Cutnell_9th_problems_ch_1_thru_10_converted

(b) The next shortest time occurs when each blade moves through an angle of
, or
rad,
between successive flashes. This time is twice the value that we found in part a, or
.
REASONING AND SOLUTION
(a) If the propeller is to appear stationary, each blade must move through an angle of
or
between
flashes. The time required is
(b) The next shortest time occurs when each blade moves through an angle of
, or
rad, between
successive flashes. This time is twice the value that we found in part a, or
.

**18. Review Conceptual Example 2 before attempting to work this problem. The moon has a
diameter of and is a distance of

from the earth. The sun has a diameter of

and is

from

the earth.

(a) Determine (in radians) the angles subtended by the moon and the sun, as measured by a person
standing on the earth.

(b) Based on your answers to part (a), decide whether a total eclipse of the sun is really “total.” Give
your reasoning.

(c) Determine the ratio, expressed as a percentage, of the apparent circular area of the moon to the
apparent circular area of the sun.

**19. The drawing shows a golf ball passing through a windmill at a miniature golf course. The
windmill has 8

blades and rotates at an angular speed of

. The opening between successive blades is equal to the

width of a blade. A golf ball

has just reached the edge of one of the rotating

blades (see the drawing). Ignoring the thickness of the blades, find the minimum linear speed with
which the ball moves along the ground, such that the ball will not be hit by the next blade.

Answer:

Section 8.3

T he

Equations

of

Rotational

Kinematics



20. A figure skater is spinning with an angular velocity of
. She then comes to a stop over a brief period of
time. During this time, her angular displacement is
. Determine
(a) her average angular acceleration and
(b) the time during which she comes to rest.
21.
A gymnast is performing a floor routine. In a tumbling run she spins through the air, increasing her
angular velocity from 3.00 to
while rotating through one-half of a revolution. How much time does this maneuver
take?
Answer:
0.125 s
REASONING The angular displacement is given as
, while the initial angular velocity is given
as
and the final angular velocity as

. Since we seek the time , we can use Equation
8.6
from the equations of rotational kinematics to obtain it.
SOLUTION Solving Equation 8.6 for the time , we find that
22. The angular speed of the rotor in a centrifuge increases from 420 to
in a time of 5.00 s.
(a) Obtain the angle through which the rotor turns.
(b) What is the magnitude of the angular acceleration?
23. A wind turbine is initially spinning at a constant angular speed. As the wind's strength gradually
increases, the turbine experiences a constant angular acceleration of
. After making 2870 revolutions, its angular speed is
.
(a) What is the initial angular velocity of the turbine?
Answer:
(b) How much time elapses while the turbine is speeding up?
Answer:

140 s

24.
A car is traveling along a road, and its engine is turning over with an angular velocity of
. The
driver steps on the accelerator, and in a time of 10.0 s the angular velocity increases to
.
(a) What would have been the angular displacement of the engine if its angular velocity had remained
constant at the initial value of
during the entire 10.0-s interval?
(b) What would have been the angular displacement if the angular velocity had been equal to its final
value of during the entire 10.0-s interval?
(c) Determine the actual value of the angular displacement during the 10.0-s interval.
25.
The wheels of a bicycle have an angular velocity of
. Then, the brakes are applied. In coming to
rest, each wheel makes an angular displacement of
.
(a) How much time does it take for the bike to come to rest?
Answer:
10.0 s
REASONING
(a) The time for the wheels to come to a halt depends on the initial and final velocities, and , and the
angular displacement
(see Equation 8.6). Solving for the time yields
(b) The angular acceleration is defined as the change in the angular velocity,
, divided by the

time :
(8.4)
SOLUTION
(a) Since the wheel comes to a rest,
. Converting 15.92 revolutions to radians

, the time for the wheel to come to rest is
(b) The angular acceleration is
(b) What is the angular acceleration (in
) of each wheel?
Answer:
REASONING
(a) The time for the wheels to come to a halt depends on the initial and final velocities, and , and the
angular displacement
(see Equation 8.6). Solving for the time yields
(b) The angular acceleration is defined as the change in the angular velocity,
, divided by the
time :
(8.4)

SOLUTION
(a) Since the wheel comes to a rest,
. Converting 15.92 revolutions to radians
, the time for the wheel to come to rest is
(b) The angular acceleration is
REASONING
(a) The time for the wheels to come to a halt depends on the initial and final velocities, and , and the
angular displacement
(see Equation 8.6). Solving for the time yields
(b) The angular acceleration is defined as the change in the angular velocity,
, divided by the time :
(8.4)
SOLUTION
(a) Since the wheel comes to a rest,
. Converting 15.92 revolutions to radians
, the
time for the wheel to come to rest is
(b) The angular acceleration is

*26.
A dentist causes the bit of a high-speed drill to accelerate from an angular speed of
to an
angular speed of
. In the process, the bit turns through
. Assuming a constant

angular acceleration, how long would it take the bit to reach its maximum speed of

, starting from

rest?

*27.

A motorcyclist is traveling along a road and accelerates for 4.50 s to pass another cyclist. The angular
acceleration of each wheel is

, and, just after passing, the angular velocity of each wheel is

, where the plus signs indicate counterclockwise directions. What is the angular displacement of each
wheel during this time?

Answer:

REASONING The equations of kinematics for rotational motion cannot be used directly to find the
angular displacement, because the final angular velocity (not the initial angular velocity), the
acceleration, and the time are known. We will combine two of the equations, Equations 8.4 and 8.6 to
obtain an expression for the angular displacement that contains the three known variables.

SOLUTION The angular displacement of each wheel is equal to the average angular velocity
multiplied by the time (8.6)

The initial angular velocity

is not known, but it can be found in terms of the angular acceleration and time, which

are known. The angular acceleration is defined as (with

)

(8.4)

Substituting this expression for

into Equation 8.6 gives

*28.

A top is a toy that is made to spin on its pointed end by pulling on a string wrapped around the body
of the top.

The string has a length of 64 cm and is wound around the top at a spot where its radius is 2.0 cm. The
thickness of the string is negligible. The top is initially at rest. Someone pulls the free end of the
string, thereby unwinding it and giving the top an angular acceleration of

. What is the final angular velocity of the top when the string is
completely unwound?
*29.
The drive propeller of a ship starts from rest and accelerates at
for
. For the
next
the propeller rotates at a constant angular speed. Then it decelerates at
until it
slows (without reversing direction) to an angular speed of
. Find the total angular displacement of the
propeller.
Answer:
*30.
The drawing shows a graph of the angular velocity of a rotating wheel as a function of time.
Although not shown in the graph, the angular velocity continues to increase at the same rate until
. What is the angular
displacement of the wheel from 0 to 8.0 s?

Problem 30

*31.

At the local swimming hole, a favorite trick is to run horizontally off a cliff that is 8.3 m above the
water. One diver runs off the edge of the cliff, tucks into a “ball,” and rotates on the way down with an
average angular speed of

. Ignore air resistance and determine the number of revolutions she makes while on the way down.

Answer:

2.1 rev

*32.

A spinning wheel on a fireworks display is initially rotating in a counterclockwise direction. The
wheel has an angular acceleration of

. Because of this acceleration, the angular velocity of the wheel changes from

its initial value to a final value of
. While this change occurs, the angular displacement of the wheel is
zero. (Note the similarity to that of a ball being thrown vertically upward, coming to a momentary
halt, and then falling downward to its initial position.) Find the time required for the change in the
angular velocity to occur.
**33.
A child, hunting for his favorite wooden horse, is running on the ground around the edge of a
stationary merry-go-round. The angular speed of the child has a constant value of
. At the instant the child spots the
horse, one-quarter of a turn away, the merry-go-round begins to move (in the direction the child is
running) with a constant angular acceleration of
. What is the shortest time it takes for the child to catch up with the
horse?
Answer:
7.37 s
REASONING The angular displacement of the child when he catches the horse is, from Equation 8.2,
. In the same time, the angular displacement of the horse is, from Equation 8.7 with
,
. If the child is to catch the horse
.
SOLUTION Using the above conditions yields
or
The quadratic formula yields
and 42.6 s; therefore, the shortest time needed to catch the horse is
.
Section 8.4 Angular Variables and Tangential Variables
34.

A fan blade is rotating with a constant angular acceleration of

. At what point on the blade, as

measured from the axis of rotation, does the magnitude of the tangential acceleration equal that of the
acceleration due to gravity?

35.

Some bacteria are propelled by biological motors that spin hair-like flagella. A typical bacterial
motor turning at a constant angular velocity has a radius of

, and a tangential speed at the rim of

.

(a) What is the angular speed (the magnitude of the angular velocity) of this bacterial motor?
Answer:
(b) How long does it take the motor to make one revolution?
Answer:
36. An auto race takes place on a circular track. A car completes one lap in a time of 18.9 s, with an
average tangential speed of
. Find
(a) the average angular speed and
(b) the radius of the track.
37.
A string trimmer is a tool for cutting grass and weeds; it utilizes a length of nylon “string” that rotates
about an axis perpendicular to one end of the string. The string rotates at an angular speed of
, and its tip has a
tangential speed of
. What is the length of the rotating string?
Answer:
0.18 m
REASONING The angular speed and tangential speed
are related by Equation 8.9
, and this
equation can be used to determine the radius . However, we must remember that this relationship is
only valid if we use radian measure. Therefore, it will be necessary to convert the given angular
speed in rev/s into rad/s.
SOLUTION Solving Equation 8.9 for the radius gives
where we have used the fact that 1 rev corresponds to

rad to convert the given angular speed from rev/s into rad/s.

38.

In 9.5 s a fisherman winds 2.6 m of fishing line onto a reel whose radius is 3.0 cm (assumed to be
constant as an approximation). The line is being reeled in at a constant speed. Determine the angular
speed of the reel.

39.

The take-up reel of a cassette tape has an average radius of 1.4 cm. Find the length of tape (in meters)
that passes around the reel in 13 s when the reel rotates at an average angular speed of

.

Answer:

0.62 m

40. The earth has a radius of

and turns on its axis once every 23.9 h.

Problem 40

(a) What is the tangential speed (in m/s) of a person living in Ecuador, a country that

lies on the equator?

(b) At what latitude (i.e., the angle in the drawing) is the tangential speed one-third that of a person
living in Ecuador?

*41.

A baseball pitcher throws a baseball horizontally at a linear speed of

. Before

being caught, the baseball travels a horizontal distance of 16.5 m and rotates through an angle of 49.0
rad. The

baseball has a radius of 3.67 cm and is rotating about an axis as it travels, much like the earth does.
What is the tangential speed of a point on the “equator” of the baseball?

Answer:

REASONING The tangential speed

of a point on the “equator” of the baseball is given by Equation 8.9 as

, where is the radius of the baseball and is its angular speed. The radius is given in the statement of the
problem. The (constant) angular speed is related to that angle through which the ball rotates by
Equation 8.2 as

, where we have assumed for convenience that

when

. Thus, the tangential speed of the

ball is

The time that the ball is in the air is equal to the distance it travels divided by its linear speed ,

, so the

tangential speed can be written as

SOLUTION The tangential speed of a point on the equator of the baseball is

*42.

A person lowers a bucket into a well by turning the hand crank, as the drawing illustrates. The crank
handle moves with a constant tangential speed of

on its circular path. The rope holding the bucket unwinds

without slipping on the barrel of the crank. Find the linear speed with which the bucket moves down
the well.

Problem 42

*43. A thin rod

is oriented vertically, with its bottom end attached to the floor by means of a

frictionless hinge. The mass of the rod may be ignored, compared to the mass of an object fixed to
the top of the rod.

The rod, starting from rest, tips over and rotates downward.

(a) What is the angular speed of the rod just before it strikes the floor?

(Hint: Consider using the principle of conservation of mechanical energy.)

Answer:

(b) What is the magnitude of the angular acceleration of the rod just before it strikes the floor?

Answer:

**44. One type of slingshot can be made from a length of rope and a leather pocket for holding the
stone. The stone can be thrown by whirling it rapidly in a horizontal circle and releasing it at the right
moment. Such a slingshot is used to throw a stone from the edge of a cliff, the point of release being
20.0 m above the base of the cliff. The stone lands on the ground below the cliff at a point X. The
horizontal distance of point X from the base of the cliff (directly beneath the point of release) is thirty

times the radius of the circle on which the stone is whirled. Determine the angular speed of the stone
at the moment of release.
Section 8.5

Centripetal
Acceleration
and
Tangential

Acceleration

45.

A racing car travels with a constant tangential speed of

around a circular track of radius 625 m. Find

(a) the magnitude of the car's total acceleration and

Answer:

REASONING Since the car is traveling with a constant speed, its tangential acceleration must be zero.
The radial or centripetal acceleration of the car can be found from Equation 5.2. Since the tangential
acceleration is zero, the total acceleration of the car is equal to its radial acceleration.

SOLUTION

(a) Using Equation 5.2, we find that the car's radial acceleration, and therefore its total acceleration, is
(b) The direction of the car's total acceleration is the same as the direction of its radial acceleration.
That is, the direction is

.

(b) the direction of its total acceleration relative to the radial direction.

Answer:

radially inward toward the center of the track

REASONING Since the car is traveling with a constant speed, its tangential acceleration must be zero.
The radial or centripetal acceleration of the car can be found from Equation 5.2. Since the tangential
acceleration is zero, the total acceleration of the car is equal to its radial acceleration.

SOLUTION

(a) Using Equation 5.2, we find that the car's radial acceleration, and therefore its total acceleration, is
(b) The direction of the car's total acceleration is the same as the direction of its radial acceleration.
That is, the direction is

.

REASONING Since the car is traveling with a constant speed, its tangential acceleration must be zero.
The radial or centripetal acceleration of the car can be found from Equation 5.2. Since the tangential
acceleration is zero, the total acceleration of the car is equal to its radial acceleration.

SOLUTION

(a) Using Equation 5.2, we find that the car's radial acceleration, and therefore its total acceleration, is

(b) The direction of the car's total acceleration is the same as the direction of its radial acceleration.
That is, the direction is

.

46.

Two Formula One racing cars are negotiating a circular turn, and they have the same centripetal
acceleration.

However, the path of car A has a radius of 48 m, while that of car B is 36 m. Determine the ratio of the
angular speed of car A to the angular speed of car B.

47.

The earth orbits the sun once a year

in a nearly circular orbit of radius

. With

respect to the sun, determine

(a) the angular speed of the earth,

Answer:

REASONING
(a) According to Equation 8.2, the average angular speed is equal to the magnitude of the angular
displacement divided by the elapsed time. The magnitude of the angular displacement is one
revolution, or
. The elapsed time is one year, expressed in seconds.
(b) The tangential speed of the earth in its orbit is equal to the product of its orbital radius and its
orbital angular speed (Equation 8.9).
(c) Since the earth is moving on a nearly circular orbit, it has a centripetal acceleration that is directed
toward the center of the orbit. The magnitude
of the centripetal acceleration is given by Equation
8.11 as
.

SOLUTION
(a) The average angular speed is
(8.2)
(b) The tangential speed of the earth in its orbit is

(8.9)
(c) The centripetal acceleration of the earth due to its circular motion around the sun is (8.11)
.
(b) the tangential speed of the earth, and
Answer:
REASONING
(a) According to Equation 8.2, the average angular speed is equal to the magnitude of the angular
displacement divided by the elapsed time. The magnitude of the angular displacement is one
revolution, or
. The elapsed time is one year, expressed in seconds.
(b) The tangential speed of the earth in its orbit is equal to the product of its orbital radius and its
orbital angular speed (Equation 8.9).
(c) Since the earth is moving on a nearly circular orbit, it has a centripetal acceleration that is directed
toward the center of the orbit. The magnitude
of the centripetal acceleration is given by Equation
8.11 as
.
SOLUTION
(a) The average angular speed is
(8.2)
(b) The tangential speed of the earth in its orbit is
(8.9)
(c) The centripetal acceleration of the earth due to its circular motion around the sun is (8.11)
.
(c) the magnitude and direction of the earth's centripetal acceleration.
Answer:
, directed toward the center of the orbit

REASONING
(a) According to Equation 8.2, the average angular speed is equal to the magnitude of the angular

displacement divided by the elapsed time. The magnitude of the angular displacement is one

revolution, or

. The elapsed time is one year, expressed in seconds.

(b) The tangential speed of the earth in its orbit is equal to the product of its orbital radius and its
orbital angular speed (Equation 8.9).

(c) Since the earth is moving on a nearly circular orbit, it has a centripetal acceleration that is directed
toward the center of the orbit. The magnitude

of the centripetal acceleration is given by Equation

8.11 as

.

SOLUTION

(a) The average angular speed is

(8.2)

(b) The tangential speed of the earth in its orbit is

(8.9)

(c) The centripetal acceleration of the earth due to its circular motion around the sun is (8.11)

.

REASONING

(a) According to Equation 8.2, the average angular speed is equal to the magnitude of the angular
displacement divided by the elapsed time. The magnitude of the angular displacement is one
revolution, or

. The

elapsed time is one year, expressed in seconds.

(b) The tangential speed of the earth in its orbit is equal to the product of its orbital radius and its
orbital angular speed (Equation 8.9).

(c) Since the earth is moving on a nearly circular orbit, it has a centripetal acceleration that is directed
toward the center of the orbit. The magnitude

of the centripetal acceleration is given by Equation 8.11 as

.

SOLUTION

(a) The average angular speed is

(8.2)

(b) The tangential speed of the earth in its orbit is

(8.9)

(c) The centripetal acceleration of the earth due to its circular motion around the sun is (8.11)

.

48.

Review Multiple-Concept Example 7 in this chapter as an aid in solving this problem. In a fast-pitch
softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm
around so that the ball in her hand moves on a circle. In one instance, the radius of the circle is 0.670
m. At one point on this circle, the ball has an angular acceleration of

and an angular speed of

.

(a) Find the magnitude of the total acceleration (centripetal plus tangential) of the ball.

(b) Determine the angle of the total acceleration relative to the radial direction.

*49.

A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly
through one corner, as the drawing shows. The centripetal acceleration measured at corner A is n
times as great as that measured at corner B. What is the ratio

of the lengths of the sides of the rectangle when

?

Answer:
0.577

*50.
Multiple-Concept Example 7 explores the approach taken in problems such as this one. The blades of
a ceiling fan have a radius of 0.380 m and are rotating about a fixed axis with an angular velocity of
. When the
switch on the fan is turned to a higher speed, the blades acquire an angular acceleration of
. After
0.500 s has elapsed since the switch was reset, what is
(a) the total acceleration (in
) of a point on the tip of a blade and
(b) the angle between the total acceleration
and the centripetal acceleration
? (See Figure 8.12 b.)
*51.
The sun has a mass of
and is moving in a circular orbit about the center of our galaxy, the Milky
Way. The radius of the orbit is
, and the angular speed of the
sun is
.
(a) Determine the tangential speed of the sun.
Answer:
REASONING
(a) The tangential speed
of the sun as it orbits about the center of the Milky Way is related to the
orbital radius and angular speed by Equation 8.9,
. Before we use this relation, however,

we must first convert to meters from light-years.
(b) The centripetal force is the net force required to keep an object, such as the sun, moving on a
circular path. According to Newton's second law of motion, the magnitude
of the centripetal force is equal
to the product of the object's mass and the magnitude
of its centripetal acceleration (see Section
5.3):
. The magnitude of the centripetal acceleration is expressed by Equation 8.11 as
, where is the radius of the circular path and is the angular speed of the object.
SOLUTION
(a) The radius of the sun's orbit about the center of the Milky Way is
The tangential speed of the sun is
(8.9)
(b The magnitude of the centripetal force that acts on the sun is
)

(b) What is the magnitude of the net force that acts on the sun to keep it moving around the center of
the Milky Way?

Answer:

REASONING
(a) The tangential speed
of the sun as it orbits about the center of the Milky Way is related to the
orbital radius and angular speed by Equation 8.9,
. Before we use this relation, however,
we must first convert to meters from light-years.
(b) The centripetal force is the net force required to keep an object, such as the sun, moving on a
circular path. According to Newton's second law of motion, the magnitude
of the centripetal force is equal
to the product of the object's mass and the magnitude
of its centripetal acceleration (see Section
5.3):
. The magnitude of the centripetal acceleration is expressed by Equation 8.11 as
, where is the radius of the circular path and is the angular speed of the object.
SOLUTION
(a) The radius of the sun's orbit about the center of the Milky Way is
The tangential speed of the sun is
(8.9)
(b The magnitude of the centripetal force that acts on the sun is
)
REASONING
(a) The tangential speed
of the sun as it orbits about the center of the Milky Way is related to the orbital radius
and angular speed by Equation 8.9,
. Before we use this relation, however, we must first convert to
meters from light-years.

(b) The centripetal force is the net force required to keep an object, such as the sun, moving on a
circular path.
According to Newton's second law of motion, the magnitude
of the centripetal force is equal to the product
of the object's mass and the magnitude
of its centripetal acceleration (see Section 5.3):
. The
magnitude of the centripetal acceleration is expressed by Equation 8.11 as
, where is the radius of the
circular path and is the angular speed of the object.
SOLUTION
(a) The radius of the sun's orbit about the center of the Milky Way is
The tangential speed of the sun is

(8.9)

(b) The magnitude of the centripetal force that acts on the sun is

**52. An electric drill starts from rest and rotates with a constant angular acceleration. After the drill
has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the
drill is twice the magnitude of the tangential acceleration. What is the angle?

Section 8.6 Rolling Motion

Note: All problems in this section assume that there is no slipping of the surfaces in contact during the
rolling motion.

53.

A motorcycle accelerates uniformly from rest and reaches a linear speed of

in a time of 9.00 s. The

radius of each tire is 0.280 m. What is the magnitude of the angular acceleration of each tire?

Answer:

REASONING AND SOLUTION From Equation 2.4, the linear acceleration of the motorcycle is Since
the tire rolls without slipping, the linear acceleration equals the tangential acceleration of a point on
the outer edge of the tire:

. Solving Equation 8.13 for gives

54. An automobile tire has a radius of 0.330 m, and its center moves forward with a linear speed of

.
(a) Determine the angular speed of the wheel.
(b) Relative to the axle, what is the tangential speed of a point located 0.175 m from the axle?
55.
A car is traveling with a speed of
along a straight horizontal road. The wheels have a radius of 0.300
m. If the car speeds up with a linear acceleration of
for 8.00 s, find the angular displacement of each
wheel during this period.
Answer:

693 rad

56. Suppose you are riding a stationary exercise bicycle, and the electronic meter indicates that the
wheel is rotating at
. The wheel has a radius of 0.45 m. If you ride the bike for 35 min, how far would you have gone if
the bike could move?
*57.
Multiple-Concept Example 8 provides useful background for part b of this problem. A motorcycle,
which has an initial linear speed of
, decelerates to a speed of
in 5.0 s. Each wheel has a radius of 0.65 m and
is rotating in a counterclockwise (positive) direction. What are
(a) the constant angular acceleration (in
) and
Answer:
(b) the angular displacement (in rad) of each wheel?
Answer:
*58.
A dragster starts from rest and accelerates down a track. Each tire has a radius of 0.320 m and rolls
without slipping. At a distance of 384 m, the angular speed of the wheels is
. Determine
(a) the linear speed of the dragster and
(b) the magnitude of the angular acceleration of its wheels.
*59.
Over the course of a multi-stage 4520-km bicycle race, the front wheel of an athlete's bicycle makes

. How many revolutions would the wheel have made during the race if its radius had been 1.2

cm larger?

Answer:

*60.

A bicycle is rolling down a circular portion of a path; this portion of the path has a radius of 9.00 m.
As the drawing illustrates, the angular displacement of the bicycle is 0.960 rad. What is the angle (in
radians) through which each bicycle wheel

rotates?

Problem 60

*61.
The penny-farthing is a bicycle that was popular between 1870 and 1890. As the drawing shows, this
type of bicycle has a large front wheel and a small rear wheel. During a ride, the front wheel
makes 276
revolutions. How many revolutions does the rear wheel
make?
Problem 61
Answer:

974 rev

REASONING As a penny-farthing moves, both of its wheels roll without slipping. This means that the
axle for each wheel moves through a linear distance (the distance through which the bicycle moves)
that equals the circular arc length measured along the outer edge of the wheel. Since both axles move
through the same linear distance, the circular arc length measured along the outer edge of the large
front wheel must equal the circular arc length measured along the outer edge of the small rear wheel.
In each case the arc length is equal to the number of revolutions times the circumference

of the wheel

.

SOLUTION Since the circular arc length measured along the outer edge of the large front wheel must
equal the circular arc length measured along the outer edge of the small rear wheel, we have

Solving for

gives

*62.

A ball of radius 0.200 m rolls with a constant linear speed of

along a horizontal table. The ball rolls off

the edge and falls a vertical distance of 2.10 m before hitting the floor. What is the angular
displacement of the ball while the ball is in the air?

**63. The differential gear of a car axle allows the wheel on the left side of a car to rotate at a
different angular speed than the wheel on the right side. A car is driving at a constant speed around a
circular track on level ground, completing each lap in 19.5 s. The distance between the tires on the left
and right sides of the car is 1.60 m, and the radius of each wheel is 0.350 m. What is the difference
between the angular speeds of the wheels on the left and right sides of the car?

Answer:

Copyright © 2012 John Wiley & Sons, Inc. Al rights reserved.

Problems

Sect io n 9.1 T he Act io n o f Fo rces and To rques o n

Rigid Objects

1.

The wheel of a car has a radius of 0.350 m. The engine of the car applies a torque of

to this

wheel, which does not slip against the road surface. Since the wheel does not slip, the road must be
applying a force of static friction to the wheel that produces a countertorque. Moreover, the car has a
constant velocity, so this countertorque balances the applied torque. What is the magnitude of the
static frictional force?

Answer:

843 N

REASONING The drawing shows the wheel as it rolls to the right, so the torque applied by the engine
is assumed to be clockwise about the axis of rotation. The force of static friction that the ground
applies to the wheel is labeled as

. This force produces a counterclockwise torque about the axis of rotation, which is given by
Equation 9.1 as

, where is the lever arm. Using this relation we can find the magnitude

of the static frictional

force.

SOLUTION The countertorque is given as

, where

is the magnitude of the static frictional force and is

the lever arm. The lever arm is the distance between the line of action of the force and the axis of
rotation; in this case the lever arm is just the radius of the tire. Solving for

gives

2. The steering wheel of a car has a radius of 0.19 m, and the steering wheel of a truck has a radius of
0.25 m. The same force is applied in the same direction to each steering wheel. What is the ratio of the
torque produced by this force in the truck to the torque produced in the car?

3.

You are installing a new spark plug in your car, and the manual specifies that it be tightened to a
torque that has a magnitude of

. Using the data in the drawing, determine the magnitude F of the force that you must exert on the
wrench.

Problem 3
Answer:
REASONING According to Equation 9.1, we have
where is the magnitude of the applied force and is the lever arm. From the figure in the text, the lever
arm is given by
. Since both the magnitude of the torque and are known, Equation 9.1 can be solved for
.
SOLUTION Solving Equation 9.1 for , we have
4.
Two children hang by their hands from the same tree branch. The branch is straight, and grows out
from the tree trunk at an angle of
above the horizontal. One child, with a mass of 44.0 kg, is hanging 1.30 m along the

branch from the tree trunk. The other child, with a mass of 35.0 kg, is hanging 2.10 m from the tree
trunk. What is the magnitude of the net torque exerted on the branch by the children? Assume that the
axis is located where the branch joins the tree trunk and is perpendicular to the plane formed by the
branch and the trunk.

5.

The drawing shows a jet engine suspended beneath the wing of an airplane. The weight

of the engine is 10 200

N and acts as shown in the drawing. In flight the engine produces a thrust

of 62 300 N that is parallel to the ground.

The rotational axis in the drawing is perpen-dicular to the plane of the paper. With respect to this axis,
find the magnitude of the torque due to

Problem 5

(a) the weight and

Answer:

REASONING To calculate the torques, we need to determine the lever arms for

each of the forces. These lever arms are shown in the following drawings:

SOLUTION
(a Using Equation 9.1, we find that the magnitude of the torque due to the weight
) is
(b Using Equation 9.1, we find that the magnitude of the torque due to the thrust is
)
(b) the thrust.
Answer:
REASONING To calculate the torques, we need to determine the lever arms for each of the forces.
These lever arms are shown in the following drawings:
SOLUTION
(a) Using Equation 9.1, we find that the magnitude of the torque due to the weight
is
(b) Using Equation 9.1, we find that the magnitude of the torque due to the thrust is
REASONING To calculate the torques, we need to determine the lever arms for each of the forces.
These lever arms are shown in the following drawings:
SOLUTION
(a) Using Equation 9.1, we find that the magnitude of the torque due to the weight

is
(b) Using Equation 9.1, we find that the magnitude of the torque due to the thrust is

6. A square, 0.40 m on a side, is mounted so that it can rotate about an axis that passes through the
center of the square.

The axis is perpendicular to the plane of the square. A force of 15 N lies in this plane and is applied to
the square. What is the magnitude of the maximum torque that such a force could produce?

*7.

A pair of forces with equal magnitudes, opposite directions, and different lines of action is called a
“couple.”

When a couple acts on a rigid object, the couple produces a torque that does not depend on the
location of the axis.

The drawing shows a couple acting on a tire wrench, each force being perpendicular to the wrench.
Determine an expression for the torque produced by the couple when the axis is perpendicular to the
tire and passes through Express your answers in terms of the magnitude F of the force and the length
L of the wrench.

(a) point ,

Answer:

REASONING AND SOLUTION The torque produced by

each force of magnitude is given by Equation 9.1,

,

where is the lever arm and the torque is positive since each force

causes a counterclockwise rotation. In each case, the torque

produced by the couple is equal to the sum of the individual

torques produced by each member of the couple.

(a) When the axis passes through point , the torque due to the

force at is zero. The lever arm for the force at is .

Therefore, taking counterclockwise as the positive direction,

we have

(b) Each force produces a counterclockwise rotation. The

magnitude of each force is and each force has a lever arm

of

. Taking counterclockwise as the positive direction,
we have
(c) When the axis passes through point , the torque due to the
force at is zero. The lever arm for the force at is .
Therefore, taking counterclockwise as the positive direction,
we have

Note that the value of the torque produced by the couple is
the same in all three cases; in other words, when the couple
acts on the tire wrench, the couple produces a torque that

does not depend on the location of the axis.

(b) point , and

Answer:

REASONING AND SOLUTION The torque produced by each force of magnitude is given by
Equation 9.1,

, where is the lever arm and the torque is positive since each force causes a

counterclockwise rotation. In each case, the torque produced by the couple is equal to the sum of the
individual torques produced by each member of the couple.

(a) When the axis passes through point , the torque due to the force at is zero. The lever arm for the
force at is . Therefore, taking counterclockwise as the positive direction, we have

(b) Each force produces a counterclockwise rotation. The magnitude of each force is and each force
has a lever arm of

. Taking counterclockwise as the positive direction, we have

(c) When the axis passes through point , the torque due to the force at is zero. The lever arm for the
force at is . Therefore, taking counterclockwise as the positive direction, we have

Note that the value of the torque produced by the couple is the same in all three cases; in other words,
when the couple acts on the tire wrench, the couple produces a torque that does not depend on the
location of the axis.

(c) point .

Answer:

REASONING AND SOLUTION The torque produced by each force of magnitude is given by
Equation 9.1,

, where is the lever arm and the torque is positive since each force causes a

counterclockwise rotation. In each case, the torque produced by the couple is equal to the sum of the
individual torques produced by each member of the couple.

(a) When the axis passes through point , the torque due to the force at is zero. The lever arm for the
force at is . Therefore, taking counterclockwise as the positive direction, we have

(b) Each force produces a counterclockwise rotation. The magnitude of each force is and each force
has a lever arm of

. Taking counterclockwise as the positive direction, we have

(c) When the axis passes through point , the torque due to the force at is zero. The lever arm for the
force at is . Therefore, taking counterclockwise as the positive direction, we have

Note that the value of the torque produced by the couple is the same in all three cases; in other words,
when the couple acts on the tire wrench, the couple produces a torque that does not depend on the
location of the axis.

REASONING AND SOLUTION The torque produced by each force of magnitude is given by
Equation 9.1,

, where is the lever arm and the torque is positive since each force causes a counterclockwise rotation.
In each case, the torque produced by the couple is equal to the sum of the individual torques produced
by each member