Conservation of angular momentum invariance

[Melian]

Homework Statement

Given a reference frame O' moving at a constant speed $\vec{V}$ in relation to another reference frame O, I want to prove that

##\vec{r_{1B}} \times m_1\vec{v_{1B}} + \vec{r_{2B}} \times m_2\vec{v_{2B}} = \vec{r_{1F}} \times m_1\vec{v_{1F}} + \vec{r_{2F}} \times m_2\vec{v_{2F}}##

in O is equal to

##\vec{r'_{1B}} \times m_1\vec{v'_{1B}} + \vec{r'_{2B}} \times m_2\vec{v'_{2B}} = \vec{r'_{1F}} \times m_1\vec{v'_{1F}} + \vec{r'_{2F}} \times m_2\vec{v'_{2F}}##

in O'. The particles 1 and 2 are colliding (elastic collision). B stands for before the collision and F, after the collision.

Homework Equations

Galilean transformation gives :

##\vec{r} = \vec{r'} + \vec{V}t## and ##\vec{v} = \vec{v'} + \vec{V}##

The Attempt at a Solution

[/B]
Substituting these expressions into the first equation and developing the cross products, I obtain, after having canceled out 8 terms (because of the conservation of linear momentum),

##\vec{r'_{1B}} \times m_1\vec{v'_{1B}} + \vec{r'_{1B}} \times m_1\vec{V} + \vec{r'_{2B}} \times m_2\vec{v'_{2B}} + \vec{r'_{2B}} \times m_2\vec{V} = \vec{r'_{1F}} \times m_1\vec{v'_{1F}} + \vec{r'_{1F}} \times m_1\vec{V} + \vec{r'_{2F}} \times m_2\vec{v'_{2F}} + \vec{r'_{2F}} \times m_2\vec{V}##

My problem is that I do not know how to cancel out the ##\vec{r'_{1B}} \times m_1\vec{V}## terms... How do I do that? Thanks!

 

[davidmoore63@y]

The sum of these terms across all particles = (total mass * position of the center of mass) X V . This quantity is preserved in collisions, since the center of mass and total mass don't change. [In addition, it is preserved across the Galilean transformation also, since r' - r is a multiple of V so contributed nothing to the cross product]