Q. 333.5( 10 Votes )

# Separation of Mot

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_{i}v_{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_{i}v_{i}’ = m_{i}v_{i} – m_{i}V

⇒ p_{i}’ = p_{i} - m_{i}V

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,

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