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=p_(i).+m_(i)V where p_(i) is the momentum of the ith 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 Sigmap_(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 (1)/(2)MV^(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 vecL=vecL.+vecRxxvec(MV) where vecL.=Sigmavec(r_(i)).xxvec(p_(i)). is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember vec(r._(i))=vec(r_(i))-vecR, rest of the notation is the velcities taken relative to the centre of mass. Remember vec(r._(i))=vec(r_(i))-vecR rest of the notation is the standard notation used in the chapter. Note vecL, and vec(MR)xxvecV can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles. (d) Show vec(dL.)/(dt)=sumvec(r_(i).)xxvec(dp.)/(dt) Further, show that vec(dL.)/(dt)=tau._(ext) where tau._(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 Thrid Law. Assume the internal forces between any two particles act along the line joining the particles.)