Question

Separating motion of a system of particles into motion of the centre of mass and motion about the centre of mass, show that K = K' + 1/2 MV²where 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 is the kinetic energy of the translation of the system as a whole (i.e., of the centre of mass motion of the system)

Answer 1

We have the relation for velocity of the ith particle as : vi = v'i + V

Or, mi vi = mi(v'i + V)

taking dot product of above equation with itself.
(mi vi).(mi vi) = mi(v'i + V).mi(v'i + V)

or, mi² vi² = mi²v'i² + mi²V² + mi²v'iV + mi²V.v'i

for the centre of mass of the system,
v'i.V = - V.vi'

So, mi²vi² = mi²V² + mi²v'i²

Or, 1/2 mi²vi² = 1/2 mi²V² + 1/2 mi²v'i²

or, K.Ei = K.E'i + 1/2 MV² [ as we assume that, sum of mass of particles , sigma mi = M ]

Hence, K = K' + 1/2 MV²

Answer 2

Answer:

Explanation:

Derive the equation for the kinetic energy and potential energy of a simple harmonic oscillator and show that the total energy of a particle in simple harmonic motion is constant at any point on its path