Separation of Motion of a system of particles into motion of the center of mass and motion about the center of mass: A. Show Where p i is the momentum of the i th particle (of mass m i ) and Note is the velocity of the i th particle relative to the canter of mass. Also, prove using the definition of the centre of mass B. Show 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 is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember; rest of the notation is the standard notation used in the chapter. Note and MR × V can be said to be angular momenta, respectively, about and of the canter of mass of the system of particles. D. Show Further, show that Where is the sum of all external torques acting on the system about the canter of mass. (Hint: Use the definition of canter of mass and Newton’s Third Law. Assume the internal forces between any two particles act along the line joining the particles.)

Question

Philoid · Accepted Answer

A. Take a system of I moving particles:

Mass of the i^th particle = m_i

Velocity of the i^th particle = v_i

∴ momentum of the i^th particle, p_i = m_iv_i

Velocity of the canter of the mass = V

Thus, velocity of the i^th particle with respect to canter of the mass, v_i’ = v_i – V …..(1)

Multiplying m_i throughout the equation (1) we get,

⇒ m_iv_i’ = m_iv_i – m_iV

⇒ p_i’ = p_i - m_iV

Where,

p_i’ = momentum of the i^th particle with respect to the canter of the mass.

Taking the summation of the momentums of the all the particle with respect to the canter the mass of the body,

…..(2)

Where,

r_i’ = position of the i^th particle with respect to the canter of mass and,

As per the definition of the canter of the mass, we have,

∴ from equation 2,

B. We have the relation for velocity of the particle as,

…..(3)

Taking the dot product with itself will give,

Here, For the canter of mass of the system of the particle,

Therefore,

C. Total angular momentum of the system of the particles,

The las two terms vanish for both contain the factor which is equal to zero from the definition of the center of the mass. Also,

So that,

D. From the previous solution,

(∵ ) Total toque,