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.
Answers
It is given that,
so,
It is also given, that,
Taking cross product of this relation with ,
where,
And
So,
In simply you can say that, L = L' + R × MV
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