Question

Separating motion of a system of particles into motion of the centre of mass and motion about the centre of mass, show L = L' + R x MVwhere L' = Σri' + pi' is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember ri' = ri - R; rest of the notation is in the standard notation used in the chapter. Note L' and MR x V can be said to be angular momentum, respectively about and of the centre of mass of the system of particles.

Answer 1

Let assume that , position vector of ith particle with respect to origin is $r_i$, position vector of ith particle with respect to centre of mass is $r'_i$ and position vector of centre of mass with respect to origin is R.
It is given that, $r'_i=r_i-R$
so, $r_i=r'_i+R$

It is also given, that,
$p_i=p'_i+m_iV$

Taking cross product of this relation with $r_i$ ,
$r_i\times p_i=r_i\times p'_i+r'_i\times m_iV\\r_i\times p_i=(r'_i+R)\times p'_i+(r'_i+R)\times m_iV\\r_i\times p_i=r'_i\times p'_i+R\times p'_i+r'_i\times m_iV+R\times m_iV\\L=L'+R\times p'_i+r'_i\times m_iV+R\times m_iV$

where, $R\times p'_i=0$
And $r'_i m_iV=0$

So, $L=L'+R\times m_iV$

In simply you can say that, L = L' + R × MV

Answer 2

Answer:

Let assume that , position vector of ith particle with respect to origin is r_ir

i

, position vector of ith particle with respect to centre of mass is r'_ir

i

′

and position vector of centre of mass with respect to origin is R.

It is given that, r'_i=r_i-Rr

i

′

=r

i

−R

so, r_i=r'_i+Rr

i

=r

i

′

+R

It is also given, that,

p_i=p'_i+m_iVp

i

=p

i

′

+m

i

V

Taking cross product of this relation with r_ir

i

,

\begin{gathered}r_i\times p_i=r_i\times p'_i+r'_i\times m_iV\\r_i\times p_i=(r'_i+R)\times p'_i+(r'_i+R)\times m_iV\\r_i\times p_i=r'_i\times p'_i+R\times p'_i+r'_i\times m_iV+R\times m_iV\\L=L'+R\times p'_i+r'_i\times m_iV+R\times m_iV\end{gathered}

r

i

×p

i

=r

i

×p

i

′

+r

i

′

×m

i

V

r

i

×p

i

=(r

i

′

+R)×p

i

′

+(r

i

′

+R)×m

i

V

r

i

×p

i

=r

i

′

×p

i

′

+R×p

i

′

+r

i

′

×m

i

V+R×m

i

V

L=L

′

+R×p

i

′

+r

i

′

×m

i

V+R×m

i

V

where, R\times p'_i=0R×p

i

′

=0

And r'_i m_iV=0r

i

′

m

i

V=0

So, L=L'+R\times m_iVL=L

′

+R×m

i

V

In simply you can say that, L = L' + R × MV