Copyright © by Holt, Rinehart and Winston. All rights reserved....244 Chapter 7 ROTATIONAL...

PHYSICS IN ACTION

When riding this spinning amusement-park

ride, people feel as if a force is pressing

them against the padding on the inside walls

of the ride.However, it is actually inertia

that causes their bodies to press against the

padding. The inertia of their bodies tends to

maintain motion in a straight-line path,

while the walls of the ride exert a force on

their bodies that makes them follow a cir-

cular path. This chapter will discuss the

force that maintains circular motion and

other rotational-motion quantities.

• How can you determine the riders’ averagelinear speed or acceleration during the ride?

• In what direction are the forces pushing orpulling the riders?

CONCEPT REVIEW

Displacement (Section 2-1)

Velocity (Section 2-1)

Acceleration (Section 2-2)

Force (Section 4-1)

CHAPTER 7

RotationalMotion and theLaw of Gravity

Rotational Motion and the Law of Gravity 243Copyright © by Holt, Rinehart and Winston. All rights reserved.

Chapter 7244

ROTATIONAL QUANTITIES

When an object spins, it is said to undergo rotational motion. Consider a

spinning Ferris wheel. The axis of rotation is the line about which the rotation

occurs. In this case, it is a line perpendicular to the side of the Ferris wheel and

passing through the wheel’s center. How can we measure the distance traveled

by an object on the edge of the Ferris wheel?

A point on an object that rotates about a single axis undergoes circular motion

around that axis. In other words, regardless of the shape of the object, any single

point on the object travels in a circle around the axis of rotation. It is difficult to

describe the motion of a point moving in a circle using only the linear quantities

introduced in Chapter 2 because the direction of motion in a circular path is

constantly changing. For this reason, circular motion is described in terms of the

angle through which the point on an object moves. When rotational motion is

described using angles, all points on a rigid rotating object, except the points on

the axis, move through the same angle during any time interval.

In Figure 7-1, a light bulb at a distance r from the center of a Ferris

wheel, like the one shown in Figure 7-2, moves about the axis in

a circle of radius r. In fact, every point on the wheel

undergoes circular motion about the center. To ana-

lyze such motion, it is convenient to set up a fixed

reference line. Let us assume that at time t = 0,

the bulb is on the reference line, as in Figure7-1(a), and that a line is drawn from the

center of the wheel to the bulb. After a

time interval ∆t, the bulb advances to a

new position, as shown in Figure 7-1(b). In this time interval, the line

from the center to the bulb (depict-

ed with a red line in both diagrams)

moved through the angle q with

respect to the reference line. Like-

wise, the bulb moved a distance s,

measured along the circumference

of the circle; s is the arc length.

7-1Measuring rotational motion

7-1 SECTION OBJECTIVES

• Relate radians to degrees.

• Calculate angular displace-ment using the arc lengthand the distance from theaxis of rotation.

• Calculate angular speed orangular acceleration.

• Solve problems using thekinematic equations for rotational motion.

Figure 7-2Any point on a Ferris wheel that spins about

a fixed axis undergoes circular motion.

Referenceline

r

Lightbulb

(a)

Referenceline

r

s

Lightbulb

(b)

θ

Figure 7-1A light bulb on a rotating Ferriswheel (a) begins at a point along areference line and (b) movesthrough an arc length s, and there-fore through the angle q .

rotational motion

motion of a body that spinsabout an axis



Radians and Arc Length

M A T E R I A L S

✔ drawing compass

✔ paper

✔ thin wire

✔ wire cutters or scissors

Use the compass to draw a circle on asheet of paper, and mark the center pointof the circle. Measure the distance fromthe center point to the outside of the cir-cle. This is the radius of the circle. Usingthe wire cutters, cut several pieces of wireequal to the length of this radius. Bend thepieces of wire, and lay them along the cir-cle you drew with your compass. Approxi-mately how many pieces of wire do you

use to go all the way around the circle?Draw lines from the center of the circle toeach end of one of the wires. Note thatthe angle between these two lines equals1 rad. How many of these angles are therein this circle? Draw a larger circle usingyour compass. How many pieces of wire(cut to the length of the radius) do you use to go all the way around this circle?

245Rotational Motion and the Law of Gravity

Angles can be measured in radians

In the situations we have encountered so far, angles have been measured in

degrees. However, in science, angles are often measured in radians (rad) rather

than in degrees. Almost all of the equations used in this chapter and the next

require that angles be measured in radians. In Figure 7-1(b), when the arc

length, s, is equal to the length of the radius, r, the angle q swept by r is equal to

1 rad. In general, any angle q measured in radians is defined by the following:

q = r

s

The radian is a pure number, with no dimensions. Because q is the ratio of

an arc length (a distance) to the length of the radius (also a distance), the

units cancel and the abbreviation rad is substituted in their place.

When the bulb on the Ferris wheel moves through an angle of

360° (one revolution of the wheel), the arc length s is equal to the

circumference of the circle, or 2pr. Substituting this value for s

in the above equation gives the corresponding angle in radians.

q = r

s =

2pr

r = 2p rad

Thus, 360° equals 2p rad, or one complete revolution. In

other words, one revolution corresponds to an angle of approxi-

mately 2(3.14) = 6.28 rad. Figure 7-3 depicts a circle marked with

both radians and degrees.

It follows that any angle in degrees can be converted to an angle in

radians by multiplying the angle measured in degrees by 2p/360°. In this

way, the degrees cancel out and the measurement is left in radians. The con-

version relationship can be simplified as follows:

q(rad) = 18

p0°q(deg)

radian

an angle whose arc length isequal to its radius, which isapproximately equal to 57.3°

x

y

360°

330°

315°300°

90°60°

45°

30°

0°

270°240°

225°210°

180°

150°135°

120°23 2

1

31

41

61

345

6ππ π

ππ

π

ππ

πππ

ππ

π

ππ

76

54 4

3 32

53

74

116

2

Figure 7-3Angular motion is measured in unitsof radians. Because there are 2pradians in a full circle, radians areoften expressed as a multiple of p.


Angular displacement describes how much an object has rotated

Just as an angle in radians is the ratio of the arc length to the radius, the angu-lar displacement traveled by the bulb on the Ferris wheel is the change in the

arc length, ∆s, divided by the distance of the bulb from the axis of rotation.

This relationship is depicted in Figure 7-4.

For the purposes of this textbook, when a rotating object is viewed from

above, the arc length, s, is considered positive when the point rotates counter-

clockwise and negative when it rotates clockwise. In other words, ∆q is posi-

tive when the object rotates counterclockwise and negative when the object

rotates clockwise.

ANGULAR DISPLACEMENT

∆q = ∆r

s

angular displacement (in radians) =change in arc lengthdistance from axis

SAMPLE PROBLEM 7A

Angular displacement

P R O B L E MWhile riding on a carousel that is rotating clockwise, a child travelsthrough an arc length of 11.5 m. If the child’s angular displacement is165�, what is the radius of the carousel?

S O L U T I O NGiven: ∆q = −165° ∆s = −11.5 m

Unknown: r = ?

First, convert the angular displacement to radians using the relationship on

page 245.

∆q(rad) = 18

p0°∆q(deg) =

18

p0°(−165°)

∆q(rad) = −2.88 rad

Use the angular displacement equation on this page. Rearrange to solve for r.

∆q = ∆r

s

r = ∆∆

qs

= −−2

1

.8

1

8

.5

r

m

ad

r = 3.99 m

CALCULATOR SOLUTION

Many calculators have a key labeledDEG � that converts from degrees toradians.

Referenceline

θ2θ1

Figure 7-4A light bulb on a rotating Ferriswheel rotates through an angulardisplacement of ∆q = q2 − q1.

angular displacement

the angle through which a point,line, or body is rotated in a speci-fied direction and about a speci-fied axis

246 Chapter 7

Copyright © by Holt, Rinehart and Winston. All rights reserved.247Rotational Motion and the Law of Gravity

Angular speed describes rate of rotation

Linear speed describes the distance traveled in a specified interval of time.

Angular speed is similarly defined. The average angular speed, wavg (w is the

Greek letter omega), of a rotating rigid object is the ratio of the angular displace-

ment, ∆q, to the time interval, ∆t, that the object takes to undergo that displace-

ment. Angular speed describes how quickly the rotation occurs.

Angular speed is given in units of radians per second (rad/s). Sometimes

angular speeds are given in revolutions per unit time. Recall that 1 rev = 2p rad.

ANGULAR SPEED

wavg = ∆∆

qt

average angular speed =angular displacement

time interval

angular speed

the rate at which a body rotatesabout an axis, usually expressedin radians per second

1. A girl sitting on a merry-go-round moves counterclockwise through an

arc length of 2.50 m. If the girl’s angular displacement is 1.67 rad, how

far is she from the center of the merry-go-round?

2. A beetle sits at the top of a bicycle wheel and flies away just before it

would be squashed. Assuming that the wheel turns clockwise, the beetle’s

angular displacement is p rad, which corresponds to an arc length of 1.2

m. What is the wheel’s radius?

3. A car on a Ferris wheel has an angular displacement of p4

rad, which cor-

responds to an arc length of 29.8 m. What is the Ferris wheel’s radius?

4. Fill in the unknown quantities in the following table:

PRACTICE 7A

Angular displacement

∆q ∆s r

? rad +0.25 m 0.10 m

+0.75 rad ? 8.5 m

? degrees −4.2 m 0.75 m

+135° +2.6 m ?

a.

b.

c.

d.


1. A car tire rotates with an average angular speed of 29 rad/s. In what time

interval will the tire rotate 3.5 times?

2. A girl ties a toy airplane to the end of a string and swings it around her

head. The plane’s average angular speed is 2.2 rad/s. In what time interval

will the plane move through an angular displacement of 3.3 rad?

3. The average angular speed of a fly moving in a circle is 7.0 rad/s. How

long does the fly take to move through 2.3 rad?


PRACTICE 7B

Angular speed

wavg ∆q ∆t

? +2.3 rad 10.0 s

+0.75 rev/s ? 0.050 s

? − 1.2 turns 1.2 s

+2p rad/s + 1.5p rad ?

SAMPLE PROBLEM 7B

Angular speed

P R O B L E MA child at an ice cream parlor spins on a stool. The child turns counter-clockwise with an average angular speed of 4.0 rad/s. In what time inter-val will the child’s feet have an angular displacement of 8.0p rad?

S O L U T I O NGiven: ∆q = 8.0p rad wavg = 4.0 rad/s

Unknown: ∆t = ?

Use the angular speed equation from page 247. Rearrange to solve for ∆t.

wavg = ∆∆

qt

∆t = w∆

a

q

vg

∆t = 4

8

.

.

0

0pra

r

d

a

/

d

s = 2.0p s

∆t = 6.3 s

Chapter 7248

a.

b.

c.

d.


Angular acceleration occurs when angular speed changes

Figure 7-5 shows a bicycle turned upside down so that a repairperson can

work on the rear wheel. The bicycle pedals are turned so that at time t1 the

wheel has angular speed w1, as shown in Figure 7-5(a), and at a later time, t2,

it has angular speed w2 , as shown in Figure 7-5(b).The average angular acceleration, aavg (a is the Greek letter alpha), of an

object is given by the relationship shown below. Angular acceleration has the

units radians per second per second (rad/s2).

ANGULAR ACCELERATION

aavg = wt2

2

−−

wt1

1 = ∆∆wt

average angular acceleration =change in angular speed

time interval

t2

w2

(b)

t1

w1

(a)

Figure 7-5An accelerating bicycle wheelrotates with (a) an angular speedw1 at time t1 and (b) an angularspeed w2 at time t2.

SAMPLE PROBLEM 7C

Angular acceleration

P R O B L E MA car’s tire rotates at an initial angular speed of 21.5 rad/s. The dri-ver accelerates, and after 3.5 s the tire’s angular speed is 28.0 rad/s. What isthe tire’s average angular acceleration during the 3.5 s time interval?

S O L U T I O NGiven: w1 = 21.5 rad/s w2 = 28.0 rad/s ∆t = 3.5 s

Unknown: aavg = ?

Use the angular acceleration equation on this page.

aavg = w2

∆−t

w1 = = 6.5

3.

r

5

ad

s

/s

aavg = 1.9 rad/s2

28.0 rad/s − 21.5 rad/s

3.5 s

angular acceleration

the time rate of change of angu-lar speed, expressed in radiansper second per second

Copyright © by Holt, Rinehart and Winston. All rights reserved.Chapter 7250

All points on a rotating rigid object have the same angularacceleration and angular speed

If a point on the rim of a bicycle wheel had an angular speed greater than a

point nearer the center, the shape of the wheel would be changing. Thus, for a

rotating object to remain rigid, as does a bicycle wheel or a Ferris wheel, every

portion of the object must have the same angular speed and the same angular

acceleration. This fact is precisely what makes angular speed and angular

acceleration so useful for describing rotational motion.

COMPARING ANGULAR AND LINEAR QUANTITIES

Compare the equations we have found thus far for rotational motion with

those we found for linear motion in Chapter 2. For example, compare the fol-

lowing defining equation for average angular speed with the defining equa-

tion for average linear speed:

wavg = qt

f

f

−−

qti

i = ∆∆

qt

vavg = x

t

f

f

−−

x

ti

i = ∆∆

x

t

The equations are similar, with q replacing x and w replacing v. Take careful

note of such similarities as you study rotational motion because nearly every

linear quantity we have encountered thus far has a corresponding twin in

rotational motion, as shown in Table 7-1.

Table 7-1Angular substitutesfor linear quantities

Linear Angular

x q

v w

a a

1. A figure skater begins spinning counterclockwise at an angular speed of

4.0p rad/s. During a 3.0 s interval, she slowly pulls her arms inward and

finally spins at 8.0p rad/s. What is her average angular acceleration dur-

ing this time interval?

2. What angular acceleration is necessary to increase the angular speed of a

fan blade from 8.5 rad/s to 15.4 rad/s in 5.2 s?


PRACTICE 7C

Angular acceleration

aavg ∆w ∆t

? + 121.5 rad/s 7.0 s

+0.75 rad/s2 ? 0.050 s

? −1.2 turns/s 1.2 s

a.

b.

c.


Use kinematic equations for constant angular acceleration

In light of the similarities between variables in linear motion and those in rota-

tional motion, it should be no surprise that the kinematic equations of rota-

tional motion are similar to the linear kinematic equations in Chapter 2. The

equations of rotational kinematics under constant angular acceleration, along

with the corresponding equations for linear motion under constant accelera-

tion, are summarized in Table 7-2. Note that the following rotational motion

equations apply only for objects rotating about a fixed axis.

Table 7-2 Rotational and linear kinematic equations

Rotational motion with con- Linear motion with constantstant angular acceleration acceleration

wf = w i + a∆t vf = vi + a∆t

∆q = w i∆t + 21a(∆t)2 ∆x = vi∆t +

21a(∆t)2

wf2 = w i

2 + 2a(∆q) vf2 = vi

2 + 2a(∆x)

∆q = 21(w i + wf)∆t ∆x =

21(vi + vf)∆t

SAMPLE PROBLEM 7D

Angular kinematics

P R O B L E MThe wheel on an upside-down bicycle moves through 11.0 rad in 2.0 s. Whatis the wheel’s angular acceleration if its initial angular speed is 2.0 rad/s?

S O L U T I O NGiven: ∆q = 11.0 rad ∆t = 2.0 s w i = 2.00 rad/s

Unknown: a = ?

Use the second angular kinematic equation from Table 7-2 to solve for a .

∆q = w i∆t + 12

a(∆t)2

a = 2(∆q − w i∆t)/(∆t)2

a = 2[11.0 rad − (2.00 rad/s)(2.0 s)]/(2.0 s)2

a = 3.5 rad/s2

Note the correlation between the rotational equations involving the angular

variables q, w, and a and the equations of linear motion involving x, v, and a.

The quantity w in these equations represents the instantaneous angular

speed of the rotating object rather than the average angular speed.

INTERACTIV

E•

T U T O RPHYSICSPHYSICS

Module 8“Angular Kinematics”provides an interactive lessonwith guided problem-solvingpractice to teach you about dif-ferent kinds of angular motion,including the types describedhere.


1. What is the angular acceleration of the upside-down bicycle wheel in

Sample Problem 7D if it rotates through 18.0 rad in 5.00 s?

2. A diver performing a double somersault spins at an angular speed of

4.0p rad/s precisely 0.50 s after leaving the platform. Assuming the diver

begins with zero initial angular speed and accelerates at a constant rate,

what is the diver’s angular acceleration during the double somersault?

3. A fish swimming behind an oil tanker gets caught in a whirlpool created

by the ship’s propellers. The fish has an angular speed of 1.0 rad/s. After

4.5 s, the fish’s angular speed is 14.5 rad/s. If the water in the whirlpool

accelerates at a constant rate, what is the angular acceleration?

4. A remote-controlled car’s wheel accelerates at 22.4 rad/s2. If the wheel

begins with an angular speed of 10.8 rad/s, what is the wheel’s angular

speed after exactly three full turns?

5. How long does the wheel in item 4 take to make the three turns?

PRACTICE 7D

Angular kinematics

Section Review

1. Convert the following angles in degrees to radians:

a. 25°b. 35°c. 128°d. 270°

2. A mosquito lands on a phonograph record 5.0 cm from the record’s cen-

ter. If the record turns clockwise so that the mosquito travels along an arc

length of 5.0 cm, what is the mosquito’s angular displacement?

3. A bicyclist rides along a circular track. If the bicyclist travels around

exactly half the track in 10.0 s, what is his average angular speed?

4. Physics in Action Find the angular acceleration of a spinning

amusement-park ride that initially travels at 0.50 rad/s then accelerates

to 0.60 rad/s during a 0.50 s time interval.

5. Physics in Action What is the instantaneous angular speed of a

spinning amusement-park ride that accelerates from 0.50 rad/s at a con-

stant angular acceleration of 0.20 rad/s2 for 1.0 s?


RELATIONSHIPS BETWEEN ANGULARAND LINEAR QUANTITIES

As described at the beginning of Section 7-1, the motion of a point on a rotat-

ing object is most easily described in terms of an angle from a fixed reference

line. In some cases, however, it is useful to understand how the angular speed

and angular acceleration of a rotating object relate to the linear speed and lin-

ear acceleration of a point on the object.

Imagine a golfer swinging a golf club. The most effective method for hit-

ting a golf ball a long distance involves swinging the club in an approximate

circle around the body. If the club head undergoes a large angular accelera-

tion, then the linear acceleration of the club head as it is swung will be large.

This large linear acceleration causes the club head to strike the ball at a high

speed and produce a significant force on the ball. This section will explore the

relationships between angular and linear quantities.

Objects in circular motion have a tangential speed

Imagine an amusement-park carousel rotating about its center. Because a

carousel is a rigid object, any two horses attached to the carousel have the same

angular speed and angular acceleration regardless of their respective distances

from the axis of rotation. However, if the two horses are different distances

from the axis of rotation, they have different tangential speeds. The tangential

speed of any point rotating about an axis is also called the instantaneous linear

speed of that point. The tangential speed of a horse on the carousel is its speed

along a line drawn tangent to its circular path. (Recall that the tangent to a cir-

cle is the line that touches the circle at one and only one point.) The tangential

speeds of two horses at different distances from the center of a carousel are rep-

resented in Figure 7-6.Note that the speed of the horse at point A is represented by a shorter arrow

than the one that represents the speed of the horse at point B; this reflects the

difference between the tangential speeds of the two horses. The horse on the

outside must travel the same angular displacement during the same amount of

time as the horse on the inside. To achieve this, the horse on the outside must

travel a greater distance, ∆s, than the horse on the inside. Thus, an object that is

farther from the axis of a rigid rotating body, such as a carousel or a Ferris

wheel, must travel at a higher tangential speed around the circular path, ∆s, to

travel the same angular displacement as would an object closer to the axis.

7-2Tangential and centripetal

acceleration


• Find the tangential speed of apoint on a rigid rotatingobject using the angularspeed and the radius.

• Solve problems involving tan-gential acceleration.

• Solve problems involvingcentripetal acceleration.

tangential speed

the instantaneous linear speed ofan object directed along the tan-gent to the object’s circular path

vt,outside

vt,inside

ω

AB

Figure 7-6Horses on a carousel move at thesame angular speed but differenttangential speeds.


How can you find the tangential speed? Again consider the rotating carousel.

If the carousel rotates through an angle ∆q, a horse rotates through an arc

length ∆s in the interval ∆t. The angular displacement of the horse is given by

the equation for angular displacement.

∆q = ∆r

s

To find the tangential speed of the horse, divide both sides of the equation

by the time the horse takes to travel the distance ∆s.

∆∆

qt

= 1

r ∆∆

s

t

From Section 7-1, you know that the left side of the equation equals wavg.

Similarly, ∆s is a linear distance, so ∆s divided by ∆t is a linear speed along an

arc length. If ∆t is very short, then ∆s is so small that it is nearly tangent to the

circle; therefore, the speed is the tangential speed.

Note that w is the instantaneous angular speed, rather than the average angu-

lar speed, because the time interval is so short. This equation is valid only when wis measured in radians per unit of time. Other measures of angular speed, such as

degrees per second and revolutions per second, must not be used in this equation.

TANGENTIAL SPEED

vt = rw

tangential speed = distance from axis × angular speed

SAMPLE PROBLEM 7E

Tangential speed

P R O B L E MThe radius of a CD in a computer is 0.0600 m. If a microbe riding on thedisc’s rim has a tangential speed of 1.88 m/s, what is the disc’s angular speed?

S O L U T I O NGiven: r = 0.0600 m vt = 1.88 m/s

Unknown: w = ?

Use the tangential speed equation on this page to solve for angular speed.

vt = rw

w = v

rt =

0

1

.

.

0

8

6

8

0

m

0 m

/s

w = 31.3 rad/s

TOPIC: Rotational motionGO TO: www.scilinks.orgsciLINKS CODE: HF2071

NSTA


Tangential acceleration is tangent to the circular path

If a carousel speeds up, the horses on it experience an angular acceleration.

The linear acceleration related to this angular acceleration is tangent to the

circular path and is called the tangential acceleration.Imagine that an object rotating about a fixed axis changes its angular speed

by ∆w in the interval ∆t. At the end of this time, the speed of a point on the

object has changed by the amount ∆vt . Using the equation for tangential

velocity on page 254 gives the following:

∆vt = r∆w

Dividing by ∆t gives ∆∆v

tt = r

∆∆wt

If the time interval ∆t is very small, then the left side of this relationship

gives the tangential acceleration of the point. The angular speed divided by

the time interval on the right side is the angular acceleration. Thus, the tan-

gential acceleration of a point on a rotating object is given by the relationship

on the next page.

1. A woman passes through a revolving door with a tangential speed of

1.8 m/s. If she is 0.80 m from the center of the door, what is the door’s

angular speed?

2. A softball pitcher throws a ball with a tangential speed of 6.93 m/s. If the

pitcher’s arm is 0.660 m long, what is the angular speed of the ball before

the pitcher releases it?

3. An athlete spins in a circle before releasing a discus with a tangential

speed of 9.0 m/s. What is the angular speed of the spinning athlete?

Assume the discus is 0.75 m from the athlete’s axis of rotation.


PRACTICE 7E

Tangential speed

vt w r

? 121.5 rad/s 0.030 m

0.75 m/s ? 0.050 m

? 1.2 turns/s 3.8 m

2.0p m/s 1.5p rad/s ?

tangential acceleration

the instantaneous linear acceler-ation of an object directed alongthe tangent to the object’s circu-lar path

a.

b.

c.

d.


SAMPLE PROBLEM 7F

Tangential acceleration

P R O B L E MA spinning ride at a carnival has an angular acceleration of 0.50 rad/s2. Howfar from the center is a rider who has a tangential acceleration of 3.3 m/s2?

S O L U T I O NGiven: a = 0.50 rad/s2 at = 3.3 m/s2

Unknown: r = ?

Use the tangential acceleration equation on this page. Rearrange to solve for r.

at = ra

r = aat =

0.

3

5

.

0

3

r

m

ad

/s

/

2

s2

r = 6.6 m

1. A dog on a merry-go-round undergoes a 1.5 m/s2 linear acceleration. If

the merry-go-round’s angular acceleration is 1.0 rad/s2, how far is the

dog from the axis of rotation?

2. A young boy swings a yo-yo horizontally above his head at an angular

acceleration of 0.35 rad/s2. If tangential acceleration of the yo-yo at the

end of the string is 0.18 m/s2, how long is the string?

3. What is a tire’s angular acceleration if the tangential acceleration at a

radius of 0.15 m is 9.4 × 10−2 m/s2?

PRACTICE 7F

Tangential acceleration

Again, the angular acceleration in this equation refers to the instantaneous

angular acceleration. This equation must use the unit radians to be valid. In

SI, angular acceleration is expressed as radians per second per second.

TANGENTIAL ACCELERATION

at = ra

tangential acceleration = distance from axis × angular acceleration

Chapter 7256


CENTRIPETAL ACCELERATION

Figure 7-7 shows a car moving in a circular path with a con-

stant tangential speed of 30 km/h. Even though the car moves

at a constant speed, it still has an acceleration. To see why this

is, consider the defining equation for acceleration.

a = v

tf

f −−

t

v

i

i

Note that acceleration depends on a change in the veloc-

ity. Because velocity is a vector, there are two ways

an acceleration can be produced: by a change in the magni-

tude of the velocity and by a change in the direction of the

velocity. For a car moving in a circular path with constant

speed, the acceleration is due to a change in direction. An

acceleration of this nature is called a centripetal (center-

seeking) acceleration. Its magnitude is given by the follow-

ing equation:

ac = v

rt2

Consider Figure 7-8(a). An object is seen first at point A, with tangential

velocity vi at time ti , and then at point B, with tangential velocity vf at a later

time, tf . Assume that vi and vf differ in direction only and their magnitudes

are the same.

The change in velocity, ∆v = vf − vi, can be determined graphically, as

shown by the vector triangle in Figure 7-8(b). Note that when ∆t is very small

(as ∆t approaches zero), vf will be almost parallel to vi and the vector ∆v will

be approximately perpendicular to them, pointing toward the center of the

circle. This means that the acceleration will also be directed toward the center

of the circle because it is in the direction of ∆v.Because the tangential speed is related to the angular speed through the

relationship vt = rw , the centripetal acceleration can be found using the angu-

lar speed as well.

CENTRIPETAL ACCELERATION

ac = v

rt2

ac = rw2

centripetal acceleration = (

d

ta

is

n

ta

g

n

en

ce

ti

f

a

r

l

o

s

m

pe

a

e

x

d

i

)

s

2

centripetal acceleration = distance from axis × (angular speed)2

Figure 7-7Although the car moves at a constant speed of 30 km/h,the car still has an acceleration because the direction ofthe velocity changes.

(a)

r

BA

r

∆svf

vi

∆θ

Figure 7-8(a) As the particle moves from A toB, the direction of the particle’svelocity vector changes. (b)Vectoraddition is used to determine thedirection of the change in velocity,∆v, which for short time intervals istoward the center of the circle.

(b)vf

−vi

∆v ∆θ

∆v = vf − vi = vf + (− vi )

centripetal acceleration

acceleration directed toward thecenter of a circular path


1. A girl sits on a tire that is attached to an overhanging tree limb by a rope.

The girl’s father pushes her so that her centripetal acceleration is 3.0 m/s2.

If the length of the rope is 2.1 m, what is the girl’s tangential speed?

2. A young boy swings a yo-yo horizontally above his head so that 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?

3. A dog sits 1.5 m from the center of a merry-go-round. If the dog under-

goes a 1.5 m/s2 centripetal acceleration, what is the dog’s linear speed?

What is the angular speed of the merry-go-round?

4. A race car moves along a circular track at an angular speed of 0.512 rad/s.

If the car’s centripetal acceleration is 15.4 m/s2, what is the distance

between the car and the center of the track?

5. A piece of clay sits 0.20 m from the center of a potter’s wheel. If the pot-

ter spins the wheel at an angular speed of 20.5 rad/s, what is the magni-

tude of the centripetal acceleration of the piece of clay on the wheel?

PRACTICE 7G

Centripetal acceleration

SAMPLE PROBLEM 7G

Centripetal acceleration

P R O B L E MA test car moves at a constant speed around a circular track. If the car is48.2 m from the track’s center and has a centripetal acceleration of8.05 m/s2, what is its tangential speed?

S O L U T I O NGiven: r = 48.2 m ac = 8.05 m/s2

Unknown: vt = ?

Use the first centripetal acceleration equation from page 257. Rearrange to

solve for vt.

ac = v

rt2

vt =√

ac�r� =√

(8�.0�5�m�/s�2)�(4�8.�2�m�)�

vt = 19.7 m/s


Tangential and centripetal accelerations are perpendicular

Centripetal and tangential acceleration are not the same. To understand why,

consider a car moving around a circular track. Because the car is moving in a

circular path, it always has a centripetal component of acceleration because its

direction of travel, and hence the direction of its velocity, is continually

changing. If the car’s speed is increasing or decreasing, the car also has a tan-

gential component of acceleration. To summarize, the tangential component

of acceleration is due to changing speed; the centripetal component of accel-

eration is due to changing direction.

Find the total acceleration using the Pythagorean theorem

When both components of acceleration exist simultaneously, the tangential

acceleration is tangent to the circular path and the centripetal acceleration

points toward the center of the circular path. Because these components of

acceleration are perpendicular to each other, the magnitude of the total accel-

eration can be found using the Pythagorean theorem, as follows:

atotal = �at�2�+� a�c2�

The direction of the total acceleration, as shown in Figure 7-9, depends on

the magnitude of each component of acceleration and can be found using the

inverse of the tangent function.

q = tan−1 a

ac

t

Section Review

1. Find the tangential speed of a ball swung at a constant angular speed of

5.0 rad/s on a rope that is 5.0 m long.

2. If an object has a tangential acceleration of 10.0 m/s2, the angular speed

will do which of the following?

a. decrease

b. stay the same

c. increase

3. Physics in Action Find the tangential acceleration of a person

standing 9.5 m from the center of a spinning amusement-park ride that

has an angular acceleration of 0.15 rad/s2.

4. Physics in Action If a spinning amusement-park ride has an angu-

lar speed of 1.2 rad/s, what is the centripetal acceleration of a person

standing 12 m from the center of the ride?

at

ac

atotalθ

Figure 7-9The direction of the total accelera-tion of a rotating object can befound using the tangent function.


FORCE THAT MAINTAINS CIRCULAR MOTION

Consider a ball of mass m tied to a string of length r that is being whirled in a

horizontal circular path, as shown in Figure 7-10. Assume that the ball moves

with constant speed. Because the velocity vector, v, changes direction con-

tinuously during the motion, the ball experiences a centripetal acceleration

directed toward the center of motion, as described in Section 7-2, with magni-

tude given by the following equation:

ac = v

rt2

The inertia of the ball tends to maintain the ball’s motion in a straight-line

path; however, the string counteracts this tendency by exerting a force on the

ball that makes the ball follow a circular path. This force is directed along the

length of the string toward the center of the circle, as shown in Figure 7-10.The magnitude of this force can be found by applying Newton’s second law

along the radial direction.

Fc = mac

The net force on an object directed toward the center of the object’s circu-

lar path is the force that maintains the object’s circular motion.

The force that maintains circular motion is measured in the SI unit of

newtons. This force is no different from any of the other forces we have

studied. For example, friction between a race car’s tires and a circular race-

track provides the force that enables the car to travel in a circular path. As

another example, the gravitational force exerted on the moon by Earth pro-

vides the force necessary to keep the moon in its orbit.

FORCE THAT MAINTAINS CIRCULAR MOTION

Fc = m

r

vt2

Fc = mrw 2

force that maintains circular motion = mass × (ta

d

n

is

g

t

e

a

n

n

t

c

i

e

al

to

sp

a

e

x

e

i

d

s

)2

force that maintains = mass × distance to axis × (angular speed)2

circular motion

7-3Causes of circular motion


• Calculate the force thatmaintains circular motion.

• Explain how the apparentexistence of an outward forcein circular motion can beexplained as inertia resistingthe force that maintains cir-cular motion.

• Apply Newton’s universal lawof gravitation to find thegravitational force betweentwo masses.

m v

r

Fc

Figure 7-10When a ball is whirled in a circle, aforce directed toward the center ofthe ball’s circular path acts on it.

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SAMPLE PROBLEM 7H

Force that maintains circular motion

P R O B L E MA pilot is flying a small plane at 30.0 m/s in a circular path with a radius of100.0 m. If a force of 635 N is needed to maintain the pilot’s circularmotion, what is the pilot’s mass?

S O L U T I O NGiven: vt = 30.0 m/s r = 100.0 m Fc = 635 N

Unknown: m = ?

Use the equation for force from page 260. Rearrange to solve for m.

Fc = m v

rt2

m = Fc v

r

t2 = 635 N

(3

1

0

0

.0

0.

m

0 m

/s)2

m = 70.6 kg

1. A girl sits in a tire that is attached to an overhanging tree limb by a

rope 2.10 m in length. The girl’s father pushes her with a tangential speed

of 2.50 m/s. If the magnitude of the force that maintains her circular

motion is 88.0 N, what is the girl’s mass?

2. A bicyclist is riding at a tangential speed of 13.2 m/s around a circular

track with a radius of 40.0 m. If the magnitude of the force that main-

tains the bike’s circular motion is 377 N, what is the combined mass of

the bicycle and rider?

3. A dog sits 1.50 m from the center of a merry-go-round with an angular

speed of 1.20 rad/s. If the magnitude of the force that maintains the dog’s

circular motion is 40.0 N, what is the dog’s mass?

4. A 905 kg test car travels around a 3.25 km circular track. If the magnitude

of the force that maintains the car’s circular motion is 2140 N, what is the

car’s tangential speed?

PRACTICE 7H

Force that maintains circular motion


A force directed toward the center is necessary for circular motion

Because the force that maintains circular motion acts at right angles to the

motion, it causes a change in the direction of the velocity. If this force van-

ishes, the object does not continue to move in its circular path. Instead, it

moves along a straight-line path tangent to the circle. To see this point, con-

sider a ball that is attached to a string and is being whirled in a vertical circle,

as shown in Figure 7-11. If the string breaks when the ball is at the position

shown in Figure 7-11(a), the force that maintains circular motion will vanish

and the ball will move vertically upward. The motion of the ball will be that of

a free-falling body. If the string breaks when the ball is at the top of its circular

path, as shown in Figure 7-11(b), the ball will fly off horizontally in a direc-

tion tangent to the path, then move in the parabolic path of a projectile.

DESCRIBING THE MOTION OFA ROTATING SYSTEM

To better understand the motion of a rotating system, consider a car

approaching a curved exit ramp to the left at high speed. 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 explana-

tion is that there must be a force that pushes the passenger outward. This force

is sometimes called the centrifugal force, but that term often creates confu-

sion, 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-line path. As the car enters the

ramp and travels along a curved path, the passenger, because of inertia, tends

to move along the original straight-line path. This is in accordance with New-

ton’s first law, which states that the natural tendency of a body is to continue

moving in a straight line. However, if a sufficiently large force that maintains

circular motion (toward the center of curvature) acts on the passenger, the

person moves in a curved path, along with the car. The origin of the force that

maintains the circular motion of the passenger is the force of friction between

the passenger and the car seat. If this frictional force is not sufficient, the pas-

senger slides across the seat as the car turns underneath. Because of inertia,

the passenger continues to move in a straight-line path. Eventually, the pas-

senger encounters the door, which provides a large enough force to enable the

passenger to follow the same curved path as the car. The passenger slides

toward the door not because of some mysterious outward force but because

the force that maintains circular motion is not great enough to enable the pas-

senger to travel along the circular path followed by the car.

(a)

Figure 7-11A ball is whirled in a vertical circu-lar path on the end of a string.When the string breaks at the posi-tion shown in (a), the ball movesvertically upward in free fall. (b)When the string breaks at the topof the ball’s path, the ball movesalong a parabolic path.

(b)

1. Pizza

Pizza makers traditionallyform the crust by throwingthe dough up in the air andspinning it. Why does thismake the pizza crust bigger?

2. Swings

The amusement-park ridepictured below spins ridersaround on swings attachedby cables from above. Whatcauses the swings to moveaway from the center of the

ride when the center

column be-gins to turn?


objects’ masses.

Gravitational force acts such that objects

are always attracted to one another. Exam-

ine the illustration of Earth and the moon

in Figure 7-12. Note that the gravitational

force between Earth and the moon is attrac-

tive, and recall that Newton’s third law

states that the force exerted on Earth by the

moon, FmE , is equal in magnitude to and in

the opposite direction of the force exerted

on the moon by Earth, FEm.

Gravitational force depends on the distance between two masses

If masses m1 and m2 are separated by distance r, the magnitude of the gravita-

tional force is given by the following equation:

FmE

FEm

Figure 7-12The gravitational force betweenEarth and the moon is attractive.According to Newton’s third law,FEm = FmE .

gravitational force

the mutual force of attractionbetween particles of matter

NEWTON’S LAW OF UNIVERSAL GRAVITATION

Fg = G m

r1m

22

gravitational force = constant ×mass 1 × mass 2

(distance between center of masses)2

G is a universal constant called the constant of universal gravitation; it can

be used to calculate gravitational forces between any two particles and has

been determined experimentally.

G = 6.673 × 10−11 N

k

•

g

m2

2

The law of universal gravitation is an example of an inverse-square law,

because the force varies as the inverse square of the separation. That is, the

force between two masses decreases as the masses move farther apart.

NEWTON’S LAW OF UNIVERSAL GRAVITATION

Note that planets move in nearly circular orbits around the sun. As mentioned

earlier, the force that keeps these planets from coasting off in a straight line is

a gravitational force. The gravitational force is a field force that always exists

between two masses, regardless of the medium that separates them. It exists

not just between large masses like the sun, Earth, and moon but between any

two masses, regardless of size or composition. For instance, desks in a class-

room have a mutual attraction because of gravitational force. The force

between the desks, however, is small relative to the force between the moon

and Earth because the gravitational force is proportional to the product of the

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The astronomer Johannes Keplerproposed that planets orbit the sunin elliptical paths. However, some sci-entists continued to believe thatEarth was the center of the solarsystem until Sir Isaac Newtonshowed that elliptical orbits could bepredicted using his laws of motion.

SAMPLE PROBLEM 7I

Gravitational force

P R O B L E MFind the distance between a 0.300 kg billiard ball and a 0.400 kg billiard ballif the magnitude of the gravitational force is 8.92 � 10�11 N.

S O L U T I O NGiven: m1 = 0.300 kg m2 = 0.400 kg Fg = 8.92 × 10−11 N

Unknown: r = ?

Use the equation for Newton’s Law of Universal Gravitation.

r2 = F

G

gm1m2 = (0.300 kg)(0.400 kg)

= 8.97 × 10−2 m2

r =√

8.�97� ×� 1�0−�2�m�2� = 3.00 × 10−1 m

6.673 × 10−11N

k

•

g

m2

2

Chapter 7264

Gravitational force is localized to the center of a spherical mass

The gravitational force exerted by a spherical mass on a particle outside the

sphere is the same as it would be if the entire mass of the sphere were concen-

trated at its center. For example, the force on an object of mass m at Earth’s

surface has the following magnitude:

Fg = G M

RE

E

m2

ME is Earth’s mass and RE is its radius. This force is directed toward the

center of Earth. Note that this force is in fact the weight of the mass, mg.

mg = G M

RE

E

m2

By substituting the actual values for the mass and radius of Earth, we can

find the value for g and compare it with the value of free-fall acceleration used

throughout this book.

Because m occurs on both sides of the equation above, these masses cancel.

g = G R

M

E

E2 = �6.673 × 10−11

N

k

•

g

m2

2

�(56.

.

9

3

8

7

××

1

1

0

0

2

6

4

m

kg

)2 = 9.83 m/s2

This value for g is approximately equal to the value used throughout this

book. The difference is due to rounding the values for Earth’s mass and radius.


1. If the mass of each ball in Sample Problem 7I is 0.800 kg, at what dis-

tance between the balls will the gravitational force between the balls have

the same magnitude as that in Sample Problem 7I? How does the change

in mass affect the magnitude of the gravitational force?

2. Mars has a mass of about 6.4 × 1023 kg, and its moon Phobos has a mass

of about 9.6 × 1015 kg. If the magnitude of the gravitational force between

the two bodies is 4.6 × 1015 N, how far apart are Mars and Phobos?

3. Find the magnitude of the gravitational force a 67.5 kg person would

experience while standing on the surface of each of the following planets:

PRACTICE 7I

Gravitational force

Planet m r

Earth 5.98 × 1024 kg 6.37 × 106 m

Mars 6.34 × 1023 kg 3.43 × 106 m

Pluto 5 × 1023 kg 4 × 105 m

Section Review

1. A roller coaster moves through a vertical loop at a constant speed and sus-

pends its passengers upside down. In what direction is the force that causes

the coaster and its passengers to move in a circle? What provides this force?

2. Identify the force that maintains the circular motion of the following:

a. a bicyclist moving around a flat circular track

b. a bicycle moving around a flat circular track

c. a bobsled turning a corner on its track

3. A 90.0 kg person stands 1.00 m from a 60.0 kg person sitting on a bench

nearby. What is the magnitude of the gravitational force between them?

4. Physics in Action A 90.0 kg person rides a spinning amusement-

park ride that has an angular speed of 1.15 rad/s. If the radius of the ride

is 11.5 m, what is the magnitude of the force that maintains the circular

motion of the person?

5. Physics in Action Calculate the mass that a planet with the same

radius as Earth would need in order to exert a gravitational force equal to the

force on the person in item 4.

a.

b.

c.


Projectiles and satellites

As explained in Chapter 3, when a ball is thrown parallel to the ground, the

motion of the ball has two components of motion: a horizontal velocity, which

remains unchanged, and a vertical acceleration, which equals free-fall accelera-

tion. Given this analysis, it may seem confusing to think of the moon and other

satellites in orbit around Earth as projectiles. But in fact, satellites are projec-

tiles. As shown in Figure 7-13, the larger the velocity parallel to Earth’s surface,

the farther the projectile moves before striking Earth. Note, however, that at

some large velocity, the projectile returns to its point of origin without moving

closer to Earth. In this case, the gravitational force between the projectile

and Earth is just great enough to keep the projectile from moving

along its inertial straight-line path. This is how satellites stay in orbit.

Escape speed

When the speed of an object, such as a rocket, is greater than the

speed required to keep it in orbit, the object can escape the gravi-

tational pull of Earth and soar off into space. The object soars off

into space when its initial speed moves it out of the range in which

the gravitational force is significant. Mathematically, the value of this

escape speed (vesc) is given by the following equation:

vesc =�2M�R�G

�Earth’s radius, R, is about 6.37 × 106 m, and its mass is approximately 5.98 × 1024 kg. Thus, the escape speed of a projectile from Earth is 1.12 × 104 m/s.(Note that this value does not depend on the mass of the projectile in question.)

As the mass of a planet or other body increases and its radius decreases, the

escape speed necessary for a projectile to escape the gravitational pull of that

body increases, as shown in Figure 7-14. If the body has a very large mass and

Figure 7-13When the speed ofa projectile is largeenough, the projec-tile orbits Earth asa satellite.


Black holes

The existence of such massive objects was first predicted in 1916 by Karl

Schwarzschild, who used his solutions to Einstein’s general-relativity

equations to predict their properties. Thus, the distance from the center of

the object to the circular orbit at which escape speed is equal to the speed of

light is called the Schwarzschild radius. Because light cannot escape from any

point within the sphere defined by the Schwarzschild radius, no information

can be obtained about events that occur in this region. Hence, the edge of the

sphere is called the event horizon, and the apparently lightless region within it

is called a black hole.

Recent observations have provided strong

evidence for the existence of black holes.

Tremendous amounts of X rays and other

radiation have been observed coming from

regions that are near visible stars, although

no stars appear to be the source of the radia-

tion. If the visible star has a black hole as a

companion, it could be losing some of its

outer atmosphere to the black hole, and those

atmospheric gases could be emitting radiation

as they accelerate closer to the black hole.

Candidates for black holes of these types

include Scorpius X-1 and Cygnus X-1.

The large amount of energy that galactic centers produce suggests to

many astrophysicists that the energy sources are supermassive black holes.

The galaxy NGC 4261, shown in Figure 7-15, is likely to have a black hole at

its center. Some astronomers believe the Milky Way, our own galaxy, contains

a black hole about the size of our solar system.


a small radius, the speed necessary for a projectile to escape the gravitational

pull of that body reaches very high values. For example, an object with a mass

three times that of the sun but with a diameter of about 10 km would require

an escape speed equal to the speed of light. In other words, the force of gravity

such an object exerts on a projectile is so great that even light does not move

fast enough to escape it.

Figure 7-14(a) A projectile that passes near any massive object will bebent from its trajectory. (b) Themore compact the object, thenearer the projectile canapproach and the greater theescape speed it needs. (c) If theobject is so small and massivethat a projectile’s escape speedat distance RS is greater than thespeed of light, then the object isclassified as a black hole.

r1r2 RS

(a) (c)(b)

Figure 7-15The gas jets and the central diskshown at right, created bycombining an optical image witha radio telescope image of NGC426 1 , show signs of a black holein this galaxy’s center.

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CHAPTER 7Summary

KEY IDEAS

Section 7-1 Measuring rotational motion• The average angular speed, wavg , of a rigid, rotating object is defined as

the ratio of the angular displacement, ∆q, to the time interval, ∆t.

• The average angular acceleration, aavg , of a rigid, rotating object is defined

as the ratio of the change in angular speed, ∆w , to the time interval, ∆t.

Section 7-2 Tangential and centripetal acceleration• A point on an object rotating about a fixed axis has a tangential speed

related to the object’s angular speed. When the object’s angular accelera-

tion changes, the tangential acceleration of a point on the object changes.

• Uniform circular motion occurs when an acceleration of constant magnitude

is perpendicular to the tangential velocity.

Section 7-3 Causes of circular motion• Any object moving in a circular path must have a net force exerted on it

that is directed toward the center of the circular path.

• Every particle in the universe attracts every other particle with a force that

is directly proportional to the product of the particles’ masses and inverse-

ly proportional to the square of the distance between the particles.

KEY TERMS

angular acceleration (p. 249)

angular displacement (p. 246)

angular speed (p. 247)

centripetal acceleration (p. 257)

gravitational force (p. 263)

radian (p. 245)

rotational motion (p. 244)

tangential acceleration (p. 255)

tangential speed (p. 253)

Variable symbols

Quantities Units

s arc length m meters

∆q angular displacement rad radians

w angular speed rad/s radians/second

a angular acceleration rad/s2 radians/second2

vt tangential speed m/s meters/second

at tangential acceleration m/s2 meters/second2

ac centripetal acceleration m/s2 meters/second2

Fc force that maintains N newtonscircular motion

Fg gravitational force N newtons

G constant of universal gravitation N

k

•

g

m2

2

new

k

t

i

o

lo

n

g

s

r

•

a

m

m

e

s2ters2

Diagram symbols

Rotationalmotion

Anglemarking


CHAPTER 7Review and Assess

EQUATIONS FOR ANGULAR MOTION

Practice problems

10. A potter’s wheel moves from rest to an angular speedof 0.20 rev/s in 30.0 s. Assuming constant angularacceleration, what is its angular acceleration in rad/s2?(See Sample Problem 7D.)

11. A drill starts from rest. After 3.20 s of constant angularacceleration, the drill turns at a rate of 2628 rad/s.

a. Find the drill’s angular acceleration.b. Determine the angle through which the drill

rotates during this period.

(See Sample Problem 7D.)

12. A tire placed on a balancing machine in a service sta-tion starts from rest and turns through 4.7 revs in1.2 s before reaching its final angular speed. Assum-ing that the angular acceleration of the wheel is con-stant, calculate the wheel’s angular acceleration.(See Sample Problem 7D.)

TANGENTIAL AND CENTRIPETALACCELERATION

Review questions

13. When a wheel rotates about a fixed axis, do all thepoints on the wheel have the same tangential speed?

14. Correct the following statement: The racing carrounds the turn at a constant velocity of 145 km/h.

15. Describe the path of a moving body whose accelera-tion is constant in magnitude at all times and is per-pendicular to the velocity.

16. An object moves in a circular path with constantspeed v.

a. Is the object’s velocity constant? Explain.b. Is its acceleration constant? Explain.

RADIANS AND ANGULAR MOTION

Review questions

1. How many degrees equal p radians? How many revolutions equal p radians?

2. What units must be used for q, w, and a in the kin-ematic equations for rotational motion listed inTable 7-2?

3. Distinguish between linear speed and angular speed.

4. When a wheel rotates about a fixed axis, do allpoints on the wheel have the same angular speed?

Practice problems

5. A car on a Ferris wheel has an angular displacementof 0.34 rad. If the car moves through an arc lengthof 12 m, what is the radius of the Ferris wheel?(See Sample Problem 7A.)

6. When a wheel is rotated through an angle of 35°, apoint on the circumference travels through an arclength of 2.5 m. When the wheel is rotated throughangles of 35 rad and 35 rev, the same point travelsthrough arc lengths of 143 m and 9.0 × 102 m,respectively. What is the radius of the wheel?(See Sample Problem 7A.)

7. How long does it take the second hand of a clock tomove through 4.00 rad?(See Sample Problem 7B.)

8. A phonograph record has an initial angular speed of33 rev/min. The record slows to 11 rev/min in 2.0 s.What is the record’s average angular accelerationduring this time interval?(See Sample Problem 7C.)

9. If a flywheel increases its average angular speed by2.7 rad/s in 1.9 s, what is its angular acceleration?(See Sample Problem 7C.)


Conceptual questions

17. Give an example of a situation in which an automo-bile driver can have a centripetal acceleration but notangential acceleration.

18. Can a car move around a circular racetrack so thatthe car has a tangential acceleration but no cen-tripetal acceleration?

19. The gas pedal and the brakes of a car accelerate anddecelerate the car. Could a steering wheel performeither of these two actions? Explain.

20. It has been suggested that rotating cylinders about16 km long and 8 km in diameter should be placedin space for future space colonies. The rotationwould simulate gravity for the inhabitants of thesecolonies. Explain the concept behind this proposal.

Practice problems

21. A small pebble breaks loose from the treads of a tirewith a radius of 32 cm. If the pebble’s tangentialspeed is 49 m/s, what is the tire’s angular speed?(See Sample Problem 7E.)

22. The Emerald Suite is a revolving restaurant at thetop of the Space Needle in Seattle, Washington. If acustomer sitting 12 m from the restaurant’s centerhas a tangential speed of 2.18 × 10−2 m/s, what isthe angular speed of the restaurant?(See Sample Problem 7E.)

23. A bicycle wheel has an angular acceleration of1.5 rad/s2. If a point on its rim has a tangential accel-eration of 48 cm/s2, what is the radius of the wheel?(See Sample Problem 7F.)

24. When the string is pulled in the correct direction on awindow shade, a lever is released and the shaft that theshade is wound around spins. If the shaft’s angularacceleration is 3.8 rad/s2 and the shade acceleratesupward at 0.086 m/s2, what is the radius of the shaft?(See Sample Problem 7F.)

25. A building superintendent twirls a set of keys in acircle at the end of a cord. If the keys have a cen-tripetal acceleration of 145 m/s2 and the cord has alength of 0.34 m, what is the tangential speed of thekeys?(See Sample Problem 7G.)

26. A sock stuck to the side of a clothes-dryer barrel hasa centripetal acceleration of 28 m/s2. If the dryerbarrel has a radius of 27 cm, what is the tangentialspeed of the sock?(See Sample Problem 7G.)

CAUSES OF CIRCULAR MOTION

Review questions

27. Imagine that you attach a heavy object to one end of aspring and then, while holding the spring’s other end,whirl the spring and object in a horizontal circle. Doesthe spring stretch? Why? Discuss your answer interms of the force that maintains circular motion.

28. Why does the water remain in a pail that is whirledin a vertical path, as shown in Figure 7-16?

29. Identify the influence of mass and distance on gravitational forces.

30. Explain the difference between centripetal accelera-tion and angular acceleration.

31. Comment on the statement, “There is no gravity inouter space.”

Conceptual questions

32. Explain why Earth is not spherical in shape and whyit bulges at the equator.

33. Because of Earth’s rotation, you would weigh slightlyless at the equator than you would at the poles. Why?

34. Why does mud fly off a rapidly turning wheel?

35. Astronauts floating around inside the space shuttleare not actually in a zero-gravity environment. Whatis the real reason astronauts seem weightless?

Figure 7-16


MIXED REVIEW PROBLEMS

41. Find the average angular speed of Earth about thesun in radians per second.(Hint: Earth orbits the sun once every 365.25 days.)

42. The tub within a washer goes into its spin cycle, start-ing from rest and reaching an angular speed of11p rad/s in 8.0 s. At this point, the lid is opened, anda safety switch turns off the washer. The tub slows torest in 12.0 s. Through how many revolutions doesthe tub turn? Assume constant angular accelerationwhile the machine is starting and stopping.

43. An airplane is flying in a horizontal circle at a speedof 105 m/s. The 80.0 kg pilot does not want the cen-tripetal acceleration to exceed 7.00 times free-fallacceleration.

a. Find the minimum radius of the plane’s path.b. At this radius, what is the net force that main-

tains circular motion exerted on the pilot bythe seat belts, the friction against the seat, andso forth?

44. A car traveling at 30.0 m/s undergoes a constantnegative acceleration of magnitude 2.00 m/s2 whenthe brakes are applied. How many revolutions doeseach tire make before the car comes to a stop,assuming that the car does not skid and that thetires have radii of 0.300 m?

45. A coin with a diameter of 2.40 cm is dropped onto ahorizontal surface. The coin starts out with an initialangular speed of 18.0 rad/s and rolls in a straight linewithout slipping. If the rotation slows with an angu-lar acceleration of magnitude 1.90 rad/s2, how fardoes the coin roll before coming to rest?

46. A mass attached to a 50.0 cm string starts from restand is rotated in a circular path exactly 40 times in1.00 min before reaching a final angular speed. Whatis the angular speed of the mass after 1.00 min?

47. A 13 500 N car traveling at 50.0 km/h rounds acurve of radius 2.00 × 102 m. Find the following:

a. the centripetal acceleration of the carb. the force that maintains centripetal accelerationc. the minimum coefficient of static friction

between the tires and the road that will allowthe car to round the curve safely

36. A girl at a state fair swings a ball in a vertical circle atthe end of a string. Is the force applied by the stringgreater than the weight of the ball at the bottom ofthe ball’s path?

Practice problems

37. A roller-coaster car speeds down a hill past point Aand then rolls up a hill past point B, as shown inFigure 7-17.

a. The car has a speed of 20.0 m/s at point A. If thetrack exerts a force on the car of 2.06 × 104 N atthis point, what is the mass of the car?

b. What is the maximum speed the car can haveat point B for the gravitational force to hold iton the track?

(See Sample Problem 7H.)

38. Tarzan tries to cross a river by swinging from onebank to the other on a vine that is 10.0 m long. Hisspeed at the bottom of the swing, just as he clearsthe surface of the river, is 8.0 m/s. Tarzan does notknow that the vine has a breaking strength of 1.0 ×103 N. What is the largest mass Tarzan can have andmake it safely across the river?(See Sample Problem 7H.)

39. The gravitational force of attraction between twostudents sitting at their desks in physics class is 3.20 × 10−8 N. If one student has a mass of 50.0 kgand the other has a mass of 60.0 kg, how far apartare the students sitting?(See Sample Problem 7I.)

40. If the gravitational force between the electron (9.11 ×10−31 kg) and the proton (1.67 × 10−27 kg) in ahydrogen atom is 1.0 × 10−47 N, how far apart are thetwo particles?(See Sample Problem 7I.)

Figure 7-17

10.0 m

15.0 m

A

B


50. Find the centripetal accelerations of the following:

a. a point on the equator of Earthb. a point at the North Pole of Earth

(See the table in the appendix for data on Earth.)

51. A copper block rests 30.0 cm from the center of asteel turntable. The coefficient of static frictionbetween the block and the surface is 0.53. Theturntable starts from rest and rotates with a constantangular acceleration of 0.50 rad/s2. After what timeinterval will the block start to slip on the turntable?(Hint: The normal force in this case equals the weightof the block.)

48. A 2.00 × 103 kg car rounds a circular turn of radius20.0 m. If the road is flat and the coefficient of staticfriction between the tires and the road is 0.70, howfast can the car go without skidding?

49. During a solar eclipse, the moon, Earth, and sun lieon the same line, with the moon between Earth andthe sun. What force is exerted on

a. the moon by the sun? b. the moon by Earth? c. Earth by the sun?

(See the table in the appendix for data on the sun,moon, and Earth.)

First be certain your graphing calculator is in

radian mode by pressing m ∂ ∂ e.

Execute “Chap7” on the p menu and press

e to begin the program. Enter the value for the

angular displacement (shown below) and press e.

The calculator will provide a graph of the angu-

lar speed versus the time interval. (If the graph is

not visible, press w and change the settings for

the graph window, then press g.)

Press ◊ and use the arrow keys to trace along

the curve. The x-value corresponds to the time

interval in seconds, and the y-value corresponds to

the angular speed in radians per second.

Determine the angular speed in the following

situations:

b. a bowl on a mixer stand that turns 2.0 rev in 3.0 s

c. the same bowl on a mixer stand that has

slowed down to 2.0 rev in 4.0 s

d. a bicycle wheel turning 2.5 rev in 0.75 s

e. the same bicycle wheel turning 2.5 rev in 0.35 s

f. The x- and y-axes are said to be asymptotic to

the curve of angular speed versus time interval.

What does this mean?

Press @ q to stop graphing. Press e to

input a new value or ı to end the program.

Graphing calculatorsRefer to Appendix B for instructions on download-

ing programs for your calculator. The program

“Chap7” allows you to analyze a graph of angular

speed versus time interval.

Angular speed, as you learned earlier in this

chapter, is described by the following equation:

wavg = ∆∆

qt

The program “Chap7” stored on your graphing cal-

culator makes use of the equation for angular speed.

Once the “Chap7” program is executed, your calcu-

lator will ask for the angular displacement in revo-

lutions. The graphing calculator will use the

following equation to create a graph of the angular

speed (Y1) versus the time interval (X). Note that

the relationships in this equation are similar to

those in the angular speed equation shown above.

Y1 = (2p q)/X

a. Why is there a factor of 2p in the equation

used by your graphing calculator?

Chapter 7272


52. An air puck of mass 0.025 kg is tied to a string andallowed to revolve in a circle of radius 1.0 m on africtionless horizontal surface. The other end of thestring passes through a hole in the center of the sur-face, and a mass of 1.0 kg is tied to it, as shown inFigure 7-19. The suspended mass remains in equi-librium while the puck revolves on the surface.

a. What is the magnitude of the force that main-tains circular motion acting on the puck?

b. What is the linear speed of the puck?

Figure 7-19

53. In a popular amusement-park ride, a cylinder ofradius 3.00 m is set in rotation at an angular speed of5.00 rad/s, as shown in Figure 7-20. The floor thendrops away, leaving the riders suspended against thewall in a vertical position. What minimum coefficientof friction between a rider’s clothing and the wall ofthe cylinder is needed to keep the rider from slipping?(Hint: Recall that Fs = msFn, where the normal forceis the force that maintains circular motion.)

Figure 7-20

Performance assessment1. Turn a bicycle upside down. Make two marks on

one spoke on the front wheel, one mark close to the

rim and another mark closer to the axle. Then spin

the front wheel. Which point seems to be moving

fastest? Have partners count the rotations of one

mark for 10 s or 20 s. Find the angular speed and

the linear speed of each point. Reassign the

observers to different points, and repeat the experi-

ment. Make graphs to analyze the relationship

between the linear and angular speeds.

2. When you ride a bicycle, the rotational motion you

create on the pedals is transmitted to the back wheel

through the primary sprocket wheel, the chain, and

the secondary sprocket wheel. Study the connection

between these components and measure how the

angular and linear speeds change from one part of

the bicycle to another. How does the velocity of the

back wheel compare with that of the pedals on a

bicycle? Demonstrate your findings in class.

Portfolio projects3. Research the historical development of the concept

of gravitational force. Find out how scientists’ ideas

about gravity have changed over time. Identify the

contributions of different scientists, such as Galileo,

Kepler, Newton, and Einstein. How did each scien-

tist’s work build on the work of earlier scientists?

Analyze, review, and critique the different scientific

explanations of gravity. Focus on each scientist’s

hypotheses and theories. What are their strengths?

What are their weaknesses? What do scientists think

about gravity now? Use scientific evidence and

other information to support your answers. Write a

report or prepare an oral presentation to share your

conclusions.

Alternative Assessment

Chapter 7274

CIRCULAR MOTIONIn this experiment, you will construct a device for measuring the tangential

speed of an object undergoing circular motion, and you will determine how

the force and the radius affect the tangential speed of the object.

PREPARATION

1. Read the entire lab, and plan what measurements you will take.

2. Prepare a data table in your lab notebook with five columns and fifteen

rows. In the first row, label the columns Trial, Hanging mass (kg), Mass of

stopper (kg), Total time (s), Radius (m). In the first column, label the sec-

ond through fifteenth rows 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, and 14.

PROCEDURE

Constant radius with varying force

3. Measure the mass of the rubber stopper, and record it in your data table.

Fasten one end of the nylon cord securely to the rubber stopper. Pass the

other end of the cord through the PVC tube and securely fasten a 100 g mass

to the other end, as shown in Figure 7-21. Leave approximately 0.75 m of

cord between the top of the tube and the rubber stopper. Attach a piece of

masking tape to the cord just below the bottom of the tube.

4. Make sure the area is clear of obstacles, and warn other students that you

are beginning your experiment. Support the 100 g mass with one hand and

hold the PVC tube in the other. Make the stopper at the end of the cord cir-

cle around the top of the tube by moving the tube in a circular motion.

5. Slowly release the 100 g mass, and adjust the speed of the stopper so that

the masking tape stays just below the bottom of the tube. Make several

practice runs before recording any data.

CHAPTER 7Laboratory Exercise

OBJECTIVES

•Examine the relationshipbetween the force thatmaintains circularmotion, the radius, andthe tangential speed of awhirling object.

MATERIALS LIST✔ 1.5 m nylon cord✔ 2-hole rubber stopper✔ masking tape✔ meterstick✔ PVC tube, about 15 cm long

and 1 cm in diameter✔ set of masses✔ stopwatch

SAFETY

• Tie back long hair, secure loose clothing, and remove loose jewelry toprevent their getting caught in moving or rotating parts.

• Wear eye protection and perform this experiment in a clear area.Swinging or dropped masses can cause serious injury.



6. When you can keep the velocity of the stopper and the position of the

masking tape relatively constant, measure the time required for 20 revo-

lutions of the stopper. Record the time interval in your data table in the

row labeled Trial 1. Repeat this trial and record the time interval in your

data table as Trial 2.

7. Place the apparatus on the lab table. Extend the cord so that it is taut and

the masking tape is in the same position it was in during the experiment.

Measure the length of the cord to the nearest millimeter from the center

of the top of the PVC tube to the center of the rubber stopper. Record this

distance in the data table as Radius for Trials 1 and 2.

8. Repeat the procedure using three different masses for Trials 3–8. Keep the

radius the same as in the first trial and use the same rubber stopper, but

increase the mass at the end of the cord each time. Do not exceed 500 g.

Attach all masses securely. Perform each trial two times, and record all

data in your data table.

Constant force with varying radius

9. For Trials 9–14, use the same stopper and the 100 g mass, and try three

different values for the radius in the range 0.50 m to 1.00 m. Make sure

that you have a clear area of at least 2.5 m in diameter to work in. Record

all data in your data table.

10. Clean up your work area. Put equipment away safely so that it is ready to

be used again.

ANALYSIS AND INTERPRETATION

Calculations and data analysis

1. Organizing data Calculate the weight of the hanging mass for each

trial. This weight is the force that maintains circular motion, Fc.

2. Organizing data For each trial, find the time necessary for one revo-

lution of the stopper by dividing the total time required for 20 revolutions

by 20.

3. Organizing data Find the tangential speed for each trial.

a. Use the equation vt = 2

∆pt

r.

b. Use the equation vt =�F

m�c r�, where r is the radius of revolution and

m is the mass of the stopper.

4. Graphing data Plot the following graphs:

a. Graph force versus tangential speed for Trials 1–8.

b. Graph tangential speed versus radius for Trials 9–14.

Figure 7-21Step 3: To attach masses, make aloop in the cord, place the massinside the loop, and secure the masswith masking tape.Step 4: You will need a clear arealarger than two times the radius.Step 5: Spin the stopper so thatthe cord makes a 90° angle with thePVC tube. Release the mass slowlywithout changing the speed.Step 7: Make sure the cord is held straight when you make yourmeasurements.


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