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CHAPTER 7
Quick Quizzes
1.
(c). For a rotation of more than 180°, the angular displacement must be larger than
  3.14 rad . The angular displacements in the three choices are (a) 6 rad  3 rad  3 rad ,
(b) 1 rad   1 rad  2 rad , (c) 5 rad  1 rad  4 rad .
2.
(b). Because all angular displacements occurred in the same time interval, the
displacement with the lowest value will be associated with the lowest average angular
speed.
3.
(b) and (d). Points on the spokes which are closer to the axis of rotation have smaller
values of tangential speed and acceleration than do points on the rim of the wheel.
4.
(e), (a), (b).
5.
(c). Both change in direction, but have constant magnitudes.
6.
(b) and (c). According to Newton’s law of universal gravitation, the force between the ball
and the Earth depends on the product of their masses, so both forces, that of the ball on
the Earth, and that of the Earth on the ball, are equal in magnitude. This follows also, of
course, from Newton’s third law. The ball has large motion compared to the Earth because
according to Newton’s second law, the force gives a much greater acceleration to the small
mass of the ball.
7.
(e).
8.
(a). From Kepler’s third law and the given period, the major axis of the comet can be
calculated. It is found to be 1.2  1011 m . Because this is smaller than the Earth-Sun
distance, the comet cannot possibly collide with the Earth.
209
C H A P T E R
7
Problem Solutions
7.1
s 60 000 mi  5280 ft 
8
(a)   

  3.2  10 rad .
r
1.0 ft  1 mi 
(b)
7.2
 1 rev 
  3.2  108 rad 
 5.0  107 rev .

 2  rad 
The distance traveled is s  r , where  is in radians.
   rad  
s  r   4.1 m  30 
   2.1 m .
  180  
For 30°,
7.3
For30 radians,
s  r   4.1 m  30 rad   1.2  102 m .
For30 revolutions,

 2  rad  
2
s  r   4.1 m  30 rev 
   7.7  10 m .

1 rev 



The Earth moves through 2 rad in one year 3.156  107 s . Thus,

2 rad
7
 1.99  10 rad s .
3.156  107 s
Alternatively, the Earth moves through 360° in one year (365.242 days).

Thus,
7.4
We use  

7.5
(a)  
 f  i
t
360
 0.986 deg day .
365.2 days
and find
0.20 rev s  0  2  rad 
-2
2

  4.2  10 rad s .
30 s
1 rev
 2.51  10

rev min  0  2 rad   1 min 
2
  821 rad s

 
3.20 s
1 rev
60.0 s
4
1
1
rad 
2
(b)   i t   t 2  0   821 2   3.20 s   4.21  103 rad
2
2
s 
210
C H A P T E R
7.6
i  100
7
rev  2 rad   1 min 
  10.47 rad s


min  1 rev   60.0 s 
(a) From  f  i  t , t 
 f  i


0  10.47 rad s
 5.24 s .
 2.00 rad s 2
  f  i   0  10.47 rad s 
t 
(b)    t  
  5.24 s   27.4 rad .
2
 2  
7.7
The final linear speed of the car is


v f  vi  at  17.0 m s  2.00 m s 2  5.00 s  27.0 m s ,
and the distance traveled in this time interval is
 v f  v i    27.0  17.0 m s 
svt 
t
  5.00 s   110 m .
2
 2  
The angular displacement of the wheel is
s
r
= 
7.8
110 m  1 rev 
 36.5 rev .
0.48 m  2  rad 
The initial angular velocity is i 
v i 24.0 m s

 50.0 rad s .
r
0.480 m
Then, from  2f  i2  2    , the angular displacement as the wheel stops is
 
 2f  i2
2
0   50.0 rad s   1 rev 

 147 rev .
2 -1.35 rad s 2  2  rad 
2


211
C H A P T E R
7.9
7
rev   2 rad   1 min 

v  r   3.80 m   450

  179 m s


min   1 rev   60 s 
Main Rotor:

m v

v =  179   sound  = 0.522 v sound

s   343 m s 
rev   2 rad   1 min 

v  r   0.510 m   4138

  221 m s


min   1 rev   60 s 
Tail Rotor:

m v

v =  221   sound  = 0.644 v sound

s   343 m s 
7.10
We will break the motion into two stages: (1) an acceleration period and (2) a
deceleration period.
The angular displacement during the acceleration period is
  5.0 rev s  2  rad 1 rev   0 
  f  i 
t
  8.0 s   126 rad ,

2
 2 


1  t  
and while decelerating,
 0   5.0 rev s  2  rad 1 rev  
  f  i 
t
 12 s   188 rad .

2
 2 


2  
The total displacement is
7.11
 1 rev 
 =1  2   126  188 rad  
 50 rev .
 2  rad 
When completely rewound, the tape is a hollow cylinder with a difference between the
inner and outer radii of ~1 cm. Let N represent the number of revolutions through which
the driving spindle turns in 30 minutes (and hence the number of layers of tape on the
spool). We can determine N from:
N
   t  1 rad s   30 min   60 s 1 min  


 286 rev
2
2
2  rad rev
Then, thickness ~
1 cm
N
~ 10-2 cm
212
C H A P T E R
7.12
7
The angular displacement of the coin while stopping is
 
 2f  i2
2

0  18.0 rad s 

2 1.90 rad s 2
2

 85.3 rad .
The linear displacement is s  r    with r  diameter 2  1.20 cm , or
s  1.20 cm  85.3 rad   102 cm  1.02 m .
7.13
The pulsar must rotate with angular velocity   15 rev s (i.e., one-half revolution per
flash). Thus, from v  r  , the maximum radius is
rmax 
v max


3.00  108 m s  1 rev 
 3.2  106 m .


15 rev s  2  rad 
With mass greater than 1.4 times the mass of the Sun and radius comparable to the
Earth, the object is thought to be a neutron star.
7.14
 1609 m 
 4.0  10 3 m . Thus, from ac  r  2 , the
The radius of the cylinder is r  2.5 mi 
 1 mi 
required angular velocity is
ac

r

7.15
9.80 m s 2
 4.9  10-2 rad s .
3
4.0  10 m
The rotational velocity of the Earth is

   2

rad   1 day 
 7.27  105 rad s ,



day   86 400 s 
and the equatorial radius of the Earth is RE  6.38  106 m .
(a) At the equator, the radius of the circular motion is r  RE .
Thus, ac  r  2   6.38  106 m  7.27  105 rad s   3.37  10-2 m s 2 .
2
(b) A point at the north pole is on the axis of rotation, so r  0 and ac  r  2  0 .
213
C H A P T E R
7.16
The speed of the plane is v 
r
7
s 2 r
. Thus, the radius of the circular path is

t
t
v  t  180 m s  120 s 

 3.44  103 m  3.44 km ,
2
2
180 m s   9.42 m s2 .
v2

and the centripetal acceleration is ac 
r 3.44  103 m
2
7.17
The final angular velocity is  f  78
rev  1 min   2 rad 


  8.17 rad s ,
min  60 s   1 rev 
 2.54 cm 
and the radius of the disk is r  5.0 in 
 12.7 cm=0.127 m .
 1 in 
(a) The tangential acceleration of the bug as the disk speeds up is
 8.17 rad s  0.35 m s 2
  
.
at  r   r 
  0.127 m  


 t 
 3.0 s 
(b) The final tangential speed of the bug is
v t  r  f   0.127 m   8.17 rad s   1.0 m s .
 8.17 rad s 
(c) At t = 1.0 s,   i   t  0  
1.0 s   2.7 rad s .
 3.0 s 
Thus,
at  r   0.35 m s 2 as above, while the radial acceleration is
ac  r  2   0.127 m   2.7 rad s   0.94 m s 2 .
2
The total acceleration is a  ac2  at2  1.0 m s 2 , and the angle this acceleration
makes with the direction of a c is
a 

  tan 1  t   tan 1 
  20° .

0.94 
a 
0.35
c
214
C H A P T E R
7.18
(a) The centripetal acceleration is ac 
vt  r ac 
v t2
. Thus, when ac  at  0.500 m s 2 , we have
r
 400 m   0.500 m s 2  
(b) At this time, t 
7
14.1 m s .
v t  vi 14.1 m s  0

 28.2 s , and the linear displacement is
at
0.500 m s 2
 v  vi   14.1 m s  0 
s  vt t   t
t
  28.2 s   200 m .
 2  
2
(c) The time is t = 28.2 s as found in part (b) above.
7.19
(a) From Fr  mac , we have
 v 2   55.0 kg  4.00 m s 
T  m t  
 1.10  103 N  1.10 kN .
0.800 m
 r
2
(b) The tension is larger than her weight by a factor of
T
1.10  103 N

 2.04 times .
mg  55.0 kg  9.80 m s 2

7.20
Since Fc  m

v t2
 m r  2 , the needed angular velocity is
r
Fc
=
mr
 =

4.0  1011 N
3.0  1016 kg  0.150 m 

 1 rev 
= 9.4  102 rad s 
= 1.5  102 rev s

 2  rad 

7.21

Friction must supply the needed centripetal force. Hence, it is necessary that
Fc   fs  max , or m
vt max 
v t2
 s  mg  and the maximum safe speed is
r
s r g   0.70 20 m   9.80 m s 2   12 m s .
215
C H A P T E R
7
2
7.22
7.23
km   1 h   1000 m  

 86.5



2


1g
vt 
h   3600 s   1 km   
ac 

2
 9.80 m s   0.966 g
r
61.0 m
(a)
ac  r  2   2.00 m   3.00 rad s   18.0 m s 2
2


(b) Fc  m ac   50.0 kg  18.0 m s 2  900 N
(c) We know the centripetal acceleration is produced by the force of friction. Therefore,
the needed static friction force is fs  900 N . Also, the normal force is
n  mg  490 N . Thus, the minimum coefficient of friction required is
s 
 fs max
n

900 N
= 1.84 .
490 N
So large a coefficient of friction is unreasonable, and he will not be able to stay on
the merry-go-round.
7.24
(a) From Fy  0 , we have
n cos   mg
(1)
where n is the normal force exerted on the car by
the ramp.
Now, require that, Fradial inward  m ac , or
n sin   m
v t2
r
(2)
tan  
Divide (2) by (1) to obtain
v t2
rg
(b) If r = 50.0 m and v t  13.4 m s , the needed bank angle is
2

13.4 m s 

  tan 
  50.0 m  9.80 m
1

2


  20.1 .

216
C H A P T E R
7.25
7
(a) Since the 1.0-kg mass is in equilibrium, the tension in the string is


T  m g  1.0 kg  9.8 m s 2  9.8 N .
(b) The tension in the string must produce the centripetal acceleration of the puck.
Hence, Fc  T  9.8 N .
r Fc
 v2 

(c) From Fc  mpuck  t  , we find v t 
mpuck
 r 
7.26
1.0 m  9.8 N 
0.25 kg
 6.3 m s .
As he passes through the bottom of his swing, the tension in the vine must equal (1) his
weight plus (2) the force needed produce the centripetal acceleration. That is,
  8.0 m s  2 
 v t2 
2
  1.4  103 N .
T  mg  m     85 kg  9.80 m s   85 kg  
10
m
 r




Since T > 1000 N, he will not make it .
7.27
(a) The centripetal acceleration is
rev   2  rad   1 min  

2
ac  r  2   9.00 m   4.00
 
   1.58 m s .
 

min
1
rev
60
s


2
(b) At the bottom of the circular path, the normal force exerted by the seat must
support the weight and also produce the centripetal acceleration.
Thus,
n  m  g  ac    40.0 kg   9.80  1.58 m s 2   455 N upward .
(c) At the top of the path, the weight must offset the normal force of the seat plus
supply the needed centripetal acceleration. Therefore, mg  n  mac , or
n  m  g  ac    40.0 kg   9.80  1.58 m s 2   329 N upward .
217
C H A P T E R
7
(d) At a point halfway up, the seat exerts an upward vertical component equal to the
child’s weight (392 N) and a component toward the center having magnitude
Fc  mac   40.0 kg  1.58 m s 2  63.2 N . The total force exerted by the seat is


 392 N  2   63.2 N  2  397 N directed inward and at
FR 
 392 N  80.8 above the horizontal

.
 63.2 N 
  tan 1 
7.28
(a) At A, the track supports the weight and supplies the centripetal acceleration.
2

20.0 m s  

v t2
2
  25 kN .
n  mg  m   500 kg  9.80 m s 
r
10 m


Thus,
(b) At B, the weight must offset the normal force exerted by the track and produce the
v2
needed centripetal acceleration, or mg  n  m t . If the car is on the verge of leaving
r
2
v
the track, then n  0 and m g=m t . Hence,
r
vt  r g 
7.29
15 m   9.80 m s 2  
12 m s .
At the half-way point the spaceship is 1.92  108 m from both bodies. The force exerted
on the ship by the Earth is directed toward the Earth and has magnitude
FE 

GmEms
r2
6.67  1011 N  m 2 kg 2 5.98  1024 kg 3.00  10 4 kg


1.92  10
8
m


2
  325 N
The force exerted on the ship by the Moon is directed toward the Moon and has a
magnitude of
FM 

Gm M ms
r2
6.67  1011 N  m 2 kg 2 7.36  10 22 kg 3.00  10 4 kg


1.92  10
8
m


2
  4.00 N
The resultant force is  325 N  4.00 N   321 N directed toward Earth .
218
C H A P T E R
7.30
7
The Sun-Earth distance is r1  1.496  1011 m , the Earth-Moon distance is
r2  3.84  108 m , and the distance from the Sun to the Moon during a solar eclipse is
r1  r2 .
(a) The force exerted on the Moon by the Sun is
FMS 

Gm M mS
 r1  r2 2
 6.67  10
11


N  m 2 kg 2 7.36  1022 kg 1.991  1030 kg
1.496  10
11
m  3.84  108 m


2
or FMS  4.39  1020 N toward the Sun .
(b) The force exerted on the Moon by the Earth is
FME 



6.67  1011 N  m 2 kg 2 7.36  1022 kg 5.98  1024 kg
Gm M mE

2
r22
3.84  108 m



= 1.99  10 N toward the Earth .
20
(c) The force exerted on the Earth by the Sun is
FES 



6.67  1011 N  m 2 kg 2 5.98  1024 kg 1.991  10 30 kg
GmEmS

2
r12
1.496  1011 m

= 3.55  1022 N toward the Sun .
219


C H A P T E R
7.31
7
The forces exerted on the 2.0-kg by the other bodies
are Fx and Fy as shown in the diagram at the right.
The magnitudes of these forces are
 6.67  10
F 
11
 4.0 m 
x
 3.3  10
and
Fy 
11
 6.67  10

N  m 2 kg 2  2.0 kg  4.0 kg 
2
N
11

N  m 2 kg 2  2.0 kg  3.0 kg 
 2.0 m 
2
 1.0  1010 N .
The resultant force exerted on the 2.0-kg is F  Fx2  Fy2  1.1  1010 N
 Fy 
directed at   tan 1    tan 1  3.0  72 above the  x  axis .
 Fx 
7.32
The equilibrium position lies between the Earth and the Sun on the line connecting their
centers. At this point, the gravitational forces exerted on the object by the Earth and Sun
have equal magnitudes and opposite directions. Let this point be located distance r from
the center of the Earth. Then, its distance from the Sun is 1.496  1011 m  r , and we

may determine the value of r by requiring that

GmE m
GmS m
, where

2
2
r
1.496  1011 m  r


mE and mS are the masses of the Earth and Sun respectively. This reduces to
1.496  10
11
r
m r

mS
 577 , or 1.496  1011 m  578r , which yields
mE
r = 2.59  108 m from center of the Earth .
7.33
(a) At the midpoint between the two masses, the forces exerted by the 200-kg and 500Gm1 m2
kg masses are oppositely directed, and from F 
we have
r2
F 
G  50.0 kg  500 kg  200 kg 
 0.200 m 
2
 2.50  105 N toward the 500-kg.
(b) At a point between the two masses and distance d from the 500-kg mass, the net
force will be zero when
G  50.0 kg  200 kg 
 0.400 m  d
2

G  50.0 kg  500 kg 
d2
220
or d = 0.245 m .
C H A P T E R
7.34
7
We know that m1  m2  5.00 kg , or m2  5.00 kg  m1
F
.
2
m1  5.00 kg  m1 

Gm1 m2
8
11 N  m 

1.00

10
N

6.67

10
2
2 

r
kg 

 0.200 m  2
 5.00 kg  m1  m
2
1
1.00  10

8

N  0.200 m 
6.67  1011 N  m 2 kg 2
2
 6.00 kg 2
Thus, m12   5.00 kg  m1  6.00 kg 2  0
or
 m1  3.00 kg  m1  2.00 kg   0
giving m1  3.00 kg, so m2  2.00 kg .
The answer m1  2.00 kg and m2  3.00 kg is physically equivalent.
7.35
(a) The gravitational force must supply the required centripetal acceleration, so
 v2 
GmE m
Gm
 m  t  . This reduces to r  2 E , which gives
2
r
 r
vt
24

N  m 2   5.98  10 kg 
r   6.67  1011
 1.596  107 m .
2
2 
kg

  5000 m s 
The altitude above the surface of the Earth is then
h  r  RE  1.596  107 m  6.38  106 m  9.58  106 m .
(b) The time required to complete one orbit is
7
circumference of orbit 2  1.596  10 m 
T

 2.00  104 s  5.57 h .
orbital speed
5000 m s
7.36
(a) The satellite moves in an orbit of radius r  2 RE and the gravitational force supplies


the required centripetal acceleration. Hence, m v t2 2 RE  GmE m  2 RE  , or
vt 



24

GmE
N  m 2  5.98  10 kg
  6.67  1011
 5.59  103 m s
2 
6
2 RE
kg
2
6.38

10
m


221

2
C H A P T E R
7
(b) The period of the satellite’s motion is
T


6
2 r 2 2 6.38  10 m 

 1.43  104 s  3.98 h .
3
vt
5.59  10 m s
(c) The gravitational force acting on the satellite is F  GmE m r 2 , or


24

N  m 2  5.98  10 kg  600 kg 
F   6.67  1011
 1.47  103 N .
2
2 
6
kg

 2 6.38  10 m 


7.37


The gravitational force exerted on Io by Jupiter provides the centripetal acceleration, so
 v2  G M m
r v t2
m t  
M

,
or
.
r2
G
 r
The orbital speed of Io is


2  4.22  108 m
2 r
vt 

 1.73  104 m s .
T
1.77 days  86 400 s day 
 4.22  10 m 1.73  10
M
8
Thus,
7.38
4
m s

2
6.67  1011 N  m 2 kg 2
27
 1.90  10 kg .
The radius of the satellite’s orbit is
r  RE  h  6.38  106 m  2.00  106 m  8.38  106 m .
(a)
PEg  
G ME m
r


2
5.98  1024 kg 100 kg 

11 N  m 
   6.67  10
  4.76  109 J
2 
6
kg 
8.38  10 m

(b) F 
7.39


2
5.98  1024 kg 100 kg 
G ME m 
11 N  m 

6.67

10
 568 N
2

r2
kg 2 
8.38  106 m


The radius of the satellite’s orbit is
r  RE  h  6.38  106 m  200  103 m  6.58  106 m .
222
C H A P T E R
7
(a) Since the gravitational force provides the centripetal acceleration,
 v 2  GmE m
m t  
, or
r2
 r




2
5.98  1024 kg

GmE
11 N  m 
vt 
  6.67  10
 7.79  103 m s .
2 
6
r
kg  6.58  10 m

Hence, the period of the orbital motion is
T


6
2  r 2  6.58  10 m

 5.31  103 s  1.48 h .
vt
7.79  103 m s
(b) The orbital speed is v t  7.79  103 m s as computed above.
(c) Assuming the satellite is launched from a point on the equator of the Earth, its
initial speed is the rotational speed of the launch point, or
vi 


6
2  RE 2  6.38  10 m

 464 m s .
1 day
86 400 s
The work-kinetic energy theorem gives the energy input required to place the

satellite in orbit as Wnc  KE  PEg
   KE  PE  , or
g
f
i
 v t2  v i2
 1 1 
GM Em   1 2 GM Em 
1
Wnc   mv t2 

mv


m
 GM E 
 


i


2
r  2
RE 
 RE r  
 2
Substitution of appropriate numeric values into this result gives the minimum
energy input as Wnc  6.43  109 J .
7.40
The gravitational force on a small parcel of material at the star’s equator supplies the
 v t2 
G Ms m
2
centripetal acceleration, or

m
 R   m Rs  .
Rs2
s


Hence,   G M s Rs3

 6.67  10
11
 


N  m 2 kg 2 2 1.99  1030 kg 
 1.63  104 rad s .
3
3
10.0  10 m

223
C H A P T E R
7.41
(a)  
7
2 rad 2  rad
5

 7.27  10 rad s
1 day
86 400 s
(b) The elapsed time is the time required for the Earth to rotate 107° on its axis, or
t 
7.42
107


107  rad 180
7.27  10
5
rad s

 7.13 h
(a) The gravitational force exerted on an object of mass m near the surface (i.e., at
distance R from the center) of a planet of mass M is
Fg 
GM m
GM m
. Equating this to Fg  m g gives m g 
, or
2
R
R2
g
GM
as the acceleration of gravity at the planet’s surface.
R2
For Mars,



2 0.1074 5.98  1024 kg 

  3.77 m s 2
11 N  m  
g   6.67  10
kg 2  0.5282 6.38  106 m 2




1
(b) From y  v i t  ay t 2 with vi  0 and ay   g , we find
2
t
7.43
2  y 
g

2  20.0 m 
 3.26 s
3.77 m s 2
The angular velocity of the ball is   0.500 rev s   rad s .
(a)
v t  r    0.800 m   rad s   2.51 m s
(b) ac 
v t2
2
 r  2   0.800 m   rad s   7.90 m s 2
r
224
C H A P T E R
7
(c) We imagine that the weight of the ball is supported by a frictionless platform. Then,
the rope tension need only produce the centripetal acceleration. The force required
to produce the needed centripetal acceleration is F  m v t2 r . Thus, if the maximum


force the rope can exert is 100 N, the maximum tangential speed of the ball is
 v t max 
7.44
r Fmax

m
 0.800 m 100 N 
5.00 kg
(a) i 
1.30 m s
vt

 56.5 rad s
ri 2.30  10-2 m
(b)  f 
1.30 m s
vt

 22.4 rad s
rf 5.80  10-2 m
 4.00 m s
(c) The duration of the recording is
t   74 min   60 s min   33 s  4473 s .
Thus,  
(d)  
 f  i
 2f  i2
2
t


 22.4  56.8
rad s
4473 s
3
2
  7.63  10 rad s
 22.4 rad s 2   56.5 rad s 2 

2 7.63  10
3
rad s
2

1.77  105 rad
(e) The track moves past the lens at a constant speed of v t  1.30 m s for 4473 seconds.
Therefore, the length of the spiral track is
s  v t  t   1.30 m s   4473 s   5.81  103 m  5.81 km
7.45
The radius of the satellite’s orbit is


r  RE  h  6.38  106 m  1.50  102 mi 1609 m 1 mi   6.62  106 m .
(a) The required centripetal acceleration is produced by the gravitational force, so
 v2  G ME m
G ME
m t  
, which gives v t 
.
2
r
r
 
r
vt 


2
5.98  1024 kg

3
11 N  m 
 7.76  10 m s
2 
6
 6.67  10
kg  6.62  10 m
225
C H A P T E R
7
(b) The time for one complete revolution is


6
2  r 2  6.62  10 m
T

 5.36  103 s  89.3 min
vt
7.76  103 m s
7.46
(a) When the car is about to slip down the incline,
the friction force, f, is directed up the incline as
shown and has the magnitude f  n . Thus,
Fy  n cos   n sin   mg  0 ,
or n 
mg
cos    sin 
(1)
 v2 
Also, Fx  n sin   n cos   m  min  ,
 R 
or v min 
nR
 sin    cos   .
m
(2)
Substituting equation (1) into (2) gives
v min 
 sin    cos  
R g

 cos    sin  
 tan    
R g
 1   tan  
If the car is about to slip up the incline, f  n is directed down the slope (opposite
to what is shown in the sketch). Then,
Fy  n cos   n sin   mg  0 , or n 
mg
cos    sin 
(3)
 v2 
Also, Fx  n sin   n cos   m  max  , or
 R 
v max 
nR
 sin    cos   .
m
(4)
Combining equations (3) and (4) gives
 sin    cos  
v max  R g 

 cos    sin  
 tan    
R g
.
 1   tan  
226
C H A P T E R
7
(b) If R  100 m, =10, and =0.10 , the lower and upper limits of safe speeds are:
v min 
and
7.47
tan 10  0.10

100 m   9.8 m s 2  

1  0.10 tan 10 
v max 
8.6 m s ,
tan 10  0.10

100 m   9.8 m s 2  

1  0.10 tan 10 
17 m s .
(a) When the car is at the top of the arc, the normal force is upward and the weight
downward. The net force directed downward, toward the center of the circular path
and hence supplying the centripetal acceleration, is Fdown  mg  n  m v t2 r .




Thus, the normal force is n  m g  v t2 r .
(b) If r  30.0 m and n  0 , then g 
vt  r g 
7.48
v t2
 0 or the speed of the car must be
r
 30.0 m   9.80 m s 2  
17.1 m s .
(a) At the lowest point on the path, the net upward force (i.e., the force directed toward
the center of the path and supplying the centripetal acceleration) is
 v2 
Fup  T  mg  m  t  , so the tension in the cable is
 r
2

3.00 m s  


v t2 
2
T  m  g     0.400 kg   9.80 m s 
  8.42 N .

r
0.800 m 


(b) Using conservation of mechanical energy, KE  PEg
   KE  PE 
f
g
from the lowest to the highest point on the path gives
v2
1
0  m g L 1  cos max    mvi2  0 , or cos max  1  i .
2gL
2
max
2


3.00 m s 


v i2 
1
 cos  1 

cos
1


  64.8

2 g L 
2 9.80 m s 2  0.800 m  

1

227

i
, as the bob goes
C H A P T E R
7
(c) At the highest point on the path, the bob is at rest and the net radial force is
 v2 
Fr  T  m g cos max  m  t   0
 r
Therefore,


T  m g cos max   0.400 kg  9.80 m s 2 cos  64.8  1.67 N
7.49
The speed the person has due to the rotation of the Earth is v t  r  where r is the
distance from the rotation axis and  is the angular velocity of rotation.
 
The person’s apparent weight, Fg
apparent
, equals the magnitude of the upward normal
 
force exerted on him by the scales. The true weight, Fg
true
 mg , is directed downward.
The net downward force produces the needed centripetal acceleration, or
 
Fdown   n  Fg
true
 
  Fg
apparent
 
(a) At the equator, r  RE , so Fg
true
 v2 
 m  t   mr  2 .
true
 r 
 
 Fg
 
 Fg
apparent
 
 m RE  2 > Fg
apparent
.
(b) At the equator, it is given that r  2  0.0340 m s 2 , so the apparent weight is
F 
g
apparent
 
 Fg
true
 mr  2   75.0 kg   9.80  0.0340 m s 2   732 N .
At either pole, r  0 (the person is on the rotation axis) and
F 
g
apparent
 
 Fg
true


 m g  75.0 kg  9.80 m s 2  735 N .
228
C H A P T E R
7.50
7
When the rope makes angle  with the vertical, the net
force directed toward the center of the circular path is
Fr  T  m g cos  as shown in the sketch. This force
supplies the needed centripetal acceleration, so
 v2 

v2 
T  m g cos   m  t  , or T  m  g cos   t  .
r
 r

Using conservation of mechanical energy, with KE  0 at   90 and PEg  0 at the
bottom of the arc, the speed when the rope is at angle  from the vertical is given by
1
mv t2  m g  r  r cos    0  m gr , or v t2  2 g r cos  . The expression for the tension in the
2
rope at angle  then reduces to T  3 m g cos  .
(a) At the beginning of the motion,   90 and T = 0 .
(b) At 1.5 m from the bottom of the arc, cos  

2.5 m 2.5 m

 0.63 and the tension is
r
4.0 m

T  3 70 kg  9.8 m s 2  0.63  1.3  103 N  1.3 kN .
(c) At the bottom of the arc,   0 and cos   1.0 , so the tension is


T  370 kg  9.8 m s 2 1.0  2.1  103 N  2.1 kN .
7.51
The normal force exerted on the person by the cylindrical wall must provide the
centripetal acceleration, so n  m r  2 .


If the minimum acceptable coefficient of friction is present, the person is on the verge of
slipping and the maximum static friction force equals the person’s weight, or
 fs max   s min n  mg .
Thus,  s  min 
mg
g
9.80 m s 2


 0.131 .
n
r  2  3.00 m   5.00 rad s  2
229
C H A P T E R
7.52
7
The horizontal component of the tension in the cord is the
only force directed toward the center of the circular path,
so it must supply the centripetal acceleration. Thus,
 v2 
 v2 
m v t2
T sin   m  t   m  t  , or T sin 2  
L
 r
 L sin  
(1)
Also, the vertical component of the tension must support
the weight of the ball, or
T cos   m g .
(2)
(a) Dividing equation (1) by (2) gives
Lg
sin 2  v t2
, or v t  sin 
.

cos  L g
cos 
(3)
With L  1.5 m s and =30 ,
v t  sin 30
1.5 m   9.8 m s 2 
cos 30
 2.1 m s .
(b) From equation (3), with sin 2   1  cos 2  , we find
 v2 
1  cos 2  v t2
or cos 2    t  cos   1  0 .

cos 
Lg
 L g
Solving this quadratic equation for cos  gives
2
 v2 
 v2 
cos     t    t   1
 2L g
 2L g
If L  1.5 m and v t  4.0 m s , this yields solutions: cos   1.7 (which is
impossible), and cos   0.59 (which is possible).
Thus,
  cos1  0.59  54 .
230
C H A P T E R
7
(c) From equation (2), when T  9.8 N and the cord is about to break, the angle is


 0.50 kg  9.8 m s 2 
 m g
1 

cos

  60 .
 T 
9.8 N


  cos 1 
Then equation (3) gives
Lg
v t  sin 
 sin 60
cos 
7.53
1.5 m   9.8 m s 2 
cos 60
 4.7 m s .
Choosing PEg  0 at the top of the hill, the speed of
the skier after dropping distance h is found using
conservation of mechanical energy as
1
mv t2  m g h  0  0 , or v t2  2 g h .
2
The net force directed toward the center of the
circular path, and providing the centripetal
acceleration, is
 v2 
Fr  m g cos   n  m  t  .
 R
Solving for the normal force, after making the substitutions v t2  2 g h and
cos  
Rh
h
 1 ,
R
R
gives
h
3h 
 2 g h


n  m g 1    m 
 m g 1   .



 R 
R
R
The skier leaves the hill when n  0 . This occurs when
1
7.54
(a)
F
R
3h
 0 or h 
3
R
GM m
r2


2 100 1.99  10 30 kg  1000 kg 


11 N  m  
  6.67  10
 1.3  1017 N .
2
3
kg 2 

10  10 m  50 m 
231
C H A P T E R
(b) F 
G M 1.0 kg 
2
front
r

G M 1.0 kg 
2
back
r
, or
7
 1
F
1 
 GM  2  2 
1.0 kg
 rfront rback 


F
N m2 
1
32

  6.67  1011
1.99

10
kg
2 
4
1.0 kg 
kg 
 10 m





2




2
4
10 m+100 m 
1

= 2.6  1012 N kg
7.55
Define the following symbols: M m  mass of moon, M e  mass of the Earth,
Rm  radius of moon, Re  radius of the Earth, and r = radius of the Moon’s orbit
around the Earth.
We interpret “lunar escape speed” to be the escape speed from the surface of a
stationary moon alone in the universe. Then,
v launch  2 v escape  2
2G M m
8G Mm
2
, or v launch
.

Rm
Rm
232
C H A P T E R
7
Applying conservation of mechanical energy from launch to impact gives
 
1
2
mvimpact
 PEg
2
2
vimpact  v launch

f
 
1
2
 mv launch
 PEg , or
i
2
   
2
PEg  PEg
i
m 
f


The needed potential energies are
 PE    G MR
g
i
m
m

m
 
G Me m
and PEg
r
f

G Me m G Mm m
.

Re
r
2
Using these potential energies and the expression for v launch
from above, the equation for
the impact speed reduces to
v impact 
 3 Mm Me  Me  Mm 
2G


.
Re
r
 Rm

With numeric values of G  6.67  1011
N m2
, M m  7.36  1022 kg , Rm  1.74  106 m ,
2
kg
Re  6.38  106 m , and r  3.84  108 m , we find
v impact  1.18  10 4 m s  11.8 km s .
7.56
The escape speed from the surface of a planet of radius R and mass M is given by
ve 
2G M
.
R
If the planet has uniform density, , the mass is given by


M    volume   4 R 3 3  4  R 3 3 .
The expression for the escape speed then becomes
ve 
2G  4  R3   8  G 

R   constant  R ,
R 
3  
3 
or the escape speed is directly proportional to the radius of the planet.
233
C H A P T E R
7.57
7
If the block will just make it through the top of the loop, the force required to produce
the centripetal acceleration at point C must equal the block’s weight, or m v c2 R  m g .

vc  R g
This gives

as the required speed of the block at point C.
We apply the work-kinetic energy theorem in the form

Wnc  KE  PEg  PEs
   KE  PE
g
f
 PEs

i
from when the block is first released until it reaches point C to obtain
 
1
1
fk AB cos180  mv c2  mg  2 R   0  0  0  kd2 .
2
2
The friction force is fk  u k  mg  , and for minimum initial compression of the spring,
v c2  Rg as found above. Thus, the work-energy equation reduces to
dmin 
dmin 
7.58
 
2 k mg AB  mRg  2 mg  2 R 
k

mg 2 k AB  5 R

k
 0.50 kg   9.8 m s 2  2  0.30 2.5 m   5 1.5 m 

78.4 N m
.
0.75 m
Choosing y  0 and PEg  0 at the level of point B, applying the work-kinetic energy
theorem to the block’s motion gives
2Wnc
1
1
Wnc  mv 2  mgy  mv 02  mg  2R  , or v 2  v 02 
 2 g  2 R  y  (1)
2
2
m
(a) At point A, y  R and Wnc  0 (no non-conservative force has done work on the
block yet). Thus, v A2  v 02  2 gR . The normal force exerted on the block by the track
must supply the centripetal acceleration at point A, so
 v 02

 v A2 
n A  m    m   2 g
 R
R

  4.0 m s  2

  0.50 kg  
 2 9.8 m s 2   15 N .
 1.5 m



234
C H A P T E R
7
At point B, y  0 and Wnc is still zero. Thus, v B2  v 02  4 gR . Here, the normal force
must supply the centripetal acceleration and support the weight of the block.
Therefore,
 v2

 v2 
n B  m  B   mg  m  0  5 g
 R
R

  4.0 m s  2

  0.50 kg  
 5 9.8 m s 2   30 N .
 1.5 m



(b) When the block reaches point C, y  2R and Wnc   fk L   k  mg  L . At this point,
the normal force is to be zero, so the weight alone must supply the centripetal
acceleration. Thus, m v c2 R  mg , or the required speed at point C is v c2  R g .


Substituting this into equation (1) yields R g  v 02  2 k gL  0 , or


2
v 02  R g  4.0 m s   1.5 m  9.8 m s
k 

 0.17 .
2 gL
2 9.8 m s 2  0.40 m 
2

7.59

(a) At point A, the weight of the coaster must be just large enough to supply the
 v2 
centripetal acceleration. Thus, m  A   mg , or v A2  Rg .
 R
Applying conservation of mechanical energy from the start to point A,
2 gh
1 2
1
 2h 
.
mv 0  mgh  mv A2  mg   , or v 02  v A2 


2
2
3
3
Using the value of v A2  Rg , so the car barely stays on track at A, gives
v0  R g 
2 gh

3
2 h

gR  

3
(b) If the speed of the coaster is to be zero at point B, conservation of mechanical energy
from the start to point B gives
2 h 
1
1  
0  mg h   mv 02  mgh  m  g  R     mgh ,
2
2  
3 
or h 
R 2h

.
2 3
235
C H A P T E R
7
Answers to Even Numbered Conceptual Questions
2.
The scratch on the rim of the disc moves in a circular path around the axis of rotation.
Whenever the disc is rotating, whether its angular speed is constant or not, the scratch has
a non-zero tangential component of velocity, but has zero radial velocity. The scratch has a
radial acceleration (the centripetal acceleration) any time the disc is rotating. It has a
tangential component of acceleration only while the disc has an angular acceleration as it
is coming up to speed.
4.
The need for a large force toward the center of the circular path on objects near the
equator will cause the Earth to bulge at the equator. A force toward the center of the
circular path is not needed at the poles, so the radius in this direction will be smaller than
at the equator.
6.
To a good first approximation, your bathroom scale reading is unaffected because you,
Earth, and the scale are all in free fall in the Sun’s gravitational field, in orbit around the
Sun. To a precise second approximation, you weight slightly less at noon and at midnight
than you do at sunrise or sunset. The Sun’s gravitational fields is a little weaker at the
center of the Earth than at the surface sub-solar point, and a little weaker still on the far
side of the planet. When the Sun is high in your sky, its gravity pulls up on you a little
more strongly than on the Earth as a whole. At midnight the Sun pulls down on you a
little less strongly than it does on the Earth below you. So you can have another doughnut
with lunch, and your bedsprings will still last a little longer.
8.
The astronaut is accelerating toward the Earth at the same rate as is the spaceship. Thus, if
the astronaut drops a wrench, it will float in space next to him. Likewise, he will float in
space next to a desk or with reference to the spaceship. Thus, he believes himself to be
weightless.
10.
Consider one end of a string connected to a spring scale and the other end connected to an
object, of true weight w. The tension T in the string will be measured by the scale and
construed as the apparent weight. We have w -T =m ac . This gives, T =w -m ac . Thus, the
apparent weight is less than the actual weight by the term m ac . At the poles the
centripetal acceleration is zero. Thus, T =w . However, at the equator the term containing
the centripetal acceleration is nonzero, and the apparent weight is less than the true
weight.
12.
If the acceleration is constant in magnitude and perpendicular to the velocity, the object is
moving in a circular path at constant speed. If the acceleration is parallel to the velocity,
the object is either speeding up, v and a in same direction, or slowing down, v and a in
opposite directions.
14.
When an object follows an orbital path, there is a force acting on it that produces the
centripetal acceleration. This force and the centripetal acceleration are always directed
toward the center of the orbit. In the case of a planet orbiting the Sun, this force is the
gravitational force exerted on the planet by the Sun, and it points toward the center of the
Sun. Hence, the center of the Sun must coincide with the center of the orbit and lie in the
plane of the orbit.
236
C H A P T E R
16.
7
Kepler’s second law says that equal areas are swept out in equal times by a line drawn
from the Sun to the planet. For this to be so, the planet must move fastest when it is closest
to the Sun. This, surprisingly, occurs during the winter.
237
C H A P T E R
7
Answers to Even Numbered Problems
2.
2.1 m, 1.2  102 m , 7.7  102 m
4.
4.2  102 rad s 2
6.
(a) 5.24 s
8.
147 rev
10.
50 rev
12.
1.02 m
14.
4.9  102 rad s
16.
3.44 km, 9.42 m s 2
18.
(a)
20.
1.5  102 rev s
22.
0.966 g
24.
(b) 20.1°
26.
The required tension in the vine is 1.4  103 N . He does not make it.
28.
(a) 25 kN
30.
(a)
(c)
32.
2.59  108 m from the center of the Earth
34.
2.00 kg and 3.00 kg
36.
(a)
5.59  103 m s
(b) 3.98 h
38.
(a)
 4.76  109 J
(b) 568 N
40.
1.63  104 rad s
42.
(a)
14.1 m s
(b) 27.4 rad
(b) 200 m
(b)
12 m s
4.39  1020 N toward the Sun
3.55  1022 N toward the Sun
3.77 m s 2
(c) 28.2 s
(b)
(b) 3.26 s
238
1.99  1020 N toward the Earth
(c)
1.47  103 N
C H A P T E R
44.
46.
(a)
56.5 rad s
(b)
(d)
1.77  105 rad
(e) 5.81 km
(a)
v min 
(b)
8.6 m s to 17 m s
 tan    
R g
, v max 
 1   tan  
7
22.4 rad s
(c)
 7.63  103 rad s 2
 tan    
R g
 1   tan  
48.
(a) 8.42 N
(b) 64.8°
(c) 1.67 N
50.
(a) 0
(b) 1.3 kN
(c) 2.1 kN
52.
(a)
2.1 m s
(b) 54°
(c)
54.
(a)
1.3  1017 N
(b)
58.
(a) 15 N, 30 N
2.6  1012 N kg
(b) 0.17
239
4.7 m s