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

**(i)** Show p = 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 with respect to the centre of mass.

Also, verify using the definition of the centre of mass that Σp’_{i} = 0

**(ii)** Prove that K = K′ + \( \dfrac{1}{2}\)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 relative to the centre of mass and MV^{2} /2 is the kinetic energy of the translation of the system as a whole.

**(iii)** Show L = L’+ R × MV

where L’ = ∑r’_{i} × p’_{i} is the angular momentum of the system about the centre of mass with velocities considered with respect to the centre of mass. Note r’_{i} = r_{i} – R, rest of the notation is the standard notation used in the lesson. 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.

**(iv)** Prove that :

\( \dfrac{dL’}{dt}\)= ∑ r’_{i} x \( \dfrac{dp’}{dt}\)

Further prove that :

\( \dfrac{dL’}{dt}\) = τ’_{ext}

Where τ’_{ext} is the sum of all external torques acting on the system about the centre of mass. ( Clue : Apply Newton’s Third Law and the definition of centre of mass . Consider that internal forces between any two particles act along the line connecting the particles.)