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SECTION 1
Preview Vocabulary
Latin Word Origins The word centripetal
is a combination of two parts, “center”
and “petal.” The second part of the word
is derived from the Latin word petere,
which means “seeking.” The word
centripetal is used to describe an
object moving in a circular path whose
acceleration is directed toward the
center of the path.
Teach
Objectives
Solve problems involving
centripetal acceleration.
Solve problems involving
centripetal force.
Explain how the apparent
existence of an outward force in
circular motion can be
explained as inertia resisting
the centripetal force.
Circular Motion
Key Term
centripetal acceleration
Centripetal Acceleration
Consider a spinning Ferris wheel, as shown in Figure 1.1. The cars on the
rotating Ferris wheel are said to be in circular motion. Any object that
revolves about a single axis undergoes circular motion. The line about
which the rotation occurs is called the axis of rotation. In this case, it is a
line perpendicular to the side of the Ferris wheel and passing through the
wheel’s center.
Tangential speed depends on distance.
FIGURE 1.1
Circular Motion Any point on a
Ferris wheel spinning about a fixed axis
undergoes circular motion.
Misconception Alert!
Tangential speed (vt) can be used to describe the speed of an object in
circular motion. The tangential speed of a car on the Ferris wheel is the
car’s speed along an imaginary line drawn tangent to the car’s circular
path. This definition can be applied to any object moving in circular
motion. When the tangential speed is constant, the motion is described
as uniform circular motion.
The tangential speed depends on the distance from the object to the
center of the circular path. For example, consider a pair of horses side-byside on a carousel. Each completes one full circle in the same time
period, but the horse on the outside covers more distance than the inside
horse does, so the outside horse has a greater tangential speed.
Some students will have difficulty with
terminology at this point because of the
previous familiarity with the term
centrifugal. It is important to emphasize
the distinction between centripetal
(center-seeking) and centrifugal
(center-fleeing). To avoid reinforcing this
misconception, avoid using the term
centrifugal.
Centripetal acceleration is due to a change in direction.
Suppose a car on a Ferris wheel is moving at a constant
speed as the wheel turns. Even though the tangential
speed is constant, the car still has an acceleration. To see
why, consider the equation that defines acceleration:
vf − vi
a=_
tf − ti
Acceleration depends on a change in the velocity.
Because velocity is a vector, acceleration can be
produced by a change in the magnitude of the
velocity, a change in the direction of the velocity,
or both. If the Ferris wheel car is moving at a
constant speed, then there is no change in the
magnitude of the velocity vector. However, think
about the way the car is moving. It is not traveling
in a straight line. The velocity vector is continuously changing direction.
Demonstration
Centripetal Acceleration
Purpose Show an example of centripetal
acceleration.
Materials battery-operated toy car,
50 cm string
Procedure Tie one end of the string to
the midpoint of the car. Hold the other
end of the string to the table or floor.
Start the car. Have students observe the
motion. Release the string and allow the
car to move in a straight line.
Point out that even though the car
was traveling at constant speed, the car
was accelerating because the direction
of motion continued to change. Show
that the change in velocity was always
perpendicular to the direction of
motion.
224 Chapter 7
©Joseph Sohm/Visions of America/Corbis
Plan and Prepare
SECTION 1
224
Chapter 7
Differentiated
Instruction
Below Level
Ask students to explain how a line of dancers
moving in a circle demonstrates that tangential
speed depends on the distance between the
object and the center of the circular path.
Also ask why jockeys maneuver their horses
during a race to get as close to the inside rail
as possible. See if students can provide other
examples of how tangential speed depends on
distance from the center.
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5/9/2011 8:14:27 AM
Untitled-27 225
The acceleration of a Ferris wheel car moving in a circular path and at
constant speed is due to a change in direction. An acceleration of this
nature is called a centripetal acceleration. The magnitude of a centripetal
acceleration is given by the following equation:
centripetal acceleration the
acceleration directed toward the
center of a circular path
FIGURE 1.2
Centripetal Acceleration
vt2
ac = _
r
centripetal acceleration =
Classroom Practice
Centripetal Acceleration
(a) As the particle moves from A
to B, the direction of the particle’s
velocity vector changes. (b) For short
time intervals, ∆v is directed toward
the center of the circle.
___
(tangential speed)2
radius of circular path
What is the direction of centripetal acceleration? To answer this
question, consider Figure 1.2(a). At time ti, an object is at point A and has
tangential velocity vi. At time tf , the object is at point B and has tangential
velocity vf. Assume that vi and vf differ in direction but have the same
magnitudes.
(a)
A
vi
B
vf
Centripetal Acceleration
The cylindrical tub of a washing machine
has a radius of 34 cm. During the spin
cycle, the wall of the tub rotates with a
tangential speed of 5.5 m/s. Calculate
the centripetal acceleration of the
clothes sitting against the tub.
Answer: 8 9 m/s2
PROBLEM guide A
The change in velocity (∆v = vf − vi) can be determined graphically,
as shown by the vector triangle in Figure 1.2(b). Note that when ∆t is very
small, vf will be almost parallel to vi. The vector ∆v will be approximately
perpendicular to vf and vi and will be pointing toward the center of the
circle. Because the acceleration is in the direction of ∆v, the acceleration
will also be directed toward the center of the circle. Centripetal acceleration is always directed toward the center of a circle. In fact, the word
centripetal means “center seeking.” This is the reason that the acceleration
of an object in uniform circular motion is called centripetal acceleration.
(b)
Use this guide to assign problems.
SE = Student Edition Textbook
PW = Sample Problem Set I (online)
PB = Sample Problem Set II (online)
Solving for:
vf
v
−v i
vt
SE Sample, 1–2;
Ch. Rvw. 8–9
PW 3–4
PB 4–6
ac
SE3; Ch. Rvw. 50a
PW 5–6
PB Sample, 1–3
r
SE 4
PW Sample, 1–2
PB 7–10
PREMIUM CONTENT
Interactive Demo
Centripetal Acceleration
HMDScience.com
Sample Problem A A test car moves at a constant speed
around a circular track. If the car is 48.2 m from the track’s center
and has a centripetal acceleration of 8.05 m/s2, what is the car’s
tangential speed?
Given:
ANALYZE
r = 48.2 m
ac =
8.05 m/s2
vt = ?
*Challenging Problem
Use the centripetal acceleration equation, and rearrange to solve for vt.
SOLVE
v 2t
ac = _
r
2
��������
vt = √a
��
cr = √(8.05 m/s ) (48.2 m)
vt = 19.7 m/s
Continued
Problem Solving
Circular Motion and Gravitation
225
Deconstructing Problems
Use the following equations for students
who are having difficulty understanding the
algebraic simplifications used to derive the
final equation.
vt2
ac = _
​  r ​  
5/9/2011 8:14:28 AM
acr = vt2
  cr ​ = ​ √
​ √
a
v  t2 ​
  cr ​ = vt
​ √a

Circular Motion and Gravitation 225
Centripetal Acceleration
Teach continued
(continued)
1. A rope attaches a tire to an overhanging tree limb. A girl swinging on the tire has
a centripetal acceleration of 3.0 m/s2. If the length of the rope is 2.1 m, what is the
girl’s tangential speed?
Answers
2. As a young boy swings a yo-yo parallel to the ground and above his head, the
yo-yo has a centripetal acceleration of 250 m/s2. If the yo-yo’s string is 0.50 m long,
what is the yo-yo’s tangential speed?
Practice A
1. 2.5 m/s
2. 11 m/s
3. 1.5 m/s2
4. 58.4 m
3. A dog sits 1.5 m from the center of a merry-go-round. The merry-go-round is set in
motion, and the dog’s tangential speed is 1.5 m/s. What is the dog’s centripetal
acceleration?
4. A racecar moving along a circular track has a centripetal acceleration of 15.4 m/s2.
If the car has a tangential speed of 30.0 m/s, what is the distance between the car
and the center of the track?
Teaching Tip
Tangential acceleration is due to a change in speed.
Tangential acceleration is covered
in more detail in the feature
“Tangential Speed and Acceleration.”
You have seen that centripetal acceleration results from a change in
direction. In circular motion, an acceleration due to a change in speed
is called tangential acceleration. To understand the difference between
centripetal and tangential acceleration, consider a car traveling in a
circular track. Because the car is moving in a circle, the car has a
centripetal component of acceleration. If the car’s speed changes,
the car also has a tangential component of acceleration.
FIGURE 1.3
Centripetal Force
Force on a Ball When a ball is
whirled in a circle, it is acted on by a
force directed toward the center of
the ball’s circular path.
Consider a ball of mass m that is tied to a string of length r and that is
being whirled in a horizontal circular path, as shown in Figure 1.3. Assume
that the ball moves with constant speed. Because the velocity vector, v,
continuously changes direction during the motion, the ball experiences
a centripetal acceleration that is directed toward the center of motion.
As seen earlier, the magnitude of this acceleration is given by the
following equation:
vt2
ac = _
r
The inertia of the ball tends to maintain the ball’s motion in a straight
path. However, the string exerts a force that overcomes this tendency.
The forces acting on the ball are gravitational force and the force exerted by
the string, as shown in Figure 1.4(a) on the next page. The force exerted by the
string has horizontal and vertical components. The vertical component is
equal and opposite to the gravitational force. Thus, the horizontal component is the net force. This net force is directed toward the center of the
circle, as shown in Figure 1.4(b). The net force that is directed toward the
center of an object’s circular path is called centripetal force. Newton’s
second law can be applied to find the magnitude of this force.
Fc = mac
226
Chapter 7
Differentiated
Instruction
Inclusion
Kinesthetic learners may need help understanding that Earth’s gravitational force causes
a satellite in a circular orbit to change its
direction but not its speed. Have students
imagine that a person is running on the side
of a road. Ask them to describe what would
happen if the runner suddenly grabbed a
lamppost or small tree trunk. Be sure students
recognize that the force was perpendicular to
the runner’s forward motion and caused the
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226 Chapter 7
runner’s direction to change. Students can try
this outdoors to feel the change in direction
for themselves.
5/9/2011 8:14:29 AM
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FIGURE 1.4
Centripetal Force The net
force on a ball whirled in a circle
(a) is directed toward the center
of the circle (b).
Fstring
Fnet = Fc
Fc
Fg
Demonstration
v
r
Centripetal Force
Purpose Show that the horizontal
component of tension in a rope
provides the centripetal force for a ball
moving in a horizontal circle.
m
(b)
(a)
The equation for centripetal acceleration can be combined with Newton’s
second law to obtain the following equation for centripetal force:
Centripetal Force
Materials 100 g mass (with hook),
30 cm string, spring scale, protractor
Procedure Attach one end of the string
to the hook on the mass and attach the
other end to the spring scale. Use the
protractor to measure the position at
which the string makes an angle of 30°
with the vertical. Then rotate the string
with a constant speed so that the mass
moves in a circular path and the string
remains at the 30° angle with the
vertical. While maintaining this motion,
have a volunteer attempt to read the
spring scale. (Maintaining a constant
angle can be difficult; you may wish to
practice this before demonstrating to
the class.)
mvt2
Fc = _
r
(tangential speed)2
centripetal force = mass × ___
radius of circular path
Centripetal force is simply the name given to the net force toward the
center of the circular path followed by an object moving in uniform
circular motion. Any type of force or combination of forces can provide
this net force. For example, friction between a racecar’s tires and a
circular track is a centripetal force that keeps the car in a circular path. As
another example, gravitational force is a centripetal force that keeps the
moon in its orbit.
PREMIUM CONTENT
Interactive Demo
Centripetal Force
HMDScience.com
Sample Problem B A pilot is flying a small plane at 56.6 m/s
in a circular path with a radius of 188.5 m. The centripetal force
needed to maintain the plane’s circular motion is 1.89 × 104 N.
What is the plane’s mass?
ANALYZE
Given:
vt = 56.6 m/s
Unknown:
m=?
r = 188.5 m
Fc = 1.89 × 104 N
Use the equation for centripetal force. Rearrange to solve for m.
SOLVE
mv 2t
Fc = _
r
Fcr ___
(1.89 × 104 N) (188.5 m)
=
m=_
2
(56.6 m/s)2
vt
m = 1110 kg
Continued
Problem Solving
Circular Motion and Gravitation
227
Deconstructing Problems
Reality Check
Use the following equations to help students
see how the equation is rearranged to solve
for mass.
mv2t
FC = _
​  r ​  
8:14:29 AM
Help students relate the equation for 5/9/2011
centripetal force to the motion of the object.
A sharper turn means a greater change in
direction, therefore an object moving in
a small circle will require a greater centripetal
force if the speed is kept constant.
rFC = mv2t
rFC mv2t
_
​  2 ​  = _
​  2 ​ 
vt
vt
rF
C
_
​  2 ​  = m
vt
Point out that the scale reading
equals the tension in the string. The
vertical component balances the weight
of the mass, while the horizontal
component of the tension holds the
mass in a circular path. Ask students to
calculate the vertical tension component and to compare it with the weight
(Fscalecosθ = mg) so that they can
confirm whether the scale reading and
angle are correct. Then have them
calculate the centripetal force from the
horizontal component of tension
(Fscalesinθ = Fc). Repeat the demonstration at angles equal to 45° and 60°. Ask
students to predict what change will
occur in the force as measured on the
spring scale.
Circular Motion and Gravitation 227
Centripetal Force
Teach continued
1. A 2.10 m rope attaches a tire to an overhanging tree limb. A girl swinging on the
tire has a tangential speed of 2.50 m/s. If the magnitude of the centripetal force is
88.0 N, what is the girl’s mass?
PROBLEM guide b
2. A bicyclist is riding at a tangential speed of 13.2 m/s around a circular track.
The magnitude of the centripetal force is 377 N, and the combined mass of
the bicycle and rider is 86.5 kg. What is the track’s radius?
Use this guide to assign problems.
SE = Student Edition Textbook
PW = Sample Problem Set I (online)
PB = Sample Problem Set II (online)
Solving for:
m
SE Sample, 1;
Ch. Rvw. 10a*, 11*
PW 3
PB 7–10
vt
SE 4 Ch. Rvw. 10b, 39*
PW Sample, 1–2
PB 5–7
r
SE 2
PW 4
PB 6–7
Fc
SE 3; Ch. Rvw. 50b, 53
PW 5
PB Sample, 1–3
(continued)
3. A dog sits 1.50 m from the center of a merry-go-round and revolves at a tangential
speed of 1.80 m/s. If the dog’s mass is 18.5 kg, what is the magnitude of the
centripetal force on the dog?
4. A 905 kg car travels around a circular track with a circumference of 3.25 km. If the
magnitude of the centripetal force is 2140 N, what is the car’s tangential speed?
Centripetal force is necessary for circular motion.
FIGURE 1.5
Because centripetal force acts at right angles to an object’s circular
motion, the force changes the direction of the object’s velocity. If this
force vanishes, the object stops moving in a circular path. Instead, the
object moves along a straight path that is tangent to the circle.
Removal of Centripetal Force
A ball that is on the end of a string is
whirled in a vertical circular path. If the
string breaks at the position shown in (a),
the ball will move vertically upward in free
fall. (b) If the string breaks at the top of
the ball’s path, the ball will move along
a parabolic path.
For example, consider a ball that is attached to a string and that is
whirled in a vertical circle, as shown in Figure 1.5. If the string breaks when
the ball is at the position shown in Figure 1.5(a), the centripetal force will
vanish. Thus, the ball will move vertically upward, as if it has been thrown
straight up in the air. If the string breaks when the ball is at the top of its
circular path, as shown in Figure 1.5(b), the ball will fly off horizontally in a
direction tangent to the path. The ball will then move in the parabolic
path of a projectile.
Describing a Rotating System
*Challenging Problem
(a)
Answers
Practice B
1. 29.6 kg
2. 40.0 m
3. 40.0 N
4. 35.0 m/s
(b)
To better understand the motion of a rotating system, consider a car
traveling at high speed and approaching an exit ramp that curves to
the left. As the driver makes the sharp left turn, the passenger slides to
the right and hits the door. At that point, the force of the door keeps the
passenger from being ejected from the car. What causes the passenger to
move toward the door? A popular explanation is that a force must push
the passenger outward. This force is sometimes called the centrifugal
force, but that term often creates confusion, so it is not used in
this textbook.
Inertia is often misinterpreted as a force.
The phenomenon is correctly explained as follows: Before the car enters the
ramp, the passenger is moving in a straight path. As the car enters the ramp
and travels along a curved path, the passenger, because of inertia, tends to
228
Chapter 7
Differentiated
Instruction
Pre-AP
Have students extend the discussion of a car
on an exit ramp by considering how changing
some of the variables would affect the
passengers. How would passengers of different
masses be affected as the moving car enters
the ramp? Who would experience greater
force, a heavier passenger or a lighter one?
How would varying the speed of the car affect
the passengers? Students should explain their
answers in terms of inertia and mass.
Untitled-27 228
228 Chapter 7
5/9/2011 8:14:30 AM
move along the original straight path. This movement is in
accordance with Newton’s first law, which states that the
natural tendency of a body is to continue moving in a
straight line.
FIGURE 1.6
Path of Car and Passenger The force of friction
on the passenger is not enough to keep the passenger
on the same curved path as the car.
However, if a sufficiently large centripetal force acts
on the passenger, the person will move along the same
curved path that the car does. The origin of the
centripetal force is the force of friction between the
passenger and the car seat. If this frictional force is not
sufficient, the passenger slides across the seat as the car
turns underneath, as shown in Figure 1.6. Eventually, the
passenger encounters the door, which provides a large
enough force to enable the passenger to follow the same
curved path as the car does. The passenger does not slide
toward the door because of some mysterious outward
force. Instead, the frictional force exerted on the
passenger by the seat is not great enough to keep the
passenger moving in the same circle as the car.
Assess and Reteach Assess Use the Formative Assessment
on this page to evaluate student
mastery of the section.
v
Fc
Reteach For students who need
additional instruction, download the
Section Study Guide.
v
Response to Intervention To reassess
students’ mastery, use the Section Quiz,
available to print or to take directly
online at HMDScience.com.
SECTION 1 FORMATIVE ASSESSMENT
Reviewing Main Ideas
1. What are three examples of circular motion?
2. A girl on a spinning amusement park ride is 12 m from the center of the
ride and has a centripetal acceleration of 17 m/s2. What is the girl’s tangential speed?
3. Use an example to describe the difference between tangential and centripetal acceleration.
4. Identify the forces that contribute to the centripetal force on the object in
each of the following examples:
a. a bicyclist moving around a flat, circular track
b. a bicycle moving around a flat, circular track
c. a racecar turning a corner on a steeply banked curve
5. A 90.0 kg person rides a spinning amusement park ride that has a radius
of 11.5 m. If the person’s tangential speed is 13.2 m/s, what is the magnitude of the centripetal force acting on the person?
6. Explain what makes a passenger in a turning car slide toward the door.
Critical Thinking
7. A roller coaster’s passengers are suspended upside down as it moves at a
constant speed through a vertical loop. What is the direction of the force
that causes the coaster and its passengers to move in a circle? What provides this force?
Answers to Section Assessment
1. Answers may vary.
H_CNLESE586694_C07S1.indd
2. 14229m/s
3. Specific examples may vary, but all should
indicate that an object moving in a circular
path undergoes centripetal acceleration
because the object changes direction.
Tangential acceleration occurs when the
object changes its speed around the path.
4.a. the forces that the bicycle seat,
handlebars, and pedals exert on the
bicyclist
Circular Motion and Gravitation
229
b. friction between the tires and the track
c. components of (1) the normal 3/26/2013
force10:53:47 PM
from the banked curve and (2) the
friction between the tires and the road
5. 1.36 × 103 N
6. When the car turns, the inertia of the
passenger keeps the passenger moving in
the same direction that the car was initially
moving.
7. toward the center of the loop; the track
and gravity
Circular Motion and Gravitation 229