Download Conservation of Linear Momentum

 2000, W. E. Haisler
Conservation of Linear Momentum
ENGR 211
Principles of Engineering I
(Conservation Principles in Engineering Mechanics)
Conservation of Linear Momentum
 Introduction
 Conservation of Linear Momentum Equations
 Linear Momentum and Newton's Laws of Motion
 Steady State
 Other Special Cases
 Reference Frame
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 2000, W. E. Haisler
2
Conservation of Linear Momentum
Introduction
The linear momentum of a system is defined to be the product
of the mass (m) and velocity ( v ) of the system. Since velocity is
a vector, then linear momentum is a vector quantity (has three
components). Because linear momentum is a conserved
property, there is no generation or consumption. Therefore, the
conservation statement becomes

















LM entering   LM leaving 
 system during - system during





time period   time period 








Accumulation of LM
within system
during time period








Accumulation of LM   LM in system  LM in system 



   at end of
within system

 -  at beginning of 










during dime period   time period   time period








 2000, W. E. Haisler
Conservation of Linear Momentum
Linear Momentum Possessed by Mass
 Linear Momentum may enter or leave the system with mass.
 Linear Momentum may also enter or leave the system at a
certain rate.
 When mass enters a system traveling at a velocity ( v ), it adds
momentum to the system.
 The rate at which mass (m) enters the system is m (where
in
dm
m  in ) and therefore the rate at which linear momentum
in dt
enters the system is given by (m v ) . Similarly, rate at which
in
the linear momentum leaves the system is (m v )out .
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 2000, W. E. Haisler
Conservation of Linear Momentum
 The proper way to look a momentum rate is as follows
momentum    momentum  mass 
time   mass  time 
If p  linear momentum mv , then the above can be written in


 (mv ) 
d
p
  m m  v m
equation form:
dt 






 If there are multiple masses entering or leaving the system (say
n of them), then we simply add them up:
n    
   mv i in
i1
and
n    
   mv i out
i1
 Note that the velocity v of a mass can be thought of as it's
linear momentum per mass.
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 2000, W. E. Haisler
Conservation of Linear Momentum
5
Linear Momentum in Transit: Forces
 Why do we refer to "Forces" as "Linear Momentum in
Transit?" Newton's 2nd Laws states that the external forces on a
system are connected to the time rate of change of linear
momentum within the system!


f ext  dp  d (mv )
dt
dt
 Linear momentum enters the system as the result of forces that
act upon the mass in the system. By Newton's second law, an
external force f ext acting on a system provides momentum to
the system at rate fext . These forces may act as surface or
contact forces on the system boundary (called tractions), or
they may act as body forces on the entire mass of the system
(for example, body forces due to gravity).
 2000, W. E. Haisler
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Conservation of Linear Momentum
Schematic for Conservation of Linear Momentum
Resultant External
Forces Acting
on System
Surroundings
Linear
Momentum
Entering
System
System
Linear
Momentum
Surroundings
Linear
Momentum
Leaving
System
System
Boundary
 2000, W. E. Haisler
Conservation of Linear Momentum
 The external forces may be due to:
 pressure acting over the surface area of the system
 drag or frictional forces acting on the surface
 supporting structure which holds the system in place (cause
forces acting on the surface)
 gravitational and electrostatic body forces acting on the
system volume
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 2000, W. E. Haisler
Conservation of Linear Momentum
Conservation of Linear Momentum Equations
We can now state the rate equation for conservation of linear
momentum as:
dpsys
 (m v ) (m v )
 f
out
ext
dt
in
The momentum of the system may change with time because of
mass that enters and leaves, and because of external forces that
act on the system.
As mass enters the system, the amount of linear momentum that
enters the system is this [mass flow rate] times the [momentum
per unit mass (i.e., the velocity) of the mass].
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 2000, W. E. Haisler
Conservation of Linear Momentum
9
We must keep in mind that conservation of linear momentum is a
vector equation! Consequently, we can write in component form.
For a Cartesian coordinate system, we have
x-direction:
d ( px)sys
 (m v x ) (m v x )
 ( f x )
out
ext
dt
in
y-direction:
d ( p y)sys
 (m v y ) (m v y )
 ( f y )
out
ext
dt
in
z-direction:
d ( pz )sys
 (m v z ) (m v z )
 ( f z )
out
ext
dt
in
 2000, W. E. Haisler
Conservation of Linear Momentum
Linear Momentum of the System
The linear momentum of the system may be written (for n
particles in the system) as


n
psys   (mv )
i
i 1
sys
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 2000, W. E. Haisler
Conservation of Linear Momentum
Rate of Change of Linear Momentum of the System
 Consider the time rate of change of linear momentum of the
system (left side of conservation equation) and use the product
rule for the derivative:
dpsys d (mv )sys
dmsys
dvsys

 vsys
 msys
dt
dt
dt
dt
 If the mass of the system is constant with time, then the time
rate of change of linear momentum reduces to
dpsys
dvsys
 msys
dt
dt
Note that the time derivative of the velocity becomes the
dvsys
 asys
acceleration:
dt
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 2000, W. E. Haisler
Conservation of Linear Momentum
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Velocity of the Center of Mass
For rigid bodies and other closed systems, it is convenient to
define a velocity that characterizes the system as a whole. This
velocity is an average velocity for the system and acts at the
center of mass. We define the velocity of the center of mass,
v , such that the system mass times the velocity of the mass
G
center is equal to system linear momentum:
n
msysv  (mv )i 
G i1

sys
and solving for the velocity of the center of mass:
 2000, W. E. Haisler
Conservation of Linear Momentum


n
 (mv )i
sys
v  i 1 m
G
sys
If the size of the elements mi becomes differentially small, the
summation process becomes an integration and the velocity of
the center of mass becomes
v  sys
G m
sys
v dm
This is sometimes called the mass-average velocity of the
system.
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 2000, W. E. Haisler
Conservation of Linear Momentum
14
Recall that the velocity of an element of mass is defined to be the
time rate of change of the position vector describing it's location.
For a particle with velocity vG whose position is refined by the
position vector rG :
_
vG
y
_
rG
x
z
dr
v  G
G dt
r  xi  yj  zk , x=x(t), etc.
G
 dy

d
dx
dz
vG  G  (xi  yj  zk )  i  j  k  vxi  vy j  vzk
dt dt
dt dt dt
dr
The center of mass can be determined in terms of the position
vectors of all of the mass elements of the system:
 2000, W. E. Haisler
15
Conservation of Linear Momentum

y
_
rG
x

n
 (mr )i
i

1
sys
r 
m
G
sys
z
As the size of mass particle becomes differentially small, we can
replace the summation by an integration to obtain
r  sys
G m
sys
r dm
This definition of the center of mass applies only for a closed
system (one in which there is no mass into or out of the system).
 2000, W. E. Haisler
Conservation of Linear Momentum
When the system is closed (no mass in or out of system), the
conservation of linear momentum becomes
dv
dpsys d (msys v )
G m ( G )m a
f


 ext dt
sys dt
sys G
dt
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 2000, W. E. Haisler
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Conservation of Linear Momentum
Integral (Finite Time Period) Equation
The rate form of the conservation of linear momentum equation
can be integrated over a finite time period to obtain


 dp

t
sys


end 
dt 


t
dt


beg 


t
end (m
 v) dt 
in
t
beg

t
end (m
 v) dt 
out
t
beg

t
end f
dt

ext
t
beg

The above equation can also be written for a finite time period as
( psys)
 ( psys)
 (m v ) t (m v ) t  ( f )t
out
ext
end
beg
in
or
( psys)
 ( psys)
 (mv ) (mv )
 ( f )t
out
ext
end
beg
in
 2000, W. E. Haisler
Conservation of Linear Momentum
18
Linear Momentum and Newton's Laws of Motion
The concept of conservation of linear momentum applied to a
rigid body, together with the realization that forces exchange
momentum between the surroundings and the body, is actually a
statement of Newton's three laws of motion.
 Newton's first law of motion - A body which is at rest will stay
at rest unless acted upon by an external force. In other words, a
body does not spontaneously generate momentum; momentum is
conserved.
 2000, W. E. Haisler
Conservation of Linear Momentum
19
 Newton's second law of motion - For a body (system) of
constant mass, the sum of the external forces is equal to the
mass times acceleration of the body. This is clearly a result of
the conservation of linear momentum when m  m out  0 so
in


 dm

sys


that by conservation of mass 
 0
dt 



 Newton's third law of motion - For every force there is an
opposite and equal reaction force. This also implies
conservation of linear momentum. A force acting on a system
by the surroundings (boundary forces or body forces) provides
an opposite and equal force exerted on the surroundings. Thus,
momentum transferred to the system must have been provided
or given up by the surroundings, implying conservation of
linear momentum.
 2000, W. E. Haisler
Conservation of Linear Momentum
Steady State
The steady state condition for a system means that the system is
not changing with time and hence the accumulation within the
system is zero for any time period.
( psys)
 ( psys)
0
end
beg
dpsys
0
or, for any instant of time:
dt
For a finite time period:
Through conservation of linear momentum, steady state implies:
0  (m v )  (m v )out   f
in
ext
dmsys
Steady state also implies that the term
 0 in the
dt
conservation of mass law.
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 2000, W. E. Haisler
Conservation of Linear Momentum
For rigid body statics (no mass enters or leaves the system and
the body is at steady state), linear momentum reduces to
0 f
ext
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 2000, W. E. Haisler
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Conservation of Linear Momentum
Reference Frame
In order to define velocity and liner momentum, we must define
a reference coordinate system. Position and velocity are then
measured with respect to the origin of the coordinate system. For
this purpose, we can choose any reference frame so long as it is
an inertial reference frame, i.e., one that is not accelerating (has
constant velocity and not rotating).
v
Consider a block sliding along a
plane at a velocity of vref with an
y
attached pendulum as shown to the
x
right. The x-y frame is fixed. The
x'-y' frame is attached to the block fixed
so that it also has a velocity of vref.
v (x, y)  v  v (x', y')
ref
y’
ref
x’
 2000, W. E. Haisler
23
Conservation of Linear Momentum
Now lets write COLM with respect to the stationary x-y
reference frame (vxy  v in x - y coordinate system)
d[mvxy]sys
 (m v xy )
 f
ext
dt
in / out
(x,y coords.) (1)
Substitute the relation between the two coordinates systems into
(1):
d[m(v v )]sys
ref x' y'
 [m (v
 v )]
 f
ext
dt
ref x' y' in / out
Taking the time derivative on the left side and rearranging, the
above can be written:
 2000, W. E. Haisler
24
Conservation of Linear Momentum
dv
d (mv )sys
dm
sys  v
x' y'
 msys ref  v
m

dt
dt
ref dt
ref
in / out
 (m v
)
x' y' in / out
 f
ext
The first underlined term is zero since v
is constant. The
ref
dmsys
  m
 0 since this is
double underlined term
in / out
dt
conservation of mass (for no mass gen/con). Thus, the COLM
equation in the x'-y' frame reduces to
d[mv ]sys
x' y'
 (m v )
 f
ext
dt
x' y' in / out
(x',y' coords.) (2)
 2000, W. E. Haisler
Conservation of Linear Momentum
Note that equation (2) is identical to equation (1) except for the
change in reference frame. Since (2) was derived from (1), we
see that COLM is independent of coordinate system as long as
we use an inertial reference frame for the velocities and the
external forces f ext are independent of the reference frame.
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 2000, W. E. Haisler
26
Conservation of Linear Momentum
Some notes on unit and radius vectors. Consider a line element
OP of length L (or a vector L ) where L  (x2  y2  z2)1/ 2. The
vector L is given by L  xi  yj  zk . Note: "O" is at the origin.
Y
P(x,y,z)
y
L
x
0
Z
z
y
X
z
x
We define the following angles:
 x  angle between x -axis and vector L
 y  angle between y -axis and vector L
 z  angle between z -axis and vector L
 2000, W. E. Haisler
Conservation of Linear Momentum
27
Now the cosine of each angle is: cos x  x , cos y  y , and
L
L
cos z  z . Note that  x is an angle in the plane containing the
L
x-axis and the line segment OP. A unit vector u in the
direction of L is given by
u  ( x )i  ( y ) j  ( z )k  (cos x)i  (cos y ) j  (cos z )k
L
L
L
where L  L  (x2  y2  z2)1/ 2 . The cosines of the three angles
are often referred to as the direction cosines of the line OP.
From geometry and trigonometry, one can prove that
cos2 x  cos2 y  cos2 z 1. Thus you only need to specify two
of the angles – the third must satisfy the trig relation. This is
similar to a unit vector – two of the components (say x and y) can
be specified, but the third (say z) must satisfy x2  y2  z2 1!!
 2000, W. E. Haisler
28
Conservation of Linear Momentum
A line segment AB of length L (or a vector) may also be
located by the coordinates of its end points: A (xbeg, ybeg, zbeg)
and B (xend,yend,zend). Then
the unit vector in the direction
of AB is given by:
u  (x)i  (y ) j  (z )k
L
L
L
where x  x
 x , etc.
end beg
Y
B(xend,yend,zend )
_
u
X
0
)
,z
,y
A(x
beg beg beg
Z
 2000, W. E. Haisler
29
Conservation of Linear Momentum
y
F
40o
o
30o 50
z
x
 2000, W. E. Haisler
Conservation of Linear Momentum
Example: Vectors P and P in x-y plane:
1
2
Y
_
P
2
_
P
1
40o
0
Z
30o
X
u  cos(30)i  cos(60) j  cos(90)k
1
P P u
1 1 1
u  cos(140)i  cos(50) j  cos(90)k
2
P P u
2 2 2
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 2000, W. E. Haisler
31
Conservation of Linear Momentum
lbm ft
Note on conversion of lb f and
in American Engineering
sec 2
units.
lbm ft
1 lb f  32.174
sec 2
lbm ft
gc  32.174
lb sec 2
f







lbm ft 
2
F
  m(lb ) g (ft/sec )
m

2
sec 

2)
g
(
ft/sec
F (lb f )  m(lbm) g
c









lb
f
Note units: gg
c lbm









 2000, W. E. Haisler
Conservation of Linear Momentum
32
Notes on determining the mass entering a system. The mass
entering a system boundary (through an area) during a period of
time may be determined or specified in many ways.
1. We can simple state that a certain amount of mass enters the
system during a specified time period, i.e., m  mass .
in time
2. However, a more common situation is when one specifies that
a fluid (of given density) flows through an area at a specified
velocity normal to the area. In this case, consider the following
dimensional analysis:
m  mass  densityvolume  m L31s  m L2 Ls  density area velocity
in time
time
L3
L3
So the rate of mass entering a system for a mass with density 
and passing through an area A with a velocity Vn normal to the
area is given by: m  AVn .
in
 2000, W. E. Haisler
Conservation of Linear Momentum
3. Another common way to describe mass flow is by the mass
flux rate, or the mass per unit area per unit time.
mass flux rate 
mass  densityvolume  densitylength  densityvelocity .
areatime
areatime
time
To obtain the mass entering a system (in terms of mass flux
rate) we write:
mass entering  mass flux rate  area time  (
mass )(area)(time period )  mass
areatime
m  mass flux rate areatime period
in
33