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Chapter 14 SYSTEMS OF PARTICLES

Chapter 14 SYSTEMS OF PARTICLES.

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Chapter 14 SYSTEMS OF PARTICLES

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  1. Chapter 14 SYSTEMS OF PARTICLES The effective force of a particle Piof a given system is the product miaiof its mass mi and its acceleration ai with respect to a newtonian frame of reference centered at O. The system of the external forces acting on the particles and the system of the effective forces of the particles are equipollent; i.e., both systems have the same resultant and the same moment resultant about O : n n S Fi =Smiai i =1 i =1 n n S (rix Fi ) =S(rixmiai) i =1 i =1 Sharif University-Aerospace Dep. Fall 2004

  2. The linear momentumLand the angular momentum Ho about point O are defined as n n L =Smivi Ho =S(rixmivi) i =1 i =1 It can be shown that . . S F = LS Mo = Ho This expresses that the resultant and the moment resultant about O of the external forces are, respectively, equal to the rates of change of the linear momentum and of the angular momentum about O of the system of particles.

  3. The mass center G of a system of particles is defined by a position vector r which satisfies the equation n mr =Smiri i =1 n where m represents the total mass S mi. Differentiating both i =1 members twice with respect to t, we obtain . L = mvL = ma where v and a are the velocity and acceleration of the mass center G. Since S F = L, we obtain . S F = ma Therefore, the mass center of a system of particles moves as if the entire mass of the system and all the external forces were concentrated at that point.

  4. Consider the motion of the particles of a system with respect to a centroidal frame Gx’y’z’ attached to the mass center G of the system and in translation with respect to the newtonian frame Oxyz. The angular momentum of the system about its mass center G is defined as the miv’i y’ y r’i Pi G x’ O x z’ z sum of the moments about G of the momenta miv’i of the particles in their motion relative to the frame Gx’y’z’. The same result is obtained by considering the moments about G of the momenta mivi of the particles in their absolute motion. Therefore n n HG = S (r’ixmivi) = S (r’ixmiv’i) i =1 i =1

  5. n n = S (r’ixmiv’i) i =1 miv’i y’ HG = S (r’ixmivi) y i =1 r’i Pi G x’ O We can derive the relation x . z’ z S MG = HG which expresses that the moment resultant about G of the external forces is equal to the rate of change of the angular momentum about G of the system of particles. When no external force acts on a system of particles, the linear momentum L and the angular momentum Ho of the system are conserved. In problems involving central forces, the angular momentum of the system about the center of force O will also be conserved.

  6. miv’i The kinetic energy T of a system of particles is defined as the sum of the kinetic energies of the particles. y’ y r’i Pi n G 1 2 T = Smivi x’ 2 O i = 1 x z’ z Using the centroidal reference frame Gx’y’z’we note that the kinetic energy of the system can also be obtained by adding the kinetic energy mv2 associated with the motion of the mass center G and the kinetic energy of the system in its motion relative to the frame Gx’y’z’: 1 2 n 1 2 1 2 T = mv 2 + Smiv’i 2 i = 1

  7. n miv’i y’ 1 2 1 2 T = mv 2 + Smiv’i 2 y i = 1 r’i Pi The principle of work and energy can be applied to a system of particles as well as to individual particles G x’ O x z’ z T1 + U1 2 = T2 where U1 2 represents the work of all the forces acting on the particles of the system, internal and external. If all the forces acting on the particles of the system are conservative, the principle of conservation of energy can be applied to the system of particles T1 + V1 = T2 + V2

  8. t2 ò SFdt t1 (mAvA)1 y y y (mBvB)2 (mAvA)2 (mCvC)2 (mBvB)1 O x O x O x t2 ò SMOdt (mCvC)1 t1 The principle of impulse and momentum for a system of particles can be expressed graphically as shown above. The momenta of the particles at timet1and the impulses of the external forces from t1 to t2 form a system of vectors equipollent to the system of the momenta of the particles at time t2 .

  9. (mAvA)1 y y (mBvB)2 (mAvA)2 (mCvC)2 (mBvB)1 O x O x (mCvC)1 If no external forces act on the system of particles, the systems of momenta shown above are equipollent and we have L1 = L2 (HO)1 = (HO)2 Many problems involving the motion of systems of particles can be solved by applying simultaneously the principle of impulse and momentum and the principle of conservation of energy or by expressing that the linear momentum, angular momentum, and energy of the system are conserved.

  10. SF Dt S SM Dt (D m)vB B B Smivi Smivi S S A A (D m)vA For variable systems of particles, first consider a steady stream of particles, such as a stream of water diverted by a fixed vane or the flow of air through a jet engine. The principle of impulse and momentum is applied to a system S of particles during a time interval Dt, including particles which enter the system at A during that time interval and those (of the same mass Dm) which leave the system at B. The system formed by the momentum (Dm)vAof the particles entering S in the time Dt and the impulses of the forces exerted on S during that time is equipollent to the momentum (Dm)vBof the particles leaving S in the same time Dt.

  11. SF Dt S SM Dt (D m)vB B B Smivi Smivi S S A A (D m)vA Equating the x components, y components, and moments about a fixed point of the vectors involved, we could obtain as many as three equations, which could be solved for the desired unknowns. From this result, we can derive the expression dm dt SF = (vB - vA) where vB - vArepresents the difference between the vectors vB and vAand where dm/dtis the mass rate of flow of the stream.

  12. v va S F Dt m Dm S mv (Dm) va S u = va - v Consider a system of particles gaining mass by continually absorbing particles or losing mass by continually expelling particles (as in the case of a rocket). Applying the principle of impulse and momentum to the system during a time interval Dt, we take care to include particles gained or lost during the time interval. The action on a system S of the particles being absorbed by S is equivalent to a thrust (m + Dm) S (m + Dm)(v + Dv) dm dt P = u

  13. v va S F Dt m Dm S mv (Dm) va S u = va - v (m + Dm) dm dt P = u S (m + Dm)(v + Dv) where dm/dtis the rate at which mass is being absorbed, and u is the velocity of the particles relative to S. In the case of particles being expelled by S , the rate dm/dt is negative and P is in a direction opposite to that in which particles are being expelled.

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