Ch 05 ISM 8e

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CHAPTER 5DYNAMICS OF UNIFORMCIRCULAR MOTION

ANSWERS TO FOCUS ON CONCEPTS QUESTIONS____________________________________________________________________________________________

1. (c) The velocity of car A has a constant magnitude (speed) and direction. Since its velocity

is constant, car A does not have an acceleration. The velocity of car B is continually

changing direction during the turn. Therefore, even though car B has a constant speed, it has

an acceleration (known as a centripetal acceleration).

2. (d) The centripetal (or center-seeking) acceleration of the car is perpendicular to its

velocity and points toward the center of the circle that the road follows.

3. (b) The magnitude of the centripetal acceleration is equal to v2/r, where v is the speed of the

object and ris the radius of the circular path. Since the radius of the track is smaller at Acompared to B, the centripetal acceleration of the car at A has a greater magnitude.

4. (a) The magnitude ac of the centripetal acceleration is given by ac = v2/r.

5. (d) The acceleration (known as the centripetal acceleration) and the net force (known as the

centripetal force) have the same direction and point toward the center of the circular path.

6. (a) According to the discussion in Example 7 in Section 5.3, the maximum speed that the

cylinder can have is given by max sv gr , where s is the coefficient of static friction, gis the acceleration due to gravity, and ris the radius of the path.

7. (d) The radius of path 1 is twice that of path 2. The tension in the cord is the centripetalforce. Since the centripetal force is inversely proportional to the radius rof the path, T

1must

be one-half ofT2.

8. (a) The centripetal force is given byFc

= mv2/r. The centripetal forces for particles 1, 2 and

3 are, respectively, 4m0v

0

2/r

0, 3m

0v

0

2/r

0, and 2m

0v

0

2/r

0.

9. (d) The centripetal force is directed along the radius and toward the center of the circularpath. The componentFN

sin of the normal force is directed along the radius and points

toward the center of the path.

10. (a) The magnitude of the centripetal force is given byFc

= mv2/r. The two cars have the

same speed v and the radius rof the turn is the same. The cars also have the same mass m,

even though they have different weights due to the different accelerations due to gravity.

Therefore, the centripetal accelerations are the same.

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244 DYNAMICS OF UNIFORM CIRCULAR MOTION

11. (e) The centripetal force acting on a satellite is provided by the gravitational force. The

magnitude of the gravitational force is inversely proportional to the radius squared (1/r2), so

if the radius is doubled, the gravitational force is one fourth as great; 1/22

= 1/4.

12. The orbital speed is 31.02 10 m/s.v

13. (b) The magnitude of the centripetal force acting on the astronaut is equal to her apparent

weight. The centripetal force is given by Equation 5.3 asFc

= mv2/r, which depends on the

square (v2) of the astronauts speed and inversely (1/r) on the radius of the ring. According

to Equation 5.1, r= vT/(2), the radius is directly proportional to the speed. Thus, thecentripetal force is directly proportional to the speed v of the astronaut. As the astronaut

walks from the inner ring to the outer ring, her speed doubles and so does her apparent

weight.

14. (d) The skier at A is speeding up, so the direction of the acceleration, and hence the net

force, must be parallel to the skiers velocity. At B the skier is momentarily traveling at aconstant speed on a circular path of radius r. The direction of the net force, called thecentripetal force, must be toward the center of the path. At C the skier is in free-fall, so the

net force, which is the gravitational force, is straight downward.

15. (b) According to Newtons second law, the net force,N

F mg , must equal the mass m

times the centripetal acceleration v2/r.

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Chapter 5 Problems 245

CHAPTER 5DYNAMICS OF UNIFORM

CIRCULAR MOTION

PROBLEMS

1. SSM REASONING The speed of the plane is given by Equation 5.1: v r T 2 / , where

Tis the period or the time required for the plane to complete one revolution.

SOLUTION Solving Equation 5.1 forTwe have

Tr

v

2 2850 m 2 (

110 m / s160 s

)

2. REASONING According to ac = v2/r (Equation 5.2), the magnitude ac of the centripetal

acceleration depends on the speed v of the object and the radius r of its circular path. In

Example 2 the object is moving on a path whose radius is infinitely large; in other words,

the object is moving along a straight line.

SOLUTION Using Equation 5.2, we find the following values for the magnitude of the

centripetal acceleration:

222

c

222

c

222

c

12 m/s290 m/s

0.50 m

35 m/s0 m/s

2.3 m/s2.9 m/s

1.8 m

va

r

va

r

va

r

Example 1

Example 2

Example 3

3. REASONING AND SOLUTIONSince the speed of the object on and off the circle

remains constant at the same value, the object always travels the same distance in equal timeintervals, both on and off the circle. Furthermore since the object travels the distance OA inthe same time it would have moved from O toPon the circle, we know that the distance OA

is equal to the distance along the arc of the circle from O toP.

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The circumference of the circle is 2 2 r (3.6 m) = 22.6 m . The arc OP subtends anangle of 25 ; therefore, since any circle contains 360, the arc OPis 25/360 or 6.9 percent of the circumference of the circle. Thus,

OP= 22.6 m 0.069 = 1.6 mb gb g

and, from the argument given above, we conclude that the distance OA is 1.6 m .

4. REASONING AND SOLUTIONLet srepresent the length of the path of the pebble

after it is released. From Conceptual

Example 2, we know that the pebble will flyoff tangentially. Therefore, the path s is per-

perpendicular to the radius r of the circle.

Thus, the distances r, s, and d form a right

triangle with hypotenuse d as shown in thefigure at the right. From the figure we see that

cos r

d

r

r10

1

10

or

= cos1

10

1 FHG

IKJ

84

Furthermore, from the figure, we see that 35 180 . Therefore,

145 145 84 61 _____________________________________________________________________________________________

5. SSM REASONING The magnitude ac of the cars centripetal acceleration is given by

Equation 5.2 as 2c

/a v r , where v is the speed of the car and r is the radius of the track.

The radius is r = 2.6 103 m. The speed can be obtained from Equation 5.1 as thecircumference (2r) of the track divided by the period Tof the motion. The period is thetime for the car to go once around the track (T= 360 s).

SOLUTION Since2

c /a v r and 2 /v r T , the magnitude of the cars centripetalacceleration is

2

2 32 22

2c 2

24 2.6 10 m4

0.79 m/s360 s

r

v rTa

r r T

______________________________________________________________________________

35

r

s

Pebble

Target

d

C

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6. REASONING Blood traveling through the aortic arch follows a circular path with a

diameter of 5.0 cm and, therefore, a radius ofr= 2.5 cm = 0.025 m. We know the speed v of

the blood flow, so the relation2

c

va

r (Equation 5.2) will give the magnitude of the

bloods centripetal acceleration.

SOLUTION With a blood flow speed of v = 0.32 m/s, the magnitude of the centripetalacceleration in the aortic arch is

22

2c

0.32 m/s4.1 m/s

0.025 m

va

r

7. REASONING The relationship between the magnitude ac of the centripetal acceleration

and the period Tof the tip of a moving clock hand can be obtained by using Equations 5.2

and 5.1:2

c

2(5.2) (5.1)

v ra v

r T

The period is the time it takes a clock hand to go once around the circle. In theseexpressions, v is the speed of the tip of the hand and ris the length of the hand. Substituting

Equation 5.1 into Equation 5.2 yields

2

2 2

c 2

2

4

r

v rTa

r r T

(1)

SOLUTION The period of the second hand is Tsecond = 60 s. The period of the minute hand

is Tminute = 1 h = 3600 s. Using Equation (1), we find that the ratio of the centripetal

acceleration of the tip of the second hand to that of the minute hand is

2

22 2c, second second minute

2 2 2c, minute second

2

minute

4

3600 s3600

4 60 s

r

a T T

a r T

T

_____________________________________________________________________________________________

8. REASONING The centripetal acceleration is given by Equation 5.2 as ac = v2/r. The value

of the radius ris given, so to determine ac we need information about the speed v. But the

speed is related to the period Tby v = (2r)/T, according to Equation 5.1. We can substitutethis expression for the speed into Equation 5.2 and see that

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av

r

r T

r

r

Tc

2

22

2

2 4 /b g

SOLUTION To use the expression obtained in the reasoning, we need a value for the

period T. The period is the time for one revolution. Since the container is turning at2.0 revolutions per second, the period is T= (1 s)/(2.0 revolutions) = 0.50 s. Thus, we find

that the centripetal acceleration is

ar

Tc2

m

sm / s

4 4 0 12

0 5019

2

2

2

2

.

.

b g

b g

_____________________________________________________________________________________________

9. SSM REASONING AND SOLUTION Since the magnitude of the centripetal

acceleration is given by Equation 5.2, aC v2

/ r, we can solve forrand find that

rv

a

2 298

C

2

.8 m / s

3.00(9.80 m / s332 m

( )

)

_____________________________________________________________________________________________

10. REASONING The magnitude of the centripetal acceleration of any point on the helicopter

blade is given by Equation 5.2, aC

v2 / r, where ris the radius of the circle on which that

point moves. From Equation 5.1: v r T 2 / . Combining these two expressions, we obtain

a rT

C 42

2

All points on the blade move with the same period T.

SOLUTION The ratio of the centripetal acceleration at the end of the blade (point 1) to thatwhich exists at a point located 3.0 m from the center of the circle (point 2) is

a

a

r T

r T

r

r

C1

C2

6.7 m

3.0 m2.2

4

4

2

1

2

2

2

2

1

2

/

/

_____________________________________________________________________________________________

11. REASONINGANDSOLUTION The sample makes one revolution in time Tas given by

T= 2r/v. The speed is

v2

= rac = (5.00 102

m)(6.25 103)(9.80 m/s2) so that v = 55.3 m/s

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Fc = mv2/r(Equation 5.3). Since the speed and the radius are the same for each car, the car

with the greater mass, which is car B, experiences the greater centripetal acceleration.

SOLUTION We find the following values for the magnitudes of the centripetalaccelerations and forces:

Car A

Car B

av

r

Fm v

r

av

r

F

m v

r

c

2

c

A

c

2

c

B

m / s

m.1 m / s (5.2)

kg m / s

mN (5.3)

m / s

m.1 m / s (5.2)

kg m / s

m N (5.3)

22

2 2

22

2 2

27

1206

1100 27

1206700

27

1206

1600 27

120 9700

b g

b gb g

b g

b gb g

_____________________________________________________________________________________________

15. SSM REASONING AND SOLUTION The magnitude of the centripetal force on the

ball is given by Equation 5.3: FC

mv2 / r. Solving forv, we have

vF r

m C

(0.028 N)(0.25 m)

0.015 kg0.68 m / s

_____________________________________________________________________________________________

16. REASONING At the maximum speed, the maximum centripetal force acts on the tires, and

static friction supplies it. The magnitude of the maximum force of static friction is specified

by Equation 4.7 as MAXs s Nf F , where s is the coefficient of static friction andFN is the

magnitude of the normal force. Our strategy, then, is to find the normal force, substitute itinto the expression for the maximum frictional force, and then equate the result to the

centripetal force, which is 2c / F mv r , according to Equation 5.3. This will lead us to an

expression for the maximum speed that we can apply to each car.

SOLUTION Since neither car accelerates in the vertical direction, we can conclude that the

cars weight mgis balanced by the normal force, soFN = mg. From Equations 4.7 and 5.3 itfollows that

2MAX

s s N s c

mv f F mg F

r

Thus, we find that2

s sor

mvmg v gr

r

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Applying this result to car A and car B gives

A s, A B s, Bandv gr v gr

In these two equations, the radius rdoes not have a subscript, since the radius is the same for

either car. Dividing the two equations and noting that the terms g and r are eliminated

algebraically, we see that

s, B s, B s, BB

B AA s, As, A s, A

0.85or 25 m/s 22 m/s

1.1

grvv v

v gr

17. REASONING AND SOLUTION Initially, the stone executes uniform circular motion in a

circle of radius rwhich is equal to the radius of the tire. At the instant that the stone flies outof the tire, the force of static friction just exceeds its maximum value f F

s

MAX

s N (see

Equation 4.7). The force of static friction that acts on the stone from one side of the tread

channel is, therefore,MAX

s 0.90(1.8 N) = 1.6 Nf

and the magnitude of the total frictional force that acts on the stone just before it flies out is

2 1.6 N 3.2 N . If we assume that only static friction supplies the centripetal force, then,Fc 3.2 N . Solving Equation 5.3 ( F mv r c

2 / ) for the radius r, we have

3 22c

6.0 10 kg (13 m/s)0.31 m

3.2 N

mvr

F

_____________________________________________________________________________________________

18. REASONINGANDSOLUTION The force P supplied by the man will

be largest when the partner is at the lowest point in the swing. The

diagram at the right shows the forces acting on the partner in thissituation. The centripetal force necessary to keep the partner swinging

along the arc of a circle is provided by the resultant of the force supplied

by the man and the weight of the partner. From the figure

P mg mv

r

2

Therefore,

Pmv

rmg

2

mg

P

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Since the weight of the partner, W, is equal to mg, it follows that m = (W/g) and

2 2 2( / ) [(475 N)/(9.80 m/s )] (4.00 m/s)(475 N) = 594 N

(6.50 m)

W g vP W

r

19. REASONING The centripetal force is the name given to the net force pointing toward the

center of the circular path. At the lowest point the net force consists of the tension in the arm

pointing upward toward the center and the weight pointing downward or away from the

center. In either case the centripetal force is given by Equation 5.3 asFc = mv2/r.

SOLUTION(a) The centripetal force is

Fmv

rc

kg .8 m / s

m

2

29 5 2

0 85

88 N.

.

b gb g

(b) Using Tto denote the tension in the arm, at the bottom of the circle we have

2

c

222 9.5 kg 2.8 m/s9.5kg 9.80 m/s 181 N

0.85 m

mv F T mg

r

mvT mg

r

20. REASONING When the penny is rotating with the disk (and not sliding relative to it), it isthe static frictional force that provides the centripetal force required to keep the penny

moving on a circular path. The magnitude MAXs

f of the maximum static frictional force is

given by MAXs s N

f F (Equation 4.7), whereFN is the magnitude of the normal force and

s is the coefficient of static friction. Solving this relation fors gives

MAXs

sN

f

F (1)

Since the maximum centripetal force that can act on the penny is the maximum static

frictional force, we have MAXc s

F f . Since Fc = mv2/r (Equation 5.3), it follows that

MAX 2s / f mv r . Substituting this expression into Equation (1) yields

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it serves as the centripetal force. The maximum centripetal force occurs when MAXc s

F f .

Therefore, the maximum speed v the car can have without slipping is related to MAXsf by

2MAX

c s

mvF f

r (2)

Substituting Equation (2) into Equation (1) yields

2

sN

mv

r

F (3)

In part a the car is subject to two downward-pointing forces, its weight W and thedownforceD. The vertical acceleration of the car is zero, so the upward normal force must

balance the two downward forces:FN = W+D. Combining this relation with Equation (3),we obtain an expression for the coefficient of static friction:

2 2

2

sN

mv mvmvr r

F W D r mg D

(4)

SOLUTIONa. Since the downforce isD = 11 000 N, Equation (4) gives the coefficient of static frictionas

22

s 2

830 kg 58 m/s0.91

160 m 830 kg 9.80 m/s 11 000 N

mv

r mg D

b. The downforce is now absent (D = 0 N). Solving Equation (4) for the speed of the car,we find that

s s s

2s

0 N

0.91 160 m 9.80 m/s 38 m/s

r mg D r mg r mg v

m m m

rg

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23. REASONING

a. The free body diagram shows the swing ride and the two forces that act on a chair: thetension T in the cable, and the weight mg of the chair and its occupant. We note that the

chair does not accelerate vertically, so the net forcey

F in the vertical direction must bezero, 0

yF . The net force consists of the upward vertical component of the tension and

the downward weight of the chair. The fact that the net force is zero will allow us to

determine the magnitude of the tension.

b. According to Newtons second law, the net forcex

F in the horizontal direction isequal to the mass m of the chair and its occupant times the centripetal acceleration

2c /a v r , so that2

c/

x F ma mv r . There is only one force in the horizontal

direction, the horizontal component of the tension, so it is the net force. We will use

Newtons second law to find the speed v of the chair.

SOLUTION

a. The vertical component of the tension is +Tcos 60.0, and the weight is mg, where we

have chosen up as the + direction. Since the chair and its occupant have no verticalacceleration, we have that 0y

F , so

cos 60.0 0

yF

T mg

Solving for the magnitude Tof the tension gives

2179 kg 9.80 m/s3510 N

cos 60.0 cos60.0

mgT

b. The horizontal component of the tension is +T sin 60.0, where we have chosen thedirection to the left in the diagram as the + direction. Since the chair and its occupant have a

centripetal acceleration in this direction, we have

60.015.0 m

r

+y

+x

T

m g

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2

csin 60.0

xF

vT ma m

r

From the drawing we see that the radius rof the circular path is r= (15.0 m) sin 60.0 =

13.0 m. Solving Equation (2) for the speed v gives

13.0 m 3510 N sin60.0sin60.014.9 m/s

179 kg

rTv

m

24. REASONING We will treat this situation asa circular turn on a banked surface, with the

angle that the rider leans serving as the

banking angle. The banking angle is

related to the speed v of the watercraft, theradius rof the curve and the magnitude gof

the acceleration due to gravity by2

tanv

rg

(Equation 5.4). If the rider is closer to the

seawall than r, she will hit the wall whilemaking the turn Therefore, the minimum

distance at which she must begin the turn is r, the minimum turn radius (see the drawing).

SOLUTION Solving Equation 5.4 forr, we obtain the minimum distance:

22

2

26 m/s170 m

tan 9.80 m/s tan22

vr

g

25. REASONING The angle at which a friction-free curve is banked depends on the radius rof the curve and the speed v with which the curve is to be negotiated, according to

Equation 5.4: tan v2 /(rg) . For known values of and r, the safe speed is

v rg tan Before we can use this result, we must determine tan for the banking of the track.

SOLUTION The drawing at the right shows a

cross-section of the track. From the drawing we have

tan . 18 m

53 m0 34

165 m 112 m = 53 m

18 m

r

r

Seawall

Beginningof turn

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a. Therefore, the smallest speed at which cars can move on this track without relying onfriction is

2min

112 m 9.80 m/s 0.34 19 m/sv

b. Similarly, the largest speed is

2max

165 m 9.80 m/s 0.34 23 m/sv

26. REASONING The relation2

tanv

r g (Equation 5.4) determines the banking angle that

a banked curve of radius r must have if a car is to travel around it at a speed v withoutrelying on friction. In this expressiongis the magnitude of the acceleration due to gravity.We will solve forv and apply the result to each curve. The fact that the radius of each curveis the same will allow us to determine the unknown speed.

SOLUTION According to Equation 5.4, we have

2

tan or tanv

v r gr g

Applying this result for the speed to each curve gives

A A B Btan and tanv r g v r g

Note that the terms r and g are the same for each curve. Therefore, these terms areeliminated algebraically when we divide the two equations. We find, then, that

BB B BB AA A AA

tan tan tan tan19or 18 m/s 22 m/s

tan tan tan13tan

rgvv v

v rg

27. REASONING From the discussion on banked curves in Section 5.4, we know that a car

can safely round a banked curve without the aid of static friction if the angle of the banked

curve is given by 2

0tan /v rg , where vo is the speed of the car and ris the radius of thecurve (see Equation 5.4). The maximum speed that a car can have when rounding an

unbanked curve is0 s

v gr (see Example 7). By combining these two relations, we can

find the angle .

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SOLUTION The angle of the banked curve is 1 20

tan /v r g . Substituting the

expression0 s

v gr into this equation gives

2

1 1 1 10 s

stan tan tan tan 0.81 39

v g r

r g r g

______________________________________________________________________________

28. REASONING The lifting force L is perpendicular to the jets

wings. When the jet banks at an angle above the horizontal,

therefore, the lifting force tilts an angle from the vertical (see thefree-body diagram). Because the jet has no vertical acceleration

during the horizontal turn, the upward vertical componentL cos of

the lifting force balances the jets weight:L cos = mg, where m isthe jets mass andgis the acceleration due to gravity. Therefore, the

magnitude of the lifting force is cos L mg .

At this point we know m and g, but not the banking angle . Sincethe jet follows a horizontal circle, the centripetal force must be

horizontal. The only horizontal force acting on the jet is the

horizontal componentL sin of the lifting force, so this must be thecentripetal force. The situation is completely analogous to that of a car driving around a

banked curve without the assistance of friction. The relation 2tan v rg (Equation 5.4),

therefore, expresses the relationship between the jets unknown banking angle , its speed v,the radius rof the turn, andg, all of which are known.

SOLUTION The magnitude of the lifting force is

5 22.00 10 kg 9.80 m/scos cos

mgL

(1)

Solving the relation 2tan v rg (Equation 5.4) for the angle , we obtain

221 1

2

123m/stan tan 22.1

3810 m 9.80 m/s

v

rg

Substituting this value for into Equation (1) for the lifting force gives

5 2 62.00 10 kg 9.80 m/s 2.12 10 Ncos cos22.1

mgL

mg

LL cos

L sin

Free-body diagram

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29. REASONING The distance d isrelated to the radius r of the circleon which the car travels by

d= r/sin 50.0 (see the drawing).

We can obtain the radius by noting

that the car experiences a centripetal

force that is directed toward the

center of the circular path. This

force is provided by the component,

FN cos 50.0, of the normal force that is parallel to the radius. Setting this force equal to themass m of the car times the centripetal acceleration 2c /a v r gives

2

N ccos50.0 / F ma mv r . Solving for the radius r and substituting it into the relation

d= r/sin 50.0 gives

2

2N

N

cos50.0

sin 50.0 sin 50.0 cos50.0 sin50.0

mvFr mv

dF

(1)

The magnitude FN of the normal force can be obtained by observing that the car has no

vertical acceleration, so the net force in the vertical direction must be zero, 0y

F . Thenet force consists of the upward vertical component of the normal force and the downward

weight of the car. The vertical component of the normal force is +FN sin 50.0, and the

weight is mg, where we have chosen the up direction as the + direction. Thus, we have

thatN

sin 50.0 0

yF

F mg

(2)

Solving this equation forFN and substituting it into the equation above will yield the

distance d.

SOLUTION Solving Equation (2) forFN and substituting the result into Equation (1) gives

2 2

N

22

2

cos50.0 sin50.0cos50.0 sin50.0

sin50.0

34.0 m/s184 m

cos50.0 9.80 m/s cos50.0

mv mv

d mgF

v

g

______________________________________________________________________________

50.0d

r

+y

+x

FN

m g

Car50.0

40.0

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30. REASONING The centripetal force Fc

required to keep an object of mass m that moves

with speed v on a circle of radius ris F mv r c

2 / (Equation 5.3). From Equation 5.1, we

know that v r T 2 / , where Tis the period or the time for the suitcase to go around once.Therefore, the centripetal force can be written as

Fm r T

r

m r

Tc

( / )2 42 2

2

(1)

This expression can be solved forT. However, we must first find the centripetal force thatacts on the suitcase.

SOLUTION Three forces act on the suitcase. They are the weight mg of the suitcase, the

force of static friction fs

MAX , and the normal force FN

exerted on the suitcase by the surface

of the carousel. The following figure shows the free body diagram for the suitcase. In this

diagram, the y axis is along the vertical direction.

The force of gravity acts, then, in the y direction.The centripetal force that causes the suitcase tomove on its circular path is provided by the net

force in the +x direction in the diagram. From the

diagram, we can see that only the forces FN

and

fs

MAX have horizontal components. Thus, we have

F f F c s

MAX

N cos sin , where the minus sign

indicates that the x component of FN

points to the

left in the diagram. Using Equation 4.7 for themaximum static frictional force, we can write this result as in equation (2).

F F F F c s N N N s

cos sin ( cos sin ) (2)

If we apply Newton's second law in they direction, we see from the diagram that

F f mg ma F F mg s

MAX

y sN N Nsin or sincos cos 0 0

where we again have used Equation 4.7 for the maximum static frictional force. Solving for

the normal force, we find

Fmg

s

N sin

cos

Using this result in equation (2), we obtain the magnitude of the centripetal force that acts

on the suitcase:

F Fmg

s

c N s

s

sin

( cos sin )

( cos sin )

cos

With this expression for the centripetal force, equation (1) becomes

+x

+y

mg

FN

fsMAX

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mg m r

Ts

( cos sin )

cos

s

sin

4 2

2

Solving for the period T, we find

2 2

2s

4 cos sin 4 (11.0 m) cos 36.0 0.760 sin 36.045 s

( cos sin ) 9.80 m/s 0.760 cos 36.0 sin 36.0

srTg

31. REASONING The speed v of a satellite in circular orbit about the earth is given by

E/v GM r (Equation 5.5), where G is the universal gravitational constant, ME is the

mass of the earth, and ris the radius of the orbit. The radius is measured from the center ofthe earth, not the surface of the earth, to the satellite. Therefore, the radius is found by

adding the height of the satellite above the surface of the earth to the radius of the earth

(6.38 106 m).

SOLUTION First we add the orbital heights to the radius of the earth to obtain the orbital

radii. Then we use Equation 5.5 to calculate the speeds.

Satellite A

Satellite B

r

vGM

r

r

vGM

r

A

E

A

2 2

A

E

A

2 2

m + 360 m = 6.74 m

N m kg kg

6.74 mm / s

m + 720 m = 7.10 m

N m kg kg

7 m/ s

6 38 10 10 10

6 67 10 5 98 10

107690

6 38 10 10 10

6 67 10 5 98 10

10 107500 m

6 3 6

11 24

6

6 3 6

11 24

6

.

. / .

.

. / .

.

c hc h

c hc h

_____________________________________________________________________________________________

32. REASONINGANDSOLUTION We have for Jupiter v2

= GMJ/r, where

r= 6.00 105 m + 7.14 107 m = 7.20 107 mThus,

11 2 2 27 47

6.67 10 N m / kg 1.90 10 kg4.20 10 m/s

7.20 10 mv

____________________________________________________________________________________________

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33. SSM WWW REASONING Equation 5.5 gives the orbital speed for a satellite in a

circular orbit around the earth. It can be modified to determine the orbital speed around any

planet P by replacing the mass of the earth ME

by the mass of the planet MP

:

v GM r P

/ .

SOLUTION The ratio of the orbital speeds is, therefore,

v

v

GM r

GM r

r

r

2

1

2

1

1

2

P

P

/

/

Solving forv2 gives

v vr

r2 11

2

1010

1010

(1.705.25

8.601.334

6

6

4m/s )m

mm / s

34. REASONINGANDSOLUTION The normal force exerted by the wall on each astronaut is

the centripetal force needed to keep him in the circular path, i.e., Fc = mv2/r. Rearranging

and lettingFc = (1/2)mgyields

r= 2v2/g= 2(35.8 m/s)

2/(9.80 m/s

2) = 262 m

35. REASONING In Section 5.5 it is shown that the period Tof a satellite in a circular orbitabout the earth is given by (see Equation 5.6)

3/ 2

E

2 rT

GM

where r is the radius of the orbit, G is the universal gravitational constant, and ME is the

mass of the earth. The ratio of the periods of satellites A and B is, then,

3/2

A

3/2EA A

3/2 3/2

BB B

E

2

2

r

GMT r

rT r

GM

We do not know the radii rA and rB. However we do know that the speed v of a satellite is

equal to the circumference (2 r) of its orbit divided by the period T, so 2 /v r T .

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SOLUTIONSolving the relation 2 /v r T forrgives / 2r vT . Substituting this valueforrinto Equation (1) yields

3/ 2 3/ 23/2

A A A AA A

3/2 3/ 23/2B B B BB B

/ 2

/ 2

v T v TT r

T r v Tv T

Squaring both sides of this equation, algebraically solving for the ratio TA/TB, and using the

fact that vA = 3vB gives

3 3

A B B

33

B A B

1= =

273

T v v

T v v

______________________________________________________________________________

36. REASONINGAND SOLUTION The period of the moon's motion (approximately the

length of a month) is given by

32 3 2 8

11 2 2 24

E

6

4 4 3.85 10 m

6.67 10 N m / kg 5.98 10 kg

2.38 10 s = 27.5 days

rT

GM

37. SSM REASONING Equation 5.2 for the centripetal acceleration applies to both the plane

and the satellite, and the centripetal acceleration is the same for each. Thus, we have

av

r

v

rv

r

rv

c

plane

2

plane

satellite

2

satellite

plane

plane

satellite

satelliteor

F

HGG

I

KJJ

The speed of the satellite can be obtained directly from Equation 5.5.

SOLUTION Using Equation 5.5, we can express the speed of the satellite as

vGm

rsatellite

E

satellite

Substituting this expression into the expression obtained in the reasoning for the speed of the

plane gives

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vr

rv

r

r

Gm

r

r Gm

r

v

plane

plane

satellite

satellite

plane

satellite

E

satellite

plane E

satellite

plane

2 2 24

6

m N m / kg 5.98 10 kg

6.7 10 m m / s

F

HGG

I

KJJ

F

HGG

I

KJJ

15 6 67 10

12

11b gc hc h.

38. REASONING The speed v of a planet orbiting a star is given by SGM

vr

(Equation 5.5), where MS is the mass of the star,11 2 26.674 10 N m / kgG is the

universal gravitational constant, and ris the orbital radius. This expression can be solved for

MS. However, the orbital radius r is not known, so we will use the relation2 r

vT

(Equation 5.1) to eliminate rin favor of the known quantities v and T(the period). Returningto Equation 5.5, we square both sides and solve for the mass of the star:

22S

SorGM rv

v Mr G

(1)

Then, solving2 r

vT

forryields

2

vTr

, which we substitute into Equation (1):

22 3

S

2

2

vTv

rv v T M G G G

(2)

We will use Equation (2) to calculate the mass of the star in part a. In part b, we will solveEquation (2) for the period Tof the faster planet, which should be shorter than that of theslower planet.

SOLUTION

a. The speed of the slower planet is v = 43.3 km/s = 43.3103 m/s. Its orbital period in

seconds is T= (7.60 yr)[(3.156 107 s)/(1 yr)] = 2.40108 s. Substituting these values intoEquation (2) yields the mass of the star:

33 83

31S 11 2 2

43.3 10 m/s 2.40 10 s4.65 10 kg

2 2 6.674 10 N m / kg

v TM

G

This is roughly 23 times the mass of the sun.

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b. Solving Equation (2) for the orbital period T, we obtain

3S

S 3

2or

2

GMv TM T

G v

(3)

The speed of the faster planet is v = 58.6 km/s = 58.6 103 m/s. Equation (3) now gives theorbital period of the faster planet in seconds:

11 2 2 31

7

33

2 6.674 10 N m / kg 4.65 10 kg9.69 10 s

58.6 10 m/s

T

Lastly, we convert the period from seconds to years:

7

9.69 10 sT 71 yr

3.156 10 s 3.07 yr

39. REASONING The satellites true weight Wwhen at rest on the surface of the planet is the

gravitational force the planet exerts on it. This force is given by 2P

W GM m r

(Equation 4.4), where G is the universal gravitational constant, MP is the mass of the planet,

m is the mass of the satellite, and ris the distance between the satellite and the center of theplanet. When the satellite is at rest on the planets surface, its distance from the planets

center isRP, the radius of the planet, so we have2

P PW GM m R . The satellites mass m is

given, as is the planets radiusRP. But we must use the relation3 2

P

2 rT

GM

(Equation 5.6)

to determine the planets mass MP in terms of the satellites orbital period T and orbital

radius r. Squaring both sides of Equation 5.6 and solving forMP, we obtain

22 2 3 2 2 3 2 3

2P2 2

PP

2 4 4or

r r rT M

GM GTGM

(1)

Substituting Equation (1) into P2P

GM mW

R (Equation 4.4), we find that

P2P

Gm GW M

R

2 3

2P

4m r

R G

2 3

2 2 2P

4 r m

T R T

(2)

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SOLUTION All of the quantities in Equation (2), except for the period T, are given in SIbase units, so we must convert the period from hours to seconds, the SI base unit for time:

T= (2.00 h)[(3600 s)/(1 h)] = 7.20103 s. The satellites orbital radius r in Equation (2) isthe distance between the satellite and the center of the orbit, which is the planets center.Therefore, the orbital radius is the sum of the planets radius RP and the satellites height h

above the planets surface: r=RP + h = 4.15106 m + 4.1105 m = 4.56106 m. We now

use Equation (2) to calculate the satellites true weight at the planets surface:

32 62 3

42 2 2 2

6 3P

4 4.56 10 m 5850 kg42.45 10 N

4.15 10 m 7.20 10 s

r mW

R T

40. REASONINGANDSOLUTIONa. The centripetal acceleration of a point on the rim of chamber A is the artificial

acceleration due to gravity,

aA = vA2/rA = 10.0 m/s

2

A point on the rim of chamber A moves with a speed vA = 2rA/T where Tis the period of

revolution, 60.0 s. Substituting the second equation into the first and rearranging yields

rA = aAT2/(4

2) = 912 m

b. Now

rB = rA/4.00 = 228 m

c. A point on the rim of chamber B has a centripetal acceleration aB = vB2/rB. The point

moves with a speed vB = 2rB/T. Substituting the second equation into the first yields

2 22B

2B 2

4 4 228 m2.50 m/s

60.0 s

ra

T

41. SSM REASONING According to Equation 5.3, themagnitude Fc of the centripetal force that acts on each

passenger is 2c

/ F mv r , where m and v are the mass and

speed of a passenger and ris the radius of the turn. From this

relation we see that the speed is given byc

/v F r m . The

centripetal force is the net force required to keep each

Passenger

r

mg

2 mg

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passenger moving on the circular path and points toward the center of the circle. With theaid of a free-body diagram, we will evaluate the net force and, hence, determine the speed.

SOLUTION The free-body diagram shows a passenger at the bottom of the circular dip.

There are two forces acting: her downward-acting weight mg and the upward-acting force

2mg that the seat exerts on her. The net force is +2mg mg = +mg, where we have takenup as the positive direction. Thus,Fc = mg. The speed of the passenger can be found byusing this result in the equation above.

SubstitutingFc = mginto the relation c /v F r m yields

2c 9.80 m/s 20.0 m 14.0 m/s

mg rF rv g r

m m

42. REASONING The riders speed v at the top of theloop is related to the centripetal force acting on her

by2

c

mvF

r (Equation 5.3). The centripetal force

Fc is the net force, which is the sum of the two

vertical forces: W (her weight) and FN (the

magnitude of the normal force exerted on her bythe electronic sensor). Both forces are illustrated in

the Top of loop free-body diagram. Because both

forces point in the same direction, the magnitude of

the centripetal force is c N F mg F . Thus, we

have that2

N

mvmg F

r . We will solve this relation to find the speed v of the rider. The

reading on the sensor at the top of the loop gives the magnitude FN = 350 N of the

downward normal force. Her weight mgis equal to the reading on the sensor when level andstationary (see the Stationary free-body diagram).

SOLUTIONSolving2

N

mvmg F

r for the speed v, we obtain

N N2 orr mg F r mg F v vm m

The only quantity not yet known is the riders mass m, so we will calculate it from her

weight W by using the relation W mg (Equation 4.5). Thus, m = W/g

= (770 N)/(9.80 m/s2) = 79 kg. The speed of the rider at the top of the loop is

mg= 770 N

FN

= 770 N

Stationary

free-body

diagram

mg

Top of loop

free-body

diagram

FN

= 350 N

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N 21 m 770 N 350 N17 m/s

79 kg

r mg F v

m

43. SSM REASONING The centripetal force is the name given to the net force pointingtoward the center of the circular path. At point 3 at the top the net force pointing toward thecenter of the circle consists of the normal force and the weight, both pointing toward the

center. At point 1 at the bottom the net force consists of the normal force pointing upward

toward the center and the weight pointing downward or away from the center. In either case

the centripetal force is given by Equation 5.3 asFc = mv2/r.

SOLUTION At point 3 we have

F F mg mv

rc N 3

2

At point 1 we have

F F mg mv

rc N 1

2

Subtracting the second equation from the first gives

2 32

1

2

mgmv

r

mv

r

Rearranging gives

v gr v3

2

1

22

Thus, we find that

v3

22 9 80 3 0 15 17 . .m / s m m / s m / s2c hb g b g

_____________________________________________________________________________________________

44. REASONING The normal force (magnitude FN) that

the pilots seat exerts on him is part of the centripetal

force that keeps him on the vertical circular path.

However, there is another contribution to the centripetalforce, as the drawing at the right shows. This additional

contribution is the pilots weight (magnitude W). Toobtain the ratioFN/W, we will apply Equation 5.3, which

specifies the centripetal force as 2c / F mv r .

SOLUTION Noting that the direction upward (toward the center of the circular path) is

positive in the drawing, we see that the centripetal force isc N

F F W . Thus, from

Equation 5.3 we have

FN

W

+

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2

c N

mv F F W

r

The weight is given by W = mg (Equation 4.5), so we can divide the expression for thecentripetal force by the expression for the weight and obtain that

2 2N N

cor 1

F W F mv vF

W mgr W gr

Solving for the ratioN

F /W, we find that

22N

2

230 m/s1 1 8.8

9.80 m/s 690 m

F v

W gr

45. REASONING The magnitude Fc of the centripetal force is2

c / F mv r (Equation 5.3).

Since the speed v, the mass m, and the radius rare fixed, the magnitude of the centripetalforce is the same at each point on the circle. When the ball is at the three oclock position,the force of gravity, acting downward, is perpendicular to the string and cannot contribute to

the centripetal force. (See Figure 5.21 in the text, point 2, for a similar situation.) At thispoint, only the tension ofT= 16 N contributes to the centripetal force. Considering that thecentripetal force is the same everywhere, we can conclude that it has a magnitude of 16 N

everywhere.

At the twelve oclock position the tension Tand the force of gravity mgboth act downward(the negative direction) toward the center of the circle, with the result that the centripetalforce at this point is Tmg. (See Figure 5.21, point 3.) At the six oclock positionthe tension points upward toward the center of the circle, while the force of gravity points

downward, with the result that the centripetal force at this point is Tmg. (See Figure 5.21,point 1.)

SOLUTION Assuming that upward is the positive direction, we find at the twelve and six

oclock positions that

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Twelve o' clock

Six o' clock

T mg

T

T mg

T

Centripetalforce

2

Centripetalforce

2

N

N kg m / s N

N

N kg m / s N

16

16 0 20 9 80 14

16

16 0 20 9 80 18

. .

. .

b gc h

b gc h

46. REASONING Because the crest of the hill is a circular arc, themotorcycles speed v is related to the centripetal forceFc acting

on the motorcycle: 2c F mv r (Equation 5.3), where m is the

mass of the motorcycle and ris the radius of the circular crest.Solving Equation 5.3 for the speed, we obtain 2

cv F r m or

cv F r m . The free-body diagram shows that two vertical

forces act on the motorcycle. One is the weight mg of themotorcycle, which points downward. The other is the normal

force FN exerted by the road. The normal force points directly

opposite the motorcycles weight. Note that the motorcycles weight must be greater than

the normal force. The reason for this is that the centripetal force is the net force produced by

mg and FN and must point toward the center of the circle, which lies below the motorcycle.

Only if the magnitude mgof the weight exceeds the magnitude FN

of the normal force will

the centripetal force point downward. Therefore, we can express the magnitude of the

centripetal force asFc = mgFN. With this identity, the relation cv F r m becomes

Nmg F r vm

(1)

SOLUTION When the motorcycle rides over the crest sufficiently fast, it loses contact withthe road. At that point, the normal force FN is zero. In that case, Equation (1) yields the

motorcycles maximum speed:

0mg r mv

m

gr

m 29.80 m/s 45.0 m 21.0 m/sgr

mg

FN

Free-body diagram of

the motorcycle

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47. REASONING When the stone is whirled in a horizontal circle, the centripetal force isprovided by the tension Th in the string and is given by Equation 5.3 as

2

h

Centripetalforce

mv

Tr

(1)

where m and v are the mass and speed of the stone, and ris the radius of thecircle. When the stone is whirled in a vertical circle, the maximum tension

occurs when the stone is at the lowest point in its path. The free-bodydiagram shows the forces that act on the stone in this situation: the tensionTvin the string and the weight mg of the stone. The centripetal force is the

net force that points toward the center of the circle. Setting the centripetal

force equal to 2 /mv r, as per Equation 5.3, we have

2

v

Centripetalforce

+ mvT mgr

(2)

Here, we have assumed upward to be the positive direction. We are given that the maximum

tension in the string in the case of vertical motion is 15.0% larger than that in the case of

horizontal motion. We can use this fact, along with Equations 1 and 2, to find the speed ofthe stone.

Solution Since the maximum tension in the string in the case of vertical motion is 15.0%

larger than that in the horizontal motion,v h

(1.000 0.150)T T . Substituting the values of

Th and Tv from Equations (1) and (2) into this relation gives

v h

2 2

1.000 0.150

1.000 0.150

T T

mv mvmg

r r

Solving this equation for the speed v of the stone yields

2

(9.80 m/s ) (1.10 m) 8.48 m/s0.150 0.150g rv

Stone

mg

Tv

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48. REASONING The drawing at the right

shows the two forces that act on a piece of

clothing just before it loses contact with thewall of the cylinder. At that instant the

centripetal force is provided by the normalforce F

Nand the radial component of the

weight. From the drawing, the radial

component of the weight is given by

mg mg mg cos = cos (90 ) = sin

Therefore, with inward taken as the positive direction, Equation 5.3 (Fc mv2

/ r) gives

F mg

mv

r

2

N sin =

At the instant that a piece of clothing loses contact with the surface of the drum, FN

0N,

and the above expression becomes

mgmv

r

2

sin =

According to Equation 5.1, v 2r/ T, and with this substitution we obtain

gr T

r

r

T

2

sin = ( / )2 4 2

2

This expression can be solved for the period T. Since the period is the required time for onerevolution, the number of revolutions per second can be found by calculating 1/T.

SOLUTION Solving for the period, we obtain

Tr

g

r

g

42 1 17

2

sin sin2

0.32 m

9.80 m/ s sin 70.0s

2c h.

Therefore, the number of revolutions per second that the cylinder should make is

1 1

1 17T

. s0.85 rev / s

FN

mg

Clothes

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49. SSM REASONING In Example 3, it was shown that the magnitudes of the centripetal

acceleration for the two cases are

Radius 33 m / sC

2 a 35 m

Radius 24 m m / sC 2 a 48

According to Newton's second law, the centripetal force is F maC C

(see Equation 5.3).

SOLUTION a. Therefore, when the sled undergoes the turn of radius 33 m,

F maC C

2350 kg)(35 m / s 1.2 ( ) 104 N

b. Similarly, when the radius of the turn is 24 m,

F maC C

2350 kg)(48 m / s 1.7 ( ) 104 N

50. REASONING Two pieces of information are provided. One is the fact that the magnitude

of the centripetal acceleration ac is 9.80 m/s2. The other is that the space station should not

rotate faster than two revolutions per minute. This rate of twice per minute corresponds to

thirty seconds per revolution, which is the minimum value for the period Tof the motion.With these data in mind, we will base our solution on Equation 5.2, which gives the

centripetal acceleration as 2c /a v r , and on Equation 5.1, which specifies that the speed v

on a circular path of radius ris 2 /v r T .

SOLUTION From Equation 5.2, we have

2 2

cc

orv v

a rr a

Substituting 2 /v r T into this result and solving for the radius gives

222 22

c2 2

c c

9.80 m/s 30.0 s2 /or 223 m

4 4

r T a T vr r

a a

51. REASONING Since the tip of the blade moves on a circular path, it experiences a

centripetal acceleration whose magnitude ac is given by Equation 5.2 as,2

c/a v r , where v

is the speed of blade tip and ris the radius of the circular path. The radius is known, and the

speed can be obtained by dividing the distance that the tip travels by the time tof travel.

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Since an angle of 90 corresponds to one fourth of the circumference of a circle, the distanceis .

SOLUTION Since 2c

/a v r and 14 2 / / 2v r t r t , the magnitude of the centripetal

acceleration of the blade tip is

2

22 22

c 2 2

0.45 m26.9 m/s

4 4 0.40 s

r

v rta

r r t

52. REASONINGANDSOLUTIONa. In terms of the period of the motion, the centripetal force is written as

Fc = 4

2

mr/T

2

= 4

2

(0.0120 kg)(0.100 m)/(0.500 s)

2

= 0.189 N

b. The centripetal force varies as the square of the speed. Thus, doubling the speed would

increase the centripetal force by a factor of 22 4 .

53. REASONING The astronaut in the chamber is subjected to a centripetal acceleration ac that

is given by 2c

/a v r (Equation 5.2). In this expression v is the speed at which the

astronaut in the chamber moves on the circular path of radius r. We can solve this relation

for the speed.

SOLUTION Using Equation 5.2, we have

2

2c c

or 7.5 9.80 m/s 15 m 33 m/sv

a v a r r

54. REASONING The person feels the centripetal force acting on his back. This force is

Fc = mv2/r, according to Equation 5.3. This expression can be solved directly to determine

the radius rof the chamber.

SOLUTION Solving Equation 5.3 for the radius rgives

rmv

F

22

83 3

560 N1 5

C

kg .2 m / sm

b gb g.

_____________________________________________________________________________________________

14 2 r

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55. SSM REASONING As the motorcycle passes over the top of the hill, it will experience

a centripetal force, the magnitude of which is given by Equation 5.3: F mv r C

2 / . The

centripetal force is provided by the net force on the cycle + driver system. At that instant,

the net force on the system is composed of the normal force, which points upward, and the

weight, which points downward. Taking the direction toward the center of the circle(downward) as the positive direction, we have F mg F C N

. This expression can be solved

forFN

, the normal force.

SOLUTIONa. The magnitude of the centripetal force is

Fmv

rC

2(342 kg)(25.0 m / s)

126 m1.70

2

103 N

b. The magnitude of the normal force is2 3 3

N C(342 kg)(9.80 m/s ) 1.70 10 N = 1.66 10 N F mg F

_____________________________________________________________________________________________

56. REASONING The speed of the satellite is given by Equation 5.1 as 2 /v r T . Since we

are given that the period is T= 1.20 104 s, it will be possible to determine the speed fromEquation 5.1 if we can determine the radius rof the orbit. To find the radius, we will use

Equation 5.6, which relates the period to the radius according to 3/ 2E

2 /T r GM , where

G is the universal gravitational constant and ME is the mass of the earth.

SOLUTION According to Equation 5.1, the orbital speed is

2 rv

T

To find a value for the radius, we begin with Equation 5.6:

3/ 2E3/ 2

E

2or

2

T GMrT r

GM

Next, we square both sides of the result forr3/2:

2

22 E3/ 2 3 E

2or

2 4

T GM T GMr r

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We can now take the cube root of both sides of the expression forr3 in order to determine r:

2

4 11 2 2 2423 7E3

2 2

1.20 10 s 6.67 10 N m /kg 5.98 10 kg1.13 10 m

4 4

T GMr

With this value for the radius, we can use Equation 5.1 to obtain the speed:

73

4

2 1.13 10 m25.92 10 m/s

1.20 10 s

rv

T

57. SSM REASONING AND SOLUTION The centripetal acceleration for any point on the

blade a distance rfrom center of the circle, according to Equation 5.2, is ac v2

/ r. From

Equation 5.1, we know that v r T 2 / where T is the period of the motion. Combiningthese two equations, we obtain

ar T

r

r

Tc

( / )2 4

2 2

2

a. Since the turbine blades rotate at 617 rev/s, all points on the blades rotate with a period

ofT (1/617) s = 1.62 103 s. Therefore, for a point with r 0.020 m, the magnitude ofthe centripetal acceleration is

ac m)(1.62

3.0 4 0 020

1010

2

2 ( . 35 2

s)m / s

b. Expressed as a multiple ofg, this centripetal acceleration is

ag

gc 2

3.01.00

9.80 m/ s3.1 F

HGIKJ

10 1045 2m / sc h

_____________________________________________________________________________________________

58. REASONING The centripetal acceleration for any point that is a distance rfrom the center

of the disc is, according to Equation 5.2, ac v2

/ r. From Equation 5.1, we know thatv r T 2 / where Tis the period of the motion. Combining these two equations, we obtain

ar T

r

r

Tc

( / )2 42 2

2

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SOLUTION Using the above expression forac , the ratio of the centripetal accelerations of

the two points in question is

a

a

r T

r T

r T

r T

2

1

2

2 2

2

2

1 1

2

2 2

2

1 1

2

4

4

/

/

/

/

Since the disc is rigid, all points on the disc must move with the same period, so T1 T2 .Making this cancellation and solving for a2 , we obtain

a ar

r2 12

1

210 FHG

IKJ

120 m/ s0.050 m

0.030 m2.0 m / s2 2c h

Note that even though T1 T2 , it is not true that v1 v2 . Thus, the simplest way to approachthis problem is to express the centripetal acceleration in terms of the period Twhich cancelsin the final step.

_____________________________________________________________________________________________

59. SSM WWW REASONING Let v0 be the initial speed of the ball as it begins its

projectile motion. Then, the centripetal force is given by Equation 5.3: FC

mv02

/ r. We

are given the values for m and r; however, we must determine the value of v0 from the

details of the projectile motion after the ball is released.

In the absence of air resistance, the x component of the projectile motion has zeroacceleration, while the y component of the motion is subject to the acceleration due togravity. The horizontal distance traveled by the ball is given by Equation 3.5a (with

ax 0 m/s2):

x v t v t x 0 0( cos )

with tequal to the flight time of the ball while it exhibits projectile motion. The time tcanbe found by considering the vertical motion. From Equation 3.3b,

v v a t y y y

0

After a time t, vy v0y . Assuming that up and to the right are the positive directions, we

have

tv

a

v

a

y

y y

2 20 0 sin

and

x vv

ay

F

HG

I

KJ( cos )

sin0

02

Using the fact that 2sin cos sin 2, we have

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xv

a

v

ay y

2

0

2

0

2cos sin sin 2(1)

Equation (1) (with upward and to the right chosen as the positive directions) can be used to

determine the speed v0 with which the ball begins its projectile motion. Then Equation 5.3

can be used to find the centripetal force.

SOLUTION Solving equation (1) forv0 , we have

vx a

y

0

86 7529 3

2

( m)(9.80 m / s )

sin 2(41 )m / s

2

sin

..

Then, from Equation 5.3,

Fmv

rC

2(7.3 kg)(29.3 m/ s)

1.8 m3500 N 0

2

_____________________________________________________________________________________________

60. REASONINGANDSOLUTION

a. The centripetal force is provided by the normal force exerted on the rider by the wall .

b. Newton's second law applied in the horizontal direction gives

FN = mv2/r= (55.0 kg)(10.0 m/s)

2/(3.30 m) = 1670 N

c. Newton's second law applied in the vertical direction gives sFN mg= 0 or

s = (mg)/FN = 0.323

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61. SSM WWW REASONING If the effects of gravity are not ignored in Example 5, the

plane will make an angle with the vertical as shown in figure A below. The figure Bshows the forces that act on the plane, and figure C shows the horizontal and verticalcomponents of these forces.

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From figure C we see that the resultant force in the horizontal direction is the horizontalcomponent of the tension in the guideline and provides the centripetal force. Therefore,

T mvr

sin =

2

From figure A, the radius r is related to the length L of the guideline by r L sin;therefore,

Tmv

L=sin

sin

2

(1)

The resultant force in the vertical direction is zero: Tcos mg 0, so that

Tcos mg (2)From equation (2) we have

Tmg

cos

(3)

Equation (3) contains two unknown, T and . First we will solve equations (1) and (3)simultaneously to determine the value(s) of the angle . Once is known, we can calculate

the tension using equation (3).

SOLUTION Substituting equation (3) into equation (1):

mg mvLcos

sin =sin

2

FHG IKJ

Thus,

sin

cos=

2

v

gL

2

(4)

Using the fact that cos2+ sin2= 1, equation (4) can be written

mg

Tcos

Tsin

mg

T

r

L L

A B C

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Ch 05 ISM 8e