__Question 7.33__: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**

_{i}=

**p**’

_{i }+

*m*

_{i}

**V**

Where

**p**

_{i}is the momentum of the

*i*

^{th}particle (of mass

*m*

_{i}) and

**p**′

_{i}=

*m*

_{i}

**v**′

_{i}. Note

**v**′

_{i}is the velocity of the

*i*

^{th}particle relative to the centre of mass.

Also, prove using the definition of the centre of mass ∑

**p**′

_{i}= 0

(b) Show

*K*=

*K*′ + ½

*MV*

^{2}

Where

*K*is the total kinetic energy of the system of particles,

*K*′ is the total kinetic energy of the system when the particle velocities are taken with respect to the centre of mass and

*MV*

^{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). The result has been used in Sec. 7.14.

(c) Show

**L**=

**L**′ +

**R**×

*M*

**V**

Where

**L**' = ∑ r' X p' is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember

**r**’

_{i }=

**r**

_{i }–

**R**; rest of the notation is the standard notation used in the chapter. Note

**L**′ and

*M*

**R**×

**V**can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.

(d) Show d

**L**' / dt = ∑

**r**' X d(

**p**')/dt

Further, show that

d

**L**' / dt =

**τ'**

_{ext}_{}

where τ'

_{ext}

_{}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.)

__Solution__:**(a)**Take a system of

*i*moving particles.

Mass of the

*i*

^{th}particle =

*m*

_{i}

Velocity of the

*i*

^{th}particle =

**v**

_{i}

Hence, momentum of the

*i*

^{th}particle,

**p**

_{i }=

*m*

_{i}

**v**

_{i}

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 }=

**v**

_{i }–

**V**… (1)

Multiplying

*m*

_{i}throughout equation (1), we get:

*m*

_{i}

**v**’

_{i }=

*m*

_{i}

**v**

_{i }–

*m*

_{i}

**V**

**p’**

_{i }=

**p**

*–*

_{i }

_{}*m*

_{i}**V**

Where,

**p**

_{i}’ =

*m*

_{i}

**v**

_{i}’ = Momentum of the

*i*

^{th}particle with respect to the centre of mass of the system

∴

**p**

_{i }=

**p’**

_{i }+

_{ }

*m*

_{i}

**V**

We have the relation:

**p’**

_{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:

∑

**p'**

_{i}= ∑ m

_{i}

**v**

_{i}' = ∑ m

_{i}(d

**r**' / dt)

_{i}Where,

**r**' = Position vector of ith particle with repsect to the centre of mass

_{i}**v**d

_{i}' =**r**' / dt

_{i}As per the definition of the centre of mass, we have

∑ m

_{i}

**r**

_{i}' = 0

∴ ∑ m

_{i}(d

**r**/ dt) = 0

_{i}'∑ p

_{i}' = 0

**(b)**We have the relation for velocity of the

*i*

^{th}particle as:

**v**

_{i }=

**v’**

_{i }+

**V**

∑ m

_{i}

**v**

_{i}= ∑ m

_{i}

**v**

_{i}' + ∑ m

_{i}

**V**....(ii)

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

∑ m

_{i}

**v**

_{i}. ∑ m

_{i}

**v**

_{i }= ∑ m

_{i}(

**v**) . ∑ m

_{i}' + V_{i}(

**v**)

_{i}' + VM

^{2}∑ v

_{i}

^{2}= M

^{2}∑ v

_{i}

^{2}+ M

^{2}∑ v

_{i}v

_{i}' + M

^{2}∑ v

_{i}' v

_{i }+ M

^{2}V

^{2}

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

_{i}v

_{i}' = -∑ v

_{i}' v

_{i}

M

^{2}∑ v

_{i}

^{2}= M

^{2}∑v

_{i}'

_{}

^{2}+ M

^{2}V

^{2}

^{}

Multiplying both sides by (1/2)

(1/2)M∑ v

_{i}

^{2}= (1/2)M∑v

_{i}'

^{2}+ (1/2)M

^{}V

^{2}

^{}K = K' + (1/2)MV

^{2}

Where K = (1/2)M∑ v

_{i}

^{2}= Total energy of the system of particles

K' = (1/2)M∑v

_{i}'

^{2}= Total kinetic energy of the system of particles with respect to the centre of mass

(1/2)MV

^{2}= inetic energy of the translation of the system as a whole

**(c)**Position vector of the

*i*

^{th}particle with respect to origin =

**r**

_{i}

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 }=

**r**

_{i }–

**R**

**r**

_{i }=

**r’**

_{i }+

**R**

We have from part

**(a)**,

**p**

_{i }=

**p’**

_{i }+

_{ }

*m*

_{i}

**V**

Taking the cross product of this relation by

**r**

_{i}, we get:

∑

**r**

_{i}X p_{i}= ∑

**r**

_{i}X p_{i}' + ∑

**r**

_{i}X m

_{i}

**V**

**L**= ∑ (

**r'**) X

_{i}+ R**p**

_{i}' + ∑ (

**r'**) X m

_{i}+ R_{i}

**V**

= ∑

**r'**'

_{i}X p_{i}+ ∑

**R X p**'

_{i}+ ∑

**r**'

_{i}X m

_{i}

**V**+ ∑

**R**X m

_{i}

**V**

=

**L'**+ ∑

**R X p**'

_{i}+ ∑

**r'**

_{i}X m

_{i}

**V**+ ∑

**R**X m

_{i}

**V**

Where,

**R X**

**∑p**'

_{i}= 0 and

( ∑

**r**'

_{i}) X M

**V**= 0

∑ m

_{i}= M

∴

**L = L**' +

**R**X M

**V**

**(d)**We have the relation:

**L' = ∑ r'**

_{i}X p'_{i}d

**L**' / dt = d(∑

**r'**'

_{i}X p_{i}) / dt

= (d/dt) ( ∑ r'

_{i}) X

**p'**

_{i}+**∑ r**'

_{i}X (d/dt)(

**p**'

_{i})

= (d/dt) ( ∑ m

_{i}

**r**'

_{i}) X

**v**'

_{i}+ ∑

**r**'

_{i}X (d/dt)(

**p**'

_{i})

Where, r'

_{i}is the position vector with respect to the centre of mass of the system of particles,

∴ ∑ m

**'**

_{i}r_{i}= 0

∴ d

**L**' / dt = ∑

**r**'

_{i}X (d/dt)(

**p**'

_{i})

We have the ralation:

d

**L**' / dt = ∑

**r**'

_{i}X (d/dt)(

**p'**

_{i})

= ∑

**r**'

_{i}X m

_{i}(d/dt) (

**v**'

_{i})

Where, (d/dt) (

**v'**

_{i}) is the rate of change of velocity of the i

^{th}particle with respect to the centre of mass of the system

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

m

_{i}(d/dt)(

**v**'

_{i}) = External force acting on the ith particle = ∑ (

**τ**'

_{i})

_{ext}

i.e. ∑

**r**'

_{i}X m

_{i}(d/dt)(

**v'**

_{i}) =**τ'**

_{ext}= External torque acting on the system as a whole

∴ d

**L**' / dt =

**τ**'

_{ext}

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