CH4_MANY_BODY

CH4_MANY_BODY - CHAPTER 4 MOTION OF MANY BODY SYSTEM 4.1...

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CHAPTER 4 MOTION OF MANY BODY SYSTEM 4.1 Newtonian Mechanics for Many Particle System In the previous chapters, we have studied the motion of single particle and the laws which govern the dynamics of single particle under the influence of a force field. In this chapter, we will extend the discussion to a system which consists of many particles. 4.1.1 Centre of mass Consider we have a system of N particles having masses of m 1 , m 2 , . ..., m N . The position vectors, velocity vectors and acceleration vectors of these N particles are denoted by ; and respectively. The centre of mass (c.m.) of this system of N particles is defined as: (4.1) where . The velocity of the c.m. and the acceleration of the c.m. can then be defined as: (4.2) and (4.3) respectively. x y z m 1 m 2 m N
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2 In triple integral form, equation (4.1) can be re-written as : (4.4) where M is the total mass of the rigid body and is the volume density (mass per unit volume) at position . Or we can write the its x, y and z coordinates as: (4.5) Likewise the centre of mass of a two-dimensional rigid body is defined as: (4.6) where is the surface density of the rigid body. If the rigid body is in the form of a wire, then the centre of mass is defined as: (4.7) where is the linear density (mass per unit length) of the wore.
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3 4.1.2 Linear momentum for system of many particles Consider one of the particle in this system, say particle m k . The total force acting on this system can be divided into two parts: (1) The resultant external force and (2) The sum of the internal forces coming from the remaining N-1 particles . Therefore, the total force experience by this particle m k is: . (4.8) where is the acceleration of the particle m k . The internal force term, we can write: (4.9) where is the force felt by the particle k exerted from the particle l . If we apply Newton's 2nd law to this single particle k: (4.10) Now we introduce the total linear momentum of the system of particles as: (4.11) m k m l
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4 where is the momentum of the single particle i in the system. From equation (A), we can modify equation (4.11) as: (4.12) Reminding you that M is the total mass of the system and V is the velocity of the c.m. of the system. Therefore we now consider the rate of the change of and try to work out what it is equal to. (4.13) Notice that the internal force term in equation (C) can be written as: (4.14) because for any term in the summation, there must exist its corresponding action- reaction counterpart such that , which will leads to a cancellation while summing up all the particles in equation (4.14). Therefore, equation (4.13) can be rewritten as: (4.15) where is the total external force exerted onto the system. Equation (4.15) is known as the Newton's second law for system of particles and it tells us that the rate of change of the total momentum of the system is equal to the total external force acting onto the system. If the total external force is zero, then the total momentum of the system will be conserved and this is known as the c onservation of momentum for system of particles .
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CH4_MANY_BODY - CHAPTER 4 MOTION OF MANY BODY SYSTEM 4.1...

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