Chapter 12
Kinetics of Particles: Newton’s Second Law
12.1 Introduction
12.2 Newton’s Second Law of Motion
If the resultant force acting on a particle is not zero, the particle will have an
acceleration proportional to the magnitude of the resultant and in the direction of this
resultant force.


F  ma or more appropriately


F  ma
Inertial frame or newtonian frame of reference – one in which Newton’s second law
equation holds.
12.3 Linear Momentum of a Particle. Rate of Change of Linear Momentum




dv d
dL 
F  m  ( mv ) 
L
dt dt
dt
12.4 Systems of Units
Reading assignment
12.5 Equations of Motion
Rectangular Components
 ( Fx î  Fy ˆj  Fz k̂ )  m( a x î  a y ˆj  a z k̂ )
or
Fx  max  mx ,
Fy  ma y  my ,
For projectile motion
mx  0
x  0
my  W
mz  0
W
y     g
z  0
m
Fz  maz  mz
Tangential and Normal Components
Ft  mat
Ft 
dv
dt
Fn  man
Fn  m
v2

12.6 Dynamic Equilibrium
See Figure 12.10

 ma  inertia vector
Addition of this vector makes the problem a static one.
Example Problems
12.7 Angular Momentum of a Particle. Rate of Change of Angular

Momentum
mv
y

HO


r



H 0  r  mv
O
H 0  rmv sin 
z
î

H0  x
mvx
ˆj
y
k̂
z
mv y
mvz
H x  m( yv z  zv y )
H y  m( zv x  xvz )
H z  m( zv y  xvx )
x
y
mv
H z  m( zv y  xvx )
For motion in x-y plane

mv
H 0  rmv sin   rmv  mr 2
In Polar Coordinates



 
 



H 0  r  mv  r  mv  v  mv  r  ma  r  ma


H 0  M o

r
mvr
x
O
12.8 Equations of Motion in Terms of Radial and Transverse Components
y
Fr  mar  m( r  r )
F  ma  m( r  2r )
Could have done it thusly
y
F
2

r
O
F r
m

r
=
x
ma
m
x
O
d
( mr 2 )  mr 2  2 mrr )
dt
F  m( r  2r )
rF 
12.9 Motion Under a Central Force. Conservation of Angular Momentum
y




H 0  0  H 0  cons tan t  r  mv

Particle moves in plane perpendicular to H 0
r0 mv0 sin 0  rmv sin 
mr 2  H
0

F
m
x
O
y
See Figures 12.17
rd
dA
dA  21 r 2 d
dA 1 2 
 r   cons tan t
dt 2
d
O
Areal velocity is constant. Example is Kepler’s second law.
12.10 Newton’s Law of Gravitation
Mm
r2
Mm
W  G 2  mg
r
F G
Example Problems

ma r

 F
r
x
12.11 Trajectory of a Particle Under a Central Force
Fr  m( r  r 2 )   F
F  m( r  2r )  0
But
r 2 
H0
 h  cons tan t
m
h
r2
dr dr d
h dr
d 1
r 

 2
 h
 
dt d dt r d
d  r 
 
r 
dr h dr
h2 d 2
 2
 2
dt r d
r d 2
Let u 

1
 
r
1
Then
r
h d 2u
h2 
Fr  m  2
 3    F
2
r 
 r d
d 2u
F
u 
2
d
mh 2 u 2
Can be solved sometimes
12.12 Applications to Space Mechanics
F  GMmu2 for satellites
d 2u
GM
u  2
2
d
h
u  c cos(   0 ) 
GM 1

r
h2
General solution + Particular solution
Choose polar axis so that  0  0
The above equation for u is a conic section, that is it is the equation for ellipses (and
circles), parabolas, or hyperbolas.
Circle
Ellipse

Eccentricity
C
GM
Parabola
h2
1 GM
 2 (1   cos )
r
h
Therefore
Three Cases:   1,   1,   1
 1
C
GM
Then 1  cos can be 0 for two 's thus r   twice.
h2
Hyperbola
 1 C 
GM
for  =1800 r  
2
h
Parabola
 1 C 
GM
ellipse
h2
Will assume that rocket burnout occurs parallel to earth’s surface.
GM GM gR 2
r0 v0  h and
 2 2  2 2
h2
r0 v0 r0 v0
1 GM
 2  C,   0
r0
h
C
1 GM
 2
r0
h
Hyperbola
For parabolic
C
If
GM 1 GM
  2
r0
h2
h
v0  vesc
1 2GM
2GM
(escape velocity)
 2 2  v0 
r0 r0 v0
r0
2GM
2 gR 2


r0
r0
hyperbolic
v0  vesc parabolic
v0  vesc
ellipse
A’
A”
A
O
Discuss
Periodic time
dA
h ab
 cons tan t  21 r 2  
dt
2


ab  area of ellipse
2ab
h
a
B
b
Discuss
A’
12.13 Kepler’s Laws of Planetary Motion
1.
2.
3.
O’ C
r1
O
A
r0
Each planet describes an ellipse, with the sun located at one of its foci.
The radius vector drawn from the sun to a planet sweeps equal areas in equal
times.
The squares of the periodic times of the planets are proportional to the cubes of
the semimajor axes of their orbits.