Separation of motion of a system of particles into motion of the centre of mass and motion about the centre of mass:
(a)n Show `vecp= vecp_(i)+m_(i)vecV` where `vecp_(i)` is the momentum of the `i^(th)` particle (of mass `m_(i))` and `vecp_(i)=m_(i)vec v_(i)`, Not `vec v_(i)` is the velocity of the `i^(th)` particle relative to the centre ofmass.
Also, prove using the definition of the centre of mass `sum vec p_(i)=0`
(b) Show `K=K'+""_(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 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)
(c) Show `vecL= vecL+ vecR xxMV`
where `vecL'=sum vec r_(i) xx vec p_(i)` is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember `vecr_(i)=vecr_(i)-vecR`, reat of the notation is the standard notation used in the chapter. Note `vecL' and M vecR xx vecV` can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.
(d) Show `(d vecL)/(dt)= sum vecr_(i)xx(d vecp')/(dt)` Further show that `(dvecL')/(dt)= vec tau_(ext)'` where Text 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 aet along the line joining the particles.)

Explanation

Refer to theory.