System of particles and Rotational motion NCERT Solutions Class 11 Physics - Solved Exercise Question 7.33

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 pi = pi + miV
Where pi is the momentum of the ith particle (of mass mi) and pi = mi vi. Note vi is the velocity of the ith particle relative to the centre of mass.
Also, prove using the definition of the centre of mass pi = 0

(b) Show K = K′ + ½MV2
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 MV2/2 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 × MV
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 ri = ri R; rest of the notation is the standard notation used in the chapter. Note L′ and MR × V can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.

(d) Show dL' / dt  =  r' X d(p')/dt
Further, show that
dL' / 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 ith particle = mi
Velocity of the ith particle = vi
Hence, momentum of the ith particle, pi = mi vi
Velocity of the centre of mass = V
The velocity of the ith particle with respect to the centre of mass of the system is given as:
vi = vi V … (1)
Multiplying mi throughout equation (1), we get:
mi vi = mi vi mi V
p’i = pi ­mi V
Where,
pi’ = mivi’ = Momentum of the ith 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:
p'i = ∑ mivi' = ∑ mi(dri' / dt)
Where,
ri' = Position vector of ith particle with repsect to the centre of mass
vi' = dri' / dt
As per the definition of the centre of mass, we have
∑ miri' = 0
∑ mi (dri' / dt)  =  0
∑ pi' = 0

(b) We have the relation for velocity of the ith particle as:
vi = v’i + V
∑ mivi = ∑ mivi' + ∑ miV    ....(ii)
Taking the dot product of equation (2) with itself, we get:
∑ mivi . ∑ miv∑ mi(vi' + V) . ∑ mi(vi' + V)
M2 ∑ vi2  =  M2 ∑ vi2 + M2 ∑ vi vi' + M2 ∑ vi' vi + M2V2
Here, for the centre of mass of the system of particles, ∑ vi vi' = -∑ vi' vi
M2∑ vi2  =  M2vi'2 + M2V2
Multiplying both sides by (1/2)
(1/2)M∑ vi2  =  (1/2)M∑vi'2 + (1/2)MV2
K = K' + (1/2)MV2
Where K = (1/2)M∑ vi2 = Total energy of the system of particles
K' = (1/2)M∑vi'2 = Total kinetic energy of the system of particles with respect to the centre of mass
(1/2)MV2 = inetic energy of the translation of the system as a whole
(c) Position vector of the ith particle with respect to origin = ri
Position vector of the ith particle with respect to the centre of mass = ri
Position vector of the centre of mass with respect to the origin = R
It is given that:
ri = 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:
ri X pi  =  ri X pi' + ri X miV
L = ∑ (r'i + R) X pi' + ∑ (r'i + R) X miV
= r'i X p'i + R X p'i + r'i X miV + R X miV
= L' + R X p'i + r'i X miV + R X miV
Where,
R X ∑p'i = 0 and 
(r'i) X MV = 0
∑ mi = M
L = L' + R X MV

(d) We have the relation:
L' = ∑ r'i X p'i
dL' / dt = d( r'i X p'i) / dt
= (d/dt) ( ∑ r'i) X p'i + ∑ r'i X (d/dt)(p'i)
= (d/dt) ( ∑ mir'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,
∑ mir'i = 0
∴ dL' / dt  = r'i X (d/dt)(p'i)
We have the ralation:
dL' / dt = r'i X (d/dt)(p'i)
= r'i X mi (d/dt) (v'i)
Where, (d/dt) (v'i) is the rate of change of velocity of the ith particle with respect to the centre of mass of the system
Therefore, according to Newton's third law of motion, we can write,
mi(d/dt)(v'i) = External force acting on the ith particle =  (τ'i )ext
i.e. r'i X mi (d/dt)(v'i) = τ'ext = External torque acting on the system as a whole
∴ dL' / dt = τ'ext

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