Motion of the Centre of Mass

A high-jumper twists in mid-air; an exploding firework scatters in all directions. Yet for every such system, one point moves with eerie regularity: the centre of mass moves only under the action of external forces, exactly as if all the mass were concentrated there.

Concept

Velocity of CM:

$\vec{V}_{\text{CM}} = \frac{1}{M}\sum_i m_i \vec{v}_i = \frac{\vec{P}}{M}$

where $\vec{P}$ is total momentum. So $\vec{P} = M\vec{V}_{\text{CM}}$.

Acceleration of CM:

$\vec{A}_{\text{CM}} = \frac{1}{M}\sum_i m_i \vec{a}_i = \frac{\vec{F}_{\text{ext}}}{M}$

Internal forces, by Newton's third law, cancel in pairs and contribute nothing.

Newton's second law for a system:

$\vec{F}_{\text{ext}} = M\vec{A}_{\text{CM}} = \frac{d\vec{P}}{dt}$

Consequence: If $\vec{F}_{\text{ext}} = 0$, total momentum is conserved and the CM moves with constant velocity (or stays at rest).

Derivation

Start with the CM definition:

$M\vec{R}_{\text{CM}} = \sum_i m_i \vec{r}_i$

Differentiate with respect to time:

$M\vec{V}_{\text{CM}} = \sum_i m_i \vec{v}_i = \vec{P}_{\text{total}}$

Differentiating again:

$M\vec{A}_{\text{CM}} = \sum_i m_i \vec{a}_i = \sum_i \vec{F}_i$

The sum $\sum_i \vec{F}_i$ over each particle's net force splits into external + internal:

$\sum_i \vec{F}_i = \vec{F}_{\text{ext}} + \sum_{i\neq j} \vec{F}_{ji}$

The internal forces $\vec{F}_{ji}$ come in third-law pairs and cancel:

$\sum_{\text{pairs}} (\vec{F}_{ji} + \vec{F}_{ij}) = 0$

Hence

$M\vec{A}_{\text{CM}} = \vec{F}_{\text{ext}}$

— Newton's second law applied to the whole system.

Worked Example

A shell of mass 5 kg is fired with velocity 100 m/s at 60° above the horizontal. At the highest point of its trajectory, it explodes into two equal fragments. One fragment falls straight down (velocity 0 horizontal, some vertical) due to a peculiar mechanism. Where does the other fragment land relative to the launch point?

Solution:

Without explosion, the projectile would land at horizontal range $R = u^2 \sin 2\theta / g = 100^2 \sin 120° / 10 \approx 866$ m.

The CM continues unaffected by internal forces, so it lands at $R = 866$ m.

The CM lands at the midpoint of the two fragments (equal masses). If fragment 1 falls straight down from the apex, it lands at $R/2 = 433$ m.

For CM to land at 866 m:

$\frac{x_1 + x_2}{2} = 866 \Rightarrow 433 + x_2 = 1732 \Rightarrow x_2 = 1299 \, \text{m}$

The other fragment lands at 1299 m, much beyond the original range.

Common Confusions

• "Internal forces can move the CM." They cannot. No matter how violently parts interact, the CM responds only to external forces.
• CM of an exploding firework still follows a parabola (until air drag becomes important).
• CM at rest stays at rest under no external force. Two skaters pushing each other apart: each moves, but the CM stays put if no friction acts.
• Mass distribution must be tracked. If parts fragment, you need their masses and positions to relate them through the CM.

Key Takeaways

• $\vec{P} = M\vec{V}_{\text{CM}}$ — total momentum lives at the CM.
• $\vec{F}_{\text{ext}} = M\vec{A}_{\text{CM}}$ — Newton's second law for the whole system.
• Internal forces cancel and don't affect CM motion.
• An exploding projectile's CM continues its original parabolic path.
• If $\vec{F}_{\text{ext}} = 0$, the CM moves with constant velocity.