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**_{i} = **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 relative to the centre of mass.

Also, prove using the definition of the centre of mass `sump'_t = 0`

b) Show K = K′ +1/2MV^{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). The result has been used in Sec. 7.14.

c)Show where `L' = sumr'_t xx p'_t` is the angular momentum of the system about the centre of mass with

velocities taken relative to the centre of mass. Remember `r'_t = r_t - 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 = sum r'_t xx (dp')/dt`

Further, show that

`(dL')/(dt) = t'_"ext"`

where `t'_"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.)