Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass

$(a)$ Show $p = p _{t}^{\prime}+m_{t} V$

where $p$, is the momentum of the the particle (of mass $m$ ) and $p_{t}^{\prime \prime}=m_{t} v_{t}$,

Note $v_{t}$, is the velocity of the particle relative to the centre of mass. Also, prove using the definition of the centre of mass $\sum p _{t}^{\prime}=0$

$(b)$ Show $K=K^{\prime}+1 / 2 M V^{2}$

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 $L = L ^{\prime}+ R \times M V$

where $L ^{\prime}=\sum r _{t}^{\prime} \times p _{t}^{\prime}$ is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember $r _{t}^{\prime}= r _{t}- R$; rest of the notation is the standard notation used in the chapter. Note $L$ ' and $M R \times V$ can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.

$(d)$ Show $\frac{d L ^{\prime}}{d t}=\sum r _{t}^{\prime} \times \frac{d p ^{\prime}}{d t}$

Further, show that

$\frac{d L ^{\prime}}{d t}=\tau_{e x t}^{\prime}$

where $\tau_{c t t}^{\prime}$ 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 third law of motion. Assume the internal forces between any two particles act along the line joining the particles.)

 

$(a)$Take a system of $i$ moving particles.

Mass of the $i^{th}$ particle $=m_{i}$

Velocity of the $i^{\text {th}}$ particle $= v _{i}$

Hence, momentum of the $i^{\text {th}}$ particle, $p _{i}=m_{i} v _{i}$

Velocity of the centre of mass $= V$

The velocity of the $i^{\text {th }}$ particle with respect to the centre of mass of the system is given

as: $v ^{\prime}_{i}= v _{i}- V \ldots(1)$

Multiplying $m_{i}$ throughout equation $(1)$, we get:

$m_{i} v ^{\prime}_{i}=m_{i} v _{i}-m_{i} V$

$p ^{\prime}_{i}= p _{i}-m_{i} V$

Where,

$p _{i}^{\prime}=m_{i} v _{i}^{\prime}=$ Momentum of the $i^{\text {th }}$ particle with respect to the centre of mass of the

system $: p _{i}= p ^{\prime} i+m_{i} V$

We have the relation: $p ^{\prime} i=m_{i} v _{i}$

Taking the summation of momentum of all the particles with respect to the centre of mass of the system, we get:

$\sum_{i} p _{i}^{\prime}=\sum_{i} m_{i} v _{i}^{\prime}=\sum_{i} m_{i} \frac{d r _{i}^{\prime}}{d t}$

Where,

$r ^{\prime}=$ Position vector of $i$ th particle with respect to the centre of mass $v _{t}^{\prime}=\frac{d r ^{\prime}}{d t}$

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

$\sum m_{i} r _{i}^{\prime}=0$

$\therefore \sum \limits _{i} m_{i} \frac{d r ^{\prime}}{d t}=0$

$\sum \limits _{i} p _{i}^{\prime}=0$

We have the relation for velocity of the $i^{\text {th }}$ particle as:

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

$\sum_{i} m_{i} v _{i}=\sum_{i} m_{i} v _{i}^{\prime}+\sum_{i} m_{i} V$

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

$\sum_{i} m_{i} v _{i}, \sum_{i} m_{i} v _{i}=\sum_{i} m_{i}\left( v _{i}^{\prime}+ v \right) \cdot \sum_{i} m_{i}\left( v _{i}^{\prime}+ v \right)$

$M^{2} \sum_{i} v_{i}^{2}=M^{2} \sum_{i} v_{i}^{2}+M^{2} \sum_{i} v _{i} v _{i}^{\prime}+M^{2} \sum_{i} v _{i}^{\prime} v _{i}+M^{2} V^{2}$

Here, for the centre of mass of the system of particles,

$\sum\limits_{i} v _{i} v ^{\prime}_i=-\sum\limits_{i} v _{i}^{\prime} v _{i}$

$M^{2} \sum_{i} v_{i}^{2}=M^{2} \sum_{i} v_{i}^{2}+M^{2} V^{2}$

$\frac{1}{2} M \sum_{i} v_{i}^{2}=\frac{1}{2} M \sum_{i} v_{i}^{\prime 2}+\frac{1}{2} M V^{2}$

$K=K^{\prime}+\frac{1}{2} M V^{2}$

Where,

$K= \frac{1}{2} M \sum_{i} v_{i}^{2}=$ Total kinetic energy of the system of particles

$K= \frac{1}{2} M \sum_{i} v_{i}^{\prime \prime 2}=$ 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 $i^{\text {th }}$ particle with respect to origin $= r _{i}$

Position vector of the $i$ "particle with respect to the centre of mass $= r ^{\prime} i$

Position vector of the centre of mass with respect to the origin $= R$

It is given that: $r ^{\prime} i= r i$

$- R r _{i}= r _{i}+ R$ We

have from part (a), pi

$= p ^{\prime} i+m i V$

Taking the cross product of this relation by $r _{i},$ we get:

$\sum_{i} r _{i} \times p _{i}=\sum_{i} r _{i} \times p _{i}^{\prime}+\sum_{i} r _{i} \times m_{i} V$

$L =\sum\left( r _{i}^{\prime}+ R \right) \times p _{i}^{\prime}+\sum\left( r _{i}^{\prime}+ R \right) \times m_{i} V$

$=\sum_{i} r _{i}^{\prime} \times p _{i}^{\prime}+\sum_{i} R \times p _{i}^{\prime}+\sum_{i} r _{i}^{\prime} \times m_{i} V +\sum_{i} R \times m_{i} V$

$= L ^{\prime}+\sum_{i} R \times p _{i}^{\prime}+\sum_{i} r _{i}^{\prime} \times m_{i} V +\sum_{i} R \times m_{i} V$

Where,

$R \times \sum_{i} p _{i}^{\prime}=0$ and

$\left(\sum_{i} r _{i}^{\prime}\right) \times M V = 0$

$\sum_{t} m_{i}=M$

$\therefore L = L ^{\prime}+R \times M V$

We have the relation:

$L ^{\prime}=\sum r ^{\prime}, \times p ^{\prime}$

$\frac{d L ^{\prime}}{d t}=\frac{d}{d t}\left(\sum r ^{\prime} \times p ^{\prime}\right)$

$=\frac{d}{d t}\left(\sum_{t} r _{t}^{\prime}\right) \times p _{i}^{\prime}+\sum_{t} r _{t}^{\prime} \times \frac{d}{d t}\left( p _{i}^{\prime}\right)$

$=\frac{d}{d t}\left(\sum_{i} m_{i} r _{i}^{\prime}\right) \times v _{i}^{\prime}+\sum_{i} r _{i}^{\prime} \times \frac{d}{d t}\left( p _{i}^{\prime}\right)$

Where, $r ^{\prime}$, is the position vector with respect to the centre of mass of the system of particles. $\therefore \sum m_{i} r _{i}^{\prime}=0$

$\therefore \frac{d L ^{\prime}}{d t}=\sum_{t} r _{t}^{\prime} \times \frac{d}{d t}\left( p _{i}^{\prime}\right)$

We have the relation:

$\frac{d L ^{\prime}}{d t}=\sum_{i} r ^{\prime}, \times \frac{d}{d t}\left( p _{i}^{\prime}\right)$

$=\sum_{i} r ^{\prime}, \times m_{i} \frac{d}{d t}\left( v ^{\prime}\right)$

Where, $\frac{d}{d t}\left( v ^{\prime}\right)$ is the rate of change of velocity of the $i$ th particle

with respect ot the centre of mass of the system

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

$m_{i} \frac{d}{d t}\left( v _{i}^{\prime}\right)=$ Extrenal force acting on the $i$ th particle $=\sum_{i}\left(\tau_{i}^{\prime}\right)$

i.e., $\sum_{i} r ^{\prime}, \times m_{i} \frac{d}{d t}\left( v _{i}^{\prime}\right)=\tau_{ ea }^{\prime}=$ External torque acting on the system as a whole

$\therefore \frac{d L ^{\prime}}{d t}=\tau^{\prime}_{ext }$