# Why is translational kinetic energy defined only for the centre of mass' velocity?

$$K_{\mathrm {translational}}= \frac{1}{2} Mv_{\mathrm {com}}^2$$

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.

• Isn't this just a matter of definition? By labelling $K$ as translational you are specifying that it is only the energy of the COM. The energy of motion of the components of your system would be counted as internal energy or heat. – John Rennie Sep 25 '17 at 16:28
• But @JohnRennie the individual particles have velocity too...Also specifying $\equiv$ defining. – Archer Sep 25 '17 at 16:29
• Yes, and we call that internal energy or heat. So total energy is the kinetic energy plus the internal energy. It isn't that we ignore the energy of the individual particles, just that we call it something else. – John Rennie Sep 25 '17 at 16:29
• What would be the formula for that @JohnRennie ? And I strongly feel that the formula should include velocity. – Archer Sep 25 '17 at 16:30
• We should take this to the chat – John Rennie Sep 25 '17 at 16:30

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).