A Firework Bursts Mid-Air, Yet One Point Keeps Flying Along the Same Arc

A spinning, tumbling cricket bat thrown into the air looks complicated — but one point on it follows a clean parabola, just like a small ball would. That point is the centre of mass.

Finding it lets us treat a car, a firework or a group of skaters as one particle for their overall motion, and deal with the spinning separately.

This part covers the centre of mass of particles, of rods and symmetric bodies, its equation of motion, and momentum conservation.

**The centre of mass is the mass-weighted average position: $X=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$ for two particles, and $\vec{R}=\frac{\sum m_i{\vec{r}}_i}{M}$ for $n$ particles of total mass $M$.

Worked example 1 — two particles.** $2$ kg at $x=0$ and $3$ kg at $x=5$ m:

$$X=\frac{2(0)+3(5)}{5}=3\text{ m}$$

The centre of mass lies nearer the heavier particle.

Worked example 2 — three particles in a plane. $1$ kg at $(0,0)$, $2$ kg at $(4,0)$ and $3$ kg at $(0,6)$ m:

$$X=\frac{0+8+0}{6}\approx 1.33\text{ m}Y=\frac{0+0+18}{6}=3\text{ m}$$

An everyday example. Carrying two buckets of water on a shoulder pole, you balance the pole at a point closer to the heavier bucket.

The substance. Taking a different origin changes the coordinates of the centre of mass, but not its physical location.

**For a uniform rod the centre of mass is at its midpoint, and for uniform symmetric bodies such as a ring, disc, sphere or cube it is at the geometric centre; continuous bodies need $X=\frac{1}{M}\int xdm$.

Uniform rod of length $L$.** With $dm=\frac{M}{L}dx$:

$$X=\frac{1}{M}{\int }_0^{L}x\frac{M}{L}dx=\frac{1}{L}\cdot \frac{L^2}{2}=\frac{L}{2}$$

Worked example 1 — two rods joined. A $1$ m rod of $2$ kg is joined end to end with a $1$ m rod of $1$ kg. Treat each as a point at its midpoint, $0.5$ m and $1.5$ m:

$$X=\frac{2(0.5)+1(1.5)}{3}\approx 0.83\text{ m}$$

Worked example 2 — an L-shape. Two identical $1$ m rods form an L along the axes; their midpoints are $(0.5,0)$ and $(0,0.5)$, so the centre of mass is at $(0.25,0.25)$ — outside the material.

Worked example 3 — a non-uniform rod. If the mass per length is $\lambda =kx$:

$$X=\frac{{\int }_0^{L}x(kx)dx}{{\int }_0^{L}kxdx}=\frac{L^3/3}{L^2/2}=\frac{2L}{3}$$

An everyday example. Balancing a 30 cm ruler on one finger works at the $15$ cm mark, the centre of the uniform strip.

The substance. A ring's centre of mass is at its centre, where there is no mass at all.

**Differentiating $M\vec{R}=\sum m_i{\vec{r}}_i$ twice gives $M{\vec{A}}_{cm}=\sum {\vec{F}}_i$, and since internal forces cancel in third-law pairs, only external forces remain: $M{\vec{A}}_{cm}={\vec{F}}_{ext}$.

Derivation.**

$$M{\vec{V}}_{cm}=\sum m_i{\vec{v}}_i=\vec{P}M{\vec{A}}_{cm}=\sum m_i{\vec{a}}_i={\vec{F}}_{ext}+\underset{=0}{\underbrace{{\vec{F}}_{int}}}$$

Worked example 1. A $1$ kg particle feels $6$ N east and a $3$ kg particle feels $2$ N west, while they also pull on each other:

$$A_{cm}=\frac{6-2}{1+3}=1{\text{ m/s}}^2\text{ east}$$

The mutual pulls do not appear at all.

Worked example 2 — an exploding shell. A shell would land $100$ m away. At the top of its path it bursts into two equal pieces; one falls straight down and lands $50$ m from the launch point. The centre of mass still lands at $100$ m:

$$100=\frac{m(50)+m(x)}{2m}\Rightarrow x=150\text{ m}$$

An everyday example. A festival firework shell bursting in the sky scatters sparks everywhere, but their centre of mass keeps following the original arc until pieces start hitting the ground.

The substance. Internal forces can change how the parts move, but never the motion of the centre of mass.

**The total momentum of a system is $\vec{P}=M{\vec{V}}_{cm}$, so when the net external force is zero the total momentum is conserved and the centre of mass moves with constant velocity — or stays at rest.

Worked example 1 — walking on a boat.** A $60$ kg person walks $4$ m along a $240$ kg boat at rest on still water. If the boat moves back by $x$, the centre of mass stays put:

$$60(4-x)=240x\Rightarrow 240=300x\Rightarrow x=0.8\text{ m}$$

The person moves $3.2$ m relative to the shore.

Worked example 2 — skaters on a rope. Skaters of $50$ kg and $70$ kg stand $12$ m apart on smooth ice and pull a rope between them. They meet at the centre of mass, which is

$$\frac{70\times 12}{120}=7\text{ m from the }50\text{ kg skater}$$

An everyday example. Stepping forward on a small boat makes the boat slide back, because the centre of mass of you and the boat cannot move.

The substance. A constant centre-of-mass velocity does not mean each particle moves steadily — the parts can speed up, slow down or spin.