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Transcript
CHAPTER 13
Kinetics of Particles:
Energy and Momentum
Methods
13.1 INTRODUCTION
Method of work and energy –
Involves relations between displacement, velocity, mass, and force.
Method of impulse and momentum –
Involves relations between mass, velocity, force, and time.
Some problems become easier to do.
13.2 WORK OF A FORCE
When a force is applied to a mass and the mass
moves through an incremental distance, the work
done by the force is
 
dU  F  dr  Fx dx  Fy dy  Fz dz
 
dU  F  dr  F (cos  )ds

F
A

r


dr A’
 
r  dr
dU  0 if   900
dU  0 if   900
dU  0 if   900
Units  ft  lb or N  m  Joule ( J )
To get the total work done along a path requires
 
  F  dr
2
U 12
1
Notice that
ds

dr
2
2
1
1
U 12   F (cos  )ds   Ft ds
When using rectangular coordinates


F
2
U 12   ( Fx dx  Fy dy  Fz dz )
1
Ft
0
s1
s2
s
The work is the area under the curve.
Work of a Constant Force in Rectilinear Motion
U 12  F (cos  )x  F(cos  )( x2  x1 )

F

1
x
2
Work of the Force of Gravity
dU  Wˆj  ( dxî  dyˆj  dzk̂ )
dU  Wdy
2
U 12    Wdy  W ( y1  y2 )
1
A2
W
U 12  Wy
dy
A
y2
A1
y
y1
Work of the Force Exerted by a Spring
Spring undeformed
x1

F
2
U 12    kxdx ( 21 kx22  21 kx12 )
x
1
x2
  12 k ( x2  x1 )( x2  x1 )
F
x
x1
F1
F2
F  kx
F  kx
x2
x
Work of a Gravitational Force
2
U 12
GMm GMm
GMm

   2 dr 
r2
r1
r
1
Forces which do not do work (ds = 0 or cos a = 0):
• reaction at frictionless pin supporting rotating body,
• reaction at frictionless surface when
body in contact moves along surface,
• reaction at a roller moving along its track, and
• weight of a body when its center of
gravity moves horizontally.
13.3 KINETIC ENERGY OF A PARTICLE.
PRINCIPLE OF WORK AND ENERGY
If the force doing work is the net force then
dv
dv ds
dv
 mv
Ft  mat  m  m
dt
ds dt
ds
2
U 12
2
2
dv
  Ft ds   mv ds   mvdv
ds
1
1
1
U 12  mv  mv
1
2
T  21 mv2
2
2
1
2
U 12  T2  T1
2
1
T1  U 12  T2
13.4 APPLICATIONS OF THE PRINCIPLE
OF WORK AND ENERGY
Work the pendulum problem in the text.
• Wish to determine velocity of
pendulum bob at A2. Consider work &
kinetic energy.
T1  U12  T2
0  Wl 
1W 2
v2
2 g
v2  2 gl
• Velocity found without determining expression for acceleration
and integrating.
• All quantities are scalars and can be added directly.
• Forces which do no work are eliminated from the problem.
• Principle of work and energy cannot be
applied to directly determine the
acceleration of the pendulum bob.
• Calculating the tension in the cord
requires supplementing the method of
work and energy with an application of
Newton’s second law.
• As the bob passes
through A2 ,
 Fn  m an
v2  2 gl
W v22
P W 
g l
W 2 gl
P W 
 3W
g l
13.5 POWER AND EFFICIENCY
Power is the rate at which work is done.
 
dU F  dr  
P

 F v
dt
dt
Units – 1 hp = 550 ft lb/s and 1 Watt = 1 J/s
Efficiency
work out power out


work in
power in
13.6 POTENTIAL ENERGY
Close to the Earth
2
U 12    Wdy  W ( y1  y2 )
1
Define
Vg  Wy
Then
U12  (Vg )1  (Vg ) 2
U 12  Vg
Not So Close to the Earth
2
U 12
GMm GMm
GMm

   2 dr 
r2
r1
r
1
Define
2
GMm
WR
Vg   r  
r
Then
U12  (Vg )1  (Vg ) 2
U 12  Vg
For a Spring
2
U 12    kxdx ( 21 kx22  21 kx12 )
1
Define
Ve  kx
1
2
2
Then
U12  (Ve )1  (Ve ) 2
U 12  Ve
Notice that the work done
by each of these three forces
is equal to a change in something that is
a function of position only.
The idea of a function of position
is valid as long as
F is conservative.
It would not work for a force like friction.
This function, V, is call the potential energy.
Potential energy is an energy of
position or
configuration.
13.7 CONSERVATIVE FORCES
Work done by conservative forces is
independent of the path
over which work is done.
U12  V ( x1 , y1 , z1 )  V ( x2 , y2 , z2 )
For short
U12  V1  V2
For any conservative force
 
 F  dr  0
An elemental work corresponding
to an elemental displacement
dU  V ( x , y , z )  V ( x  dx , y  dy , z  dz )
dU  dV ( x, y, z )
dU  dV ( x, y, z )
 V


V

V
Fx dx  Fy dy  Fz dz  
dx 
dy 
dz 
y
z 
 x
V
Fx  
x
V
Fy  
y
V
Fz  
z


F  Fx î  Fy ˆj  Fz k̂   V î  V ˆj  V
y
z
 x

k̂ 

The vector in the parentheses is known as the
gradient of the scalar function V.

F   grad V

F  V
13.8 CONSERVATION OF ENERGY
Kinetic plus Potential
If the only forces doing work on a system of particles are
conservative, then the total mechanical energy is conserved.
U 12  V  T
V1  V2  T2  T1
T1  V1  T2  V2
Pendulum Motion
l
T1  V1  T2  V2
0  Wl  mv  0
1
2
v  2 gl
2
13.9 MOTION UNDER A CONSERVATIVE
CENTRAL FORCE. APPLICATION
TO SPACE MECHANICS
The gravitational attractive force is conservative.
So, in space mechanics both energy and angular momentum
are conserved since this force is
both conservative and central.

v
P

r
O

r0 mv0 sin 0  rmv sin 

v0
T0  V0  T  V


r0
0
P0
1
2
2
GMm
1
mv02  GMm

mv

2
r
r
0
13.10 PRINCIPLE OF IMPULSE AND
MOMENTUM
 d

F  ( mv )
dt
t2 


mv2  mv1   Fdt  Imp 12
t1
t2 


mv1   Fdt  mv2
t1
For a system of particles
external impulses are considered only
(remember Newton’s third law)
If no external forces act on the particles then


 mv2   mv1
13.11 IMPULSE MOTION
Non-impulsive forces can be neglected for they are small
in comparison (usually) to the impulsive forces.
If it is not known for sure that the forces are small,
then include them.
Impulsive Motion
• Force acting on a particle during a very
short time interval that is large enough
to cause a significant change in
momentum is called an impulsive force.
• When impulsive forces act on a
particle,



mv1   Ft  mv2
• When a baseball is struck by a bat,
contact occurs over a short time
interval but force is large enough to
change sense of ball motion.
• Nonimpulsive forces are forces for
which

 F t is small and therefore, may be
neglected.
13.12 IMPACT
• Impact: Collision between two bodies
which occurs during a small time interval
and during which the bodies exert large
forces on each other.
• Line of Impact: Common normal to the
surfaces in contact during impact.
Direct Central
Impact
• Central Impact: Impact for which the
mass centers of the two bodies lie on the
line of impact; otherwise, it is an
eccentric impact.
• Direct Impact: Impact for which the
velocities of the two bodies are directed
along the line of impact.
Oblique Central
Impact
• Oblique Impact: Impact for which one or
both of the bodies move along a line other
than the line of impact.
• Block constrained to move along
horizontal surface.


• Impulses from internal forces F and  F
along the
 n axis and from external
force Fext
exerted by horizontal surface and
directed along the vertical to the
surface.
• Final velocity of ball unknown in direction
and magnitude and unknown final block
velocity magnitude. Three equations
required.
13.13 DIRECT CENTRAL IMPACT
• Bodies moving in the same straight
line, vA > vB .
• Upon impact the bodies undergo a
period of deformation, at the end of
which, they are in contact and moving
at a common velocity.
• A period of restitution follows
during which the bodies either
regain their original shape or remain
permanently deformed.
vA
vA
• Wish to determine the final
velocities of the two bodies. The
total momentum of the two body
system is preserved,
mAvA  mB vB  mAvA  mB vB
• A second relation between the final
velocities is required.
e  coefficient of restitution
• Period of
deformation:
• Period of
restitution:
mAv A   Pdt  mAu
mAu   Rdt  mAvA
• A similar analysis of particle B yields
• Combining the relations leads to the
desired second relation between the
final velocities.
• Perfectly plastic impact, e = 0:
• Perfectly elastic impact, e = 1:
Total energy and total momentum
conserved.
Rdt

e
 Pdt
e
u  vA

vA  u
vB  u
u  vB
vB  u u  vA vB  vA

e

u  vB v A  u v A  vB
vB  vA
e
v A  vB
vB  vA  evA  vB 
0 e1
vB  vA  v
mAvA  mB vB  mA  mB v
v  v  v  v
B
A
A
B
13.14 OBLIQUE CENTRAL IMPACT
• No tangential impulse
component; tangential
component of momentum for
each particle is conserved.
• Normal component of total
momentum of the two particles
is conserved.
• Normal components of relative
velocities before and after
impact are related by the
coefficient of restitution.
• Final velocities are
unknown in
magnitude and
direction. Four
equations are
required.
v   v 
A t
A t
v   v 
B
t
B
t
mA vA n  mB vB n  mA vA n  mB vB n
v   v 
B
n
A
n
 evA n  vB n 
v   v 
• Tangential momentum of ball is
conserved.
• Total horizontal momentum of
block and ball is conserved.
• Normal component of relative
velocities of block and ball are
related by coefficient of
restitution.
B
B
t
t
mA vA   mB vB x  mA vA   mB vB x
v   v 
B
n
A
n
 evA n  vB n 
• Note: Validity of last expression does not follow from previous
relation for the coefficient of restitution. A similar but separate
derivation is required.
13.15 PROBLEMS INVOLVING ENERGY
AND MOMENTUM
• Three methods for the analysis of kinetics
problems:
- Direct application of Newton’s second law
- Method of work and energy
- Method of impulse and momentum
• Select the method best suited for the problem or part of a
problem under consideration.