Question

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

Show p_i = p’i + miV

Where p_i is the momentum of the i^th particle (of mass mi) 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

Show K = K′ + ½MV²

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 MV²/2 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.

Show L = L′ + R × MV

Where

is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember r’i = r_i – 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.

where τ′_ext is the sum of all external torques acting on the system about the centre of mass.

Answer 1

The velocity of the i^th particle with respect to the centre of mass of the system is given

p_i’ = m_iv_i’ = Momentum of the i^th particle with respect to the centre of mass of the

Taking the summation of momentum of all the particles with respect to the centre of mass of the system, we get:

Position vector of the i^th particle with respect to the centre of mass = r’i
Position vector of thecentre of mass with respect to the origin = R

We have the relation: