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148 PHYSICS

of Fig. 7.11 had different masses. How will you Thus, the total mass of a system of particles
then determine the centre of mass of the lamina? times the acceleration of its centre of mass is
t the vector sum of all the forces acting on the
system of particles.
7.3 MOTION OF CENTRE OF MASS
Note when we talk of the force F on the first
1
Equipped with the definition of the centre of particle, it is not a single force, but the vector
mass, we are now in a position to discuss its sum of all the forces on the first particle; likewise
physical importance for a system of n particles. for the second particle etc. Among these forces
We may rewrite Eq.(7.4d) as on each particle there will be external forces
+
MR = ∑ m r = m r + m r + ... m r (7.7) exerted by bodies outside the system and also
internal forces exerted by the particles on one
n n
1 1
2 2
i i
Differentiating the two sides of the equation another. We know from Newton’s third law that
with respect to time we get these internal forces occur in equal and opposite
pairs and in the sum of forces of Eq. (7.10),
dR dr dr dr n
M = m 1 1 + m 2 2 + ... m+ n their contribution is zero. Only the external
dt dt dt dt
forces contribute to the equation. We can then
rewrite Eq. (7.10) as
or
MA = F ext (7.11)
+
M V = m v + m v + ... m v (7.8)
1 1 2 2 n n
where F ext represents the sum of all external
where v 1 ( d /dt= r 1 ) is the velocity of the first forces acting on the particles of the system.
particle v 2 ( d= r 2 dt ) is the velocity of the Eq. (7.11) states that the centre of mass
second particle etc. and V = d /dt is the of a system of particles moves as if all the
R
mass of the system was concentrated at the
velocity of the centre of mass. Note that we centre of mass and all the external forces
assumed the masses m , m , ... etc. do not
1 2 were applied at that point.
change in time. We have therefore, treated them Notice, to determine the motion of the centre
as constants in differentiating the equations of mass no knowledge of internal forces of the
with respect to time. system of particles is required; for this purpose
Differentiating Eq.(7.8) with respect to time, we need to know only the external forces.
we obtain To obtain Eq. (7.11) we did not need to
specify the nature of the system of particles.
dV dv dv dv
+
M = m 1 1 + m 2 2 + ... m n n The system may be a collection of particles in
dt dt dt dt which there may be all kinds of internal
or motions, or it may be a rigid body which has
either pure translational motion or a
+
MA = m a + m a + ... m a (7.9)
1 1 2 2 n n combination of translational and rotational
where a ( d= v /dt ) is the acceleration of the motion. Whatever is the system and the motion
1 1
of its individual particles, the centre of mass
first particle, ( d=a 2 v 2 /dt ) is the acceleration moves according to Eq. (7.11).
Instead of treating extended bodies as single
of the second particle etc. and ( d /dt=A V ) is
particles as we have done in earlier chapters,
the acceleration of the centre of mass of the we can now treat them as systems of particles.
system of particles. We can obtain the translational component of
Now, from Newton’s second law, the force their motion, i.e. the motion of the centre of mass
acting on the first particle is given by F = m a . of the system, by taking the mass of the whole
1 1 1
The force acting on the second particle is given system to be concentrated at the centre of mass
and all the external forces on the system to be
by F = m a and so on. Eq. (7.9) may be written
2 2 2 acting at the centre of mass.
as This is the procedure that we followed earlier
MA = F + F + ... + F (7.10) in analysing forces on bodies and solving
1 2 n

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