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 =iS=1miai
i =1
n
n
S (ri x Fi ) =iS=1(ri x miai)
i =1
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The linear momentum L and the angular momentum Ho about
point O are defined as
n
L = S mivi
i =1
It can be shown that
.
S F=L
n
Ho = S (ri x mivi)
i =1
.
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.
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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
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concentrated at that point.
Consider the motion of the particles
of a system with respect to a
y
centroidal frame Gx’y’z’ attached to
r’i
Pi
the mass center G of the system and
G
in translation with respect to the
x’
O
newtonian frame Oxyz. The angular
x
momentum of the system about its
z’
z
mass center G is defined as the
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
y’
miv’i
n
n
i =1
i =1
HG = S (r’i x mivi) = S (r’i x miv’i)
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y’
miv’i
y
r’i
G
x’
z
z’
x
HG = S (r’i x mivi)
i =1
n
Pi
O
n
= S (r’i x miv’i)
i =1
We can derive the relation
.
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.
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y’
miv’i
y
r’i
G
Pi
x’
O
z’
x
The kinetic energy T of a system of
particles is defined as the sum of
the kinetic energies of the particles.
n
1
2
T = 2 S mivi
i=1
Using the centroidal reference
frame Gx’y’z’ we note that the
kinetic energy of the system can also be obtained by adding the
1
kinetic energy 2 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’ :
z
1
2
T = mv 2 +
n
2
1
miv’i
2 iS
=1
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y’
miv’i
y
r’i
G
Pi
x’
O
z
z’
x
n
1
2
2
T = mv + 2 S miv’i
i=1
The principle of work and energy
can be applied to a system of
particles as well as to individual
particles
1
2
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
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y
(mAvA)1
y
t2
S ٍ Fdt
y
(mAvA)2
t1
(mCvC)2
(mBvB)1
O
x
(mCvC)1
(mBvB)2
O
t2
Sٍ
t1
x
MOdt
O
x
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 .
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y
(mAvA)1
y
(mBvB)2
(mAvA)2
(mCvC)2
(mBvB)1
O
x
(mCvC)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 = 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.
12 - 9
Smivi
B
S
A
(D m)vB
S F Dt
S
S M Dt
B
Smivi
S
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.
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Smivi
B
S
A
(D m)vB
S F Dt
S
S M Dt
B
Smivi
S
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
SF =
(vB - vA)
dt
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.
12 - 11
v
va
m
mv
S F Dt
Dm
S
S (Dm) va
u = va - v
Consider a system of particles gaining
(m + Dm)
mass by continually absorbing particles
S
or losing mass by continually expelling
(m + Dm)(v + Dv)
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
dm
P = dt u
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v
va
m
mv
S F Dt
Dm
S
S (Dm) va
u = va - v
dm
P = dt u
(m + Dm)
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.
12 - 13
Problem 1
x
A 30-g bullet is fired with a
480 m/s
velocity of 480 m/s into block A,
A
which has a mass of 5 kg. The
coefficient of kinetic friction
B
C
between block A and cart BC is
0.05. Knowing that the cart has
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.
12 - 14
Problem 1
x
A 30-g bullet is fired with a
velocity of 480 m/s into block A,
A
which has a mass of 5 kg. The
B
C
coefficient of kinetic friction
between block A and cart BC is
0.05. Knowing that the cart has
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.
480 m/s
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.
12 - 15
Problem 1
x
A 30-g bullet is fired with a
velocity of 480 m/s into block A,
A
which has a mass of 5 kg. The
B
C
coefficient of kinetic friction
between block A and cart BC is
0.05. Knowing that the cart has
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.
480 m/s
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
1
computed from T = 2 S mivi2.
12 - 16
Problem 1
x
A 30-g bullet is fired with a
velocity of 480 m/s into block A,
A
which has a mass of 5 kg. The
B
C
coefficient of kinetic friction
between block A and cart BC is
0.05. Knowing that the cart has
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.
480 m/s
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.
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Problem 1 Solution
(mO + mA + mC) vf
mOvO
A
B
A
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) = (0.03 + 5 + 4) vf
vf = 1.595 m/s
12 - 18
Problem 1 Solution
(mO + mA) v’
mOvO
A
A
Conservation of linear
momentum during impact
is used to determine the
kinetic energy immediately
after impact.
Conservation of linear mementum:
mO vO = (mO + mA) v’
0.03(480) = (0.03 + 5) v’
v’ = 2.86 m/s
Kinetic energy after impact = T’ :
1
T ’= (mO + mA)(v’)2 = 0.5(5.03)(2.86)2 = 20.61 N-m
2
12 - 19
vf = 1.595 m/s
Problem 1 Solution
mg
F = m mg
x
N = mg
The work-energy principle is
applied to determine how far
the block slides.
T ’= 20.61 N-m
Final kinetic energy= Tf:
1
Tf = 2 (mO + mA + mC)(vf )2 = 0.5(9.03)(1.595)2 = 11.48 N-m
The only force to do work is the friction force F.
T ’+ U1
2=
Tf :
20.61 - m(mg)(x) = 11.48
20.61 - 0.5(5.03)(9.81)(x) = 11.48
x = 0.370 m
12 - 20
Problem 2
An 80-Mg railroad
20 Mg
engine A coasting at
B
A
C 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.
6.5 km/h
30 Mg
12 - 21
Problem 2
An 80-Mg railroad
6.5 km/h
20 Mg
engine A coasting at
B
A
C 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.
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.
12 - 22
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
F = mkN
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
N
0 + 0 = mL ( vL )1
( vL )1 = 0
12 - 23
Problem 2 Solution
(a) Velocity immediately after impact
We apply the principle of conservation of linear momentum to
the entire system.
mEv1
mEv0
mL(vL)1= 0
LO = L1: mE vO = (mE + mC) v1
mE
v1 = m + m
E
C
mcv1
80
vO =
(6.5 km/h)
80 + 20
v1 = 5.2 km/h
12 - 24
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:
mEv2
mEv0
LO = L1: mE vO = (mE + mC + mL) v2
mE
v2 =
mE + mC + mL
mL(vL)2
mcv2
80
vO =
(6.5 km/h)
80 + 20 + 30
v2 = 4 km/h
12 - 25
Problem 3
60 lb
A
B
q
The 40-lb block B is suspended
from a 6-ft cord attached to the 60lb 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.
40 lb
12 - 26
Problem 3
60 lb
A
B
40 lb
q
The 40-lb block B is suspended
from a 6-ft cord attached to the 60lb 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.
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.
12 - 27
Problem 3
60 lb
A
B
40 lb
q
The 40-lb block B is suspended
from a 6-ft cord attached to the 60lb 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.
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.
12 - 28
Problem 3 Solution
60 lb
A
q
B
40 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
mAvA
B
mBvB
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 = 0
vA = -(mB/mA)vB
vA = -[(40/32.2)/(60/32.2)]vB
vA = -(4/6)vB
12 - 29
Problem 3 Solution
Conservation of energy is used.
60 lb
A
vA = -(4/6)vB
A
vA
6 cos 35o
B
6 ft
40 lb
Datum
B
Conservation of
energy:
T0 + V0 = T + V
vB
T0 = 0
V=0
V0 = mg(6 - 6 cos 35o) = 40(6)(1 - cos 35o) = 43.4
1
1
2
2
T = 2 mAvA + mBvB = 0.5(60/g)(vA)2 + 0.5(40/g)(vB)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
12 - 30
Problem 3 Solution
60 lb
T0 = 0
A
A
vA
6 cos 35o
B
6 ft
40 lb
Datum
V0 = 43.4
B
V=0
T = 1.035(vB)2
vB
T0 + V0 = T + V : 0 + 43.4 = 1.035vB + 0
2
vB = 6.48 ft/s
vA = -(4/6)vB = -(4/6)(6.476)
vA = 4.32 ft/s
12 - 31
Problem 4
B
A
v
C h
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.
12 - 32
Problem 4
B
A
v
C h
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.
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.
12 - 33
B
B
S(Dm)v
Dmv0 = 0
h
r
r
S(Dm)v
A
Dx
Problem 4 Solution
B
C
Dx
r
A
A
S(Dm)(v + D v)
C
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.)
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B
B
S(Dm)v
Dmv0 = 0
h
r
r
S(Dm)v
A
Dx
Problem 4 Solution
B
C
Dx
r
A
A
S(Dm)(v + D v)
C
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
Dv
Dx
-v
+ gh = L Dt
Dt
12 - 35
B
B
S(Dm)v
Dmv0 = 0
h
-v
r
r
S(Dm)v
A
Dx
C
r
A
A
S(Dm)(v + D v)
Dx
C
C
mg( h/L)Dt
Dx
Dv
+ gh = L
Dt
Dt
Letting Dt
Problem 4 Solution
B
0, and noting that (dx/dt) = v ,
gh -
v2=
dv
L
dt
If the chain is to keep moving at its initial speed, dv/dt = 0, and
h = v2/g
12 - 36