Transcript Chapter 9

Chapter 9
Center of Mass and Linear Momentum
Linear momentum
• Linear momentum (or, simply momentum)
point-like object (particle) is

p
of a


p  mv
• SI unit of linear momentum is kg*m/s
• Momentum is a vector, its direction coincides with
the direction of velocity
Newton’s Second Law revisited
• Originally, Newton formulated his Second Law in a
more general form

Fnet

dp

dt
• The rate of change of the momentum of an object is
equal to the net force acting on the object
• For a constant mass

Fnet



dp d (mv )
dv



m
 ma
dt
dt
dt
Center of mass
• In a certain reference frame we consider a system of
particles, each of which can be described by a mass
and a position vector
• For this system we can define a center of mass:

rcom 

 mi ri
i
m
i
i


 mi ri
i
M
Center of mass of two particles
• A system consists of two particles on the x axis
• Then the center of mass is
xcom
m1 x1  m2 x2

m1  m2
• Changing the reference frame
xcom
m2 d

m1  m2

rcom 

m
r
 ii
i
m
i
i


m
r
 ii
i
M
Newton’s Second Law for a system of
particles
• For a system of particles, the center of mass is

rcom 

 mi ri
i
m

i
• Then
i


Mrcom   mi ri

Macom
i


d   mi ri 
  i 2 
dt
2

d ri
  mi 2
dt
i
2

 mi ri
i
M


2
d rcom d ( Mrcom )
M

2
dt
dt 2
2

  mi ai
i

  Fi
i
Newton’s Second Law for a system of
particles
• From the previous slide:


Macom   Fi
i

• Here Fi is a resultant force on particle i
• According to the Newton’s Third Law, the forces that
particles of the system exert on each other (internal
forces) should cancel:


Macom  Fnet

• Here Fnet is the net force of all external forces that
act on the system (assuming the mass of the system
does not change)
Newton’s Second Law for a system of
particles


Macom  Fnet
Linear momentum for a system of
particles
• We define a total momentum of a system as:



P   pi   mi vi
i
i
• Using the definition of the center of mass


dri


P   mi vi   mi
dt
i
i

d  mi ri
i
dt


drcom
 Mvcom
M
dt
• The linear momentum of a system of particles is
equal to the product of the total mass of the system


m
r
m
r
and the velocity of the center of mass
 ii  ii

rcom 
i
m
i
i

i
M
Linear momentum for a system of
particles
• Total momentum of a system:


P  Mvcom
• Taking a time derivative




dvcom
dP
 Macom  Fnet
M
dt
dt
• Alternative form of the Newton’s Second Law for a
system of particles

dP 
 Fnet
dt
Conservation of linear momentum
• From the Newton’s Second Law

dP 
 Fnet
dt
• If the net force acting on a system is zero, then

dP
0
dt

P  const
• If no net external force acts on a system of particles,
the total linear momentum of the system is conserved
(constant)
• This rule applies independently to all components
Fnet , x  0  Px  const
Center of mass of a rigid body
• For a system of individual particles we have

rcom 

 mi ri
i
m
i
i
• For a rigid body (continuous assembly of matter)
with volume V and density ρ(V) we generalize a
definition of a center of mass:

rcom 


volume

r dV
volume
dV


volume

r dV
M
Chapter 9
Problem 82
Impulse
• During a collision, an object is acted upon by a
force exerted on it by other objects participating in
the collision
 
dp
 Fnet (t )
dt
 
dp  Fnet (t )dt

tf
ti
tf 

dp   Fnet (t )dt
• We define impulse as:

tf 
J   Fnet (t ) dt
ti
• Then (momentum-impulse theorem)



p f  pi  J
ti
Elastic and inelastic collisions
• During a collision, the total linear momentum is always
conserved if the system is isolated (no external force)
• It may not necessarily apply to the total kinetic energy
• If the total kinetic energy is conserved during the
collision, then such a collision is called elastic
• If the total kinetic energy is not conserved during the
collision, then such a collision is called inelastic
• If the total kinetic energy loss during the collision is a
maximum (the objects stick together), then such a
collision is called completely inelastic
Elastic collision in 1D
K  const
2
2
m1v12i m2 v22i m1v1 f m2 v2 f



2
2
2
2
m1 (v1i  v1 f )(v1i  v1 f ) 
 m2 (v2i  v2 f )(v2i  v2 f )
(m1  m2 )
2m2
v1 f 
v1i 
v2 i
(m1  m2 )
(m1  m2 )
v2 f
2m1
(m1  m2 )

v1i 
v2 i
(m1  m2 )
(m1  m2 )
P  const
m1v1i  m2 v2i  m1v1 f  m2 v2 f
m1 (v1i  v1 f )  m2 (v2i  v2 f )
Elastic collision in 1D: stationary target
(m1  m2 )
2m2
v1 f 
v1i 
v2 i
(m1  m2 )
(m1  m2 )
2m1
(m1  m2 )
v2 f 
v1i 
v2 i
(m1  m2 )
(m1  m2 )
• Stationary target: v2i = 0
• Then
(m1  m2 )
v1 f 
v1i
(m1  m2 )
2m1
v2 f 
v1i
(m1  m2 )
Chapter 9
Problem 58
Completely inelastic collision in 1D
P  const
m1v1i  (m1  m2 )V
m1
V
v1i
(m1  m2 )
Chapter 9
Problem 62
Answers to the even-numbered problems
Chapter 9:
Problem 2
(a) −1.50 m;
(b) −1.43 m
Answers to the even-numbered problems
Chapter 9:
Problem 10
6.2 m
Answers to the even-numbered problems
Chapter 9:
Problem 18
4.9 kg · m/s
Answers to the even-numbered problems
Chapter 9:
Problem 26
(a) 42 N · s;
(b) 2.1 kN
Answers to the even-numbered problems
Chapter 9:
Problem 46
3.1 × 102 m/s
Answers to the even-numbered problems
Chapter 9:
Problem 66
(a)(10 m/s)ˆi + (15 m/s)ˆj;
(b) −500 J
Answers to the even-numbered problems
Chapter 9:
Problem 70
(a) 2.7;
(b) 7.4