# UY1: Motion Of A System Of Particles

We have calculated the centre of mass of a system of particles. How does it related to the movement of a system of particles?

The centre of mass of a system of particles:

$$\vec{r}_{CM} = \frac{\sum\limits_{i} m_{i}\vec{r}_{i}}{M}$$

The velocity of the centre of mass will be:

\begin{aligned} \vec{v}_{CM} &= \frac{d \vec{r}_{CM}}{dt} \\ &= \frac{1}{M} \sum\limits_{i} m_{i} \frac{d\vec{r}_{i}}{dt} \\ &= \frac{\sum\limits_{i} m_{i} \vec{v}_{i}}{M} \end{aligned}

The total momentum of a system of particles will then be:

\begin{aligned} M \vec{v}_{CM} &= \sum\limits_{i} m_{i} \vec{v}_{i} \\ &= \sum\limits_{i} \vec{p}_{i} \\ &= \vec{p}_{\text{tot}} \end{aligned}

From the above equation, we can see that the total linear momentum of a system of particles is equal to that of a single particle of mass $M$ moving with a velocity $\vec{v}_{CM}$.

Let’s find the acceleration of the centre of mass of the system of particles:

\begin{aligned} \vec{a}_{CM} &= \frac{d \vec{v}_{CM}}{dt} \\ &= \frac{1}{M} \sum\limits_{i} m_{i} \frac{d\vec{v}_{i}}{dt} \\ &= \frac{1}{M} \sum\limits_{i} m_{i} \vec{a}_{i} \end{aligned}

From Newton’s second law for a system of particles, we can see that:

$$M\vec{a}_{CM} = \sum\limits_{i} m_{i}\vec{a}_{i} = \sum\limits_{i} \vec{F}_{i}$$

Since $\vec{F}_{i}$ includes both internal and external forces, the internal forces will cancel out, leaving the external forces. We will have:

$$\sum \vec{F}_{\text{ext}} = M \vec{a}_{CM} = \frac{d \vec{p}_{tot}}{dt}$$

From the above equation, we can see that the centre of mass of a system of particles moves like an imaginary particle of mass $M$ under the influence of the external resultant force on the system.

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