Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass:
(a) Show $\vec{p}=\vec{p}_i+m_i \vec{V}$ where $\vec{p}_i$ is the momentum of the $\boldsymbol{i}^{\text {th }}$ particle (of mass $m_i$ ) and $\vec{p}_i^{\prime}=m_i \vec{v}_i^{\prime}$. Note $\vec{v}_i^{\prime}$ is the velocity of the $i^{\text {th }}$ particle relative to the centre of mass.
Also, prove using the definition of the centre of $\operatorname{mass} \sum \vec{p}_i^{\prime}=0$
(b) Show $K=K^{\prime}+\frac{1}{2} M V^2$
where $K$ is the total kinetic energy of the system of particles, $K^{\prime}$ is the total kinetic energy of the system when the particle velocities are taken with respect to the centre of mass and $M V^2 / 2$ is the kinetic energy of the translation of the system as a whole (i.e., of the centre of mass motion of the system).
(c) Show $\vec{L}=\vec{L}^{\prime}+\vec{R} \times \vec{M} \vec{V}$
where $\vec{L}^{\prime}=\sum \vec{r}_i^{\prime} \times \vec{p}_i^{\prime}$ is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember $\vec{r}_i^{\prime}=\vec{r}_i-\vec{R}$; rest of the notation is the standard notation used in the chapter. Note $\vec{L}^{\prime}$ and $\overrightarrow{M P} \times \vec{V}$ can be said to be angular momenta respectively, about and of the centre of mass of the system of particles.
(d) Show $\frac{d \vec{L}^{\prime}}{d t}=\sum \vec{r}_i^{\prime} \times \frac{d \vec{p}^{\prime}}{d t}$ Further, show that: $\frac{d \vec{L}^{\prime}}{d t}=\tau_{e x t}^{\prime}$ where $\tau_{\text {ext }}^{\prime}$ is the sum of all external torques acting on the system about the centre of mass.
[Hint: Use the definition of centre of mass and Newton's Third Law. Assume the internal forces between any two particles act along the line joining the particles.]

Options:

None

Solution:
Here $r_i=r_i^{\prime}+R$ and $V_i=V_i^{\prime}+V$ where $r_i^{\prime}$ and $v_i^{\prime}$ denote the radius vector and velocity of the $l^{\text {th }}$ particle referred to centre of mass $O^{\prime}$ as the new origin and $V=R^{\prime}$ is the velocity of centre of mass relative to $O$.
(a) Momentum of $i^{\text {th }}$ particle
$$
\begin{aligned}
&\quad P=m_i V_i^{\prime} \\
&=m_i\left(V_i^{\prime}+V\right)\left(\text { since } V_i=v_i+V\right) \\
&\text { or } P=m_i V_i^{\prime}+m_i V \\
&P=P_i+m_i V
\end{aligned}
$$
(b) Kinetic energy of the system of particles.
$$
\begin{aligned}
&K=\frac{1}{2} \sum m_i v=\frac{1}{2} \sum_i m_i V_i V_i \\
&=\frac{1}{2} \sum_i m_i\left(V_i^{\prime}+V\right)\left(V_i+V\right) \\
&=\frac{1}{2} \sum_i m_i\left(V_i^2+V^2+V_i V\right) \\
&=\frac{1}{2} \sum_i m_i V^2+\frac{1}{2} \sum_i m_i V_i^{\prime 2}+\sum_i m_i V_i \cdot V \\
&=\frac{1}{2} M V^2+K^{\prime}
\end{aligned}
$$
where $M=\sum m_i=$ total mass of the system
$$
K^{\prime}=\frac{1}{2} \sum_i m_i v_1^{\prime 2}
$$
$=$ kinetic energy of motion about the centre of mass or $\frac{1}{2} M v^2=$ kinetic energy of motion of centre of mass.
(Proved)
Since $\sum_i m_i V_i^{\prime} \cdot V=\sum_i m_i \frac{d V_i}{d t} V$
$$
=\frac{d}{d t}\left(\sum m_i V_i^{\prime}\right) \cdot V=0
$$
(c) Total angular momentum of the system of particles.
$$
\begin{array}{r}
L=r_i \times p=\left(r_i^{\prime}+R\right) \times \sum_i m_i V_i \\
=\left(r_i^{\prime}+R\right) \times \sum_i m_i\left(V_i^{\prime}+V\right) \\
=\sum_i\left(R \times m_i V\right)+\sum_i r_i^{\prime} \times m_i V_i^{\prime}+\left(\sum_i m_i r_i^{\prime}\right) \\
\times V+R \times \sum_i m_i V_i \\
=\sum_i\left(R \times m_i V\right)+\sum_i r_i^{\prime} \times m_i V_i+\left(\sum_i m_i r_i^{\prime}\right) \\
\times V+R \times \frac{d}{d t}\left(\sum_i m_i r_i^{\prime}\right)
\end{array}
$$
The last two terms vanish for both contain the factor $\sum m_i r_i^{\prime}$ which is equal to
$\sum_i m_i r_i=\sum_i m_i\left(r_i^{\prime}-R\right)=M R-M R=0$
from the definition of centre of mass. Also
$\sum_i\left(R \times m_i V\right)=R \times M V$
so that $L=R \times M V+r_{\mathrm{i}}^{\prime} P_{\mathrm{i}}$
or $L=R \times M V+L^{\prime}$
where $L^{\prime}=\sum_i r_i^{\prime} \times P_i$
(d) From previous solution
$$
\begin{aligned}
&\mathrm{L}^{\prime}=\sum_{\mathrm{i}} r_i^{\prime} \times p_i \\
&\frac{d L^{\prime}}{d t}=\sum r_i^{\prime} \times \frac{d P_i}{d t}+\sum \frac{d r_i^{\prime}}{d t} \times P_i
\end{aligned}
$$
$\quad=\sum r_i^{\prime} \times \frac{d P_i}{d t}=\sum \mathrm{r}_{\mathrm{i}} \times F_i^{e x t}=\tau^{\prime}$ ext $^{\text {ext }}$
Since $\sum \frac{d r_i^{\prime}}{d t} \times P_i=\sum \frac{d r_i^{\prime}}{d t} \times m v_i=0$
Total torque $\tau=\sum r_i \times F_i^{e x t}$
$=\sum\left(r_i+R\right) \times F_i^{e x t}$
$=\sum r_i^{\prime} \times F_i^{e x t}+R \times \sum F_i^{e x t}$
$=\tau_{\text {ext }}^{\prime}+\tau_0^{(\text {ext })}$
where $\tau_{\text {ext }}$ is the total torque about the centre of mass as
origin and $\tau_0^{\text {ext }}$, that about the origin $\mathrm{O}$.
$\tau_{\text {ext }}^{\prime}=\sum r_i^{\prime} \times F_i^{\text {ext }}$
$=\sum r_i \times \frac{d P_i^{\prime}}{d t}=\frac{d}{d t} \sum\left(r_i \times P_i\right)=\frac{d L^{\prime}}{d t}$

Topic: Physics, System of Particles and Rotational Motion