Separation of motion of a system of particles into motion of the centre of mass and motion about the centre of mass:
$(a)$ Show $p = \sum p_{i}^{\prime} + M V$,where $p$ is the total momentum of the system,$p_{i}^{\prime} = m_{i} v_{i}^{\prime}$,and $v_{i}^{\prime}$ is the velocity of the $i^{th}$ particle relative to the centre of mass. Also,prove using the definition of the centre of mass that $\sum p_{i}^{\prime} = 0$.
$(b)$ Show $K = K^{\prime} + \frac{1}{2} M V^{2}$,where $K$ is the total kinetic energy of the system,$K^{\prime}$ is the total kinetic energy of the system relative to the centre of mass,and $\frac{1}{2} M V^{2}$ is the kinetic energy of the translation of the system as a whole.
$(c)$ Show $L = L^{\prime} + R \times M V$,where $L^{\prime} = \sum r_{i}^{\prime} \times p_{i}^{\prime}$ is the angular momentum of the system about the centre of mass. Note $r_{i}^{\prime} = r_{i} - R$.
$(d)$ Show $\frac{d L^{\prime}}{d t} = \sum r_{i}^{\prime} \times \frac{d p_{i}^{\prime}}{d t}$. Further,show that $\frac{d L^{\prime}}{d t} = \tau_{ext}^{\prime}$,where $\tau_{ext}^{\prime}$ is the sum of all external torques acting on the system about the centre of mass.

(A) Let $m_{i}$ be the mass and $v_{i}$ be the velocity of the $i^{th}$ particle. The velocity relative to the centre of mass is $v_{i}^{\prime} = v_{i} - V$,so $v_{i} = v_{i}^{\prime} + V$. The total momentum $p = \sum m_{i} v_{i} = \sum m_{i}(v_{i}^{\prime} + V) = \sum m_{i} v_{i}^{\prime} + V \sum m_{i} = \sum p_{i}^{\prime} + MV$. Since $\sum m_{i} r_{i}^{\prime} = 0$,differentiating gives $\sum m_{i} v_{i}^{\prime} = \sum p_{i}^{\prime} = 0$.
$(b)$ $K = \sum \frac{1}{2} m_{i} v_{i}^{2} = \sum \frac{1}{2} m_{i} (v_{i}^{\prime} + V) \cdot (v_{i}^{\prime} + V) = \sum \frac{1}{2} m_{i} v_{i}^{\prime 2} + \sum m_{i} v_{i}^{\prime} \cdot V + \sum \frac{1}{2} m_{i} V^{2} = K^{\prime} + V \cdot (\sum p_{i}^{\prime}) + \frac{1}{2} M V^{2}$. Since $\sum p_{i}^{\prime} = 0$,$K = K^{\prime} + \frac{1}{2} M V^{2}$.
$(c)$ $L = \sum r_{i} \times p_{i} = \sum (R + r_{i}^{\prime}) \times m_{i} (V + v_{i}^{\prime}) = \sum (R \times m_{i} V + R \times m_{i} v_{i}^{\prime} + r_{i}^{\prime} \times m_{i} V + r_{i}^{\prime} \times m_{i} v_{i}^{\prime}) = R \times MV + R \times (\sum p_{i}^{\prime}) + (\sum m_{i} r_{i}^{\prime}) \times V + L^{\prime}$. Since $\sum p_{i}^{\prime} = 0$ and $\sum m_{i} r_{i}^{\prime} = 0$,$L = L^{\prime} + R \times MV$.
$(d)$ $\frac{d L^{\prime}}{d t} = \frac{d}{d t} \sum (r_{i}^{\prime} \times p_{i}^{\prime}) = \sum (v_{i}^{\prime} \times p_{i}^{\prime}) + \sum (r_{i}^{\prime} \times \frac{d p_{i}^{\prime}}{d t})$. Since $v_{i}^{\prime} \times (m_{i} v_{i}^{\prime}) = 0$,$\frac{d L^{\prime}}{d t} = \sum r_{i}^{\prime} \times F_{i}^{\prime} = \tau_{ext}^{\prime}$.