(a)Take a system of i moving particles.
Mass of the i
th particle = mi
Velocity of the i
th particle = vi
Hence, momentum of the i
th particle, pi = mi vi
Velocity of the centre of mass = V
The velocity of the i
th particle with respect to the centre of mass of the system is given
as: v’i = vi – V … (1)
Multiplying mi throughout equation (1), we get:
mi v’i = mi vi – mi V
p’i = pi – mi V
Where,

pi’ = mivi’ = Momentum of the i
th particle with respect to the centre of mass of the
system pi = p’i +mi V
We have the relation: p’i = mivi’
Taking the summation of momentum of all the particles with respect to the centre of mass of the system, we get:

$\sum p_i^{\mathrm{\prime }}=\sum m_i{v}_i^{\mathrm{\prime }}=\sum m_j\frac{d{r}_i^{\mathrm{\prime }}}{dt}$

$\begin{array}{l}\text{ Where, }\\ {\mathrm{r}}^{\mathrm{\prime }}=\text{ Position vector of ith particle with respect to the centre of mass }\\ {\mathrm{v}}_{\mathrm{i}}^{\mathrm{\prime }}=\frac{{\mathrm{dr}}_{\mathrm{i}}^{\mathrm{\prime }}}{\mathrm{dt}}\end{array}$

As per the definition of the center of mass, we have:

$\begin{array}{l}\sum {\mathrm{m}}_{\mathrm{i}}{\mathrm{r}}_{\mathrm{i}}^{\mathrm{\prime }}=0\\ \therefore \sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{j}}\frac{{\mathrm{dr}}_{\mathrm{j}}^{\mathrm{\prime }}}{\mathrm{dt}}=0\\ \sum {\mathrm{p}}_{\mathrm{i}}^{\mathrm{\prime }}=0\end{array}$

We have the relation for velocity of the ith particle as:

$v_{\mathit{i}}\mathit{=}{v}_{\mathit{i}}^{\mathit{\prime }}\mathit{+}V$

$\sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}{\mathrm{v}}_{\mathrm{i}}=\sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}{\mathrm{v}}_{\mathrm{i}}^{\mathrm{\prime }}+\sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}\mathrm{V}$                        ...(2)

Taking the dot product of equation (2) with itself, we get:

$\begin{array}{l}\sum _{\mathrm{j}}{\mathrm{m}}_{\mathrm{i}}{\mathrm{v}}_{\mathrm{i}}\sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}{\mathrm{v}}_{\mathrm{i}}=\sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}({\mathrm{v}}_{\mathrm{i}}^{\mathrm{\prime }}+\mathrm{V})\cdot \sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}({\mathrm{v}}_{\mathrm{i}}^{\mathrm{\prime }}+\mathrm{V})\\ {\mathrm{M}}^2\sum {\mathrm{v}}_{\mathrm{i}}^2={\mathrm{M}}^2\sum {\mathrm{v}}_{\mathrm{i}}^{\mathrm{\prime }2}+{\mathrm{M}}^2\sum {\mathrm{v}}_{\mathrm{i}}{\mathrm{v}}^{\mathrm{\prime }}+{\mathrm{M}}^2\sum {\mathrm{v}}^{\mathrm{\prime }}{\mathrm{v}}_{\mathrm{i}}+{\mathrm{M}}^2{\mathrm{V}}^2\end{array}$

Here, for the centre of mass ofthe system of particles. $\sum {\mathrm{v}}_{\mathrm{i}}{\mathrm{v}}_{\mathrm{i}}^{\mathrm{\prime }}=-\sum _{\mathrm{i}}{\mathrm{v}}_{\mathrm{i}}^{\mathrm{\prime }}\mathrm{v}$

$\begin{array}{l}{\mathrm{M}}^2\sum _{\mathrm{i}}{\mathrm{v}}_{\mathrm{i}}^2={\mathrm{M}}^2\sum _{\mathrm{i}}{\mathrm{v}}_{\mathrm{i}}^{\mathrm{\prime }2}+{\mathrm{M}}^2{\mathrm{V}}^2\\ \frac{1}{2}\mathrm{M}\sum _{\mathrm{i}}{\mathrm{v}}_{\mathrm{i}}^2=\frac{1}{2}\mathrm{M}\sum {\mathrm{v}}_1^{\mathrm{\prime }2}+\frac{1}{2}{\mathrm{MV}}^2\\ \mathrm{K}={\mathrm{K}}^{\mathrm{\prime }}+\frac{1}{2}{\mathrm{MV}}^2\end{array}$

Where,
$K\mathit{=}\frac{\mathit{1}}{\mathit{2}}M\mathit{\sum }{v}_{\mathit{i}}^{\mathit{2}}$ = Total kinetic energy of the system of particles
$K=\frac{1}{2}M\sum v_i^{\mathrm{\prime }2}$ = Total kinetic energy of the system of particles with respect to the
centre of mass
$\frac{1}{2}M{V}^2$ = Kinetic energy of the translation of the system as a whole Position vector of the ith particle with respect to origin = ri

Position vector of the i
th particle with respect to the centre of mass = r’i
Position vector of the centre of mass with respect to the origin = R
It is given that: r’i = ri
– R ri = r’i + R We
have from part (a), pi
= p’i +mi V
Taking the cross product of this relation by ri, we get:

$\begin{array}{l}\sum _{\mathrm{i}}{\mathrm{r}}_{\mathrm{i}}\times {\mathrm{p}}_{\mathrm{i}}=\sum _{\mathrm{i}}{\mathrm{r}}_{\mathrm{i}}\times {\mathrm{p}}_{\mathrm{i}}^{\mathrm{\prime }}+\sum _{\mathrm{i}}{\mathrm{r}}_{\mathrm{i}}\times {\mathrm{m}}_{\mathrm{i}}\mathrm{V}\\ \mathrm{L}=\sum _{\mathrm{i}}({\mathrm{r}}_{\mathrm{i}}^{\mathrm{\prime }}+\mathrm{R})\times {\mathrm{p}}_{\mathrm{i}}^{\mathrm{\prime }}+\sum _{\mathrm{i}}({\mathrm{r}}_{\mathrm{i}}^{\mathrm{\prime }}+\mathrm{R})\times {\mathrm{m}}_{\mathrm{i}}\mathrm{V}\\ =\sum _{\mathrm{i}}{\mathrm{r}}_{\mathrm{i}}^{\mathrm{\prime }}\times {\mathrm{p}}_{\mathrm{i}}^{\mathrm{\prime }}+\sum _{\mathrm{i}}\mathrm{R}\times {\mathrm{p}}_{\mathrm{i}}^{\mathrm{\prime }}+\sum _{\mathrm{i}}{\mathrm{r}}_{\mathrm{i}}^{\mathrm{\prime }}\times {\mathrm{m}}_{\mathrm{i}}\mathrm{V}+\sum _{\mathrm{i}}\mathrm{R}\times {\mathrm{m}}_{\mathrm{i}}\mathrm{V}\\ ={\mathrm{L}}^{\mathrm{\prime }}+\sum \mathrm{R}\times {\mathrm{p}}_{\mathrm{i}}^{\mathrm{\prime }}+\sum {\mathrm{r}}_{\mathrm{i}}^{\mathrm{\prime }}\times {\mathrm{m}}_{\mathrm{i}}\mathrm{V}+\sum \mathrm{R}\times {\mathrm{m}}_{\mathrm{i}}\mathrm{V}\end{array}$

Where,

$\begin{array}{l}\mathrm{R}\times \sum _{\mathrm{i}}{\mathrm{p}}_{\mathrm{i}}^{\mathrm{\prime }}=0\text{ and }\\ \left(\sum _{\mathrm{i}}{\mathrm{r}}_{\mathrm{i}}^{\mathrm{\prime }}\right)\times \mathrm{MV}=0\\ \sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}=\mathrm{M}\\ \therefore \mathrm{L}={\mathrm{L}}^{\mathrm{\prime }}+\mathrm{R}\times \mathrm{MV}\end{array}$

We have the relation:

$\begin{array}{l}{\mathrm{L}}^{\mathrm{\prime }}=\sum _{\mathrm{i}}{\mathrm{r}}_{\mathrm{i}}^{\mathrm{\prime }}\times {\mathrm{p}}_{\mathrm{i}}^{\mathrm{\prime }}\\ \frac{{\mathrm{dL}}^{\mathrm{\prime }}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\sum {\mathrm{r}}_{\mathrm{i}}^{\mathrm{\prime }}\times {\mathrm{p}}_1^{\mathrm{\prime }})\end{array}$
$\begin{array}{l}=\frac{\mathrm{d}}{\mathrm{dt}}(\sum {\mathrm{r}}_{\mathrm{i}}^{\mathrm{\prime }})\times {\mathrm{p}}_{\mathrm{i}}^{\mathrm{\prime }}+\sum _{\mathrm{i}}{\mathrm{r}}^{\mathrm{\prime }},\times \frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{p}}_{\mathrm{i}}^{\mathrm{\prime }}\right)\\ =\frac{\mathrm{d}}{\mathrm{dt}}\left(\sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}{\mathrm{r}}_{\mathrm{i}}^{\mathrm{\prime }}\right)\times {\mathrm{v}}_{\mathrm{i}}^{\mathrm{\prime }}+\sum _{\mathrm{j}}{\mathrm{r}}_{\mathrm{i}}^{\mathrm{\prime }}\times \frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{p}}_{\mathrm{i}}^{\mathrm{\prime }}\right)\end{array}$

Where. r', is the position vector with respect to the center of mass of the system of particles.

$\begin{array}{l}\therefore \sum {\mathrm{m}}_{\mathrm{i}}{\mathrm{r}}_{\mathrm{i}}^{\mathrm{\prime }}=0\\ \therefore \frac{{\mathrm{dL}}^{\mathrm{\prime }}}{\mathrm{dt}}=\sum {\mathrm{r}}_{\mathrm{i}}^{\mathrm{\prime }}\times \frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{p}}_{\mathrm{i}}^{\mathrm{\prime }}\right)\end{array}$

We have the relation:

$\begin{array}{l}\frac{{\mathrm{dL}}^{\mathrm{\prime }}}{\mathrm{dt}}=\sum _{\mathrm{i}}{\mathrm{r}}_{\mathrm{i}}^{\mathrm{\prime }}\times \frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{p}}_{\mathrm{i}}^{\mathrm{\prime }}\right)\\ =\sum _{\mathrm{i}}{\mathrm{r}}_{\mathrm{i}}^{\mathrm{\prime }}\times {\mathrm{m}}_{\mathrm{i}}\frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{v}}_{\mathrm{i}}^{\mathrm{\prime }}\right)\end{array}$

Where, $\frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{v}}_{\mathrm{i}}^{\mathrm{\prime }}\right)$ is the rate of change of velocity of the ith particle with respect to the center of mass of the system

Therefore, according to Newton's third law of motion, we can write:

$\begin{array}{l}{\mathit{m}}_{\mathit{i}}\frac{\mathit{d}}{\mathit{d}\mathit{t}}\mathit{\left(}{\mathit{v}}_{\mathit{i}}^{\mathit{\prime }}\mathit{\right)}\mathit{=}\mathit{\text{ Extrenal force acting on the }}\mathit{i}\mathit{\text{ th particle }}\mathit{=}\mathit{\sum }_{\mathit{i}}\mathit{\left(}{\mathit{\tau }}_{\mathit{i}}^{\mathit{\prime }}\mathit{\right)}\\ \mathit{\text{ i.e., }}\mathit{\sum }_{\mathit{i}}{\mathit{r}}_{\mathit{i}}^{\mathit{\prime }}\mathit{\times }{\mathit{m}}_{\mathit{i}}\frac{\mathit{d}}{\mathit{d}\mathit{t}}\mathit{\left(}{\mathit{v}}_{\mathit{i}}^{\mathit{\prime }}\mathit{\right)}\mathit{=}{\mathit{\tau }}_{\mathit{\text{ et }}}^{\mathit{\prime }}\mathit{=}\mathit{\text{ External torque acting on the system as a whole }}\end{array}$
$\mathit{\therefore }\frac{\mathit{d}{\mathit{L}}^{\mathit{\prime }}}{\mathit{d}\mathit{t}}\mathit{=}{\tau }_{\mathit{e}\mathit{x}\mathit{f}}^{\mathit{\prime }}$