase - Rotational part of angular momentum is I(CM)*Angular velocity of body. That means that it is independent of where the point is etc. Why is it so? $\endgroup$ – Shodai Nov 16 '15 at 14:42
  • $\begingroup$ For question 1...I is the moment of inertia around its axis of rotation which may or may not be the CM. omega is the number of revolutions done by the body in 1 second and its same even with respect to the fixed point. And yes this means that the angular momentum of a body will be same with respect to every point at equal perpendicular (to linear momentum) distance $\endgroup$ – manshu Nov 16 '15 at 14:53

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