$$K_{\mathrm {translational}}= \frac{1}{2} Mv_{\mathrm {com}}^2$$
Why does the term for translational kinetic energy include only the velocity of the centre of mass of a rigid body? How can we ignore the velocity of the different particles constituting the system?
Can someone prove this to me? I tried finding it on net but couldn't as at most places there was derivation of translational $K$ for ideal gases given.
The notion of 'translational' and 'rotational' kinetic energy comes by starting from the kinetic energy of a point particle (which is only translational), and building up a notion of the properties of systems of particles.
It turns out that the energy factors into a portion that has the $\frac{1}{2}Mv_\text{com}^2$ form (where $M$ is the total mass and $v_\text{com}$ is the velocity of the center of mass) and a portion due to motion of the parts relative the center of mass (which for a rigid body is the rotational kinetic energy).
This development is purely algebraic so you might be able to work it yourself with no more hint than I've given you, but it is also shown in every serious mechanics book.
