Angular momentum of a system relative to a moving reference frame.

[Davidllerenav]

Homework Statement

Demostrate that the angular momentum $\vec M$ of the system of particles relative to a point $O$ of the reference frame $K$ can be presented as

$\vec M=\vec M'+\vec r_c\times\vec p$

where $\vec M'$ is the proper angular momentum (in the reference frame moving translationally and fixed to the centre of inertia), $\vec r_c$ is the radius vector of the centre of inertia relative to the point $O$, $\vec p$ is the total momentum of the system of particles in the reference frame $K$.

Relevant Equations

$\vec M=\vec r\times\vec p$

I don't have too much of a clue of how to begin the problem.
I first wrote the angular moementum of the system of particles: →M=∑mi(→ri×→vi)M→=∑mi(r→i×v→i). Then I know that the angular momentum from of the moving reference frame would have the velocity as the sum of the velocity of the frame plus the velocity of each particle and the radius vector as the sum of the the radius vector relative to the center of inertia and the radius vector relative to the moving frame: →M′=∑mi[(→rc+→ri)×(→vi+→vc)]M→′=∑mi[(r→c+r→i)×(v→i+v→c)]. After that I don't know what to do.

 

[PeroK]

I don't have too much of a clue of how to begin the problem.
I first wrote the angular moementum of the system of particles: →M=∑mi(→ri×→vi)M→=∑mi(r→i×v→i). Then I know that the angular momentum from of the moving reference frame would have the velocity as the sum of the velocity of the frame plus the velocity of each particle and the radius vector as the sum of the the radius vector relative to the center of inertia and the radius vector relative to the moving frame: →M′=∑mi[(→rc+→ri)×(→vi+→vc)]M→′=∑mi[(r→c+r→i)×(v→i+v→c)]. After that I don't know what to do.

That should give you four separate terms to work on.

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