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Circular Motion, Orbits,
and Gravity
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Looking Ahead
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Looking Back
You have previously been introduced to basic concepts of circular motion
and gravity. The goal of Chapter 6 is to learn about these concepts in more
detail. In this chapter you will learn to

Describe the kinematics of circular motion in terms of relevant
variables.

Understand the forces that produce circular motion.
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Analyze a wide range of physical phenomena as circular motion
problems.

Further develop your understanding of the force of gravity as a longrange force.
This chapter uses what you have learned about one-dimensional motion,
circular motion, Newton’s laws and the concept of appararent weight. Please
review:
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Sections 1.1-1.2: Coordinate systems and motion variables
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Sections 2.2-2.5: Equations for one-dimensional motion
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Section 3.6: Circular motion
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Section 4.3: The force of gravity
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Sections 4.6-4.8: Newton’s laws

Section 5.4: Apparent weight
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How are the motorcyclists in the “Globe of Death” able to ride their bikes up the sides
of the globe and even upside down over the top? This is one of the questions we will
address in this chapter.
Engines revving, the motorcyclists build up speed until they are riding in
excess of 30 miles per hour inside the spherical steel frame. As their speeds
increase, they begin to ride up the sides, culminating in complete vertical
loops around the top of the frame that take them completely upside down. In
this chapter we will consider circular motion problems that range from such
extreme examples to more prosaic situations such as cars going around
corners.
Circular motion is accelerated motion, as the velocity is always changing.
We will develop a new set of variables to describe circular motion and then
look at its dynamics, the forces that produce the motion. Finally, we will
consider a particular kind of circular motion—the orbital motion of the moon
about the earth and the planets about the sun—that will lead us to consider
the nature of gravity, the force that keeps things in orbit.
This is a very practical chapter, in which we will learn how to compute the
orbital speeds of satellites, why water stays in a bucket when you swing it
over your head, and why you need to drive more slowly on icy roads.
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6.1 Circular Motion: Constant Speed But Changing
Velocity
In Chapter 3, we considered the motion of a car on a Ferris wheel. The car
moves in a circle at a constant speed, but in a constantly changing direction.
A motion diagram for the Ferris wheel showed that such motion implied a
constant acceleration that was directed toward the center of the circle, which
we called a centripetal acceleration. We review the details of such motion in
Figure 6.1.
Figure 6.1 Velocity and acceleration vectors for circular motion.
We derived the details of circular motion for the particular example of a
Ferris wheel, but the motion any object moving in a circle can be described
in exactly the same way. Such circular motions are quite common; several
examples are illustrated in the table below. In each case, the velocity is
parallel to the circular path and the acceleration is directed toward the center.
In Chapter 4 we learned that an acceleration requires a force. For each of the
examples of motion in the table, there must be a force that produces the
centripetal acceleration. The force must be directed toward the center of
the circle, as this is the direction of the acceleration.
Circular Motion
A car going around a corner is
traveling in a circular arc. The
velocity is parallel to the circle,
and is constantly changing
direction. There must be a force
toward the center of the circle to
produce this centripetal
acceleration. This force is
provided by the friction between
the tires and the road.
KJF: College Physics Ch. 6, Draft 1 1/16/04
A ball at the end of a string is
swung in a vertical circle. The
tension in the string attached to the
ball provides the necessary force
to produce the centripetal
acceleration. The force is directed
toward the center of the circle,
along the length of the string.
The moon goes in an orbit around
the earth that is nearly circular.
There must be a force directed
toward the center of the circle to
produce the centripetal
acceleration. The force is provided
by the gravitational attraction
between the earth and the moon.
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The force the produces the centripetal acceleration of circular motion could
be provided by tension in a string, friction, gravity—or any of the other
forces we have seen.
We will begin our discussion of circular motion by defining variables that we
use to describe circular motion. After this, we will consider the dynamics of
such motion.
6.2 Describing Angular Motion
When a particle moves at constant speed around a circle of radius r, we call
this uniform circular motion. Figure 6.2 shows a particle moving in a
uniform circular motion. The particle might be a satellite moving in an orbit,
a ball on the end of a string, or even just a dot painted on the side of a wheel.
Regardless of what the particle represents, its velocity vector is always
parallel to the circular path; we say that the velocity vector is tangent to the
circle. The particle’s speed v is constant, so the vector’s length stays
constant as the particle moves around the circle.
KJF: College Physics Ch. 6, Draft 1 1/16/04
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Figure 6.2 A particle in uniform circular motion.
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We will now define an appropriate set of coordinates and variables to
characterize circular motion, much as we did for one-dimensional motion in
Chapter 1.
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Period
The time interval it takes the particle to go around the circle once,
completing one revolution (abbreviated rev), is called the period of the
motion. Period is represented by the symbol T. This is a logical symbol,
because period is an interval of time, but there’s a risk of confusing the
period T with the symbol T for tension or, even worse, the identical symbol
T for the magnitude of the tension force.
The number of symbols used in science and engineering far
exceeds the number of letters in the English alphabet. Even after we’ve
borrowed from the Greek alphabet, it’s inevitable that some letters are
used several times to represent entirely different quantities. The use of T
is the first time we’ve run into this problem, but it won’t be the last. You
must be alert to the context of a symbol’s use in order to deduce its
meaning. b
NOTE c
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The period of the orbit of the earth around the sun is one year.
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Rather than specify the time for one revolution, circular motion may be
specified by its frequency, the number of revolutions per second, for which
we use the symbol f. The frequency is the inverse of the period:
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f
1
T
KJF: College Physics Ch. 6, Draft 1 1/16/04
(6.1)
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The units of frequency are inverse seconds, or s-1 , for which we also use the
name hertz (Hz). Frequency may also be given in revolutions per minute or
another time interval.
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EXAMPLE 6.1 Period of an audio CD
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An audio CD spins at a frequency of up to 540 rpm (revolutions per minute).
At 540 rpm, how much time is required for one revolution of the CD?
Prepare First, we convert units from minutes to seconds:
revolutions 1 minute
rev

 9.0
minute
60 seconds
s
A “rev” is not a unit, so we can simplify to
f  540
1
 9.0 s-1  9.0 Hz
s
Solve We rearrange Equation (6.1) to read
f  9.0
T
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1
f
and then compute
T
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 0.111 s
f 9.0 s-1
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Assess 0.111 second is 1/9 of a second, which makes sense, as the
frequency is 9 rev/s.
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Angular Position
Rather than using xy-coordinates, it will be more convenient to describe the
position of the particle by its distance r from the center of the circle and its
angle  from the positive x-axis. This is shown in Figure 6.3. The angle 
is the angular position of the particle.
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Figure 6.3 A particle’s position is described by distance r and angle θ.
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We can distinguish a position above the x-axis from a position an equal angle
below the x-axis by defining  to be positive when measured
counterclockwise from the positive x-axis. An angle measured clockwise
from the positive x-axis has a negative value. “Clockwise” and
“counterclockwise” in circular motion are analogous, respectively, to “left of
the origin” and “right of the origin” in linear motion, which we associated
with negative and positive values of x.
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Rather than measure angles in degrees, mathematicians and scientists usually
measure the angle  in the angular unit of radians. In Figure 6.3, we have
noted the arc length s that the particle has traveled along the edge of the
circle of radius r. We define the particle’s angle  in radians in terms of this
arc length and the radius of the circle:
s
(6.2)
r
The radian, which is abbreviated rad, is the SI unit of an angle. An angle of 1
rad has an arc length s exactly equal to the radius r.
 (radians) 
The arc length completely around a circle is the circle’s circumference 2 r.
Thus the angle of a full circle is
2 r
 2 rad
r
We can use this to define conversion factors among revolutions, radians and
degrees:
1 rev  360  2 rad
full circle 
360
 57.3  60
2 rad
We will often specify angles in degrees, but keep in mind that the SI unit is
the radian.
1 rad  1 rad 
An important consequence of Equation (6.2) is that the arc length spanning
the angle  is
s  r
(6.3)
This is valid only if  is measured in radians and not in degrees. This very
simple relationship between angle and arc length is one of the primary
motivations for using radians.
While the concept of measuring an angle is simple, units of
angle are often troublesome. Unlike the kilogram or the second, for
which we have standards, the radian is a defined unit. Further, its
definition as a ratio of two lengths makes it a pure number without
dimensions. Thus the unit of angle, be it radians or degrees or
revolutions, is really just a name to remind us that we’re dealing with an
angle. The practical implication is that the radian unit sometimes appears
or disappears without warning. This seems rather mysterious until you
get used to it. This textbook will call your attention to such behavior the
first few times it occurs. With a little practice, you’ll soon learn when the
rad unit is needed and when it’s not. b
NOTE c
Angular Velocity
Figure 6.4 shows a particle moving in a circle from an initial angular position
 i at time t i to a final angular position  f at a later time t f . The change
   f   i is called the angular displacement. We can measure the
particle’s circular motion in terms of the rate of change of  just as we
measured the particle’s linear motion in terms of the rate of change of its
position x.
KJF: College Physics Ch. 6, Draft 1 1/16/04
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FIGURE 6.4 A particle moves from
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In analogy with linear motion, we define the angular velocity to be
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i
to
f
with angular velocity
.
angular displacement 

time interval
t
Angular velocity of a particle in uniform circular motion

(6.4)
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The symbol  is a lowercase Greek omega, not an ordinary w. The SI unit
of angular velocity is rad/s.
Angular velocity is the rate at which a particle’s angular position is changing.
A particle that starts from   0 rad with an angular velocity of 0.5 rad/s will
be at angle   0.5 rad after 1 s, at   1.0 rad after 2 s, at   1.5 rad after
3 s, and so on. Its angular position is increasing at the rate of 0.5 radians per
second. In analogy with uniform linear motion, which you studied in Chapter
2, uniform circular motion is motion in which the angle increases at a
constant rate: A particle moves with uniform circular motion if and only
if its angular velocity ω is constant and unchanging.
Angular velocity, like the velocity v of one-dimensional motion, can be
positive or negative. The signs shown in Figure 6.5 are based on the fact that
 was defined to be positive for a counterclockwise rotation.
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Figure 6.5 Positive and negative angular velocities.
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Circular motion is analogous to linear motion with angular variables
replacing linear variables. Much of what you learned about linear kinematics
and dynamics carries over to circular motion; we will illustrate with an
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example. For motion with constant velocity, Equation 2.6 gave us a formula
for computing a displacement during a time interval:
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We can write down a similar equation for angular motion, substituting angle
 for position x and angular velocity  for linear velocity v:
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Angular displacement for uniform circular motion
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xf  xi  x  vt
 i   f    t
We will often be able to deduce equations for angular motion from their
linear equivalents.
The angular velocity  is closely related to the period T and the frequency f
of the motion. As a particle goes around a circle one time, its angular
displacement is   2 rad during the interval T. Thus, using the definition
of angular velocity, we find
2 rad
T
We can also express the angular velocity in terms of the frequency:
 
  (2 rad) f
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(6.5)
(6.6)
(6.7)
Knowing the period or the frequency alone gives only the absolute value of
 . To determine the sign of  you need to know the direction of motion.
You can use the following table to deduce angular velocity in rad/s given the
frequency in rev/s or rpm.
TABLE 6.6.1 Frequency and angular velocity conversions
Frequency
Equivalent angular velocity
1.00 rev/s
1.00 rpm
1
/
2


 rev/s  .159 rev/s
30 /   rpm  9.55 rpm
2 rad/s  6.28 rad/s
 / 30  rad/s  .0105 rad/s
1.00 rad/s
1.00 rad/s
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EXAMPLE 25.2 At the roulette wheel
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Prepare Assume the ball is in uniform circular motion.
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A small, steel roulette ball rolls around the inside of a 30-cm-diameter
roulette wheel. The ball completes 2 rev in 1.20 s.
a. What is the ball’s angular velocity?
b. What is the ball’s position at t  2.0 s? Assume  i  0.
Solve a. The period of the ball’s motion, the time for 1 rev, is T  0.60 s.
Thus

2 rad 2 rad

 10.47 rad/s
T
0.60 s
b. The ball starts at  i  0 rad. After t  2.0 s, its position is given by
Equation (6.5):
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f  i  t  0 rad  10.47 rad/s2.0 s  20.94 rad
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Assess Although this is a mathematically acceptable answer, an observer
would say that the ball is always located somewhere between 0° and 360°.
Thus it is common practice to subtract off an integer multiple of 2 ,
representing the completed revolutions. Because 20.94 / 2  3.333, we can
write
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 f  20.94 rad  3.333  2 rad
 3  2 rad  0.33  2 rad
 3  2 rad  2.09 rad
In other words, at the ball has completed 3 rev and is 2.09 rad  120 into
its fourth revolution. An observer would say that the ball’s position is
  120.
Angular Acceleration
Most of the cases we will consider in this chapter involve constant angular
velocity, at least during the time of the motion we will consider. But there are
many examples of motion for which the angular velocity is changing. For
instance, in the above example of the roulette wheel, the ball will eventually
slow down and stop, meaning the angular velocity has decreased to zero.
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Figure 6.6 A particle moves in a circle with changing angular velocity.
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Figure 6.6 shows the motion of a particle with an increasing angular velocity.
At time t i the angular velocity is  i ; at the later time t f the angular velocity
is  f . The change in angular velocity during this time interval is
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   f   i
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The rate of change of velocity is acceleration; we define the rate of change of
angular velocity as angular acceleration. In analogy with linear motion, we
define the angular acceleration to be
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change in angular velocity 
(6.8)

time interval
t
Angular acceleration for a particle in circular motion
The symbol for angular acceleration is the Greek letter  ; the units are
rad/s2. The sign of  depends on the sign of the change in angular
velocity—again in analogy with the one-dimensional motion situation.
As we have seen, the definitions of angular velocity and angular acceleration
can be easily deduced by analogy with the related variable for onedimensional motion. Table 6.2 lists the variables for one-dimensional motion
and angular motion and their respective definitions side-by-side.

TABLE 6.2 Linear and angular motion variables
1D motion
variable
Postion
x
Velocity
v
Acceleration
a
Angular
motion
variable
1D motion definition
Angular motion definition
Angle
Displacement ∆x:
Angular displacement  :

x  xf  xi
   f   i
Angular
velocity
v
x
t


t
Angular
acceleration
a
v
t


t


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All of the equations for one-dimensional motion have rotational analogs as
well. We saw one example above; Equation (6.5) was obtained from a onedimensional motion equation by substituting angle for position and angular
velocity for velocity. Table 6.3 lists some one-dimensional motion equations
and the analogous equations for angular motion.
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TABLE 6.3 Linear and angular motion equations
1D motion equation
Angular motion equation
Displacement of object moving
at constant speed:
Angular displacement of
object moving at constant
angular speed:
x  vt
Change in velocity of object
undergoing constant
acceleration:
  t
Change in angular velocity of
object undergoing constant
angular acceleration:
v  at
  t
Displacement of object
undergoing constant
acceleration:
Angular displacement of
object undergoing constant
angular acceleration:
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x  v0 t  at 2
2
  0   0t   t 2
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EXAMPLE 6.3 Spinning up a disk drive platter
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The platter in a computer disk drive spins up to a steady 5400 rpm in a time
of 2.0 seconds. What is the angular acceleration? At the end of 2.0 seconds,
how many revolutions has the platter made?
Prepare We convert 5400 rpm into angular velocity:
5400 rev 1 min 2 rad


 565 rad/s
min
60 s
1 rev
Solve Using the definition of angular acceleration in Table 6.2, we compute:
 565 rad/s

 283 rad/s2
t
2.0 s
During the time of this angular acceleration, we can compute the angular
displacement using an equation from Table 6.3:



1
1
2
   0 t   t 2  0 rad/s   283 rad/s2 2.0 s 
2
2
 566 rad
Each revolution corresponds to an angular displacement of 2π, so we
compute:
566 rad
 90 revolutions
2 rad/revolution
The platter completes 90 complete revolutions during the first two seconds.
Assess As you might expect, the solution to this problem is exactly
analogous to the solution of a one-dimensional motion problem.
Number of revolutions 
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In the rest of the chapter, we will consider situations in which the angular
velocity is constant; there will be no angular acceleration. When we speak of
acceleration, we will mean the centripetal acceleration.
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6.3 Velocity and Acceleration in Uniform Circular
Motion
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Velocity
A person swings a ball on the end of a string in a
horizontal circle once every second. For each of the following quantities, tell
whether they are zero, constant (but not zero) or changing.
STOP TO THINK 6.1
A. Period
B. Velocity
C. Angular velocity
D. Acceleration
E. Angular Acceleration
In the previous section we described uniform circular motion in terms of
angular variables. In Chapter 3, we looked at a description of uniform
circular motion in terms of velocity and acceleration. In this section we will
describe the relationship between these two different ways of describing the
motion.
It’s easy to relate the particle’s period T to its speed v, as we saw at the end
of Chapter 3. For a particle moving with constant speed, speed is simply
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distance/time. In one period, the particle moves once around a circle of
radius r and travels the circumference 2 r. Thus
v
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(6.9)
If we combine this result with Equation (6.6) for the angular velocity, we can
deduce a relationship between the speed v and the angular velocity  :
v
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1 circumference 2 r

1 period
T
2 r  2 
   r  r
 T 
T
(6.10)
In Equation (6.10),  must be in units of rad/s. If you are given
a frequency in rev/s or rpm, you must convert to an angular velocity in
rad/s. b
NOTE c
EXAMPLE 6.4 Speed at the edge of a CD
The diameter of an audio compact disc is 12.0 cm. When the disk is spinning
at 540 rpm, as in Example 6.1, at what speed is a point moving a) at the
outside edge of the disk, and b) at a distance 3.0 cm from the axis?
Prepare At 540 rpm, the period is 0.111 s, from Example 6.1. We compute
the angular velocity using Equation (6.6):
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2 rad 2 rad

 56.6 rad/s
T
0.111 s
The disk as a whole turns at this angular velocity. But different points on the
disk move at different speeds, as we see in Figure 6.7.
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Figure 6.7 The rotation of an audio compact disc
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
Solve Now that we have  in the correct units, we compute the speed for
each point using Equation (6.10):
a) At the outside edge of the disc, r  6.0 cm  0.060 m and we
find:
KJF: College Physics Ch. 6, Draft 1 1/16/04
v   r  56.6 rad/s.060 m   3.4 m/s
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b) At r  3.0 cm  0.030 m, we compute:
v   r  56.6 rad/s.030 m   1.7 m/s
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to two significant figures.
Assess The velocities are a few m/s, which seems reasonable. The point
farther from the axis is moving at a higher speed, as we expect.
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CONCEPTUAL EXAMPLE 6.1 Spin rate of a CD
As an audio CD plays, a laser reads the data from the inside of the disc
toward the outside. At the position where the data is being read, the disc must
be moving at a constant speed. This means that the angular velocity of the
disc must change as different parts of the disc are read. As the laser reads
tracks farther from the center of the disk, does the angular velocity increase
or decrease?
Reason We consider the relationship of different variables given in
Equation (6.10)
As the laser reads points farther out from the center, at larger values of r, the
angular velocity must decrease.
Assess In fact, the angular speed of an audio compact disc does change as it
is read. At the very start, when the laser is near the center of the disc, the
frequency is about 540 rpm; by the time the laser is reading tracks near the
outside of the disc, it is just over 200 rpm.
Acceleration
In Chapter 3, we derived a relationship between the velocity and acceleration
for uniform circular motion:
v2
r
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We can combine this relationship with the relationship between v and  in
Equation (6.10) to obtain a relationship between a and  :
v2  r 
(6.11)

  2r
r
r
Centripetal acceleration for uniform circular motion
Recall that this is a centripetal acceleration, which is directed toward the
center of the circle. The angular velocity is constant, but there is an
acceleration as the direction of the velocity is changing. A larger angular
velocity will correspond to a greater acceleration.
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EXAMPLE 6.5 Maximum rotation frequency of a carnival ride
In the “Quasar” carnival ride shown in Figure 6.8, passengers travel in a
horizontal circle of radius 5.0 meters. For safe operation, the maximum
sustained acceleration that riders may experience is 2 g. What is the
maximum frequency of this ride in rpm?
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Figure 6.8 The Quasar
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Prepare We begin with the pictorial representation in Figure 6.9, a top view
of the ride showing all relevant variables.
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Figure 6.9 Pictorial representation for the Quasar carnival ride.
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Solve Running the ride at its maximum frequency will produce the
maximum acceleration. Therefore, we can simply solve for the angular
velocity and frequency at the maximum acceleration. We compute the
maximum possible angular velocity from the maximum acceleration by
rearranging Equation (6.11):
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 max 
amax
19.6 m/s 2

 1.98 rad/s
r
5.0 m
We can convert this to a frequency in rpm by using the conversions from
Table 6.1. 1.00 rad/s is equivalent to 9.55 rpm:
 9.55 rpm 
fmax  1.98 rad/s 
 18.9 rpm
 1.00 rad/s 
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Assess 18.9 rpm is one rotation in just over three seconds; this seems
reasonable for a pretty zippy carnival ride. In fact, the maximum rated rpm
for the ride is 16 rpm, a bit less than the maximum safe value.
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STOP TO THINK 6.2 Rank in order, from largest to smallest, the
centripetal accelerations of particles a to d.
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6.4 Dynamics of Circular Motion
A particle in uniform circular motion is clearly not traveling at constant
velocity in a straight line. Consequently, according to Newton’s first law, the
particle must have a net force acting on it. We’ve already determined the
acceleration of a particle in uniform circular motion—the centripetal
acceleration of Equation (3.9) and Equation (6.11). Newton’s second law
tells us exactly how much net force is needed to cause this acceleration:

r  mv 2
Fnet  ma  
, toward center of circle 
 r

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(6.12)
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Net force producing the centripetal acceleration of circular motion
In other words, a particle of mass m moving at constant speed v around a
circle of radius r must have a net force of magnitude mv2 / r pointing toward
the center of the circle, as in Figure 6.10. Without such a force, the particle
would move off in a straight line tangent to the circle.
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Figure 6.10 Net force for circular motion.
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The force described by Equation (6.12) is not a new force. Our
rules for identifying forces have not changed. What we are saying is that
a particle moves with uniform circular motion if and only if a net force
always points toward the center of the circle. The force itself must have
an identifiable agent and will be one of our familiar forces, such as
tension, friction, or the normal force. Equation (6.12) simply tells us how
NOTE c
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the force needs to act—how strong and in which direction—to cause the
particle to move with speed v in a circle of radius r. b
We can clarify these ideas by looking at some examples of circular motion
and how it is produced by familiar forces. First, we will sum up what we
have learned in a Problem-Solving Strategy.
PROBLEM-SOLVING STRATEGY 6.1 Circular dynamics problems
Prepare We have one basic equation for circular dynamics problems:
Equation (6.12). This is just a version of Newton’s second law; the right side
is a net force, the left a mass times an acceleration, the centripetal
acceleration. Circular dynamics problems thus tend to break out into two
different types:
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Details of the motion are given, and questions asked about the forces
that produce the motion.

Details of the forces are given, and questions are asked about the
resulting motion.
In each case, it is a good idea to begin with a sketch. For completeness, you
may wish to draw both a pictorial and a physical representation:
Pictorial Representation: Sketch the motion, showing relevant
physical dimensions. Note all relevant variables, and record values of
all known quantities. List quantities to be determined.
Physical Representation: The main component will be a free-body
diagram, showing all of the forces involved.
Solve The solution will use Equation (6.12). For the two types of problem,
you should either

Begin with motion calculations, apply Equation (6.12) to determine
the net force, and complete an analysis of forces, or

Begin with force calculations, determine a net force, and then use
Equation (6.12) to compute a velocity, from which other motion
variables can be determined.
Assess A few things that are easy to check for this kind of problem:

Make sure that your net force points toward the center of the circle.

Consider the values for forces, velocities, periods. Are they
reasonable?
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EXAMPLE 6.6 Spinning in a circle
An energetic father places his 20 kg child on a 5.0 kg cart to which a 2.0-mlong rope is attached. He then holds the end of the rope and spins the cart and
child around in a circle, keeping the rope parallel to the ground. If the tension
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in the rope is 100 N, how much time does it take for the cart to make one
rotation? Rolling friction between the cart’s wheels and the ground is
negligible.
Prepare We proceed according to the steps of Problem-Solving Strategy 6.1.
Figure 6.11 shows the pictorial and physical representations of the problem.
The main idea of the pictorial representation is to illustrate the relevant
geometry and to define the symbols that will be used. A circular dynamics
problem usually does not have starting and ending points like a projectile
problem, so numerical subscripts such as x1 or y2 are usually not needed.
Here we need to define the cart’s speed v and the radius r of the circle.
Notice that there are two quantities for which we use the symbol T: the
tension and the period. We will include additional information when
necessary to distinguish the two.
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Figure 6.11 Pictorial and physical representations of the cart spinning in a circle.
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A motion diagram is not needed for uniform circular motion because we
know the acceleration a points to the center of the circle. The essential part
of the physical representation is the free-body diagram. For uniform
circular motion we’ll draw the free-body diagram looking at the edge of
the circle, because this is the plane of the forces. There are three forces
acting on the cart: The weight force w, the normal force of the ground n
and the tension force of the rope T. There is no vertical motion, so w and
n must be equal and opposite; the net force is just the tension force T.
Solve We begin by using Equation (6.12) and noting that the net force is
equal to the tension:
Fnet  T 
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mv 2
r
We know the mass, the radius of the circle, and the tension, so we can solve
for v:
v
Tr

m
100 N 2.0 m 
25 kg
 2.83 m/s
From this, we compute the period using Equation (6.9):
v
2 r
2 r 2 2.0 m 
T 

 4.4 s
T
v
2.83 m/s
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Assess Once around the circle in just over 4 seconds at a speed of about 6
miles per hour sounds reasonable; this is a fast ride, but not so fast as to be
scary!
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This has been a fairly typical circular dynamics problem. You might want to
think about how the solution would change if the rope is not parallel to the
ground.
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EXAMPLE 6.7 Turning the corner
What is the maximum speed with which a 1500 kg car can make a left turn
around a curve of radius 20 m on a level (unbanked) road without sliding?
(This radius turn is about what you might expect at a major intersection in a
city.)
Prepare We will start with diagrams, the pictorial and physical
representations of Figure 6.12. The car moves along a circular arc at constant
speed for the quarter-circle necessary to complete the turn. The motion
before and after the turn is not relevant to the problem. The more interesting
issue is how a car turns a corner. What force or forces can we identify that
cause the direction of the velocity vector to change? Imagine you are driving
a car on a completely frictionless road, such as a very icy road. You would
not be able to turn a corner. Turning the steering wheel would be of no use;
the car would slide straight ahead, in accordance with both Newton’s first
law and the experience of anyone who has ever driven on ice! So it must be
friction that somehow allows the car to turn.
Figure 6.12 Pictorial and physical representations of a car turning a corner.
The force-identification section of Figure 6.12 shows the top view of a tire as
it turns a corner. If the road surface were frictionless, the tire would slide
straight ahead. The force that prevents an object from sliding across a surface
is static friction. Static friction fs pushes sideways on the tire, toward the
center of the circle. How do we know the direction is sideways? If fs had a
component either parallel to v or opposite to v, it would cause the car to
speed up or slow down. Because the car changes direction but not speed,
static friction must be perpendicular to v Thus fs causes the centripetal
acceleration of circular motion around the curve. With this in mind, the freebody diagram, drawn from behind the car, shows the static friction force
pointing toward the center of the circle. There are three forces acting on the
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car: The weight force w, the normal force n, and the force of static friction
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and the net force is the force of static friction. Recall from Equation 5.10 in
Chapter 5 that static friction has a maximum possible value:
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Solve The next step is to use Equation (6.12), noting that the net force is the
force of static friction.
fs . As in the previous example, because there is no vertical motion the
weight force and the normal force must be equal and opposite. The
magnitude of the normal force is just equal to the weight:
n  w  mg
fs max  s n
mv 2
r
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Fnet  fs 
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The only difference between this example and the previous one is that the
tension force toward the center has been replaced by a static friction force
toward the center.
Because the static friction force has a maximum value, there will be a
maximum speed with which a car can turn without sliding. The maximum
speed is reached when the static friction force reaches its maximum
fs max  s n If the car enters the curve at a speed higher than the maximum,
static friction will not be large enough to provide the necessary centripetal
acceleration and the car will slide. The maximum speed occurs at the
maximum value of the force of static friction:
fs max 
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2
mvmax
r
We can put in what we know about static friction to find:
2
mvmax
 fs max  s n  s w  s mg
r
2
vmax
 s gr
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vmax  s gr
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For rubber tires on pavement, we can find the relevant value of s in Table
5.1; it is 1.0. We can then complete our calculation:
vmax  s gr 
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1.0 9.8 m/s2 20 m   14.0 m/s
Assess 14.0 m/s  30 mph, which seems like a reasonable upper limit for
the speed at which a car can turn a corner. There are a few other things to
note about the solution:

The car’s mass canceled out. The maximum speed does not depend
on the mass of the vehicle. This makes sense.

Consider the final expression for the vmax ; it does depend on the
coefficient of friction and the radius of the turn. vmax decreases if 
does (meaning the maximum value of friction is less) or if r is less
(meaning a tighter turn.) Both of these make sense as well.
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Because vmax depends on s and because s depends on road conditions, the
maximum safe speed through turns can vary dramatically. Wet or icy roads
lower the value of s and thus lower the maximum speed of turns. A car that
handles normally while driving straight ahead on a wet road can suddenly
slide out of control when turning a corner. Icy conditions are even worse. If
your lower the value of the coefficient of friction in Example 6.7 from 1.0
(dry pavement) to 0.1 (icy pavement), the maximum speed for the turn is 4.4
m/s—about 10 mph!
Cars with wings Race cars may have wings, not to give them lift, but to
provide an additional force pushing the car into the pavement. The wings
deflect air upward which means the air pushes down on the car. This down
force may be twice the weight of the car. The down force increases the
normal force from the pavement, increasing the maximum friction force,
making faster turns possible.
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Banked turns If a road is banked, a component of the normal force of the
road will be directed toward the center of the turn. This additional force will
allow for a larger centripetal acceleration than would be possible with
friction alone.
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CONCEPTUAL EXAMPLE 6.2 Will the swing break?
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A man has used a rope to make a swing hung from a tree branch. The rope is
just strong enough to support his weight, but no more. He pulls himself back
on the swing and lets go. Will the rope break?
Reason As we saw in Chapter 3, the motion of a person in a swing is
circular motion: The swing moves in an arc at a constant distance from a
central point. Consider the swing at the lowest point of its arc. Figure 6.13
show pictorial and physical representations at this point. The tension in the
rope must be pulling upward on the man with a force greater than his weight,
as there is a net upward force on the man necessary to keep him moving in a
circle. The rope cannot provide this force, and will break.
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Figure 6.13 Motion of a person in a swing.
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Assess Note that the swing is in circular motion even though it does not
travel through a complete circle; it travels through an arc of a circle.
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Until now, we have considered circular motion from the point of view of a
stationary observer. In the next section of the chapter we will also consider
the acceleration of circular motion from the point of view of the person or
object undergoing the motion.
STOP TO THINK 6.3 An athlete is performing the track and field event called
the hammer throw. She swings the hammer (a large weight at the end of a
chain) in a circle and the releases it so that it will fly through the air. The
figure shows a top view of the circular motion; the hammer is moving
counterclockwise. At which point in the circle should she release the hammer
so that it will travel in the noted direction?
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6.5 Apparent Forces in Circular Motion
We have seen that motion in a circle implies an acceleration. Suppose that
several people were standing in a room that could be rotated around its center
on an axle. Everyone in the room would be going in a circle, meaning
everyone would be experiencing an acceleration. The direction of the
acceleration, and the direction of the force producing this acceleration, would
be directed toward the center of the circle, toward the center of the room. But
what would the people in the room feel?
Figure 6.14 shows the a carnival ride that works just this way; it is a room
that rotates around its center. The people in the second panel are inside this
rotating room. We know that they are being pushed inward, toward the center
of the circle. But you can see from the photo that the people feel an outward
force; they feel as though they are pushed outward, toward the wall. In the
next section, we will explore how to explain this difference in perception.
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Figure 6.14 Outside and inside views of the Gravitron, a rotating circular room.
Centrifugal Force?
If you are riding in a car that turns a sharp corner, you may feel as if a force
pushes you toward the door. But there really is no such force. You can not
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identify any agent that does the pushing; there is no physical force pushing
you toward the door.
We will introduce a new concept here, the reference frame. When you are
riding in a car, this is your reference frame. When you are traveling down the
highway at a constant velocity, everything in the car moves along with you.
If you were to toss a ball in the air, it would go up and come back down in
your hand. When the car is moving at a constant speed, we will call it an
inertial reference frame. In an inertial reference frame, the laws of physics
work just like they do in the laboratory—which is, to a good approximation,
an inertial reference frame.
But a car going around a corner is accelerating, and so is not an inertial
reference frame. In such a noninertial reference frame Newton’s laws are
not valid. If you were to toss a ball in the air in the car, when it was going
around the corner, it wouldn’t come back down in your hand. In a noninertial
frame, odd things happen; objects follow unusual paths, and strange forces
seem to exist. But we don’t need any new physics to explain this—just a
different point of view. We must consider the situation from the point of
view of someone in an inertial reference frame.
If you are a passenger in a car that turns a corner quickly, you feel a force
pushing you against the door, as we noted. But is there really such a force?
Let’s look at the situation from the point of view of someone in an inertial
reference frame, someone suspended above the road with a bird’s-eye view
of you riding in a car as it makes a left turn. This person will see you going in
a circle, which requires a force. From their point of view, it is the normal
force of the door, pointing inward toward the center of the curve, that is
causing you to turn the corner. You are not pushed into the door; the door
pushes into you! The bird’s-eye view, from an inertial reference frame, gives
the proper perspective of what actually happens.
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Figure 6.15 Bird’s-eye view of a passenger in a car turning a corner.
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Now, let’s revisit the example of the rotating room carnival ride. In this case,
passengers in the room spin with it. A person watching from above would
see people in the room moving in a circle, with the walls providing the
necessary inward force to cause them to move in a circular path.
But, for a person in the room, the situation appears quite different: There
appears to be a very strong outward force, as we saw in Figure 6.14. You
may have heard this referred to as centrifugal force. But viewed from above,
from the point of view of an inertial reference frame, there is no force
pushing the person out. The centrifugal force does not exist; it is not a real
force. The centrifugal force is what we refer to as a fictitious force. It is an
apparent force that arises only because the frame of reference is accelerating.
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A centrifugal force will never appear on a free-body diagram and never
be included in Newton’s laws.
In Chapter 5, we explored another example of an apparent force that results
from the acceleration of a reference frame, apparent weight. A person in an
elevator that is accelerating will sense a change in her weight. But recall that
the sensation of weight is really due to contact forces. In the rotating room of
the Gravitron, the contact force of the wall, which is pushing in, will give a
person in the room a sensation of a force that is pushing out.
We will avoid using the concept of centrifugal force to solve problems, as it
is not a real force. We will talk about the apparent force than an object in
circular motion feels, as we will see in examples later in the chapter; our
treatment will be analogous to that of apparent weight that you saw in
Chapter 5.
TRY IT YOURSELF
Catch on a curve A good place to explore a non-inertial reference frame is
on a merry-go-round at the playground. As you spin, note what forces you
feel. Try tossing a ball up in the air so that you can catch it again; this is
surprisingly hard! This is a wonderful way to see fictitious forces in action.
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Why Does the Water Stay in the Bucket?
Imagine swinging a bucket of water over your head. If you swing the bucket
quickly, the water stays in. But you’ll get a shower if you swing too slowly.
Why does the water stay in the bucket? You might have said that there was a
centrifugal force holding the water in, but we have seen that there really isn’t
a centrifugal force. Analyzing this question will tell us a lot about forces in
general and circular motion in particular.
Suppose you swing a bucket of water fast enough for the water to stay in.
Figure 6.16 shows the bucket at the top of a circle of radius r. At this
moment, the water has a velocity that is tangent to the circle. If the bucket
suddenly disappeared, the water wouldn’t fall straight down. Instead, it
would fall along a parabolic trajectory like a ball that is thrown horizontally.
This is the motion of an object acted on only by gravity.
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Figure 6.16 Water in a bucket moving in a circle.
When you swing the bucket, you are making the water follow a path with
with more curvature than the parabolic path, so it needs more force than just
its weight. This extra force is provided by the bottom of the bucket pushing
on the water. As long as the bucket is pushing on the water, the water stays in
the bucket.
Notice the similarity to the car making the left turn in Figure 6.15. The
passenger feels like he’s being “hurled” into the door by a centrifugal force,
but it’s actually the contact force from the door, pushing inward toward the
center of the circle, that causes the passenger to turn the corner instead of
moving straight ahead. Here it seems like the water is being “pinned” against
the bottom of the bucket by a centrifugal force, but it’s really the pushing
force from the bottom of the bucket that causes the water to move in a circle
instead of following a free-fall parabola.
You know that there will be a certain critical angular velocity  c for
swinging the bucket. For angular velocity    c , the water stays in the
bucket; for    c , it won’t. At an angular velocity    c the bucket is
going just fast enough. Figure 6.17 shows a series of diagrams of the motion
of the bucket for these three cases.
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Figure 6.17 The forces on the water in a bucket.
For    c , at the top of the arc, there are two forces acting on the water:
The weight force, which is directed down, and the normal force of the
bucket, which is directed toward center of the circle. The magnitude of the
net force necessary to keep the water moving in the circle is w+n.
If you gradually slow the speed of the bucket, the normal force of the bucket
on the water gets smaller and smaller. There comes a point, as the angular
velocity  decreases, when n reaches zero at the top point of the circle, and
the magnitude of the net force is just w. At this point, the bucket is not
pushing against the water. Instead, the water is able to complete the circle
because the weight force alone provides sufficient centripetal acceleration.
The critical angular velocity v c is the angular velocity at which the
weight force alone is sufficient to cause circular motion at the top.
Circular motion with an angular velocity less than  c isn’t possible because
there is too much downward force. If you attempt to swing the bucket with a
smaller angular velocity, the normal force drops to zero before the water
reaches the top. “No normal force” means “no contact.” The water leaves the
bucket when n becomes zero, becoming a projectile moving under the
influence of only the weight force. That’s when you get wet!
We can calculate the critical angular velocity by using Equation (6.12),
noting that the net force is equal to the weight, and replacing the velocity
with the angular velocity:
mv 2 m  c r 
Fnet  w  mg 

 m c2 r
r
r
2
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Solving for the critical angular velocity, we find:
c 
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g
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(6.13)
At the beginning of the chapter, we considered the case of a motorcyclist
riding in a spherical cage. It is possible to ride upside down in the cage for
exactly the same reason that the water stays in the bucket—as long as you
ride fast enough! The velocity must be high enough that, at the top of the arc,
the force necessary to keep the motorcycle moving in a circle is greater than
the weight. In this case, there will be a normal force, and the tires will stay in
contact with the cage.
EXAMPLE 6.8 How slow can you go?
A motorcyclist is riding in the Globe of Death, as pictured at the start of the
chapter. The radius is 2.2 m. The rider is executing a vertical loop. In order to
keep control of the bike, the rider wants to keep the normal force between the
tires and the cage equal to the weight at the top of the loop. What is the
minimum speed at which the rider can take the loop?
Prepare The physical and pictorial representations for the motion are shown
in Figure 6.18. We can use the above diagrams for the bucket as a guide.
Figure 6.18 Riding in a vertical loop.
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At the top of the loop, at the minimum speed, the normal force and the
weight force are equal in magnitude and direction, so we can say that
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Solve The net force is the force that keeps the motorcyclist moving in a
circle, so we can say that
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Fnet 
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Fnet  w  n  2w
Equating these two expressions for Fnet and solving for v we get
2w 
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v
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mv 2
r
mv 2
r
2wr

m


2mgr
 2gr  2 9.8 m/s 2 2.2 m   6.6 m/s  15 mph
m
Assess This isn’t all that fast; the bikes can easily reach this speed. The big
challenge is to keep all riders moving at this speed in synchrony; at this
speed in this globe, the period of the motion will be T  2 r / v  1 s!
KJF: College Physics Ch. 6, Draft 1 1/16/04
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6.6 Carnival Rides and Centrifuges
We know that centrifugal force isn’t a real force. But moving in a circle
causes some real effects. Passengers on carnival rides going in circles
experience accelerations that produce real physical effects. And even though
there is no such thing as centrifugal force, there is such a thing as a
centrifuge, a machine that can spin fast enough to cause particles to settle
out of a fluid. In this section we will explore these effects.
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A centrifuge spins tubes of liquids at high rotational velocities. If the liquid is blood,
this motion will cause it to separate into different fractions.
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Apparent Weight on a Roller Coaster
Figure 6.19a shows a roller coaster car going around a vertical loop-the-loop
of radius r. If you’ve ever ridden a roller coaster, you know that your
sensation of weight changes as you go over the crests and through the dips.
To understand why, let’s look at the forces on passengers going through the
loop.
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Figure 6.19 A roller coaster car going around a loop-the-loop.
Figure 6.19b shows a passenger’s free-body diagram at the bottom of the
loop. The only forces acting on the passenger are her weight w and the
normal force n of the seat pushing up on her. Recall, from Chapter 5, that
the passenger’s apparent weight, her sensation of weight, is the magnitude of
the force supporting her. Here the seat is supporting her with the normal
force n, so her apparent weight is wapp  n Based on our understanding of
circular motion, we can say that:

She’s moving in a circle, so there must be a net force toward the
center of the circle—above her head—to provide the centripetal
acceleration.

The net force points upward, so it must be the case that n > w

Her apparent weight is wapp  n, so her apparent weight is larger


than her true weight wapp  w  Thus she “feels heavy” at the
bottom of the circle.
In short, the normal force has to exceed the weight force to provide the net
force she needs to “turn the corner” at the bottom of the circle. To analyze
the situation quantitatively, we note that the magnitude of the net force is
Fnet  n  w
As this is the force that provides the centripetal acceleration, we can say that
KJF: College Physics Ch. 6, Draft 1 1/16/04
Fnet  n  w 
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2
mv 2
r
(6.14)
From Equation (6.14) we find her apparent weight to be:
wapp  n  w 
3
mv 2
r
(6.15)
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The passenger’s apparent weight at the bottom is larger than her true weight
w, which agrees with your experience when you go through a dip or a valley.
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we can compute
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Now let’s look at the roller coaster car as it crosses the top of the loop. Figure
6.19c shows the car’s free-body diagram at the top of the loop. The car is still
moving in a circle, so there must be a net force toward the center of the circle
to provide the centripetal acceleration; this net force now points downward.
As the net force is given by
Fnet  n  w
Fnet  n  w 
mv 2
r
(6.16)
And from Equation (6.16) we find
wapp  n 
mv 2
w
r
(6.17)
An especially interesting case is when the car is moving at the critical
speed—just fast enough to make the loop. At that point, n  0 , and the track
is not pushing against the car. The car is able to complete the circle because
the weight force alone provides sufficient centripetal acceleration.
As the critical speed vc is when n  0 we can compute:
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mvc2
 w  mg
r
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vc  rg
(6.18)
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At this critical speed, we can also say that
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The passengers in the car feel weightless at this point—they are in free fall.
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CONCEPTUAL EXAMPLE 6.3 Vertical loop in a real roller coaster
wapp  n  0
When you ride a coaster, you want some variation in your apparent weight;
this is exciting. But if there is too much variation over a short period of time,
it can be painful. To make for a more comfortable ride, the vertical loops of
roller coasters aren’t really circles. If you look at a vertical loop, it has a
sharper curvature at the top than at the sides or the bottom, as you can see in
Figure 6.20. The radius of the circle that matches the track at the top of the
loop is about quite a bit smaller than that of a matching circle at other places
on the track. Explain why this shape would lead to a more comfortable ride
for passengers on the coaster.
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Fig_CoasterLoop Figure 6.20 Variable radius of the loop on a real coaster.
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Reason Looking at the analysis of the vertical loop above, we see that at the
bottom of the loop a passenger’s apparent weight is the sum of two
quantities:
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wapp 
mv 2
w
r
At the top of the loop, the apparent weight is the difference of these two
quantities:
wapp 
mv 2
w
r
Increasing r at the bottom of the loop decreases the apparent weight at this
point; decreasing r at the top of the loop increases the apparent weight here.
Varying the radius makes the apparent weight more nearly equal in different
parts of the loop, making for a much more comfortable ride.
Assess Many coasters have vertical loops in which the apparent weight is
quite nearly equal throughout the loop. This produces the rather disorienting
experience of turning upside down while feeling no apparent change in the
forces acting on the body.
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KJF: College Physics Ch. 6, Draft 1 1/16/04
A fast-spinning world Saturn is quite a bit larger than the earth, has no solid
surface, and rotates in just under 11 hours. This rapid rotation decreases the
apparent weight at the equator enough to distort the fluid surface; the planet
is noticeably out of round, as the red circle shows. The diameter at the
equator is 11% greater than the diameter at the poles.
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Centrifuges
We finished the previous section by looking at the apparent weight of a
person moving in a circular trajectory. If the angular velocity of circular
motion is very large, the apparent weight can be quite large; this is the
principle behind a centrifuge.
Suppose you have a particle that is more dense than water that is suspended
in water. Gravity will exert a downward force; it will begin to move toward
the bottom of its container. But if the particle size is small, the terminal
velocity can be minuscule, and it can take a very long time for this settling to
occur.
A faster separation would require the force of gravity to be increased. This is
not possible, but it is possible to increase the apparent weight. If the particle
in water is placed in a tube in a centrifuge that is spun at a high angular
velocity, the apparent weight can be increased significantly, which will make
the particle settle out much more rapidly.
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Figure 6.21 A spinning centrifuge.
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For a very rapid rotation, the normal force that keeps the liquid going in a
circle is much larger than any other force, and so we can say that
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wapp  n  ma 
mv 2
 m 2 r
r
Equation (6.11) gives the acceleration as a   2 r , so we can write
wapp  m 2 r  ma
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A typical centrifuge is characterized by its centripetal acceleration in units of
g, the acceleration of gravity.
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EXAMPLE 6.9 The ultracentrifuge
An ultracentrifuge is designed to produce very large accelerations. An
ultracentrifuge with diameter 18 cm produces centripetal accelerations of
250,000 g’s. What is the frequency in rpm? If the centrifuge is spinning a
sample with a mass of 300 g, what is its apparent weight?
Prepare First, we compute the acceleration in m/s2:

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
a  250, 000 9.8 m/s2  2.45 105 m/s2
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The radius is half the diameter, 9.0 cm, or 0.090 m.
Solve We can use Equation (6.11) to compute the angular velocity:
a   2r   
a

r
2.45  10 5 m/s 2
 1.65  10 3 rad/s
.090 m
Expressed as a frequency in rpm, this is
 9.55 rpm 
f  1.65 10 3 rad/s 
 1.57 10 4 rpm

 1 rad/s 


or over 15,000 revolutions per minute. At this rotation rate, the 300 g mass
will have an apparent weight of


wapp  ma  0.300 kg 2.45 105 m/s2  7.35 104 N
Assess This is the weight of a mass of over 7,000 kg. The forces in the
ultracentrifuge are very large, so great care must be taken in the design.
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TRY IT YOURSELF
Before spinning
After spinning
TIY_HandBefore TIY_HandAfter
Human centrifuge If your spin your arm in a vertical circle, the motion will
produce an effect like that in a centrifuge. The forces that produce the
acceleration will result in a significant increase in the blood pressure in your
hand; you will be able to see (and feel) the difference quite easily.
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STOP TO THINK 6.4
A car is rolling over the top of a hill at speed v. At
this instant,
A. n  w.
B. n  w.
C. n  w.
D. We can’t tell about n without knowing v.
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6.7 Circular Orbits and Weightlessness
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Orbital Motion
How does a satellite orbit the earth? What forces act on it? Why does it move
in a circle? To answer these important questions, let’s return, for a moment,
to projectile motion. Projectile motion occurs when the only force on an
object is gravity. Our analysis of projectiles made an implicit assumption that
the earth is flat and that the acceleration due to gravity is everywhere straight
down. This is an acceptable approximation for projectiles of limited range,
such as baseballs or cannon balls, but there comes a point where we can no
longer ignore the curvature of the earth.
Figure 6.22 shows a perfectly smooth, spherical, airless planet with one
tower of height h. A projectile is launched from this tower parallel to the
ground   0 with speed v0  If v0 is very small, as in trajectory A, the
“flat-earth approximation” is valid and the problem is identical to Example
3.3 in which a car drove off a cliff. The projectile simply falls to the ground
along a parabolic trajectory.
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Figure 6.22 Projectiles being launched at increasing speeds from height h on a
smooth, airless planet.
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As the initial speed v0 is increased, it seems to the projectile that the ground
is curving out from beneath it. It is still falling the entire time, always getting
closer to the ground, but the distance that the projectile travels before finally
reaching the ground—that is, its range—increases because the projectile
must “catch up” with the ground that is curving away from it. Trajectories B
and C are of this type.
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If the launch speed v0 is sufficiently large, there comes a point at which the
curve of the trajectory and the curve of the earth are parallel. In this case, the
projectile “falls” but it never gets any closer to the ground! This is the
situation for trajectory D. The projectile returns to the point from which it
was launched, at the same speed at which it was launched, making a closed
trajectory. Such a closed trajectory around a planet or star is called an orbit.
The most important point of this qualitative analysis is that an orbiting
projectile is in free fall. This is, admittedly, a strange idea, but one worth
careful thought. An orbiting projectile is really no different from a thrown
baseball or a car driving off a cliff. The only force acting on it is gravity, but
its tangential velocity is so large that the curvature of its trajectory matches
the curvature of the earth. When this happens, the projectile “falls” under the
influence of gravity but never gets any closer to the surface, which curves
away beneath it.
When we first discussed the force of gravity in Chapter 2, we said that it was
always directed vertically downward. As we see in Figure 6.23 , this is due to
our perspective; “downward” really means “toward the center of the earth.”
For a projectile in orbit, the direction of the force of gravity changes, always
pointing toward the center of the earth.
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Figure 6.23 The force of gravity is really directed toward the center of the earth.
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As you have learned, a force of constant magnitude that always points toward
the center of a circle causes the centripetal acceleration of uniform circular
motion. Since the only force acting on the orbiting projectile in Figure 6.23 is
gravity, we can write
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a
Fnet w mg
 
g
m m m
(6.20)
An object moving in a circle of radius r at speed vorbit will have this
centripetal acceleration if
2
vorbit 

a
g
r
(6.21)
That is, if an object moves parallel to the surface with the speed
vorbit  rg
KJF: College Physics Ch. 6, Draft 1 1/16/04
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then the acceleration due to gravity provides exactly the centripetal
acceleration needed for a circular orbit of radius r. An object with any other
speed will not follow a circular orbit.
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The earth’s radius is r  Re  637 106 m The orbital speed of a projectile
just skimming the surface of an airless, bald earth is
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vorbit  rg 
637 10 m980 m/s  7900 m/s  16,000 mph
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Even if there were no trees and mountains, a real projectile moving at this
speed would burn up from the friction of air resistance.
Suppose, however, that we launched the projectile from a tower of height
h  200 mi  32  10 5 m, above most of the earth’s atmosphere. This is
approximately the height of low-earth-orbit satellites, such as the Space
Shuttle. Note that h = Re , so the radius of the orbit
r  Re  h  669 106 m is only 5% greater than the earth’s radius. Many
people have a mental image that satellites orbit far above the earth, but in fact
most satellites come pretty close to skimming the surface. Our calculation of
vorbit thus turns out to be quite a good estimate of the speed of a satellite in
low earth orbit.
We can use vorbit to calculate the period of a satellite orbit:
T
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2r
r
 2
vorbit
g
(6.23)
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For a low earth orbit, with r  Re  200 miles, we find T  5192 s  87 min
The period of the Space Shuttle at an altitude of 200 mi is, indeed, just about
87 minutes.
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Weightlessness in Orbit
When we discussed weightlessness in Chapter 5, we discovered that it occurs
during free fall. We asked the question, at the end of Section 5.4, whether
astronauts and their spacecraft were in free fall. We can now give an
affirmative answer: They are, indeed, in free fall. They are falling
continuously around the earth, under the influence of only the gravitational
force, but never getting any closer to the ground because the earth’s surface
curves beneath them. Weightlessness in space is no different from the
weightlessness in a free-falling elevator. It does not occur from an absence of
weight or an absence of gravity. Instead, the astronaut, the spacecraft, and
everything in it are “weightless” because they are all falling together.
KJF: College Physics Ch. 6, Draft 1 1/16/04
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The orbiting space shuttle, and everything it in, including this astronaut and all of the
equipment in the photo, are all in free fall. The astronaut feels weightless; her
apparent weight is zero.
Rotating space stations The weightlessness astronauts experience in orbit
has serious physiological consequences. Astronauts who spend time in
weightless environments lose bone and muscle mass and suffer other adverse
effects. The solution is to introduce “artificial gravity.” The easiest way to do
this is to make the station rotate, producing an apparent weight. The
designers of this space station model for 2001: A Space Odyssey made it
rotate for just that reason.
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The Orbit of the Moon
If a satellite is simply “falling” around the earth, with the gravitational force
causing a centripetal acceleration, then what about the moon? Is it obeying
the same laws of physics? Or do celestial objects obey laws that we cannot
discover by experiments here on earth?
The radius of the moon’s orbit around the earth is r  Rm  384 108 m If
we use Equation (6.23) to calculate the period of the moon’s orbit, the time it
takes the moon to circle the earth once, we get
KJF: College Physics Ch. 6, Draft 1 1/16/04
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T  2
r
384  10 8 m
 2
 655 min  11 hours
g
980 m/s2
(6.24)
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This is clearly wrong; the period of the moon’s orbit is approximately one
month. For the moon, T  273 days  236 106 s, a factor of 60 longer
than we calculated it to be.
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6.8 Universal Gravitation
Newton believed that the laws of motion he had discovered were universal
and so should apply to the motion of the moon as well as to the motion of
objects in the laboratory. But why should we assume that the acceleration
due to gravity g is the same at the distance of the moon as it is on or near the
earth’s surface? If gravity is the force of the earth pulling on an object, it
seems plausible that the size of that force, and thus the size of g, should
diminish with increasing distance from the earth.
Newton proposed the idea that the earth’s force of gravity decreases
inversely with the square of the distance from the earth. This is the basis of
Newton’s law of gravity, a topic we will study in the next section. Gravity is
less at the distance of the moon, and it has exactly the strength needed to
make the moon orbit at the observed rate. The moon, just like the space
shuttle, is simply “falling” around the earth!
Isaac Newton was born to a poor farming family in 1642, the year of
Galileo’s death. He entered Trinity College at Cambridge University at age
19 as a “subsizar,” a poor student who had to work his way through school.
Newton graduated in 1665, at age 23, just as an outbreak of the plague in
England forced the universities to close for two years. He returned to his
family farm for that period, during which he made important experimental
discoveries in optics, laid the foundations for his theories of mechanics and
gravitation, and made major progress toward his invention of calculus as a
whole new branch of mathematics.
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Isaac Newton, 1642–1727
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A popular image has Newton thinking of the idea of gravity after an apple
fell on his head. This amusing story is at least close to the truth. Newton
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himself said that the “notion of gravitation” came to him as he “sat in a
contemplative mood” and “was occasioned by the fall of an apple.” It
occurred to him that, perhaps, the apple was attracted to the center of the
earth but was prevented from getting there by the earth’s surface. And if the
apple was so attracted, why not the moon? Newton’s genius was his sudden
realization that the force of the sun on the planets was identical to the
force of the earth on the apple. In other words, gravitation is a universal
force between all objects in the universe! This is not shocking today, but no
one before Newton had ever thought that the mundane motion of objects on
earth had any connection at all with the stately motion of the planets around
the sun.
In order to be responsible for the motion of the moon around the earth, the
force of the earth’s gravity at the position of the moon must be less than it
value at the surface of the earth, as we saw in the previous section; gravity
must decrease with distance.
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Newton’s Law of Gravity
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These two notions about gravity—that it is universal, and that it decreases
with distance—form the basis for Newton’s Law of Gravity.
Newton proposed that every object in the universe attracts every other object
with a force that has the following properties:
1. The force is inversely proportional to the square of the distance
between the objects.
2. The force is directly proportional to the product of the masses of the
two objects.
Figure 6.24 shows two spherical masses m1 and m2 separated by distance r.
Each mass exerts an attractive force on the other, a force that we call the
gravitational force. These two forces form an action/reaction pair, so F1 on 2
is equal and opposite to F2 on 1. The magnitude of the forces is given by
Newton’s law of gravity.
Figure 6.24 The gravitational forces on masses
m1 and m2 .
Newton’s law of gravity
If two objects with masses m1 and m1 are a distance r apart, the objects exert
attractive forces on each other of magnitude:
Gm1m2
r2
The forces are directed along the line joining the two objects.
F1 on 2  F2 on 1 
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The constant G is called the gravitational constant. In the SI system of units,
G has the value
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G  6.67 1011 Nm2 / kg2
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The gravitational force decreases as one over the square of the distance
between the objects; it is what we call an inverse square law force. This
mathematical form is one we will see again, so it is worth our while to
explore it in more detail.
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Math Model: Inverse square relationships
The force between two spherical objects is the distance between
the centers of the objects. Proving this was one of Newton’s big
accomplishments; we will not attempt to duplicate it. b
NOTE c
Two quantities have an inverse
square relationship if
increasing one by a certain
factor results in a decrease of the
other one by the square of that
factor. In symbols, we can write
this relationship as
k
y 2
x
y equals k over x squared
Limits: as x becomes large, y becomes
very small; as x becomes small, y
becomes very large.
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As an example of the variation of the gravitational force, let’s consider the
force of the earth’s gravity at the distance of the moon.
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EXAMPLE 6.10 Earth’s gravity at the moon
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For a satellite in low earth orbit, g has a value of approximately 9.8 m/s2. The
moon is at a distance of 3.84 108 m from the earth, about 60 times as far
away from the center of the earth as a satellite in a low earth orbit. What
approximate value would g, the acceleration of the earth’s gravity, have at
the position of the moon?
Solve By considering the math model for inverse square relationships, we
see that the force (and thus the acceleration) will decrease with the square of
the distance—increasing the distance by a factor of 60 will decrease the force
by a factor of 60 squared, or 3600. This means estimate g at the position of
the moon to be:
gat position of moon 
gnear surface of earth 9.8 m/s2

 .0027 m/s 2
3600
3600
(6.26)
Assess We can do a check on our calculation by using this value for g in
Equation (6.24), in which we calculated the period of the moon. When we
used the value of g at the surface of the earth to calculate this period, we
obtained a value that was clearly incorrect. Using the value from Equation
(6.26) in Equation (6.24), we obtain:
T  2
r
384  10 8 m
 2
 39, 500 min  27 days
g
.0027 m/s2
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This is the correct orbital period for the moon, meaning our calculation of the
acceleration of gravity at the position of the moon must be correct as well.
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Newton found that if he assumed that gravity decreased as an inverse square,
the acceleration of the earth’s gravity at the position of the moon was exactly
what was needed to produce the observed centripetal acceleration of the
moon as it orbits the earth—as we have seen.
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If all objects in the universe attract all other objects, why don’t you sense the
attractive force between yourself and this book, or feel a small gravitational
tug toward a car driving by? We will consider this issue in the next example.
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EXAMPLE 6.11 Gravitational force between two people
You are seated in your physics lecture class next to another student who is a
distance of 0.6 meters away. What is the magnitude of the gravitational force
between you? Assume that you each have a mass of 65 kg.
Prepare We will assume that we can model each of you as a sphere; this is
not a particularly good model, but it will serve our purposes, as we will see.
In this case, the gravitational force will have the form of Equation (6.25). We
will take the 0.6 m as the distance between your centers.
Solve Substituting relevant numbers in Equation (6.25), we get
F(you) on (other student) 

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Gmyou mother student
r2
6.67  10 11 Nm 2 /kg 2 65 kg 65 kg 


0.6 m 2
 7.8  10 7 N
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to two significant figures.
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The gravitational force between two ordinary-sized objects very small
because the masses are; this is the reason that we are not aware of it. Only
when one (or both) of the masses is exceptionally large—planet-size—does
the force of gravity become important.
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EXAMPLE 6.12 Force of the earth on a person
Assess The force is quite small, on the order of the weight of one hair on
your head. The force is too small to be of practical import, so our original
spherical approximation was justified. Making a more careful analysis of the
forces would lead to a result that was a bit different, but still insignificant;
there is no need to make a more accurate calculation.
Suppose a person has a mass of 60 kg. What is the magnitude of the
gravitational force of the earth on this person? Compute the result by using
Newton’s law of gravity and data for the earth.
Prepare First, we draw a sketch showing the forces on the person due to the
Earth. There are two opposing forces, as we see in Figure 6.25.
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Fig_WeightOnEarth
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Figure 6.25 Forces on a person on the surface of the earth.
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Solve The force of gravity on the person due to the earth can be computed
using Equation (6.25). The masses in the equation are the mass of the earth
and the mass of the person. The distance that we use in the equation is the
distance of the person from the center of the earth—the radius of the earth Re.
Data for the mass and radius of the earth can be obtained from Table 6.4.
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FMe on m 



6.67  10 11 Nkg 2 /m 2 5.98  10 24 kg 60 kg 
GM e m

2
Re2
6.37  10 6 m


=590 N
Assess This force is the same as we would calculate using the formula for
the weight force, w=mg. The weight of an object is simply the “force of
gravity” acting on it. We can see that the person’s apparent weight, wapp=n, is
equal to this force as well.
The force of gravitational attraction between the earth and you is responsible
for your weight. If you were to venture to another planet, your mass would
be the same, but your weight would vary, as we discussed in Chapter 5. We
will explore this concept in more detail in the next section.
Variable gravity The proof that we can use the distance to the center of the
earth in our gravitation calculations assumes that the earth’s composition is
uniform. The earth’s composition actually varies, and so there is a very small
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variation in gravity at the surface of the earth. If the materials of the earth’s
crust are quite dense in a region, gravity will be a bit stronger here. Satellite
orbits and other information may be used to make maps of the earth’s surface
gravity, useful tools in searching for mineral and oil deposits.
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Gravity on other Worlds
When astronauts ventured to the moon, the television images sent back
showed them walking—and even jumping and skipping—with some ease,
even though they were wearing bulky spacesuits and life support systems that
had a mass of over 80 kg. This was a visible reminder that the weight of
objects is less on the moon. Let’s consider why this is so.
Apollo astronaut on the moon.
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Figure 6.26 shows an astronaut on the moon, who is weighing a rock of mass
m. When we compute the weight of an object on the surface of the earth, we
use the formula w  mg. We can do the same calculation for a mass on the
moon, as long as we use the value of g on the moon.
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This is the “little g” perspective. Falling body experiments done on the moon
would give the value of gmoon to be 1.62 m/s2.
w  mgmoon
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(6.27)
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Figure 6.26 An astronaut weighing a mass on the moon.
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But we can also take a “big G” perspective. We know that the weight of the
rock comes from the gravitational attraction of the moon. We can compute
this value using Equation (6.25). The distance r will be the radius of the
moon, as it is the distance to the center of the moon.
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FM moon on m 
GM moon m
Rmoon 2
(6.28)
Because Equations (6.27) and (6.28) are two names and two expressions for
the same force, we can equate the right-hand sides to find that
gmoon 
GM moon
Rmoon 2
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We have done this calculation for an object on the moon, but the result is
completely general. At the surface of a planet (or a star) the acceleration of
gravity, the value of g can be computed as
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gplanet 
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GM planet
Rplanet 2
(6.29)
Acceleration of gravity on the surface of a planet
If we use values for the mass and the radius of the moon, we compute a value
for gmoon of 1.62 m/s2.
Both perspectives lead to the same result, that the value of g is 1.62 m/s2.
This means that an object will weigh less on the moon than it would on the
earth, where g is 9.8 m/s2. A 70 kg astronaut wearing an 80 kg spacesuit and
backpack will weigh over 330 pounds on the earth, but only 54 pounds on the
moon, making jumping and skipping quite easy.
EXAMPLE 25.13 Gravity on Saturn
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Saturn, at 5.68 1026 kg, has nearly 100 times the mass of the earth. It is
also much larger, with a radius of 5.85 107 m. What is the value of g on the
surface of Saturn?
Solve We can use Equation (6.29) to compute the value of gSaturn, using
values for Saturn’s mass and radius:
gSaturn



6.67  10 11 Nm 2 / kg 2 5.68  10 26 kg
GM Saturn


 11.1 m/s 2
2
2
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RSaturn
5.85  10 m


Assess Even though Saturn is much more massive than the Earth, its larger
radius gives it a surface gravity that is not markedly different than that of the
Earth. If Saturn had a solid surface, you could walk and move around quite
normally.
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EXAMPLE 25.14 Orbiting Phobos
Mars has two moons, each of which is much smaller than the Earth’s moon.
The smaller of these two bodies, Deimos, has an average diameter of only
6.3 km, and a mass of 1.8 1015 kg. At what speed would a projectile move
in a very low orbit around Deimos?
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Deimos, the smaller of Mars’ two moons.
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Prepare Equation (6.22) shows that a low orbital velocity depends on two
things: Surface gravity and the radius. To prepare, we will calculate surface
gravity.
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gDeimos




6.67  10 11 Nm 2 /kg 2 1.8  1015 kg
GM Deimos


 .0030 m/s2
2
2
3
RDeimos
6.3  10 m

Solve We can now use Equation (6.22) to calculate the orbital speed:
vorbit  gr 
.0030 m/s 6.3  10 m  4.3 m/s  10 mph
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Assess This is quite modest. If you were to run and jump, you could easily
launch yourself into orbit on Deimos.
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Orbital Motions in the Solar System
The planets of the solar system orbit the sun due to the influence of the sun’s
gravity. As the planets are all at different distances from the sun, the strength
of the gravitational force of the will vary among the planets, with Mercury,
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the closest planet experiencing the largest acceleration due to the sun’s
gravity and Pluto, the most distant planet, the smallest.
Figure 6.27 shows a massive body of mass M, such as the earth or the sun,
with a lighter body of mass m orbiting it. The lighter body is called a
satellite, even though it may be a planet orbiting the sun. Newton’s second
law for the satellite is
FM on m 
GMm
mv 2

ma

r2
r
(6.30)
Thus the speed of a satellite in a circular orbit is
GM
r
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v
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Speed of a satellite in a circular orbit of radius r about a sun or planet of mass
M
(6.31)
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Figure 6.27 The orbital motion of a satellite due to the force of gravity.
A satellite must have this specific speed in order to maintain a circular orbit
of radius r about the larger mass M. If the velocity differs from this value, the
orbit will become elliptical rather than circular. Notice that the orbital speed
does not depend on the satellite’s mass m. This is consistent with our
previous discoveries, that free fall and projectile motion due to gravity is
independent of the mass.
For a planet orbiting the sun, the period T is the time to complete one full
orbit. The relationship among speed, radius, and period is the same as for any
2 r
circular motion, v 
. Combining this with the value of v for a circular
T
orbit from Equation (6.31) give:
GM 2 r

r
T
If we square both sides and solve for T, we find
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 4 2  3
T2 
r
 GM 
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Relationship of orbital period T and radius r for circular orbits
In other words, the square of the period of the orbit is proportional to the
cube of the radius of the orbit. This law of orbital motion for planets orbiting
the sun is known as Kepler’s third law. Kepler was an astronomer who did a
serious study of the orbital motion of Mars and the other planets. He
formulated three laws of orbital motion of the planets that were an important
test case for Newton’s laws to explain. You can see that Kepler’s third law is
a direct consequence of Newton’s law of gravity and Newton’s second law;
his other laws can similarly be deduced from Newton’s laws of motion and
gravity.
Table 12.2 contains astronomical information about the sun and the planets
of the solar system. Note that the orbits of planets farther from the sun have
longer periods.
TABLE 6.4 Astronomical data
Planetary body
Mean distance
from sun (m)
Period (years)
Mass (kg)
Mean radius
(m)
Sun
–
–
Mercury
579  1010
108 1011
150 1011
228  1011
778  1011
143 1012
287  1012
450  1012
5.911012
0.241
199  10 30
318  10 23
488 1024
598 1024
642  10 23
190  10 27
568 1026
868 1025
1031026
1.31022
696  10 8
243  10 6
606  10 6
637  10 6
337 106
699 10 7
585 10 7
233  10 7
221 10 7
1.15 106
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
Pluto
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(6.32)
0.615
1.00
1.88
11.9
29.5
84.0
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STOP TO THINK 6.5 Two planets orbit a star. Planet 1 has orbital radius r1
and planet 2 has r2  4r1  Planet 1 orbits with period T1  Planet 2 orbits with
period
a.
T2  8T1 
b.
T2  4T1 
c.
T2  2T1 
d.
T2  12 T1 
e.
T2  41 T1 
f.
T2  18 T1 
6.9 Circular Motion and Gravity on a Grand Scale
Although weak, gravity is a long-range force. No matter how far apart two
objects may be, there is a gravitational attraction between them.
Consequently, gravity is the most ubiquitous force in the universe. It not only
keeps your feet on the ground, but is also at work at a much larger scale. The
Milky Way galaxy, the collection of stars of which our sun is a part, is held
together by gravity. Why doesn’t this attractive force of gravity simply pull
all of the stars together?
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The answer is that all of the stars in the galaxy are in orbit around the center;
the gravitational attraction keeps the starts moving in circular orbits around
the center of the galaxy rather than making them fall toward it. In the nearly
5 billion years that our solar system has existed, it has orbited the center of
the galaxy approximately 20 times.
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A spiral galaxy, similar to our Milky Way galaxy.
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All of the stars in a galaxy revolve with different periods; the characteristic
spiral structure of galaxies like ours results from the complicated dynamics
of this rotation.
In the next chapter, we will consider the simpler case of extended objects that
rotate as a whole, rigid objects that keep their shape as they rotate.
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