# S Fi = S miai S (ri x Fi ) = S (ri x miai) SYSTEMS OF PARTICLES

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S Fi = S miai S (ri x Fi ) = S (ri x miai) SYSTEMS OF PARTICLES
The effective force of a particle Pi of a given system is the product miai of 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 = S miai i =1 i =1 n n S (ri x Fi ) = S (ri x miai) i =1 i =1

. . S F = L S Mo = Ho L = S mivi Ho = S (ri x mivi)
The linear momentum L and the angular momentum Ho about point O are defined as n n L = S mivi Ho = S (ri x mivi) i =1 i =1 It can be shown that . . S F = L S 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.

. S F = ma mr = S miri L = mv L = ma .
The mass center G of a system of particles is defined by a position vector r which satisfies the equation n mr = S miri i =1 n where m represents the total mass S mi. Differentiating both i =1 members twice with respect to t, we obtain . L = mv L = 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.

HG = S (r’i x mivi) = S (r’i x miv’i)
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 y’ miv’i 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’i x mivi) = S (r’i x miv’i) i =1 i =1

. S MG = HG HG = S (r’i x mivi) = S (r’i x miv’i) n y’ miv’i y i =1 n
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.

T = S mivi 2 T = mv 2 + S miv’i 2 y’ miv’i
The kinetic energy T of a system of particles is defined as the sum of the kinetic energies of the particles. y r’i Pi n G 1 2 T = S mivi 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 S miv’i 2 i = 1

T = mv 2 + S miv’i 2 T1 + U1 2 = T2 T1 + V1 = T2 + V2 n y’ miv’i 1 1 2
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 + U = T2 where U 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

ٍ ٍ (mAvA)1 y y S Fdt y (mBvB)2 (mAvA)2 (mCvC)2 (mBvB)1 O O x x O x
S MOdt ٍ (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 time t1 and 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 .

L1 = L2 (HO)1 = (HO)2 (mAvA)1 y y (mBvB)2 (mAvA)2 (mCvC)2 (mBvB)1 O x
If no external forces act on the system of particles, the systems of momenta shown above are equipollent and we have L1 = L (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.

(D m)vB S S F Dt S M Dt B Smivi B 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)vA of 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)vB of the particles leaving S in the same time Dt.

dm SF = (vB - vA) dt (D m)vB S F Dt B B Smivi Smivi S S S S M Dt 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 - vA represents the difference between the vectors vB and vA and where dm/dt is the mass rate of flow of the stream.

dm dt P = u Consider a system of particles gaining
v va m Dm S F Dt 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

v va m Dm S F Dt S mv (Dm) va S u = va - v dm dt (m + Dm) P = u S (m + Dm)(v + Dv) where dm/dt is 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.

Problem 1 x A 30-g bullet is fired with a velocity of 480 m/s into block A, which has a mass of 5 kg. The coefficient of kinetic friction between block A and cart BC is Knowing that the cart has 480 m/s A B C a mass of 4 kg and can roll freely, determine (a) the final velocity of the cart and block, (b) the final position of the block on the cart.

1. Conservation of linear momentum of a system of particles is
Problem 1 x A 30-g bullet is fired with a velocity of 480 m/s into block A, which has a mass of 5 kg. The coefficient of kinetic friction between block A and cart BC is Knowing that the cart has 480 m/s A B C a mass of 4 kg and can roll freely, determine (a) the final velocity of the cart and block, (b) the final position of the block on the cart. 1. Conservation of linear momentum of a system of particles is used to determine the final velocity of the system of particles. Conservation of linear momentum occurs when the resultant of the external forces acting on the particles of the system is zero.

2. Conservation of linear momentum during impact is used to
Problem 1 x A 30-g bullet is fired with a velocity of 480 m/s into block A, which has a mass of 5 kg. The coefficient of kinetic friction between block A and cart BC is Knowing that the cart has 480 m/s A B C a mass of 4 kg and can roll freely, determine (a) the final velocity of the cart and block, (b) the final position of the block on the cart. 2. Conservation of linear momentum during impact is used to determine the kinetic energy immediately after impact. The kinetic energy T ‘ immeditely after the collision is computed from T = S mivi2. 1 2

3. The work-energy principle is applied to determine how far
Problem 1 x A 30-g bullet is fired with a velocity of 480 m/s into block A, which has a mass of 5 kg. The coefficient of kinetic friction between block A and cart BC is Knowing that the cart has 480 m/s A B C a mass of 4 kg and can roll freely, determine (a) the final velocity of the cart and block, (b) the final position of the block on the cart. 3. The work-energy principle is applied to determine how far the block slides. The final kinetic energy of the system Tf is determined knowing the final velocity of the system of particles (from step 1). The work is done by the friction force.

mO vO = (mO + mA + mC) vf 0.03(480) = (0.03 + 5 + 4) vf vf = 1.595 m/s
Problem 1 Solution (mO + mA + mC) vf mOvO A A B C B C Conservation of linear momentum of a system of particles is used to determine the final velocity of the system of particles. mO vO = (mO + mA + mC) vf 0.03(480) = ( ) vf vf = m/s

T ’= (mO + mA)(v’)2 = 0.5(5.03)(2.86)2 = 20.61 N-m
Problem 1 Solution (mO + mA) v’ mOvO Conservation of linear momentum during impact is used to determine the kinetic energy immediately after impact. A A Conservation of linear mementum: mO vO = (mO + mA) v’ 0.03(480) = ( ) v’ v’ = 2.86 m/s Kinetic energy after impact = T’ : 1 2 T ’= (mO + mA)(v’)2 = 0.5(5.03)(2.86)2 = N-m

Tf = (mO + mA + mC)(vf )2 = 0.5(9.03)(1.595)2 = 11.48 N-m
vf = m/s Problem 1 Solution mg The work-energy principle is applied to determine how far the block slides. F = m mg x N = mg T ’= N-m Final kinetic energy= Tf: 1 2 Tf = (mO + mA + mC)(vf )2 = 0.5(9.03)(1.595)2 = N-m The only force to do work is the friction force F. T ’+ U = Tf : m(mg)(x) = 11.48 (5.03)(9.81)(x) = 11.48 x = m

Problem 2 An 80-Mg railroad engine A coasting at
6.5 km/h strikes a 20-Mg flatcar C carrying a 30-Mg load B which can slide along the floor of the car (mk =0.25). Knowing that the car was at rest with its brakes released and that it automatically coupled with the engine upon impact, determine the velocity of the car (a) immediately after impact, (b) after the load has slid to a stop relative to the car. C A B 6.5 km/h 20 Mg 30 Mg

flatcar C carrying a 30-Mg
Problem 2 An 80-Mg railroad engine A coasting at 6.5 km/h strikes a 20-Mg flatcar C carrying a 30-Mg load B which can slide along the floor of the car (mk =0.25). Knowing that the car was at rest with its brakes released and that it automatically coupled with the engine upon impact, determine the velocity of the car (a) immediately after impact, (b) after the load has slid to a stop relative to the car. C A B 6.5 km/h 20 Mg 30 Mg Conservation of linear momentum of a system of particles is used to determine the final velocity of the system of particles immediately after coupling and after the load slides to a stop.

mL ( vL )O + FDt = mL ( vL )1 0 + 0 = mL ( vL )1 ( vL )1 = 0
Problem 2 Solution (a) Velocity immediately after impact Conservation of linear momentum of a system of particles is used to determine the final velocity of the system of particles. W First consider the load. We have F = mkN = 0.20N. Since coupling occurs in Dt 0 : F Dt 0 mL ( vL )O + FDt = mL ( vL )1 F = mkN N 0 + 0 = mL ( vL )1 ( vL )1 = 0

mcv1 LO = L1: mE vO = (mE + mC) v1 mE 80 mE + mC v1 = vO = (6.5 km/h)
Problem 2 Solution (a) Velocity immediately after impact We apply the principle of conservation of linear momentum to the entire system. mEv0 mEv1 mL(vL)1= 0 mcv1 LO = L1: mE vO = (mE + mC) v1 mE mE + mC 80 v1 = vO = (6.5 km/h) v1 = 5.2 km/h

LO = L1: mE vO = (mE + mC + mL) v2
Problem 2 Solution (b) Velocity after load has stopped moving The engine, car, and load have the same velocity v2. Using conservation of linear momentum for the entire system: mEv0 mEv2 mL(vL)2 mcv2 LO = L1: mE vO = (mE + mC + mL) v2 mE mE + mC + mL 80 v2 = vO = (6.5 km/h) v2 = 4 km/h

Problem 3 60 lb The 40-lb block B is suspended from a 6-ft cord attached to the 60-lb cart A which can roll freely on a frictionless horizontal track. If the system is released from rest when q = 35o, determine the velocities of A and B when q = 0. A q B 40 lb

Problem 3 60 lb The 40-lb block B is suspended from a 6-ft cord attached to the 60-lb cart A which can roll freely on a frictionless horizontal track. If the system is released from rest when q = 35o, determine the velocities of A and B when q = 0. A q B 40 lb Conservation of linear momentum and conservation of energy. In problems involving two-dimensional motion, the initial and final linear momentum of the system are used to determine a relationship between the velocities of the two particles. Equating the initial total energy of the system of particles (including potential energy as well as kinetic energy) to its final total energy yields an additional equation.

1. Conservation of linear momentum of a system of particles is
Problem 3 60 lb The 40-lb block B is suspended from a 6-ft cord attached to the 60-lb cart A which can roll freely on a frictionless horizontal track. If the system is released from rest when q = 35o, determine the velocities of A and B when q = 0. A q B 40 lb 1. Conservation of linear momentum of a system of particles is used to determine the first equation relating the final velocities of the particles. 2. Conservation of energy is used to determine the second equation relating the final velocities of the particles.

mAvA + mBvB = 0 vA = -(mB/mA)vB
Problem 3 Solution 60 lb Conservation of linear momentum of a system of particles is used to determine the the first equation relating the final velocities of the particles. A A mAvA q B mBvB B 40 lb The block and cart A are initially at rest, so L0 = 0 When q = 0, B is directly under A and L = mAvA + mBvB = 0 + mAvA + mBvB = vA = -(mB/mA)vB vA = -[(40/32.2)/(60/32.2)]vB vA = -(4/6)vB

V0 = mg(6 - 6 cos 35o) = 40(6)(1 - cos 35o) = 43.4
Problem 3 Solution Conservation of energy is used. A vA = -(4/6)vB A 60 lb vA Conservation of energy: 6 cos 35o 6 ft 40 lb T0 + V0 = T + V B T0 = 0 V = 0 B vB Datum V0 = mg(6 - 6 cos 35o) = 40(6)(1 - cos 35o) = 43.4 1 2 1 2 T = mAvA + mBvB = 0.5(60/g)(vA) (40/g)(vB)2 2 2 = (30/g)[(4/6)vB]2 + (20/g)(vB)2 = (33.33/g)(vB)2 = (33.33/32.2)(vB)2 = 1.035(vB)2

Problem 3 Solution 60 lb T0 = 0 A A vA V0 = 43.4 6 cos 35o 6 ft V = 0 40 lb B T = 1.035(vB)2 B vB Datum T0 + V0 = T + V : = 1.035vB + 0 2 vB = 6.48 ft/s vA = -(4/6)vB = -(4/6)(6.476) vA = 4.32 ft/s

Problem 4 B The ends of a chain lie in piles at A and C. When given an initial speed v, the chain keeps moving freely at that speed over the pulley at B. Neglecting friction, determine the required value of h. v A C h

Problem 4 B The ends of a chain lie in piles at A and C. When given an initial speed v, the chain keeps moving freely at that speed over the pulley at B. Neglecting friction, determine the required value of h. v A C h The motion of a variable system of particles, i.e. a system which is continually gaining or losing particles or doing both at the same time involves (1) steady streams of particles and (2) systems gaining or losing mass. To solve problems involving a variable system of particles, the principle of impulse and momentum is used.

Problem 4 Solution B B B S(Dm)v r r r S(Dm)v Dmv0 = 0 A A A h S(Dm)(v + D v) Dx C C Dx C mg( h/L)Dt To solve problems involving a variable system of particles, the principle of impulse and momentum is used. We apply the principle of impulse and momentum to the portion of the chain of mass m in motion at t + Dt. Let L be the length and m be the mass of the portion of the chain in motion at t + Dt. Of this portion of chain, an element at A of length Dx and mass Dm =(m /L)Dx is not in motion at time t. (The extra element at C is not part of the system considered here.)

r(m - Dm)v + rmg(h/L)Dt = rm(v + Dv)
Problem 4 Solution B B B S(Dm)v r r r S(Dm)v Dmv0 = 0 A A A h S(Dm)(v + D v) Dx C C Dx C mg( h/L)Dt Equating moments about O : + r(m - Dm)v + rmg(h/L)Dt = rm(v + Dv) -(Dm)v + mg(h/L)Dt = m(Dv) Substituting Dm = (m/L)Dx and dividing by (m/L)Dt Dx Dt Dv Dt -v gh = L

dv gh - v2= L dt h = v2/g -v + gh = L Dx Dt Dv
Problem 4 Solution B B B S(Dm)v r r r S(Dm)v Dmv0 = 0 A A A h S(Dm)(v + D v) Dx C C Dx C mg( h/L)Dt -v gh = L Dx Dt Dv Letting Dt , and noting that (dx/dt) = v , dv dt gh - v2= L If the chain is to keep moving at its initial speed, dv/dt = 0, and h = v2/g

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