sm chapter6

6Circular Motion and Other Applications of Newton’s Laws

CHAPTER OUTLINE

6.1 Newton’s Second Law Applied to Uniform Circular Motion

6.2 Nonuniform Circular Motion6.3 Motion in Accelerated Frames6.4 Motion in the Presence of

Resistive Forces

ANSWERS TO QUESTIONS

*Q6.1 (i) nonzero. Its direction of motion is changing. (ii) zero.Its speed is not changing. (iii) zero: when v = 0, v2�r = 0.(iv) nonzero: its velocity is changing from, say 0.1 m �s north to 0.1 m �s south.

Q6.2 (a) The object will move in a circle at a constant speed. (b) The object will move in a straight line at a changing

speed.

Q6.3 The speed changes. The tangential force component causes tangential acceleration.

*Q6.4 (a) A > C = D > B = E. At constant speed, centripetal acceleration is largest when radius is smallest.A straight path has infi nite radius of curvature. (b) Velocity is north at A, west at B, and south at C.(c) Acceleration is west at A, nonexistent at B, and east at C, to be radially inward.

*Q6.5 (a) yes, point C. Total acceleration here is centripetal acceleration, straight up. (b) yes, point A. Total acceleration here is tangential acceleration, to the right and downward perpendicular to the cord. (c) No. (d) yes, point B. Total acceleration here is to the right and upward.

Q6.6 I would not accept that statement for two reasons. First, to be “beyond the pull of gravity,” one would have to be infi nitely far away from all other matter. Second, astronauts in orbit are moving in a circular path. It is the gravitational pull of Earth on the astronauts that keeps them in orbit. In the space shuttle, just above the atmosphere, gravity is only slightly weaker than at the Earth’s surface. Gravity does its job most clearly on an orbiting spacecraft, because the craft feels no other forces and is in free fall.

Q6.7 This is the same principle as the centrifuge. All the material inside the cylinder tends to move along a straight-line path, but the walls of the cylinder exert an inward force to keep everything moving around in a circular path.

Q6.8 The water has inertia. The water tends to move along a straight line, but the bucket pulls it in and around in a circle.

Q6.9 Blood pressure cannot supply the force necessary both to balance the gravitational force and to provide the centripetal acceleration, to keep blood fl owing up to the pilot’s brain.

*Q6.10 (a) The keys shift backward relative to the student’s hand. The cord then pulls the keys upward and forward, to make them gain speed horizontally forward along with the airplane.

(b) The angle stays constant while the plane has constant acceleration. This experiment is described in the book Science from Your Airplane Window by Elizabeth Wood.

127

13794_06_ch06_p127-152.indd 12713794_06_ch06_p127-152.indd 127 12/2/06 1:58:33 PM12/2/06 1:58:33 PM

128 Chapter 6

Q6.11 The person in the elevator is in an accelerating reference frame. The apparent acceleration due to gravity, “g,” is changed inside the elevator. “g” = ±g a

Q6.12 From the proportionality of the drag force to the speed squared and from Newton’s second law, we derive the equation that describes the motion of the skydiver:

md

dtmg

D Ayy

vv= − ρ

22

where D is the coeffi cient of drag of the parachutist, and A is the projected area of the parachu-tist’s body. At terminal speed,

ad

dtyy= =

v0

and vT

mg

D A=

⎛⎝⎜

⎞⎠⎟

21 2

ρ

When the parachute opens, the coeffi cient of drag D and the effective area A both increase, thus reducing the speed of the skydiver.

Modern parachutes also add a third term, lift, to change the equation to

md

dtmg

D A L Ayy x

vv v= − −ρ ρ

2 22 2

where vy is the vertical velocity, and vx is the horizontal velocity. The effect of lift is clearly seen

in the “paraplane,” an ultralight airplane made from a fan, a chair, and a parachute.

Q6.13 (a) Static friction exerted by the roadway where it meets the rubber tires accelerates the car forward and then maintains its speed by counterbalancing resistance forces. (b) The air around the propeller pushes forward on its blades. Evidence is that the propeller blade pushes the air toward the back of the plane. (c) The water pushes the blade of the oar toward the bow. Evi-dence is that the blade of the oar pushes the water toward the stern.

Q6.14 The larger drop has higher terminal speed. In the case of spheres, the text demonstrates that terminal speed is proportional to the square root of radius. When moving with terminal speed, an object is in equilibrium and has zero acceleration.

*Q6.15 (a) Speed increases, before she reaches terminal speed. (b) The magnitude of acceleration decreases, as the air resistance force increases to counterbalance more and more of the gravita-tional force.

Q6.16 The thesis is false. The moment of decay of a radioactive atomic nucleus (for example) cannot be predicted. Quantum mechanics implies that the future is indeterminate. On the other hand, our sense of free will, of being able to make choices for ourselves that can appear to be random, may be an illusion. It may have nothing to do with the subatomic randomness described by quantum mechanics.


Circular Motion and Other Applications of Newton’s Laws 129

SOLUTIONS TO PROBLEMS

Section 6.1 Newton’s Second Law Applied to Uniform Circular Motion

P6.1 m = 3 00. kg, r = 0 800. m. The string will break if the tension exceeds the weight corresponding to 25.0 kg, so

T Mgmax . .= = ( ) =25 0 9 80 245 N

When the 3.00 kg mass rotates in a horizontal circle, the tension causes the centripetal acceleration,

so Tm

r= = ( )v v2 23 00

0 800

.

.

Then

v2 0 800

3 00

0 800

3 00

0 800 24= = ( )≤ ( )

=rT

m

T T.

.

.

.

.max 55

3 0065 3

( )=

.. m s2 2

and 0 65 3≤ ≤v .

or 0 8 08≤ ≤v . m s

P6.2 In F mr∑ = v2

, both m and r are unknown but remain constant. Therefore, F∑ is proportional

to v2 and increases by a factor of 18 0

14 0

2.

.⎛⎝

⎞⎠ as v increases from 14.0 m �s to 18.0 m �s. The total

force at the higher speed is then

Ffast N N∑ = ⎛⎝

⎞⎠ ( ) =18 0

14 0130 215

2.

.

Symbolically, write Fm

rslow m s∑ = ⎛⎝

⎞⎠ ( )14 0

2. and F

m

rfast m s∑ = ⎛⎝

⎞⎠ ( )18 0

2. .

Dividing gives F

Ffast

slow

∑∑

= ⎛⎝

⎞⎠

18 0

14 0

2.

., or

F Ffast slow∑ ∑= ⎛⎝

⎞⎠ = ⎛

⎝⎞⎠

18 0

14 0

18 0

14 0130

2 2.

.

.

.N N( ) = 215

This force must be horizontally inward to produce the driver’s centripetal acceleration.

P6.3 (a) Fm

r= =

×( ) ×( )×

−v2 31 6 29 11 10 2 20 10

0 530 1

. .

.

kg m s

008 32 1010

8−

−= ×m

N inward.

(b) ar

= =×( )

×= ×−

v2 6 2

10222 20 10

0 530 109 13 10

.

..

m s

mm s inward2

P6.4 (a) F may y∑ = , mgm

rmoon down down= v2

v = = ( ) × + ×( ) =g rmoon2m s m m1 52 1 7 10 100 10 16 3. . .665 103× m s

(b) v = 2πr

T, T =

×( )×

= × =2 1 8 10

1 65 106 84 10 1 90

6

33π .

.. .

m

m ss hh

FIG. P6.1


130 Chapter 6

FIG. P6.9

P6.5 (a) static friction

(b) ma f n mgˆ ˆ ˆ ˆi i j j= + + −( ) F n mgy∑ = = −0

thus n mg= and F mr

f n mgr∑ = = = =v2

µ µ .

Then µ = =( )

( )( ) =v2 250 0

30 0 9800 085

rg

.

..

cm s

cm cm s2 00 .

P6.6 Neglecting relativistic effects. F mam

rc= = v2

F = × ×( ) ×( )(

−2 1 661 102 998 10

0 48027

7 2

..

.kg

m s

m)) = × −6 22 10 12. N

P6.7 Standing on the inner surface of the rim, and moving with it, each person will feel a normal force exerted by the rim. This inward force causes the 3 00. m s2 centripetal acceleration:

a rc = v2/ so v = = ( )( ) =a rc 3 00 60 0 13 4. . .m s m m s2

The period of rotation comes from v = 2πr

T: T

r= = ( )=2 2 60 0

13 428 1

π πv

.

..

m

m ss

so the frequency of rotation is fT

= = = ⎛⎝

⎞⎠ =1 1

28 1

1

28 1

602 14

. ..

s s

s

1 minrev miin .

P6.8 T mgcos . . .5 00 80 0 9 80° = = ( )( )kg m s2

(a) T = 787 N: �T i j= ( ) + ( )68 6 784. ˆ ˆN N

(b) T macsin .5 00° = : ac = 0 857. m s2 toward the center of the circle.

The length of the wire is unnecessary information. We could, on the other hand, use it to fi nd the radius of the circle, the speed of the bob, and the period of the motion.

P6.9 n mg= since ay = 0

The force causing the centripetal acceleration is the frictional force f.

From Newton’s second law f mam

rc= = v2

.

But the friction condition is f ns≤ µ

i.e., m

rmgs

v2

≤ µ

v ≤ = ( )( )µsrg 0 600 35 0 9 80. . .m m s2

v ≤ 14 3. m s

FIG. P6.8



P6.10 (a) v = =2356 53

m

36.0 sm s.

The radius is given by 1

42 235πr = m so r = 150 m

(b) �ar r

=⎛⎝⎜

⎞⎠⎟

=( )

v2

26 53

150

toward center

m s

m

.at 35.0 north of west

m s2

°

°= ( )0 285 35 0. cos . −−( ) +( )= − +

ˆ sin . ˆ

. ˆ . ˆ

i j

i

35 0

0 233 0 163

°

m s m s2 2 jj

(c) �

� �a

v v

j i

a g

f i

tv =−( )

=−( )6 53 6 53

36 0

. ˆ . ˆ

.

m s m s

ss

m s m s2 2= − +0 181 0 181. ˆ . ˆi j

P6.11 F mgg = = ( )( ) =4 9 8 39 2kg m s N2. .

sin.

.

θ

θ

=

=

1 5

48 6

m

2 m°

r = ( ) =2 48 6 1 32m mcos . .°

F mam

r

T T

x x

a b

∑ = =

+ =( )

v2

48 6 48 64 6

cos . cos .° °kg m ss

mN

N

( )

+ = =

2

1 32109

48 6165

.

cos .T Ta b °

F ma

T T

T T

y y

a b

a b

∑ =

+ − − =

−

sin . sin . .48 6 48 6 39 2 0° ° N

== =39 2

48 652 3

.

sin ..

NN

°

(a) To solve simultaneously, we add the equations in Ta and Tb:

T T T Ta b a b+ + − = +165 52 3N N.

Ta = =217108

N

2N

(b) T Tb a= − = − =165 165 108 56 2N N N N.

FIG. P6.11

θ

39.2 N

T a

Tb

forces

motion

vac


132 Chapter 6

Section 6.2 Nonuniform Circular Motion

P6.12 (a) arc = =

( ) =v2 24 00

1 33.

.m s

12.0 mm s2

(b) a a ac t= +2 2

a = ( ) + ( ) =1 33 1 20 1 792 2. . . m s2

at an angleθ =⎛⎝⎜

⎞⎠⎟

=−tan .1 48 0a

ac

t

° inward

P6.13 M = 40 0. kg, R = 3 00. m, T = 350 N

(a) F T MgM

R∑ = − =22v

v2 2= −( )⎛⎝

⎞⎠T Mg

R

M

v2 700 40 0 9 803 00

40 023 1= − ( )( )[ ]⎛

⎝⎞⎠ =. .

.

.. m s2 2(( )

v = 4 81. m s

(b) n Mg FM

R− = = v2

n MgM

R= + = +⎛

⎝⎞⎠ =v2

40 0 9 8023 1

3 00700. .

.

.N

P6.14 (a) v = 20 0. m s,

n = force of track on roller coaster, and

R = 10 0. m.

F

M

Rn Mg∑ = = −v2

From this we fi nd

n MgM

R= + = ( )( ) +

( )v2

500 9 80500 20 0

kg m skg m2.

. ss

m

N N N

2( )

= + = ×10 0

4 900 20 000 2 49 104

.

.n

(b) At B, n MgM

R− = − v2

The maximum speed at B corresponds to

n

MgM

RRg

=

− = − ⇒ = = ( ) =

0

15 0 9 80 12 12v

vmaxmax . . . m s

FIG. P6.12

FIG. P6.13(a)

T T

Mg

child + seat

FIG. P6.13(b)

Mg

child alone

n

FIG. P6.14

10 m 15 m

B

A

C



continued on next page

FIG. P6.15

P6.15 Let the tension at the lowest point be T.

F ma T mg mam

r

T m gr

T

c∑ = − = =

= +⎛⎝⎜

⎞⎠⎟

= (

:

.

v

v

2

2

85 0 kg)) +( )⎡

⎣⎢

⎤

⎦⎥ = >9 80

8 00

10 01 38 1

2

..

..m s

m s

mkN2 0000 N

He doesn’t make it across the river because the vine breaks.

*P6.16 (a) Consider radial forces on the object, taking inward as positive.

ΣF ma

T r

r r=

− = =

:

( . )( . ) cos .0 5 9 8 20 0 5kg °m s2 2mv kkg m

N N N

( )

. . .

8 2

4 60 16 0 20 6

2m s

T = + =

(b) We already found the radial component of acceleration,

( ) . .8 2 32 0m s m s2 2m inward=

Consider tangential forces on the object.

ΣF ma

a

a

t t

t

t

=

=

=

:

( . )( . )sin .0 5 9 8 20 0 5kg m s kg2

33 35. tanm s2 downward gent to the circle

(c) a = +[ . ]32 3 352 2 1 2 2m s inward and below the cord at angle tan ( . )−1 3 35 32

= 32 2. m s2 inward and below the cord at 5.98°

(d) No change. If the object is swinging down it is gaining speed. If the object is swinging up it is losing speed but its acceleration is the same size and its direction can be described in the same terms.

P6.17 Fm

rmg ny∑ = = +v2

But n = 0 at this minimum speed condition, so

m

rmg gr

vv

2

9 80 1 00 3 13= ⇒ = = ( )( ) =. . .m s m m s2

P6.18 (a) arc = v2

rac

= =( )( ) =v2 213 0

2 9 808 62

.

..

m s

m sm2

(b) Let n be the force exerted by the rail.Newton’s second law gives

Mg n

M

r+ = v2

n Mr

g M g g Mg= −⎛

⎝⎜⎞⎠⎟

= −( ) =v2

2 , downward

FIG. P6.17

FIG. P6.18


134 Chapter 6

(c) arc = v2

ac =( ) =13 0

20 08 45

2.

..

m s

mm s2

If the force exerted by the rail is n1

then n MgM

rMac1

2

+ = =v

n M a gc1 = −( ) which is < 0 since ac = 8 45. m s2

Thus, the normal force would have to point away from the center of the curve. Unless they have belts, the riders will fall from the cars.

To be safe we must require n1 to be positive. Then a gc > . We need

v2

rg> or v > = ( )( )rg 20 0 9 80. .m m s2 , v > 14 0. m s

Section 6.3 Motion in Accelerated Frames

P6.19 (a) F Max∑ = ,

aT

M= = =18 0

3 60.

.N

5.00 kgm s2 to the

right.

(b) If v = const, a = 0, so T = 0

(This is also an equilibrium situation.)

(c) Someone in the car (noninertial observer) claims that the forces on the mass along x are T and a fi ctitious force (–Ma). Someone at rest outside the car (inertial observer) claims that T is the only force on M in the x-direction.

P6.20 The water moves at speed

v = = ( )=2 2 0 12

7 250 104

π πr

T

.

..

m

sm s.

The top layer of water feels a downward force of gravity mg and an outward fi ctitious force in the turntable frame of reference,

m

r

mm

v2 220 104

0 129 01 10=

( ) = × −.

..

m s

mm s2

It behaves as if it were stationary in a gravity fi eld pointing downward and outward at

tan.

..− =1 0 090 1

9 80 527

m s

m s

2

2 °

Its surface slopes upward toward the outside, making this angle with the horizontal.

FIG. P6.19

5.00 kg



P6.21 The only forces acting on the suspended object are the force of gravity m�g and the

force of tension T forward and upward at angle θ with the vertical, as shown in the free-body diagram. Applying Newton’s second law in the x and y directions,

F T max∑ = =sinθ (1)

F T mgy∑ = − =cosθ 0

or T mgcosθ = (2)

(a) Dividing equation (1) by (2) gives

tan.

..θ = = =a

g

3 00

9 800 306

m s

m s

2

2

Solving for θ, θ = 17 0. °

(b) From Equation (1),

Tma= =

( )( )( ) =

sin

. .

sin ..

θ0 500 3 00

17 05 1

kg m s2

°22 N

P6.22 Consider forces on the backpack as it slides in the Earth frame of reference.

F ma n mg ma n m g a f m g a

Fy y k k

x

∑∑

= + − = = +( ) = +( )=

: , , µmma m g a max k x: − +( ) =µ

The motion across the fl oor is described by L t a t t g a tx k= + = − +( )v v1

2

1

22 2µ .

We solve for µk : vt L g a tk− = +( )1

22µ , 2

2

vt L

g a t k

−( )+( ) = µ .

P6.23 F F magmax = + = 591 N

F F magmin = − = 391 N

(a) Adding, 2 982Fg = N, Fg = 491 N

(b) Since F mgg = , m = =49150 1

N

9.80 m skg2 .

(c) Subtracting the above equations,

2 200ma = N ∴ =a 2 00. m s2

FIG. P6.21

T cos

T sinmg

θ

θ


136 Chapter 6

P6.24 In an inertial reference frame, the girl is accelerating horizontally inward at

v2 25 70

2 4013 5

r=

( ) =.

..

m s

mm s2

In her own non-inertial frame, her head feels a horizontally outward fi ctitious force equal to its mass times this acceleration. Together this force and the weight of her head add to have a magni-tude equal to the mass of her head times an acceleration of

gr

22 2

2 29 80 13 5 16 7+⎛⎝⎜

⎞⎠⎟

= ( ) + ( ) =v. . .m s m s2 2

This is larger than g by a factor of 16 7

9 801 71

.

..= .

Thus, the force required to lift her head is larger by this factor, or the required force is

F = ( ) =1 71 55 0 93 8. . .N N .

P6.25 aR

Tre=

⎛⎝⎜

⎞⎠⎟

=4

35 0 0 027 62

2

πcos . .° m s2

We take the y axis along the local vertical.

a a

a

y r y

x

net2

net

m s

m

( ) = − ( ) =

( ) =

9 80 9 77

0 015 8

. .

. ss2

θ = =arctan .a

ax

y

0 092 8°

Section 6.4 Motion in the Presence of Resistive Forces

P6.26 m = 80 0. kg, vT = 50 0. m s, mgD A D A mgT

T

= ∴ = =ρ ρvv

2

22 20 314. kg m

(a) At v = 30 0. m s

a gD A

m= − = − ( )( )

=ρ v2 229 80

0 314 30 0

80 06 27

/.

. .

.. m ss downward2

(b) At v = 50 0. m s, terminal velocity has been reached.

F mg R

R mg

y∑ = = −

⇒ = = ( )( ) =

0

80 0 9 80 784. .kg m s N d2 iirected up

(c) At v = 30 0. m s

D Aρ v2

2

20 314 30 0 283= ( )( ) =. . N upward

FIG. P6.25

N

ar

g0

anet

Equator35.0°

θ

35.0°(exaggerated size)



P6.27 (a) a g b= − v

When v v= T , a = 0 and g b T= v bg

T

=v

The Styrofoam falls 1.50 m at constant speed vT in 5.00 s.

Thus,

vT

y

t= = =1 50

0 300.

.m

5.00 sm s

Then

b = = −9 80

0 30032 7 1.

..

m s

m ss

2

(b) At t = 0, v = 0 and a g= = 9 80. m s2 down

(c) When v = 0 150. m s, a g b= − = − ( )( ) =−v 9 80 32 7 0 150 4 901. . . .m s s m s m2 ss2 down

P6.28 (a) ρ = m

V, A = 0 020 1. m2, R AD mgT= 1 =

22ρair v

m V= = ( )⎡⎣⎢

⎤⎦⎥

=ρ πbead3g cm cm0 830

4

38 00 1 783. . . kg

Assuming a drag coeffi cient of D = 0 500. for this spherical object, and taking the density of air at 20°C from the endpapers, we have

vT =( )( )( )

2 1 78 9 80

0 500 1 20 0 02

. .

. . .

kg m s

kg m

2

3 00 153 8

mm s2( ) = .

(b) v vf i gh gh2 2 2 0 2= + = + : hgf= =

( )( ) =

v2 2

2

53 8

2 9 80148

.

.

m s

m sm2

P6.29 Since the upward velocity is constant, the resultant force on the ball is zero. Thus, the upward applied force equals the sum of the gravitational and drag forces (both downward):

F mg b= + v.

The mass of the copper ball is

mr= = ⎛

⎝⎞⎠ ×( ) ×( −4

3

4

38 92 10 2 00 10

33 2πρ π . .kg m m3 )) =

30 299. kg

The applied force is then

F mg b= + = ( )( ) + ( ) ×( ) =−v 0 299 9 80 0 950 9 00 10 32. . . . .001 N

P6.30 The resistive force is

R D A= = ( )( )( )1

2

1

20 250 1 20 2 20 27 82ρ v . . . .kg m m3 2 m s

N

N

1200 kgm s

( )=

= − = − = −

2

255

2550 212

R

aR

m. 22


138 Chapter 6

P6.31 (a) At terminal velocity, R b mgT= =v

∴ = =×( )( )

×

−

−bmg

Tv

3 00 10 9 80

2 00 10

3

2

. .

.

kg m s

m

2

ssN s m= ⋅1 47.

(b) In the equation describing the time variation of the velocity, we have

v v= −( )−T

bt me1 v v= 0 632. T when e bt m− = 0 368.

or at time tm

b= − ⎛

⎝⎞⎠ ( ) = × −ln . .0 368 2 04 10 3 s

(c) At terminal velocity, R b mgT= = = × −v 2 94 10 2. N

*P6.32 (a) Since the window is vertical, the normal force is horizontal. n = 4 N f = n =k kµ 0.9(4 N) = 3.6 N upward, to oppose downward motion

ΣF may y= : + − + =3 6 0 16 9 8 0. ( . ) ( . )N kg m s2 Py Py = − =2 03 2 03. .N N down

(b) ΣF may y= : + − − =3 6 0 16 9 8 1 25 2 03 0 16. ( . ) ( . ) . ( . ) .N kg N kgm s2 ay

ay = − = − =0 508 0 16 3 18 3 18. . . .N kg downm s m s2 2

(c) At terminal velocity, ΣF may y= : + ⋅ − − =( ) ( . )( . ) . ( . )20 0 16 9 8 1 25 2 03 0N s m kg NvT m s2

vT = ⋅ =4 11 20 0 205. ( ) .N N s m downm s

P6.33 v = ⎛⎝

⎞⎠ − −⎛

⎝⎞⎠

⎡⎣⎢

⎤⎦⎥

mg

b

bt

m1 exp where exp x ex( ) = is the exponential function.

At t → ∞ v v→ =T

mg

b

At t = 5 54. s 0 500 15 54

9 00. exp

.

.v vT T

b= − − ( )⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦

s

kg ⎥⎥

exp.

.. ;

.

− ( )⎛⎝⎜

⎞⎠⎟

=

− ( )

b

b

5 54

9 000 500

5 54

s

kg

s

99 000 500 0 693

9 00 0 693

5

.ln . . ;

. .

kg

kg

= = −

=( )( )

b..

.54

1 13s

kg s=

(a) vT

mg

b= vT =

( )( )=

9 00 9 80

1 1378 3

. .

..

kg m s

kg sm s

2

(b) 0 750 11 13

9 00. exp

.

.v vT T

t= − −⎛⎝

⎞⎠

⎡⎣⎢

⎤⎦⎥s

exp.

..

−⎛⎝

⎞⎠ =1 13

9 000 250

t

s

t = ( )−

=9 00 0 250

1 1311 1

. ln .

..s s




(c) dx

dt

mg

b

bt

m= ⎛

⎝⎞⎠ − −⎛

⎝⎞⎠

⎡⎣⎢

⎤⎦⎥

1 exp ; dxmg

b

bt

mdt

x

x t

0

10

∫ ∫= ⎛⎝

⎞⎠ − −⎛

⎝⎞⎠

⎡⎣⎢

⎤⎦⎥

exp

x xmgt

b

m g

b

bt

m

mgt

b

m gt

− = +⎛⎝⎜

⎞⎠⎟

−⎛⎝

⎞⎠ = +0

2

20

2

expbb

bt

m2 1⎛⎝⎜

⎞⎠⎟

−⎛⎝

⎞⎠ −⎡

⎣⎢⎤⎦⎥

exp

At t = 5 54. s,

x = ( ) +( )

9 005 54 9 00

.. .

kg 9.80 m ss

1.13 kg s

kg2

22

2

9 80

1 130 693 1

.

.exp .

m s

kg s

2( )( )

⎛

⎝⎜

⎞

⎠⎟ −( ) −[ ]]

x = + −( ) =434 626 0 500 121m m m.

P6.34 F ma∑ =

− =

− =

− =

− −(

∫ ∫ −

km md

dt

kdtd

k dt d

k t

t

vv

vv

v vv

v

2

2

0

2

0

0)) =−

= − +

= + =+

=+

−vv v

v vv

v

vv

v

v1

0

0

0

0

0

1

1 1

1 1 1

1

0

ktkt

vv0kt

P6.35 (a) From Problem 34,

vvv

vv

v

= =+

=+

=∫ ∫

dx

dt kt

dxdt

kt k

kdtx t

0

0

0

000

0

1

1

1

1++

= +( )

− = +( )

∫ v

v

v

00

0 0 0

0

11

01

1

kt

xk

kt

xk

kt

t

x tln

ln −−⎡⎣ ⎤⎦

= +( )

ln

ln

1

11 0x

kktv

(b) We have ln 1 0+( ) =v kt kx

1 0+ =v kt ekx so vvv

vv v=

+= = =−0

0

001 kt eekx

kx


140 Chapter 6

P6.36 We write − = −km D Av v2 21

2ρ so

k

D A

m= =

( ) ×( )−ρ2

0 305 1 20 4 2 10

2 0 145

3. . .

.

kg m m3 2

kgm

m s

( ) = ×

= = ( )

−

− − × −

5 3 10

40 2

3

0

5 3 10

.

..v v e ekx

33 18 336 5

m mm s( )( ) =.

.

P6.37 (a) v vt eict( ) = − v v20 0 5 00 20 0. . .s( ) = = −

ice , vi = 10 0. m s

So 5 00 10 0 20 0. . .= −e c and − = ⎛⎝

⎞⎠20 0

1

2. lnc c = − ( ) = × − −ln

..

12 2 1

20 03 47 10 s

(b) At t = 40 0. s v = ( ) = ( )( ) =−10 0 10 0 0 250 2 5040 0. . . ..m s m s m se c

(c) v v= −i

cte ad

dtc e ci

ct= = − = −−vv v

P6.38 In R D A= 1

22ρ v , we estimate that D = 1 00. , ρ = 1 20. kg m3,

A = ( )( ) = × −0 100 0 160 1 60 10 2. . .m m m2 and v = 27 0. m s. The resistance force is then

R = ( )( ) ×( )(−1

21 00 1 20 1 60 10 27 02. . . .kg m m m s3 2 )) =2

7 00. N

or

R ~ 101 N

Additional Problems

*P6.39 Let v0 represent the speed of the object at time 0. We have

d b

mdt

b

mt

tvv

vv

v

v

v

0 00 0∫ ∫= − = −ln tt

b

mt

bt

m

e

ln ln /

/

v v v v

v v

− = − −( ) ( ) = −

=

0 0

0

0 ln

−− −=bt m bt me/ /v v0

From its original value, the speed decreases rapidly at fi rst and then more and more slowly, asymptotically approaching zero.

In this model the object keeps losing speed forever. It travels a fi nite distance in stopping.

The distance it travels is given by

dr e dt

rm

be

b

mdt

rbt m

t

bt mt

0 0 0

0 0

∫ ∫

∫

=

= − −⎛⎝

⎞

−

−

v

v

/

/

⎠⎠ = − = − −( ) = −− − −m

be

m

be

m

bebt m t bt m btv v

v0 0 0

01 1/ / //m

t

( )

As goes to infinity, the distance appproachesm

bm b

vv0

01 0−( ) = /

FIG. P6.39

t0

0



P6.40 At the top of the vertical circle,

T mR

mg= −v2

or T = ( ) ( )− ( )( ) =0 400

4 00

0 5000 400 9 80 8 88

2

..

.. . . N

P6.41 (a) The speed of the bag is 2 7 46

381 23

π ..

m

sm s

( )= .

The total force on it must add to

mac =( )( ) =3 1 23

7 466 12

20 kg m s

mN

.

..

F ma f n

F ma f

x x s

y y s

∑∑

= − =

=

: cos sin .

: sin

20 20 6 12

2

N

00 20 30 9 8 0

20 6 12

+ − ( )( ) =

= −

n

nfs

cos .

cos .

kg m s2

NN

sin 20

Substitute:

f fs ssincos

sin.

cos

sin20

20

206 12

20

2029

2

+ − ( ) =N 44

2 92 294 16 8

106

N

N N

N

f

f

s

s

. .( ) = +

=

(b) v = ( )=2 7 94

341 47

π ..

m

sm s

mac =( )( ) =30 1 47

7 948 13

2kg m s

mN

.

..

f n

f n

n

s

s

cos sin .

sin cos

20 20 8 13

20 20 294

− =+ =

=

N

N

ff

f f

s

s s

cos .

sin

sincos

sin

20 8 13

20

2020

208

2

−

+ −

N

..cos

sin. .

1320

20294

2 92 294 22 4

N N

N

( ) =

( ) = +fs N

N

N NN

f

n

s =

= ( ) − =

108

108 20 8 13

20273

cos .

sin

µµssf

n= = =108

2730 396

N

N.

FIG. P6.41

n

mg

fs

ac x

y


142 Chapter 6

P6.42 When the cloth is at a lower angle θ, the radial component of F ma∑ =

reads

n mg

m

r+ =sinθ v2

At θ = 68 0. °, the normal force drops to

zero and gr

sin 682

° = v.

v = = ( )( ) =rg sin . . sin .68 0 33 9 8 68 1 73° °m m s m s2

The rate of revolution is

angular speed = ( )⎛⎝

⎞⎠ ( )

⎛⎝⎜

⎞⎠⎟

=1 731

2

2

2 0 330.

.m s

rev

mππ

πr

r.. .835 50 1rev s rev min=

P6.43 (a) v = ( )⎛⎝⎜

⎞⎠⎟

⎛⎝

⎞⎠ =30

1 1 0008 3km h

h

3 600 s

m

1 km. 33 m s

F may y∑ = : + − = −n mgm

r

v2

n m gr

= −⎛⎝⎜

⎞⎠⎟

= −( )v2 2

1 800 9 88 33

20kg m s

m s2..

.44

1 15 104

m

N up

⎡

⎣⎢

⎤

⎦⎥

= ×.

(b) Take n = 0. Then mgm

r= v2

.

v = = ( )( ) = =gr 9 8 20 4 14 1 50 9. . . .m s m m s km h2

P6.44 (a) F mam

Ry y∑ = = v2

mg nm

R− = v2

n mgm

R= − v2

(b) When n = 0 mgm

R= v2

Then,

v = gR

A more gently curved bump, with larger radius, allows the car to have a higher speed with-out leaving the road. This speed is proportional to the square root of the radius.

P6.45 (a) slope = − =0 160 0

9 90 016 2

.

..

N

m skg m2 2

(b) slope = = =R D AD A

vv

v2

12

2

2

1

2

ρ ρ

FIG. P6.43

n

mg

FIG. P6.42

R68°

p

mg

p

mg cos 68°

mg sin 68°




(c) 1

20 016 2D Aρ = . kg m

D =

( )( ) ( )

=2 0 016 2

1 20 0 1050 7782

.

. ..

kg m

kg m m3 π

(d) From the table, the eighth point is at force mg = ×( )( ) =−8 1 64 10 9 8 0 1293. . .kg m s N2

and horizontal coordinate 2 802

. m s( ) . The vertical coordinate of the line is here

0 016 2 2 8 0 127

2. . .kg m m s N( )( ) = . The scatter percentage is 0 129 0 127

1 5. .

. %N N

0.127 N

− = .

(e) The interpretation of the graph can be stated thus:

For stacked coffee fi lters falling at terminal speed, a graph of air resistance force as a function of squared speed demonstrates that the force is proportional to the speed squared within the experimental uncertainty estimated as 2%. This proportionality agrees with that

described by the theoretical equation R D A= 1

22ρ v . The value of the constant slope of the

graph implies that the drag coeffi cient for coffee fi lters is D = ±0 78 2. %.

*P6.46 (a) The forces acting on the ice cube are the Earth’s gravitational force, straight down, and the basin’s normal force, upward and inward at 35° with the vertical. We choose the x and y axes to be horizontal and vertical, so that the acceleration is purely in the x direction. Then

∑Fx = ma

x: n sin 35° = mv2�R

∑Fy = ma

y: n cos 35° − mg = 0

Dividing eliminates the normal force: n sin 35°� n cos 35° = mv2�Rmg

tan 35° = v2�Rg v = = ( )Rg Rtan . .35 0 6 86° m s2

(b) The mass is unnecessary.

(c) The answer to (a) indicates that the speed is proportional to the square root of the radius, so

doubling the radius will make the required speed increase by 2 times .

(d) The period of revolution is given by TR R

RgR= = = ( )2 2

35 02 40

π πv tan .

.°

s m

When the radius doubles, the period increases by 2 times .

(e) On the larger circle the ice cube moves 2 times faster but also takes longer to get around, because the distance it must travel is 2 times larger. Its period is also proportional to the square root of the radius.

P6.47 Take x-axis up the hill

F ma T mg ma

aT

mg

F m

x x

y

∑

∑

= + − =

= −

=

: sin sin

sin sin

θ φ

θ φ

aa T mg

Tmg

ag

y: cos cos

cos

coscos sin

c

+ − =

=

=

θ φφ

θφ θ

0

oossin

cos tan sin

θφ

φ θ φ

−

= −( )

g

a g


144 Chapter 6

P6.48 (a) v = ( )⎛⎝⎜

⎞⎠⎟

=30088 0

60 0440mi h

ft s

mi hft s

.

.

At the lowest point, his seat exerts an upward force; therefore, his weight seems to increase. His apparent weight is

′ = + = + ⎛⎝

⎞⎠

( )=F mg m

rg

v2 2

160160

32 0

440

1 200967

.lbb

(b) At the highest point, the force of the seat on the pilot is directed down and

′ = − = −F mg mrg

v2

647 lb

Since the plane is upside down, the seat exerts this downward force as a normal force.

(c) When ′ =Fg 0, then mgm

R= v2

. If we vary the aircraft’s R and v such that this equation is

satisfi ed, then the pilot feels weightless.

P6.49 (a) Since the centripetal acceleration of a person is downward (toward the axis of the earth), it is equivalent to the effect of a falling elevator. Therefore,

′ = −F Fm

rg g

v2

or F Fg g> ′

(b) At the poles v = 0 and ′ = = = ( ) =F F mgg g 75 0 9 80 735. . N down.

At the equator, ′ = − = − ( ) =F F mag g c 735 75 0 0 033 7 732N N N. . down.

*P6.50 (a) Since the object of mass m2 is in equilibrium, F T m gy∑ = − =2 0

or

T m g= 2

(b) The tension in the string provides the required centripetal acceleration of the puck.

Thus,

F T m gc = = 2

(c) From

Fm

Rc = 12v

we have

v = =⎛⎝⎜

⎞⎠⎟

RF

m

m

mgRc

1

2

1

(d) The puck will spiral inward,gainingspeed as it doess so. It gains speed because the extra-

large string tension produces forward tangential acceleration as well as inward radial acceleration of the puck, pulling at an angle of less than 90° to the direction of the inward-spiraling velocity.

(e) The puck will spiral outward,slowing down as it doees so.

FIG. P6.49



*P6.51

(a) The only horizontal force on the car is the force of friction, with a maximum value determined by the surface roughness (described by the coeffi cient of static friction) and the normal force (here equal to the gravitational force on the car).

(b) ∑Fx = ma

x −f = ma a = −f �m = (v2 − v

02)�2(x − x

0)

x − x0 = (v2 − v

02)m�2f = (02 − [20 m�s]2)1200 kg�2(−7000 N) = 34 3. m

(c) Now f = mv2�r r = mv2�f = 1200 kg [20 m�s]2�7000 N = 68 6. m

A top view shows that you can avoid running into the wall by turning through a quarter-circle, if you start at least this far away from the wall.

(d) Braking is better. You should not turn the wheel. If you used any of the available fric-tion force to change the direction of the car, it would be unavailable to slow the car, and the stopping distance would be longer.

(e) The conclusion is true in general. The radius of the curve you can barely make is twice your minimum stopping distance.

P6.52 v = = ( )( ) =2 2 9 00

15 03 77

π πr

T

.

..

m

sm s

(a) arr = =v2

1 58. m s2

(b) F m g arlow N= +( ) = 455

(c) F m g arhigh N= −( ) = 328

(d) F m g armid N upward and= + =2 2 397 at θ = = =− −tan tan.

..1 1 1 58

9 89 15

a

gr ° inward .

P6.53 (a) The mass at the end of the chain is in vertical equilibrium. Thus T mgcosθ = .

Horizontally T mam

rrsinθ = = v2

r

r

= +( )= +( ) =

2 50 4 00

2 50 28 0 4 00

. sin .

. sin . .

θ m

m° 55 17. m

Then ar = v2

5 17. m.

By division tan.

θ = =a

g gr v2

5 17

v

v

2 5 17 5 17 9 80 28 0

5

= = ( )( )( )=

. tan . . tan .g θ ° m s2 2

..19 m s

(b) T mgcosθ =

Tmg= =

( )( )°

=cos

. .

cos .θ50 0 9 80

28 0555

kg m sN

2

FIG. P6.53

TR = 4.00 m

θl = 2.50 m

r

mg


146 Chapter 6

P6.54 (a) The putty, when dislodged, rises and returns to the original level in time t. To fi nd t, we use

v vf i at= + : i.e., − = + −v v gt or tg

= 2vwhere v is the speed of a point on the rim of the

wheel.

If R is the radius of the wheel, v = 2π R

t, so t

g

R= =2 2vvπ

.

Thus, v2 = π Rg and v = π Rg .

(b) The putty is dislodged when F, the force holding it to the wheel, is

Fm

Rm g= =v2

π

*P6.55 (a) nm

R= v2

f mg− = 0

f ns= µ v = 2π R

T

TR

gs= 4 2π µ

(b) T = 2 54. s

# .rev

min

rev

2.54 s

s

min

rev

min= ⎛

⎝⎞⎠ =1 60

23 6

(c) The gravitational and frictional forces remain constant. The normal force increases. The person remains in motion with the wall.

(d) The gravitational force remains constant. The normal and frictional forces decrease. The person slides relative to the wall and downward into the pit.

P6.56 Let the x-axis point eastward, the y-axis upward, and the z-axis point southward.

(a) The range is Zg

i i= v2 2sin θ

The initial speed of the ball is therefore

vi

i

gZ= = ( )( )°

=sin

.

sin ..

2

9 80 285

96 053 0

θm s

The time the ball is in the air is found from ∆y t a tiy y= +v1

22 as

0 53 0 48 0 4 90 2= ( )( ) − ( ). sin . .m s m s2° t t

giving t = 8 04. s .


FIG. P6.55

f

mg

nn



(b) vixe iR= =

×( ) °2

86 400

2 6 37 10 35 06π φ πcos . cos .

s

m

886 400379

sm s=

(c) 360° of latitude corresponds to a distance of 2π Re, so 285 m is a change in latitude of

∆φπ π

=⎛⎝⎜

⎞⎠⎟

( ) =×( )

⎛

⎝⎜

S

Re2360

285

6 37 106°m

2 m.

⎞⎞

⎠⎟ ( ) = × −360 2 56 10 3° . degrees

The fi nal latitude is then φ φ φf i= − = − =∆ 35 0 0 002 56 34 997 4. . .° ° °.

The cup is moving eastward at a speed v fxe fR

=2

86 400

π φcos

s, which is larger than the eastward

velocity of the tee by

∆v v vx fx fie

f ieR R= − = −⎡⎣ ⎤⎦ =2

86 400

2π φ φ πs

cos cos886 400

2

86 400

s

s

cos cos

cos

φ φ φ

π

i i

eR

−( ) −⎡⎣ ⎤⎦

=

∆

φφ φ φ φ φi i icos sin sin cos∆ ∆+ −[ ]

Since ∆φ is such a small angle, cos ∆φ ≈ 1and ∆ ∆vxe

i

R≈ 2

86 400

π φ φs

sin sin .

∆vx ≈×( )2 6 37 10

86 40035 0 0 002 56

6π .sin . sin .

m

s° °° = × −1 19 10 2. m s

(d) ∆ ∆x tx= ( ) = ×( )( ) = =−v 1 19 10 8 04 0 095 5 92. . .m s s m ..55 cm

P6.57 (a) If the car is about to slip down the incline, f is directed up the incline.

F n f mgy∑ = + − =cos sinθ θ 0 where f ns= µ gives

n

mg

s

=+( )cos tanθ µ θ1

and fmgs

s

=+( )

µθ µ θcos tan1

Then, F n f mRx∑ = − =sin cos minθ θ v2

yields

vmin

tan

tan=

−( )+

Rg s

s

θ µµ θ1

When the car is about to slip up the incline, f is directed

down the incline. Then, F n f mgy∑ = − − =cos sinθ θ 0

with f ns= µ yields

nmg

s

=−( )cos tanθ µ θ1

and fmgs

s

=−( )

µθ µ θcos tan1

In this case, F n f mRx∑ = + =sin cos maxθ θ v2

, which gives

vmax

tan

tan=

+( )−

Rg s

s

θ µµ θ1

(b) If vmin

tan

tan=

−( )+

=Rg s

s

θ µµ θ1

0, then µ θs = tan .



148 Chapter 6

(c) vmin

. tan . .

.=

( )( ) −( )+

100 9 80 10 0 0 100

1 0 1

m m s2 °

000 10 08 57

( ) =tan .

.°

m s

vmax

. tan . .

.=

( )( ) +( )−

100 9 80 10 0 0 100

1 0 1

m m s2 °

000 10 016 6

( ) =tan .

.°

m s

*P6.58 (a) We let R represent the radius of the hoop and T represent the period of its rotation. The bead moves in a circle with radius v = R sinθ at a speed of

v = =2 2π π θr

T

R

T

sin

The normal force has an inward radial component of n sinθ and an upward component of n cosθ

F ma n mgy y∑ = − =: cosθ 0

or

nmg=

cosθ

Then F n mrx∑ = =sinθ v2

becomes mg m

R

R

Tcossin

sin

sin

θθ

θπ θ⎛

⎝⎞⎠ = ⎛

⎝⎞⎠

2 2

which reduces to g R

T

sin

cos

sinθθ

π θ= 4 2

2

This has two solutions: sinθ θ= ⇒ = °0 0 (1)

and cosθπ

= gT

R

2

24 (2)

If R = 15 0. cm and T = 0 450. s, the second solution yields

cos. .

..θ

π=

( )( )( ) =

9 80 0 450

4 0 1500 335

2

2

m s s

m

2

and θ = 70 4. °

Thus, in this case, the bead can ride at two positions θ = 70 4. ° and θ = 0° .

(b) At this slower rotation, solution (2) above becomes

cos. .

..θ

π=

( )( )( ) =

9 80 0 850

4 0 1501 20

2

2

m s s

m

2

, which is impossible.

In this case, the bead can ride only at the bottom of the loop, θ = 0° .

(c) The equation that the angle must satisfy has two solutions whenever 4π2R > gT 2 but only the solution 0° otherwise. The loop’s rotation must be faster than a certain threshold value in order for the bead to move away from the lowest position. Zero is always a solution for the angle. There are never more than two solutions.

FIG. P6.58(a)

FIG. P6.57



P6.59 At terminal velocity, the accelerating force of gravity is balanced by frictional drag: mg ar br= +v v2 2

(a) mg = ×( ) + ×( )− −3 10 10 0 870 109 10 2. .v v

For water, m = = ( )⎡⎣⎢

⎤⎦⎥

−ρ πV 1 0004

310 5 3

kg m m3

4 11 10 3 10 10 0 870 1011 9 10 2. . .× = ×( ) + ×( )− − −v v

Assuming v is small, ignore the second term on the right hand side: v = 0 013 2. m s .

(b) mg = ×( ) + ×( )− −3 10 10 0 870 108 8 2. .v v Here we cannot ignore the second term because the coeffi cients are of nearly equal magnitude.

4 11 10 3 10 10 0 870 10

3 1

8 8 8 2. . .

.

× = ×( ) + ×( )

=−

− − −v v

v00 3 10 4 0 870 4 11

2 0 8701 03

2± ( ) + ( )( )( ) =

. . .

.. m s

(c) mg = ×( ) + ×( )− −3 10 10 0 870 107 6 2. .v v

Assuming v > 1 m s, and ignoring the fi rst term:

4 11 10 0 870 105 6 2. .× = ×( )− − v v = 6 87. m s

P6.60 (a) t ds m

.

.

.

.

.

.

.

.

.

( ) ( )1 00

2 00

3 00

4 00

5 00

6 00

7 00

8 00

9 000

10 0

11 0

12 0

13 0

14 0

15 0

16 0

17 0

18 0

19 0

2

.

.

.

.

.

.

.

.

.

.

00 0

4 88

18 9

42 1

73 8

112

154

199

246

296

347

399

45

.

.

.

.

.

22

505

558

611

664

717

770

823

876

(b)

(c) A straight line fi ts the points from t = 11 0. s to 20.0 s quite precisely. Its slope is the termi-nal speed.

vT = = −

−=slope

m m

20.0 s sm s

876 399

11 053 0

..

(s)

(m)

900

800

700

600

500

400

300

200

100

00 2 4 6 8 10 12 14 16 18 20

t

d


150 Chapter 6

P6.61 F L T mg L T may y y y∑ = − − = − − =cos . sin . .20 0 20 0 7 35° ° N ==

= + = + ° =

=

∑

0

20 0 20 0

0

2

2

F L T L T mr

mr

x x x sin . cos .°v

v..

.

. cos ..750

35 0

60 0 20 016 3

2

kgm s

mN

We

( )( ) =

°

have the simultaneous equations

L sin .20 0° + TT

L T

cos . .

cos . sin . .

20 0 16 3

20 0 20 0 7 35

°

° °

=− =

N

N

LL T

L T

+ =

−

cos .

sin .

.

sin .sin

20 0

20 0

16 3

20 020

°

° °

N

..

cos .

.

cot . tan .

0

20 0

7 35

20 0 20

°

° °

°

=

+

N

cos20.0

T 0016 3

20 0

7 35

20 03 11

°° °

( ) = −

( ) =

.

sin .

.

cos ..

N N

T 339 8

12 8

.

.

N

NT =

*P6.62 (a) v v= +i kx implies the acceleration is ad

dtk

dx

dtk= = + = +vv0

Then the total force is F ma m k∑ = = +( )v

As a vector, the force is parallel or antiparallel to the velocity: � �F v∑ = km .

(b) For k positive, some feedback mechanism could be used to impose such a force on an object for a while. The object’s speed rises exponentially. Riding on such an object would be more scary than riding on a skyrocket. It would be a good opportunity for learning about exponential growth in population or in energy use.

(c) For k negative, think of a duck landing on a lake, where the water exerts a resistive force on the duck proportional to its speed.

ANSWERS TO EVEN PROBLEMS

P6.2 215 N horizontally inward

P6.4 (a) 1 65. km s (b) 6 84 103. × s

P6.6 6 22 10 12. × − N

P6.8 (a) 68.6 N toward the center of the circle and 784 N up (b) 0.857 m�s2

P6.10 (a) − +( )0 233 0 163. ˆ . ˆ m s2i j (b) 6 53. m s (c) − +( )0 181 0 181. ˆ . ˆ m s2i j

P6.12 (a) 1 33. m s2 (b) 1 79. m s2 forward and 48.0° inward

P6.14 (a) 2.49 × 104 N up (b) 12.1 m�s

FIG. P6.61



P6.16 (a) 20.6 N (b) 3.35 m �s2 downward tangent to the circle; 32.0 m �s2 radially inward (c) 32.2 m �s2 at 5.98° to the cord, pointing toward a location below the center of the circle. (d) No change. If the object is swinging down it is gaining speed. If it is swinging up it is losing speed but its acceleration is the same size and its direction can be described in the same terms.

P6.18 (a) 8.62 m (b) Mg downward (c) 8 45. m s2 Unless they are belted in, the riders will fall from the cars.

P6.20 0.527°

P6.22 µk

t L

g a t= −( )

+( )2

2

v

P6.24 93.8 N

P6.26 (a) 6 27. m s downward2 (b) 784 N up (c) 283 N up

P6.28 (a) 53 8. m s (b) 148 m

P6.30 −0 212. m s2

P6.32 (a) 2.03 N down (b) 3.18 m �s2 down (c) 0.205 m �s down

P6.34 see the solution

P6.36 36 5. m s

P6.38 ~101 N

P6.40 8.88 N

P6.42 0 835. rev s

P6.44 (a) mgm

R− v2

upward (b) v = gR

P6.46 (a) v = = ( )Rg Rtan . .35 0 6 86° m/s2 (b) The mass is unnecessary. (c) Increase by 2

times (d) Increase by 2 times (e) On the larger circle the ice cube moves 2 times faster but also takes longer to get around, because the distance it must travel is 2 times larger. Its period is

described by TR R

RgR= = = ( )2 2

35 02 40

πv

πtan .

.°

s m/ .

P6.48 (a) The seat exerts 967 lb up on the pilot. (b) The seat exerts 647 lb down on the pilot. (c) If the plane goes over the top of a section of a circle with v2 = Rg, the pilot will feel weightless.

P6.50 (a) m g2 (b) m g2 (c) m

mgR2

1

⎛⎝⎜

⎞⎠⎟

(d) The puck will move inward along a spiral, gaining

speed as it does so. (e) The puck will move outward along a spiral as it slows down.

P6.52 (a) 1.58 m /s2 (b) 455 N (c) 329 N (d) 397 N upward and 9.15° inward

P6.54 (a) v = π Rg (b) m gπ

P6.56 (a) 8.04 s (b) 379 m s (c) 1 19. cm s (d) 9.55 cm


152 Chapter 6

P6.58 (a) either 70.4° or 0° (b) 0° (c) The equation that the angle must satisfy has two solutions whenever 4π2R > gT 2 but only the solution 0° otherwise. (Here R and T are the radius and period of the hoop.) Zero is always a solution for the angle. There are never more than two solutions.

P6.60 (a) and (b) see the solution (c) 53 0. m s

P6.62 (a) Σ� �F v= mk (b) For k positive, some feedback mechanism could be used to impose such a

force on an object for a while. The object’s speed rises exponentially. Riding on such an object would be more scary than riding on a skyrocket. It would be a good opportunity for learning about exponential growth in population or in energy use. (c) For k negative, think of a duck landing on a lake, where the water exerts a resistive force on the duck proportional to its speed.


sm chapter6

Education