Download ƒ A S ƒ ƒ B

CHAPTER
1
SUMMARY
Physical quantities and units: Three fundamental physical quantities are mass, length, and time.
The corresponding basic SI units are the kilogram, the meter, and the second. Derived units for
other physical quantities are products or quotients of the basic units. Equations must be dimensionally consistent; two terms can be added only when they have the same units. (See Examples 1.1
and 1.2.)
Significant figures: The accuracy of a measurement can be indicated by the number of significant
figures or by a stated uncertainty. The result of a calculation usually has no more significant figures
than the input data. When only crude estimates are available for input data, we can often make useful order-of-magnitude estimates. (See Examples 1.3 and 1.4.)
Significant figures in magenta
C
0.424 m
5
5 3.14
2r
2(0.06750 m)
p5
123.62 1 8.9 5 132.5
Scalars, vectors, and vector addition: Scalar quantities are numbers and combine with the usual
rules of arithmetic. Vector quantities have direction as well as magnitude and combine according
to the rules of vector addition. The negative of a vector has the same magnitude but points in the
opposite direction. (See Example 1.5.)
Vector components and vector addition: Vector addition can be carried out
using
components
of vectors.
S
S
S
The x-componentSof R !
is
the
sum
of the
A
"
B
S
x-components of A and B, and likewise for the y- and
z-components. (See Examples 1.6–1.8.)
Rx = Ax + Bx
Ry = Ay + By
Rz = Az + Bz
S
S
A
1
S
A1B
S
B
S
A
5
S
B
y
(1.10)
S
R
By
Ry
S
B
Ay
S
A
Ax Bx
O
x
Rx
Unit vectors: Unit vectors describe directions in space.
A unit vector has a magnitude of 1, with no units. The
unit vectors Nı, n≥ , and kN , aligned with the x-, y-, and
z-axes of a rectangular coordinate system, are especially useful. (See Example 1.9.)
Scalar product:
The scalar product C = A # B of two
S
S
vectors A and B is a scalar quantity.
It
can be expressed
S
S
in terms of the magnitudes of A and B and the angle f
between
the
two vectors, or in terms of the components
S
S
of
and
The
scalar product is commutative;
A
.
B
S S
S S
A # B ! B # A. The scalar product of two perpendicular
vectors is zero. (See Examples 1.10 and 1.11.)
S
S
S
S
S
Vector product:
The vector productS C ! A : B of two
S
S
vectors
A and B is another vector C. The
magnitude
of
S
S
S
S
A : B depends on the magnitudes of A and B and
the
angle
f between the two vectors. The direction of
S
S
A : B is perpendicular to the plane of the two vectors
being multiplied,
as given
by the right-hand rule. The
S
S
S
components of C ! AS : B S
can be expressed in terms
of the components
of
and
vector product is not
A
. The
B
S
S
S
S
commutative; A : B ! # B : A. The vector product
of two parallel or antiparallel vectors is zero. (See
Example 1.12.)
26
S
A ! Ax nı " Ay n≥ " Az kN
(1.16)
y
Ay j^
S
A 5 Ax i^ 1 Ay j^
j^
O
A # B = AB cos f = ƒ A ƒ ƒ B ƒ cos f (1.18)
S
S
S
S
A # B = Ax Bx + Ay By + Az Bz
S
S
x
Ax i^
i^
Scalar product A # B 5 AB cos f
S
(1.21)
S
S
B
f
S
A
C = AB sin f
(1.22)
Cx = Ay Bz - Az By
Cy = Az Bx - Ax Bz
Cz = Ax By - Ay Bx
S
S
S
A $ B is perpendicular
S
S
to the plane of A and B.
S
A$B
(1.27)
S
B
S
A
f
S
S
(Magnitude of A $ B) 5 AB sin f
SUMMARY
dx
¢x
=
¢t
dt
(2.3)
vx = lim
¢tS 0
x
p2
x2
av
p1
x1
v2x - v1x
¢vx
=
t2 - t1
¢t
(2.4)
¢vx
dvx
=
¢t
dt
(2.5)
ax = lim
¢tS 0
p2
-x
e5
op
Sl
t1
t 5 Dt
vx 2 = v0x2 + 2ax 1x - x 02
(2.13)
t 5 2Dt
(2.14)
t 5 3Dt
x - x0 = a
v0x + vx
bt
2
t 5 4Dt
x
a
v
x
0
v
a
x
0
v
0
a
x
ay 5 2g
5 29.80 m s2
/
vx = v0x +
x = x0 +
L0
L0
t
ax dt
(2.17)
t
vx dt
(2.18)
ax
aav-x
O
58
a
0
Freely falling bodies: Free fall is a case of motion with
constant acceleration. The magnitude of the acceleration
due to gravity is a positive quantity, g. The acceleration
of a body in free fall is always downward. (See Examples 2.6–2.8.)
Straight-line motion with varying acceleration: When the
acceleration is not constant but is a known function of
time, we can find the velocity and position as functions
of time by integrating the acceleration function. (See
Example 2.9.)
x
v
(2.12)
t
t2
a
0
x = x 0 + v0x t + 12 ax t 2
ax
Dt 5 t2 2 t1
v
t50
a av
e5
Slop
p1
Constant x-acceleration only:
(2.8)
t
t2
!t 5 t2 2 t1
v2x
v1x
vx = v0x + ax t
vx
vx
O
Straight-line motion with constant acceleration: When
the x-acceleration is constant, four equations relate the
position x and the x-velocity vx at any time t to the
initial position x 0 , the initial x-velocity v0x (both
measured at time t = 0), and the x-acceleration ax .
(See Examples 2.4 and 2.5.)
pe 5
Slo
t1
O
aav-x =
!x 5 x2 2 x1
(2.2)
-x
x2 - x1
¢x
=
t2 - t1
¢t
Dvx 5 v2x 2 v1x
Average and instantaneous x-acceleration: The average
x-acceleration aav-x during a time interval ¢t is equal
to the change in velocity ¢vx = v2x - v1x during
that time interval divided by ¢t. The instantaneous
x-acceleration ax is the limit of aav-x as ¢t goes to zero,
or the derivative of vx with respect to t. (See Examples
2.2 and 2.3.)
vav-x =
v
Straight-line motion, average and instantaneous
x-velocity: When a particle moves along a straight line,
we describe its position with respect to an origin O by
means of a coordinate such as x. The particle’s average
x-velocity vav-x during a time interval ¢t = t 2 - t 1 is
equal to its displacement ¢x = x 2 - x 1 divided by ¢t.
The instantaneous x-velocity vx at any time t is equal to
the average x-velocity for the time interval from t to
t + ¢t in the limit that ¢t goes to zero. Equivalently, vx
is the derivative of the position function with respect to
time. (See Example 2.1.)
e5
2
Sl
op
CHAPTER
t1
Dt
t2
t
CHAPTER
3
SUMMARY
Position, velocity, and acceleration vectors: The position
S
vector r of a point P in space is the vector from the
origin to P. Its components are the coordinates x, y, and z.
S
The average velocity vector vav during the time
S
interval ¢t is the displacement ¢ r (the change in the
S
position vector r ) divided by ¢t. The instantaneous
S
S
velocity vector v is the time derivative of r , and its
components are the time derivatives of x, y, and z. The
S
instantaneous speed is the magnitude of v. The velocity
S
v of a particle is always tangent to the particle’s path.
(See Example 3.1.)
S
The average acceleration vector a av during the time
S
interval ¢t equals ¢v (the change in the velocity vector
S
v2 divided by ¢t. The instantaneous acceleration vector
S
S
a is the time derivative of v, and its components are the
time derivatives of vx , vy , and vz . (See Example 3.2.)
The component of acceleration parallel to the
direction of the instantaneous velocity affects the speed,
S
S
while the component of a perpendicular to v affects the
direction of motion. (See Examples 3.3 and 3.4.)
r ! x Nı # y ≥N # z kN
S
S
S
r2 " r1
¢r
S
!
vav !
t2 - t1
¢t
S
S
¢r
dr
S
v ! lim
!
¢tS0 ¢t
dt
dy
dx
dz
vy =
vz =
vx =
dt
dt
dt
S
S
S
v2 " v1
¢v
S
!
a av !
t2 - t1
¢t
S
a ! lim
S
¢t 0
ax =
ay =
az =
y1
(3.3)
S
vav 5 Dr
Dt
S
(3.2)
S
Dr
S
r1
Dy
y2
(3.4)
S
O
(3.8)
r2
x1
x
x2
Dx
S
S
¢v
dv
!
¢t
dt
S
y
(3.1)
dvx
dt
dvy
(3.9)
S
v2
y
S
v1
S
(3.10)
dt
dvz
S
v1
dt
S
v2
x
O
Projectile motion: In projectile motion with no air
resistance, ax = 0 and ay = - g. The coordinates and
velocity components are simple functions of time, and
the shape of the path is always a parabola. We usually
choose the origin to be at the initial position of the
projectile. (See Examples 3.5–3.10.)
Uniform and nonuniform circular motion: When a particle
moves in a circular path of radius R with constant speed v
S
(uniform circular motion), its acceleration a is directed
S
toward the center of the circle and perpendicular to v.
The magnitude arad of the acceleration can be expressed
in terms of v and R or in terms of R and the period T
(the time for one revolution), where v = 2pR/T. (See
Examples 3.11 and 3.12.)
If the speed is not constant in circular motion
(nonuniform circular motion), there is still a radial
S
component of a given by Eq. (3.28) or (3.30), but there
S
is also a component of a parallel (tangential) to the
path. This tangential component is equal to the rate of
change of speed, dv/dt.
Relative velocity: When a body P moves relative to a
body (or reference frame) B, and B moves relative to A,
S
we denote the velocity of P relative to B by vP>B , the
S
velocity of P relative to A by vP>A , and the velocity of B
S
relative to A by vB>A . If these velocities are all along the
same line, their components along that line are related
by Eq. (3.33). More generally, these velocities are
related by Eq. (3.36). (See Examples 3.13–3.15.)
x = 1v0 cos a02t
(3.20)
y = 1v0 sin a02t - 12 gt2
(3.21)
vx = v0 cos a0
(3.22)
vy = v0 sin a0 - gt
(3.23)
arad =
arad =
S
y
vy
v
vy
v
vx
S
v
vy
ay 5 2g
vx
x
O
S
v2
R
4p2R
v
(3.28)
(3.30)
T2
S
v
S
arad
S
arad
S
arad
S
v
S
S
arad
S
S
v
vP/A-x = vP/B-x + vB/A-x
(relative velocity along a line)
S
S
v
vx
S
arad
S
(relative velocity in space)
S
S
v
S
vB/A
(3.33)
S
vP/A
S
vP>A ! vP>B # vB>A
v
arad
(3.36)
S
S
S
vP/A 5 vP/B 1 vB/A
S
vP/B
P (plane)
B (moving air)
A (ground
observer)
94
S
aav 5 Dv
Dt
S
Dv
CHAPTER
4
SUMMARY
Force as a vector: Force is a quantitative measure
of the interaction between two bodies. It is a vector
quantity. When several forces act on a body, the
effect on its motion is the same as when a single
force, equal to the vector sum (resultant) of the
forces, acts on the body. (See Example 4.1.)
The net force on a body and Newton’s first law:
Newton’s first law states that when the vector sum
of all forces acting on a body (the net force) is
zero, the body is in equilibrium and has zero
acceleration. If the body is initially at rest, it
remains at rest; if it is initially in motion, it
continues to move with constant velocity. This
law is valid only in inertial frames of reference.
(See Examples 4.2 and 4.3.)
Mass, acceleration, and Newton’s second law: The
inertial properties of a body are characterized by its
mass. The acceleration of a body under the action
of a given set of forces is directly proportional to
the vector sum of the forces (the net force) and
inversely proportional to the mass of the body. This
relationship is Newton’s second law. Like Newton’s
first law, this law is valid only in inertial frames of
reference. The unit of force is defined in terms of
the units of mass and acceleration. In SI units, the
unit of force is the newton (N), equal to 1 kg # m>s2.
(See Examples 4.4 and 4.5.)
S
S
S
S
R ! F1 # F2 # F3 #
Á
! a F (4.1)
S
S
aF ! 0
S
(4.3)
S
v 5 constant
S
S
F1
S
F2 5 2F1
S
SF 5 0
a F ! ma
S
S
(4.7)
a Fx = max
SF
Weight: The weight w of a body is the gravitational
force exerted on it by the earth. Weight is a vector
quantity. The magnitude of the weight of a body
at any specific location is equal to the product
of its mass m and the magnitude of the acceleration
due to gravity g at that location. While the weight
of a body depends on its location, the mass is
independent of location. (See Examples 4.6
and 4.7.)
w = mg
(4.9)
Newton’s third law and action–reaction pairs:
Newton’s third law states that when two bodies
interact, they exert forces on each other that at each
instant are equal in magnitude and opposite in
direction. These forces are called action and reaction forces. Each of these two forces acts on only
one of the two bodies; they never act on the same
body. (See Examples 4.8–4.11.)
FA on B ! " FB on A
S
/
S
a 5 SF m
S
a Fz = maz
Mass m
S
w 5 mg
S
F1
Mass m
S
S
S
S
F2
(4.8)
126
R
Fx
a Fy = may
S
S
S
Fy
S
g
B
(4.11)
S
FA on B
A
S
FB on A
CHAPTER
5
SUMMARY
Using Newton’s first law: When a body is in equilibrium
in an inertial frame of reference—that is, either at rest or
moving with constant velocity—the vector sum of
forces acting on it must be zero (Newton’s first law).
Free-body diagrams are essential in identifying the
forces that act on the body being considered.
Newton’s third law (action and reaction) is also frequently needed in equilibrium problems. The two forces
in an action–reaction pair never act on the same body.
(See Examples 5.1–5.5.)
The normal force exerted on a body by a surface is not
always equal to the body’s weight. (See Example 5.3.)
Using Newton’s second law: If the vector sum of forces
on a body is not zero, the body accelerates. The acceleration is related to the net force by Newton’s second law.
Just as for equilibrium problems, free-body diagrams
are essential for solving problems involving Newton’s
second law, and the normal force exerted on a body is
not always equal to its weight. (See Examples
5.6–5.12.)
aF ! 0
S
(vector form)
a Fx = 0
a Fy = 0
y
(5.1)
n
n
(component form)
(5.2)
T
a
w sin a
T
w cos a
x
a
w
w
Vector form:
y
a F ! ma
S
S
Component form:
a Fx = max
(5.3)
a
a Fy = may
n
m
(5.4)
Forces in circular motion: In uniform circular motion,
the acceleration vector is directed toward the center of
the circle.
The motion is governed by Newton’s second
S
S
law, gF ! ma . (See Examples 5.19–5.23.)
Acceleration in uniform circular motion:
ƒk = mk n
(5.5)
Magnitude of static friction force:
ƒs … ms n
w
Static
friction
f
1 fs 2max
Kinetic
friction
fk
(5.6)
T
O
arad
v2
4p2R
=
=
R
T2
x
a
w
Magnitude of kinetic friction force:
w sin a
T
w cos a
a
Friction and fluid resistance: The contact force between
two bodies can always be represented in terms of a norS
mal force n perpendicular
to the surface of contact and a
S
friction force ƒ parallel to the surface.
When a body is sliding over the surface, the friction
force is called kinetic friction. Its magnitude ƒk is
approximately equal to the normal force magnitude n
multiplied by the coefficient of kinetic friction mk .
When a body is not moving relative to a surface, the
friction force is called static friction. The maximum possible static friction force is approximately equal to the
magnitude n of the normal force multiplied by the coefficient of static friction ms . The actual static friction
force may be anything from zero to this maximum
value, depending on the situation. Usually ms is greater
than mk for a given pair of surfaces in contact. (See
Examples 5.13–5.17.)
Rolling friction is similar to kinetic friction, but the
force of fluid resistance depends on the speed of an
object through a fluid. (See Example 5.18.)
n
ax
T
S
v
S
(5.14), (5.16)
SF
S
v
S
arad
S
arad
S
SF
S
SF
S
arad
S
v
161
CHAPTER
6
SUMMARY
S
#
S
S
Work done by a force: When a constant force F acts on
S
a particle that undergoes a straight-line displacement s ,
the work done by the force on the particle is defined to
S
S
be the scalar product of F and s . The unit of work in
SI units is 1 joule = 1 newton-meter 11 J = 1 N # m2.
Work is a scalar quantity; it can be positive or negative,
but it has no direction in space. (See Examples 6.1
and 6.2.)
W = F s = Fs cos f
Kinetic energy: The kinetic energy K of a particle equals
the amount of work required to accelerate the particle
from rest to speed v. It is also equal to the amount of
work the particle can do in the process of being brought
to rest. Kinetic energy is a scalar that has no direction in
space; it is always positive or zero. Its units are the same
as the units of work: 1 J = 1 N # m = 1 kg # m2>s2.
K = 12 mv2
The work–energy theorem: When forces act on a particle while it undergoes a displacement, the particle’s
kinetic energy changes by an amount equal to the total
work done on the particle by all the forces. This relationship, called the work–energy theorem, is valid
whether the forces are constant or varying and whether
the particle moves along a straight or curved path. It is
applicable only to bodies that can be treated as particles.
(See Examples 6.3–6.5.)
Wtot = K2 - K1 = ¢K
Work done by a varying force or on a curved path: When
a force varies during a straight-line displacement, the
work done by the force is given by an integral, Eq. (6.7).
(See Examples 6.6 and 6.7.) When a particleSfollows a
curved path, the work done on it by a force F is given
by an integral that involves the angle f between the
force and the displacement. This expression is valid
even if the force magnitude and the angle f vary during
the displacement. (See Example 6.8.)
Power: Power is the time rate of doing work. The average power Pav is the amount of work ¢W done in time
¢t divided by that time. The instantaneous power is the
limit of
the average power as ¢t goes to zero. When a
S
S
force F acts on a particle moving with velocity v, the
instantaneous power (the rate at
which the force does
S
S
work) is the scalar product of F and v. Like work and
kinetic energy, power is a scalar quantity. The SI unit of
power is 1 watt = 1 joule>second 11 W = 1 J>s2. (See
Examples 6.9 and 6.10.)
196
S
S
(6.2), (6.3)
S
F
F'
f = angle between F and s
W 5 Fis
5 (F cosf)s
f
Fi 5 F cos f
m
(6.5)
2m
S
v
S
v
Doubling m doubles K.
m
m
S
v
S
2v
Doubling v quadruples K.
(6.6)
m
K1 5
1
2
v1
Wtot 5 Total work done on
particle along path
m
mv12
K2 5
1
2
mv22 5 K1 1 Wtot
x2
Lx1
W =
Fx dx
P2
LP1
W =
LP1
F # dl
S
S
¢W
¢t
¢W
dW
P = lim
=
¢t S 0 ¢t
dt
Pav =
S
#
S
P = F v
Area 5 Work done by
force during displacement
Fx
P2
F cos f dl =
P2
=
(6.7)
LP1
FΠdl
(6.14)
(6.15)
O
x1
t55s
(6.16)
(6.19)
t50
v2
x2
Work you do on the
box to lift it in 5 s:
W 5 100 J
Your power output:
100 J
W
5
P5
t
5s
5 20 W
x
CHAPTER
7
SUMMARY
Gravitational potential energy and elastic potential
energy: The work done on a particle by a constant
gravitational force can be represented as a change
in the gravitational potential energy Ugrav = mgy.
This energy is a shared property of the particle and
the earth. A potential energy is also associated with
the elastic force Fx = - kx exerted by an ideal
spring, where x is the amount of stretch or compression. The work done by this force can be represented as a change in the elastic potential energy
of the spring, Uel = 12 kx 2.
Wgrav = mgy1 - mgy2
= Ugrav,1 - Ugrav,2
= - ¢Ugrav
When total mechanical energy is conserved:
The total potential energy U is the sum of the
gravitational and elastic potential energy:
U = Ugrav + Uel . If no forces other than the
gravitational and elastic forces do work on a
particle, the sum of kinetic and potential energy
is conserved. This sum E = K + U is called the
total mechanical energy. (See Examples 7.1, 7.3,
7.4, and 7.7.)
K 1 + U1 = K 2 + U2
When total mechanical energy is not conserved:
When forces other than the gravitational and elastic
forces do work on a particle, the work Wother done
by these other forces equals the change in total
mechanical energy (kinetic energy plus total
potential energy). (See Examples 7.2, 7.5, 7.6,
7.8, and 7.9.)
K 1 + U1 + Wother = K 2 + U2
Conservative forces, nonconservative forces, and the
law of conservation of energy: All forces are either
conservative or nonconservative. A conservative
force is one for which the work–kinetic energy
relationship is completely reversible. The work of a
conservative force can always be represented by a
potential-energy function, but the work of a nonconservative force cannot. The work done by nonconservative forces manifests itself as changes in
the internal energy of bodies. The sum of kinetic,
potential, and internal energy is always conserved.
(See Examples 7.10–7.12.)
¢K + ¢U + ¢Uint = 0
Uel 5
1
2
kx2
x
Wel = 12 kx 12 - 12 kx 22
= Uel, 1 - Uel, 2 = - ¢Uel
x50
(7.10)
x
Ugrav,2 5 mgy2
O
(7.4), (7.11)
y
At y 5 h
E 5K 1Ugrav
h
zero
At y 5 0
x
O
(7.14)
At point 1
E 5K 1Ugrav
Point 1 f 5 0
n50
w
E5K 1Ugrav
R
f
zero
At point 2
E 5K 1Ugrav
n
f
w
Point 2
w
zero
zero
v
n
zero
(7.15)
E5K1Ugrav
E5K1 Ugrav
v50
As friction slows block,
mechanical energy is converted
to internal energy of block and ramp.
Fx 1x2 = -
0U
0x
0U
Fz = 0z
Fx = -
S
F ! "a
230
Ugrav,1 5 mgy1
(7.1), (7.3)
zero
Determining force from potential energy: For motion
along a straight line, a conservative force Fx 1x2 is
the negative derivative of its associated potentialenergy function U. In three dimensions, the components of a conservative force are negative partial
derivatives of U. (See Examples 7.13 and 7.14.)
y
dU1x2
(7.16)
dx
Fy = -
0U
0y
0U
0U
0U n
nı #
n≥ #
kb
0x
0y
0z
U
Unstable equilibria
(7.17)
O
(7.18)
x
Stable equilibria
CHAPTER
8
SUMMARY
S
Momentum of a particle: The momentum p of a particle
is a vector quantity equal to the product of the particle’s
S
mass m and velocity v. Newton’s second law says that
the net force on a particle is equal to the rate of change
of the particle’s momentum.
S
S
p ! mv
y
(8.2)
dp
gF !
dt
S
S
S
(8.4)
S
p 5 mv
py
S
v
px
m
x
O
Impulse and momentum: If a constant net force g F acts
on a particle for a time interval ¢t from t 1 to t 2 , the
S
impulse J of the net force isSthe product of the net
force
S
and the time interval. If gF varies with time, J is the
integral of the net force over the time interval. In any
case, the change in a particle’s momentum during a time
interval equals the impulse of the net force that acted on
the particle during that interval. The momentum of a particle equals the impulse that accelerated it from rest to its
present speed. (See Examples 8.1–8.3.)
S
Conservation of momentum: An internal force is a force
exerted by one part of a system on another. An external
force is a force exerted on any part of a system by something outside the system. If the net external force on
a
S
system is zero, the total momentum of the system P (the
vector sum of the momenta of the individual particles
that make up the system) is constant, or conserved. Each
component of total momentum is separately conserved.
(See Examples 8.4–8.6.)
J ! gF1t 2 - t 12 ! g F ¢t
S
S
J!
Lt1
S
S
S
t2
S
g F dt
S
S
J ! p2 " p1
(8.6)
(Fav)x
t1
O
S
S
Á
S
P ! pA # pB #
S
S
! m A vA # m B vB #
Á
If gF ! 0, then P ! constant.
S
(8.14)
t2
A
B
S
S
FB on A
S
y
y
S
FA on B
x
S
S
P 5 pA 1 pB 5 constant
S
S
r cm !
S
S
m1 r1 # m2 r2 # m3 r3 #
m1 + m2 + m3 + Á
g im i r i
!
g im i
Á
S
A
vA1
S
S
S
! M vcm
g Fext ! M a cm
S
S
B
S
vA2
Shell explodes
B
S
vB2
cm
cm
cm
(8.29)
S
P ! m 1 v1 # m 2 v2 # m 3 v3 #
S
S
vB1
A B
A
S
Á
(8.32)
(8.34)
Rocket propulsion: In rocket propulsion, the mass of a rocket changes as the fuel is used up
and ejected from the rocket. Analysis of the motion of the rocket must include the momentum
carried away by the spent fuel as well as the momentum of the rocket itself. (See Examples 8.15
and 8.16.)
266
Jx 5 (Fav)x(t2 2 t1)
(8.7)
Collisions: In collisions of all kinds, the initial and final total momenta are equal. In an elastic collision between two bodies, the initial and final total kinetic energies are also equal, and the initial and
final relative velocities have the same magnitude. In an inelastic two-body collision, the total
kinetic energy is less after the collision than before. If the two bodies have the same final velocity,
the collision is completely inelastic. (See Examples 8.7–8.12.)
Center of mass: The position vector of the center of
S
mass of a system of particles, r cm , is a weighted averS
S
age of the positions r 1 , r 2 , Á of the individual partiS
cles. The total momentum P of a system equals its total
mass M multiplied by the velocity of its center of mass,
S
vcm . The center of mass moves as though all the mass
M were concentrated at that point. If the net external
force on the system is zero, the center-of-mass velocity
S
vcm is constant. If the net external force is not zero, the
center of mass accelerates as though it were a particle
of mass M being acted on by the same net external
force. (See Examples 8.13 and 8.14.)
Fx
(8.5)
1x-direction
vfuel 5 v 2 vex
v 1 dv
2dm
m 1 dm
x
t
CHAPTER
9
SUMMARY
Rotational kinematics: When a rigid body rotates about
a stationary axis (usually called the z-axis), its position
is described by an angular coordinate u. The angular
velocity vz is the time derivative of u, and the angular
acceleration az is the time derivative of vz or the second
derivative of u. (See Examples 9.1 and 9.2.) If the angular acceleration is constant, then u, vz , and az are related
by simple kinematic equations analogous to those for
straight-line motion with constant linear acceleration.
(See Example 9.3.)
¢u
du
=
¢t
dt
¢vz
dvz
d 2u
az = limS
=
= 2
¢t 0 ¢t
dt
dt
vz = limS
¢t 0
du
dt
At t2
(9.3)
y
vz 5
dvz
dt
At t1
Du
(9.5), (9.6)
u = u0 + v0z t + 12 az t 2
az 5
u2
u1
O
x
(9.11)
(constant az only)
u - u0 = 12 1v0z + vz2t
(9.10)
(constant az only)
vz = v0z + az t
(constant az only)
Relating linear and angular kinematics: The angular
speed v of a rigid body is the magnitude of its angular
velocity. The rate of change of v is a = dv>dt. For a
particle in the body a distance r from the rotation axis,
S
the speed v and the components of the acceleration a
are related to v and a. (See Examples 9.4 and 9.5.)
(9.7)
vz2 = v0z2 + 2az 1u - u02
(constant az only)
(9.12)
v = rv
dv
dv
= r
= ra
atan =
dt
dt
v2
arad =
= v 2r
r
(9.13)
(9.14)
(9.15)
y
v
atan 5 ra
v 5 rv
S
a
Linear
acceleration
of point P
r
P
arad 5 v2r
s
u
x
O
Moment of inertia and rotational kinetic energy: The
moment of inertia I of a body about a given axis is a
measure of its rotational inertia: The greater the value
of I, the more difficult it is to change the state of the
body’s rotation. The moment of inertia can be expressed
as a sum over the particles m i that make up the body,
each of which is at its own perpendicular distance ri
from the axis. The rotational kinetic energy of a rigid
body rotating about a fixed axis depends on the angular
speed v and the moment of inertia I for that rotation
axis. (See Examples 9.6–9.8.)
Calculating the moment of inertia: The parallel-axis
theorem relates the moments of inertia of a rigid body
of mass M about two parallel axes: an axis through the
center of mass (moment of inertia Icm) and a parallel
axis a distance d from the first axis (moment of inertia
IP). (See Example 9.9.) If the body has a continuous
mass distribution, the moment of inertia can be calculated by integration. (See Examples 9.10 and 9.11.)
I = m 1 r 12 + m 2 r 22 + Á
= a m ir i 2
(9.16)
i
K = 12 Iv2
(9.17)
Axis of
rotation
m1
v
r2
m2
I 5 S miri2
i
1
r1
K 5 2 Iv2
r3
IP = Icm + Md 2
m3
(9.19)
d
cm
Mass M
P
Icm
IP 5 Icm 1
Md 2
297
CHAPTER
10
SUMMARY
S
Torque: When a force F acts on a body, the torque of
that force with respect to a point O has a magnitude
given by the product of the force magnitude F and the
S
lever arm l. More generally, torque is a vector T equal to
S
the vector product of r (the position
vector of the point
S
at which the force acts) and F . (See Example 10.1.)
Rotational dynamics: The rotational analog of
Newton’s second law says that the net torque acting
on a body equals the product of the body’s moment of
inertia and its angular acceleration. (See Examples 10.2
and 10.3.)
Combined translation and rotation: If a rigid body is
both moving through space and rotating, its motion can
be regarded as translational motion of the center of mass
plus rotational motion about an axis through the center
of mass. Thus the kinetic energy is a sum of translational
and rotational kinetic energies. For dynamics, Newton’s
second law describes the motion of the center of mass,
and the rotational equivalent of Newton’s second law
describes rotation about the center of mass. In the case of
rolling without slipping, there is a special relationship
between the motion of the center of mass and the rotational motion. (See Examples 10.4–10.7.)
Work done by a torque: A torque that acts on a rigid
body as it rotates does work on that body. The work can
be expressed as an integral of the torque. The work–
energy theorem says that the total rotational work done
on a rigid body is equal to the change in rotational
kinetic energy. The power, or rate at which the torque
does work, is the product of the torque and the angular
velocity (See Example 10.8.)
Angular momentum: The angular momentum of a par-
ticle with respect to point O is the vector product of the
S
particle’s position vector r relative to O and its momenS
S
tum p ! mv. When a symmetrical body rotates about a
stationary axis of symmetry, its angular momentum is
the product of its moment of inertia and its angular
S
velocity vector V . If the body is not symmetrical or the
rotation 1z2 axis is not an axis of symmetry, the component of angular momentum along the rotation axis is
Ivz . (See Example 10.9.)
Rotational dynamics and angular momentum: The net
external torque on a system is equal to the rate of
change of its angular momentum. If the net external
torque on a system is zero, the total angular momentum
of the system is constant (conserved). (See Examples
10.10–10.13.)
t = Fl
S
(10.2)
S
S
T! r : F
(10.3)
Frad 5 F cos f
S
F
f
l 5 r sin f
5 lever arm
f
S
r
Ftan 5 F sin f
S
S
S
t5r3F
a tz = Iaz
(10.7)
F
y
F
R
O
n
R
x
M
Mg
K = 12 Mvcm2 + 12 Icm v2
a Fext ! M a cm
a tz = Icm az
S
S
(10.8)
R
(10.12)
vcm = Rv
(rolling without slipping)
(10.11)
vcm 5 0
v50
M
1
(10.13)
h
v
2
vcm
u2
W =
Lu1
tz du
(10.20)
S
Ftan
W = tz1u2 - u12 = tz ¢u
(constant torque only)
(10.21)
Wtot =
(10.22)
1
1
2
2
2 Iv2 - 2 Iv1
P = tz vz
S
S
S
O
(10.23)
S
S
S
(10.24)
L ! IV
(rigid body rotating
about axis of symmetry)
(10.28)
L ! r : p ! r : mv
(particle)
S
Ftan
ds
du
R
R
S
dL
S
a T ! dt
S
L
S
v
S
(10.29)
331
CHAPTER
11
SUMMARY
Conditions for equilibrium: For a rigid body to be in
equilibrium, two conditions must be satisfied. First, the
vector sum of forces must be zero. Second, the sum of
torques about any point must be zero. The torque due to
the weight of a body can be found by assuming the
entire weight is concentrated at the center of gravity,
S
which is at the same point as the center of mass if g has
the same value at all points. (See Examples 11.1–11.4.)
a Fx = 0
a Fy = 0
a T ! 0 about any point
S
S
r cm !
a Fz = 0
(11.1)
T
(11.2)
S
S
S
m1 r1 " m2 r2 " m3 r3 " Á
m1 + m2 + m3 + Á
w
E
y
(11.4)
T
Ty
Ex
Tx
w
Stress, strain, and Hooke’s law: Hooke’s law states that
in elastic deformations, stress (force per unit area) is
proportional to strain (fractional deformation). The proportionality constant is called the elastic modulus.
Tensile and compressive stress: Tensile stress is tensile
force per unit area, F! >A. Tensile strain is fractional
change in length, ¢l>l 0 . The elastic modulus is called
Young’s modulus Y. Compressive stress and strain are
defined in the same way. (See Example 11.5.)
Stress
= Elastic modulus
Strain
Y =
x
Ey
(11.7)
F! >A
F! l 0
Tensile stress
=
=
Tensile strain
¢l>l 0
A ¢l
Initial
A state
(11.10)
l0
Dl
A F'
F'
l
Bulk stress: Pressure in a fluid is force per unit area.
Bulk stress is pressure change, ¢p, and bulk strain is
fractional volume change, ¢V>V0 . The elastic modulus
is called the bulk modulus, B. Compressibility, k, is
the reciprocal of bulk modulus: k = 1>B. (See
Example 11.6.)
p =
F!
A
¢p
Bulk stress
= B =
Bulk strain
¢V>V0
(11.11)
(11.13)
F'
F'
Pressure 5 p
5 p0 1 Dp
Shear stress: Shear stress is force per unit area, FŒ>A,
for a force applied tangent to a surface. Shear strain is
the displacement x of one side divided by the transverse
dimension h. The elastic modulus is called the shear
modulus, S. (See Example 11.7.)
S =
Volume
V0
Pressure 5 p0
FŒ>A
FΠh
Shear stress
=
=
Shear strain
x>h
A x
F'
A
x
Volume
V
F'
F'
Initial
state
h
(11.17)
F'
A
F||
F||
The limits of Hooke’s law: The proportional limit is the maximum stress for which stress and strain
are proportional. Beyond the proportional limit, Hooke’s law is not valid. The elastic limit is the
stress beyond which irreversible deformation occurs. The breaking stress, or ultimate strength, is
the stress at which the material breaks.
359
CHAPTER
12
SUMMARY
m
V
dF!
p =
dA
Density and pressure: Density is mass per unit volume.
If a mass m of homogeneous material has volume V, its
density r is the ratio m>V. Specific gravity is the ratio of
the density of a material to the density of water. (See
Example 12.1.)
Pressure is normal force per unit area. Pascal’s law
states that pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid.
Absolute pressure is the total pressure in a fluid; gauge
pressure is the difference between absolute pressure and
atmospheric pressure. The SI unit of pressure is the pascal (Pa): 1 Pa = 1 N>m2. (See Example 12.2.)
r =
Pressures in a fluid at rest: The pressure difference
between points 1 and 2 in a static fluid of uniform density r (an incompressible fluid) is proportional to the
difference between the elevations y1 and y2 . If the pressure at the surface of an incompressible liquid at rest is
p0 , then the pressure at a depth h is greater by an
amount rgh. (See Examples 12.3 and 12.4.)
p2 - p1 = - rg1y2 - y12
(pressure in a fluid
of uniform density)
(12.5)
p = p0 + rgh
(pressure in a fluid
of uniform density)
(12.6)
(12.1)
(12.2)
dF!
392
dF!
dA
Equal normal forces exerted on
both sides by surrounding fluid
Fluid, density r
p2 5 p0
Buoyancy: Archimedes’s principle states that when a
body is immersed in a fluid, the fluid exerts an upward
buoyant force on the body equal to the weight of the
fluid that the body displaces. (See Example 12.5.)
Fluid flow: An ideal fluid is incompressible and has no
viscosity (no internal friction). A flow line is the path of
a fluid particle; a streamline is a curve tangent at each
point to the velocity vector at that point. A flow tube is a
tube bounded at its sides by flow lines. In laminar flow,
layers of fluid slide smoothly past each other. In turbulent flow, there is great disorder and a constantly changing flow pattern.
Conservation of mass in an incompressible fluid is
expressed by the continuity equation, which relates the
flow speeds v1 and v2 for two cross sections A1 and A2
in a flow tube. The product Av equals the volume flow
rate, dV>dt, the rate at which volume crosses a section
of the tube. (See Example 12.6.)
Bernoulli’s equation relates the pressure p, flow speed
v, and elevation y for any two points, assuming steady
flow in an ideal fluid. (See Examples 12.7–12.10.)
Small area dA within fluid at rest
2
y2 2 y1 5 h
p1 5 p
y2
1 y1
dF'
B
wbody
A1 v1 = A2 v2
(continuity equation,
incompressible fluid)
(12.10)
cg
Fluid element
replaced with
solid body of
the same size
and shape
v2
d p2A2
c
A2 dV
ds2
dV
= Av
dt
(volume flow rate)
(12.11)
p1 + rgy1 + 12 rv12 = p2 + rgy2 + 12 rv22
(Bernoulli’s equation)
(12.17)
Flow
v1 b
y2
a
dV
A
p1A1 1
ds1
y1
CHAPTER
14
SUMMARY
1
T
Periodic motion: Periodic motion is motion that repeats
itself in a definite cycle. It occurs whenever a body has a
stable equilibrium position and a restoring force that
acts when it is displaced from equilibrium. Period T is
the time for one cycle. Frequency ƒ is the number of
cycles per unit time. Angular frequency v is 2p times
the frequency. (See Example 14.1.)
ƒ =
Simple harmonic motion: If the restoring force Fx in
periodic motion is directly proportional to the displacement x, the motion is called simple harmonic motion
(SHM). In many cases this condition is satisfied if the
displacement from equilibrium is small. The angular
frequency, frequency, and period in SHM do not depend
on the amplitude, but only on the mass m and force constant k. The displacement, velocity, and acceleration in
SHM are sinusoidal functions of time; the amplitude A
and phase angle f of the oscillation are determined by
the initial position and velocity of the body. (See Examples 14.2, 14.3, 14.6, and 14.7.)
Fx = - kx
T =
1
ƒ
2p
v = 2pƒ =
T
(14.2)
x = 2A
x=0
x=A
x,0
x.0
y
y ax
ax y
n
Fx
n
ax =
v =
ƒ =
T =
x
(14.3)
A
Fx
k
= - x
m
m
(14.4)
O
2A
k
Am
(14.10)
1
k
v
=
2p
2p A m
(14.11)
1
m
= 2p
ƒ
Ak
(14.12)
(14.13)
Energy
Simple pendulum: A simple pendulum consists of a point
mass m at the end of a massless string of length L. Its
motion is approximately simple harmonic for sufficiently small amplitude; the angular frequency, frequency, and period then depend only on g and L, not on
the mass or amplitude. (See Example 14.8.)
Physical pendulum: A physical pendulum is any body
suspended from an axis of rotation. The angular frequency and period for small-amplitude oscillations are
independent of amplitude, but depend on the mass m,
distance d from the axis of rotation to the center of gravity, and moment of inertia I about the axis. (See Examples 14.9 and 14.10.)
v =
and
ƒ =
g
AL
g
v
1
ƒ =
=
2p
2p A L
2p
1
L
T =
= = 2p
v
ƒ
Ag
v =
mgd
B I
I
T = 2p
A mgd
k
1
2p A I
(14.24)
E5K1U
U
(14.21)
k
AI
t
2T
T
E = 12 mvx2 + 12 kx 2 = 12 kA2 = constant
v =
x
mg
mg
K
O
2A
Angular simple harmonic motion: In angular SHM, the
frequency and angular frequency are related to the
moment of inertia I and the torsion constant k.
n
Fx
x
x
mg
x = A cos1vt + f2
Energy in simple harmonic motion: Energy is conserved
in SHM. The total energy can be expressed in terms of
the force constant k and amplitude A. (See Examples
14.4 and 14.5.)
x
(14.1)
x
A
Balance wheel Spring
tz
u
Spring torque tz opposes
angular displacement u.
(14.32)
L
(14.33)
u
T
(14.34)
O
z
(14.38)
u d
d sin u
(14.39)
mg cos u
mg sin u
mg
mg sin u
cg
mg cos u
mg
461
462
CHAPTER 14 Periodic Motion
Damped oscillations: When a force Fx = - bvx proportional to velocity is added to a simple harmonic oscillator, the motion is called a damped oscillation. If
b 6 2 2km (called underdamping), the system oscillates with a decaying amplitude and an angular frequency v¿ that is lower than it would be without
damping. If b = 21km (called critical damping) or
b 7 2 1km (called overdamping), when the system
is displaced it returns to equilibrium without
oscillating.
Driven oscillations and resonance: When a sinusoidally
varying driving force is added to a damped harmonic
oscillator, the resulting motion is called a forced oscillation. The amplitude is a function of the driving frequency vd and reaches a peak at a driving frequency
close to the natural frequency of the system. This behavior is called resonance.
BRIDGING PROBLEM
x = Ae -1b>2m2t cos 1v¿t + f2
k
b2
v¿ =
Bm
4m 2
(14.42)
A
x
Ae2(b /2m)t
(14.43)
t
O
T0
2A
A =
Fmax
21k - mvd2 22 + b 2vd2
(14.46)
2T0 3T0 4T0 5T0
b 5 0.1!km
b 5 0.4!km
5Fmax/k
4Fmax/k
3Fmax/k
2Fmax/k
Fmax/k
A
0
b 5 0.2!km
b 5 0.4!km
b 5 0.7!km
b 5 1.0!km
b 5 2.0!km
v v
0.5 1.0 1.5 2.0 d /
Oscillating and Rolling
Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest on a
horizontal tabletop (Fig. 14.29). A frictionless ring at the center of
the rod is attached to a spring with force constant k; the other end
of the spring is fixed. The cylinders are pulled to the left a distance
x, stretching the spring, and then released from rest. Due to friction
between the tabletop and the cylinders, the cylinders roll without
slipping as they oscillate. Show that the motion of the center of
mass of the cylinders is simple harmonic, and find its period.
SOLUTION GUIDE
See MasteringPhysics® study area for a Video Tutor solution.
IDENTIFY and SET UP
1. What condition must be satisfied for the motion of the center of
mass of the cylinders to be simple harmonic? (Hint: See Section 14.2.)
2. Which equations should you use to describe the translational
and rotational motions of the cylinders? Which equation should
you use to describe the condition that the cylinders roll without
slipping? (Hint: See Section 10.3.)
3. Sketch the situation and choose a coordinate system. Make a
list of the unknown quantities and decide which is the target
variable.
14.29
M
x
R
k
EXECUTE
4. Draw a free-body diagram for the cylinders when they are
displaced a distance x from equilibrium.
5. Solve the equations to find an expression for the acceleration of
the center of mass of the cylinders. What does this expression
tell you?
6. Use your result from step 5 to find the period of oscillation of
the center of mass of the cylinders.
EVALUATE
7. What would be the period of oscillation if there were no friction and the cylinders didn’t roll? Is this period larger or
smaller than your result from step 6? Is this reasonable?
CHAPTER
15
SUMMARY
Waves and their properties: A wave is any disturbance
that propagates from one region to another. A mechanical wave travels within some material called the
medium. The wave speed v depends on the type of
wave and the properties of the medium.
In a periodic wave, the motion of each point of the
medium is periodic with frequency f and period T. The
wavelength l is the distance over which the wave pattern repeats, and the amplitude A is the maximum displacement of a particle in the medium. The product of l
and ƒ equals the wave speed. A sinusoidal wave is a
special periodic wave in which each point moves in
simple harmonic motion. (See Example 15.1.)
Wave functions and wave dynamics: The wave function
y1x, t2 describes the displacements of individual particles in the medium. Equations (15.3), (15.4), and (15.7)
give the wave equation for a sinusoidal wave traveling
in the +x-direction. If the wave is moving in the
-x-direction, the minus signs in the cosine functions are
replaced by plus signs. (See Example 15.2.)
The wave function obeys a partial differential equation called the wave equation, Eq. (15.12).
The speed of transverse waves on a string depends
on the tension F and mass per unit length m. (See
Example 15.3.)
v = lf
Wave
speed
v
Wavelength l
Each particle of
Amplitude A rope oscillates
in SHM.
y1x, t2 = A cos cv a
y
x
- tb d
v
A
x
= A cos 2pƒ a - t b
v
y1x, t2 = A cos 2p a
x
A
(15.3)
Wavelength l
x
t
- b
l
T
y
(15.4)
A
y1x, t2 = A cos1kx - vt2
(15.7)
t
A
where k = 2p>l and v = 2pf = vk
0 2y1x, t2
0 x2
v=
Wave power: Wave motion conveys energy from one
region to another. For a sinusoidal mechanical wave,
the average power Pav is proportional to the square
of the wave amplitude and the square of the frequency.
For waves that spread out in three dimensions, the
wave intensity I is inversely proportional to the
square of the distance from the source. (See
Examples 15.4 and 15.5.)
(15.1)
F
Am
Pav =
1
2
=
2
1 0 y1x, t2
v2
0t2
Period T
(15.12)
(waves on a string) (15.13)
2mF v2A2
Wave power versus time t
at coordinate x 5 0
(15.25)
(average power, sinusoidal wave)
r 22
I1
= 2
(15.26)
I2
r1
(inverse-square law for intensity)
Wave superposition: A wave reflects when it reaches a
boundary of its medium. At any point where two or
more waves overlap, the total displacement is the sum
of the displacements of the individual waves (principle
of superposition).
y1x, t2 = y1 1x, t2 + y2 1x, t2
(principle of superposition)
Standing waves on a string: When a sinusoidal wave is
reflected from a fixed or free end of a stretched string,
the incident and reflected waves combine to form a
standing sinusoidal wave with nodes and antinodes.
Adjacent nodes are spaced a distance l>2 apart, as are
adjacent antinodes. (See Example 15.6.)
When both ends of a string with length L are held
fixed, standing waves can occur only when L is an integer multiple of l>2. Each frequency with its associated
vibration pattern is called a normal mode. (See
Examples 15.7 and 15.8.)
(15.28)
y1x, t2 = 1ASW sin kx2 sin vt
(standing wave on a string,
fixed end at x = 0)
v
= nƒ1 1n = 1, 2, 3, Á 2
ƒn = n
2L
P
Pmax
Pav 5 12 Pmax
0
t
Period T
(15.27)
O
(15.33)
1
F
2L A m
(string fixed at both ends)
ƒ1 =
(15.35)
N
N
A
N
N
N
A
l
2
A
A
N
2
l
2
3
l
2
N
N
5L
5L
A
A
4
A
N
N
5L
N
l
2
N
A
A
N
A
N
5L
499
CHAPTER
16
SUMMARY
Sound waves: Sound consists of longitudinal waves in a
medium. A sinusoidal sound wave is characterized by its
frequency ƒ and wavelength l (or angular frequency v
and wave number k) and by its displacement amplitude
A. The pressure amplitude pmax is directly proportional
to the displacement amplitude, the wave number, and
the bulk modulus B of the wave medium. (See
Examples 16.1 and 16.2.)
The speed of a sound wave in a fluid depends on the
bulk modulus B and density r. If the fluid is an ideal
gas, the speed can be expressed in terms of the temperature T, molar mass M, and ratio of heat capacities g of
the gas. The speed of longitudinal waves in a solid rod
depends on the density and Young’s modulus Y. (See
Examples 16.3 and 16.4.)
pmax = BkA
1sinusoidal sound wave2
(16.5)
v =
(16.7)
B
Ar
(longitudinal wave in a fluid)
gRT
A M
(sound wave in an ideal gas)
v =
y
Wavelength l
y.0
A
y.0
x
(16.10)
!A
y,0
y,0
Rarefaction
p
Compression
pmax
Y
(16.8)
v =
Ar
(longitudinal wave in a solid rod)
x
!pmax
Intensity and sound intensity level: The intensity I of a
sound wave is the time average rate at which energy is
transported by the wave, per unit area. For a sinusoidal
wave, the intensity can be expressed in terms of the displacement amplitude A or the pressure amplitude pmax .
(See Examples 16.5–16.7.)
The sound intensity level b of a sound wave is a logarithmic measure of its intensity. It is measured relative
to I0 , an arbitrary intensity defined to be 10 -12 W>m2.
Sound intensity levels are expressed in decibels (dB).
(See Examples 16.8 and 16.9.)
Standing sound waves: Standing sound waves can be set
up in a pipe or tube. A closed end is a displacement
node and a pressure antinode; an open end is a displacement antinode and a pressure node. For a pipe of length
L open at both ends, the normal-mode frequencies are
integer multiples of the sound speed divided by 2L.
For a stopped pipe (one that is open at only one end),
the normal-mode frequencies are the odd multiples of
the sound speed divided by 4L. (See Examples 16.10
and 16.11.)
A pipe or other system with normal-mode frequencies can be driven to oscillate at any frequency. A maximum response, or resonance, occurs if the driving
frequency is close to one of the normal-mode frequencies of the system. (See Example 16.12.)
I =
=
1
2
2rB v2A2 =
p max2
p max2
2rv
Point source
P1
(16.12), (16.14)
2 2rB
P2
(intensity of a sinusoidal sound wave)
b = 110 dB2 log
I
I0
(16.15)
(definition of sound intensity level)
nv
1n = 1, 2, 3, Á2 (16.18)
2L
1open pipe2
ƒn =
nv
1n = 1, 3, 5, Á2 (16.22)
ƒn =
4L
1stopped pipe2
Interference: When two or more waves overlap in the same region of space, the resulting effects
are called interference. The resulting amplitude can be either larger or smaller than the amplitude of
each individual wave, depending on whether the waves are in phase (constructive interference) or
out of phase (destructive interference). (See Example 16.13.)
A
Open
pipe
l
2
A
N
A
N
A
N
A
f1 5
v
f2 5 2 2L
5 2f1
v
2L
N
Stopped
pipe
N
A
l
4
N
A
A
f1 5
Waves
arrive in
phase.
v
f3 5 3 4L
5 3f1
v
4L
d2 1
d1
l
2
d1
d2
Q
Waves
arrive 12
cycle out
of phase.
P
541
542
CHAPTER 16 Sound and Hearing
Beats: Beats are heard when two tones with slightly different frequencies ƒa and ƒb are sounded together. The
beat frequency ƒbeat is the difference between ƒa and fb .
ƒbeat = fa - ƒb
(16.24)
Displacement
(beat frequency)
t
t
Beat
Doppler effect: The Doppler effect for sound is the frequency shift that occurs when there is motion of a
source of sound, a listener, or both, relative to the
medium. The source and listener frequencies ƒS and ƒL
are related by the source and listener velocities vS and
vL relative to the medium and to the speed of sound v.
(See Examples 16.14–16.18.)
Shock waves: A sound source moving with a speed vS
greater than the speed of sound v creates a shock wave.
The wave front is a cone with angle a. (See Example
16.19.)
BRIDGING PROBLEM
ƒL =
v + vL
ƒS
v + vS
(16.29)
L to S
!
(Doppler effect, moving source
and moving listener)
v
v
v
vL v
l
L
a vS b l
S
S
vS
v
v
sin a =
v
vS
(shock wave)
v
(16.31)
vS . v
a
Shock wave
Loudspeaker Interference
Loudspeakers A and B are 7.00 m apart and vibrate in phase at
172 Hz. They radiate sound uniformly in all directions. Their acoustic
power outputs are 8.00 × 10-4 W and 6.00 " 10-5 W, respectively.
The air temperature is 20°C. (a) Determine the difference in phase of
the two signals at a point C along the line joining A and B, 3.00 m
from B and 4.00 m from A. (b) Determine the intensity and sound
intensity level at C from speaker A alone (with B turned off) and from
speaker B alone (with A turned off). (c) Determine the intensity and
sound intensity level at C from both speakers together.
SOLUTION GUIDE
See MasteringPhysics® study area for a Video Tutor solution.
IDENTIFY and SET UP
1. Sketch the situation and label the distances between A, B, and C.
2. Choose the equations that relate power, distance from the
source, intensity, pressure amplitude, and sound intensity level.
3. Decide how you will determine the phase difference in part
(a). Once you have found the phase difference, how can you
use it to find the amplitude of the combined wave at C due to
both sources?
4. List the unknown quantities for each part of the problem and
identify your target variables.
EXECUTE
5. Determine the phase difference at point C.
6. Find the intensity, sound intensity level, and pressure amplitude at C due to each speaker alone.
7. Use your results from steps 5 and 6 to find the pressure amplitude at C due to both loudspeakers together.
8. Use your result from step 7 to find the intensity and sound
intensity level at C due to both loudspeakers together.
EVALUATE
9. How do your results from part (c) for intensity and sound
intensity level at C compare to those from part (b)? Does this
make sense?
10. What result would you have gotten in part (c) if you had
(incorrectly) combined the intensities from A and B directly,
rather than (correctly) combining the pressure amplitudes as
you did in step 7?
CHAPTER
17
SUMMARY
Temperature and temperature scales: Two bodies in thermal equilibrium must have the same temperature. A conducting material between
two bodies permits them to interact and come to thermal equilibrium;
an insulating material impedes this interaction.
The Celsius and Fahrenheit temperature scales are based on the
freezing 10°C = 32°F2 and boiling 1100°C = 212°F2 temperatures of water. One Celsius degree equals 95 Fahrenheit degrees.
(See Example 17.1.)
The Kelvin scale has its zero at the extrapolated zero-pressure
temperature for a gas thermometer, -273.15°C = 0 K. In the gasthermometer scale, the ratio of two temperatures T1 and T2 is defined
to be equal to the ratio of the two corresponding gas-thermometer
pressures p1 and p2 .
Thermal expansion and thermal stress: A temperature change ¢T
causes a change in any linear dimension L 0 of a solid body. The
change ¢L is approximately proportional to L 0 and ¢T. Similarly, a
temperature change causes a change ¢V in the volume V0 of any solid
or liquid; ¢V is approximately proportional to V0 and ¢T. The quantities a and b are the coefficients of linear expansion and volume expansion, respectively. For solids, b = 3a. (See Examples 17.2 and 17.3.)
When a material is cooled or heated and held so it cannot contract
or expand, it is under a tensile stress F> A. (See Example 17.4.)
Heat, phase changes, and calorimetry: Heat is energy in transit from
one body to another as a result of a temperature difference. Equations
(17.13) and (17.18) give the quantity of heat Q required to cause a
temperature change ∆T in a quantity of material with mass m and
specific heat c (alternatively, with number of moles n and molar heat
capacity C ! Mc, where M is the molar mass and m ! nM). When
heat is added to a body, Q is positive; when it is removed, Q is
negative. (See Examples 17.5 and 17.6.)
To change a mass m of a material to a different phase at the same
temperature (such as liquid to vapor), a quantity of heat given by
Eq. (17.20) must be added or subtracted. Here L is the heat of fusion,
vaporization, or sublimation.
In an isolated system whose parts interact by heat exchange, the
algebraic sum of the Q’s for all parts of the system must be zero. (See
Examples 17.7–17.10.)
Conduction, convection, and radiation: Conduction is the transfer of
heat within materials without bulk motion of the materials. The heat
current H depends on the area A through which the heat flows, the
length L of the heat-flow path, the temperature difference 1TH - TC2,
and the thermal conductivity k of the material. (See Examples
17.11–17.13.)
Convection is a complex heat-transfer process that involves mass
motion from one region to another.
Radiation is energy transfer through electromagnetic radiation.
The radiation heat current H depends on the surface area A, the emissivity e of the surface (a pure number between 0 and 1), and the
Kelvin temperature T. Here σ is the Stefan–Boltzmann constant. The
net radiation heat current Hnet from a body at temperature T to its
surroundings at temperature Ts depends on both T and Ts. (See
Examples 17.14 and 17.15.)
578
TF = 95 TC + 32°
(17.1)
If systems A and B are each in
thermal equilibrium with system C …
TC = 59 1TF - 32°2 (17.2)
Insulator
p2
T2
=
p1
T1
C
Conductor
TK = TC + 273.15 (17.3)
(17.4)
A
A
B
B
C
… then systems A and B are in
thermal equilibrium with each other.
¢L = aL 0 ¢T
(17.6)
¢V = bV0 ¢T
(17.8)
L 5 L 0 1 DL
5 L 0 (1 1 a DT )
T0
T0 1 DT
F
= - Ya ¢T
A
(17.12)
Q = mc ¢T
(17.13)
Q = nC ¢T
(17.18)
Q = "mL
(17.20)
L0
DL
Phase changes, temperature is constant:
Q 5 1mL
T (°C)
100
Boiling point
Melting point
0
t
Temperature rises, phase does not change:
Q 5 mcDT
H =
dQ
TH - TC
= kA
(17.21)
dt
L
H = AesT 4
Heat current H
TH
(17.25)
A
Heat current H 5 kA
Hnet = Aes1T 4 - T s42 (17.26)
TC
L
TH 2 TC
L
CHAPTER
25
SUMMARY
Current and current density: Current is the amount of
charge flowing through a specified area, per unit time.
The SI unit of current is the ampere 11 A = 1 C>s2. The
current I through an area A depends on the concentration n and charge q of the charge carriers, as well as on
S
the magnitude of their drift velocity vd . The current
density is current per unit cross-sectional area. Current
is usually described in terms of a flow of positive
charge, even when the charges are actually negative or
of both signs. (See Example 25.1.)
I =
dQ
= n ƒ q ƒ vd A
dt
S
S
J ! nqvd
S
vd
S
S
vd
vd
S
E
J
Resistors: The potential difference V across a sample of
material that obeys Ohm’s law is proportional to the
current I through the sample. The ratio V>I = R is the
resistance of the sample. The SI unit of resistance is the
ohm 11 Æ = 1 V>A2. The resistance of a cylindrical
conductor is related to its resistivity r, length L, and
cross-sectional area A. (See Examples 25.2 and 25.3.)
V = IR
(25.11)
rL
A
(25.10)
(25.6)
vd
r
(25.5)
r1T2 = r0 31 + a1T - T024
Slope 5 r0a
r0
O
T0
Metal: r increases with
increasing T.
R =
Higher
potential
A
(25.15)
Vab = E - Ir
(source with internal resistance)
L
S
S
V
+
R
(power into a resistor)
(25.17)
Conduction in metals: The microscopic basis of conduction in metals is the motion of electrons that
move freely through the metallic crystal, bumping into ion cores in the crystal. In a crude classical
model of this motion, the resistivity of the material can be related to the electron mass, charge,
speed of random motion, density, and mean free time between collisions. (See Example 25.11.)
R 5 4V
I
b!
Vb
Va
Circuit
element
I
(25.18)
b
r 5 2 V, E 5 12 V A
a!
P = Vab I = I R =
I
Vab 5 Va!b!
I
2
J
V
I
a
V ab2
T
Lower
potential
E
P = Vab I
(general circuit element)
S
E
S
vd
r =
Energy and power in circuits: A circuit element with a
potential difference Va - Vb = Vab and a current I puts
energy into a circuit if the current direction is from
lower to higher potential in the device, and it takes
energy out of the circuit if the current is opposite. The
power P equals the product of the potential difference
and the current. A resistor always takes electrical energy
out of a circuit. (See Examples 25.8–25.10.)
S
vd
(25.4)
Resistivity: The resistivity r of a material is the ratio of
the magnitudes of electric field and current density.
Good conductors have small resistivity; good insulators
have large resistivity. Ohm’s law, obeyed approximately
by many materials, states that r is a constant independent of the value of E. Resistivity usually increases with
temperature; for small temperature changes this variation is represented approximately by Eq. (25.6), where
a is the temperature coefficient of resistivity.
Circuits and emf: A complete circuit has a continuous
current-carrying path. A complete circuit carrying a
steady current must contain a source of electromotive
force (emf) E. The SI unit of electromotive force is the
volt (1 V). Every real source of emf has some internal
resistance r, so its terminal potential difference Vab
depends on current. (See Examples 25.4–25.7.)
I
(25.2)
a
I
b
S
E
Net displacement
841
SUMMARY
1
1
1
1
=
+
+
+ Á
Req
R1
R2
R3
(resistors in parallel)
(26.1)
Resistors in series
R1
a
(26.2)
R2
x
I
(junction rule)
(loop rule)
b
R3
I
Junction
(26.5)
(26.6)
At any junction:
SI 5 0
I1
I2
I1 1 I2
Loop 1
+
Loop 2
E
R
Loop 3
E
Around any loop: SV 5 0
Ammeter
|||||||||||||
||
Voltmeter
|
|
|||
I
–
q = CE A 1 - e
= Qf A 1 - e
-t/RC
-t/RC
dq
E
= e -t/RC
dt
R
= I0 e -t/RC
i =
B
B
Q 0 -t/RC
dq
= e
dt
RC
-t/RC
= I0 e
i =
||
|
a R sh b
–
a
I
b
Va Circuit Vb
element
I
+ E
(26.12)
i
R
(26.13)
Capacitor discharging:
q = Q 0 e -t/RC
||| ||||||||||
+
I
Capacitor charging:
|||
Rc
Rs
Rc
+
||
||
||
||
Electrical measuring instruments: In a d’Arsonval galvanometer, the deflection is proportional to
the current in the coil. For a larger current range, a shunt resistor is added, so some of the current
bypasses the meter coil. Such an instrument is called an ammeter. If the coil and any additional
series resistance included obey Ohm’s law, the meter can also be calibrated to read potential difference or voltage. The instrument is then called a voltmeter. A good ammeter has very low resistance;
a good voltmeter has very high resistance. (See Examples 26.8–26.11.)
R-C circuits: When a capacitor is charged by a battery in
series with a resistor, the current and capacitor charge are
not constant. The charge approaches its final value asymptotically and the current approaches zero asymptotically.
The charge and current in the circuit are given by
Eqs. (26.12) and (26.13). After a time t = RC, the charge
has approached within 1>e of its final value. This time is
called the time constant or relaxation time of the circuit.
When the capacitor discharges, the charge and current are
given as functions of time by Eqs. (26.16) and (26.17).
The time constant is the same for charging and discharging. (See Examples 26.12 and 26.13.)
b
I
R2
I
aI = 0
aV = 0
R3
y
R1
Resistors
in parallel
a
+
Kirchhoff’s rules: Kirchhoff’s junction rule is based on
conservation of charge. It states that the algebraic sum
of the currents into any junction must be zero. Kirchhoff’s loop rule is based on conservation of energy and
the conservative nature of electrostatic fields. It states
that the algebraic sum of potential differences around
any loop must be zero. Careful use of consistent sign
rules is essential in applying Kirchhoff’s rules. (See
Examples 26.3–26.7.)
Req = R1 + R2 + R3 + Á
(resistors in series)
||
Resistors in series and parallel: When several resistors
R1 , R2 , R3 , Á are connected in series, the equivalent
resistance Req is the sum of the individual resistances.
The same current flows through all the resistors in a
series connection. When several resistors are connected
in parallel, the reciprocal of the equivalent resistance
Req is the sum of the reciprocals of the individual resistances. All resistors in a parallel connection have the
same potential difference between their terminals. (See
Examples 26.1 and 26.2.)
|
26
||
CHAPTER
i, q
O
i
1q
2q
C
q versus t
i versus t
t
(26.16)
(26.17)
Household wiring: In household wiring systems, the various electrical devices are connected in
parallel across the power line, which consists of a pair of conductors, one “hot” and the other
“neutral.” An additional “ground” wire is included for safety. The maximum permissible current
in a circuit is determined by the size of the wires and the maximum temperature they can tolerate.
Protection against excessive current and the resulting fire hazard is provided by fuses or circuit
breakers. (See Example 26.14.)
873
CHAPTER
33
SUMMARY
Light and its properties: Light is an electromagnetic
wave. When emitted or absorbed, it also shows particle
properties. It is emitted by accelerated electric charges.
A wave front is a surface of constant phase; wave
fronts move with a speed equal to the propagation speed
of the wave. A ray is a line along the direction of propagation, perpendicular to the wave fronts.
When light is transmitted from one material to
another, the frequency of the light is unchanged, but the
wavelength and wave speed can change. The index of
refraction n of a material is the ratio of the speed of
light in vacuum c to the speed v in the material. If l0 is
the wavelength in vacuum, the same wave has a shorter
wavelength l in a medium with index of refraction n.
(See Example 33.2.)
n =
c
v
(33.1)
l =
l0
n
(33.5)
Reflection and refraction: At a smooth interface between
two optical materials, the incident, reflected, and
refracted rays and the normal to the interface all lie in a
single plane called the plane of incidence. The law of
reflection states that the angles of incidence and reflection are equal. The law of refraction relates the angles of
incidence and refraction to the indexes of refraction of
the materials. (See Examples 33.1 and 33.3.)
ur = ua
(law of reflection)
(33.2)
n a sin ua = n b sin ub
(law of refraction)
(33.4)
Total internal reflection: When a ray travels in a material
of greater index of refraction n a toward a material of
smaller index n b , total internal reflection occurs at the
interface when the angle of incidence exceeds a critical
angle ucrit . (See Example 33.4.)
sin ucrit =
Polarization of light: The direction of polarization of a
linearlyS polarized electromagnetic wave is the direction
of the E field. A polarizing filter passes waves that are
linearly polarized along its polarizing axis and blocks
waves polarized perpendicularly to that axis. When polarized light of intensity Imax is incident on a polarizing filter
used as an analyzer, the intensity I of the light transmitted
through the analyzer depends on the angle f between the
polarization direction of the incident light and the polarizing axis of the analyzer. (See Example 33.5.)
I = Imax cos2f
(Malus’s law)
Polarization by reflection: When unpolarized light strikes
an interface between two materials, Brewster’s law states
that the reflected light is completely polarized perpendicular to the plane of incidence (parallel to the interface)
if the angle of incidence equals the polarizing angle up .
(See Example 33.6.)
nb
na
(Brewster’s law)
Rays
Source
Wave fronts
Incident
ua
Reflected
nb
na
(33.6)
Normal
ub
Refracted
ur
na , nb
Material a
nb
na
ucrit
(33.7)
Material b
Incident
natural
light
f
. ucrit
E cos f
E cos f
S
E
f
Photocell
Analyzer
Polarizer
tan up =
Normal
(33.8)
Huygens’s principle: Huygens’s principle states that if the position of a wave front at one instant is
known, then the position of the front at a later time can be constructed by imagining the front as a
source of secondary wavelets. Huygens’s principle can be used to derive the laws of reflection and
refraction.
up
up
na
nb
ub
r 5 vt
B
A
A! B!
1105