Motion Of Centre Of Mass

Equipped with the definition of the centre of mass, we are now in a position to discuss its physical importance for a system of n particles. We may rewrite Eq.(7.4d) as

$M\mathbf{R}=\Sigma m_i \mathbf{r_i}=m_1\mathbf{r_1}+m_2\mathbf{r_2}+...+m_n\mathbf{r_n}$

Differentiating the two sides of the equation with respect to time we get

$M\frac {d\mathbf{R}}{dt}=\Sigma m_i \mathbf{r_i}=m_1\frac{d\mathbf{r_1}}{dt}+m_2\frac{d\mathbf{r_2}}{dt}+...+m_n\frac{\mathbf{r_n}}{dt}$

or

$M\mathbf{V}=m_1\mathbf{v_1}+m_2\mathbf{v_2}+...+m_n\mathbf{v_n}$

where $\mathbf{v_1}(=\frac{d\mathbf{r_1}}{dt})$  is the velocity of the first particle $\mathbf{v_2}(=\frac{d\mathbf{r_2}}{dt})$ is the velocity of the second particle etc. and $\mathbf{V}=\frac{d\mathbf{R}}{dt})$  is the velocity of the centre of mass. Note that we assumed the masses $m_1$, $m_2$, ... etc. do not change in time. We have therefore, treated them as constants in differentiating the equations with respect to time.

Differentiating Eq.(7.8) with respect to time, we obtain

$M\frac{d\mathbf{V}}{dt}=m_1\frac{d\mathbf{v_1}}{dt}+m_2\frac{d\mathbf{v_2}}{dt}+...+m_n\frac{d\mathbf{v_n}}{dt}$

or

$M\mathbf{A}=m_1\mathbf{a_1}+m_2\mathbf{a_2}+...+m_n\mathbf{a_n}$

where $\mathbf{a_1}(=\frac{d\mathbf{v_1}}{dt})$ is the acceleration of the first particle, $\mathbf{a_2}(=\frac{d\mathbf{v_2}}{dt})$ is the acceleration of the second particle etc. and $\mathbf{A}=\frac{d\mathbf{V}}{dt})$ is the acceleration of the centre of mass of the system of particles.

Now, from Newton’s second law, the force acting on the first particle is given by $\mathbf{F_1} = m_1 \mathbf{a_1}$. The force acting on the second particle is given by $\mathbf{F_2} = m_2 \mathbf{a_2}$ and so on. Eq. (7.9) may be written as

$M\mathbf{A}=\mathbf{F_1}+\mathbf{F_2}+...+\mathbf{F_n}$

Thus, the total mass of a system of particles times the acceleration of its centre of mass is the vector sum of all the forces acting on the system of particles.

Note when we talk of the force $\mathbf{F_1}$ on the first particle, it is not a single force, but the vector sum of all the forces on the first particle; likewise for the second particle etc. Among these forces on each particle there will be external forces exerted by bodies outside the system and also internal forces exerted by the particles on one another. We know from Newton’s third law that these internal forces occur in equal and opposite pairs and in the sum of forces of Eq. (7.10), their contribution is zero. Only the external forces contribute to the equation. We can then rewrite Eq. (7.10) as

$M\mathbf{A}=\mathbf{F_{ext}}$

where $\mathbf{F_{ext}}$ represents the sum of all external forces acting on the particles of the system.

Eq. (7.11) states that the centre of mass of a system of particles moves as if all the mass of the system was concentrated at the centre of mass and all the external forces were applied at that point.

Notice, to determine the motion of the centre of mass no knowledge of internal forces of the system of particles is required; for this purpose we need to know only the external forces.

To obtain Eq. (7.11) we did not need to specify the nature of the system of particles. The system may be a collection of particles in which there may be all kinds of internal motions, or it may be a rigid body which has either pure translational motion or a combination of translational and rotational motion. Whatever is the system and the motion of its individual particles, the centre of mass moves according to Eq. (7.11).

Instead of treating extended bodies as single particles as we have done in earlier chapters, we can now treat them as systems of particles. We can obtain the translational component of their motion, i.e. the motion of the centre of mass of the system, by taking the mass of the whole system to be concentrated at the centre of mass and all the external forces on the system to be acting at the centre of mass.

This is the procedure that we followed earlier in analysing forces on bodies and solving problems without explicitly outlining and justifying the procedure. We now realise that in earlier studies we assumed, without saying so, that rotational motion and/or internal motion of the particles were either absent or negligible. We no longer need to do this. We have not only found the justification of the procedure we followed earlier; but we also have found how to describe and separate the translational motion of (1) a rigid body which may be rotating as well, or (2) a system of particles with all kinds of internal motion.

Fig. 1

Fig.1 (The centre of mass of the fragments of the projectile continues along the same parabolic path which it would have followed if there were no explosion.)

Figure 1 is a good illustration of Eq. (7.11). A projectile, following the usual parabolic trajectory, explodes into fragments midway in air. The forces leading to the explosion are internal forces. They contribute nothing to the motion of the centre of mass. The total external force, namely, the force of gravity acting on the body, is the same before and after the explosion. The centre of mass under the influence of the external force continues, therefore, along the same parabolic trajectory as it would have followed if there were no explosion.