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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 99 of Fig. 6.11 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 ⊳ the vector sum of all the forces acting on the system of particles. 6.3 MOTION OF CENTRE OF MASS Note when we talk of the force F1on the first 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.(6.4d) as on each particle there will be external forces exerted by bodies outside the system and also MR = ∑ m i im r1 1 r +2 2+m r n n (6.7) internal forces exerted by the particles on one 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 dR dr dr dr pairs and in the sum of forces of Eq. (6.10), their M = m1 1 +m 2 2 +...+m n n contribution is zero. Only the external forces dt dt dt dt contribute to the equation. We can then rewrite Eq. (6.10) as or M V = m v +m v +...+m v (6.8) MA = F ext (6.11) 1 1 2 2 n n where F represents the sum of all external where v =(dr /dt ) is the velocity of the first ext 1 1 forces acting on the particles of the system. Eq. (6.11) states that the centre of mass of particle v 2(= dr 2 dt)is the velocity of the second particle etc. and V = dR/dt is the a system of particles moves as if all the mass of the system was concentrated at the centre velocity of the centre of mass. Note that we of mass and all the external forces were assumed the masses m , m , ... etc. do not applied at that point. 1 2 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.(6.8) with respect to time, we need to know only the external forces. we obtain To obtain Eq. (6.11) we did not need to specify the nature of the system of particles. The system dV dv1 dv 2 dv n may be a collection of particles in which there M = m1 + m 2 + ... m n dt dt dt dt may be all kinds of internal motions, or it may or be a rigid body which has either pure translational motion or a combination of MA = m a1 1 a +2..2+m a n n (6.9) translational and rotational motion. Whatever where a1( dv 1dt ) is the acceleration of the is the system and the motion of its individual particles, the centre of mass moves according first particle,a (= dv /dt ) is the acceleration 2 2 to Eq. (6.11). of the second particle etc. and is Instead of treating extended bodies as single A ( dV/dt ) 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 =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 2 m a2 2 and so on. Eq. (6.9) may be written acting at the centre of mass. as This is the procedure that we followed earlier MA = F 1 F +2..+ F n (6.10) in analysing forces on bodies and solving Reprint 2026-27