· CHAPTER 5 DYNAMICS OF UNIFORM CIRCULAR MOTION ANSWERS TO FOCUS ON CONCEPTS QUESTIONS

CHAPTER 5 DYNAMICS OF UNIFORM CIRCULAR 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 r is the radius of the circular path. Since the radius of the track is smaller at A compared 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 grµ= , where µs is the coefficient of static friction, g is the acceleration due to gravity, and r is 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 centripetal force. Since the centripetal force is inversely proportional to the radius r of the path, T1 must be one-half of T2.

8. (a) The centripetal force is given by Fc = mv2/r. The centripetal forces for particles 1, 2 and 3 are, respectively, 4m0v0

2/r0, 3m0v02/r0, and 2m0v0

2/r0.

9. (d) The centripetal force is directed along the radius and toward the center of the circular path. The component FN sin θ 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 by Fc = mv2/r. The two cars have the same speed v and the radius r of 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.

Chapter 5 Answers to Focus on Concepts Questions 237

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 as Fc = mv2/r, which depends on the square (v2) of the astronaut’s speed and inversely (1/r) on the radius of the ring. According to Equation 5.1, r = vT /(2π), the radius is directly proportional to the speed. Thus, the centripetal 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 skier’s velocity. At B the skier is momentarily traveling at a constant speed on a circular path of radius r. The direction of the net force, called the centripetal 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 Newton’s second law, the net force, NF mg− , must equal the mass m

times the centripetal acceleration v2/r.

238 DYNAMICS OF UNIFORM CIRCULAR MOTION

CHAPTER 5 DYNAMICS OF UNIFORM

CIRCULAR MOTION

PROBLEMS 1. SSM REASONING The magnitude ac of the car’s 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 the circumference (2π r) of the track divided by the period T of the motion. The period is the time for the car to go once around the track (T = 360 s).

SOLUTION Since 2

c /a v r= and ( )2 /v r Tπ= , the magnitude of the car’s centripetal acceleration is

( )( )

2

2 32 22

2c 2

24 2.6 10 m4 0.79 m/s

360 s

rv rTar r T

πππ

⎛ ⎞⎜ ⎟ ×⎝ ⎠= = = = =

______________________________________________________________________________ 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/s 290 m/s

0.50 m

35 m/s 0 m/s

2.3 m/s 2.9 m/s

1.8 m

var

var

var

= = =

= = =∞

= = =

Example 1

Example 2

Example 3

Chapter 5 Problems 239

3. REASONING AND SOLUTION Let s represent the length of the path of the pebble after it is released. From Conceptual Example 2, we know that the pebble will fly off 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 the figure at the right. From the figure we see that

35° θ

α r

s

Pebble

Target

d

C –11 1cos or =cos 84

10 10 10r rd r

α α ⎛ ⎞= = = = °⎜ ⎟⎝ ⎠

Furthermore, from the figure, we see that α +θ +35°=180°. Therefore,

145 145 84 61θ α= °− = °− ° = ° _____________________________________________________________________________________________

4. REASONING In each case, the magnitude of the centripetal acceleration is given by 2

c /a v r= (Equation 5.2). Therefore, we will apply this expression to each boat and set the centripetal accelerations equal. The resulting equation can be solved for the desired ratio.

SOLUTION Using Equation 5.2 for the centripetal acceleration of each boat, we have

2 2A B

cA cBA B

and v v

a ar r

= =

Setting the two centripetal accelerations equal gives

2 2A B

A B

v vr r

=

Solving for the ratio of the speeds gives

A A

B B

120 m 0.71240 m

v rv r

= = =

_____________________________________________________________________________________________ 5. SSM REASONING The speed of the plane is given by Equation 5.1: v r T= 2π / , where

T is the period or the time required for the plane to complete one revolution. SOLUTION Solving Equation 5.1 for T we have

Trv

= = =2 2850 mπ π2 (

110 m / s160 s)


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

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

the blood flow, so the relation 2

cvar

= (Equation 5.2) will give the magnitude of the

blood’s centripetal acceleration.

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

( )222

c0.32 m/s

4.1 m/s0.025 m

var

= = = 7. 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, 2c /a v r= , where v

is the speed of blade tip and r is 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 t of travel. Since an angle of 90° corresponds to one fourth of the circumference of a circle, the distance is ( )1

4 2 rπ . SOLUTION Since 2

c /a v r= and ( ) ( )14 2 / / 2v r t r tπ π= = , the magnitude of the centripetal

acceleration of the blade tip is

( )( )

2

22 22

c 2 20.45 m2 6.9 m/s

4 4 0.40 s

rv rtar r t

πππ

⎛ ⎞⎜ ⎟⎝ ⎠= = = = =

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

of the radius r is given, so to determine ac we need information about the speed v. But the speed is related to the period T by v = (2π r)/T, according to Equation 5.1. We can substitute this expression for the speed into Equation 5.2 and see that

ac =v2

r=2π r /T( )2

r= 4π

2rT 2

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 at 2.0 revolutions per second, the period is T = (1 s)/(2.0 revolutions) = 0.50 s. Thus, we find that the centripetal acceleration is

ac =4π 2rT 2 =

4π 2 0.12 m( )0.50 s( )2 = 19 m/s2


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

acceleration is given by Equation 5.2, aC = v2 / r, we can solve for r and find that

r va

= = =2 298

C2

.8 m / s3.00(9.80 m / s

332 m( ))

_____________________________________________________________________________________________ 10. REASONING The centripetal acceleration for any point that is a distance r from the center

of the disc is, according to Equation 5.2, ac = v2 / r . From Equation 5.1, we know that

v r T= 2π / where T is the period of the motion. Combining these two equations, we obtain

ar Tr

rTc = =

( / )2 42 2

2

π π

SOLUTION Using the above expression for ac, the ratio of the centripetal accelerations of

the two points in question is

aa

r Tr T

r Tr T

2

1

22 2

2

21 1

22 2

2

1 12

44

= =ππ

//

//

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

a2 = a1r2r1= 120 m/s2( ) 0.050 m

0.030 m⎛⎝⎜

⎞⎠⎟ = 2.0 × 102 m/s2

Note that even though T1 = T2, it is not true that v1 = v2. Thus, the simplest way to approach

this problem is to express the centripetal acceleration in terms of the period T which cancels in the final step.

_____________________________________________________________________________________________ 11. REASONING AND SOLUTION The sample makes one revolution in time T as given by

T = 2π r/v. The speed is v2 = rac = (5.00 × 10–2 m)(6.25 × 103)(9.80 m/s2) so that v = 55.3 m/s

The period is

T = 2π (5.00 × 10–2 m)/(55.3 m/s) = 5.68 × 10–3 s = 9.47 × 10–5 min The number of revolutions per minute = 1/T = 10 600 rev/min . _____________________________________________________________________________________________


12. REASONING AND SOLUTION a. At the equator a person travels in a circle whose radius equals the radius of the earth,

r = Re = 6.38 × 106 m, and whose period of rotation is T = 1 day = 86 400 s. We have

v = 2πRe/T = 464 m/s The centripetal acceleration is

( )22–2 2

c 6

464 m/s3.37 10 m/s

6.38 10 mvar

= = = ××

b. At 30.0° latitude a person travels in a circle of radius,

r = Re cos 30.0° = 5.53 × 106 m Thus,

v = 2π r/T = 402 m/s and ac = v2/r = 2.92 × 10–2 m/s2 _____________________________________________________________________________________________ 13. 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 C2350 kg)(35 m/ s 1.2= = = ×( ) 104 N

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

F maC C2350 kg)(48 m/ s 1.7= = = ×( ) 104 N

14. 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 r of the chamber.

SOLUTION Solving Equation 5.3 for the radius r gives

r = mv2

FC

=83 kg( ) 3.2 m/s( )2

560 N= 1.5 m

_____________________________________________________________________________________________


15. 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 MAX

s s Nf Fµ= , where µs is the coefficient of static friction and FN is the magnitude of the normal force. Our strategy, then, is to find the normal force, substitute it into the expression for the maximum frictional force, and then equate the result to the centripetal force, which is 2

c /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

car’s weight mg is balanced by the normal force, so FN = mg. From Equations 4.7 and 5.3 it follows that

2MAXs s N s c

mvf F mg Fr

µ µ= = = =

Thus, we find that

2

s s or mvmg v grr

µ µ= =

Applying this result to car A and car B gives

A s, A B s, B and v gr v grµ µ= = In these two equations, the radius r does 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, BBB A

A s, As, A s, A

0.85 or 25 m/s 22 m/s1.1

grvv v

v gr

µ µ µµµ µ

= = = = =

16. REASONING The centripetal force that acts on the skater is Fc = mv2/r (Equation 5.3). This expression can be solved directly to determine the mass m.

SOLUTION Solving Equation 5.3 for the mass m gives

( )( )( )

c2 2

460 N 31 m73 kg

14 m/s

F rm

v= = =

_____________________________________________________________________________________________


17. REASONING AND SOLUTION 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 this situation. 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 diagram, we see that

P mg mvr

− =2

Therefore,

P mvr

mg= +2

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 Wr

= + = +

18. REASONING The centripetal force Fc that keeps the car (mass = m, speed = v) on the

curve (radius = r) is 2c /F mv r= (Equation 5.3). The maximum force of static friction

MAXsF provides this centripetal force. Thus, we know that MAX 2

s /F mv r= , which can be

solved for the speed to show that MAXs /v rF m= . We can apply this result to both the dry

road and the wet road and, in so doing, obtain the desired wet-road speed. SOLUTION Applying the expression MAX

s /v rF m= to both road conditions gives

MAX MAXs, dry s, wet

dry wet and rF rF

v vm m

= =

We divide the two equations in order to eliminate the unknown mass m and unknown radius r algebraically, and we remember that MAX MAX1

s, wet s, dry3F F= :

MAX

MAX

MAXMAX

s, wets, wetwet

dry s, drys, dry

13

rFFv m

v FrFm

= = =

Solving for vwet, we obtain

drywet

21 m / s 12 m / s3 3

vv = = =

_____________________________________________________________________________________________


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 as Fc = mv2/r.

SOLUTION (a) The centripetal force is

Fc =mv2

r=

9.5 kg( ) 2.8 m/s( )2

0.85 m= 88 N

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

( )( ) ( )( )

2

c

222 9.5 kg 2.8 m/s

9.5kg 9.80 m/s 181 N0.85 m

mvF T mgr

mvT mgr

= − =

= + = + =

20. REASONING When the penny is rotating with the disk (and not sliding relative to it), it is

the static frictional force that provides the centripetal force required to keep the penny moving on a circular path. The magnitude MAX

sf of the maximum static frictional force is

given by MAXs s Nf Fµ= (Equation 4.7), where FN is the magnitude of the normal force and

µs is the coefficient of static friction. Solving this relation for µs gives

MAXs

s N

fF

µ = (1)

Since the maximum centripetal force that can act on the penny is the maximum static frictional force, we have MAX

c sF f= . Since Fc = mv2/r (Equation 5.3), it follows that MAX 2s /f mv r= . Substituting this expression into Equation (1) yields

2

MAXs

s N N

mvf rF F

µ = = (2)

The speed of the penny can be determined from the period T of the motion and the radius r according to v = 2π r/T (Equation 5.1). Furthermore, since the penny does not accelerate in the vertical direction, the upward normal force must be balanced by the downward-pointing weight, so that FN = mg, where g is the acceleration due to gravity. Substituting these two expressions for v and FN into Equation (2) gives


( )

2

2 2

s 2 N

24

rmmv rTr F r mg gT

ππµ

⎛ ⎞⎜ ⎟⎝ ⎠= = = (3)

SOLUTION Using Equation (3), we find that the coefficient of static friction required to keep the penny rotating on the disk is

( )( )( )

22

s 2 22

4 0.150 m4 0.1879.80 m/s 1.80 s

rgT

ππµ = = =

______________________________________________________________________________ 21. SSM 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 zero acceleration, while the y component of the motion is subject to the acceleration due to gravity. The horizontal distance traveled by the ball is given by Equation 3.5a (with ax = 0m/s2):

x v t v tx= =0 0( cos )θ with t equal to the flight time of the ball while it exhibits projectile motion. The time t can

be found by considering the vertical motion. From Equation 3.3b,

v v a ty y y= +0 After a time t, vy = −v0y . Assuming that up and to the right are the positive directions, we

have

tv

ava

y

y y

=−

=−2 20 0 sinθ

and

x = (v0 cosθ )−2v0 sinθ

ay

⎛

⎝⎜⎞

⎠⎟

Using the fact that 2sinθ cosθ = sin 2θ , we have

xv

avay y

= − = −2 0

202cos sin sinθ θ θ 2

(1)

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

determine the speed v0with which the ball begins its projectile motion. Then Equation 5.3 can be used to find the centripetal force.


SOLUTION Solving equation (1) for v0 , we have

vx a y

086 75 29 3=

−= −

°=

2( m)(–9.80 m / s )

sin 2(41 ) m / s

2

sin. .

θ

Then, from Equation 5.3,

FmvrC

2(7.3 kg)(29.3 m / s)1.8 m

3500 N= = =02

_____________________________________________________________________________________________ 22. REASONING The coefficient µs of static friction is related to the magnitude MAX

sf of the maximum static frictional force and the magnitude FN of the normal force acting on the car by MAX

s s Nf Fµ= (Equation 4.7), so that:

MAXs

sN

fF

µ = (1)

The car is going around an unbanked curve, so the centripetal force 2

cmvFr

=

(Equation 5.3) must be horizontal. The static frictional force is the only horizontal force, so it serves as the centripetal force. The maximum centripetal force occurs when MAX

c sF f= .

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

2

MAXc s

mvF fr

= = (2)

Substituting Equation (2) into Equation (1) yields

2

sN

mvrF

µ = (3)

In part a the car is subject to two downward-pointing forces, its weight W and the

downforce D. 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 22

sN

mv mvmvr r

F W D r mg Dµ = = =

+ + (4)

SOLUTION a. Since the downforce is D = 11 000 N, Equation (4) gives the coefficient of static friction as


( )( )( )

( ) ( )( )22

s 2

830 kg 58 m/s0.91

160 m 830 kg 9.80 m/s 11 000 N

mvr mg D

µ = = =+ ⎡ ⎤+⎣ ⎦

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 mgv

m m m

rg

µ µ µ

µ

+ += = =

= = =

23. REASONING a. The free body diagram shows the swing ride and the two forces that act on a chair: the

tension 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 force yF∑ in the vertical direction must be

zero, 0yF =∑ . 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 Newton’s second law, the net force xF∑ in the horizontal direction is

equal to the mass m of the chair and its occupant times the centripetal acceleration ( )2c /a v r= , so that 2

c /xF 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 Newton’s second law to find the speed v of the chair.

SOLUTION a. The vertical component of the tension is +T cos 60.0°, and the weight is −mg, where we

have chosen “up” as the + direction. Since the chair and its occupant have no vertical acceleration, we have that 0yF =∑ , so

60.0° 15.0 m

r

+y

+x

T

m g


+T cos60.0°− mgFy∑

= 0 (1)

Solving for the magnitude T of the tension gives

( )( )2179 kg 9.80 m/s3510 N

cos60.0 cos60.0mgT = = =

° °

b. The horizontal component of the tension is +T sin 60.0°, where we have chosen the

direction to the left in the diagram as the + direction. Since the chair and its occupant have a centripetal acceleration in this direction, we have

T sin60.0°Fx∑

= mac= m

v2

r⎛⎝⎜

⎞⎠⎟

(2)

From the drawing we see that the radius r of 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 sin 60.0sin 60.0 14.9 m/s179 kg

rTvm

°°= = =

24. REASONING The angle θ at which a friction-free curve is banked depends on the radius r

of the curve and the speed v with which the curve is to be negotiated, according to Equation 5.4: tanθ = v2 /(rg). For known values of θ and r, the safe speed is

v rg= tanθ

Before we can use this result, we must determine tan θ 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 .θ = =18 m53 m

0 34

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

friction 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 = =

25. 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 θ of the banked curve is given by ( )2

0tan /v rgθ = , where vo is the speed of the car and r is the radius of the curve (see Equation 5.4). The maximum speed that a car can have when rounding an unbanked curve is 0 sv grµ= (see Example 7). By combining these two relations, we can

find the angle θ. SOLUTION The angle of the banked curve is ( )1 2

0tan /v rgθ − ⎡ ⎤= ⎣ ⎦ . Substituting the

expression 0 sv grµ= into this equation gives

( ) ( )2

1 1 1 10 sstan tan tan tan 0.81 39

v g rrg rg

µθ µ− − − −⎛ ⎞ ⎛ ⎞= = = = = °⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

______________________________________________________________________________

26. REASONING We will treat this situation as

a circular turn on a banked surface, with the angle θ that the rider leans serving as the banking angle. The banking angle θ is related to the speed v of the watercraft, the radius r of the curve and the magnitude g of the

acceleration due to gravity by 2

tan vrg

θ =

(Equation 5.4). If the rider is closer to the seawall than r, she will hit the wall while making 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 for r, we obtain the minimum distance:

( )( )

22

2

26 m/s170 m

tan 9.80 m/s tan 22vr

g θ= = =

o

27. REASONING The relation 2

tan vrg

θ = (Equation 5.4) determines the banking angle θ that

a banked curve of radius r must have if a car is to travel around it at a speed v without

r

r

Seawall

Beginning of turn


relying on friction. In this expression g is the magnitude of the acceleration due to gravity. We will solve for v and apply the result to each curve. The fact that the radius of each curve is the same will allow us to determine the unknown speed.

SOLUTION According to Equation 5.4, we have

2tan or tanv v rg

rgθ θ= =

Applying this result for the speed to each curve gives

A A B Btan and tanv rg v rgθ θ= = Note that the terms r and g are the same for each curve. Therefore, these terms are

eliminated algebraically when we divide the two equations. We find, then, that

( )BB B BB A

A A AA

tan tan tan tan19 or 18 m/s 22 m/stan tan tan13tan

rgvv v

v rg

θ θ θθ θθ

°= = = = =°

28. REASONING The distance d is

related to the radius r of the circle on 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 the mass m of the car times the centripetal acceleration ( )2

c /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

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

mvFr mvd

F°

= = =° ° ° °

(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, 0yF =∑ . The net 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 that

50.0° d

r

+y

+x

FN

m g

Car 50.0°

40.0°


+ FN

sin50.0°− mg

Fy∑

= 0 (2)

Solving this equation for FN and substituting it into the equation above will yield the distance d.

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

( )( ) ( )( )

( )( )

2 2

N

22

2

cos50.0 sin 50.0 cos50.0 sin 50.0sin 50.0

34.0 m/s184 m

cos50.0 9.80 m/s cos50.0

mv mvdmgF

vg

= =⎛ ⎞° ° ° °⎜ ⎟°⎝ ⎠

= = =° °

______________________________________________________________________________

29. REASONING The lifting force L is perpendicular to the jet’s

wings. When the jet banks at an angle θ above the horizontal, therefore, the lifting force tilts an angle θ from the vertical (see the free-body diagram). Because the jet has no vertical acceleration during the horizontal turn, the upward vertical component L cos θ of the lifting force balances the jet’s weight: L cos θ = mg, where m is the jet’s mass and g is the acceleration due to gravity. Therefore, the magnitude of the lifting force is cosL mg θ= .

At this point we know m and g, but not the banking angle θ. Since the jet follows a horizontal circle, the centripetal force must be horizontal. The only horizontal force acting on the jet is the horizontal component L sin θ of the lifting force, so this must be the centripetal 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 rgθ = (Equation 5.4), therefore, expresses the relationship between the jet’s unknown banking angle θ, its speed v, the radius r of the turn, and g, all of which are known.

SOLUTION The magnitude of the lifting force is

( )( )5 22.00 10 kg 9.80 m/s

cos cosmgLθ θ

×= = (1)

Solving the relation ( )2tan v rgθ = (Equation 5.4) for the angle θ, we obtain

mg

L L cos θ

L sin θ

θ

Free-body diagram


( )( )( )

221 1

2123 m/s

tan tan 22.13810 m 9.80 m/s

vrg

θ − −⎡ ⎤⎛ ⎞ ⎢ ⎥= = =⎜ ⎟⎢ ⎥⎝ ⎠ ⎣ ⎦

o

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

( )( )5 262.00 10 kg 9.80 m/s

2.12 10 Ncos cos22.1mgLθ

×= = = ×

°

30. REASONING The centripetal force Fc required to keep an object of mass m that moves

with speed v on a circle of radius r is F mv rc =2 / (Equation 5.3). From Equation 5.1, we

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

Fm r T

rm rTc = =

( / )2 42 2

2

π π (1) This expression can be solved for T. However, we must first find the centripetal force that

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

force of static friction f sMAX , and the normal force FN exerted on the suitcase by the

surface of the carousel. The following drawing 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 to move 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 f s

MAX have horizontal components. Thus, we have F f Fc s

MAXN= cos – sinθ θ , where the

minus sign indicates that the x component of FN points to the left in the diagram. Using Equation

+x

+y

mg

FN

θ

θθ

fsMAX

4.7 for the maximum static frictional force, we can write this result as in Equation (2).

F F F Fc s N N N s= =µ θ θ µ θ θcos – sin ( cos – sin ) (2) If we apply Newton's second law in the y direction, we see from the diagram that

F f mg ma F F mgsMAX

y sN N Nsin or sincos – cos –θ θ θ µ θ+ = = + =0 0 where we again have used Equation 4.7 for the maximum static frictional force. Solving for

the normal force, we find


Fmg

sN sin=

+cosθ µ θ Using this result in Equation (2), we obtain the magnitude of the centripetal force that acts

on the suitcase:

F F

mg

sc N s

s

sin= =

+( cos – sin )

( cos – sin )cos

µ θ θµ θ θθ µ θ

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

mg m r

Ts

( cos – sin )cosµ θ θθ µ θ

πs

sin+= 4

2

2

Solving for the period T, we find

( ) ( )( )( )

2 2

2s

4 cos sin 4 (11.0 m) cos36.0 0.760 sin 36.045 s

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

Tgπ θ µ θ π

µ θ θ+ °+ °

= = =° °

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 r is the radius of the orbit. The radius is measured from the center of the 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 rA= 6.38×106 m+360×103 m=6.74×106 m

v =GM

E

rA

=6.67 ×10−11 N ⋅m2 / kg2( ) 5.98×1024 kg( )

6.74×106 m= 7690 m/s

Satellite B rA= 6.38×106 m+720×103 m=7.10×106 m

v =GM

E

rA

=6.67 ×10−11 N ⋅m2 / kg2( ) 5.98×1024 kg( )

7.10×106 m= 7500 m/s

_____________________________________________________________________________________________


32. REASONING The speed v of a satellite in a circular orbit of radius r about the earth is given by E /v GM r= (Equation 5.5). In this expression 11 2 26.67 10 N m / kgG −= × ⋅ is

the universal gravitational constant and 24E 5.98 10 kgM = × is the mass of the earth. The

orbital radius of a synchronous satellite in earth orbit is calculated in Example 11 to be 74.23 10 mr = × .

SOLUTION Using Equation 5.5, we find that the speed of a synchronous satellite in earth

orbit is ( )( )11 2 2 24

E7

6.67 10 N m / kg 5.98 10 kg3070 m/s

4.23 10 mGM

vr

−× ⋅ ×= = =

×

_____________________________________________________________________________________________ 33. SSM 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,

vv

GM r

GM r

rr

2

1

2

1

1

2

= =P

P

/

/

Solving for v2 gives

v vrr2 1

1

2

10 1010

10= = × ××

= ×(1.70 5.258.60

1.3346

64 m / s) m

m m / s

34. REASONING AND SOLUTION 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 letting Fc = (1/2)mg yields

r = 2v2/g = 2(35.8 m/s)2/(9.80 m/s2) = 262 m

35. REASONING The speed of the satellite is given by Equation 5.1 as 2 /v r Tπ= . Since we

are given that the period is T = 1.20 × 104 s, it will be possible to determine the speed from Equation 5.1 if we can determine the radius r of the orbit. To find the radius, we will use Equation 5.6, which relates the period to the radius according to 3/2

E2 /T r GMπ= , 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 rvTπ=

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

3/ 2E3/ 2

E

2 or 2

T GMrT rGMπ

π= =

Next, we square both sides of the result for r3/2:

( )2 22 E3/ 2 3 E

2 or 2 4

T GM T GMr r

π π

⎛ ⎞⎜ ⎟= =⎜ ⎟⎝ ⎠

We can now take the cube root of both sides of the expression for r3 in order to determine r:

( ) ( )( )24 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 4T GM

rπ π

−× × ⋅ ×= = = ×

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

( )7

34

2 1.13 10 m2 5.92 10 m/s1.20 10 s

rvT

ππ ×= = = ×

×

36. REASONING The period of a satellite is given by 3/2

E2 /T r GMπ= (Equation 5.6),

where 11 2 26.67 10 N m / kgG −= × ⋅ is the universal gravitational constant. SOLUTION Using Equation 5.6, we find that the period is

( )( )( )

3/23/2 64

11 2 2 24E

2 2 2 6.38 10 m 1.43 10 s6.67 10 N m / kg 5.98 10 kg

rTGMπ π

−

⎡ ⎤×⎣ ⎦= = = ×× ⋅ ×

_____________________________________________________________________________________________ 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

ac=

vplane2

rplane

=v

satellite2

rsatellite

or vplane

=r

plane

rsatellite

⎛

⎝⎜⎜

⎞

⎠⎟⎟

vsatellite

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

vGmrsatellite

E

satellite

=

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

plane gives

vplane

=r

plane

rsatellite

⎛

⎝⎜⎜

⎞

⎠⎟⎟

vsatellite

=r

plane

rsatellite

⎛

⎝⎜⎜

⎞

⎠⎟⎟

GmE

rsatellite

=r

planeGm

E

rsatellite

vplane

=15 m( ) 6.67 ×10−11 N ⋅m2 /kg2( ) 5.98×1024 kg( )

6.7 ×106 m= 12 m/s

38. REASONING The satellite’s true weight W when at rest on the surface of the planet is the

gravitational force the planet exerts on it. This force is given by 2PW 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 r is the distance between the satellite and the center of the planet. When the satellite is at rest on the planet’s surface, its distance from the planet’s center is RP, the radius of the planet, so we have 2

P PW GM m R= . The satellite’s mass m is

given, as is the planet’s radius RP. But we must use the relation 3 2

P

2 rTGMπ= (Equation 5.6)

to determine the planet’s mass MP in terms of the satellite’s orbital period T and orbital radius r. Squaring both sides of Equation 5.6 and solving for MP, we obtain

( )

( )

22 2 3 2 2 3 2 32

P2 2PP

2 4 4 or r r rT M

GM GTGM

π π π= = = (1)

Substituting Equation (1) into P2P

GM mW

R= (Equation 4.4), we find that

( )P2P

Gm GW MR

= =2 3

2P

4m rR G

π 2 3

2 2 2P

4 r mT R T

π⎛ ⎞=⎜ ⎟⎜ ⎟⎝ ⎠

(2)

SOLUTION All of the quantities in Equation (2), except for the period T, are given in SI base 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)] = 7.20×103 s. The satellite’s orbital radius r in Equation (2) is the distance between the satellite and the center of the orbit, which is the planet’s center. Therefore, the orbital radius is the sum of the planet’s radius RP and the satellite’s height h


above the planet’s surface: r = RP + h = 4.15×106 m + 4.1×105 m = 4.56×106 m. We now use Equation (2) to calculate the satellite’s true weight at the planet’s surface:

( ) ( )

( ) ( )

32 62 34

2 2 2 26 3P

4 4.56 10 m 5850 kg4 2.45 10 N4.15 10 m 7.20 10 s

r mWR T

ππ ×= = = ×

× ×

39. REASONING The speed v of a planet orbiting a star is given by SGMv

r=

(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 r is the orbital radius. This expression can be solved for

MS. However, the orbital radius r is not known, so we will use the relation 2 rvTπ=

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

2

2SS or

GM rvv Mr G

= = (1)

Then, solving 2 rvTπ= for r yields

2vTrπ

= , which we substitute into Equation (1):

22 3

S2

2

vT vrv v TMG G G

ππ

⎛ ⎞⎜ ⎟⎝ ⎠= = =

(2)

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

SOLUTION a. The speed of the slower planet is v = 43.3 km/s = 43.3×103 m/s. Its orbital period in seconds is T = (7.60 yr)[(3.156 × 107 s)/(1 yr)] = 2.40×108 s. Substituting these values into Equation (2) yields the mass of the star:

( ) ( )( )

33 8331

S 11 2 2

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

2 2 6.674 10 N m / kgv TMGπ π −

× ×= = = ×

× ⋅

This is roughly 23 times the mass of the sun.

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

3

SS 3

2 or

2GMv T M T

G vπ

π= = (3)


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

( )( )

( )11 2 2 31

733

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

58.6 10 m/sT

π −× ⋅ ×= = ×

×

Lastly, we convert the period from seconds to years:

79.69 10 sT = ×( ) 7

1 yr3.156 10 s×

3.07 yr⎛ ⎞

=⎜ ⎟⎝ ⎠

40. REASONING AND SOLUTION a. The centripetal acceleration of a point on the rim of chamber A is the artificial

acceleration due to gravity, aA = vA

2/rA = 10.0 m/s2 A point on the rim of chamber A moves with a speed vA = 2π rA/T where T is the period of

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

rA = aAT2/(4π2) = 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 = 2π rB/T. Substituting the second equation into the first yields

( )

( )

2 22B

2B 2

4 4 228 m 2.50 m/s60.0 s

ra

Tπ π= = =

41. 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 rC = 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 FC N= − . This expression can be solved for FN , the normal force.

SOLUTION a. The magnitude of the centripetal force is


F mvrC

2(342 kg)(25.0 m / s)126 m

1.70= = = ×2

103 N

b. The magnitude of the normal force is

2 3 3N C (342 kg)(9.80 m/s ) 1.70 10 N = 1.66 10 NF mg F= − = − × ×

_____________________________________________________________________________________________ 42. REASONING The normal force (magnitude FN) that

the pilot’s 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 centripetal force, as the drawing at the right shows. This additional contribution is the pilot’s weight (magnitude W). To obtain the ratio FN/W, we will apply Equation 5.3, which specifies the centripetal force as 2

c /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 is c NF F W= − . Thus, from Equation 5.3 we have 2

c NmvF F Wr

= − = The weight is given by W = mg (Equation 4.5), so we can divide the expression for the

centripetal force by the expression for the weight and obtain that

2 2N N

c or 1F W Fmv vFW mgr W gr−

= = − = Solving for the ratio NF /W, we find that

( )( )( )

22N

2

230 m/s1 1 8.8

9.80 m/s 690 m

F vW gr

= + = + =

43. SSM REASONING The centripetal force is the name given to the net force pointing

toward the center of the circular path. At point 3 at the top the net force pointing toward the center 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 as Fc = mv2/r.

SOLUTION At point 3 we have

FN

W

+

−


F F mgmvrc N= + = 3

2

At point 1 we have

F F mgmvrc N= − = 1

2

Subtracting the second equation from the first gives

2 32

12

mgmvr

mvr

= −

Rearranging gives v gr v32

122= +

Thus, we find that

v

3= 2 9.80 m/s2( ) 3.0 m( ) + 15 m/s( )2

= 17 m/s _______________________________________________________

______________________________________ 44. REASONING The rider’s speed v at the top of the

loop is related to the centripetal force acting on her

by 2

cmvFr

= (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 by the 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 NF mg F= + . Thus, we have that

2

Nmvmg Fr

+ = . 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 mg is equal to the reading on the sensor when level and stationary (see the “Stationary” free-body diagram).

SOLUTION Solving 2

Nmvmg Fr

+ = for the speed v, we obtain

( ) ( )N N2 or r mg F r mg F

v vm m+ +

= =

The only quantity not yet known is the rider’s mass m, so we will calculate it from her weight W by using the relation W mg= (Equation 4.5). Thus, we find that 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


( ) ( )( )N 21 m 770 N 350 N17 m/s

79 kgr mg F

vm+ +

= = =

45. REASONING The magnitude Fc of the centripetal force is 2

c /F mv r= (Equation 5.3). Since the speed v, the mass m, and the radius r are fixed, the magnitude of the centripetal force is the same at each point on the circle. When the ball is at the three o’clock position, the force of gravity, acting downward, is perpendicular to the stick and cannot contribute to the centripetal force. (See Figure 5.20 in the text, point 2, for a similar situation.) At this point, only the tension of T = 16 N contributes to the centripetal force. Considering that the centripetal force is the same everywhere, we can conclude that it has a magnitude of 16 N everywhere.

At the twelve o’clock position the tension T and the force of gravity mg both act downward

(the negative direction) toward the center of the circle, with the result that the centripetal force at this point is –T – mg. (See Figure 5.20, point 3.) At the six o’clock position the 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 T – mg. (See Figure 5.20, point 1.)

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

o’clock positions that

Twelve o'clock −T − mgCentripetal

force

= −16 N

T = 16 N − 0.20 kg( ) 9.80 m/s2( ) = 14 N

Six o'clock T − mgCentripetal

force

= 16 N

T = 16 N + 0.20 kg( ) 9.80 m/s2( ) = 18 N

46. REASONING When the stone is whirled in a horizontal circle, the centripetal force is

provided by the tension Th in the string and is given by Equation 5.3 as

+Th

Centripetalforce

= mv2

r

(1)

where m and v are the mass and speed of the stone, and r is the radius of the

circle. When the stone is whirled in a vertical circle, the maximum tension

Stone

mg

Tv


occurs when the stone is at the lowest point in its path. The free-body diagram shows the forces that act on the stone in this situation: the tension Tv in 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

+Tv− mg

Centripetalforce

= mv2

r

(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 of the 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 mvmgr 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 = = =

47. REASONING Because the crest of the hill is a circular arc, the

motorcycle’s speed v is related to the centripetal force Fc acting on the motorcycle: 2

cF mv r= (Equation 5.3), where m is the mass of the motorcycle and r is 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 the motorcycle, which points downward. The other is the normal force FN exerted by the road. The normal force points directly opposite the motorcycle’s weight. Note that the motorcycle’s 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 mg of 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 as Fc = mg − FN. With this identity, the relation

cv F r m= becomes

mg

FN

Free-body diagram of the motorcycle


( )Nmg F r

vm−

= (1)

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

( )0mg r mvm−

= = grm ( )( )29.80 m/s 45.0 m 21.0 m/sgr= = =

48. REASONING The drawing at the right

shows the two forces that act on a piece of clothing just before it loses contact with the wall of the cylinder. At that instant the centripetal force is provided by the normal force FN and the radial component of the weight. From the drawing, the radial component of the weight is given by

mg mg mg cos = cos (90 – ) = sinφ θ θ°

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

2 / r) gives

F mg mvr

2

N sin =+ θ At the instant that a piece of clothing loses contact with the surface of the drum, FN = 0N,

and the above expression becomes

mg mvr

2

sin =θ According to Equation 5.1, v = 2πr /T , and with this substitution we obtain

gr Tr

rT

2

sin =θπ π( / )2 4 2

2= This expression can be solved for the period T. Since the period is the required time for one

revolution, the number of revolutions per second can be found by calculating 1/T. SOLUTION Solving for the period, we obtain

T = 4π 2rg sinθ

= 2π rg sinθ

= 2π 0.32 m9.80 m/s2( ) sin 70.0°

= 1.17 s


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

1 1

1 17T= =

. s0.85 rev / s

49. SSM REASONING The magnitude Fc of the centripetal force that acts on the skater is

given by Equation 5.3 as 2c /F mv r= , where m and v are the mass and speed of the skater,

and r is the distance of the skater from the pivot. Since all of these variables are known, we can find the magnitude of the centripetal force.

SOLUTION The magnitude of the centripetal force is

( )( )22

c

80.0 kg 6.80 m/s606 N

6.10 mmvFr

= = =

______________________________________________________________________________ 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 T of the motion. With these data in mind, we will base our solution on Equation 5.2, which gives the centripetal acceleration as 2

c /a v r= , and on Equation 5.1, which specifies that the speed v on a circular path of radius r is 2 /v r Tπ= .

SOLUTION From Equation 5.2, we have

2 2

cc

or v va rr a

= =

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

( ) ( )( )222 22c

2 2c c

9.80 m/s 30.0 s2 / or 223 m

4 4r T a Tvr r

a aπ

π π= = = = =

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

and the period T of the tip of a moving clock hand can be obtained by using Equations 5.2 and 5.1:

2

c2(5.2) (5.1)v ra v

r Tπ= =


The period is the time it takes a clock hand to go once around the circle. In these expressions, v is the speed of the tip of the hand and r is the length of the hand. Substituting Equation 5.1 into Equation 5.2 yields

2

2 2

c 2

24

rv rTar r T

ππ

⎛ ⎞⎜ ⎟⎝ ⎠= = = (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

2minute

43600 s

36004 60 s

ra T Ta r T

T

π

π= = = =

_____________________________________________________________________________________________ 52. REASONING AND SOLUTION a. In terms of the period of the motion, the centripetal force is written as

Fc = 4π2mr/T2 = 4π2 (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 In Section 5.5 it is shown that the period T of a satellite in a circular orbit

about the earth is given by (see Equation 5.6) 3/ 2

E

2 rTGMπ=

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

A3/2

EA A3/2 3/2

BB B

E

2

2

rGMT rrT rGM

π

π= = (1)


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 (2π r) of its orbit divided by the period T, so 2 /v r Tπ= .

SOLUTION Solving the relation 2 /v r Tπ= for r gives / 2r vT π= . Substituting this value

for r into Equation (1) yields ( )( )

( )( )

3/ 2 3/ 23/2A A A AA A

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

/ 2

/ 2

v T v TT rT r v Tv T

π

π

⎡ ⎤⎣ ⎦= = =⎡ ⎤⎣ ⎦

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 B33

B A B

1 = = 273

T v vT v v

=

______________________________________________________________________________ 54. REASONING The centripetal acceleration depends on the speed v and the radius r of the

curve, according to ac = v2/r (Equation 5.2). The speeds of the cars are the same, and since they are negotiating the same curve, the radius is the same. Therefore, the cars have the same centripetal acceleration. However, the magnitude Fc of the centripetal force depends on the mass m of the car, as well as the speed and the radius of the curve, according to 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 centripetal

accelerations and forces:

Car A ac =v2

r=

27 m/s( )2

120 m= 6.1 m/s2 (5.2)

Fc =mAv

2

r=

1100 kg( ) 27 m/s( )2

120 m= 6700 N (5.3)

Car B ac =v2

r=

27 m/s( )2

120 m= 6.1 m/s2 (5.2)

Fc =mBv

2

r=

1600 kg( ) 27 m/s( )2

120 m= 9700 N (5.3)

_____________________________________________________________________________________________


55. SSM REASONING According to Equation 5.3, the magnitude Fc of the centripetal force that acts on each passenger is 2

c /F mv r= , where m and v are the mass and speed of a passenger and r is the radius of the turn. From this relation we see that the speed is given by c /v F r m= . The centripetal force is the net force required to keep each passenger moving on the circular path and points toward the center of the circle. With the aid 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 taken “up” as the positive direction. Thus, Fc = mg. The speed of the passenger can be found by using this result in the equation above.

Substituting Fc = mg into the relation c /v F r m= yields

( ) ( )( )2c 9.80 m/s 20.0 m 14.0 m/smg rF r

v g rm m

= = = = =

56. 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/sva v a r

r⎡ ⎤= = = =⎣ ⎦

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

blade a distance r from center of the circle, according to Equation 5.2, is ac = v2 / r . From

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

ar Tr

rTc = =

( / )2 42 2

2

π π

Passenger

r

mg

2 mg


a. Since the turbine blades rotate at 617 rev/s, all points on the blades rotate with a period of T = (1/617) s = 1.62 ×10œ3 s. Therefore, for a point with r = 0.020 m, the magnitude of the centripetal acceleration is

ac m)

(1.623.0=

×= ×4 0 020

1010

2

2

π ( .–3

5 2

s) m / s

b. Expressed as a multiple of g, this centripetal acceleration is

ac = 3.0 ×105 m / s2( ) 1.00 g

9.80 m/s2⎛⎝⎜

⎞⎠⎟ = 3.1×104 g

____________________________________________________________________________________ 58. REASONING The magnitude of the centripetal acceleration of any point on the helicopter

blade is given by Equation 5.2, aC = v2 / r, where r is the radius of the circle on which that point moves. From Equation 5.1: v r T= 2π / . Combining these two expressions, we obtain

a rTC = 42

2

π 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 that

which exists at a point located 3.0 m from the center of the circle (point 2) is

aa

r Tr T

rr

C1

C2

6.7 m3.0 m

2.2= = = =44

21

2

22

21

2

ππ

//

_____________________________________________________________________________________________

59. REASONING The drawing shows the block (mass m) hanging as it does when the van goes around the curve. Two forces act on the block, the tension T in the string and its weight mg. The upward vertical component of the tension is T cos θ. The horizontal component of the tension is T sin θ and points to the left, toward the center of the circular curve. Since the block does not accelerate in the vertical direction, the upward vertical component of the tension balances the downward-directed weight. The horizontal component of the tension provides the centripetal force that keeps the block moving on its circular path in the horizontal plane.

SOLUTION Since the horizontal component of the tension provides the centripetal force,

Equation 5.3 can be written as follows:

T cos θ

T sin θ

T θ

θ

mg


T sinθCentripetal

force

= mv2

r (1)

where v is the speed of the van and r is the radius of the curve. Since the vertical component

of the tension balances the weight, we have cosT mgθ = (2)

Dividing Equation (1) by Equation (2), we obtain

2 2sin / sin or tancos cosT mv r vT mg rg

θ θ θθ θ= = =

This result indicates that

( )( )( )

221 1

228 m / s

tan tan 28150 m 9.80 m/s

vrg

θ − −⎡ ⎤⎛ ⎞ ⎢ ⎥= = = °⎜ ⎟⎢ ⎥⎝ ⎠ ⎣ ⎦

60. REASONING We seek the ratio m1/m2of the masses. The mass appears in the centripetal

force expression 2c /F mv r= (Equation 5.3), so we begin by applying it to each of the

particles. The centripetal force Fc arises because of the tension in each section of the rod. The tension in the inner section is greater than the tension in the outer section because the inner section not only keeps particle 1 on its circular path, but also helps to keep particle 2 on its circular path. We note the following three items for use in our solution: • The drawing shows the rod and the particles. The tension in the inner section has a

magnitude of TI, whereas the tension in the outer section has a magnitude of TO. The tension in the inner section causes a leftward-pointing force of magnitude TI to be exerted on particle 1. Similarly, the tension in the outer section causes a leftward-pointing force of magnitude TO to be exerted on particle 2. However, the tension in the outer section also causes a rightward-pointing force of magnitude TO to be exerted on particle 1.

• Since particle 1 is located at the rod’s center and particle 2 is located at the rod’s end, the distances r1 and r2 of the particles from the end (axis) about which the rod is rotating are related according to 1 2/ 1/ 2r r = .

• The speeds v1 and v2 of the particles are also related, since the rod and the particles rotate as a rigid unit. Consider that each particle travels the circumference of its circle in the same time. Since the circumference is 2 rπ and 1 2/ 1/ 2r r = , it follows that

1 2/ 1/ 2v v = .

Axis TO TO TI

m2 m1


SOLUTION Applying Equation 5.3 to each particle, we obtain

TI −TO

Centripetalforce

=

m1v12

r1 and TO

Centripetalforce

=

m2v22

r2

Dividing the left equation by the right equation gives

2 2 2I O 1 1 1 1 1 2 I 1 1 2

2 2 2O O2 2 2 2 2 1 2 2 1

/ or 1

/T T m v r m v r T m v rT Tm v r m v r m v r−

= = − =

Into this result we can now substitute I o/ 3T T = and 1 2/ 1/ 2v v = and 1 2/ 1/ 2r r = :

21 1 1

2 2 2

1 2 13 1 or 2 or 42 1 2

m m mm m m

⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

_____________________________________________________________________________________________

61. SSM REASONING If the effects of gravity are not ignored in Example 5, the plane will make an angle θ with the vertical as shown in figure A below. The figure B shows the forces that act on the plane, and figure C shows the horizontal and vertical components of these forces.

mg

T cos θ

T sin θ

mg

T θ

θ

r

L L

A B C From figure C we see that the resultant force in the horizontal direction is the horizontal

component of the tension in the guideline and provides the centripetal force. Therefore,

T mvr

sin = θ2

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

therefore,

T mvL

=

sinsin

θθ

2

(1)

The resultant force in the vertical direction is zero: T cosθ −mg = 0, so that


T cosθ =mg (2) From equation (2) we have

Tmg

=cos θ

(3)

Equation (3) contains two unknown, T and θ . First we will solve equations (1) and (3)

simultaneously to determine the value(s) of the angle θ . Once θ is known, we can calculate the tension using equation (3).

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

mgcos θ

⎛⎝⎜

⎞⎠⎟ sin θ = mv2

L sin θ

Thus,

sin cos

=

2 θθ

vgL

2

(4)

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

1 – cos cos

= 2 θθ

vgL

2

or 1

cos – cos =

θθ v

gL

2

This can be put in the form of an equation that is quadratic in cos θ. Multiplying both sides

by cos θ and rearranging yields:

cos + cos – 1 = 02 θ θv

gL

2

(5)

Equation (5) is of the form ax bx c2 0+ + = (6)

with x = cos θ, a = 1, b = v2/(gL), and c = –1. The solution to equation (6) is found from the

quadratic formula:

x b b aca

= – ± −2 42

When v = 19.0 m/s, b = 2.17. The positive root from the quadratic formula gives

x = cos θ = 0.391. Substitution into equation (3) yields


Tmg

= = =cos

)

(0.900 kg)(9.80 m / s0.391

2

θ23 N

When v = 38.0 m/s, b = 8.67. The positive root from the quadratic formula gives

x = cos θ = 0.114. Substitution into equation (3) yields

Tmg

= = =cos

)

(0.900 kg)(9.80 m / s0.114

2

θ77 N

_____________________________________________________________________________________________ 62. REASONING The value given for either acceleration corresponds to the centripetal

acceleration ac = v2 / r (Equation 5.2) in the corresponding ring. This expression can be

solved for the radius r. However, we are given no direct information about the speed v. Instead, it is stated that the period of rotation T is chosen so that the outer ring simulates gravity on earth. Although no value is given for T, we know that the laboratory is a rigid structure, so that all points on it make one revolution in the same time. This is an important observation, because it means that both rings have the same value for T. Thus, we will be able to use the fact that v = 2πr /T (Equation 5.1) together with ac = v

2 / r in order to find the radius of the inner ring.

SOLUTION According to Equation 5.2, the centripetal acceleration of the inner ring is ac,I = vI

2 / rI , which can be solved for the radius r1:

rI =vI2

ac,I

The speed vI is given by vI = 2πrI /T (Equation 5.1). With this substitution, the expression for the inner radius becomes:

rI =(2πrI /T )

2

ac,I

Solving for rI gives the following: rI =ac,IT

2

4π 2 . The term T2 in this expression is unknown,

but we can solve for it using the information about the outer ring. Both the inner and outer rings have the same period. Repeating the same sequence of steps for the outer

ring, we find: rO =ac,OT

2

4π 2 , which we can then solve for T2: T 2 = 4π2rO

ac,O. We can now

substitute this result into the expression for rI:


rI =ac,IT

2

4π 2 =ac,I(4π

2rO / ac,O )4π 2 =

ac,IrOac,O

= (3.72 m/s2 )(2150 m)9.80 m/s2 = 816 m

_____________________________________________________________________________ 63. SSM CONCEPTS (i) Acceleration is a vector and, for it to be constant, both its

magnitude and direction must be constant. Automobile A has a constant acceleration because the magnitude of its accelerations is constant (3.5 m/s2), and its direction always points forward along the straight road. Automobile B has an acceleration with a constant magnitude (again, 3.5 m/s2), but the acceleration does not have a constant direction. Automobile B is in uniform circular motion, so it has a centripetal acceleration which points towards the center of the circle at every instant. Therefore, for automobile B, the direction of the acceleration vector continually changes, and the vector is not constant. (ii) The equations of kinematics apply for automobile A because it has a constant acceleration, which must be the case when you apply these equations. They do not apply to automobile B because it does not have a constant a constant acceleration. CALCULATIONS To determine the speed of automobile A at t = 2.0 s we use Equation 2.4 from the equations of kinematics: To determine the speed of automobile B, we note that this car is in uniform circular motion and, therefore, has a constant speed as it goes around the turn. At a time of t = 2.0 s its speed is the same as at t = 0 s , or

_____________________________________________________________________________ 64. CONCEPTS (i) As part (b) of the figure illustrates, only a single tension force of magnitude

TA acts on ball A. It points to the left in the drawing and is due to the tension in the rod between the two balls. This force provides the centripetal force keeping ball A on its circular path of radius 2L. (ii) Part (c) of the figure shows two tension forces acting on ball B. One has a magnitude TB and points to the left in the drawing. The other has a magnitude TA and points to the right. It is due to the tension in the right half of the rod. The centripetal force acting on ball B points toward the center of the circle and is the vector sum of these two forces, or TB −TA . (Note here that we are treating the vectors in one dimension, so we

v = v0 + at = 18 m/s( ) + 3.5 m/s2( ) 2.0 s( ) = 25 m/s

v = 18 m/s


omit the vector signs, however, the sign indicates the direction). (iii) No, it is not. The reason is that ball A travels farther than ball B in the same time. Consider a time of one period, which is the same for either ball, since the arrangement is a rigid unit. The period is the time for either ball to travel once around its circular path. In this

time, ball A travels a distance equal to the circumference of its path, which is 2π 2L( ) . In contrast, the circumference for ball B is just2πL . Therefore, the speed of ball B is one-half the speed of ball A, or vB = 2.5 m/s . CALCULATIONS Applying Equation 5.3 to ball A, we have: And for ball B: The tension in the right half of the rod follows directly from the first equation: The tension in the left half of the rod is obtained by adding the equations for both balls:

_____________________________________________________________________________

TA

Centripetal Force, Fc

= mvA2

2L

TB −TA

Centripetal Force, Fc

= mvB2

L

TA = mvA2

2L=

0.50 kg( ) 5.0 m/s2( )2

2 0.40 m( ) = 16 N

TB = mvB2

L+ mvA

2

2L=

0.50 kg( ) 2.5 m/s2( )2

0.40 m+

0.50 kg( ) 5.0 m/s2( )2

2 0.40 m( ) = 23 N

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