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Chapter 14
Systems of Particles
1
Introduction
• In the current chapter, you will study the motion of 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 of
external forces acting on a system of particles is equipollent
with the system of effective forces of the system.
• The mass center of a system of particles will be defined
and its motion described.
• Application of the work-energy principle and the
impulse-momentum principle to a system of particles will
be described. Result obtained are also applicable to a
system of rigidly connected particles, i.e., a rigid body.
• Analysis methods will be presented for variable systems
of particles, i.e., systems in which the particles included
in the system change.
2
Application of Newton’s Laws. Effective Forces
• Newton’s second law for each particle Pi
in a system of n particles,

Fi 
n

j 1
 
ri  Fi 


fij  mi ai
 


ri  fij  ri  mi ai

n
j 1



Fi  external force
fij  internal forces

mi ai  effective force
• The system of external and internal forces on
a particle is equivalent to the effective force
of the particle.
• The system of external and internal forces
acting on the entire system of particles is
equivalent to the system of effective forces.
3
Application of Newton’s Laws. Effective Forces
• Summing over all the elements,
n 
n n 
n

F

f

m
a
 i   ij  i i
i 1
i 1 j 1
i 1
n n 
n 

 






r

F

r

f

r

m
a
 i i   i ij  i i i 
n
i 1
i 1 j 1
i 1
• Since the internal forces occur in equal
and opposite collinear pairs, the resultant
force and couple due to the internal
forces are zero,


 Fi   mi ai
 





r

F

r

m
a
 i i  i i i
• The system of external forces and the
system of effective forces are
equipollent by not equivalent.
4
Linear & Angular Momentum
• Linear momentum of the system of
particles,
 n 
L   mi vi

L
i 1
n
n


 mi vi   mi ai
i 1
i 1
• Resultant of the external forces is
equal to rate of change of linear
momentum of the system of
particles,
 
FL

• Angular momentum about fixed point O
of system of particles,
n 


H O   ri  mi vi 
i 1
n
n 




H O   ri  mi vi    ri  mi vi 
i 1
n
i 1


  ri  mi ai 
i 1
• Moment resultant about fixed point O of
the external forces is equal to the rate of
change of angular momentum of the
system of particles,


M H

O
O
5
Motion of the Mass Center of a System of Particles
• Mass center G of system of particles is defined

by position vector rG which satisfies
n


mrG   mi ri
i 1
• Differentiating twice,
n


mrG   mi ri

mvG 
i 1
n


 mi vi  L
i 1



maG  L   F
• The mass center moves as if the entire mass and
all of the external forces were concentrated at
that point.
6
Angular Momentum About the Mass
• The angular momentum of the system of
Center
particles about the mass center,
n 


   ri  mi vi 
HG

 
HG
 

ai  aG  ai
• Consider the centroidal frame
of reference Gx’y’z’, which
translates with respect to the
Newtonian frame Oxyz.
• The centroidal frame is not,
in general, a Newtonian
frame.

i 1
n
n 


 
 ri  mi ai    ri  mi ai  aG 
i 1
n
i 1
n





 ri  mi ai     mi r    aG

n 



  ri  mi ai    ri  Fi 
i 1
i 1

  MG
i 1
n
 i 1
• The moment resultant about G of the external
forces is equal to the rate of change of angular
momentum about G of the system of particles.
7
Angular Momentum About the Mass Center
• Angular momentum about G of particles in
their absolute motion relative to the
Newtonian Oxyz frame of reference.
n 


H G   ri mi vi 

 

vi  vG  vG
i 1
n






r

m
v

v
 i i G i 
i 1
 n
n 
 

   mi ri  vG   ri mi vi 
 i 1

i 1



   MG
HG  HG
• Angular momentum about G of
the particles in their motion
relative to the centroidal Gx’y’z’
frame of reference,
• Angular momentum about G of the particle
momenta can be calculated with respect to
n 


H G   ri mi vi 
either the Newtonian or centroidal frames of
i 1
reference.
8
Conservation of Momentum
• If no external forces act on the
particles of a system, then the linear
momentum and angular momentum
about the fixed point O are
conserved.




L  F  0
HO   M O  0


L  constant
H O  constant
• Concept of conservation of momentum
also applies to the analysis of the mass
center motion,




HG   M G  0
L  F  0


L  mvG  constant


vG  constant
H G  constant
• In some applications, such as
problems involving central forces,




L  F  0
HO   M O  0


L  constant
H O  constant
9
Sample Problem 14.2
SOLUTION:
• Since there are no external forces, the
linear momentum of the system is
conserved.
• Write separate component equations
for the conservation of linear
momentum.
A 20 N projectile is moving with a
velocity of 100 m/s when it explodes into
5 and 15 N fragments. Immediately after
the explosion, the fragments travel in the
directions qA = 45o and qB = 30o.
• Solve the equations simultaneously
for the fragment velocities.
Determine the velocity of each fragment.
10
Sample Problem 14.2
SOLUTION:
• Since there are no external forces, the
linear momentum of the system is
conserved.
• Write separate component equations for
the conservation of linear momentum.



mAv A  mB vB  mv0
5 g vA  15 g vB  20 g v0
x components:
5vA cos 45  15vB cos 30  20100
y components:
y
5vA sin 45  15vB sin 30  0
x
• Solve the equations simultaneously for the
fragment velocities.
v A  207 m s
v B  97.6 m s
11
Kinetic Energy
• Kinetic energy of a system of particles,
n
 
1
T   mi vi  vi   2  mi vi2
1
2
n
i 1
i 1
• Expressing the velocity in terms of the
centroidal reference frame,
T

 

vi  vG  vi

n




1

m
v

v

v

v

i G
i
G
i
2
i 1
n
n
 1 n
1  m v 2  v 
   mi vi2
m
v


i
G
G
i
i


2
2
 i 1 
i 1
i 1
n
1 mv 2  1
2
m
v

G
i
i
2
2
i 1
 
 

• Kinetic energy is equal to kinetic energy of
mass center plus kinetic energy relative to
the centroidal frame.
12
Work-Energy Principle. Conservation of Energy
• Principle of work and energy can be applied to each particle Pi ,
T1  U12  T2

where U12 represents the workdone by the internal forces f ij
and the resultant external force Fi acting on Pi .
• Principle of work and energy can be applied to the entire system by
adding the kinetic energies of all particles and considering the work
done by all external and internal forces.


• Although f ij and f ji are equal and opposite, the work of these
forces will not, in general, cancel out.
• If the forces acting on the particles are conservative, the work is
equal to the change in potential energy and
T1  V1  T2  V2
which expresses the principle of conservation of energy for the
system of particles.
13
Principle of Impulse and Momentum
 
F  L
t2



  Fdt  L2  L1
t1


 M O  HO
t2



  M O dt  H 2  H1
t1
t2 


L1    Fdt  L2
t1
t2 


H1    M O dt  H 2
t1
• The momenta of the particles at time t1 and the impulse of the forces
from t1 to t2 form a system of vectors equipollent to the system of
momenta of the particles at time t2 .
14
Sample Problem 14.4
SOLUTION:
Ball B, of mass mB, is suspended from
a cord, of length l, attached to cart A,
of mass mA, which can roll freely on a
frictionless horizontal tract. While the
cart is at rest, the ball is given an initial
velocity
v0  2 gl .
Determine (a) the velocity of B as it
reaches it maximum elevation, and (b)
the maximum vertical distance h
through which B will rise.
• With no external horizontal forces, it
follows from the impulse-momentum
principle that the horizontal component of momentum is conserved. This
relation can be solved for the velocity
of B at its maximum elevation.
• The conservation of energy principle
can be applied to relate the initial
kinetic energy to the maximum potential
energy. The maximum vertical distance
is determined from this relation.
15
Sample Problem 14.4
SOLUTION:
• With no external horizontal forces, it follows from the
impulse-momentum principle that the horizontal component of momentum is conserved. This relation can be
solved for the velocity of B at its maximum elevation.
t2 


L1    Fdt  L2
t1
y
x
x component equation:
m Av A,1  mB vB,1  m Av A,2  mB vB,2
Velocities at positions 1 and 2 are
v A,1  0 vB,1  v0
vB,2  v A,2  vB
A, 2
 v A,2 (velocity of B relative to
A is zero at position 2)
mB v0  m A  mB v A, 2
v A,2  vB,2 
mB
v0
m A  mB
16
Sample Problem 14.4
• The conservation of energy principle can be applied to relate
the initial kinetic energy to the maximum potential energy.
T1  V1  T2  V2
Position 1 - Potential Energy: V1  mA gl
Kinetic Energy:
T1  12 mBv02
Position 2 - Potential Energy: V2  mA gl  mB gh
Kinetic Energy:
1 m v2
2 B 0
T2  12 mA  mB v 2A,2
 mA gl  12 mA  mB v 2A,2  mA gl  mB gh
2

v02 m A  mB v A, 2 v02 m A  mB  mB

h



v0 
2g
mB
2g 2g
2 g mB  m A  mB 
v02
mB
v02
h

2 g mA  mB 2 g
2
mA v02
h
mA  mB 2 g
17
Sample Problem 14.5
SOLUTION:
• There are four unknowns: vA, vB,x, vB,y,
and vC.
Ball A has initial velocity v0 = 10 m /s
parallel to the axis of the table. It hits
ball B and then ball C which are both at
rest. Balls A and C hit the sides of the
table squarely at A’ and C’ and ball B
hits obliquely at B’.
• Solution requires four equations:
conservation principles for linear
momentum (two component equations),
angular momentum, and energy.
• Write the conservation equations in
terms of the unknown velocities and
solve simultaneously.
Assuming perfectly elastic collisions,
determine velocities vA, vB, and vC with
which the balls hit the sides of the table.
18
Sample Problem 14.5
SOLUTION:
• There are four unknowns: vA,
vB,x, vB,y, and vC.


vA  vA j



vB  vB , x i  vB , y j


vC  vC i
• The conservation of momentum and energy equations,



L1    Fdt  L2
mv0  mvB, x  mvC
0  mv A  mvB, y



H O,1    M O dt  H O, 2
 2 m mv0  8 m mv A  7 m mvB , y  3 m mvC
T1  V1  T2  V2
1 mv 2
0
2


 12 mv 2A  12 m vB2 , x  vB2 , y  12 mvC2
Solving the first three equations in terms of vC,
v A  vB, y  3vC  20
vB, x  10  vC
Substituting into the energy equation,
23vC  20 2  10  vC 2  vC2  100
20vC2  260vC  800  0
y
x
v A  4 m/s vC  8 m s



v B  2i  4 j  m s v B  4.47 m s
19