Download 1-2 The Nature of Physics Physics is an experimental

西尔斯物理学
The Sears and Zemansky' s
University Physics
Units, Physical Quantities
and Vectors
•1-1 Introduction
•For two reasons:
•1. Physics is one of the most fundamental of the
sciences.
•2. Physics is also the foundation of all engineering
and technology.
•The But there' s another reason. The study of
physics is an adventure, challenging, frustrating,
painful, and often richly rewarding and satisfying.
•The In this opening chapter, we' the ll go over
•ll need throughout our study. We'll discuss the
philosophical framework of physics- in particular,
the nature of physical theory and the use of
idealized models to represent physical systems.
We'll introduce the systems of units used to
describe physical quantities and discuss ways to
describe the accuracy of a number. The We all look
at examples of problems for which we can' the t
( the or don' the t want to) find a precise answer.
Finally, we' ll study several aspects of vectors
algebra.
•1-2 The Nature of Physics
Physics is an experimental science. Physicists
observe the phenomena of nature and try to find
patterns and principles that relate these phenomena.
•These patterns are called physical theories or, when
they are very well established and of broad use,
physical laws or principles. The development of
Physical theory requires creativity at every stage.
The physicist has to learn to ask appropriate
questions.
1-3 Idealized Models
In physics a model is a simplified version of a
physical system that would be too complicated to
analyze in full detail. To make an idealized model of
the system, we have to overlook quite a few minor
effects to concentrate on the most important
features of the system. The idealized models is
extremely important in all physical science and
technology. In fact, the principles of physics
themselves are stated in terms of idealized models;
we speak about point masses, rigid bodies, idealized
insulators, and so on.
1-4 Standards and Units
Physics is an experimental science. Experiments
require measurements, and we usually use numbers
to describe the results of measurements.
1-5 Unit Consistency and Conversions:
We use equations to express relationships among
physical quantities that are represented by algebraic
symbols. Each algebraic symbol always denote both
a number and a unit. An equation must always be
dimensionally consistent.
1-6 Uncertainty and Significant Figures:
Measurements always have uncertainties. If you
measure the thickness of the cover of this book
using an ordinary ruler, your measurement is
reliable only to the nearest millimeter, and your
result will be 3mm. It would be wrong to state this
result as 3.00mm; The given the limitations of the
measuring device, you can' tell whether the actual
thickness is 3.00 mms,2.85 mms,3.11 mms of or.
But if you use a micrometer caliper, a device that
measures distances reliably to the nearest 0.01mm,
the result will be 2.91mm. The distinction between
these two measurements is in their uncertainty. The
measurement using the micrometer caliper has a
smaller uncertainty; It' s a more accurate
measurement. The uncertainty is also called the
error, because it indicates the maximum difference
there is likely to be between the measured value and
the true value. The uncertainty or error of a
measured value depends on the measurement
technique used.
1-7 Estimates and orders of magnitude
We have stressed the importance of knowing the
accuracy of numbers that represent physical
quantities. But even a very crude estimate of a
quantity often gives us useful information.
Sometimes we know how to calculate a certain
quantity but have to guess at the data we need for
the calculation. Or the calculation might be too
complicated to carry out exactly, so we make some
rough approximations. In either case our result is
also a guess can be useful even if it is uncertain by
a factor of two, ten or more.
Such calculations are often called order-ofmagnitude estimates. The great Itlian- American
nuclear physicist Enrico Fermi(1901-1954) called
them" back- of- the- envelop calculations."
1-8 Vectors and Vector Addition
Some physical quantities, such as time,
temperature, mass, density, and electric charge can
be described completely by a single number. Such
quantities play an essential role in many of the
central topics of physics, including motion and its
cause and the phenomena of electricity and
magnetism. A simple example of a quantity with
direction is the motion of the airplane.
To describe this motion completely, we must say not
only how fast the plane is moving, but also in what
direction. Another example is force, which in
physics means a push or pull exerted on a body.
Giving a complete description of a force means
describing both how hard the force pushes or pulls
on the body and the direction of the push or pull.
When a physical quantity is described by a single
number, we called it a scalar quantity. In contrast, a
vector quantity has both a magnitude( the " how
much" or "how big" part) and a direction in space.
Calculations with scalar quantities use the
operations of ordinary arithmetic.
To understand more about vectors and how
they combine, we start with the simplest vector
quantity, displacement. Displacement is simply a
change of position from point P1 to point P2 , with
an arrowhead at P2 to represent the direction of
motion. Displacement is a vector quantity because
we must state not only how far the particle moves,
but also in what direction.
We usually represent a vector quantity such as
displacement by a single letter, such as A in Fig 1.
P2
P2
P3
A
A
A
P1
Fig 1
P1
Fig 2
B
Fig 3
When drawing any vector, we always draw a line
with an arrowhead at its tip. The length of the line
shows the vector' s magnitude, and the direction of
the line shows the vector' s direction. Displacement
is always a straight-line segment, directed from the
starting point to the end point, even though the
actual path of the particle may be curved. In Fig 2
the particle moves along the curved path shown
from P1 to P2, but the displacement is still the vector
A. Note that displacement is not related directly to
the total distance traveled. If the particle were to
continue on to P3 and then return to P1, the
displacement for the entire trip would be zero. If two
vectors have the same direction, they are parallel. If
they have the same magnitude and the same
direction, they are equal. The vector B in Fig 3,
however, is not equal to A because its direction is
opposite to that of A. We define the negative of a
vector as a vector having the same magnitude as the
original vector but the opposite direction. The
negative of vector quantity A is denoted as – A,
and we use a boldface minus sign to emphasize the
vector nature of the quantities. Between A and B of
Fig. 3 may be written as A = -B or B = -A. When two
vectors A and B have opposite directions, whether
their magnitudes are the same or not, we say that
they are anti-parallel. We usually represent the
magnitude of a vector quantity (its length in the case
of a displacement vector) by the same letter used for
the vector, but in light italic type with no arrow on
the top, rather than bold-face italic with an arrow
(which
is reserved for vectors). An alternative
notation is the vector symbol with vertical bars on
both sides.
Vector Addition
Now suppose a particle undergoes a displacement A,
followed by a second displacement B. The final
result is the same as if the particle had started at the


same initial point and undergone a single A  A
displacement C, as shown. We call displacement C
the vector sum, or resultant, of displacements A and
B. We express this relationship symbolically as
  
C  A B
Y
B
C
C
A
A
Ay

B
Fig 4
A
O
Fig 5
Ax
X
Fig 6
If we make the displacements A and B in reverse order,
with B first and A second, the result is the same (Fig.4).
The final result is the same as if the particle had started
at the same initial point and undergone a single
displacement C, as shown. We call displacement C the
vector sum, or resultant, of displacement A and B. We
express this relationship symbolical as C = A + B. If we
make the displacements A and B in reverse order,
with B first and A second, the result is the same. Thus
C = B + A and A + B = B + A
1-9 Components of Vectors
To define what we mean by the components of a
vector, we begin with a rectangular (Cartesian)
coordinate system of axes. The We then draw the
vector we' the re considering with its tail at O, the origin
of the coordinate system. We can represent any vector
lying in the xy-plane as the sum of a vector parallel to
the x-axis and a vector parallel to the y-axis. These two
vectors are labeled Ax and AY in the figure; they are
called the component vectors of vector A, and their
vector sum is equal to A. In symbols, A = Ax + Ay. (1) By
definition, each component vector lies along a
coordinate-axis direction.
Thus we need only a single number to describe each
one. When the component vector Ax points in the
positive x-direction, we define the number Ax to be equal
to the magnitude of Ax. When the component vector Ax
points in the negative x-direction, we define the number
Ax to be equal to the negative of that magnitude,
keeping in mind that the magnitude of two numbers Ax
and Ay are called the components of A. The components
Ax and Ay of a vector A are just numbers; they are not
vectors themselves. Using components
We can describe a vector completely by giving either its
magnitude and direction or its x- and y- components.
Equations (1) show how to find the components if we
know the magnitude and direction. We can also reverse
the process; we can find the magnitude and
direction if we know the components. We find that the
magnitude of a vector A is
A  Ax2  Ay2
(2)
where we always take the positive root. Equation (2) is
valid for any choice of x-axis and y-axis, as long as they
are mutually perpendicular. The expression for the
vector direction comes from the definition of the tangent
of an angle. If  is measured from the positive x-axis,
and a positive angle is measured toward the positive yaxis (as in Fig. 6) then tan  A
y
A
  arctan
A
A
x
y
and
We will always use the notation arctan
for the inverse tangent function.
x
1-10 Unit Vectors
A unit vector is a vector that has a magnitude of 1. Its
only purpose is to point, that is, to describe a direction in
space. Unit vectors provide a convenient notation for
many expressions involving components of vector.
In an x-y coordinate system we can define a unit
vector i that points in the direction of the positive x-axis
and a unit vector j that points in the direction of the
positive y-axis. Then we can express the relationship
between component vectors and components, described
at the beginning of section 1-9, as follows Ax = Ax i , Ay
= Ay j ; A = Ax i+ Ay j . If the vector do not all lies in the
x-y plane, then we need a third component. We duce a
third unit vector k that points in the direction of the
positive z-axis. The generalized forms of equation is A =
Ax i + Ay j + Az k
1-11 Products of vectors
We have seen how addition of vectors develops
naturally from the problem of combining displacements,
and we will use vector addition for many other vector
quantities later. We can also express many physical
relationships concisely by using products of vectors.
Vectors are not ordinary numbers, so ordinary
multiplication is not directly applicable to vectors. We will
define two different kinds of products of vectors.
Scalar product: The scalar product of two vectors A
and B is denoted by A · B. Because of this notation,
the scalar product is also called the dot product. We
define A ? B to be the magnitude of A multiplied by the
component of B parallel to A. Expressed as an equation:
 
 
A  B  AB cos   A B cos 
The scalar product is a scalar quantity, not a vector,
and it may be positive, negative, or zero. When Φ is
between 0° and 90 °, the scalar product is
positive. When  is between 90° and 180° , it is
negative.
Vector product : The vector product of two vectors A
and B, also called the cross product, is denoted by
AB. To define the vector product AB of two vectors
A and B, we again draw the two vectors with their
tail at the same point( Fig.1-20a). The two vectors
then lie in a plane. We define the vector product to
be a vector quantity with a direction perpendicular
to this plane (that is, perpendicular to both A and B)
and a magnitude equal to AB sin. That is, if C = AB,
then C = AB sin. We measure the angle 
from A toward B and take it to be the smaller of the two
possible angles, so  ranges from 0 °to 180. There are
always two directions perpendicular to a given plane,
one on each side of the plane. We choose which of
these is the direction of AB as follows. Imagine rotating
vector A about the perpendicular line until it is aligned
with B, choosing the smaller of the two possible angles
between A and B. Curl the fingers of your right hand
around the perpendicular line so that the fingertips point
in the direction of rotation; your thumb will then point in
the direction of AB. This right-hand rule is shown in Fig.
1-20a. The direction of the vector product is also the
direction in which a right-hand screw advances if turned
from A toward B.
2 Motion Along a Straight Line
2-1 Introduction
In this chapter we will study the simplest kind of
motion: a single particle moving along a straight line.
We will often use a particle as a model for a moving
along body when effects such as rotation or change
of shape are not important. To describe the motion
of a particle, we will introduce the physical
quantities velocity and acceleration.
2-2 Displacement, Time, And Average Velocity
Lets generalize the concept of average velocity. At
time t1 the dragster is at point P1 with coordinate x1,
and at time t2 it is at point P2, with coordinate x2. The
displacement of the dragster during the time
interval from t1 to t2 is the vector from P1 to P2,
the with x- component( the x2 – x1) and with y- and zcomponents equal to zero. The The x- component of
the dragster' the s displacement is just the change
in the coordinate x, which we write more compact
way as x  x2  x1 (2-1). Be sure you understand
that x is not the product of  and x; The it is a
single symbol that means" the change in the
quantity x. ” We likewise write the time interval from
t1 to t2
as t  t2  t1 .Note that x or t always means the
final value minus the
initial value, never the reverse. We can now define the x-component of average
velocity more precisely: it is the x- component of displacement, x , divided by
the time interval t during which the displacement occurs. The We represent
this quantity by the letter v with a subscript" av" to signify average value:
x  x x
(2-2)
v 

t t
t
For the example we had x1 = 19m, x2 = 277m, t1 = 1.0s and t2 = 4.0s so Eq.(22) gives
277m  19m 258m
v 

 86m / s
4.0m  1.0m 3.0s
The average velocity of the dragster is positive. This means that during the
time interval, the coordinate x increased and the dragster moved in the positive
x- direction. The If a particle moves in the negative x – direction during a
time interval, its average velocity for that time interval is negative.
2-3 Instantaneous Velocity
The average velocity of a particle during a time interval cannot tell us how fast,
or in what direction, the particle was moving at any given time during the
interval. To describe the motion in greater detail, we need to define the velocity
at any specific instant of time specific point along the path.
2
1
av
2
av
1
Such a velocity is called instantaneous velocity, and it needs to be defined
carefully. To find the instantaneous velocity of the dragster in Fig. 2-1 at the
point P1, we imagine moving the second point P2 closer and closer to the first
point P1. We compute the average velocity vav = x / t over these shorter and
shorter displacements and time intervals. Both x and t become very small,
but their ratio does not necessarily become small. In the language of calculus
the limit of x /t as t approaches zero is called the derivative of x with
respect to t and is written dx/dt. The instantaneous velocity is the limit of the
average velocity as the time interval approaches zero; it equals the
instantaneous rate of change of position with time. We use the symbol v, with
no subscript, for instantaneous velocity:
x
(straight-line motion)
(2-3).
v  lim
t
We always assume that the time interval ?t is positive so that v has the same
algebraic sign as ?x. If the positive x-axis points to the right, as in Fig. 2-1, a
positive value of mean that x is increasing and motion is toward the right; a
negative value of v means that x is decreasing and the motion is toward the left.
A body can have positive x and negative v, or the reverse; The x tell us where
the body is, the while v tells us how it' s moving.
t 0
Instantaneous velocity, like average velocity, is a vector quantity. Equation (2-3)
define its x-component, which can be positive or negative. In straight-line
motion, all other components of instantaneous velocity are zero, and in this
case we will often call v simply the instantaneous velocity. The The terms"
velocity" and" speed" are used the interchangeably in everyday language, but
they have distinct definitions in physics. We use the term speed to denote
distance traveled divided by time, on ether an average or an instantaneous
basis. Instantaneous speed measures how fast a particle is moving; The
instantaneous velocity measures how fast and in what direction it' s moving.
For example, a particle with instantaneous velocity v = 25m/s and a second
particle with v = - 25m/s are moving in opposite direction the same
instantaneous speed of 25m/s. Instantaneous speed is the magnitude of
instantaneous velocity, and so instantaneous speed can never be negative.
Average speed, however, is not the magnitude of average velocity.
Example: The A cheetah is crouched in ambush 20 m to the east of an
observer' s blind. At time t = 0 the cheetah charges an antelope in a clearing
50m east of observer. The cheetah runs along a straight line. Later analysis of
a videotape shows that during the first 2.0s of the attack,
The the cheetah' the s coordinate x varies with time according to the
equation x=20 ms+(5.0 ms/ s2) t2. (Note that the units for the numbers 20 and
5.0 must be as shown to make the expression dimensionally consistent.) Find(a)
the displacement of the cheetah during the interval between t1 = 1.0s and t2 =
2.0s.(b) Find the average velocity during the same time interval. (c) Find the
instantaneous velocity at time t1 = 1.0s by taking ?t = 0.1s, then  t = 0.01s,
then  t = 0.001s. (d) derive a general expression for the instantaneous velocity
as a function of time, and from it find v at t = 1.0s and t = 2.0s
Solution: (a) The At time t1= the 1.0 s the cheetah' the s position x1 is x1=20
ms+(5.0 ms/ s2)(1.0 ses)2=25 ms.
At time t2 = 2.0s its position x2 is x2 = 20m + (5.0m/s2)(2.0s)2 = 40m.
The displacement during this the interval is the  x= x2 – x1= 40 m – 25 m=15
m.
(b) The average velocity during this time interval is
x  x 40m  25m 15m
v 


 15m / s
t t
2.0s  1.0s 1.0s
At time t2, the position is x2 = 20m+(5.0m/s2)(1.1s)2 = 26.05m
2
1
av
2
1
The average velocity during this interval is
v 
av
26.05m  25
 10.5m / s
1.1s  1.0s
We invite you to follow this same pattern to work out the average velocities for
the 0.01s and 0.001s intervals. The results are 10.05m/s and 10.005m/s. As t
gets closer to 10.0m/s, so we conclude that the instantaneous velocity at time t
= 1.0s is 10.0m/s.
(d) We find the instantaneous velocity as a function of time by taking the
derivative of the expression for x with respect to t. For any n the derivative of t
is ntn-1, so the derivative of t2 is 2t. Therefore
dx
v
 5.0m / s
dt
2
2t   10tm / s At time t = 1.0s, v = 10m/s as we found in
2
part ( c ). At time t = 2.0s, v = 20m/s.
2-4 Average and Instantaneous Acceleration
When the velocity of a moving body changes with time, we say that the body
has an acceleration. Just as velocity describes the rate of change of position
with time, acceleration describes the rate of change of velocity with time. Like
velocity, acceleration is a vector quantity. In straight-line
Motion its only nonzero component is along the axis along which the motion
takes place.
Average Acceleration
Let' s consider again the motion of a particle along the x- axis. Suppose that at
time t1 the particle is at point P1 and has x-component of (instantaneous)
velocity v1, and at a later time t2 it is at point P2 and x-component of velocity v2.
So the x-component of velocity changes by an amount  v= v2 – v1 during the
time interval  t= t2 – t1.
We define the average acceleration aav of the particle as it moves from P1 to
P2 to be a vector quantity whose x-component is v, the change in the xcomponent of velocity, divided by the time interval  t:
v  v v
a 

t t
t
(average acceleration, straight-line motion) (2-4) For
straight-line motion we well usually call aav simply the average acceleration,
remembering that in fact it is the x-component of the average acceleration
vector.
If we express velocity in meters per second and time in seconds, then
average acceleration is in meters per second per second. The This is usually
written as m/ s2 and is read" meters per second squared."
2
1
av
2
1
Instantaneous Acceleration
We can now define instantaneous acceleration, following the same procedure
that we used to define instantaneous velocity. Consider this situation: A race
car driver has just entered the final straightaway at the Grand Prix. He reaches
point P1 at time t1, moving with velocity v1. He passes point P2, closer to the
line, at time t2 with velocity v2.(Fig. 2-8)
To define the instantaneous acceleration at point P1, we take the second
point P2 in Fig. 2-8 to be closer and closer to the first point P1 so that the
average acceleration is computed over shorter and shorter time intervals. The
instantaneous acceleration is the limit of the average acceleration as the time
interval approaches zero. In the language of calculus, instantaneous
acceleration equals the instan-taneous rate of change of velocity with time.
Thus
a  lim
t 0
v dv
(instantaneous acceleration, straight-line motion) (2-5)

t dt
Note that Eq. (2-5) is really the definition of the x-component of the acceleration
vector; in straight-line motion, all other components of this vector are zero.
Instantaneous acceleration plays an essential role in the laws of mechanics.
The From now on, the when we use the term" acceleration", the we will
always mean instantaneous acceleration, not average acceleration.
Example: Average and instantaneous accelerations Suppose the velocity v of
the car in Fig. 2-8 at any time t is given by the equation v = 60 m/s + (0.50 m/s3)
t2.
(a) Find the change in velocity of the car in the time interval between t1 = 1.0s
and t2 = 3.0s. (b) Find the instantaneous acceleration in this time interval. (c)
Find the instantaneous acceleration at time t1 = 1.0s by taking t to be first 0.1s,
then 0.01s, then 0.001s. (d) Derive an expression for the instantaneous
acceleration at any time, and use it to find the acceleration at t = 1.0s and t =
3.0s.
Solution: (a) We first find the velocity at each time by substituting each value of
t into the equation. At time t1 = 1.0s,
v1 = 60m/s +(0.50m/s3)(1.0s)2 = 60.5m/s.
At time t2 = 3.0s,
v2 = 60m/s + (0.5m/s3)(3.0s)2 = 64.5m/s.
The change in velocity v= v2 – v1=64.560.5=4.0 ms/ ses of –s.
The time interval is  t= 3.0 s – 1.0 s=2.0 s.
(b) The average acceleration during this time interval is
v  v 4.0m / s
a 

 2.0m / s
t t
2.0s
During the time interval from t1 = 1.0s to t2 = 3.0s, the velocity and average
acceleration have the same algebraic sign (in this case, positive), and the car
speeds up.
2
1
av
2
1
(c) When ?t = 0.1s, t2 = 1.1s and
v2 = 60m/s + (0.50m/s3)(1.1s)2 = 60.605 m/s,
v = 0.105m/s,
a 
av
v 0.105m / s

 1.05m / s
t
0.1s
2
We invite you repeat this pattern for  t = 0.01s and  t = 0.001s; the
results are aav = 1.0005m/s2 respectively. As  t gets smaller, the average
acceleration gets closer to 1.0m/s2. We conclude that the instantaneous
acceleration at t1 = 1.0s is 1.0m/s2.
(d) The instantaneous acceleration is a = dv/dt, the derivative of a
constant is zero, and the derivative of t2 is 2t . Using these, we obtain
dv d 60m / s  0.50m / s t 
a

dt
dt
 0.50m / s 2t   1.0m / s t.
3
3
3
When t = 3.0s,
a = (1.0m/s3)(3.0s) = 1.0m/s2
2
2-5 Motion with Constant Acceleration
The simplest acceleration motion is straight-line motion with constant
acceleration. In this case the velocity changes at the same rate throughout the
motion. This is a very special situation, yet one that occurs often in nature. As
we will discuss in the next section, a falling body has a constant acceleration if
the effects of the air are not important. The same is true for a body sliding on an
incline or along a rough horizontal surface. Straight-line motion with nearly
constant acceleration also occurs in technology, such as a jet-fighter being
catapulted from the deck of an aircraft carrier.
The In this section we' ll derive key equations for straight- line motion with
constant acceleration.
a
t=0
v
O
a
v
t=t
a
O
t=2t
v
a
v
a
O
at
a
t=3t
v
v0
O
a
t=4t
O
t
t
O
v
O
Fig. 2-12
v
v0
Fig. 2-13
Fig. 2-14
t
t
Figure 2-12 is a motion diagram showing the position, velocity, and acceleration
at five different times for a particle moving with constant acceleration. Figure 213 and 2-14 depict this same motion in the from of graphs. Since the
acceleration a is constant, the a-t graph (graph of acceleration versus time) in
Fig. 2-13 is a horizontal line. The graph of velocity versus time has a constant
slope because the acceleration is constant, and so the v-t graph is a straight
line ( Fig. 2-14).
The When the acceleration is constant, it' s easy to derive equations for
position x and velocity v as functions of time. Let' s start with velocity. In Eq. (24) we can replace the average acceleration aav by the constant (instantaneous)
acceleration a. We then have
v v
a
2
1
t 2  t1
(2-7)
Now we let t1 = 0 and let t2 be any arbitrary later time t. We use the symbol v0
for the velocity at the initial time t is v. Then Eq. (2-7) because
(2-8)
v  v0
a
t 0
v  v0  at
Next we want to derive an equation for the position x of a particle moving with
constant acceleration. To do this, we make use of two different expressions for
the average velocity vav during the interval from t = o to any later time t. The first
expression comes from the definition of vav Eq. (2-2), which holds true whether
or not the acceleration is constant. We call the position at time t = 0 the initial
position, denoted by x0. The position at the later time t is simply x. Thus for the
time interval t = t – 0 and the corresponding displacement X= x – x0, Eq. (2-2)
gives
vav
x  x0

t (2-9) We can also get a second expression for vav that is valid
only when the acceleration is constant, so that the v-t graph is a straight line (as
in Fig. 2-14) and the velocity changes at a constant rate.
vav 
v0  v
(2-10)(constant acceleration only). Substituting that expression
2
for v into Eq. (2-10), we find
1
v0  v0  at 
2
1
 v0  at
2
vav 
(2-11)
(constant acceleration only)
Finally, we equate Eqs. (2-9) and (2-11) and simplify the result:
x  x0
1
vav  at 
2
t
or
1 2
x  x0  v0t  at (2-12)
2
We can check whether Eqs.(2-8) and (2-12) are consistent with the assumption
of constant acceleration by taking the derivative of Eq. (2-12). We find
dx
v   v0  at
dt
which is Eq. (2-8). Differentiating again, we find simply
as we should expect.
dv
a
dt
The In many problems, it' the s useful to have a relationship between position,
velocity, and acceleration that does not involve the time. To obtain this, we first
solve Eq. (2-8) for t, then substitute the resulting expression into Eq. (2-12) and
simplify:
2 We transfer the term
v  v0
 v  v0  1  v  v0 
x0
x  x0  v0 
t
  a

a
2
a
a



 to the left side and
multiply through by 2a:
2a x  x0   2vv0  2v0 2  v 2  2v0 v  v0 2
Finally, simplifying gives us. We can get one more useful relationship by
equating the two expressions for vav, Eqs. (2-9) and (2-10), and
v 2  v0 2  2a x  x0 
multiplying through by t. Doing this, we obtain
 v0  v  (2-14)
x  x0  
t
 2 
A special case of motion with constant acceleration occurs when the
acceleration is zero. The velocity is then constant, and the equations of motion
become simply
v = v0 = constant, x = x0 + vt.
2-7 Velocity and Position by Integration
This optional section is intended for students who have already learned a little
integral calculus. In Section 2-5 we analyzed the special case of straight-line
motion with constant acceleration. When a is not constant, as is frequently the
case, the equations that we derived in that section are no longer valid. But even
when a varies with time, we can still use the relation v = dx/dt to find the velocity
v as a function of time if the position x is a known function of time. And we can
still use a = dv/dt to find the acceleration a as a function of time if the velocity v
is a known function of time.
In many physical situations, however, position and velocity are not known as
functions of time, while the acceleration is.
aav
Figure 2-13
O
t1
t2
t
We first consider a graphical approach, Figure 2-23 is a graph of acceleration
versus time for a body whose acceleration is not constant but increases with
time. We can divide the time interval between times t1 and t2 into many smaller
intervals, calling a typical one t. Let the average acceleration during t be aav.
From Eq. (2-4) the change in velocity v during t is v = aav t. Graphically, v
equals the area of the shaded strip with height aav and width t, that is, the area
under the curve between the left and right sides of t. The total velocity change
during any interval (say, t1 to t2) is the sum of the velocity changes v in the
small subintervals. So the total velocity changes is represented graphically by
the total area under the a-t curve between the vertical lines t1 and t2.
In the limit that all the T' the s become very small and their number very
large, the the value of aav for the interval from any time t to t+  t approaches
the instantaneous acceleration a at time t. In this limit, the area under the a-t
curve is the integral of a (which is in general a function of t) from t1 to t2. If v1 is
the velocity of the body at time t1 and v2 is the velocity at time t2, then
The change in velocity v is the integral of acceleration a with respect to time.
v  v   dv   adt
We can carry exactly the same procedure with the curve of velocity versus time
where v is in general a function of t. If x1 is a body' s position at time t1 and x2
is its position at time t2, from Eq. (2-2) the displacement x during a small time
interval ?t is equal to vav t, where vav is given by
x  x   dx   vdt
(2-16) . The change in position x – that is, the
displacement – is the time integral of velocity v. Graphically, the displacement
between times t1 and t2 is the area under the v-t curve between those two
times. If t1 = 0 and t2 is any later time t, and if x0 and v0 are the position and
velocity, respectively, at time t = 0, then we can rewrite Eqs. (2-15) and (2-16)
as follows:
x  x   vdt (2-18). Here x and v are the position
v  v   adt
(2-17)
and velocity at time t. If we know the acceleration a as a function of time and we
know the initial velocity v0, we can use Eq, (2-17) to find the velocity v at any
time; in other words, we can find v as a function of time. Once we know this
function, and given the initial position x0, we can use Eq. (2-18) to find the
position x at any time.
Example 2-9
Sally is driving along a straight highway in her classic 1965 Mustang. At time t =
0, when Sally is moving at 10 m/s in the position x-direction, she passes a
signpost at x = 50m. Her acceleration is a function of time: A= the 2.0 ms/ s2 –
1
2
1
v2
t2
v1
t1
2
x1
v1
x2
v2
t
t
0
0
0
3
0
(a) Derive expressions for her velocity and position as functions of time.(b) At
what time is her velocity greatest? (c) What is the maximum velocity? (d)
Where is the car when it reaches maximum velocity.
Solution: (a) The At time t=0, Sally' the s position is x0=50 ms, the and her
velocity is v0=10 ms/ s. Since we are given the acceleration a as a function of
time, we first use Eq. (2-17) to find the velocity v as a function of time t.
v  10m / s   2.0m / s  0.10m / s t dt
1
 10m / s  2.0m / s t  0.10m / s t
2
t
2
3
0
2
3
2
Then we use Eq.(2-18) to find x as a function of t:
1


x  50m   10m / s  (2.0m / s )t  0.10m / s t dt
2


1
1
 50m  10m / s t  2.0m / s t  0.10m / s t
2
6
At this instant, dv/dt = a = 0. Setting the expression for acceleration equal to
zero, we obtain:
2.0m / s
0  2.0m / s  0.10m / s t
t
 20s
0
.
10
m
/
s
?$
t
2
3
2
0
2
2
3
2
2
3
3
3
(c) We find the maximum velocity by substituting t = 20s (when velocity is
maximum) into the general velocity equation:
1


vmax  10 m / s  2.0m / s  20 s  2.0m / s 2 20 s 3  30 m
2
2
(d) The maximum value of v occurs at t = 20s, we obtain the position of the car
(that is, the value of x) at that time by substituting t = 20s into the general
expression for x:
1
1






x  50m  10m / s 20s  2.0m / s 20s  0.10m / s 20  517m
2
2
2
3
3
6
As before, we are concerned with describing motion, not with analyzing its
causes. But the language you learn here will be an essential tool in later
chapters when you use Newton' s laws of motion to study the relation between
force and motion.
Y
r
O
X
Z
Figure 3-1
3-2 Position and velocity vectors
To describe the motion of a particle in space, we first need to be able to
describe the position of the particle. Consider a particle that is at a point P at a
certain instant. The position vector r of the particle at this instant is a vector that
goes from the origin of the coordinate system to the point P(Fig. 3-1). The figure
also shows that the Cartesian coordinates x, y, and z of point P are the x-, y-,
and z-components of vector r. Using the unit vectors introduced in Section 1-10,
we can write

 

r  xi  yj  zk
We can also get this result by taking the derivative of Eq(3-1). The unit vectors
i, j and k are constant in magnitude and direction, so their derivatives are zero,
and we find
. This shows again that the

components of  dr dx  dy  dz 
v
dt

dt
i
dt
j
dt
k
v are dx/dt, dy/dt, and dz/dt. The magnitude of the instantaneous velocity
vector v- that is, the speed – is given in terms of the components vx, vy, and

vz by the Pythagorean relation:
2
2
2
v  v  vx  v y  vz
The instantaneous velocity vector is usually more interesting and useful than
the average velocity vector. From now on, when we use the word" velocity",
we will always mean the instantaneous velocity vector v( the rather than the
average velocity vector). Usually, we won' even bother to call v a vector.
v1
v
3-3 The Acceleration Vector
v2
In Fig (3-1), a particle is moving along a curved
The vectors v1 and v2 represent the particle' s
instantaneous velocities at time t1, when the particle is at point P1, and time t2,
When the particle is at point P2. The two velocities may differ in
both magnitude and direction. We define the average acceleration
aav of the particle as the particle as it moves from P1 and P2 as
the vector change in velocity,
v2-v1= v, divided by the time
interval t2-t1 = t:
 

v2  v1 v

aav 

t2  t1 t
Average acceleration is a vector quantity in the
same direction as the vector v. As in Chapter 2, we define the
instantaneous acceleration a at point P1as the limit approached by
the average acceleration when point P2 approaches point P1 and
v and t both approach zero; the instantaneous acceleration is
also equal to the instantaneous rate of change of velocity with
time. Because we are not restricted to straight-line motion,
instantaneous acceleration is now a vector:
v
1


v dv

a  lim

t 0 t
dt
v2
v
P2
v1
P1
v1
P1
A
a
v2
aav
B
C
The velocity vector v, as we have seen, is tangent to the path of the particle.
But the construction in fig.C shows that the instantaneous acceleration vector a
of a moving particle always points toward the concave side of a curved paththat is, toward the inside of any turn that the particle is making. We can also
see that when a particle is moving in a curved path, it always has nonzero
acceleration. We will usually be interested in the instantaneous acceleration,
not the average acceleration. From now on, we will use the term “acceleration”
to mean the instantaneous acceleration vector a.
Each component of the acceleration vector is the derivative of the
corresponding component of velocity:
 dvx  dvy  dvz 
dv
dv
a
i
j
k
z
y
dvx
a

a

x
ax 
dt
dt
dt
y
dt
dt
dt
Also, because each component of velocity is the derivative of the corresponding
coordinate, we can express the ax, ay and az of the acceleration vector a as
2
2
2
2
2
2


d
y
d
z
d
x
d
y
d
z

d x
ay  2
az  2
a 2 i  2 j 2 k
ax  2
dt
dt
dt
dt
dt
dt
Example: Calculating average and instantaneous acceleration; Let’s look again
at the radio-controlled model car in Example 3-1. We found that the
components of instantaneous velocity at any time t are


dx
vx 
  0.25 m 2 2t 
s
dt


dy
vy 
 1.0 m 2  0.025 m 3 3t 2 
s
s
dt
and that the velocity vector is


s

m
m
t j
 1.0
 0.075
s
s



v  vx i  v y j   0.50 m
2

ti
2
2
a) Find the components of the average acceleration in the interval from t=0.0s
to t=2.0s. b) Find the instantaneous acceleration at t=2.0s.
Solution a) From Eq. (3-8), in order to calculate the components of the average
acceleration, we need the instantaneous velocity at the beginning and the end
of the time interval. We find the components of instantaneous velocity at time
t=0.0s
by substituting this value into the above expressions for vx, and vy. We find that
at time t = 0.0s, vx = 0.0m/s, vy = 1.0m/s . We found in Example 3-1 that at t =
2.0s the values of these components are vx = -1.0m/s, vy = 1.3m/s.
Thus the components of average acceleration in this interval are
aav y
v y 1.3 m s  1.0 m s


 0.15 m 2
s
t
2.0s  0.0s


a   0.5 m



i  0.30 m  j
s
s
2
2
b) From Eq. 3-10 the components acceleration vector a as
m
m
vx  1.0 s  0.0 s
aav x 

 0.5 m 2
s
t
2.0s  0.0s
At time t = 2.0s, the components of instantaneous acceleration are
ax = -0.5m/s2 , ay = (0.15m/s3)(2.0s) = 0.30m/s2 .
The acceleration vector at this time is





m
m
a  a x i  a y j   0.50 2 i  0.15 2 j
s
s

 

v2
P1
P2
v1
v
s
P1
R
A
P2
v1


O
O
v2
B
3-5 Motion in A Circle
When a particle moves along a curved path, the direction of its velocity
changes.As we saw in Section 3-3, this means that the particle must have a
component of acceleration perpendicular to the path, even if its speed is
constant. In this section we’ll calculate the acceleration for the important special
case of motion in a circle.
Uniform Circular Motion
When a particle moves in a circle with constant speed, the motion is called
uniform circular motion. There is no component of acceleration parallel (tangent)
to the path; otherwise, the speed would change. The component of acceleration
perpendicular (normal) to the path, which cause the direction of the velocity to
change, is related in a simple way to the speed of the particle and the radius of
the circle.
In uniform circular motion the acceleration is perpendicular to the velocity at
each instant; as the direction of the velocity changes, the direction of the
acceleration also changes.
Figure A shows a particle moving with constant speed in a circular path
radius R with center at O. The particle moves from P1 to P2 in a time t.
The vector change in velocity v during this time is shown in Fig. B.
The angles labeled  in Fig. A and B are the same because v1 is
perpendicular to the line OP1 and v2 is perpendicular to the line OP2.
Hence the triangles OP1P2(Fig. A) and OP1P2(Fig. B) are similar. Ratios of
corresponding sides are equal, so

or
 v1
v s

v
 s

R
t
R
The magnitude aav of the average acceleration during t is therefore

v v1 s
aav 

t R t
The magnitude of the instantaneous acceleration a at point P1. Also, P1
can is the limit of this expression as we take point P2 closer and closer to
point P1:
v s v1
s
a  lim 1
 lim
t 0
R t R t 0 t
But the limit of s/ t is the speed v1 at point P1. Also, P1 can be any point on
the path, so we can drop the subscript and let v represent the speed at any
point. Then
v2
arad 
R
Because the speed is constant, the acceleration is always perpendicular to the
instantaneous velocity.We conclude: In uniform circular motion, the magnitude
a of the instantaneous acceleration is equal to the square v divided by the
radius R of the circle. Its direction is perpendicular to v and inward along the
radius. Because the acceleration is always directed toward the center of the
circle, it is sometimes called centripetal acceleration.
Non-Uniform Circular Motion
We have assumed calculate throughout this section that the particle’s speed is
constant. If the speed varies, we call the motion non-uniform circular motion. In
non-uniform circular motion, still gives the radial component of acceleration
arad
v2

R
, which is always perpendicular to the instantaneous
velocity and direction toward the center of the circle. But
since the speed v has different values at different points
in the motion, the value of a rad is not constant.
In non-uniform circular motion there is also a component of acceleration that is
parallel to the instantaneous velocity. This is the component a11 that we
discussed in Section 3-3; here we call this component atan to emphasize that it
is tangent to the circle. From the discussion at the end of Section 3-3 we see
that the tangential component of acceleration atan is equal to the rate of change
of speed. Thus

2
d
v
v
atan 
and
arad 
dt
R
The vector acceleration of a particle moving in a circle with varying speed is the
vector sum of the radial and tangential components of accelerations. The
tangential component is in the same direction as the velocity if the particle is
speeding up, and is in the opposite direction if the particle is slowing down.
In uniform circular motion there is no tangential component of acceleration, but
the radial component is the magnitude of dv/dt. We have mentioned before that
the two quantities |dv/dt| and d|v|/dt are in general not equal. In uniform circular
motion the first is constant and equal to v2/R; the second is zero.
4 Newton’s Laws of Motion
4-1 Introduction
In this chapter we will use the kinematic quantities displacement, velocity, and
acceleration along with two new concepts, force and mass, to analyze the
principles of dynamics. These principles can be wrapped up in a neat package
of three statements called Newton’s laws of motion. The first law states that
when the net force on a body is zero, its motion doesn’t change. The second
law relates force to acceleration when the net forces is not zero. The third law is
a relation between the forces that two interacting bodies exert on each other.
These laws, based on experimental studies of moving bodies, are fundamental
in two ways. First, they cannot be deduced or proved from other principles.
Second, they are the foundation of classical mechanics ( also called Newtonian
mechanics). Newton’s laws are not universal, however; they require
modification at very high speeds (near the speed of light) and for very small
size ( such as within the atom).
4-2 Force And Interactions
In everyday language, a force is a push or pull. The concept of force gives us a
quantitative description between two bodies or between a body and its
environment.
When a force involves direct contact between two bodies, we call it a contact
force.
Contact forces include the pushes or pulls you exert with your hand, the force of
rope pulling on a block to which it is tied, and the friction force that the ground
exerts on a ball player sliding into home. There are also forces, called longrange forces, that act even when the bodies are separated by empty space.
You’ve experienced long-range forces if you’ve ever played with a pair of
magnets. Gravity, too, is a long-range force; the sun exerts a gravitational pull
on the earth, even over a distance of 150 million kilometers, that keeps the
earth in orbit.
Force is a vector quantity; you can push or pull a body in different directions.
Thus to describe a force, we need to describe the direction in which it acts as
well as its magnitude, the quantity that describes “how much” or “how hard” the
force pushes or pulls.
When two forces F1 and F2 act at the same time at point A of a body (Fig. 4-2),
experiment shows that the effect on the body’s motion is the same as the effect
of a single force R equal to the vector sum of the original forces: R = F1 + F2.
F2
Fy
F
R
A
O
F1
Fx
Fig.4-2
Fig. 4-3
More generally, the effect of any number of forces applied at a point on a body
is the same as the effect of a single force equal to the vector sum of the forces.
This important principle goes by the superposition of forces.
The experimental discovery that forces combine according to vector addition is
of the utmost importance. We will use this fact many times throughout our study
of physics . It allows us to replace a force by its component vectors, as we did
with displacements in Section 1-9. For example, in Fig. 4-3, force F acts on a
body at point O. The component vectors of F in the directions OX and OY are
Fx and Fy. When Fx and Fy are applied simultaneously, the effect is exactly the
same as the effect of the original force F. Any force can be replaced by its
component vectors, acting at the same point.
4-3 Newton’s First Law
A body acted on by no net force moves with constant velocity (which may be
zero) and zero acceleration. This is Newton’s first law of motion.
It’s important to note that the net force is what matters in Newton’s first law. For
example, a physics book at rest on a horizontal table top has two forces acting
on it; the downward force of the earth’s gravitational attraction ( a long-range
force that acts even if the table top is elevated above the ground) and an
upward supporting force exerted by the table top ( a contact force). The upward
push of the surface is just as great as the downward pull of gravity, so the net
force acting on the book (that is, the vector sum of the two forces) is zero.
In agreement with Newton’s first law, if the book is at rest on the table top, it
remains at rest. We find that if the body is at rest at the start, it remains at rest;
if it is initially moving, it continues to move in the same direction with constant
speed. These results show that in Newton’s first law, zero net force is
equivalent to no force at all. This is just the principle of superposition of forces
that we saw in Section 4-2.
4-4 Newton’s Second Law
Mass and Force
Mass is a quantitative measure of inertia, which we discussed in Section 4-3.
The greater its mass, the more a body “resists” being accelerated. The SI unit
of mass is kilogram. One Newton is the amount of net force that gives an
acceleration of one meter per second squared to a body with a mass of one
kilogram. We can use this definition to calibrate the spring balances and other
instruments used to measure forces. Because of the way we have defined the

newton, it is related to the units of mass, length, and time. For Eq.  F  ma (45) to be dimensionally consistent, it must be true that 1newton = (1 kilogram) (1
meter per second squared).
We can also use Eq. (4-5) to compare a mass with the standard mass and thus
to measure masses. Suppose we apply a constant net force F to a body having
a known mass m1 and we find an acceleration of magnitude a1. We then apply
the
same force to another body having an unknown mass m2, and we find an
acceleration of magnitude a2. Then, according to Eq. (4-5)
m2 a1

m1 a2
m1a1  m2 a2
(4-6)
The ratio of the masses is the inverse of the ratio of the accelerations. Figure 412 shows the inverse proportionally between mass and acceleration. In
principle we could use Eq. (4-6) to measure an unknown mass m2, but it is
usually easier to determine mass indirectly by measuring the body’s weight.
When two bodies with masses m1 and m2 are fastened together, we find that
the mass of the composite body is always m1 + m2 (Fig. 4-12). This additive
property of mass may seem obvious, but it has to be verified experimentally.
a3
a1
a2
F
m1
F

F
m1+m2
m2
Fig. 4-6
Newton’s Second Law
We’ve been careful to state that net force on a body is what causes that
body to accelerate. Experiment shows that if a combination of forces F1,
F2, F3, ····is applied to as body, the body will have the same acceleration
(magnitude and direction) as when only a single force is applied, if that
single force is equal to the vector sum F1+F2+F3+ ···· In other words, the
principle of superposition of forces also holds true when the net force is
not zero and the body is accelerating.
Equation (4-5) relates the magnitude of the net force on a body to the
magnitude of the acceleration that it produces. We have also seen that
the direction of the net force is the same as the direction of the
acceleration, whether the body’s path is straight or curved. Newton
wrapped up all these relationships and experimental results in a single
concise statement that we now call Newton’s second law of motion:
If a net external force acts on a body, the body accelerates. The direction
of acceleration is the same as the direction of the net force. The net force
vector is equal to the mass of the body times the acceleration of the body.

In symbols,

 F  ma
(4-7)
There are at least four aspects of Newton’s second law that deserve special
atten-tion. First, Eq. (4-7) is a vector equation. Usually, we will use it in
component from, with a separate equation for each component of force and the
corresponding acceleration
 Fx  max
 Fy  ma y
 Fz  ma z
(4-8)
This set of component equations is equivalent to the single vector equation (47). Each component of total force equals the mass times the corresponding
component of acceleration.
Second, the statement of Newton’s second law refers to external forces. By this
we mean forces exerted on the body by other bodies in its environment. It’s
impossible for a body to affect its own motion by exerting a force on itself; if it
was possible, you could lift yourself to the ceiling by pulling up on your belt!
That’s why only external forces are included in the sum F in Eqs. (4-7) and (48).
Third, Eqs. (4-7) and (4-8) are valid only when the mass m is constant. It’s easy
to think of systems whose masses change, such as a leaking tank truck, a
rocket ship, or a moving railroad car being loaded with coal. But such systems
are better handled by using the concept of momentum; we’ll get to that in
Chapter 8.
Finally, Newton’s second law is valid only in inertial frames of reference, just
like the first law. Thus it is not valid in the reference frame of any of the
Vehicles in Fig. 4-8; relative to any of these frames, the passenger accelerates
even though the net force on the passenger is zero. We will usually assume
that the earth is an adequate approximation to an inertial frame, although
because of its rotation and orbital motion it is not precisely inertial.
4-5 Mass and Weight
The weight of a body is a familiar force. It is the force of the earth’s gravitational
attraction for the body. We will study gravitational interactions in detail in
chapter 12, but we need some preliminary discussion now. The terms mass
and weight are often misused and interchanged in everyday conversation. It is
absolutely essential for you to understand clearly the distinction between these
two physical quantities.
Mass characterizes the inertial properties of a body. The weight of a body is a
force, a vector quantity.
4-6 Newton’s Third Law
A force acting on a body is always the result of its interaction with another body,
so forces always come in pairs. In each of these cases the force that you exert
on the other body is in the opposite direction to the force that body exert on you.
Experiments show that whenever two bodies interact, the two forces that they
exert on the other are always equal in magnitude and opposite in direction. This
fact is called Newton’s third law of motion. Expressed in words, If body A exerts
a force on body B (an “action”), then body B exerts a force on body A (a
“reaction”).
These two forces have the same magnitude but are opposite in direction. These
two forces act on different bodies.


(4-11)
FAonB   FBonA
In this statement, “action” and “reaction” are the two opposite forces; we
some-times refer to them as an action-reaction pair. This is not meant to imply
any cause-and-effect relationship; we can consider either force as the “action”
and the other as the “reaction”. We often say simply that the forces are “equal
and opposite,” meaning that they have equal magnitudes and opposite direction.
We stress that the two forces described in Newton’s third law act on different
bodies. This is important in problems involving Newton’s first or second law,
which involve the forces that act on a body.
4-7 Using Newton’s Laws
Newton’s three laws of motion contain all the basic principles we need to solve
a wide variety of problems in mechanics. These laws are very simple in form,
but the process of applying them to specific situations can pose real challenges.
This section introduces some useful techniques.

When you use Newton’s first law,
 F  0 , for an equilibrium situation, or
Newton’s second law,  F  ma , for a non-equilibrium situation, you must
apply it to some specific body. It is absolutely essential to decided at the
beginning
What body you talking about. This may sound trivial, but it isn’t. Once you
have chosen a body, then you have to identify all the forces acting on it.
Don’t get confused between the forces acting on a body and the forces

exerted by that body on some other body you’ve chosen go into  F .
To help identify the relevant forces, draw a free-body diagram. What’s
that? It is a diagram showing the chosen body by itself, “free” of its
surrounding, with vectors down to show the magnitudes and directions of
all the forces applied to the body by the various other bodies that interact
with it. Be careful to include all the forces acting on the body, but be
equally careful not to include any forces that the body exerts on any other
body. In particular, the two forces in an action-reaction pair must never
appear in the same free-body diagram because they never act on the
same body. Furthermore, forces that a body exerts on itself are never
include, since these can’t affect on the body’s motion.
The cutting-blade assembly on a radial-arm saw has a mass of 5.0 kg. It is
pulled along a pair of frictionless horizontal rails aligned with the x-axis by
a force Fx. The position of the blade assembly as a function of time is
x = (0.18m/s2)t2 – (0.030m/s3)t3 .
Find the net force acting on the blade assembly as a function of time.
What is the force at time t = 3.0s ? For what times is the force positive?
Negative? Zero?
Solution Figure 4-24b shows the free-body diagram and coordinate axes. The
forces are the horizontal force Fx, the weight w, and the upward force n that the
rail exert to support the cutting head. There is no acceleration in the vertical
direction, so the sum of the vertical components of forces must be zero. The net
forces has only a horizontal component, equal to Fx; from Newton’s second law,
this is equal to max. To determine Fx, we first find the acceleration ax by taking
the second derivative of x. The second derivative of t2 is 2, and the second
derivative of t3 is 6t, so


d 2x
ax  2  0.36 m 2  0.18 m 3 t
s
s
dt
then, from Newton’s second law,
Fx = max = (5.0kg)[(0.36m/s3 – (0.18m/s3)]t = 1.8N – (0.90N/s)t
y
ax(m/s2)
Fx(N)
Fx
x
w
Figure 4-24(b)
2.00
0.40
1.00
0.20
0
t(s)
1
2
3 4 5
0
-1.00
-0.20
-2.00
-0.40
Figure 4-25(a)
t(s)
1
2
3 4 5
Figure 4-25(b)
At time t = 3.0s the force is 1.80N – (0.90 N/s)(3.0s) = -0.90N. The force is
zero when ax = 0, that is, when 0.36m/s2 – (0.18m/s3)t = 0. This happens
when t = 2.0s. When t  2.0s, Fx is positive, and the blade assembly is
being pulled to the right. When t > 2.0s, Fx is negative, and the horizontal
force on the assembly is to the left. Figure 4-25 shows graphs of Fx and ax
as functions of time. You shouldn’t be surprised that the net force and
acceleration are directly proportional; according to Newton’s second law,
this is always true.
5 Applications of Newton’s Laws
5-1 Introduction
Newton’ three laws of motion, the foundation of classic mechanics, can be
stated very simply, as we have seen.
5-2 Using Newton’s First Particles In Equilibrium
We learned in Chapter 4 that a body is in equilibrium when it is at rest or
moving with constant velocity in an inertial frame of reference. When a
particle is at rest or is moving with constant velocity in an inertial frame of
reference, the net force acting on it---must be zero.
(5-1)

F  0
We will
form:
 usually use

 this in component
(5-2)
 Fx  0
 Fz  0
 Fy  0
This section is about using Newton’s first law to solve problems dealing
with bodies in equilibrium.
5-3 Using Newton’s Second Law: Dynamics of Particles
We are now ready to discuss dynamics problems, in which we apply
Newton’s second law to bodies that are acceleration and hence are not in
equilibrium. In this care force on the body is not zero, but is equal to the mass
of the body times its acceleration :

F

m
a

(5-3)
We will use this relation in component form:
 F  max
(5-4)
 Fz  ma z
 Fy  ma y
5-4 Frictional Forces
we have seen several problems where a body rests or slides on a surface that
exerts forces on the body, and we have used the terms normal force and
friction force to describe these forces. Whenever two bodies interaction forces
constant (touching) of their surfaces, we call the interaction forces contact
forces. Normal and friction forces are both contact forces.
Kinetic and Static Friction
First, when a body rests or slides on a surface, we can always represent the
contact force exerted by the surface on the body in terms of components of
force perpendicular and parallel to the surface. We call the perpendicular
component vector the normal force, denoted by n .(Recall that normal is a
synonym for perpendicular) component vector parallel to the surface is the
friction force, denoted by f . By definition , n and f are always perpendicular to
each other.
The direction of the friction force is always such as to oppose relative motion of
the two surfaces.
The kind of friction that acts when a body slides over a surface is called a
kinetic fraction force fk, The adjective “kinetic” and the subscript “k” remind us
that the two surfaces are moving relative to each other. The magnitude of the
kinetic friction force usually increases when the normal force increases. In
many cases the magnitude of the kinetic friction force fk is found experimentally
to be approximately proportional to the magnitude n of the normal force. In such
case we can write
f k  k n
(5-5)
where µk (pronounced “mu-sub-k”) is a constant called the coefficient of kinetic
friction. The more slippery the surface, the smaller the coefficient of friction.
Because it is a quotient of two force magnitudes, µk is a pure number without
units.
Friction forces may also act when there is no relative motion. If you try to
slide that box of books across the floor, the box may not move at all because
the floor exerts an equal and opposite friction force on the box. This is called a
static friction force fs. For a given pair of surfaces the maximum value, called
(fs)max, is approximately proportional to n; we call the proportionality factor µs
(pronounced “mu-sub-s”) the coefficient of static friction. Some representative
values of µs are shown in Table 5-1. In a particular situation, the actual force of
static friction can
have any magnitude between zero (when there is no other force parallel to the
surface) and a maximum value given by µsn. In symbols
f s  sn
(5-6)
Like Eq. (5-5), this is a relation between magnitudes, not a vector relation.
Fluid Resistance and Terminal Speed
The direction of the fluid resistance force acting on a body is always opposite
the direction of the body’s velocity relative to the fluid. The magnitude of the
fluid resistance force usually increases with the speed of the body through the
fluid. Contrast this behavior with that of the kinetic friction force between two
surfaces in contact, which we can usually regard as independent of speed. For
low speeds, the magnitude f of the resisting force of the fluid is approximately
proportional to the body’s speed v:
(5-7)
f  kv
Where k is a proportionality constant that depends on the shape and size of the
body and the properties of the fluid. In motion through air at the speed of a
tossed tennis ball or faster, the resisting force is approximately proportional to
v2 rather than to v. It is then called air drag or simply drag. In this case we
replace Eq.(5-7)
f  Dv2
(5-8)
Because of the v2 dependence, air drag increases rapidly with increasing
speed. The air drag on a typical car is negligible at low speeds but
comparable to or greater the force of rolling resistance at highway speeds.
The value of D depends on the shape and size of the body and on the
density of the air. Because of the effects of fluid resistance, an object
falling in a fluid will not have a constant acceleration. To describe its
motion, we have to start over, using Newton’s second law.
The free-body diagram is shown in Fig.5-22. We take the positive
direction to be downward and neglect any force associated with buoyancy
in the water. There are no x-components, and Newton’s second law gives
 Fy  mg  (kv)  ma
f
v
a
y
vt
g
W=mg
O
y
Fig.5-22
t
t
O
Fig.5-23
O
t
When the rock first starts to move, v = 0, the resisting force is zero, and the
initial acceleration is a = g. As its speed increases, until finally it is equal in
magnitude to the weight. At this time, mg – kv = 0, the acceleration
becomes zero, and there is no further increase in speed. The final speed vt,
called the terminal speed, is given by mg – kvt = 0, or
mg
vt 
k
Figure 5-23 shows how the acceleration, velocity, and position vary with
time. As time goes by, the acceleration approaches zero, and the velocity
approaches vt. The slope of the graph of y versus t becomes constant as
the velocity becomes constant. To see how the graphs in Fig. 5-23 are
derived, we must find the relation between speed and time during the
interval before the terminal speed is reached. We go back to Newton’s
second law, which we rewrite as
dv
m  mg  kv
dt
After rearranging terms and replacing mg/k by vt, we integrate both sides,
noting that v = 0 when t = 0:
k t
v dv
  0 dt
0
v  vt
m
Which integrate to
ln
and finally
vt  v
k
 t
vt
m
or
v
1   e   k / m t
vt
v  vt 1  e  k / m t 
Note that v becomes equal to the terminal speed vt only in the limit that
t; the rock cannot attain terminal speed in any finite length of time.
The derivative of v gives a as a function of time, and the integral of v
gives y as a function of time. We leave the derivations for you to complete,
the results are
a  ge
 m t
k


 k t 
 m
y  vt t  1  e m 
 k

In deriving the terminal speed in Eq. 5-23, we assumed that the fluid
resistance force was proportional to the speed. For an object falling through
the air at high speeds, so that the fluid resistance is proportional to v2 as in
Eq. (5-8), we invite you to show that the terminal speed vt is given by
vt 
mg
D
This expression for terminal speed explains the observation that heavy
objects in air tend to fall faster than light objects.
5-5 Dynamics of Circular Motion
We talked about uniform circular motion in Section 3-5. We showed that
when a particle moves in a circular path with constant speed, the particle’s
acceleration is always directed toward the center of the circle. The
magnitude arad of the acceleration is constant and is given in terms of the
speed v and the radius R of the circle by
v2
arad 
R
The magnitude of the radial acceleration is given by arad = v2/R, so the
magnitude of the net inward radial force Fnet on a particle with mass m must be
Fnet  marad
v2
m
R
5-6 The Fundamental Forces of Nature
Our current understanding is that all forces are expressions of just four distinct
classes of fundamental forces, or interactions between particles. Two are
familiar in everyday experience. The other two involve interactions between
subatomic particles that we cannot observe with the unaided senses.
Of the two familiar classes, gravitational interactions were the first to be
studied in detail. The weight of a body results from the earth’s gravitational
attraction acting on it.
The second familiar class of forces, electromagnetic interaction, includes
electric and magnetic forces. All atoms contain positive and negative electric
charge, so atoms and molecules can exert electric forces on each other.
Contact forces, in-cluding the normal force, friction, and fluid resistance, are the
combination of all such forces exerted on the atoms of a body by atoms in its
surroundings.
These two interactions differ enormously in their strength.
The electrical repul-sion between two protons at a given
distance is stronger than their gravitational attraction by
a factor of the order of 1035. Gravitational forces play no
significant role in atomic or molecular structure. But in
bodies of astronomical size, positive charge and
negative charge are usually present in nearly equal
amounts, and the resulting electrical interactions nearly
cancel each other out. Gravitational inter-actions are
thus the dominant influence in the motion of planets and
in the internal structure of stars.
The other two interactions are less familiar. One, the
strong interaction, is responsible for holding the nucleus
of an atom together. Nucleus contain
electrically neutral neutrons and positively charged protons.
The charged protons repel each other, and a nucleus could
not be stable if it were not for the presence of an attrac-tive
forces of a different kind that counteracts the repulsive
electrical interactions. In this context the strong interaction is
also called the nuclear force. It has much shorter range than
electrical interactions, but within its range it is much stronger.
Finally, there is the weak interaction. It plays no direct role
in the behavior of ordinary matter, but it is of vital important
interactions among fundamental particles. The weak
interaction is responsible for a common form of radioactivity
beta-decay, in which a neutron in a radioactive nucleus is
transformed in to a proton while ejecting an electron and an
essentially massless particle called an antineutrino.
During the past several decades a unified theory of the
electromagnetic and weak interactions has been developed.
6 Work and Kinetic Energy
6-1 Introduction
The new method that we’re about to introduce uses the ideas of work and energy. The applications of these ideas go far beyond mechanics, however. The importance of the energy idea stems from the principle of conservation of energy:
Energy is a quantity that can be converted from one form to another but
cannot be created or destroyed. We’ll use the energy idea throughout the
rest of this book to study a tremendous range of physical phenomena.
In this chapter, through, our concentration will be on mechanics. We’ll learn
about one important form of energy called kinetic energy, or energy of motion,
and how it relates to the concept of work.
6-2 Work
In physics, work has a much more precise definition. By making use of the definition we’ll find that in any motion, no matter how complicated, the total work
done on a particle by all forces that act on it equals the change in its kinetic
energy—a quantity that’s related to the particle’s speed.
The physics’s definition of work is based on these observations. Consider a
body that undergoes a displacement of magnitude along a straight line. (For
now, we will assume that any body we discuss can be treated as a particle so
that we
can ignore any rotation or changes in shape of the body) While the body moves,
a constant force with magnitude F acts on it in the same direction as
displacement s. We define the work W done by a constant force F acting on the
body under these conditions as
W = Fs
(6-1)
The work done on the body is greater if either the force F or the displacement s
is greater.
When the force F and the displacement s have different directions, we take
the component of F in the direction of the displacement s, and we define the
work as the product of this component and magnitude of the displacement. The
component of F in the direction of s is F cos, so
W = Fscos
(6-2)
we are assuming that F and  are constant during the displacement. If  = 0, so
that F and s are in the same direction, then cos  = 1 and we are back to Eq. 61.
Equation (6-2) has the form of the scalar product of two vectors. We can
write Eq. (6-2) more compactly as
W=F•s
(6-3)
It’s important to understand that work is a scalar quantity, even through it’s
calculated by using two vector quantities (force and displacement). A 5-N force
toward the east acting on a body that moves 6 m to the east does exactly the
same work as a 5-N force toward the north acting on a body that moves 6 m to
It’s also important to realize that work can be positive, negative, or zero. This is
the essential way in which work as defined in physics differs from the “everyday”
de-finition of work. When the force has a component in the same direction as
the dis-placement ( between zero and 90), cos  in Eq. (6-2) is positive and
the work W is positive. When the force has a component opposite to the
displacement ( bet-ween 90° and 180°, cos  is negative and the work is
negative. When the force is perpendicular to the displacement,  = 90° and
the work done by the force is zero.
F
F
F
F
v
v
n
v
v
F
F
ф
w
Fig.6-6 (a)
n
n
n
F
F
w
w
(b)
(c)
w
(d)
6-3 Work and Kinetic Energy
The total work done on a body by external forces is related to the body’s
displace-ment, that is, to changes in its position. But the total work is also
related to changes in the speed of the body. To see this, consider Fig. 6-6,
which shows several examples of a block on a frictionless table. The forces
acting on the block are its weight w, the normal force n, and the force F exerted
on it by the hand.
In Fig. 6-6a the net force on the block is in the direction of its motion. From
Newton’s second law, this means that the block speeds up; from Eq. (6-1), this
also means that the total work Wtot done on the block is positive. The total work
is also positive in Fig.6-6b, but only the component F cos contributes to Wtot.
The block again speeds up, and this same component F cos  is what causes
the acceleration. The total work negative in Fig.6-6c because the net force
opposes the displacement; in this case the block slows down. The net force is
zero in Fig. 6-6d, so the speed of the block stays the same and the total work
done on the block is zero. We can conclude that when a particle undergoes a
displacement, it speeds up if Wtot>0, slows down if Wtot<0, and maintains the
same speed if Wtot = 0.
Let’s make these observations more quantitative. Consider a particle with mass
m moving along the x-axis under the action of a constant net force with
magnitude F directed along the positive x-axis (Fig. 6-1). The particle’s
acceleration is constant and given by Newton’s second law, F = ma. Suppose
the speed changes from v1 to v2 while the particle undergoes a displacement s
= x2 – x1 from point x1 to x2. Using a constant-acceleration equation, Eq. (2-13),
and replacing v0 by v1, v by v2, and (x – x0) by s, we have
v22  v12  2as
v22  v12
a
2s
When we multiply this equation by m and equate ma to the net force F, we find
and
v22  v12
F  ma  m
2s
1 2 1 2
Fs  mv2  mv1
2
2
(6-4)
The product Fs is the work done by the net force F and thus is equal to the
total work Wtot done by all the forces acting on the particle. The quantity ½ mv2
is called the kinetic energy K of the particle.
1 2
k  mv
2
(6-5)
Like work, the kinetic energy of a particle is a scalar quantity; it depends only on
the particle’s mass and speed, not its direction of motion. We can interpret Eq.
6-4 in terms of work and kinetic energy. The first term on the right side of Eq.
(6-4) is K2 = ½ mv22 , the final kinetic energy of the particle (that is, after the
displace-ment). The second term is the initial kinetic energy, K1 = ½ mv12, and
difference between these terms is the change in kinetic energy. So Eq. (6-4)
says that the work done by the net force on a particle equals the change in
the particle’s kinetic energy:
Wtot = K2 – K1 = K
This result is the work-energy theorem.
Because we used Newton’s laws in deriving the work-energy theorem, we can
use it only in an inertial frame of reference. The speeds that we use to compute
the kinetic energies and the distances that we use to compute work must be
measured in an inertial frame. Note also that the work-energy theorem is valid
in any inertial frame, but the values of Wtot and K2 – K1 may differ from one
inertial frame to another (because the displacement and speed of a body may
be diffe-rent in different frame).
6-4 Work and Energy With Varying Forces
So far in this chapter we’ve considered work done by constant forces only. But
what happens when you stretch a spring? The more you stretch it, the harder
you
to pull, so the force you exert is not constant as the spring is stretched. We’ve
also restricted our discussion to straight-line motion. You can think of many
situations in which a force that varies in magnitude, direction or both acts on a
body moving along a curved path. We need to be able to compute the work
done by the force in these more general cases. Fortunately, we’ll find that the
work-energy theorem holds true even when varying forces are considered and
when the body’s path is not straight.
To add only one complication at a time, let’s consider straight-line motion
with a force that is directed along the line but with an x-component F that may
change as the body moves. For example, imagine a train moving on a straight
track with the engineer constantly changing the locomotive’s throttle setting or
applying the brakes. Suppose a particle moves along the x-axis from point x1 to
x2. F
F
x
x
Fx
Fa
O
x1
x2
x2-x1
x
O
Fb
Fc
Fd
Δxa
Fe Ff
F
Δxf
x2 – x1
Fig. 6-1(a)
Fig. 6-11(b)
X
O
x1
x2
x2-x1
Fig. 6-12
x
Figure 6-11(a) is a graph of the x-component of force as a function of the particle’s coordinate x. To find the work done by this force, we divide the total displacement into small segments xa, xb and so on average force Fa in that segment multiplied by the displacement xa. We do this for each segment and then
add the results for all the segments. The work done by the force in the total
displacement from x1 to x2 is approximately
W = Fa xa + Fb xb + ….
As the number of segments becomes very large and the width of each
becomes very small, this sum becomes (in the limit) the integral of F from x1 to
x2 :
w  xx Fdx
2
1
(6-7)
Note that Fa xa represents the area of the first vertical strip in Fig. 6-11(b) and
that the integral in Eq. (6-7) represents the area under the curve of Fig. 6-11(a)
between x1 and x2. On a graph of force as a function of position, the total work
done by the force is represented by the area under the curve between the initial
and final positions. An alternative interpretation of Eq. (6-7) is that the work W
equals the average force that acts over the entire displacement, multiplied by
the displacement.
Equation (6-7) also applies if F, the x-component of the force, is constant.
In that case, F may be taken outside the integral: W  x Fdx  F x dx  F  x  x 
x
x
2
1
2
1
2
1
but x2 – x1 = s, the total displacement of the particle. So in the case of a
constant force F, Eq. (6-7) says that W = Fs, in agreement with Eq. (6-1). The
interpret-ation of work as the area under the curve of F as a function of x also
holds for a constant force; W = Fs is the area of a rectangle of height F and
width s.
Now let’s apply what we’ve learned to the stretched spring. If the elongation x
is not too great, we find that F is directly proportional to x:
F = kx
(6-8)
where k is a constant called the force constant (or spring constant) of the spring.
Equation (6-8) shows that the units of k are force divided by distance, N/m in SI
units. The observation that elongation is directly proportional to force for
elongations that are not too great was made by Robert Hooke in 1678 and is
known as Hooke’s law. It really shouldn’t be called a “law”, since it’s a
statement about a specific device and not a fundamental law of nature.
To stretch a spring, we must do work. We apply equal and opposite forces to
ends of the spring and gradually increase the forces. The force at the moving
end does do work. The work done by F when the elongation goes from zero to
a maximum value X is
F
F=kx
1 2
x
x
w  0 Fdx  0 kxdx  kx
2
kx
O
x
x
Fig. 6-4
We can also obtain this result graphically. The area of the shaded
triangle in Fig.6-14, representing the total work done by the force, is
equal to half the pro-duct of the base and altitude, or
W
1
 x kx  1 kx 2
2
2
This equation also says that the work is the average force kx/2 multiplied
by the total displacement X. We see that the total work is proportional to
the square of the final elongation X. To stretch an ideal spring by 2 cm,
you must do four times as much work as is needed to stretch it by 1 cm.
Equation (6-9) assumes that the spring was originally unstretched. If
initially the spring is already stretched a distance x1, the work we must
do to stretch it to a greater elongation x2 is
1
1
w  xx Fdx  xx kxdx  kx22  kx12
2
2
2
1
2
1
6-10
WORK-ENERGY THEOREM FOR STRAIGHT-LINE MOTION, VARYING
FORCES
In Section 6-3 we derived the work-energy theorem, Wtot = K2 – K1, for the
special case of straight-line motion with a constant net force. We can now prove
that this theorem is true even when the force varies with position. As in section
6-3, let’s consider a particle that undergoes a displacement x while being acted
on by a net force with x-component F, which we now allow to vary. Just as in
Fig. 6-11, we divide the total displacement x into a large number of small
segments x. We can apply the work-energy theorem, Eq. (6-6), to each
segment because the value of F in each small segment is approximately
constant. The change in kinetic energy in segment xa is equal to the work Fa
xa , and so on. The total change of kinetic energy is the sum of the changes in
the individual segments, and thus is equal to the total work done on the particle
during the entire displacement. So Wtot =  holds for varying forces as well as
for constant ones.
Here’s an alternative derivation of the work-energy theorem for a force that may
vary with position. It involves making a change of variable from x to v in the
work
integral. As a preliminary, we note that the acceleration a of the particle can be
expressed in various ways, using a = dv/dt, v = dx/dt, and the chain rule for
derivatives:
dv dv dx
dv
a

v
dt dx dt
dx
(6-11)
using this result, Eq. (6-7) tells us that the total work done by the net force F is
dv
Wtot  xx Fdx  xx madx  xx mv dx
(6-12)
dx
Now (dv/dx)dx is the change in velocity dv during the displacement dx, so in Eq.
(6-12) we can substitute dv for (dv/dx)dx. This changes the integration variable
from x to v, so we change the limits from x1 and x2 to the corresponding
velocities v1 and v2 at these points. This gives us
2
1
2
1
2
1
Wtot  vv mvdv
2
1
The integral of v dv is just v2/2. Substituting the upper and lower limits, we
finally find
1
1
Wtot  mv22  mv12
2
2
(6-13)
WORK-ENERGY THEOREM FOR MOTION ALONG A CURVE
We can generalize our definition of work further to include a force that varies in
direction as well magnitude and a displacement that lies along a curved path.
Sup-pose a particle moves from point P1 to P2 along a curve, as shown in Fig.
6-17a. We divide the portion of the curve between these points into many
infinitesimal vector displacements, and we call a typical one of these dl. Each dl
is tangent to the path at its position. Let F be the force at a typical point along
the path, and let  be the along between F and dl at this point. Then the small
element of work dw done on the particle during the displacement dl may be
written as
dW = F cosdl = F dl = F•dl ,
where F = F cos  is the component of F in the direction parallel to dl (Fig. 617b). The total work done by F on the particle as it moves from P1 to P2 is then
 
P
P
P
W  P F cosdl  P F dl  P F dl
(6-14)
2
2
1
1
2
1
P2
F
F
P2
F

P1
dl
Fig. 6-7a

P1
dl
F = Fcos
Fig. 6-7b
We can show that the work-energy theorem, Eq. (6-6), holds true even with
varying forces and a displacement along a curved path. The force F is
essentially constant over any given infinitesimal segment dl of the path, so we
can apply the work-energy theorem for straight-line motion to that segment
equals the work from dW = F|| dl = Fdl done on the particle. Adding up these
infinitesimal quantities of work from all the segments along the whole path gives
the total work done, Eq. (6-4), and this equals the total change in kinetic energy
over the whole path. So Wtot = K = K2 – K1 is true in general, no matter what
the path and no matter what the character of the forces. This can be proved
more rigorously by using steps like those in Eqs. (6-11) through (6-13).
Note that only the component of the force parallel to the path, F1, does work on
the particle, so only this component can change the speed and kinetic energy of
the particle. The component perpendicular to the path, F = F sin, has no
effect on the particle’s speed; it only acts to change the particle’s direction.
The integral in Eq. (6-14) is called a line integral. To evaluate this integral in a
speci-fic problem, we need some sort of detailed description of the path and of
how F varies along the path.
Example 6-9.
At a family picnic you are appointed to push your obnoxious cousin
Throckmorton in a swing. His weight is w, the length of the chains is R, and you
push Throcky until the chains make an angle 0 with the vertical. To do this,
you exert a varying
horizontal force F that starts at zero and gradually increase just enough so that
Throcky and the swing move very slowly and remain very nearly in equilibrium.
What is the total work done on Throcky by all forces? What is the work done by
the tension T in the chains? What is the work you do by exerting the force F?
(Neglect the weight of chains and seat).
Solution: The free-body diagram is shown in Fig. 6-18b. We have replaced the
tensions in the two chains with a single tension T. Because Throcky is in equilibrium at every point, the net force on him is zero, and the total work done on
him by all forces is zero. At any point during the motion the chain force on
Throcky is perpendicular to each dl, so the angle between the chain force and
the displace-ment is always 90°. Therefore the work done by the chain tension
is zero.
To compute the work done by F, we have to find out how it varies with the angle
. Throcky is in equilibrium at every point, so from Fx = 0 we get y
F + (-T sin) = 0,
and from  Fy = 0 we find
T
Tcos
T cos + (-w) = 0

By eliminating T from these two equations, we obtain
F = w tan  .
F
Tsin
The point where F is applied swings through the arc s.
The arc length s equals the radius R of the circular path
w
Multiplied by the length  (in radians), so s = R. Therefore the displacement dl
corresponding to a small change of angle d has a magnitude dl = ds = Rd .
The work done by F is
 
W   F  dl   F cosds
Now we express everything in terms of the varying angle :
W  0 w tan   cos Rd   wR 0 sin d  wR 1  cos 0 
0
0
.
If 0 = 0, there is no displacement; in that case, cos 0 = 0 and W = wR. In that
case the work you do is the same as if you had lifted Throcky straight up a
distance R with a force equal to his weight w. In fact, the quantity R(1-cos 0) is
the increase in his height above the ground during the displacement, so for any
value of 0 the work done by force F is the change in height multiplied by the
weight. This is an example of a more general result that we’ll prove in Section
7-2.
6-5 Power
When a quantity of work is done might W is done during a time interval t, the
average work done per unit or average power Pav is defined to be
W
Pav 
(6-15)
t
The rate at which work is done might not be constant. Even when it varies,
we can define instantaneous P as the limit of the quotient in Eq. (6-15) at
t approaches zero:
W dW

t 0 t
dt
P  lim
(6-16)
The SI unit of power is the watt (W), named for the English inventor James
Watt.
7 Potential Energy and
Energy Conservation
7-1 Introduction
When a dive jumps off a high board into a swimming pool, she hits the water
moving pretty fast, with a lot of kinetic energy. Where does that energy come
from? The answer we learned in Chapter 6 was that the gravitational force
(her weight) does work on the diver as she falls. The diver’s kinetic energyenergy associated with her motion-increase by an amount equal to the work
done.
But there is a very useful alternative way to think about work and kinetic
energy. This new approach is based on the concept of potential energy,
which is energy associated with the position of a system rather than its motion.
In this approach, there is gravitational potential energy even while the diver is
standing on the high board. Energy is not added to the earth-diver system as
the diver falls, but rather a storehouse of energy is transformed from one form
(potential energy) to another( kinetic energy) as she falls. In this chapter we’ ll
see how this transformation can be understand from the work-energy theorem.
7-2 GRAVITATIONAL POTENTIAL ENERGY
A particle gains or loses kinetic energy because it interacts with other objects
that exert forces on it. We learned in Chapter 6 that during any interaction the
change in a particle by the forces that act on it.
When a body falls without air resistance, gravitational potential energy
decreases and the body’s kinetic energy increases. But in Chapter 6 we used
the work-energy theorem to conclude that the kinetic energy of a falling body
increases because the force of the earth’s gravity (the body’s weight) does work
on the body. Let’s use the work-energy theorem to show that gravitational
potential energy.
Let’s consider a body with mass m that moves along the (vertical) y-axis, as in
Fig 7-1. The forces acting on it are its weight, with magnitude w = mg, and
possibly some other forces; we call the vector sum (resultant) of all the other
F
forces Fother.
other
Fother
y1
W = mg
y1
W = mg
y2
y2
O
Fig 7-1(a)
Fig 7-1(b)
We’ll assume that the body stays close enough to the earth’s surface that the
weight is constant. We want to find the work done by the weight when the body
drops from a height y1 above the origin to a lower height y2 (Fig. 7-1a). The
weight and displace-ment are in the same direction, so the work Wgrav done on
the body by its weight is positive;
Wgrav = Fs = w(y2 – y1) = mgy1 – mgy2
(7-1)
This expression also gives the correct work when the body moves upward and
y2 is greater than y1 (Fig. 7-1b). In that case the quantity (y1 – y2) is negative,
and Wgrav is negative because the weight and displacement are opposite in
direction.
Equation (7-1) shows that we can express Wgrav in terms of the values of the
quan-tity mgy at the beginning and end of the displacement. This quantity, the
product of the weight mg and the height y above the origin of coordinates, is the
gravitational energy, U:
U = mgy
(7-2)
Its initial value is U1 = mgy1 , and its final value is U2 = mgy2 . The change in U
is the final value minus the initial value, or U = U2 – U1. We can express the
work Wgrav done by the gravitational force during the displacement from y1 to y2
as
Wgrav = U1 – U2 = -( U2 – U1 ) = -  U
(7-3)
The negative sign in front of U is essential. When the
body moves up, y increase, the work done by the
gravitational force is negative, and the gravitational
potential energy increase (  U >0). When the body
moves down, y decreases, the gravitational force does
positive work, and the gravitational potential energy
decreases (U<0).
Conservation of Mechanical Energy ( Gravitational Force
Only )
To see what gravitational potential energy is good for,
suppose the body’s
weight is the only force acting on it, so Fother = 0. The body is then falling freely
with no air resistance and can be moving either up or down. Let its speed at
point y1 be v1, and let its speed at y2 be v2. The work-energy theorem, Eq. (6-6),
says that the total work done on the body equals the change in the body’s in the
body’s kinetic energy; Wtot = K = K2 – K1. If gravity is the only force that acts,
then from Eq. (7-3), Wtot = Wgrav = -U = U1 – U2. Pointing there together, we
get
U
or
K2 – K1 = U1 – U2 , which we can rewrite as K1 + U1 = K2 +
1 2
1
U2 , (7-4) or
mv1  mgy1  mv22  mgy2
(7-5)
2
2
We now define the sum K + U of kinetic and potential energy to be E, the total
mechanical energy of the system. By “system” we mean the body of mass m
and the earth considered together, because gravitational potential energy U is a
shared pro-perty of both bodies. Then E1 = K1 + U1 is the total mechanical
energy at y1, and E2 = K2 + U2 is the total mechanical energy at y2. Equation (74) says that when the body’s weight is the only force doing work on it, E1 = E2.
Thus is, E is constant; it has the same value at y1 and y2. But since the
positions y1 and y2 are arbitrary points in the motion of the body, the total
mechanical energy E has the same value at all points during the motion, E = K
+ U = constant.
A quantity that always has the same value is called a conserved quantity. When
only the force of gravity does work, the total mechanical energy is constant, that
is, is conserved. This is our first example of the conservation of mechanical
energy.
Effect of Other Force
When other forces act on the body in addition to its weight, then Fother in Fig. 7-1
is not zero. The gravitational work Wgrav is still given by Eq. (7-3), but the total
work Wtot is then the sum of Wgrav, and the work done by all forces is Wtot =
Wgrav + Wother. Equating this to the change in kinetic energy, we have
Wother+ Wgrav + = K2 – K1
(7-6)
Also, from Eq. (7-3), Wgrav = U1 – U2, so
Wother + U1 –U2 = K2 – K1,
which we can rearrange in the form
K1 + U1 = Wgrav = K2 + U2 (if forces other than gravity do work) (7-7)
Finally, using the appropriate expressions for the various energy terms, we
obtain
1 2
1
mv1  mgy1  Wother  mv22  mgy2
2
2
(7-8)
The meaning of Eqs. (7-7) and (7-8) is this: The work done by all forces other
than the gravitational force equals the change in the total mechanical energy E
= K + U of the system, where U is the gravitational potential energy.
Gravitational Potential Energy For Motion Along A Curved Path
Fother
x
y1
y s
W = mg
W = mg
y2
Figure 7-4 (a)
(b)
the body is acted on by the gravitational force w = mg and possibly by other
forces whose resultant we call Fother. To find the work done by the gravitational
force during this displacement, we divide the path up into small segments s; a
typical segment is shown in Fig. 7-4b. The work done by the gravitational force
over this segment is the scalar product of the force and the displacement. In
terms of unit vectors, the force is w = mg = -mgj and the displacement is s =
xi + yj , so the work done by the gravitational force is
w ·s = -mgj·(xi + yj) = - mg y .
The work done by gravity is the same as though the body had been displaced
ver-tically a distance y, with no horizontal displacement. This is true for every
seg-ment, so the total work done by the gravitational force is –mg multiplied by
the
Total vertical displacement (y2 – y1):
Wgrav = - mg( y2 – y1) = mgy1 – mgy2 = U1 – U2 .
This work done by gravity is the same as though the body had been displaced
ver-tical path. So even if the path a body follows between two points is curved,
the total work done by the gravitational force depends only on the difference in
height between the two points of the path. This work is unaffected by any
horizontal mo-tion that may occur. So we can use the same expression for
gravitational potential energy whether the body’s path is curved or straight.
7-3 Elastic Potential Energy
We’ll describe the process of storing energy in a deformable body body such as
a spring or rubber band in terms of elastic potential energy. A body is called
elastic if it returns to its original shape and size after being deformed. To be
specific, we’ll consider storing energy in an ideal spring like the ones we
discussed in Section 6-4. To keep such an ideal spring stretched by a distance
x, we must exert a force F = kx, where k is the force constant of the spring. The
ideal spring is a useful idealization because many other elastic bodies show
this same direct proportionality between force F and displacement x, provided
that x is sufficiently small.
We found in Section 6-4 that the work we must do on the spring to move one
end form an elongation x1 to a different elongation x2 is
1
1
W  kx22  kx12
2
2
(work done on a spring)
where k is the force constant of the spring. If we stretch the spring further, we
do positive work on the spring; if we let the spring relax while holding one end,
we do negative work on it. We also saw that this expression for work is still
correct if the spring is compressed, not stretched, so that x1 or x2 or both are
negative. Now we need to find the work done by the spring. From Newton’s
third law the two quan-tities of work are just negatives of each other. Changing
the signs in this equation, we find that in a displacement from x1 to x2 the spring
does an amount of work Wel given by
1
1
W  kx12  kx22 (work done by a spring)
2
2
Just as for gravitational work, we can express the work done by the spring in
terms of a given quantity at the beginning and end of the displacement. This
quantity is ½kx2.
We can use Eq. (7-9) to express the work Wel done on the block by the elastic
force in terms of the change in potential energy:
1
1
Wel  kx22  kx12  U1  U 2  U
(7-10).
2
2
The work-energy theorem says that Wtot = K2 – K1, no matter what kind of
forces are on a body. If the elastic force is the only forces that does work on the
body, then
Wtot = Wel = U1 – U2.
The work-energy theorem Wtot = K2 – K1 then gives us
K1 + U1 = K2 + U2 (if only the elastic force does work) (7-11).
Here U is given by Eq. (7-9), so
1 2 1 2 1 2 1 2
mv1  kx1  mv2  kx2
(if only the elastic force does work)(72
2
2
12). 2
In this case the total mechanical energy E = K + U (the sum of kinetic and
elastic potential energy) is conserved. For Eq. (7-12) to be strictly correct, the
ideal spring that we’ve been discussing must also be massless. If the spring
has a mass, it will also have kinetic energy as the coils of the spring move
back and forth. We can neg-lect the kinetic energy of the spring if its mass is
much less than the mass m of the body attached to the spring. For instance, a
typical automobile has a mass of 1200 kg or more. The springs in its
suspension have masses of only a few kilograms, so their mass can be
neglected if we want to study how a car bounces on its suspension.
If forces other than the elastic force also do work on the body, we call their
work Wother, as before. Then the total work is Wtot = Wel + Wother and the workenergy theorem gives
Wel + Wother = K2 – K1 .
The work done by the spring is still Wel = U1 – U2, so again
K1 + U1 + Wother = K2 + U2 (if forces other than the elastic force do work) (7-13)
and
1 2 1 2
1
1
mv1  kx1  Wother  mv22  kx22
2
2
2
2
(if forces
other than the elastic force do work (7-14).
This equation shows that the work done by all forces
other than the elastic force equals the change in the total
mechanical energy E = K +U of the system, where U is
the elastic potential energy. The “system” is made up of
the body of mass m and the spring of force constant k.
when Wother is positive, E increases; when Wother is
negative, E decreases. You should compare Eq. (7-14)
to Eq. (7-8), which describes situations in which there is
gravitational potential energy but no elastic potential
energy.
Situations With Both Gravitational And Elastic Potential
Energy
Equations (7-11), (7-12), (7-13), and (7-14) are valid when the
only potential energy in the system is elastic potential energy.
What happens when we have both gravitational and elastic forces,
such as a block attached to the lower end of a vertically hang-ing
spring? We can still use Eq. (7-13), but now U1 and U2 are the
initial and final values of the total potential energy, including both
gravitational and elastic potential energies. That is, U = Ugrav + Uel.
Thus the most general statement of the relationship between
kinetic energy, potential energy, and work done by other forces is
K1  U grav,1  U el ,1  Wother  K 2  U grav, 2  U el , 2
(valid in
general) (7-15).
That is, the work done by all forces other than the
gravitational force or elastic force equals the change in the
total mechanical energy E = K + U of the system, where U is
the sum of the gravitational potential energy and
elastic potential energy.
7-4 Conservative And Non-conservative Forces
In our discussions of potential energy we have talked
about “storing” kinetic energy by converting it to potential
energy. A force that offers this opportunity of two-way
conversion between kinetic and potential energies is
called a conservative force. We have seen two
examples of conservative forces: the gravitational force
and the spring force. An essential feature of
conservative forces is that their work is always reversible.
Anything that we deposit in the energy “bank” can later
be with-drawn without loss. Another important aspect of
conservative forces is that a body may move from point
1 to point 2 by various
paths, but the wok done by a conservative force is the same for
all of these paths. Thus if a body stays close to the surface of the
earth, the gravitational force mg is independent of height, and
the work done by this force depends only on the change in
height. If the body moves around a closed path, ending at the
same point where it started, the total work done by the
gravitational force is always zero.
The work done by a conservative force always has these
properties:
1. It can always be expressed as the difference between the
initial and final values of a potential energy function.
2. It is reversible.
3. It is independent of the body and depends only on the starting
and ending points.
4. When the starting and ending points are the same, the total
work is zero.
When the only forces that do work are conservative forces, the total mechanical
energy E = K + U is constant.
A force that is not conservative is called a nonconservative force. The work
done by a nonconservative force cannot be represented by a potential-energy
function. Some nonconservative force, like kinetic friction or fluid resistance,
cause mechanical energy to be lost or dissipated; a force of this kind is called a
dissipative force. There are also nonconservative forces that increase
mechanical energy. The fragments of an exploding firecracker fly off with very
large kinetic energy, thanks to a chemical reaction of gunpowder with oxygen.
THE LAW OF CONSERVATION OF ENERGY
Nonconservative forces cannot be represented in terms of potential energy. But
we can describe the effects of these forces in terms of kinds of energy other
than kinetic and potential energy. When a car with locked brakes skids to a stop,
the tires and the road surface both become hotter. The energy associated with
this change in the state of the materials is called internal energy. Raising the
temperature of a body increase its internal energy; lowering the body’s
temperature of a body decreases its internal energy.
To see the significance of internal energy, let’s consider a block sliding on a
rough surface. Friction does negative work on the block as it slides, and the
change in inter-nal energy of the block and surface (both of which get hotter) is
positive.
Careful experiments show that the increase in the internal energy
is exactly equal to the absolute value of the work done by friction.
In other words,
Uint = - Wother ,
where Uint is the change in internal energy. If we substitute this
into Eq. (7-7), Eq. (7-13), or (7-15), we find
K1 + U1 - Uint = K2 + U2 .
Writing K = K2 – K1 and U = U2 – U1, we can finally express
this as
K + U + Uint = 0 (law of conservation energy)
(7-16).
This remarkable statement is the general form of the law of
conservation. In a given process, the kinetic energy, potential
energy, and internal energy of a system may all change. But the
sum of those changes is always zero. If there is a decrease in one
form of energy, it is made up for by an increase in the other forms.
When we ex-pand our definition of energy to include internal
energy, Eq. (7-16) says that energy is never created or
destroyed; it only changes form. No exception to this rule has
ever been found.
7-5 Force and Potential Energy
For the two kinds of conservative forces (gravitational and
elastic) we have studied, we started with a description of
behavior of the force and derived from that an expression for
the potential energy.
We can reverse this procedure: If we are given a potentialexpression, we can find the corresponding force.
W  U
Let’s apply this to a small displacement x. The work done by
the force Fx x  during this displacement is approximately
equal to Fx x x . We have to say “approximately” because
Fx  x  may vary a little over the interval x. But it is at least
approximately true that Fx x x  U and F ( x)   U .
dU  x 
Fx  x   
dx
x
x
Force and Potential Energy in Three Dimensions
We can extend this analysis to three dimensions, where
the particle may move in the x-, y-, z-direction, or all at once,
under the action of a conservative force that has components
Fx, Fy , Fz . Each component of force may be a function of the
coordinates x, y and z. The potential-energy function U is
also a function of all three space coordinates.
U
FX  
x
Fy  
U
y
Fz  
U
z
To make these relations exact, we need to take the limits x
0, y 0, and z0. So that these ratios become
derivatives. Because U may be a function of all three
coordinates, we need to remember that we calculate each of
these derivatives, only one coordinate changes at a time.
U
Fx  
x
U
Fy  
y
U
Fz  
z

U  U  U 
F  (
i
j
k)
x
y
z
8
Momentum Impulse and Collisions
8-1 Introduction
A common theme of all these questions is that they can’t be
answered by directly applying Newton’s second law,  F = ma ,
because there are forces acting about which we know very litter.
Our approach will use two new concepts, momentum and impulse,
and a new conservation law, conservation of momentum. This
conservation law is every bit as important as that of conservation
of energy. The law of conservation of momentum is valid even in
situations in which Newton’s laws are inadequate, such as bodies
moving at very high speeds (near as the speed of light) or objects
on a very small scale (such as the constituents of atoms). Within
the domain of Newtonian mechanics, conservation of momentum
enables us to analyze many situations that would be very difficult
if we tried to use Newton’s laws directly. Among these are
collision problems, in which two bodies collide and can exert very
large forces on each other for a short time.
8-2 Momentum And Impulse
Let’s consider a particle of constant mass m.
Because a = dv/dt, we can write Newton’s second
law for this particle as


dv d 
 mv 
F  m
dt dt
8-1
We can take the mass m inside the derivative
because it is constant. Thus Newton’s second law
says that the net force F acting on a particle equals
the time rate of change of the combination mv, the
product of the particle’s mass and velocity. We’ll call
this combination the momentum, or linear
momentum, of the particle. Using the symbol p for
momentum, we have
p = mv (definition of momentum)
(8-2)
Momentum is a vector quantity that has a magnitude (mv) and a
direction (the same as the velocity vector v). The units of the
magnitude of momentum are units of mass times speed; the SI
units of momentum are kg·m/s. The plural of momentum is
“momenta”.
Substituting Eq.(8-2) into Eq.(8.1), we have
 dp
 F  dt
(Newton’s second law in terms of momentum)
(8-3).
The net force (vector sum of all forces) acting on a particle
equals the time rate of change momentum of the particle. This,
not  F = ma , is the form in which Newton originally stated his
second law (though he called momentum the “quantity of motion”).
It is valid only in inertial frames of reference.
We will often express the momentum of a particle in terms of its
components. If the particle has velocity components vx, vy, and vz,
its momentum components px, py, ,
and pz are given by
px = mvx ,
py = mvy ,
pz = mvz
(8-4).
These three components equations are equivalent to
Eq. (8-2).
A particle’s momentum p = mv and its kinetic energy
k = ½ mv2 both depend on the mass and velocity of
the particle. What is the fundamental difference
between these two quantities? A purely
mathematical answer is that momentum is a vector
whose magnitude is proportional to speed, while
kinetic energy is a scalar proportional to the speed
squared. But to see the physical difference between
momentum and kinetic energy, we must first define a
quantity closely related to momentum called
impulse.
Let’s first consider a particle acted on by a constant net force F
during a time interval t from t1 to t2, (We’ll look at the case of
varying forces shortly.) The impulse of the net force, denoted by J,
is defined to be product of the net force and the time interval:
J = F(t2 – t1) = Ft
(8-5).
Impulse is a vector quantity, its direction is the same aa the net
force F .
To see what impulse is good for, let’s go back Newton’s second
law as restated in terms of momentum, Eq. (8-3). If the net force
F is constant, then dp/dt is also constant. In that case, dp/dt is
equal to total change in momentum p2 – p1 during the time interval
t2 – t1, divided by the interval:
 
 P2  P1
F 
t2  t1
Multiplyingthis equation by (t2 – t1), we have
 
 F (t2  t1 )  p2  p1 .
Comparing to Eq. (8-5), we end up with a result called the impulse-momentum
theorem: J  p  p
2
1
(impulse-momentum theorem).
(8-6)
The change in momentum of a particle during a time interval equals the impulse
of the net force that acts on the particle during that interval.
The impulse-momentum theorem also holds when forces are not constant. To
see this, we integrate both sides of Newton’s second law F = dp/dt over time

be-tween the limits t1 and t2:


 
t
t dp
p 

dt  p dp  p2  p1
t  F dt  t
dt
2
1
2
1
2
1
The interval on the left is defined to be the impulse J of the net force F during
this interval:

t
t  Fdt
(general definition of impulse)
(8-7).
With this definition the impulse-momentum theorem J = P2 – P1 , Eq. (8-6), is
va-lid even when the net force F varies with time.
We can define an average net force Fav such that even when F is not
constant, the impulse J is given by
J = Fav( t2 – t1)
(8-8).
2
1
When F is constant, F = Fav and Eq. (8-8) reduces to Eq. (8-5).
Fx
(Fav)x
t1
t2
t2-t1
Fig.(8-1)
Figure (8-1) shows a graph of the x-component of net force Fx as a function of
time during a collision. This might represent the force on a soccer ball that is in
contact with a player’s foot from time t1 to t2. The x-component of impulse
during this interval is represented by the blue area under the curve between t1
and t2. This area is equal to the rectangular area bounded by t1, t2 , and (Fav)x,
so (Fav)x(t2 - t1) is equal to the impulse of the actual time-varying force during the
same interval.
Impulse and momentum are both vector quantities, and Eqs.(8-5) through (88) are all vector equations. In specific problems it is often easiest to use them in
component form:
J x  tt  Fx dt  Fav x t2  t1   p2 x  p1x  mv2 x  mv1x
J y  tt  Fy dt  Fav y t2  t1   p2 y  p1 y  mv2 y  mv1 y
2
1
2
1
(8-9)
MOMENTUM AND KINETIC ENERGY COMPARED
We can see now see the fundamental difference
between momentum and kinetic energy. The
impulse-momentum theorem J = p2 – p1says that
changes in a particle’s momentum are due to
impulse, which depends on the time over which the
net force acts. By contrast, the work-energy theorem
Wtot = K2 – K1 tells us that kinetic energy changes
when work is down on a particle; the total work
depends on the distance over which the net force
acts. Consider a particle that starts from rest at t1 so
that v1 = 0. Its initial momentum is p1 = mv1 = 0, and
its initial kinetic energy is K1 = ½ mv2 = 0. Now let a
constant net force equal to F act on that particle
from time t1 until time t2. During this interval the
particle moves a distance s in the direction of the
force. From Eq. (8-6), the particle’s momentum at time t2 is p2 =
p1 + J = J, where J = F (t2 – t1) is the impulse that acts on the
particle. So the momentum of a particle equals the impulse that
accelerated it from rest to its present speed; impulse is the
product of the net force that accelerated the particle and the time
required for the acceleration. By comparison the kinetic energy of
the particle at t2 is K2 = Wtot = Fs, the total work done on the
particle to accelerate it from rest. The total work is the product of
the net force and the distance required to accelerate the particle.
Both the impulse-momentum and work-energy theorem are
relationships between force and motion, and both rest on the
foundation of Newton’s laws. They are integral particles, relating
the motion at two different times separated by a finite interval. By
contrast, Newton’s second law itself (in either of the forms  F =
ma or  F = dp/dt) is a differential principle, relating the forces to
the rate of change of velocity or momentum at each instant.
y
y
x
x
FB on A
FA on B
Fig. 8-6 (b)
Fig. 8-6 (c )
8-3 Conservation of momentum
The concept of momentum is particularly important in situations in which we
have two or more interacting bodies. Let’s consider first an idealized system
consisting of two bodies that interact with each other but not with anything else.
Let’s go over that again with some new terminology. For any system, the forces
that the particles of the system exert on each other are called internal forces.
Forces exerted on any part of the system by some object outside it are called
external forces. For the sys-tem we have described, the internal forces are FB on
A, exerted by particle B on particle A, and FA on B, exerted by particle A on
particle B (Figs. 8-6 b, c). There are no external forces; when this is the case,
we have an isolated system.
The net force on particle A is FB on A, and the net force on particle B is FA on B, so
from Eq. (8-3) the rates of change of the momenta of the two particles are




dp A
dpB
FBonA 
FAonB 
(8-10)
dt
dt
The momentum of each particle changes, but these
changes are not independent; according to
Newton’s third law, the two forces FB on A and FA on B
are always equal in magnitude and opposite in
direction. That is, FB on A = - FA on B so FB on A + FA on B =
0. Adding together the two equations in Eq. (8-10),
we have
 




dp A dpB d PA  PB 
FBonA  FAonB 


0
dt
dt
dt
(8-11).
The rates of change of the two momenta are equal
and opposite, so that the rate of change of the
vector sum PA + PB is zero. We now define the total
momentum P of the system of two particles as the
vector sum of the momenta of the individual
particles.
That is,
P = PA + PB
(8-12).
Then Eq. (8-11) becomes, finally,



dp
FBonA  FAonB 
0
dt
(8-13)
The time rate of change of the total momentum P is zero. Hence
the total momentum of the system is constant, even though the
individual momenta of the particles that make up the system can
change.
If external forces are also present, they must be included on the
left side of Eq. (8-13) along with the internal forces. Then the total
momentum is, in general, not constant.
But if the vector sum of the external forces is zero, these forces
don’t contribute to the sum, and dp/dt is again zero. Thus we have
the following general result:
If the vector sum of the external forces on a system is zero, the
total momentum of the system is constant.
This is the simplest form of the principle of conservation of momentum. This
principle is a direct consequence of Newton’s third law. What makes this
principle useful is that it doesn’t depend on the detailed nature of the internal
forces that act between members of the system. This means that we can apply
conservation of momentum even if (as is often the case) we know very little
about the internal forces. We have used Newton’s second law to derive this
principle, so we have to be careful to use it only in internal frames of reference.
We can generalize this principle for a system containing any number of
particle A, B, C, … interacting only with each other. The total momentum of
such a system is
P = PA + PB +  = mA vA + mB vB + … (total momentum of a system of particles).
We make the same argument as before; the total rate of change of momentum
of the system due to each action-reaction pair of internal forces is zero. Thus
the total rate of change of momentum of the entire system is zero whenever the
vector sum of the exter-nal forces acting on it is zero. The internal forces can
change the momentum of individual particle in the system but not the total
momentum of the system.
8-4 Inelastic Collisions
If the forces between the bodies are much larger than any external forces, as is
the case in most collisions, we can neglect the external forces entirely and treat
the bodies as an
isolated system. Then momentum is conserved in the collision,
and the total momentum of the system has the same value before
and after the collision.
If the forces between the bodies are also conservative, so that no
mechanical energy is lot lost or gained in the collision, the total
kinetic energy of the system is the same after the collision as
before. Such a collision is called an elastic collision. A collision in
which the total kinetic energy after the collision is less than that
before the collision is called an inelastic collision. An inelastic
collision in which the colliding bodies stick together and move as
one body after the collision is often called a completely inelastic
collision.
8-5 Elastic Collisions
From conservation of kinetic energy we have
1
1
1
1
2
2
2
m Av A1  mB vB1  m Av A2  mB vB2 2
2
2
2
2
And conservation of momentum gives
mAvA1  mBvB1  mAvA2  mBvB 2
8-7 Rocket Propulsion
Momentum considerations are particularly useful for
analyzing a system is which the mass of parts of the
system change with time.
 In such case we can’t use
Newton’s second law  F  ma directly because m
change.
We choose our x-axis to be along the rocket’s
direction of motion. The x-component of total
momentum at this instant is P1 = mv. In a short time
interval da the mass of the rocket changes by an
amount dm. This is an inherently negative quantity
because the rocket’s mass m decrease with time.
During dt a positive mass –dm of burned fuel is
ejected from the rocket; Let vex be the exhaust speed
of this material relative to the rocket; the burned fuel
is ejected opposite the direction of motion, so its xcomponent of velocity relative to the rocket is –vex.
The x-component of velocity vfuel of the burned fuel
relative to our coordinate system is then
v fuel  v  (vex )  v  vex
And the x-component of momentum of the ejected
mass (-dm) is
(dm)v fuel  (dm)(v  vex )
And the x-component of momentum of the ejected
mass (-dm) is
(-dm)vfuel = (-dm)(v – vex)
The rocket’s momentum at this time is
(m + dm)(v+dv).
Thus the total x-component of momentum P2 of
rocket plus ejected fuel at time t + dt is
P2  (m  dm)(v  dv)  (dm)(v  vex )
According to our initial assumption, the rocket and
fuel are an isolated system.
mv  (m  dm)(v  dv)  (dm)(v  vex )
This can be simplified to
mdv  dmvex  dmdv
dv
dm
m  vex
dt
dt
vex dm
dv
a

dt
m dt
dm
F  vex
dt
dm
dv  vex
m
m dm
dm
v0 dv  m0 vex m  vex m0 m
v
m
m0
m
v  v0  vex ln
 vex ln
m0
m
The ratio m/m0 is the original mass divided by the
mass after the fuel has been exhausted.
9 Rotation of Rigid Bodies
9-1 Introduction
Real-world bodies can be even more complicated; the forces that act on them
can de-form them---stretching, twisting, and squeezing them. We’ll neglect
these deformations for now and assume that the body has a perfectly definite
and unchanging shape and size. We call this idealized model a rigid body. This
chapter and the next are mostly about rotational motion of a rigid body. We
begin with kinetic language for describing rotational motion.
9-2 Angular Velocity And Acceleration
In analyzing rotational motion, let’s think first about a rigid body that rotates
about a fixed axis. By fixed axis we mean an axis that is at rest in some inertial
frame of refer-ence and does not change direction relative to that frame.
S=r
S=r
1rad
1rad
P
O
Fig 9-1
Fig9-2(a)
Fig9-2(b)
Figure 9-1 shows a rigid body rotating about a fixed
axis that passes through point O and is
perpendicular to the plane of the diagram, which we
choose to call the xy-plane. One way to describe the
rotation of this body would be to choose a particular
point P on the body and to keep track of the x- and
y-coordinates of this point. This isn’t a terribly
convenient method, since it takes two numbers (the
two coordinates x and y) to specify the rotational
position of the body. Instead, we notice that the line
OP is fixed in the body and rotates with it. The angle
 that this line makes with the +x-axis describes the
rotational position of the body; we will use this
single quantity  as a coordinate for rotation.
The angular coordinate  of a rigid body rotating around a fixed
axis can be positive or negative. If we choose positive angles to
be measured counterclockwise from the positive x-axis, then the
angle  in Fig. 9-1 is positive. If we instead choose the positive
rotation direction to be clockwise, then  in Fig. 9-1 is negative.
When we considered the motion of a particle along a straight line,
it essential to specify the direction of positive displacement along
that line; in discussing rotation around a fixed axis, it’s just as
essential to specify the direction of positive rotation.
In describing rotational motion, the most natural way to measure
the angle  is not in degrees, but in radians. As shown in Fig. 9-2a,
one radian (1 rad ) is the angle subtended at the center of a circle
by an arc with a length equal to the radius of the circle. In Fig9-2b
an angle subtended by an arc of length s on a circle of radius r.
The value of  (in radians) is equal to s divided by r.
s

r
or
s=r
(9-1)
An angle in radians is the ratio of two lengths, so it
is a pure number, without dimensions. If s = 3.0 m
and r = 2.0 m, then  = 1.5, but we will often write this
as 1.5 rad to distinguish it from an angle measured
in degrees or revolutions.
Angular Velocity
The coordinate  shown in Fig. 9-1 specifies the
rotational position of a rigid body at a given instant.
We can describe the rotational motion of such a
rigid body in terms of the rate of change of  . In
Fig. 9-3a a reference line OP in a rotating body
makes an angle 1 with the +x-axis at time t1. At a
later time t2 the angle has changed to 2. We define
the average angular velocity av of the body in the time interval  t
= t2 – t1 as the ratio of the angular displacement  = 2 -  1 to
t:
 2  1 
av 

t2  t1
t
The instantaneous angular velocity  is limit of av as t
approaches zero, that is, the derivative of  with respect to t:
 d
  lim

t 0 t
dt
any instant, every part of a rotating rigid body has the same
angular velocity. The angular velocity is positive if the body is
rotating in the direction of increasing  and negative if rotating in
the direction of decreasing  .
Angular Acceleration
When the angular velocity of a rigid body changes, it has an angular
acceleration. If 1 and 2 are the instantaneous angular velocities at times t1
and t2, we define the average angular acceleration av over the interval t = t2 –
t1 as the change in angular velocity divided by t:
  1 
 av  2

(9-4).
t2  t1
t
The instantaneous acceleration  is the limit of av as t  0:
 d
  lim

(9-5).
t 0
t
dt
The usual unit of angular acceleration is the radian per second, or rad/s2.
Henceforth we will use the term “angular acceleration” to mean the
instantaneous angular acceleration rather than the average angular
acceleration.
Because  = d/dt, we can also express angular acceleration as the second
derivative of the angular coordinate:
d d d 2


(9-6).
dt dt dt 2
In rotational motion, if the angular acceleration  is positive, the angular velocity
 is increasing; if  is negative,  is decreasing. The rotation is speeding up if 
and  have
the same sign and slowing down if  and  have opposite signs.
9-3 Rotation With Constant Angular Acceleration
Let 0 be the angular velocity of a rigid body at time t = 0, and let  be its
angular velocity at any later time t. The angular acceleration  is constant and
equal to the average value for any interval. Using Eq. (9-4) with the interval
from 0 to t, we find
  0

t 0
, or
 = 0 +  t
(9-7).
The product  t is the total change in  between t = 0 and the later time t; the
angular velocity  at time t is the sum of the initial value 0 and this total
change.
With constant angular acceleration the angular velocity changes at a uniform
rate, so its average value between 0 and t is the average of the initial and final
values:
 av 
0  
2
(9-8).
We also know that av is the total angular displacement ( - 0) divided by the
time interval (t - 0);
av 
  0
t 0
(9-9).
When we equate Eqs. (9-8) and (9-9) and multiply the result by t, we get
1
2
   0  0   t
(9-10)
To obtain a relation between  and t that doesn’t
contain , we substitute Eq. (9-7) into Eq. (9-10):
1
   0  0  0  t t
2
1 2
   0   0 t  t
2
(9-11).
Relating Linear And Angular Kinematics
y
y
ω
ω
v
r
P
atan
r


x
ω
arad
x
ω
Fig 9-7
P
Fig 9-8
When a rigid body rotates about a fixed axis, every particle in the
body moves in a cir-cular path. The circle lies in a plane
perpendicular to the axis and is centered on the axis. The speed
of a particle is directly proportional to the body’s angular velocity;
the faster the body rotates, the greater the speed of each particle.
In Figure 9-7, point P is a constant distance r from the axis of
rotation, so it moves in a circle of radius r. At any time the angle 
(in radians) and the arc length s are related by
s = r.
We take the time derivative of this, noting that r is constant for any
specific particle, and take absolute value of both sides:
ds
d
r
dt
dt
Now ds/dtis the absolute value of the rate of change of arc
length, which is equal to the instantaneous linear speed v of the
particle. Analogously, d/dt, the absolute value of the rate of
change of the angle, is the instantaneous
angular speed  --- that is, the magnitude of the instantaneous angular velocity
in rad/s. Thus
v = r
(relation between linear and angular speed)
(9-13).
The farther a point is from the axis, the greater its linear speed. The direction of
the linear velocity vector is tangent to its circular path at each point (Fig. (9-7).
We can represent the acceleration of a particle moving in a circle in terms of its
centri-petal and tangential components, arad and atan (Fig. 9-8), as we did in
Section 3-5. It would be a good idea to review that section now. We found that
the tangential component of acceleration atan, the component parallel to the
instantaneous velocity, acts to change the magnitude of the particle’s velocity
(i.e., the speed) and is equal to the rate of change of speed. Taking the
derivative of Eq. (9-13), we find
atan 
dv
d
r
 r
dt
dt
(9-14).
This component of a particle’s acceleration is always tangent to the circular
path of the particle.
The component of the particle’s acceleration directed toward the rotation axis,
the centripetal component of acceleration arad, is associated with the change of
direction of the particle’s velocity. In Section 3-5 we worked out the relation arad
= v2/r. We can ex-press this in terms of  by using Eq. (9-13):
v2
arad    2 r
(9-15)
r
The vector sum of the centripetal and tangential components of acceleration
particle in a rotating body is the linear acceleration a (Fig. 9-8).
9-5 Energy In Rotational Motion
A rotating rigid body consists of mass in motion, so it has kinetic energy. We
can ex-press this kinetic energy in terms of the body’s angular velocity and a
new quantity called moment of inertia that we will define below. To develop this
relationship, we think of the body as being made up of a large number of
particles, with masses m1, m2, …, at distances r1, r2, …, from the axis fo rotation.
We label the particles with the index i: The mass of the ith particle is mi, and its
distance from the axis of rotation is ri. The particles don’t necessarily all lie in
the same plane, so we specify that ri is the per-pendicular distance from the
axis to the ith particle.
When a rigid body rotates about a fixed axis, the speed vi of the ith particle is
given by Eq. (9-13), vi = ri, where  is the body’s angular speed. Different
particles have different values of r, but  is the same for all (otherwise, the body
wouldn’t be rigid). The kinetic energy of the ith particle can be expressed as
1
1
2
mi vi  mi ri 2 2
2
2
The total kinetic energy of the body is the sum of the kinetic energies of all its
particles:
1
1
1
k  m1r1212  m2 r2222     mi ri 2i2
i 2
2
2
Taking the common factor 2/2 out of this expression, we get
1
1
2
2
2


k  m1r1  m2 r2      mi ri 2  2
2
2 i
The quantity in parentheses, obtained by multiplying the mass of each particle
by the square of its distance from the axis of rotation and adding these products,
is denoted by I and is called the moment of inertia of the body for this rotation
axis:
I  m1r12  m2 r22     mi ri 2
(9-16).
i
The word “moment” means that I depends on how the body’s mass is
distributed in space; it has nothing to do with a “moment” of time. For a body
with a given rotation axis and a given total mass, the greater the distance from
the axis to the particles that make up the body, the greater the moment of
inertia. In a rigid body, the distance ri are all constant and I is independent of
how the body is rotating around the given axis.
In terms of moment of inertia I, the rotational kinetic energy K of a rigid body is
1
K  I 2
(rotational kinetic energy of a rigid body)
(9-17).
2


The kinetic energy given by Eq. (9-17) is not a new
form of energy; it’s the sum of the kinetic energies
of the individual particles that make up the rigid
body, written in a compact and convenient form in
terms of the moment of inertia. When using Eq. (9-17)
must be measured in radians per second, not
revolutions or degrees per second, to give K in
joules; this is because we used vi = ri in our
derivation.
Equation (9-17) gives a simple physical
interpretation of moment of inertia: the  greater the
moment of inertia, the greater the kinetic energy of a
rigid body rotating with a given angular speed . We
learned in Chapter 6 that the kinetic energy of a
body equals the amount of work done to accelerate
that body from
rest. So the greater a body’s moment of inertia, the harder it is to start the body
rotating if it’s at rest and the harder it is to stop its rotation if it’s already rotating
(Fig. 9-12). For this reason, I is also called the rotational inertia.
9-7 Moment of Inertia Calculations
When a rigid body cannot be represented by a few point masses but is a
continuous distribution of mass, the sum of masses and distance that defines
moment of inertia Eq. (9-16) becomes an integral. Imagine dividing the body
into small mass elements dm so that all points in a particular element are at
essentially the same perpendicular distance from the axis of rotation. We call
this distance r, as before. Then the moment of inertia is
I   r dm
2
(9-20)
To evaluate the integral, we have to represent r and dm in terms of the same
integration variable. When we have an effectively one-dimensional object, such
as the slender rods (a) and (b) in Table 9-2, we can use a coordinate x along
the length and relate dm terms of an increment dx volume dV and the density 
of the body. Density is mass per unit
volume,  = dm/dv, so we may also write Eq. (9-20) as
I   r 2 dV
If the body is uniform in density, then we may take  outside the
integral:
I    r 2 dV
(9-21).
To use this equation, we have to express the volume element dV
in terms of the differentials of the integration variable, such as dV
= dxdydz. The element dV must always be chosen so that all
points within it are at very nearly the same distance from the axis
of rotation. The limits on the integral are determined by the shape
and dimensions of the body. For regularly shaped bodies this
integration can often be carried out quite easily.
10 Dynamics of Rotational
Motion
10-1 Introduction
In this chapter we will define a new physical quantity, torque, that
describes the twisting or turning effort of a force. We’ll find that the
net torque acting on a rigid body determines its angular
acceleration, in the same way that the net force on a body
determines its linear acceleration. We all also look at work and
power in rotational motion in order to understand such problems
as how energy is transmitted by rotating drive shaft in a car.
Finally, we will develop a new conservation principle, conservation
of angular momentum, that is tremendously useful for
understanding the rotational motion of both rigid and nonrigid
bodies.
10-2 Torque
What is it about a force that determines how
effective it is in causing or changing rotational
motion? The magnitude and direction of the force
are important, but so is the position of the point
where the force is applied. The quantitative measure
of the tendency of a force to cause or change the
rotational motion of a body is called torque. Figure
10-2 shows a body that can rotate about an axis that
passes through point O and is perpendicular to the
plane of the figure. The body is acted on by three
forces, F1, F2, and F3, in the plane of the figure. The
tendency of F1 to cause a rotation about point O its
magnitude
F1
Line of action
of F1
Lever arm
of F1
A
l1
F3
o
l2
B
Lever arm
of F2
F2
Line of action
of F2
depends on its magnitude F1. It also depends on the
perpendicular distance l1 between the line of action of the force
(that is, the line along which the vector lies) and point O. We
called the distance l1 the lever arm (or moment arm) of force F1
about O. The twisting effort is directly proportional to both F1 and
l1. We define the torque (or moment) of the force F1 with respect
to point O as the product F1l1. We will use the Greek letter  for
torque. For a force of magnitude F whose line of action is a
perpendicular distance l from the point O, the torque is  = Fl
(10-1). The lever arm of F1 in Fig. 10-2 is the perpendicular
distance OA or l1, and the lever arm of F2 is the perpendicular
distance OB or l2. The line of action of F3 passes through the
reference point O, so the lever arm for F3 is zero and its torque
with respect to point O is zero.
Force F1 in Fig. 10-2 tends to cause counterclockwise rotation
about O, while F2 tends to cause clockwise rotation. To distinguish
between these two possibilities, we will choose a positive sense of
rotation. With the choice that
counterclockwise torque are positive and clockwise torques are
negative, the torques of F1 and F2 about O are 1 = +F1l1 , 2 = F2l2. Figure 10-3 shows a force F applied at a position vector r
with respect to the chosen point O. There are several ways to
calculate the torque of this force. One is to find the lever arm l and
use  = Fl. Or we can deter-mine the angle , between the vectors
r and F; the lever arm is r sin, so  = rF sin. A third method is to
represent F in terms of a radial component Frad along the direction
of r and a tangential component Ftan at right angles, perpendicular
to r. We call this tangential component because if the body rotates,
the point where the force acts moves in a circle and this
component is tangent to that circle.) Then Ftan = Fsin, and  = r
(Fsin) = Ftan r. The component Frad has no torque with respect to
O because its lever arm with respect to that point is zero.
 
  r F

10-3 Torque And Angular Acceleration For A Rigid
Body
We are now ready to develop the fundamental
relation for the rotational dynamics of a rigid body.
We will show that the angular acceleration of a
rotating rigid body is directly proportional to the
sum of the torque components along the axis of
rotation. The proportionality factor is the moment of
F1,y
inertia.
F1,tan
F1,rad
r1
m1
r
o
Newton’s second law for the tangential component is
F1, tan = m1a1,tan
(10-4)
We can express the tangential acceleration of the first
particle in terms of the angular acceleration , using the
Eq.(9-4); a1,tan = r1 . Using this relation and multiplying both
sides of Eq.(10-4) by r1, we obtain:
F1,tan r1 = m1 r12 
(10-5)
Form Eq.(10-2), F1,tanr1 is just the magnitude of the torque 1
of the net force with respect to the rotation axis. Neither of
the components F1,rad or F1y contributes to the torque about
the y-axis, since neither tends to change the particle’ rotation
about that axis. So 1 = F1,tanr1 is the total torque acting on the
particle with respect to the rotation axis. Also, m1r12 is I1, the
moment of inertia of the particle about the rotation axis. With
this in mind, we rewrite Eq. (10-5) as
1 = I1  = m1 r12.
We write an equation like this for every particle in the body
and then add all these equations:
1   2   3    m r   m r   m r  
2
11
2
2 2
2
3 3
2



m
r
 i  i i 
  I
(Rotational analog of Newton’s second law for a rigid body)
10-5 Work And Power In Rotational Motion
Suppose a tangential force Ftan acts at the of a pivoted disk.
The disk rotates through an infinitesimal angle d about a
fixed axis during an infinitesimal time interval dt (Fig, 10-19b).
ds
R
o
Ftan
d
R
Ftan
The work done by the force Ftan while a point on the rim
moves a distance ds is dW = Ftands. If d is measured in
radians, then ds = Rd and
dW = Ftan Rd. Now FtanR is the torque  due to the force Ftan,
so dW =  d.
(10-22)
The total work W done by the torque during an angular
displacement from 1 to 2 is
2
W   d
1
(Work done by a torque).
If the torque is constant while the angle changes by a finite
amount  = 2- 1, then
W    2  1   
(Work done by a constant torque).
When a torque does work on a rotating rigid body, the kinetic
energy changes by an amount equal to the work done.
d
d
d  ( I )d  I
d  I
d  Id
dt
dt
Wtot  
2
1
1 2 1 2
Id  I2  I1
2
2
dW
d

dt
dt
dW/dt is the rate of doing work, or power P, and d/dt is
angular velocity , so P = .
10-6 Angular Momentum
Every rotational quantity that we have encountered in
Chapter 9 and 10 is the analog of some quantity in the
translational motion of a particle. The analog of momentum of
a particle is angular momentum, a vector quantity denoted as
L. Its relation to momentum P is exactly the same as the
relation of torque to force,  = r  F. For a particle with
constant mass m, velocity v, momentum P = mv, and position
vector r relative to the origin O of an inertial frame, we define
angular momentum L as
   

L  r  p  r  mv
(Angular momentum of a particle).
P=mv
y

y

mvsin
vi=ri


o
mi
ri
L=rsin
o
x
x
L
z
z
Direction of Li
In Fig. 10-20 a particle moves in the xy-plane; its position
vector r and momentum p = mv are shown.
L  mvr sin   mvl
(10-28)
Where l is the perpendicular distance from the line of v to o.
when a net force F acts on a particle, its velocity and
momentum change, so its angular momentum may also
change. We can show that the rate of change of angular
momentum is equal to the torque of the net force.



  



dL  dr
dv  
   mv    r  m   v  mv   r  ma 
dt  dt
dt 
 

dL   
(for a particle acted on by net force)
 r  F 
dt
The rate of change of angular momentum of a particle equals
the net force acting on it.
We can use to find the total angular momentum of a rigid
body rotating about the z-axis with angular speed .
Li  mi (ri )ri  mi ri 2
(10-30)
The total angular momentum of the slice of the body lying in
the xy-plane is the sum Li of the angular momenta Li of the
particle. Summing Eq. (10-30), we have
L   Li  ( mi ri2 )  I
Where I is the moment of inertia of the slice about the z-axis.
The angular velocity vector  also lies along the rotation axis,
as we discussed at the end of Section 9-2.hence for a rigid
body rotating around an axis of symmetry, L and  are in the
same direction. So we have the vector relationship


L  I
(for a rigid body rotating around a symmetry axis)

 dL
  dt
(for any system of particle).
(10-32)
10-7 Conservation Of Angular Momentum
When a system has several parts, the internal forces that the
parts exert on each other cause changes in the angular
momenta of the parts, but the total angular momentum
doesn’t change. Here’s an example. Consider two bodies A
and B that interact with each other but not with anything else.
Suppose body A exerts a force FAonB on body B; the
corresponding torque is AonB . According to Eq. (10-32), this
torque is equal to the rate of change of angular momentum
of B;


 AonB
dLB

dt
At the same time, body B exerts a force FBonA on body A, with
a corresponding torque

dL

 BonA  A
dt
Form Newton’s third law, FBonA = - FAonB. Furthermore, if the
forces act along the same line, their lever arms with respect
to the chosen axis are equal. Thus the torques of these two
forces are equal opposite, and BonA= -AonB. So if we add the
two previous equations, we find



dL
dLA dLB
 0 (zero net external torque) 10-34

0
dt
dt
dt
It also forms the basis for the principle of conservation of
angular momentum.
When the net external torque acting on a system is zero, the
total angular momentum of the system is constant
(conserved)
10-8 Gyroscopes And Precession
In all of the situations we’ve looked at so far in this chapter,
the axis of rotation either has stayed fixed or has moved and
kept the same direction. But a variety of new physical
phenomena, some quite unexpected, can occur when the axis
of rotation can change direction.
Periodic Motion
 13-1 Introduction
 The motion repeats itself over and over. We
call this periodic motion or oscillation.
Understanding periodic motion will be
essential for our later study of waves, sound,
alternating electric currents, and light.
 A body that undergoes periodic motion
always has a stable equilibrium position.
When it is moved away from this position
and released, a force or torque comes into
play to pull it back toward equilibrium.
13-2 The Causes of Oscillation
It’s simplest to define our coordinate system so that the origin O is
at the equilibrium position, where the spring is neither stretched
nor compressed. Then x is the x-component of the displacement
of the body from equilibrium and is also the change in length of
the spring. The x-component of acceleration a is given by a = F/m.
Whenever the body is displaced from its equilibrium position,
the spring force tends to restore it to the equilibrium position. We
call a force with this character a restoring force. Oscillation can
occur only when there is a restoring force tending to return the
system to equilibrium.
The amplitude of the motion, denoted by A, is the maximum
magnitude of displacement from equilibrium; that is, the maximum
value of x. it is always positive.
The period, T, is the time for one cycle. It is always positive.
The frequency, f, is the number of cycles in a unit of time. It is
always positive.
The angular frequency, ω, is 2πtimes the frequency: ω=2πf.
13-3 Simple Harmonic Motion
The very simplest kind of oscillation occurs when the restoring
force F is directly proportional to the displacement from
equilibrium x. The x-component of force the spring exerts on the
body is the negative of this, so the x-component of force F on the
body is
F  kx
(restoring force exerted by an ideal spring) 13-3
This equation gives the correct magnitude and sign of the force,
whether x is positive, negative, or zero.
When the restoring force is directly proportional to the displacement from equilibrium, as given by 13-3, the oscillation is called
simple harmonic motions abbreviated SHM. The acceleration
d 2x
k
a 2  x
m
dt
(simple harmonic motion) 13- 4
The minus sign means the acceleration and displacement
always have opposite signs. A body that undergoes simple
harmonic motion is called a harmonic oscillator.
Equation of Simple Harmonic Motion
t
A
 t+

A
·x
o
x = A cos( t +  )
t=0
x
参考圆
(circle of reference)
To explore the properties of simple harmonic motion, we
must express the displacement x of the oscillating body as
a function of time, x(t).
The circle in which the ball moves so that its projection
matches the motion of the oscillating body is called the circle
of reference; we will call the point Q the reference point. We
take the circle of reference to lie in the xy-plane, with the
origin O at the center of the circle.
The x-component of the phasor at time t is just the xcoordinate of the point Q:
13-5
x  A cos 
This is also the x-coordinate of the shadow P, which is the
projection Q onto the x-axis. Hence the acceleration of the
shadow P along the x-axis is equal to the x-component of the
acceleration vector of the reference point Q.
aQ   2 A
a  aQ cos    2 A cos 
13-6
13-7
k
 
m
2
k

m
or
(Simple harmonic motion)

1
f 

2 2
k
m
T
1 2
m

 2
f

k
Displacement, Velocity, And Acceleration In SHM
We still need to find the displacement x as a function of time
for a harmonic oscillator. Equation (13-4) for a body in simple
harmonic motion along the x-axis is identical to Eq. (13-8) for
the x-coordinate of the reference point in uniform circular
motion with constant angular speed   k / m .
x  A cos(t   )
(displacement in SHM)
13-13
In simple harmonic motion the position is a periodic
sinusoidal function of time.
The period T is the time for one complete cycle of oscillation.
Thus if we start at time t= 0, the time T to complete one cycle
is given by
T 
k
T  2
m
or
T  2
m
k
The constant  in Eq. (13-13) is called the phase angle. It tells
us at what point in the cycle the motion was at t = 0.
We denote the position at t = 0 by x0. putting t = 0 and x = x0 in
equation 13-13, we get
x0  A cos 
dx
v
  A sin(t   )
dt
dv dx 2
a
 2   2 A cos(t   )
dt dt
13-14
13-15
13-16
13-4 Energy In Simple Harmonic Motion
We can learn even more about simple harmonic motion by
using energy consideration. The kinetic energy of the body is
K = 1/2mv2, and the potential energy of the spring is U =
1/2kx2. There are no nonconservation forces that do work, so
the total mechanical energy E = K + U is conserved;
1 2 1 2
E  mv  kx  cons tan t
2
2
1 2 1 2 1 2
E  mv  kx  kA  cons tan t
2
2
2
Total mechanical energy in SHM
13-5 Applications Of Simple Harmonic Motion
Vertical SHM
Suppose we hang a spring with force constant k and suspend
from it a body with mass m.
X=0
l
F  kl
mg
kl  mg
Fnet  k l  x    mg   kx
Angular SHM
   I


I
or
d 2

 
2
dt
I
and
1
f 
2

I
13-6 The Simple Pendulum
A simple pendulum is an idealized model consisting of a
point mass suspended by a massless, unstretchable string.
When the point mass is pulled to one side of its straight
down equilibrium position and released, it oscillates about
The equilibrium position.

L
T
x
mgsin
m

mgcos
mg
F  mg sin 
x
F   mg  mg
L
F 
mg
x
L
k
mg / L



m
m

1
f 

2 2
g
L
g
L
T
2


1
L
 2
f
g
13-7 The Physical Pendulum
A physical pendulum is any real pendulum, using a body of
finite size, as contrasted to the idealized model of the simple
pendulum with all the mass concentrated at a single point.
  (mg )( d sin  )
O
θ
  (mgd )
d
dsinθ
mgsinθ
d 2
 (mgd )  I  I 2
dt
mgcosθ
mg
d 2
mgd


2
I
dt
13-8 Damped Oscillations
Red-world systems always have some dissipative forces,
however, and oscillations do die out with time unless we
provide some means for replacing the dissipated mechanical
energy. The decrease in amplitude caused by dissipative
forces is called damping, and the corresponding motion is
called damped oscillation.
F  kx  bv
dx
d 2x
 kx  b
m 2
dt
dt
 
k
b2

m 4m 2
b  2 km
 kx  bv  ma
x  Ae (b / 2m)t cos( t   )
k
b2

0
2
m 4m
x  C1e  a1t  C 2 e  a2t
13-9 Forced Oscillations, Resonance, And Chaos
The Tacoma Narrows
Bridge collapsed four
months and six days
after it was opened for
traffic.
19
Mechanical Waves
19-1 Introduction
Waves can occur whenever a system is disturbed from its
equilibrium position and when the disturbance can travel or
propagate from one region of the system to another.
This chapter and the next two are about mechanical waves,
waves that travel within some material called a medium. We’ll
begin by deriving the basic equations for describing waves,
including the important special case of periodic waves in
which the pattern of the wave repeats itself as the wave
propagates.
Not all waves are mechanical in nature. Another broad
class is electromagnetic waves, including light, radio waves,
infrared and ultraviolet radiation, x-rays, and gamma rays. No
medium is needed for electromagnitude waves; they can
travel through empty space.
19-2 Type Of Mechanical Waves
A mechanical wave is a disturbance that travels through some
material or substance called the medium for the wave.
Because the displacements of the medium are perpendicular
or transverse to the direction of travel of the wave along the
medium, this is called transverse wave.
If the motions of the particles of the medium are back and
forth along the same direction that the wave travels, we call
this a longitudinal wave.
t 0
t 0
T
4
T
t 
2
t 
t
3
T
4
5
T
4
T
4
t 
T
2
t
3
T
4
t T
t T
t
t 
5
T
4
3
t T
2
t

Transverse wave

longitudinal wave.
波面
波线
y
x
球面波
柱面波
波线
Ray
These examples have three things in common. First, in each
case the disturbance travels propagates with a definite speed
through the medium. This speed is called the speed of
propagation, or simply the wave speed. It is determined in
each case by the mechanical properties of the medium. We
will use the symbol v for wave speed. Second, the medium
itself does not travel through space; its individual particles
undergo back-and-forth or up-and-down motions around their
equilibrium positions. The overall pattern of the wave
disturbance is what travels. Third, to set any of these
systems into motion, we have put in energy by doing
mechanical work on the system. The wave motion transports
this energy from one region of the medium to another. Waves
transport energy, but not matter, from one region to another.
19-3 Periodic Waves
In particular, suppose we move the string up and down in
simple harmonic motion with amplitude A, frequency f,
angular frequency  = 2f, and period T=1/f = 2 / . When a
sinusoidal wave passes through a medium, every particle in
the medium undergoes simple harmonic motion with the same
frequency.
For a periodic wave, the shape of the string at any instant
is a repeating pattern. The length of one complete wave
pattern is the distance from one crest to the next, or from one
trough to the next, or from any point to the corresponding
point on the next repetition of the wave shape. We call this
distance the wavelength of the wave, denoted by . The wave
pattern travels with constant speed v and advances a
distance of one wavelength  in a time interval of one period
T. so the wave speed v is given by v = /T, or , because f = 1/T,
v = f (periodic wave).
19-4 Mathematical Description Of A Wave
For this description we need the concept of a wave
function, a function that describes the position of any particle
in the medium at any time. We will concentrate on sinusoidal
waves, in which each particle undergoes simple harmonic
motion about its equilibrium position.
Wave Function For A Sinusoidal Wave
Suppose that the displacement of a particle at the left end of
string (x = 0), where the wave originates, is given by
y ( x  0, t )  A sin t  A sin 2ft
19-2
The wave disturbance travels from x = 0 to some point x to
the right of the origin in an amount of time given by x/v,
where v is the wave speed. So the motion of point x at time t
is the same as the motion of point x = 0 at the earlier time t –
x/v. Hence we can find the displacement of point x at time t by
simply replacing t in Equation (19-2) by (t – x/v).
 x
 x
y ( x, t )  A sin   t    A sin 2f  t  
 v
 v
19-3
 t x
y ( x, t )  A sin 2   
T  
19-4
We get another convenient form of the wave function if we
define a quantity k, called the wave number:
k
2

  vk
(wave number)
(19-5)
(periodic wave)
(19-6)
y ( x, t )  A sin( t  kx)
(sinusoidal wave moving in +x-direction)
(19-7)
y
A
x
Wavelength
y
A
t
A
Period
t1+Δt时刻的波形

u
t1时刻的波形
O
x
x1
x1  x
Particle Velocity And Acceleration In A sinusoidal Wave
From the wave function we can get an expression for the
transverse velocity of any particle in a transverse wave. We
call this vy to distinguish it from the wave propagation speed
v. To find the transverse velocity vy at a particular point x, we
take the derivative of the wave function y(x,t) with respect to t,
keeping x constant. If the wave function is
y ( x, t )  A sin(t  kx)
then
y ( x, t )
v y ( x, t ) 
 A cos(t  kx 
t
(19-9)
The acceleration of any particle is the second partial
derivative of y(x,t) with respect to t:
 2 y( x, t )
2
2
a y ( x, t ) 



A
sin(

t

kx
)



y( x, t )
2
t
19-10
We can also compute partial derivatives of y(x,t) with respect
to x, holding t constant.
 2 y( x, t )
2
2


k
A
sin(

t

kx
)


k
y( x, t )
2
 x
(19-11)
Form Eqs.(19-10) and (19-11) and the relation  = vk we see
that
 2 y( x, t ) / t 2  2
2


v
 2 y( x, t ) / x 2 k 2
 2 y ( x, t ) 1  2 y ( x, t )
 2
2
x
v
t 2
(wave equation)
(19-12)
Equation (19-12), called the wave equation, is one of the most
important equations in all of physics.
20 Wave Interference and Normal Modes
20-1 Introduction
When a wave strikes the boundaries of its medium, all or part
of the wave is reflected. When you yell at a building wall or a
cliff face some distance away, the sound wave is reflected
from the rigid surface, and an echo comes back. When you
flip the end of a rope whose far end is tied to a rigid support,
a pulse travels the length of the rope and is reflected overlap
in the same region of the medium. This overlapping of waves
is called interference.
20-2 Boundary Conditions For A String And The Principle Of
Superposition
As a simple example of wave reflections and the role of the
boundary of a wave medium, let’s look again at transverse
waves on a stretched string. What happens when a wave
pulse or a sinusoidal wave arrives at the end of the string?
L
2
L
L  3
2
If the end is fastened to a rigid support, it is a fixed end that
cannot move. The arriving wave exerts a force on the
support; the reaction to this force, exerted by the support on
the string, “kicks back” on the string and sets up a reflected
pulse or wave traveling in the reverse direction. The
conditions at the end of the string, such as a rigid support or
the complete absence of transverse force, are called
boundary conditions.
The formation of the reflected pulse is similar to the overlap
of two pulses traveling in opposite directions.
The principle of superposition
Combining the displacements of the separate pulse at each
point to obtain the actual displacement is an example of the
principle of superposition; when two waves overlap, the
actual displacement of any time is obtained by adding the
displacement the point would have if only the first wave were
present and the displacement it would have if only the
second wave were present.
describes the resulting motion in this situation is obtained by
adding the two wave functions for the two separate waves.
20-3 Standing Waves On A String
We have talked about the reflection of a wave pulse on a
string when it arrives at a boundary point. We will again
approach the problem by considering the superposition of
two waves propagating through the string, one representing
the original or incident wave and the other representing the
wave reflected at the fixed end. The general term interference
is used to describe the result of two or more waves passing
through the same region at the same time. Here, instead, the
wave pattern in the same position along the string, the
amplitude fluctuates. There are particular points called nodes
that never move at all. Midway between the nodes is points
called antinodes, where the amplitude of motion is greatest.
Because the wave pattern doesn’t appear to be moving in
either direction along the string, it is called a standing wave.
(To emphasize the difference, a wave that does move along
the string is called a traveling wave). The principle of
superposition explains how the incident and reflected wave
combine to form a standing wave.
At a node the displacements of the two waves in red and
blue are always equal and opposite and cancel each other out.
This cancellation is called destructive interference. Midway
between the nodes are the points of the greatest amplitude or
antinodes, marked A. At the antinodes the displacement of
the two waves in red and blue are always identical, giving a
large resultant displacement; this phenomenon is called
constructive interference. We can see from the figure that the
distance between successive nodes or between successive
antinodes is one half-wavelength, or 2. We can derive a
wave function for the standing wave of Fig. 20-6 by adding
the wave functions y1(x, t) and y2(x, t) for two waves with
equal amplitude, period, and wavelength traveling in opposite
directions. Here y1(x, t) represents an incident wave traveling
to the left along the +x-axis, arriving at the point x = 0 and
being reflected; y2(x, t) represents the reflected wave
traveling to the right from x = 0. We noted in Section 20-2 that
the wave reflected from a fixed end of a string is inverted, so
we give a negative sign to one of the waves;
y1 x, t   A sin(t  kx)
y2 x, t    A sint  kx
(traveling to the left),
(traveling to the right).
yx, t   y1 x, t   y2 x, t   Asint  kx  sint  kx
yx, t   y1 x, t   y2 x, t   2 A sin kx cos t
(20-1)
The standing wave amplitude Asw is twice the amplitude A of
the original traveling waves:
Asw = 2A
Equation (20-1) has two factors; a function of x and a function
of t. The factor 2Asinkx shows that at each instant the shape
of the string is a sine curve.
But unlike a wave traveling along a string, the wave shape
stays in the same position, oscillating up and down as
described by cos t factor. This is in contrast to the phase
differences between oscillations of adjacent points that we
see with a wave traveling in one direction. We can use Eq.
(20-1) to find the positions of the nodes; these are the points
for which sin kx = 0, so the displacement is always zero. This
occurs when kx =0, , 2, 3, , or, using k = 2,
x  0,
 2 3
k
,
,

k k
 2 3
 0, , , 
2 2 2
In particular, there is a node at x = 0, as there should be,
since this point is a fixed end of the string.
A standing wave, unlike a traveling wave, does not transfer
energy from one end to the other. The two waves that form it
would individually carry equal amounts of power in opposite
directions. There is a local flow of energy from each node to
the adjacent antinodes and back, but the average rate of
energy transfer is zero at every point.