Velocity of Centre of Mass

Derivation

Differentiate the CM position with respect to time:

$\vec{r}_{cm} = \frac{\sum m_i \vec{r}_i}{M}$

$\frac{d\vec{r}_{cm}}{dt} = \frac{\sum m_i \frac{d\vec{r}_i}{dt}}{M}$

$\boxed{\vec{v}_{cm} = \frac{\sum m_i \vec{v}_i}{M} = \frac{\vec{p}_{total}}{M}}$

The velocity of the CM equals the total momentum of the system divided by the total mass.

Momentum interpretation

$M\vec{v}_{cm} = \sum m_i \vec{v}_i = \vec{p}_{total}$

The total momentum of the system equals the momentum of a single particle of mass $M$ moving with the CM velocity. This is why the CM is so important — it captures the entire momentum of the system.

Conservation of momentum and CM motion

If $\vec{F}_{ext} = 0$: total momentum is conserved → $\vec{v}_{cm}$ is constant.

A system with no external forces has its CM moving at constant velocity (or at rest). Whatever the internal interactions — explosions, collisions, rotations — the CM moves in a straight line at constant speed.

Example: A grenade flying through the air explodes. Each fragment follows its own parabolic path. But the CM of all fragments continues on the original parabolic path as if the explosion never happened — because gravity is the only external force, same as before.

Example: Two bodies

$m_1 = 3$ kg at $v_1 = 4$ m/s (right) and $m_2 = 1$ kg at $v_2 = -2$ m/s (left):

$v_{cm} = \frac{3(4) + 1(-2)}{4} = \frac{12-2}{4} = \frac{10}{4} = 2.5 \text{ m/s (right)}$

CM velocity and kinetic energy

$KE = \frac{1}{2}Mv_{cm}^2 + KE_{internal}$

Total KE = KE of CM motion + KE of motion relative to CM.

The CM contribution $\frac{1}{2}Mv_{cm}^2$ is the "translational" KE. The internal KE is the KE in the CM frame.