systems of particles and rotational motion


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 + miV

Where pi is the momentum of the ith particle (of mass mi ) and p’i = mivi‘.  Note v’i is the velocity of the ith 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′ + \( \dfrac{1}{2}\)MV2

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 MV2 /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 = ri – 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.)