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CHAPTER 7 -
~
W?,WfA,v'
~
Rotational
Motion and the
Law of Gravity'
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 bodles 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 circular path. This chapter will discuss theforce that maintains circular motion and
other rotational-motion
quantities .
• How conyou determine the riders' average
linear speed or acceleration during the ride?
• In what direction are the forces pushing or
pulling the riders?
CONCEPT REVIEW
Displacement
Velocity
(Section 2·1)
Acceleration
Force
Bot~tional
(Section 2·1)
(Section 2·2)
(Section 4·1)
Motion and the Law of Gravity
243
7-1
Measuring rotational motion
7-1 SECTION OBJECTIVES
ROTATIONAL QUANTITIES
• Relate radians to degrees.
Figure 7-1
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 wh~
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 analyze 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 Figure
7-1(a), and that a line is drawn from the
center of the wheel to the bulb. After a
time interval i1t, 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 (depicted with a red line in both diagrams)
moved through the angle () with
respect to the reference line. Likewise, the bulb moved a distance s,
measured along the circumference
of the circle; s is the arc length.
A light bulb on a rotating Ferris
wheel (a) begins at a point along a
reference line and (b) moves
through an arc length s, and therefore through the angle e.
Any point on a Ferris wheel that spins about
a fixed axis undergoes circular motion.
• Calculate angular displacement using the arc length
and the distance from the
axis of rotation.
• Calculate angular speed or
angular acceleration.
• Solve problems using the
kinematic equations for
rotational motion.
rotational motion
motion of a body that spins
about an axis
Reference
line
(a)
(b)
244
Chapter7
Figure 7-2
•
-
•
..
Angles can be measured inradians
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, 1~the angle () swept by r is equal to
1 rad. In general, any angle ()measured in radians is defined by the following:
radian
an angle whose arc length is
equal to its radius, which is
approximately equal to 57.3°
..
(}=!_
r
•
The radian is a pure number, with no dimensions. Because ()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 5 is equal to the
circumference of the circle, or 27rr. Substituting this value for 5
in the above equation gives the corresponding angle in radians.
..
5
27rr
r
r
(}=-=-=
2nrad
Thus, 360° equals zn rad, or one complete revolution. In
other words, one revolution corresponds to an angle of approximately 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 27r/360°. In this
way, the degrees cancel out and the measurement is left in radians. The conversion relationship can be simplified as follows:
n
(}(rad) =-(}(deg)
180°
.J
Radians and Arc Length
MATERIALS
v' drawing compass
V paper
V thin wire
v' wire cutters or scissors
y
Figure 7-3
Angular motion is measured in units
of radians. Because there are 211:
radians in a full circle, radians are
often expressed as a multiple of 11:.
Use the compass to draw a circle on a
sheet of paper-and mark the center point
use to go all the way around the circle?
of the circle. Measure the distance from
the center point to the outside of the circle. This is the radius of the circle. Using
each end of one of the wires. Note that
the angle between these two lines equals
1 rad. How many of these angles are there
Draw lines from the center of the circle to
the wire cutters, cut several pieces of wire
in this circle? Draw a larger circle using
equal to the length of this radius. Bend the
pieces of wire, and lay them along the circle you drew with your compass. Approxi-
your compass. HoY!'many pieces of wire
(cut to the length of the radius) do you
use to go all the way around this circle?
mately how many pieces of wire do you
Rotational Motion and the Law of Gravity
245
Angular displacement describes how much an object has rotated
angular displacement
Just as an angle in radians is the ratio of the arc length to the radius, the angu-
the angle through which a point,
line, or body is rotated in a specified direction and about a specified axis
lar displacement traveled by the bulb on the Ferris wheel is the change in the
arc length, Lls,divided by the distance of the bulb from the axis of rotation.
This relationship
is depicted in Figure 7-4.
ANGULAR DISPLACEMENT
angular, displacement.Iin
radians)
changein
arc length
distance from axis
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 counterFigure 7-4
clockwise and negative when it rotates clockwise. In other words, Ll() is posi-
A light bulb on a rotating Ferris
wheel rotates through an angular
displacement of /,;(J = (J2 - (J,.
tive when the object rotates counterclockwise
and negative when the object
rotates clockwise.
SAMPLE PROBLEM 7A
Angular displacement
PROBLEM
While riding on a carousel that is rotating clockwise, a child travels
through an arc length of 11.5 m. If the child's angular displacement is
165°, what is the radius of the carousel?
SOLUTION
Given:
Unknown:
Ll()=-165°
Lls=-11.5 m
r;;?
First, convert the angular displacement
to radians using the relationship
on
page 245.
"
L18(rad) =-- -Ll()(deg)
1800
"
=-(-165°)
180°
L1()(rad) = -2.88 rad
Use the angular displacement
equation on this page. Rearrange to solve for r.
L1s
L18=-
r
L1s
r=-=----
-u.s m
L1() -2.88 rad
r= 3.99 m
246
CALCULATOR SOLUTION
Many calculators have a key labeled
DEG~ that converts from degrees to
radians.
Chapter 7
L-_------__L-C'I--- _"
PRACTICE 7A
Angular displacement
1.
A girl sitting on a merry-go-round
moves counterclockwise
arc length of, 2.50 m. If the girl's angular displacement
through an
-,
4
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 tt rad, which corresponds
to an arc length of
1.2
m. What is the wheel's radius?
3.
A car'o~ a Ferris wheel has an angular displacement
n
of'4 rad, which zor-
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: ""
118
115
? rad
a.
r
+0.25m
h. +0.75 tad
c. ? degrees
d.
-4.2m
0
+2.6m
+135
J
, !
Angular speed describes rate of rotation
Linear speed describes the distance traveled in a specified interval of time.
if
Angular speed is similarly defined. The average angular speed,
,
Greek letter omega), of a rotating rigid object is the ratio of the angular displace-
OJavg (OJ
is the
ment, 118, to the time interval, M, that the object takes to undergo that displace-
angular speed
the rate at which a body rotates
about an axis, usually expressed
in radians per second
ment. Angular speed describes how quickly the rotation occurs.
ANGULAR SPEED
118
OJavg=-
~;,
average.angurar
spee
d
M
= angular..
displacement
time mterval
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 = 2n rad.
Rotational Motion and the Law of Gravity
247
SAMPLE PROBLEM 7B
Angular speed
PROBLEM
A child at an ice cream parlor spins on a stool. The child turns counterclockwise with an average angular speed of 4.0 rad/s, In what time interval will the child's feet have an angular displacement of S.On:radt
SOLUTION
Given:
118 = 8.0n rad
Unknown:
M= ?
(1)avg= 4.0 rad/s
- Use the angular speed equation from page 247. Rearrange to solve for I1t.
118
(1)avg= I1t
I1t=-
118
(1)avg
8.0n rad
4.0 rad/s
2.0ns
I M=6.3s I
(
PRACTICE 7B
Angular speed
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
--::_Jong does the
4. _Eiil
fly take to move through 2.3 rad?
in the unknown quantities in the following table:
{j)avg
a.
118
+2.3rad
h. +0.75 revls
c.
?
d. +2n:radls
248
Chapter7
M
10>0s
0.050 s
-1.2 turns
+1.5nrad
1.2 s
Figure 7-5
An accelerating bicycle wheel
rotates with (a) an angular speed
(1)/ at time t/ and (b) an angular
speed (1)2 at time t2·
Angular acceleration
occurs when angular speed changes
Figure 7-5 shows a bicycle turned
upside down so that a repairperson
work on the rear wheel. The bicycle pedals are turned
wheel has angular speed
(1)1'
can
ts the
so that at time
as shown in Figure 7-5(a), and at a later time,
it has angular speed (1)2> 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
/
t2,
has the
units radians per second per second (rad/s'').
angular acceilration
the time rate of change of sngular speed, expressed in radians
per second per second
ANGULAR ACC~LERATION
SAMPLE PROBLEM
7C
Angular acceleration
PROBLEM
A car's tire rotates at an initial angular speed of 21.5 rad/s. The driver accelerates, and after 3.5 s the tire's angular speed is 28.0 rad/s. What is
the tire's average angular acceleration during the 3.5 s time interval?
-L
SOLUTION
21.5 rad/s
Given:
(1)1 =
Unknown:
aavg--7.
Use the angular acceleration
= 28.0
rad/s
!J.t = 3.5 s
equation on this page.
(1)2 -
aavg=
(1)2
At
Ll
(1)1
28.0 rad/s - 21.5 rad/s
3.:J S
r-
6.5 rad/s
3.5 s
Rotational Motion and the Law of Gravity
249
PRACTICE 7C
Angular acceleration
1. A figure skater begins spinning counterclockwise at an angular speed of
4.0:n:rad/s. During a 3.0 s interval, she slowly pulls her arms inward and
finally spins at 8.0:n:rad/s.What is her average.angular acceleration during this time interval?
..
2. What angular acce1eration is necessary to increase the angular speed of a
fan blade from 8.§_ra~s to 15.4 rad/s in 5.2 s?
3. .Eill in the unknown quantities in the following table:
a.
b. +0.75 rad/s2
c.
+ 121.5 rad/s
7.0 s
0.050 S
1.28
All points on a rotating rigid object have the same angular
acceleration 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
Table 7·1
Angular substitutes
for linear quantities
Linear
Angular
x
e
v
OJ
a
a
250
Chapter7
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 following defining equation for average angular speed with the defining equation for average linear speed:
The equations are similar, with ()replacing x and (J) 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.
Use kinematic
equations
for const~t
angular acceleration
In light of the similarities between variables in linear motion and those in rotational motion, it should be no surprise that the kinematic equations of rotational 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 acceleration, 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 kinem,atic equations
Linear motion with constant
acceleration
Rotational motion with constant angular acceleration
(l)f = au
+ aAt
Vf= Vj+ aAt
AO = d/At
+ 2!txCLit)2
1
.
(1)/ = (I)? + 2a(LlO)
AO
Ax = VjLlt
+ ~a(At)2
~1£!!_.q~
\
= ~«(I)i + (l)f)At
PlWSIC!
+ 2a(Ax)
v/ = v?
Llx = ~(Vi
~o .. •
+ Vf1M
Module
"Angular Kinematics"
provides an interactive lesson
with guided problem-solving
practice to teach you about different kinds of angular motion,
including the types described
here.
Note the correlation between the rotational equations involving the angular
variables (I), and a and the equations of linear motion involving x, u, and a.
The quantity OJ in these equations represents the instantaneous angular
speed of the rotating object rather than the average angular speed.
e,
J
B
SAMPLE PROBLEM 7D
Angular kinematics
'I
PROBLEM
The wheel on an upside-down bicycle moves through 11.0 rad in 2.0 s, What
is the wheel's angular acceleration if its initial angular speed is 2.0 rad/s!
SOLUTION
Given:
Unknown:
•
I
= 11.0 rad
a=?
M= 2.0 s
Lle
oi, = 2.00
rad/s
Use the second angular kinematic equation from Table 7-2 to solve for
1
Lle = OJjM+ 2:a(M)
a
= 2(Lle
a.
2
- OJjM)j(M)2
a = 2[11.0 rad - (2.00 rad/s)(2.0 s)]/(2.0
5)2
a = 3.5 rad/s2
Rotational Motion and the Law of Gravity
251
Use kinematic equations for constant angular acceleration
In light of the similarities between variables in linear motion and those in rotational motion, it should be no surprise that the kinematic equations of rotational 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 acceleration, are summarized in Table 7-2. Note that the following rotational motion
equations apply only for objects rotating about a fixed axis.
Rotational and linear kinematic equations '
Table 7·2
Rotational motion with constant angular acceleration
Linear motion with constant
acceleration
OJf= OJi+aM
Vf = Vj + a!1t
I
!18;:::OJjM + 'ia(M)
OJ/ =
2
m? + 2a(Ae)
M) = ~(OJi
+ ~a(M)2
!1x = ViM
\
p"'r~!
~.
+ 2a(L\x)
v/ = v?
I
+ OJf)M
~'(6~
!1x = "2(Vi
+ vf)M
ute 8
gular Kinematics"
provides an interactive lesson
with guided problem-solving
practice to teach you about different kinds of angular motion,
including the types described
here.
Note the correlation between the rotational equations involving the angular
variables (), (1), and aand the equations of linear motion involving x, v, and a.
The quantity (1) in these equations represents the instantaneous angular
speed of the rotating object rather than the average angular speed.
SAMPLE PROBLEM 70
Angular kinematics
PROBLEM
The wheel on an upside-down bicycle moves through 11.0 rad in 2.0 s, What
is the wheel's angular acceleration if its initial angular speed is 2.0 rad/s?
SOLUTION
Given:
!1()= 11.0 rad
Unknown:
a=?
M= 2.0 s
OJi = 2.00 rad/s
Use the second angular kinematic equation from Table 7-2 to solve for a.
•
I ( M) 2
!1();:::(1)j!1t+ 'ia
a
= 2(!1()-
(1)jM)/(M)2
a = 2[11.0 rad - (2.00 rad/s)(2.0 s)]/(2.0 s)2
a = 3.5 rad/s2
Rotational Motion and the Law of Gravity
251
-
/,f
PRACTICE 70
Angular kinematics
1.
2.
1.
Convert the following angles in degrees to radians:
a.25°
h. 35°
c. 128°
d. 270°
2. A mosquito lands on a phonograph
record 5.0 em from the record's cen-
ter. If the record turns clockwise so that the mosquito travels along an arc
length of 5.0 em, 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
amusement-park
Find the angular acceleration
of a spinning
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
spinning amusement-park
What is the instantaneous
2
stant angular acceleration of 0.20 rad/s
252
Chapter 7
angular speed of a
ride that accelerates from 0.50 rad/s at a confor 1.0 s?
7-2
Tangential and centripetal
acceleration
RELATIONSHIPS BETWEEN ANGULAR
AND LINEAR QUANTITIES
7-2 SECTION OBJECTIVES .
As described at the beginning of Section 7-1, the motion of a point on a rotating 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 linear acceleration of a point on the object.
Imagine a golfer swinging a golf club. The most effective method for hitting 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 acceleration, 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.
•
Find the tangential speed of a
point on a rigid rotating
object using the angular
speed and the radius.
•
Solve problems involving tangential acceleration.
•
Solve problems involving
centripetal acceleration.
Objects in circular motion have a tangential speed
1
1
I
I
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 circle 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 represented in Figure 7-6.
Note that the speed of the horse at point A is represented by a longer 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.
tangential speed
the instantaneous linear speed of
an object directed along the tengent to the object'S circular path
Figure 7·6
Horses on a carousel move at the
same angular speed but different
tangential speeds.
Rotational Motion and the Law of Gravity
253
IlJ internet connect
How can you find the tangential speed? Again consider the rotating carousel.
If the carousel rotates through
an angle !::..(), a horse rotates through
length !::..s in the interval !::..t. The angular displacement
an arc
of the horse is given by
the equation for angular displacement.
TOPIC: Rotational motion
GO TO: www.scilinks.org
sciLiNKS COOE: HF2071
!::..()= !::..s
r
To find the tangential
speed of the horse, divide both sides of the equation
by the time the horse takes to travel th7 distance Ss.
!::..() l!::..s
-=-M r St
From Section 7-1, you know that the left side of the equation equals
(.t)avg.
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.
TANGENTIAL SPEED
tangential
speed
= distance
Note that to is the instantaneous
from axis x angular speed
angular speed, rather than the average angu-
lar speed, because the time interval is so short. This equation is valid only when
(.t)
is 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.
SAMPLE PROBLEM 7E
Tangential speed
PROBLEM
The radius of a CD in a computer is 0.0600 m. H a microbe riding on the
disc's rim has a tangential speed of 1.88 m/s, what is the disc's angular speed?
SOLUTION
Given:
r= 0.0600 m
Unlrnown:
(.t)
Vt= 1.88
mls
=?
Use the tangential speed equation on this page to solve for angular speed.
Vt=roi
vt
1.88 mls
r
0.0600 m
{.t)=-=----
I
254
(.t)
= 31.3 radls
I
Chapter 7
~---------------------------------------------------------------
PRACTICE 7E
Tangential speed
1.
2.
I
-I
Tangential acceleration is tangent to the circular path
If a carousel speeds up, the horses on it experience an angular acceleration.
f
1
I
,
•
•
•
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 D..OJ in the interval M. At the end of this time, the speed of a point on the
object has changed by the amount D..Vt. Using the equation for tangential
velocity on page 254 gives the following:
=
!J..Vt
tangential acceleration
the instantaneous linear acceleration of an object directed along
the tengent.to the object's circular path
rD..OJ
!J..Vt
D..OJ
M
D..t
Dividing by Ar gives -=r-
If the time interval M is very small, then the lett 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 tangential acceleration of a point on a rotating object is given by the relationship
on the next page.
Rotational Motion and the Law of Gravity
255
TANGENTIAL
ACCELERATION
tangential acceleration
= distance from axis x angular acceleration
_l
Again, the angular acceleration
angular acceleration.
This equation
in this equation refers to the instantaneous
is valid only when the angular accelera-
tion is expressed in units of radians per second per second.
SAMPLE PROBLEM 7F
Tangential acceleration
PROBLEM
A spinning ride at a carnival has an angular acceleration of 0.50 radls2• How
far from the center is a rider who has a tangential acceleration of 3.3 m/s2?
SOLUTION
Given:
a = 0.50 rad/s2
Unknown:
r=?
Use the tangential acceleration equation on this page. Rearrange to solve for r.
at= ra
at
r=-=
a
r=6.6
3.3 m/s2
0.50 rad/s
2
I
m
PRACTICE 7F
Tangential acceleration
1. A dog on a merry-go-round
the merry-go-round's
undergoes a 1.5 m/s2linear
acceleration. If
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
If tangential acceleration
of the yo-yo at the
end of the string is 0.18 m/s2, how long is the string?
256
of 0.35 rad/i.
3. What is a tire's angular acceleration if the tangential acceleration at a
radius of 0.15 m.is 9.4 x 10-2 m/s2?
Chapter 7
CENTRIPETAL
ACCELERATION
Figure 7-7 shows a car moving in a circular path with a constant 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.
Vf-Vi
a=--
'r: ti
Note that acceleration
ity. Because
velocity
an acceleration
depends on a change in the velocis a vector,
there
are
two
ways
by a change in the magni-
can be produced:
tude of the velocity and by a change in the direction of the
velocity. For a car moving in a circular path with constant
is due to a change in direction. An
speed, the acceleration
I
acceleration
of this nature is called a centripetal
seeking) acceleration.
Its magnitude
(center-
is given by the follow-
Figure 7·7
Although the car moves at a constant speed of 30 km/h,
the car still has an acceleration because the direction of
the velocity changes.
\
ing equation:
centripetal
1
Consider Figure 7-8(a). An object is seen first at point A, with tangential
1
velocity
time,
_I
Vi
at time
tr Assume
ti'
that
and then at point B, with tangential velocity
Vi
and
Vf
differ in direction
Vf
acceleration
acceleration directed toward the
center of a circular path
at a later
only and their magnitudes
are the same.
(a)
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 Mis very small
•
(as M approaches
zero),
be approximately
perpendicular
Vf
will be almost parallel to
to them, pointing
Vi
and the vector
~V
will
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
relationship
speed is related to the angular speed through
the
vt= rto, the centripetal acceleration can be found using the angu-
u
lar speed as well.
u
CENTRIPETAL
(b)
ACCELERAl'lON
~V
= vf
-
Vi
= vr+
(- Vi)
Figure 7·8
a
(a) As the particle moves from A to
.
centripetal
centripetal
.
(tangential speed)2
acceleration = --::..._--~-distance from axis
acceleration = distance.from
axis x (angular speed)2
B,the direction of the particle's
velocity vector changes. (b) Vector
addition is used to determine the
direction of the change in velocity,
/:;v, which for short time intervals is
toward the center of the circle.
Rotational Motion and the Law of Gravity
257
SAMPLE PROBLEM 7G
Centripetal acceleration
PROBLEM
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 its tangential speed?
SOLUTION
Given:
r=48.2 m
Unknown:
vt=?
Use the first centripetal
solve for
ac= 8.05 m/s 2
acceleration
equation from page 257. Rearrange to
Vt'
2
v
ac=- t
r
Vt=;;;=
2
.)(8.05 m/s )(48.2
!v
t= 19.7 ro/s
m)
I
PRACTICE 7G
Centripetal acceleration
1. A girl sits on a tire that is attached to an overhanging tree lim:6 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
2
of 250 m/s . If the yo-yo's string is 0.50 rn
long, what is the yo-yo's tangential sReed?
3. A dog sits 1.5 m from the center of a merry-gotround: If the dog undergoes 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 ear's centripetal acceleration is 15.4
2
ill/S , what
is the distance
between the car and the center of the track?
5. A piece of clay sits 0.20
ill
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 magnitude of the centripetal
258
Chapter7
acceleration
of the piece of day on the wheel?
Tangential and centripetal accelerations are perpendicular
Centripetal
/
I
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
direction
component
of travel, and hence the direction
of acceleration because its
of its velocity, is continually
changing. If the car's speed is increasing or decreasing, the car also has a tangential component
1
of acceleration
of acceleration.
To summarize,
the tangential
is due to changing speed; the centripetal
component
component
of accel-
eration is due to changing direction.
Find the total acceleration using the Pythagorean theorem
When both components
acceleration
of acceleration
exist simultaneously,
the tangential
is tangent to the circular path and the centripetal
acceleration
points toward the center of the circular path. Because these components
acceleration
are perpendicular
to each other, the magnitude
of
of the total accel-
eration can be found using the Pythagorean theorem, as follows:
"'~
atotal = V
at + at
Figure 7-9
The direction of the total acceleration of a rotating object can be
found using the tangent function .
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 .
•
J
I
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
h. 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
has an angular acceleration of 0.15 rad/s2.
4. Physics in Action If a spinning amusement-park
ride that
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?
1
Rotational Motion and the Law of Gravity
1<~
_
259
7-3
I~
Causes of circular motion
.
7-3 SECTION OBJECTIVES
FORCE THAT MAINTAINS CIRCULAR MOTION
• Calculate the 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 continuously during the motion, the ball experiences a centripetal acceleration
directed toward the center of motion, as described in Section 7-2, with magnitude given by the following equation:
• Explain how the apparent
existence of an outward force
in circular motion can be
explained as inertia resisting
the force that maintains circular motion.
• Apply Newton's universal law
of gravitation to find the
gravitational force between
two masses.
,,~-
...
- _--------
.....
~,
,
~~Fc
"
"r
m
fIII";
v
----------
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.
...~
Figure 7-10
When a ball is whirled in a circle. a
force directed toward the center of
the ball's circular path acts on it.
The net force on an object directed toward the center of the object's circular path is the force that maintains the object's circular motion.
FORCE THAT MAINTAINS CIRCULAR MOTION
IEJ internetconnect
(tangential speed) '2
force that 1l1aintainscircular motion =mass x ---=:-_--'=---"-distance to axis
TOPIC: Circular motion
GO TO: www.scilinks.org
sciLlNKS CODE: HF2072
for~ethat mail~tains =. mass x distance to axis x (angular speed)2
CIrcularmotion
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 racetrack 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 provides the force necessary to keep the moon in its orbit.
260
Chapter 7
•
•
SAMPLE PROBLEM 7H
•
force that maintains circular motion
II
•
PROBLEM
A pilot is flying a small plane at 30.0 mls in a circular path with a radius of
100.0 m. If a force of 635 N is needed to maintain the pilot's circular
motion, what is the pilot's mass?
•1
•
SOLUTION
I
= 30.0
Given:
Vt
Unknown:
m=?
mls
Fe=635 N
r= 100.0 m
Use the equation for force from page 260. Rearrange to solve for m.
I-
Vt
•
•
•,
•
•
Fe=m-
2
r
r
m = Fel
= 635 N
Vt
100.0 m
(30.0 m/s)
2
I m=70.6kg I
I
PRACtiCE
7H
III
,
Force that maintains circular motion
•
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 ml 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 combinedmass
of
the bicycle and rider?
A dog sits 1.50 m from the center of a merry-go-round
speed of 1.20 rad/s. ITthe magnitude
with an angular
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?
Rotational Motion and the Law of Gravity
261
~I
,
,,
,,,
,I
A force directed toward the center is necessary for circular motion
,
....
/
\
~1''',
\
,,
I
,
/
(a)
1/
,
"'-1-'-_===-
'\
T
I
f
\
......... "-
I"
,""
\
I
....
/
....
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 vanishes, 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, consider 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 direction tangent to the path, then move in the parabolic path of a projectile.
".
(b)
Figure 7·11
A ball is whirled in a vertical circular path on the end of a string.
When the string breaks at the position shown in (a), the ball moves
vertically upward in free fall. (b)
When the string breaks at the top
of the bali's path. the ball moves
along a parabolic path.
----..ceplual
Challenge
1. Pizza
Pizza
makers
traditionally
form the crust by throwing
the dough up in the air and
spinning it. Why does this
make the pizza crust bigger?
2. Swings
The
amusement-park
pictured
ride
below spins riders
around on swings attached
by cables from above. What
causes the swings to move
away from the center of the
ride when
the center
DESCRIBING THE MOTION OF
A 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 explanation 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 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-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 Newton'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 passenger slides across the seat as the car turns underneath. Because of inertia,
the passenger continues to move in a straight-line path. 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. 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 passenger to travel along the circular path followed by the car.
NEWTON'S UNIVERSAL
LAW OF 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
that separates them. It exists
between two masses, regardless of the medium
not just between large masses like the sun, Earth, and mOOD but between any
two masses, regardless of size or composition.
room
have a mutual
attraction
because
gravitational force
the mutual force of attraction
between particles of matter
For instance, desks in a class-
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
objects'masses.
Gravitational
force acts such that objects
are always attracted
ine the illustration
to one another. Examof Earth and the mOOD
in Figure 7-12. Note that the gravitational
force between Earth and the moon is attractive, and recall that
Newton's
third
law
states that the force exerted on Earth by the
moon, FIDE' is equal in magnitude
the opposite
direction
Figure 7·12
to and in
The gravitational force between
Earth and the moon is attractive.
According to Newton's third law,
of the force exerted
on the moon by Earth, FEm•
FEm
_j
= FmE·
Gravitational force depends on the distance between two masses
If masses m1 and
mz are
separated by distance r, the magnitude
of the gravita-
tional force is given by the following equation:
NEWTON'S
UN1VERSAL
r:J internetconneet
LAW OF GRAVITATION
I
,-'
,
TOPIC: Law of gravitation
GO TO: www.scilinks.org
sciliNKS CODE: HF2073
•
•
•
G is a universal constant called the constant of universal gravitation; it can
be used to calculate gravitational
been determined
forces between any two particles and has
experimentally.
G = 6.673
N
0
x 10-11_
_
m2
-
2
kg
The universal law of 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.
Rotational Motion and the Law of Gravity
263
Gravitational force is localized to the center of a spherical mass
Did you know?
The gravitational
force exerted by a spherical mass on a particle outside the
Sir Isaac Newton knew from his first
sphere is the same as it would be if the entire mass of the sphere were concen-
law that a net force had to be acting
on the moon. Otherwise, the moon'
would move in a straight-line path
trated at its center. For example, the force on an object of mass m at Earth's
surface has the following magnitude:
rather than in an elliptical orbit. He
MEm
F =C
reasoned 1:hat this force arises as a
result of an attractive field force
betwe~n the moon and Earth and
that a force of the same origin causes
Ri
g
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.
an apple to fall from a tree to Earth.
MEm
mg=C--2
RE
By substituting
the actual values for the mass and radius of Earth, we can
find the value for gand compare it with the value of free-fall acceleration
throughout
used
this book.
Because m occurs on both sides of the equation above, these masses cancel.
g=
ME (
G-1
= 6.673 x 10-11
RE
2
N .m
kg
--2-
This value for g is approximately
)
24
5.98 X 10 kg
2
6
2 = 9.83 mls
(6.37 x 10 m)
equal to the value used throughout
this
book. The difference is due to rounding the values for Earth's mass and radius.
SAMPLE PROBLEM 7I
Gravitational force
PROBLEM
Find the distance between a 0.300 kg billiard ball and a 0.400 kg billiard ball
if the magnitude of the gravitational force is 8.92 x 10-11 N.
SOLUTION
Given:
ml
Unknown:
r
= 0.300kg
Fg= 8.92
X
10-11 N
=?
Use the equation for Newton's Universal Law of Gravitation.
N.m2
r2
=.E..m1m2=
Fg
6.673 x 10-11--2-
----.,..,1l-k.=...g-(0.300
8.92 x 10 N
= 8.97 X 10-2 m2
r
264
Chapter 7
= ";8.97 X 10-2 m2 = 3.00 x 10-1 m
kg)(OAOO kg)
I(_
I
PRACTICE 7I
!
U
<
II
J
i
iii
JlI
L
1
I
1. A roller coaster moves through a vertical loop at a constant speed and suspends its passengers upside down. In what direction is the force that causes
I
the coaster and its passengers to move in a circle? What provides this force?
r.
•
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
I.
a
nearby. What is the magnitude
4.
Physics in Action
of the gravitational
force between them?
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
III
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.
Rotational Motion and the Law of Gravity
265
CHAPTER 7
Summary
KEY IDEAS
KEY TERMS
angular acceleration
(p. 249)
angular displacement
(p.246)
angular speed (p. 247)
centripetal
(p.257)
acceleration
Section 7-1 Measuring rotational
• The average angular speed,
motion
wavg, of a rigid, rotating object is defined as
the ratio of the angular displacement,
• The average angular acceleration,
tl.() to the time interval, tl.t.
aavg' of a rigid, rotating
as the ratio of the change in angular speed,
Section 7-2 Tangential and centripetal
object is defmed
Sea, to the time interval, M.
acceleration
• A point on an object rotating about a fixed axis has a tangential speed
- gravitational
force (p. 263)
related to the object's angular speed. When the object's angular acceleration changes, the tangential acceleration of a point on the object changes.
radian (p.245)
rotational
--...tangential
(p.255)
tangential
motion (p.244)
acceleration
• 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
speed (p.253)
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
ly proportional
Diagram symbols
to the product of the particles' masses and inverse-
to the square of the distance between the particles.
Variable symbols
Rotational
motion
!1(j
Angle
marking
m
asc Ie-ngt~
5
~,
angular displacement
':.£
r..;:",.
radians
~.radians/second
angular speed
cr'
angular acceleration
rap./52
radians/second2
vt".
t~ngential speed
m/s
meters/second
ra.d/s
2
at
tangential acceleration
ill/5
meters/s~cond}
a(
"'centripetal acceleration'
mls4
meters/second2
Fe;~
,force that maintains
circular motion
N
newtons
G
/',
'.
N·
N.m2
const~nt of ~niv~rsal gravitation
kg2
';:l'
Chapter 7
'",
rad
(l)
gravitational force.
268
meters
».
~;:.,
..
, newtons,s
7.~~
newtons- meters2
kilograms2
"0/~
~.
..
CHAPTER 7
Review and Assess
RADIANS AND ANGULAR MOTION
EQUATIONS FOR ANGULAR MOTION
Review questions
Practice problems
1. How many degrees equal tt radians? How many
10. A potter's wheel moves from rest to an angular speed
revolutions equal n radians?
of 0.20 rev/s in 30.0 s. Assuming constant angular
acceleration, what is its angular acceleration in rad/s2?
(See Sample Problem 7D.)
e,
(1), and a in the kinematic equations for rotational motion listed in
2. What units must be used for
Table7-2?
n. A drill starts from rest. After 3.20 s of constant angular
acceleration,the drill turns at a rate of2628 rad/s.
3. Distinguish between linear speed and angular speed.
•
4. When a wheel rotates about a fixed axis, do all
points on the wheel have the same angular speed?
..
Practice problems
5. A car on a Ferris wheel has an angular displacement
of 0.34 rad. If the car moves through an arc length
of 12 m, what is the radius of the Ferris wheel?
(See Sample Problem 7A.)
... •
a. Find the drill's angular acceleration .
"
h. 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 in
1.2 s before reaching its final angular speed.Assuming that the angular acceleration of the wheel is constant, calculate the wheel's angular acceleration.
(See Sample Problem 7D.)
6. When a wheel is rotated through an angle of 35°, a
point on the circumference travels through an arc
length of 2.5 m. When the wheel is rotated through
angles of 35 rad and 35 rev, the same point travels
through arc lengths of 143 m and 9.0 x 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 to
move through 4.00 rad?
(See Sample Problem 7B.)
8. A phonograph record has an initial angular speed of
33 rev/min. The record slows to 11 rev/min in 2.0 s.
What is the record's average angular acceleration
during this time interval?
(See Sample Problem 7C.)
9. If a flywheel increases its average angular speed by
2.7 rad/s in 1.9 s, what is its angular acceleration?
(See Sample Problem 7C.)
TANGENTIAL AND CENTRIPETAL
ACCELERATION
Review questions
13. When a wheel rotates about a fixed axis, do all the
points on the wheel have the same tangential speed?
14. Correct the following statement: The racing car
rounds 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 perpendicular to the velocity.
16. An object moves in a circular path with constant
speed v.
a. Is the object's velocity constant? Explain.
b. Is its acceleration constant? Explain.
Rotational Motion and the Law of Gravity
269
Conceptual questions
17. Give an example of a situation in which an automobile driver can have a centripetal acceleration but no
tangential acceleration.
18. Can a car move around
, - a circular racetrack so that
the car has a tangential acceleration but no centripetal acceleration? •
19. The gas pedal and-the brakes of a car accelerate and
decelerate the car. Could a steering wheel perform
either of these two actions? Explain.
20. It has been suggested that rotating cylinders about
16 km long and 8 km in diameter should be placed
in space for future space colonies. The rotation
would simulate gravity for the inhabitants of these
colonies. Explain the concept behind this proposal.
26. A sock stuck
a centripetal
barrel has a
speed of the
(See Sample
to the side of a clothes-dryer barrel has
acceleration of 28 m/s2. If the dryer
radius of 27 cm, what is the tangential
sock?
Problem 7G.)
CAUSES OF CIRCULAR MOTION
Review questions
27. Imagine that you attach a heavy object to one end of a
spring and then, while holding the spring's other end,
whirl the spring and object in a horizontal circle. Does
the spring stretch? Why? Discuss your answer ill
terms of the force that maintains circular motion.
28. Why does the water remain in a pail that is whirled
in a vertical path, as shown in Figure 7-16?
Practice problems
21. A small pebble breaks loose from the treads of a tire
with a radius of 32 ern. If the pebble's tangential
speed is 49 mis, what is the tire's angular speed?
(See Sample Problem 7E.)
22. The Emerald Suite is a revolving restaurant at the
top of the Space Needle in Seattle, Washington. If a
customer sitting 12 m from the restaurant's center
has a tangential speed of 2.18 x 10-2 mis, what is
the angular speed of the restaurant?
(See Sample Problem 7E.)
23. A bicycle wheel has an angular acceleration of
1.5 rad/s2. If a point on its rim has a tangential acceleration of 48 cm/52, what is the radius of the wheel?
(See Sample Problem 7F.)
24. When the string is pulled in the correct direction on a
window shade, a lever is released and the shaft that the
shade is wound around spins. If the shaft's angular
acceleration is 3.8 rad/s2 and the shade accelerates
upward 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 a
circle at the end of a cord. If the keys have a centripetal acceleration of 145 m/s2 and the cord has a
length of 0.34 m, what is the tangential speed of the
keys?
(See Sample Problem 7G.)
270
Chapter7
Figure 7·16
29. Identify the influence
gravitational forces.
of mass and distance
30. Explain the difference between centripetal
tion and angular acceleration.
on
accelera-
31. Comment on the statement, "There is no gravity in
outer space."
Conceptual questions
32. Explain why Earth is not spherical in sbape and why
it bulges at the equator.
33. Because of Earth's rotation, you would weigh slightly
less 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 shuttle
are not actually in a zero-gravity environment. What
is the real reason astronauts seem weightless?
36. A girl at a state fair swings a ball in a vertical circle at
the end of a string. Is the force applied by the string
greater than the weight of the ball at the bottom of
the ball's path?
Practice problems
37. A roller-coaster
car speeds down a hill past point A
and then rolls up a hill past point B, as shown in
I
l
Figure 7-17.
a. The car has a speed of20.0 mls at point A. If the
track exerts a force on the car of 2.06 x 104 N at
this point, what is the mass of the car?
b. What is the maximum speed the car can have
at point B for the gravitational force to hold it
on the track? (See Sample Problem 7H.)
.. .. - ...
B
\
IS.0m
,
I
I
I
MIXED REVIEW PROBLEMS
41. Find the average angular speed of Earth about the
sun 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, starting from rest and reaching an angular speed of
lIn rad/s in 8.0 s. At this point, the lid is opened, and
a safety switch turns off the washer. The tub slows to
rest in 12.0 s. Through hQ'Y many revolutions does
the tub turn? Assume constant angular acceleration
while the machine is starting and stopping.
43. An airplane is flying in a horizontal circle at a speed
of 105 m/s. The 80.0 kg pilot does not want the centripetal acceleration to exceed 7.00 times free-fall
acceleration.
a. Find the minimum radius of the plane's path .
b. At this radius, what is the net force that maintains circular motion exerted on the pilot by.,
the seat belts, the friction against the seat, and
so forth?
\
A
Figure 7-17
38. Tarzan tries to cross a river by swinging from one
bank to the other on a vine that is 10.0 ill long. His
speed at the bottom of the swing, just as he clears
the surface of the river, is 8.0 m/s. Tarzan does not
know that the vine has a breaking strength of 1.0 x
103 N. What is the largest mass Tarzan can have and
make it safely across the river?
(See Sample Problem 7H.)
39. The gravitational force of attraction between two
students sitting at their desks in physics class is
3.20 x 10-8 N. If one student has a mass of 50.0 kg
and the other has a mass of 60.0 kg, how far apart
are the students sitting?
(See Sample Problem 71.)
40. If the gravitational force between the electron (9.11 X
10-31 kg) and the proton (1.67 x 10-27 kg) in a
hydrogen atom is 1.0 x 10-47 N, how far apart are the
two particles?
(See Sample Problem 71.)
44. A car traveling at 30.0 mls undergoes a constant
negative acceleration of magnitude 2.00 m/s2 when
the brakes are applied. How many revolutions does
each tire make before the car comes to a stop,
assuming that the car does not skid and that the
tires have radii of 0.300 m?
45. A coin with a diameter of 2.40 em is dropped onto a
horizontal surface. The coin starts out with an initial
angular speed of 18.0 rad/s and rolls in a straight line
without slipping. If the rotation slows with an angular acceleration of magnitude 1.90 rad/s'', how far
does the coin roll before coming to rest?
46. A mass attached to a 50.0 cm string starts from rest
and is rotated in a circular path exactly 40 times in
1.00 min before reaching a final angular speed. What
is the angular speed of the mass after 1.00 min?
47. A 13500 N car traveling at 50.0 km/h rounds
curve of radius 2.00 X 102 m. Find the following:
a
a. the centripetal acceleration of the car
b. the force that maintains centripetal acceleration
c. the minimum
coefficient of static friction
between the tires and the road that will allow
the car to round the curve safely
Rotational
Motion and the Law of Gravity
271
48. A 2.00 X 103 kg car rounds a circular turn of radius
50. Find the centripetal accelerations of the following:
20.0 m. If the road is flat and the coefficient of static
friction between the tires and the road is 0.70, how
fast can the car go without skidding?
a. a point on the equator of Earth
b. a point at the North Pole of Earth
(See the table in the appendix for data on Earth.)
49. During a solar eclipse, the moon, Earth, and sun lie
51. A copper block rests 30.0 em from the center of a
on the same line, with the moon between Earth and
the SUll. What force is exerted on
a. the moon by the sun?
b. the moon by Earth?
c. Earth by the sun?
steel turntable. The coefficient of static friction
between the block and the surface is 0.53. The
turntable starts from rest and rotates with a constant
angular acceleration of 0.50 rad/s'. After what time.
interval will the block start to slip on the turntable?
(Hint: The normal force in this case equals the weight
of the block.)
(See the table in the appendix for data on the sun,
moon, and Earth.)
ing
Techno.
First be certain your graphing calculator is in
radian mode by pressing I
llIllIll l.
Execute"Chap7" on the I
I menu and press
I I to begin the program. Enter the value for the
angular displacement (shown below) and press I
I.
The calculator will provide a graph of the angular speed versus the time interval. (If the graph is
not visible, press I
I and change the settings for
the graph window, then press IGRAPH!.)
Press I
1 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 B 8 to stop graphing. Press I
I to
input a new value or I
I to end the program.
MODE
Graphing calculators
EmER
PRGM
Refer to Appendix B for instructions on downloading programs for your calculator. The program
"Chap 7" 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:
ENTER
ENTER
WINDOW
TRACE
!J.()
(l)avg=-
!J.t
The program "Chap7" stored on your graphing calculator makes use of the equation for angular speed.
Once the "Chap7" program is executed, your calculator will ask for the angular displacement in revolutions. 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.
a. Why is there a factor of 27rin the equation
used by your graphing calculator?
272
Chapter 7
EmER
CLEAR
52. An air puck of mass 0.025 kg is tied to a string and
allowed to revolve in a circle of radius 1.0 m on a
frictionless horizontal surface. The other end of the
string passes through a hole in the center of the surface, and a mass of 1.0 kg is tied to it, as shown in
Figure 7-19. The suspended mass remains in equilibrium while the puck revolves on the surface.
a. What is the magnitude of the force that maintains circular motion acting on the puck?
h. What is the linear speed of the puck?
53. In a popular amusement-park ride, a cylinder of
radius 3.00 m is set in rotation at an angular speed of
5.00 rad/s, as shown in Figure 7-20. The floor then
drops away,leaving the riders suspended against the
wall in a vertical position. What minimum coefficient
of friction bern\-eena rider'~ clothing and the wall of
the cylinder is needed to keep the rider from slipping?
(Hint: Recall that P, = flsFn' where the normal force
is the force that maintains circular motion.)
.,
Figure 7·20
Figure 7·19
Performance assessment
..
...
I
..
1. 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 experiment. 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 projects
3. 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 scientist'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.
Rotational Motion and the Law of Gravity
273