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REVIEW AND SUMMARY
This chapter was devoted to Newtons second law and its application
to the analysis of the motion of particles.
Newtons second law
Denoting by m the mass of a particle, by oF the sum, or resultant,
of the forces acting on the particle, and by a the acceleration of the
particle relative to a newtonian frame of reference [Sec. 12.2], we
wrote
(12.2)
oF 5 ma
Linear momentum
Introducing the linear momentum of a particle, L 5 mv [Sec. 12.3],
we saw that Newtons second law can also be written in the form
oF 5 L
(12.5)
which expresses that the resultant of the forces acting on a particle is
equal to the rate of change of the linear momentum of the particle.
Consistent systems of units
Equation (12.2) holds only if a consistent system of units is used. With
SI units, the forces should be expressed in newtons, the masses in
kilograms, and the accelerations in m/s2; with U.S. customary units,
the forces should be expressed in pounds, the masses in lb s2/ft (also
referred to as slugs), and the accelerations in ft/s2 [Sec. 12.4].
Equations of motion for a particle
To solve a problem involving the motion of a particle, Eq. (12.2) should
be replaced by equations containing scalar quantities [Sec. 12.5]. Using
rectangular components of F and a, we wrote
oFx 5 max
oFy 5 may
oFz 5 maz
(12.8)
Using tangential and normal components, we had
Ft 5 m
Dynamic equilibrium
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dv
dt
Fn 5 m
v2
r
(12.99)
We also noted [Sec. 12.6] that the equations of motion of a particle
can be replaced by equations similar to the equilibrium equations
used in statics if a vector 2ma of magnitude ma but of sense opposite to that of the acceleration is added to the forces applied to the
particle; the particle is then said to be in dynamic equilibrium. For
the sake of uniformity, however, all the Sample Problems were solved
by using the equations of motion, first with rectangular components
[Sample Probs. 12.1 through 12.4], then with tangential and normal
components [Sample Probs. 12.5 and 12.6].
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In the second part of the chapter, we defined the angular momentum
HO of a particle about a point O as the moment about O of the linear
momentum mv of that particle [Sec. 12.7]. We wrote
and noted that HO is a vector perpendicular to the plane containing
r and mv (Fig. 12.24) and of magnitude
HO 5 r mv sin f
j
y
mvy
k
z
mvz
y
HO
mv
(12.13)
Resolving the vectors r and mv into rectangular components, we
expressed the angular momentum HO in the determinant form
(12.14)
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Angular momentum
(12.12)
HO 5 r 3 mv
i
HO 5 x
mvx
Review and Summary
f
P
O
r
x
z
Fig. 12.24
In the case of a particle moving in the xy plane, we have z 5
vz 5 0. The angular momentum is perpendicular to the xy plane and
is completely defined by its magnitude. We wrote
HO 5 Hz 5 m(xvy 2 yvx)
(12.16)
.
Computing the rate of change H O of the angular momentum HO,
and applying Newtons second law, we wrote the equation
.
oMO 5 H O
(12.19)
Rate of change of angular
momentum
which states that the sum of the moments about O of the forces acting
on a particle is equal to the rate of change of the angular momentum
of the particle about O.
In many problems involving the plane motion of a particle, it is found
convenient to use radial and transverse components [Sec. 12.8, Sample Prob. 12.7] and to write the equations
(12.21)
oFr 5 m(r 2 ru2)
1 2ru)
(12.22)
oFu 5 m(ru
Radial and transverse components
When the only force acting on a particle P is a force F directed
toward or away from a fixed point O, the particle is said to be moving
under a central force [Sec. 12.9]. Since oMO 5 0 at any given instant,
.
it follows from Eq. (12.19) that H O 5 0 for all values of t and, thus,
that
Motion under a central force
HO 5 constant
(12.23)
We concluded that the momentum angular of a particle moving under
a central force is constant, both in magnitude and direction, and that
the particle moves in a plane perpendicular to the vector HO.
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r mv sin f 5 r0 mv0 sin f0
mv
P
mv0
where h is a constant representing the angular momentum per unit
mass, HO /m, of the particle. We observed (Fig. 12.26) that the infinitesimal area dA swept by the radius vector OP as it rotates
through du is equal to 1 r2du and, thus, that the left-hand mem2
ber of Eq. (12.27) represents twice the areal velocity dA/dt of the
particle. Therefore, the areal velocity of a particle moving under a
central force is constant.
f0
r0
(12.25)
for the motion of any particle under a central force (Fig. 12.25).
Using polar coordinates and recalling Eq. (12.18), we also had
r2u 5 h
(12.27)
f
O
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Recalling Eq. (12.13), we wrote the relation
Kinetics of Particles: Newtons Second Law
r
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P0
Fig. 12.25
r dq
dA
P
dq
r
O
F
q
Fig. 12.26
Newtons law of universal
gravitation
m
r
F
An important application of the motion under a central force is provided by the orbital motion of bodies under gravitational attraction
[Sec. 12.10]. According to Newtons law of universal gravitation, two
particles at a distance r from each other and of masses M and m,
respectively, attract each other with equal and opposite forces F and
2F directed along the line joining the particles (Fig. 12.27). The
common magnitude F of the two forces is
F
M
Fig. 12.27
F5G
Mm
r2
(12.28)
where G is the constant of gravitation. In the case of a body of mass m
subjected to the gravitational attraction of the earth, the product GM,
where M is the mass of the earth, can be expressed as
GM 5 gR 2
2
(12.30)
2
where g 5 9.81 m/s 5 32.2 ft/s and R is the radius of the earth.
Orbital motion
It was shown in Sec. 12.11 that a particle moving under a central
force describes a trajectory defined by the differential equation
F
d 2u
1u5
du2
mh2u2
(12.37)
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Review and Summary
where F . 0 corresponds to an attractive force and u 5 1/r. In the
case of a particle moving under a force of gravitational attraction
[Sec. 12.12], we substituted for F the expression given in Eq. (12.28).
Measuring u from the axis OA joining the focus O to the point A of
the trajectory closest to O (Fig. 12.28), we found that the solution
to Eq. (12.37) was
r
q
1
GM
5 u 5 2 1 C cos u
r
h
(12.39)
This is the equation of a conic of eccentricity 5 Ch2/GM. The
conic is an ellipse if , 1, a parabola if 5 1, and a hyperbola if
. 1. The constants C and h can be determined from the initial
conditions; if the particle is projected from point A (u 5 0, r 5 r0)
with an initial velocity v0 perpendicular to OA, we have h 5 r0v0
[Sample Prob. 12.9].
It was also shown that the values of the initial velocity corresponding,
respectively, to a parabolic and a circular trajectory were
vesc 5
2GM
A r0
GM
A r0
Fig. 12.28
Escape velocity
(12.43)
vcirc 5
O
(12.44)
and that the first of these values, called the escape velocity, is the
smallest value of v0 for which the particle will not return to its starting
point.
The periodic time t of a planet or satellite was defined as the time
required by that body to describe its orbit. It was shown that
t5
2pab
h
Periodic time
(12.45)
where h 5 r0v0 and where a and b represent the semimajor and
semiminor axes of the orbit. It was further shown that these semiaxes
are respectively equal to the arithmetic and geometric means of the
maximum and minimum values of the radius vector r.
The last section of the chapter [Sec. 12.13] presented Keplers laws
of planetary motion and showed that these empirical laws, obtained
from early astronomical observations, confirm Newtons laws of
motion as well as his law of gravitation.
Keplers laws
A
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