CHAPTER 14 Systems of Particles The effective force of a particle is defined as the product of it mass and acceleration. It will be shown that the system.

CHAPTER 14CHAPTER 14

Systems of ParticlesSystems of Particles

• The The effective forceeffective force of a particle is defined as of a particle is defined as the product of it mass and acceleration. It will the product of it mass and acceleration. It will be shown that the be shown that the system of external forcessystem of external forces acting on a system of particles is acting on a system of particles is equipollentequipollent with the with the system of effective forcessystem of effective forces of the of the system.system.

14.1. INTRODUCTION14.1. INTRODUCTION

• In the current chapter, you will study the In the current chapter, you will study the motion of motion of systems of particlessystems of particles..

• The The mass centermass center of a system of particles will of a system of particles will be defined and its motion described.be defined and its motion described.

• Application of the Application of the work-energy principlework-energy principle and and the the impulse-momentum principleimpulse-momentum principle to a system to a system of particles will be described. Results of particles will be described. Results obtained are also applicable to a system of obtained are also applicable to a system of rigidly connected particles, i.e., a rigidly connected particles, i.e., a rigid bodyrigid body..

• Analysis methods will be presented for Analysis methods will be presented for variable systems of particlesvariable systems of particles, i.e., systems in , i.e., systems in which the particles included in the system which the particles included in the system change.change.

14.2. APPLICATION OF NEWTON’S LAWS TO THE 14.2. APPLICATION OF NEWTON’S LAWS TO THE MOTION OF A SYSTEM OF PARTICLES. MOTION OF A SYSTEM OF PARTICLES.

EFFECTIVE FORCESEFFECTIVE FORCES• Newton’s second law for each Newton’s second law for each

particle particle PPii in a system of in a system of n n particles,particles,

ii

n

1jiji amfF

• The system of external and internal The system of external and internal forces on a particle is forces on a particle is equivalentequivalent to to the effective force of the particle.the effective force of the particle.

• The system of external and The system of external and internal forces acting on the entire internal forces acting on the entire system of particles is system of particles is equivalentequivalent to to the system of effective forces.the system of effective forces.

forceeffective iiam

iii

n

1jijiii amrfrFr

forces internal force external iji fF

• Summing over all the elements,Summing over all the elements,

n

1iii

n

1i

n

1jij

n

1ii amfF

• Since the internal forces occur in Since the internal forces occur in equal and opposite collinear equal and opposite collinear pairs, the resultant force and pairs, the resultant force and couple due to the internal forces couple due to the internal forces are zero,are zero,

iii amF

• The system of external forces The system of external forces and the system of effective and the system of effective forces are forces are equipollentequipollent by not by not equivalentequivalent..

n

1iiii

n

1i

n

1jiji

n

1iii amrfrFr

iiiii amrFr

14.3. LINEAR AND ANGULAR MOMENTUM OF A 14.3. LINEAR AND ANGULAR MOMENTUM OF A SYSTEM OF PARTICLESSYSTEM OF PARTICLES

• Linear momentum of the Linear momentum of the system of particles,system of particles,

n

1iiivmL

• Angular momentum about fixed Angular momentum about fixed point point O of system of particles, of system of particles,

n

1iiiiO vmrH

• Resultant of the external Resultant of the external forces is equal to rate of forces is equal to rate of change of linear change of linear momentum of the system momentum of the system of particles,of particles,

LF

OO HM

• Moment resultant about fixed Moment resultant about fixed point point O of the external forces is of the external forces is equal to the rate of change of equal to the rate of change of angular momentum of the angular momentum of the system of particles,system of particles,

n

1iii

n

1iii amvmL

n

1iiii

n

1iiiiO vmrvmrH

n

1iiiiO amrH

14.4. MOTION OF THE MASS CENTER OF A SYSTEM OF 14.4. MOTION OF THE MASS CENTER OF A SYSTEM OF PARTICLESPARTICLES

• Mass center Mass center G of system of particles is of system of particles is defined by position vector which defined by position vector which satisfiessatisfies

Gr

n

1iiiG rmrm

• Differentiating twice,Differentiating twice,

n

1iiiG rmrm

• The mass center moves as if the The mass center moves as if the entire mass and all of the external entire mass and all of the external forces were concentrated at that forces were concentrated at that point.point.

Lvmvmn

1iiiG

FLam G

GG MH

14.5. ANGULAR MOMENTUM OF A SYSTEM OF 14.5. ANGULAR MOMENTUM OF A SYSTEM OF PARTICLES ABOUT ITS MASS CENTERPARTICLES ABOUT ITS MASS CENTER

n

1iiiiG vmrH

• The angular momentum of the system The angular momentum of the system of particles about the mass center,of particles about the mass center,

• The moment resultant about The moment resultant about G of the of the external forces is equal to the rate of external forces is equal to the rate of change of angular momentum about change of angular momentum about G of the system of particles.of the system of particles.

• The centroidal frame is The centroidal frame is not, in general, a not, in general, a Newtonian frame.Newtonian frame.

• Consider the centroidal Consider the centroidal frame of reference frame of reference Gx’y’z’, which translates which translates with respect to the with respect to the Newtonian frame Newtonian frame Oxyz..

iGi aaa

n

1iGiii

n

1iiiiG aamramrH

G

n

1iii

n

1iiiiG armamrH

n

1iii

n

1iiiiG FramrH

• Angular momentum about Angular momentum about GG of of particles in their absolute motion particles in their absolute motion relative to the Newtonian relative to the Newtonian Oxyz frame frame of reference.of reference.

GGG MHH

• Angular momentum about Angular momentum about G of the particles in their of the particles in their motion relative to the motion relative to the centroidal centroidal Gx’y’z’, frame of frame of reference,reference,

iGi vvv

• Angular momentum about Angular momentum about G of the of the particle momenta can be calculated particle momenta can be calculated with respect to either the Newtonian with respect to either the Newtonian or centroidal frames of reference.or centroidal frames of reference.

n

1iiiiG vmrH

n

1iiGiiG vvmrH

n

1iiiiG

n

1iiiG vmrvrmH

n

1iiiiG vmrH

14.6. CONSERVATION OF MOMENTUM FOR A SYSTEM 14.6. CONSERVATION OF MOMENTUM FOR A SYSTEM OF PARTICLESOF PARTICLES

• If no external forces act on If no external forces act on the particles of a system, the particles of a system, then the linear momentum then the linear momentum and angular momentum and angular momentum about the fixed point about the fixed point O are are conserved.conserved.

constant constant

O

OO

HL

0MH0FL

• In some applications, such as In some applications, such as problems involving central problems involving central forces,forces,

constant constant

O

OO

HL

0MH0FL

• Concept of conservation of Concept of conservation of momentum also applies to the momentum also applies to the analysis of the mass center analysis of the mass center motion,motion,

constant constant

constant

GG

G

GG

Hv

vmL

0MH0FL

14.7. KINETIC ENERGY OF A SYSTEM OF PARTICLES14.7. KINETIC ENERGY OF A SYSTEM OF PARTICLES

• Kinetic energy of a system of Kinetic energy of a system of particles,particles,

n

1i

2ii2

1n

1iiii2

1 vmvvmT

iGi vvv

• Expressing the velocity in terms Expressing the velocity in terms of the centroidal reference frame,of the centroidal reference frame,

n

1iiGiGi2

1 vvvvmT

• Kinetic energy is equal to kinetic Kinetic energy is equal to kinetic energy of mass center plus kinetic energy of mass center plus kinetic energy relative to the centroidal energy relative to the centroidal frame.frame.

n

1i

2ii2

1n

1iiiG

2G

n

1ii2

1 vmvmvvmT

n

1i

2ii2

12G2

1 vmmvT

14.8. WORK-ENERGY PRINCIPLE. CONSERVATION OF 14.8. WORK-ENERGY PRINCIPLE. CONSERVATION OF ENERGY FOR A SYSTEM OF PARTICLESENERGY FOR A SYSTEM OF PARTICLES

• Principle of work and energy can be applied to each particle Principle of work and energy can be applied to each particle Pi ,,

2211 TUT

where represents the work done by the internal where represents the work done by the internal forces and the resultant external force acting on forces and the resultant external force acting on Pi ..

ijf

iF21U

• Principle of work and energy can be applied to the entire Principle of work and energy can be applied to the entire system by adding the kinetic energies of all particles and system by adding the kinetic energies of all particles and considering the work done by all external and internal considering the work done by all external and internal forces.forces.

• Although and are equal and opposite, the work of Although and are equal and opposite, the work of these forces will not, in general, cancel out.these forces will not, in general, cancel out.

ijf

• If the forces acting on the particles are conservative, the If the forces acting on the particles are conservative, the work is equal to the change in potential energy andwork is equal to the change in potential energy and

2211 VTVT

which expresses the principle of conservation of energy for which expresses the principle of conservation of energy for the system of particles.the system of particles.

jif

2

t

t

O1 HdtMH2

1

LF

14.9. PRINCIPLE OF IMPULSE AND MOMENTUM FOR A 14.9. PRINCIPLE OF IMPULSE AND MOMENTUM FOR A SYSTEM OF PARTICLESSYSTEM OF PARTICLES

• The momenta of the particles at time The momenta of the particles at time t1 and the impulse and the impulse of the forces from of the forces from t11 to to t2 form a system of vectors form a system of vectors equipollentequipollent to the system of momenta of the particles to the system of momenta of the particles at time at time t2 . .

12

t

t

LLdtF2

1

2

t

t

1 LdtFL2

1

OO HM

12

t

t

O HHdtM2

1

14.10. VARIABLE SYSTEMS OF PARTICLES14.10. VARIABLE SYSTEMS OF PARTICLES

• Kinetics principles established so far were derived for Kinetics principles established so far were derived for constant systems of particles, i.e., systems which neither constant systems of particles, i.e., systems which neither gain nor lose particles.gain nor lose particles.

• A large number of physics and engineering applications A large number of physics and engineering applications require the consideration of variable systems of particles, require the consideration of variable systems of particles, e.g., hydraulic turbine, rocket engine, etc.e.g., hydraulic turbine, rocket engine, etc.

• For analyses, consider auxiliary systems which consist of For analyses, consider auxiliary systems which consist of the particles instantaneously within the system plus the the particles instantaneously within the system plus the particles that enter or leave the system during a short time particles that enter or leave the system during a short time interval. The auxiliary systems, thus defined, are constant interval. The auxiliary systems, thus defined, are constant systems of particles.systems of particles.

14.11. STEADY STREAM OF PARTICLES14.11. STEADY STREAM OF PARTICLES

•System consists of a steady stream System consists of a steady stream of particles against a vane or of particles against a vane or through a duct.through a duct.

BiiAii

vmvmtFvmvm

• The auxiliary system is a constant The auxiliary system is a constant system of particles over system of particles over t..

• Define auxiliary system which Define auxiliary system which includes particles which flow in and includes particles which flow in and out over out over t. .

AB

vvdtdmF

2

2t

1t

1LdtFL

Steady Stream of Particles. ApplicationsSteady Stream of Particles. Applications

• Fluid Flowing Through a Fluid Flowing Through a PipePipe

• Jet Jet EngineEngine

• FanFan

•Fluid Stream Diverted by Fluid Stream Diverted by Vane or DuctVane or Duct

• HelicopterHelicopter

14.12. STREAMS GAINING OR LOSING MASS14.12. STREAMS GAINING OR LOSING MASS

• Define auxiliary system to include Define auxiliary system to include particles of mass particles of mass m within system within system at time at time t plus the particles of mass plus the particles of mass m which enter the system over which enter the system over time interval time interval t. .

• The auxiliary system is a constant The auxiliary system is a constant system of particles.system of particles.

udtdmFam

vmvvmvmtF a

2

t

t

1 LdtFL2

1

vvmmtFvmvm a

udtdm

dtvdmF

Work Some Example Work Some Example ProblemsProblems

14.92 A chain of length 14.92 A chain of length l and mass and mass m falls through a small hole in a falls through a small hole in a plate. Initially, when plate. Initially, when y is very small, the chain is at rest. In is very small, the chain is at rest. In each case shown on this slide and a later slide, determine (a) each case shown on this slide and a later slide, determine (a) the acceleration of the first link as a function of the acceleration of the first link as a function of y and (b) the and (b) the velocity of the chain as the last link passes through the hole. velocity of the chain as the last link passes through the hole. In this first case assume that the individual links are at rest In this first case assume that the individual links are at rest until they fall through the hole and in the later case assume until they fall through the hole and in the later case assume that at any instant all links have the same speed. Ignore the that at any instant all links have the same speed. Ignore the effect of friction in both cases.effect of friction in both cases.

y

y

Define a mass/lengthDefine a mass/length

l/m

The mass of length The mass of length y is is

'm y

tgy vy

vy

)vv( vyy

y

Apply the principle of momentum-impulse.Apply the principle of momentum-impulse.Direction is positive downward.Direction is positive downward.

tgy

dtdy

vdtdvygy

dt)yv(d

gy

)yy(

vy vy yv

Can you see what weCan you see what we can do to simplify this?can do to simplify this?

y

y

dt

)vy(dgy

)vy(ddtgy

Remember that Remember that

dtdy

v

oorr

vdy

dt

ThereforeTherefore

)vy(dv

dygy

)vy(d)vy(dyyg 2

In general this integrates toIn general this integrates to

23

23 )vy(yg

y

y

23

23 )vy(yg

ygv32

When When y = l

lgv32dy

dvva

yg

g

dydv

32

3

3g

a WOWWOW

y

y

y

y

)vv(m vm tgy

tgyl

m vm

gl

y

dt

dv

dy

dvv

y

y

gl

ydy

dvv

yl

ga

gl

ydy

dvv

gl

y

2

2

2

2v

yl

gv

ForFor y = l lgv

A 1200 lb spacecraft is designed to include a two stage rocket A 1200 lb spacecraft is designed to include a two stage rocket with stages A and B, each weighing 21,300 lb, including with stages A and B, each weighing 21,300 lb, including 20,000 lb of fuel. The fuel is consumed at a rate of 500 lb/s 20,000 lb of fuel. The fuel is consumed at a rate of 500 lb/s and ejected with a relative velocity of 12,000 ft/s. Knowing and ejected with a relative velocity of 12,000 ft/s. Knowing that when stage A expels its last particle of fuel, its casing is that when stage A expels its last particle of fuel, its casing is released and jettisoned, determine the altitude at which (a) released and jettisoned, determine the altitude at which (a) stage A of the rocket is released and (b) the fuel of both stage A of the rocket is released and (b) the fuel of both stages has been consumed.stages has been consumed.

Given:Given:

Total WeightTotal Weight

)lb300,21(2lb1200W

qgs/lb500dt

dW

Velocity of the exhaustVelocity of the exhaust

j)uv(jve

s/ft000,12u

Time to burn one tank of fuelTime to burn one tank of fuel

s40s/lb500

lb000,20t

mglb800,43W

tg)qtm(

)uv(tq

emv

y

y

Apply the principle of momentum-impulse.Apply the principle of momentum-impulse.Direction is positive upward.Direction is positive upward.

There is a correspondingThere is a corresponding

v

v)qtm( tg)qtm( ))tt(qm( )vv(

v)qtm( v)qtm(

tqv vtq

v)qtm(

g)qtm(dtdv)qtm( quqvqv

g)qtm(dtdv)qtm( qu

dt}g)qtm(

qu{dv

dt}g)qtm(

qu{dv

t

0

v

0

dt}g)qtm(

qu{dv

gt]}mln[)]qtm{ln[(uv

gt]m

)qtm(ln[uv

At At t = 40 s

s/ft6031vA

This is needed later.This is needed later.

dtdy

gt]m

)qtm(ln[uv

dydt}gt]m

)qtm(ln[u{

t

0

y

0

dt}gt]m

)qtm(ln[u{dy

s40

0

2

Agt

21}

m)qtm(

]m

)qtm(ln[

m)qtm(

{qmuy

mi0.20yA

dtdy

vgt]m

)qtm(ln[uv

A

To get the height at the second stage burnout useTo get the height at the second stage burnout use

You would again integrate from You would again integrate from t = 0 to to t = 40 sand use and use

)lb300,21(lb1200W

A

s40

0

A

2 ytvgt21}

m)qtm(

]m

)qtm(ln[

m)qtm(

{qmuy

mi8.126y

mglb500,22W