3-7 The coordinates of a bird flying in t

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```					               EXERCISES AND SOLUTIONS OF UNIVERSITY PHYSICS

2008-4-10
CHAPTER 3
3-7 The coordinates of a bird flying in the xy -plane are given by x(t )  t , and

y (t )  3.0m  t 2 , where   2.4m / s and   1.2m / s 2 . a) Find the trajectory equation
of the bird. b) Calculate the velocity and acceleration vectors of the bird as function of time. c)
Calculate the magnitude and direction of the bird’s velocity and acceleration vectors at t  2.0s .
At this instant, is the bird speeding up, is it slowing down, or is its speed instantaneously not
changing? Is the bird turning? If so, in what direction?

3-8 A particle moves along a path as shown in Fig. 3-31. Between points B and D , the path is
a straight line. Sketch the acceleration vectors at A , C , and E in the cases in which a) the
particle moves with a constant speed; b) the particle moves with steadily increasing speed; c) the
particle moves with a steadily decreasing speed.


                           v
v                                                           
v
E                            E
E

                 D                          D                              D
v                              v                                 v
C                              C
C
B                  
B                                                 v             B
                           
v                           v
A                              A                                A

v

3-55 A basketball player is fouled and knocked to the floor during a layup attempt. The player is
awarded two free throws. The center of the basket is a horizontal distance of 4.21m from the
foul line and is a height of 3.05m above the floor. On the first free-throw attempt, he shoots the

above the horizontal and with a speed of v 0  4.88 m / s . The ball is
0
ball at an angle of 35

released 1.83m above the floor. This shot misses badly. Ignore air resistance, a) What is the
maximum height reached by the ball? b) At what distance along the floor from the free-throw line
does the ball land? c) For the second throw, the ball goes through the center of the basket. For this
0
second free throw, the player again shoots the ball at 35            above the horizontal and releases it

1.83m above the floor. What initial speed does the player give the ball on this second attempt? d)

1
For the second throw, what is the maximum height reached by the ball? At this point, how far
horizontally is the ball from the basket?

3.05m

1.83m

4.21m

4-30 The position of a           2.75  10 5 kg training helicopter under test is given by

                                                               ˆ
ˆ
r  (0.020 m / s 3 )t 3 i  (2.2m / s )tˆ  (0.060 m / s 2 )t 2 k . Find the net force on the helicopter at
j
t  5.0s
4-47 The two blocks in Fig. 4-31 are connected by a heavy uniform rope with a mass of 4.0kg .

An upward force of 200N is applied as shown. a) Draw three free-body diagrams, one for the

6.00kg block, one for the 4.00kg rope, and another one for the 5.00kg block. For each

force, indicate what body exerts that force. b) What is the acceleration of the system? c) What is
the tension at the top of the heavy rope? d) What is the tension at the midpoint of the rope?

F  200N

6.00kg

4.00kg

5.00kg

Chapter 5
5-80 Two blocks connected by a cord passing over a small frictionless pulley rest on frictionless
planes. A) Which way will the system move when the blocks are released from rest? B) What is
the acceleration of the blocks? C) What is the tension in the cord?

100kg                             50kg
2

30.0 0                 53.10
5-95 A rock with mass m  3.00kg falls from rest in a viscous medium. The rock is acted on by

a net constant downward force of 18.0N (a combination of gravity and the buoyant force

exerted by the medium) and by a fluid resistance force f  kv where v is the speed in m / s

and k  2.20N  s / m . a) Find the initial acceleration, a 0 . b) Find the acceleration when the

speed is 3.00m / s . c) Find the speed when the acceleration equals 0.1a 0 . d) Find the terminal

speed, v t . e) Find the coordinate, speed, and acceleration at time t after the start of the motion.

5-96 The 4.00 -kg block in Fig. 5-65 is attached to a vertical rod by means of two strings. When
the system rotates about the axis of the rod, the strings are extended as shown in the diagram and
the tension in the upper string is 80.0N . a) What is the tension in the lower cord? b) How many
revolutions per minute does the system make? c) Find the number of revolutions per minute at
which the lower cord just goes slack. d) Explain what happens if the number of revolutions per
minutes is less than in part (c).

1.25m

2.00m                           4.00kg

1.25m

5-109 A wedge with mass M rests on a frictionless horizontal table top. A block with mass m

is placed on the wedge, and a horizontal force F is applied to the wedge. What must be the

magnitue of F if the block is to remain at a constant height above the table top?

m

F
M           


6-24 A child applies a force F parallel to the x -axis to a 10.0-kg sled moving on the frozen

surface ofa small pond. As the child controls the speed of the sled, the x component of the force

3
she applies varies with the x -coordiante of the sled as shown in Fig. 6-21. Callculate the work
done by the force when the sled moves a) from x  0 to x  8.0m ; b) from x  8.0m to
x  12.0m ; c) from x  0 to 12.0m .
Fx (N )

10

5

4           8          12      x(m)

6-57 An object that can move along the x -axis is attracted toward the origin with a force of

magnitude F  x , where   4.00 N / m . What is the force F when the object is a) at
3                          3

x  1.00m ? b) at x  2.00m ? c) How much work is done by the force F when the object
moves from x  1.00m to x  2.00m ? Is this work positive or negative?

6-59 A small block with a mass of 0.120kg is attached to a cord passing through a hole in a

horizontal frictionless surface. The block is originally revolving at a distance of 0.40m from
below, shortening the radius of the circle in which the block revolves to 0.10m . At this new
distance the speed of the block is observed to be 2.80m / s . a) What is the tension in the cord in
the original situation when the block has speed v  0.70m / s ? b) What is the tension in the cord
in the final situation when the block has speed v  2.80m / s ? c) How much work was done by
the person who pulled on the cord?


F

6-73 Consider the system shown in Fig. 6-26. The rope and pulley have negligible mass, and the
pulley is frictionless. Initially the 6.00-kg block is moving downward and the 8.00-kg block is
moving to the right, both with a speed of 0.900m/s. the blocks come to rest after moving 2.00m.
use the work-energy theorem to calculate the coefficient of kinetic friction between the 8.00-kg
block and the table top.                                8.00kg

4
6.00kg
CHAPTER 7
7-33 An experimental device moving in the xy -plane is acted on by a conservative force

described by the potential-energy function U ( x, y )  k ( x  y )  k ' xy whre k and k '
2     2

are positive constants. Derive an expression for the force expressed in terms of the unit vectors

iˆ and ˆ .
j
7-51 A skier starts at the top of a very large frictionless snowball, with a very small initial speed,
and skis straight down the side. At what point does she lose contact with the snowball and fly off
at a tangent? That is, at the instant she loses contact with the snowball, what angle  does a
radial line from the center of the snowball to the skier make with the vertical?



7-55 A certain spring is found not to obey Hooke’s law; it exerts a restoring force

Fx  x   x 2      if   it   is       streched       or   compressed,   where     60.0N / m       and

  18 .0 N / m 2 . The mass of the spring is negligible. a) Calculate the potential energy function

U (x) for this spring. Let U  0 when x  0 . b) An objetc with a mass of 0.900kg on a

frictionless horizontal surface is attatched to this spring, pulled a distance 1.00m to the right to
stretch the spring, and released. What is the speed of the object when it is 0.50m to the right of
the x  0 equilibrium position?

7-67 A cutting tool under microprocessor control has several forces acting on it. One force is

F  xy 2 ˆ , a force in the negative y -direction whose magnitude depends on the position of
j

the tool. The constant is   2.50 N / m . Consider the displacement of the tool from the
3


origin to the point x  3.00m , y  3.00m . a) Calculate the work done on the tool by F if

this displacement is along the straight line y  x that connects these two points. b) Calculate

the work done on the tool by F if the tool is first moved out along the x -axis to the point

x  3.00m , y  0 and then moved parallel to the y -axis to x  3.00m , y  3.00m . c)

5
                               
Compare the work done by F along these two paths. Is F conservative or nonconservative?

Explain.

CHAPTER 8
8-30 At the intersection of Texas Avenue and University Drive, a blue subcompact car of mass

950kg traveling east on University collides with a maroon pickup truck of mass 1900kg that

is traveling north on Texas and ran a red light. The two vehicles stick together as a result of the
0
collision, and after the collison the wreckage is sliding at 16.0m / s in the direciton 24.0

east of north. Calculate the speed of each vehicle before the collision. The collision occurs during
a heavy rainstorm; fricition forces between the vehicles and the wet road can be neglected.
y (north)

24.0 0

x(east )

8-33 A 0.150 -kg glider is moving to the right on a horizontal frictionless air track with a speed
of 0.80m / s . It makes a head-on collision with a 0.300 -kg glider that is moving to the left
with a speed of 2.20m / s . Find the final velocity (magnitude and direction) of each glider if the
collision is elastic.

8-57 The net force acting on a 2.00 -kg discus while it is being thrown is (t )i  (   t ) ˆ ,
2 ˆ
j

where   25.0 N / s ,   30.0 N and   5.0 N / s . If the discus was originally at rest,
2

what is its velocity after the net force has acted for 0.500s ? Express your answer in terms of the
ˆ
unit vectors i and ˆ .
j
8-81 The objects in Fig. 8-41 are constructed of uniform wire bent into the shapes shown. Find the
position of the center of mass of each.

6

L                  L
L                              L

L

L                                          L                   L

L                                      L

8-83 You are standing on a concrete slab, whch in turn is resting on a frozen lake. Assume that
there is no friction between the slab and the ice. The slab has a weight five times your weight. If
you begin walking forward at 2.00m / s , relative to the ice, with what speed relative to the ice
does the slab move?
CHAPTER 9
9-33 Four small spheres, each of which can be regarded as a point of mass 0.200kg , are

arranged in a square 0.400m on a side and connected by light rods. Find the moment of inertia
of the system about an axis a) through the center of the square (point O ), perpendicular to its
plane; b) bisecting two opposite sides of the square (an axis along the line AB ); c) that passes
through the centers of the upper left and lower right spheres and through point O .

0.400m              0.200kg

A                       O                            B

9-35 A wagon wheel is constructed as in Fig. 9-22. The radius of the wheel is 0.300m , and the

rim has a mass of 1.40kg . Each of the eight spokes, which lie along a diameter and are

0.300m long, has a mass of 0.280kg . What is the moment of inertia of the wheel about an
axis through its center perpendicular to the plane of the wheel?                 0.600m

7
9-57 A roller in pringting press turns through an angle  (t ) given by  (t )  t   t , where
2      3

  3.20 rad / s 2 and   0.500 rad / s 3 .a) Calculate the angular velocity of the roller as a
function of time. b) Calculate the angular acceleration of the roller as a function of time.c) What is
the maximum positive angular velocity, and at what value of t does it occur?
9-71 The pulley in Fig. 9-24 has radius R and moment of inertia I . The rope does not slip over
the pulley, and the pulley spins on a frictionless axle. The coefficient of kinetic friction between

block A and the table top is  k . The system is released from rest, and block B descends.

Block A has mass m A and block B has mass m B . Use energy methods to calculate the

speed of block B as a function of the distance d that it has descended.

A
I

B

9-77 A thin flat disk hass mass M and radius R . A circular hole of radius R / 4 , centered at a
point R / 2 from the disk’s center, is then punched in the disk. a) Find the moment of inertia of
the disk with the hole about an axis through the original center of the disk, perpendicular to the
plane of the disk. b) Find the moment of inertia of the disk with the hole about an axis through
the center of the hole, perpendicular to the plane of the disk.

R/4
R

CHAPTER 10

ˆ
10-5 One force acting on a machine part is F  (5.00 N )i  (4.00 N ) ˆ . The vector from the
j

            ˆ
origin to the point where the force is applied is r  (0.450 m)i  (0.150 m) ˆ . a) In a sketch
j

    
show r , F , and the origin. b) Use the right-hand rule to determine the direction of the torque.

8
c) Calculate the vector torque produced by this force. Verify that the direction of the torque is the
same as you obtained in (b).
10-29 A 2.00 -kg rock has a horizontal velocity of magnitude 12.0m / s when it is at point P .
A) At this instant, what is the magnitude and direction of its angluar momentum relative to point
O ? b) If the only force acting on the rock is its weight, what is the rate of change of its angular
momentum at this instant?

v  12.0m / s

8.00m

36.9 0
O
10-33 A small block on a frictionless horizontal surface has a mass of 0.0250kg . It is attached

to a massless cord passing through a hole in the surface. The block is originally revolving at a
distance of 0.300m from the hole with an angular speed of 1.75rad / s . The cord is then
pulled from below, shortening the radius of the circle in which the block revolves to 0.150m .
The block may be treated as a particle. a) Is angular momentum conserved? Why or why not? b)
What is the new angular speed? c) Find the change in kinetic energy of the block. d) How much
work was done in pulling the cord?

10-34 The outstretched hands and arms of a figure skater preparing for a spin can be considered a
slender rod pivoting about an axis through its center. When his hands and arms are brought in
and wrapped around his body to execute the spin, the hands and arms can be considered a

thin-walled hollow cylinder. His hands and arms have a combined mass of 8.0kg . When

outstretched, they span 1.8m ; when wrapped, they form a cylinder of radius 25cm . The
moment of inertia about the rotation axis of the remainder of his body is constant and equal to

0.40 kg  m 2 . If his original angular speed is 0.40rev / s , what is his final angular speed?

10-38 A solid wood door 1.00m wide and 2.00m high is hinged along one side and has a

total mass of 40.0kg . Initially open and at rest, the door is struck at its center by a handful of

sticky mud of mass 0.500kg , traveling perpendicular to the door at 12.0m / s just before

impact. Find the final angular speed of the door. Does the mud make a significant contribution to

9
the momentum of inertia?

10-55 A block with mass m  5.00kg slides down a surface inclined 36.9
0
to the horizontal.

The coefficient of kinetic friction is 0.25 . A string attached to the block is wrapped around a

flywheel on a fixed axis at O . The flywheel has mass 25.0kg and moment of inertia

0.500 kg  m 2 with respect to the axis of rotation. The string pulls without slipping at a
perpendicular distance of 0.200m from that axis. a) What is the acceleration of the block
down the plane? b) What is the tension in the string?

5.00kg

36.9 0

10-75 Disks A and B are mounted on a shaft SS and may be connected or disconnected by
a clutch C . Disk A is made of lighter material than disk B so that the moment of inertia of
disk A about the shaft is one-third that of disk B . The moments of inertia of the shaft and the

clutch are negligible. With the clutch disconnected, A is brought up to an angular speed  0 .

The accelerating torque is then removed from A , and A is coupled to disk B by the clutch.
(Bearing friction may be neglected.) It is found that 2400J of thermal energy is developed in
the clutch when the ocnnection is made. What was the original kinetic energy of disk A ?

S                                                     S

C

A                        B

CHAPTER 22
22-60 Three point charges are placed on the y -axis: a charge q at y  a , a charge  2q at

the origin, and a charge q at y  a . Such an arrangement is called an electric quadrupole. a)
Find the magnitude and direction of the electric field at points on the y -axis for y  a . b) Use
the binomial expansion to show that very far from the quadupole, so that y  a , the electric
4
field is proportional to y        . Contrast this behavior to that of the electric field of a point charge

and that of the electric field of a dipole.

22-69 Positive charge Q is distributed uniformly along the x -axis from x  0 to x  a . A

10
positive point charge q is located on the positive x -axis at x  a  r , a distance r to the

right of the end of Q . a) Calculate the x - and y -components of the electric field produced by

the charge distribution Q at points on the positive x -axis where x  a . b) Calculate the

force that the charge distribution Q exerts on q . c) Show that if r  a , the magnitude of

the force in part (b) is approximately Qq / 4 0 r . Explain why this result is obtained.
2

y

q
Q
x
a                            r

22-77 Negative charge  Q is distributed uniformly around a quarter-circle of radius a that

lies in the first quadrant, with the center of curvature at the origin. Find the x - and
y –components of the net electric field at the origin.

2
22-81 Three large parallel insulating sheets have surface charge densities 0.0200C / m ,

0.0100C / m 2 , and  0.0200C / m 2 , respectively. Adjacent sheets are a distance of 0.300m
from each other. Calculate the net electric field due to all three sheets at a) point P ; b) point R ;
c) point S ; d) point T .

 0.0200C / m 2                         0.0200C / m 2

0.150m

P                 R                    S                T

 0.0100C / m 2

11
CHAPTER 23
23-3 A cube has sides of length L . It is placed with one corner at the origin as shown in Fig.

ˆ         ˆ
23-28. The electric field is uniform and given by E   Bi  Cˆ  Dk , where B , C , and
j

D are positive constants. a) Find the electric flux through each of the six cube faces S 1 , S 2 ,

S 3 , S 4 , S 5 , and S 6 . b) Find the electric flux through the entire cube.
S 2 (top)

S 6 (back)
L

L
S1                                       S 3 (right side)
(left side)

L
S5                   S 4 (bottom)
(front)

23-6 The three small spheres shown in Fig. 23-30 carry charges q1  4.00 nC , q 2  7.80 nC ,

and q . Find the net electric flux through each of the following closed surfaces shown in cross

section in the figure: a) S 1 ; b) S 2 ; c) S 3 ; d) S 4 ; e) S 5 ; f) Do your answers to parts (a)

through (e) depend on how the charge is distributed over each small sphere? Why or why not?

S3

q2
S1                         S2
q1

S4                 q3
S5

23-25 A solid conducting sphere of radius R that carries positive charge Q is concentric with a

very thin insulating shell of radius 2R that also carries charge Q . The charge Q is

distributed uniformly over the insulating shell. a) Find the electric field in each of the regions
0  r  R , R  r  2R , and r  2R . b) Graph the electric field magnitude as a function of
r.

12
23-26 A conducting spherical shell with inner radius a and outer radius b has a positive point

charge Q located at its center. The total charge on the shell is  3Q , and it is insulated from

its surroundings. a) Derive expressions for the electric field magnitue in terms of the distance
from the center for the regions r  a , a  r  b , and r  b . b) What is the surface charge
density on the inner surface of the conducting shell? c) What is the surface charge density on the
outer surface of the conducting shell? d) Sketch the electric field lines and the location of all
charges. e) graph the electric field magnitude as a function of r .

a

b
Q

 3Q

23-33 A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer
coaxial cylinder with inner radius b and outer radius c . The outer cylinder is mounted on
insulating supports and has no net charge. The inner cylinder has a uniform positive charge per
unit length  . Calculate th elelectric field a) at any point between the cylinders, a distance r
from the axis; b) at any point outside the outer cylinder. C) Graph the magnitude of the electric
field as a function of the distance r from the axis of the cable, from r  0 to r  2c . d)
Find the charge per unit length on the inner surface and on the outer surface of the outer cylinder.

23-37 A single, isolated, large conducting plate has a charge per unit area  on its surface.
Because the plate is a conductor, the electric field at its surface is perpendicular to the surface

and has magnitude E   /  0 . a) In Example 23-7 it was shown that the field caused by a large,

uniformly charged sheet with charge per unit area  has magnitude E   / 2 0 , exactly half

as much as for a charged conducting plate. Why is there a difference? b) By regarding the charge
distribution on the conducting plate as being two sheets of charge(one on each surface), each
with charge per unit are  , use the result of Exampke 23-7 and the principle of superposition to

show that E  0 inside the plate and E   /  0 outside the plate.
     

E   /0                                      E   /0

13
CHAPTER 24
24-31 In a certain region of space, the electric potential is V ( x, y, z )  Axy  Bx  Cy , where
2

A , B , and C are positive constants. a) Calculate the x -, y -, and z -components of the
electric field. b) At what points is the electric field equal to zero?
24-66 A thin insulating rod is bent into a semicircular arc of radius a , and a total electric charge

Q is distributed uniformly along the rod. Calculate the potential at the center of curvature of the

arc if the potential is assumed to be zero at infinity.

24-68 a) From the expression for E (r ) obtained in Example 23-9, find the expression for the

electric potential V (r ) as a function of r both inside and outside the uniformly charged

sphere. Assume that V  0 at infinity. b) Graph V and E as function of r from r  0
to r  3R .
24-69 Use the electric field calculated in Problem 23-25 to calculate the potential difference
between the solid conducting sphere and the thin insulating shell.

24-73 Electric charge is distributed uniformly along a thin rod of length a , with total charge Q .

Take the potential to be zero at infinity. Find the potential at the following points: a) point P , a

distance x to the right of the rod; b) point R , a distance y above the right-hand end of the

rod. c) In part (a) and pat (b), what does your result reduce to as x or y becomes much larger
than a ?
R

y

x         P
a

CHAPTER 25
25-50 An air capacitor is made by using two flat plates, each with area A , separated by a distance
d . Then a metal slab having thickness a (less that d ) and the same shape and size as the
plates is inserted between them, parallel to the plates and not touching either plate. a) What is the
capacitance of this arrangement? b) Express the capacitance as a multiple of the capacitance
when the metal slab is not present. C) Discuss what happens to the capacitance in the limits
a  0 and a  d .

d                                      a

14
25-53 A uniformly charged sphere of radius R has total charge Q , as described in Example

23-9. Calculate the electric-field energy density at a point a distance r from the center of the
sphere for a) r  R ; b) r  R . c) Calculate the total electric-field energy.
25-55 A parallel-plate capacitor has the space between the plates filled with two slabs of dielectric,

one with constant K 1 and one with constant K 2 . Each slab has thickness d / 2 , where d is

2 0 A  K1 K 2 
the plate separation. Show that the capacitance is C                      
d  K1  K 2 
        

K1                                  d /2

K2                                   d /2

CHAPTER 28
3                     8
28-4 A particle with a mass of 1.81  10 kg and a charge of 1.22  10 C has at a given


instant a velocity v  (3.00  10 m / s ) . What are the magnitude and direction of the particle’s
4


ˆ
acceleration produced by a uniform magnetic field B  (1.63T )i  (0.980 T ) ˆ ?
j
28-5 Each of the lettered points at the corners of the cube in Fig. 28-37 represents a positive
charge q moving with a velocity v of magnitude in the direction indicated . The region in the

figure is in a uniform magnetic field B , parallel to the x –axis and directed toward the right.

Copy the figure, find the magnitude and direction of the ofrce on each charge, and show the force

b
c
d

B

x
a

e

z
28-23 The electric field between the plates of the velocity selector in a Bainbridge mass

15
spectrometer is 1.12  10 V / m , and the magnetic field in both regions is 0.540T . A stream
5

of singly charged selenium ions moves in a circular path with a radius of 31.0cm in the
magnetic field. Determine the mass of one selenium ion and the mass number of this selenium
isotope in atomic mass units, rounded to the nearest integer.
28-51 The cube in Fig. 28-45, 75.0cm on a side, is in a uniform magnetic field of 0.860T

parallel to the x -axis. The wire abcdef carries a current of 6.58A in the direction

indicated a) Determine the magnitude and direction of the force acting on each of the segments

ab , bc , cd , de , and ef . b) What are the magnitude and direction of the total force on the
y
wire?

c

b                                   B

e
f                         x
a
d

z

CHAPTER 29
29-3 A pair of point charges q  4.00C and q'  1.50C , are moving in a reference

frame as shown in Fig. 29-29. At this instant, what are the magnitude and direction of the

magnetic field produced at the origin? Take v  2.00  10 m / s and v'  8.00  10 m / s .
5                           5

y


q           v


v'

0.400m                          x
q'
29-9 Two long, straight wires, one above the other, are separated by a distance 2a and are
parallel to the x -axis. Let the  y -axis be in the plane of the wires in the direction from the
lower wire to the upper wire. Each wire carries current I in the  x - direction. What are the

16
magnitude and direction of the net magnetic field of the two wires at a point in the plane of the
wires a) midway between them? a) at a distance a above the upper wire? C) at a distance a
below the lower wire?
29-19 Calculate the magnitude of the magnetic field at point P of Fig. 29-36 in terms of R ,

I 1 , and I 2 . What does your expression give when I1  I 2 ?

I1                                         I1
P

I2                R                    I2

29-23 Figure 29-37 shows, in cross section, several conductors that carry curents through the

plane of the figure. The currents have the magnitude I 1  4.0 A , I 2  6.0 A , and I 3  2.0 A ,

and the directions shown. Four paths, labled a through d , are shown. What is the line integral
 
 B  dl for each path? Each integral involves going around the path in the counterclockwise

a

I1

b                           I3

c
I2

d

29-25 Repeat Exercise 29-24 for the case in which the current in the central solid conductor is I 1 ,

that in the tube is I 2 , and these currents are in the same direction rather than in opposite

directions.
29-57 A long, straight wire with circular cross section of radius R carries a current I . Assume
that the current density is not constant across the cross section of the wire but rather varies as
J  r , where  is a constant. a) By the requirement that J integrated over the cross section
of the wire gives the total current I , calculate the constant  in terms of I and R . b) Use
Ampere’s law to calculate the magnetic field for i) r  R , ii) r  R . Express your answers in
terms of I .

17
CHAPTER 30
30-34 Suppose the loop in Fig. 30-28 is a) rotated about the y –axis; b) roatated about the
x -axis; c) rotated about an edge parallel to the z -axis. What is the maximum induced emf in

each case if A  600cm ,
2
  35.0rad / s , and B  0.450T ?
z

A                y



B
x

30-36 The current in the long, straight wire AB of Fig. 30-30 is upward and is increasing
steadily at a rate di / dt . a) At an instant when the current is i , what are the magnitude and

direction of the field B at a distance r to the right of the wire? b) What is the flux d B

through the narrow shaded strip? c) What is the total flux through the loop? e) Evaluate the
numerical value of the induced emf if a  12.0cm , b  36.0cm , L  24.0cm , and
di / dt  9.60 A / s .
B
r           dr

a                            L

b
A
SOLUTIONS:
3-7
1 2
a) x  2.4t , y  3  1.2t , y  3.0 
2
x
4.8
        
 dx ˆ dy ˆ                             dv
b) v     i           ˆ
j  2.4i  2.4tˆ(m / s) , a 
j                  2.4 j (m / s 2 )
dt    dt                               dt
c) v    2.4 2  (2.4t ) 2  2.4 2  (2.4  2) 2  5.36m / s , tg  2 ,

18
  3600  63.4 0  296.6 0

a  2.4m / s 2 , tg   ,   270 0
The particle is speeding up

a av                             
a av                       
3-8                                                                                                                 a av

E                                E
           E
                                                                        a av
a av                                     a av
D                                D                             D

C                                         C                            C
B                 
B                                                         v              B
                                
v                                v
A                                A                                    A

3-55
1 2
a) y  y 0  v0 sin t       gt , v y  v0 sin   gt .
2
v 0 sin 
At t  t 0 , v y  v 0 sin   gt 0  0 , So t 0                          .
g

1 2 (v0 sin  ) 2
y  y 0  v0 sin t 0      gt 0            y0
2        2g
(4.88  sin 350 ) 2
                      1.83  2.23(m)
2  9.8
1 2
b) At t  t1 , y  y 0  v0 sin t1          gt1  0 . Therefore t1  0.96 s .
2
x1  v 0 cos t1  4.88  cos 35 0  0.96  3.84 (m)

1 2
c) From y 2  y 0  v' 0 sin t 2       gt 2 and x 2  v' 0 cos t 2
2
2
1  x2       
 v' cos   x 2 tg  y 0
we get y 2   g          
2  0        

2 cos2   2
1
v' 0  [( y 2  y0  x2 tg )          2
]  8.65m / s
gx2

                  
       d 2r                             ˆ            ˆ                ˆ        ˆ
4-30 F  ma  m           2
 2.75  10 5  (6  0.020 ti  2  0.06 k )         1.65i  0.33 k ( N )
dt                                                t 5.0

19
4-47
a) Free body diagrams are shown as folows:

Tt
Tt
F                                  Tb

m1 g
Tt
m1 g                                            m1 g
Tm         m2 g / 2

b) F  m1 g  Tt  m1 a , Tt  m2 g  Tb  m2 a , Tb  m3 g  m3 a .

F  (m1  m 2  m3 ) g
a                           3.53 m / s 2 ,
m1  m 2  m3

Tt  F  m1 ( g  a)  120 ( N ) ,

m2
Tm  Tt       ( g  a)  93.34( N )
2
5-80 Free body diagrams of two particles
y1
                                                                    y2
n                                     x2       T             
T         x1                            n2

a

a
2
1                                                   
m2 g

m1 g

Newton’s secon law for each particles:

m1 g sin 1  T  m1a

T  m2 g sin  2  m2 a

m1 sin  1  m 2 sin  2       m m (sin  1  sin  2 )
Therefore we find a                                g, T  1 2
m1  m 2                     m1  m 2

20
5-95 Free body diagram:                      
 kv

18 .0 Nˆ
j

dv ˆ
a) Newton’s second law at time t : 18.0 ˆ  kvˆ  m
j     j                  j.
dt
dv 18.0
At t  0 , v  0 , so            6m / s 2 .
dt    3.0
b) When v  3.0m / s , for k  2.2N  s / m given in the text,
dv 18.0  2.2  3.0
a                        3.8m / s 2 .
dt         3.0
18.0  2.2v
c) When a  0.1a0  0.1  6                 , v  7.36m / s
3.0
18.0
d) vt        8.18m / s
k
dv     k
e) From       (v  vt ) ,where v t is given above, we have
dt     m
dv               k
vv    t
 
m
dt . The integration gives

k
k                             t
ln(v  vt )         t  C . That is v  vt  Ce m . At t  0 , v  0 , C  vt , we get
m
k
 t
v  vt (1  e    m
).

5-96                                                                         Tu
a) Newton’s second law gives
l
Tu cos   Tl cos   m 2 l sin 


Tu sin   Tl sin   mg                                     L

2                                                             
L                                                                       mg
l   2
Tl
2
sin           ,
l

Tu sin   mg
Tl 
sin 

where the values of l , L, Tu are all given in the text. Substitute these values , we can get the

21
numerical result of T l .

(Tu  Tl ) cos 
b)  
ml sin 

Tu cos 
c) When Tl  0 ,  
ml sin 
5-109

N                         
Free body diagram is shown on the right.                                                                  n

F
According to Newton’s secon law:
F  n sin   Ma                                                                       
n sin   ma                                                            
n
n cos  mg                                                                                         
Mg                    mg
a                                         F
We get tg        and F  (m  M )a . Therefore tg             .
g                                     (m  M ) g
6-24
8  10                    4  10
a) W             40( J ) , b) W          20( J ) ; c) W  40  20  60( J )
2                         2
6-57 F  x  4.0 x
3                   3

a) x  1.00m F  x  4.0 x  4.0( N )
3                3

b) x  2.00m F  x  4.0  2  32 .0( N )
3                    3

x2               x2                        2.0
x4
c) W12   Fdx    x dx                                 15 .0( J ) . The work is negative.
3

x1         x1
4                      1.0

v12
6-59 a) F1  m            0.147( N )
r1
2
v2
b) F2  m     9.408( N )
r2
1
c) W12  E K 2  E K 1                  m(v2  v12 )  0.441( J )
2

2
1
m2 gs  (m1  m2 )v12
6-73  k         2               0.786
m1 gs

7-33 From U ( x, y )  k ( x  y )  k ' xy , we get
2       2

22
U                               U                           U
Fx              (2kx  k ' y) , and Fy       (2ky  k ' x) , Fz      0.
x                               y                           z

v2
7-51 mg cos   m                        ,
R                                                                
1
mgR  mgR cos  mv 2
2
2
cos 
3
7-55
x                  x
U (0)  U ( x)   Fx dx   (x  x 2 )dx
a)                                0                  0

              
           x2            x3
2              3
            
Because, U (0)  0 , U ( x)                                  x2         x3 .
2            3
1 2                          2(U (1)  U (0.5))
b) U (1)  U (0.5)                     mv , v                                        7.85 m / s .
2                                    m
y
7-67
3.0

3.0                                                   (3.0,3.0)
a) W        y dy                                        50.625( J )
3                       4
y
0
4           0

3 3
3.0                                                3.0

b) W    3y dy                                                 67.5( J )
2
y
0
3                                     0

c) Non-conservative. Because the work is dependent on the path.
x
8-30
       
In the text, m A  950 kg , m B  1900 kg , v A1  v A1i , v B1  v B1 ˆ ,
ˆ               j

v  v(sin 24 0  cos 24 0 ) , and v  16.0m / s .
From the conservation of linear momentum
                                       ˆ
m A v A1  m B v B1  (m A  m B )v(sin 24 0 i  cos 24 0 ˆ)
j

(m A  mB )v sin 240              (m A  mB )v cos240
We have v A1                                 , and v B1 
mA                                mB

Substitute all the values of m A , mB , v , we can get the values of v A1 , v B1 .

8-33 m A  0.150 kg , mB  0.300 kg , v A1  0.80 m / s , v B1  2.20 m / s .

23
2m B (v A1  v B1 )                               2m A (v A1  v B1 )
v A 2  v A1                          2.67 m / s , v B 2  v B1                       1.27 m / s .
m A  mB                                          m A  mB
8-57
 0.5      0.5                                                     0.5
ˆ]dt  1 t 3i  ( t  1 t 2 ) ˆ
J   Fdt   [t i  (   t ) j
2ˆ                           ˆ                   j
0        0
3                2          0


 J       ˆ
v   0.53i  7.81 ˆ(m / s )
j                                                                                     y
m                                                                                  x
8-81

a) x cm  0                                            L                L
L                               L
L
2m cos
2        L
y cm              cos                                      y                                         L
2m       2                         y
y
b) x cm  0

L
2m
y cm    2 L                                  L                                                 L                 L
3m   3
L
m                                                      L                                              L
c) xcm    2 L
x                                   x
2m 4

L
m
L
y cm    2 
2m 4

L 3
2m
d) x cm  0 y cm             2 2  3L .
3m     6
         
8-83 m A v A  mB v B  0
mA                
vA
     m        1
vB   A v A   v A
mB        5
             mB
vB
1
9-33 a) I O  0.8            2  0.16  0.064kgm2
4
b) I AB  0.20  (0.20 )  4  0.032 kgm
2                    2

1
c) I  0.20  2           2  0.16  0.032kgm2
4

24
8
9-35 I  MR               mD 2  0.1932kgm2
2

12
9-57
d
a)       2t  3t 2 (rad / s)
dt
d
b)       2  6t (rad / s 2 )
dt
                        2
c) When   0 ,    m ax ,  t 0           2.13( s ) .  m ax      6.81(rad / s )
3                        3
9-71
1                    1
W  W g  W f    k Ns  m B gs , E K 1  0 E K 2         (m A v 2  mB v 2  Mv 2 )
2                    2
2 gs(m B   k m A )
v2 
1
m A  mB  M
2
9-77
The moment of inertia of solid disk is
1
IO        MR 2 ,
2
1              3
The moment of inertia of hole (if it is treated as a disk) is I '      mR 2  mR 2  mR 2 , where
2              2
M
m         .
4
1
The total moment of inertia is I  I O  I '      MR2 .
8

10-5
           
ˆ                   ˆ                  ˆ
c)   r  F  (0.450 i  0.150 ˆ)  (5.0i  4.0 ˆ)  1.05 k ( Nm)
j                 j
10-29
             ˆ
a) L  r  p  r mv k  19 .2kgm / s
2


dL                          ˆ
b)       r  F  r  mg  125 .4k (kgm2 / s 2 )
dt

10-33 m  0.025kg , r1  0.30 m , r2  0.15 m , 1  1.75 rad / s

             
  r  F  0 , L  constant vector, r1mv1  r2 mv 2 ,

v2 r121
b)   2           2  7(rad / s)
r2   r2

25
1
c) E K       ( I 2 2  I 112 )  0.0103( J )
2

2
10-34 m  8.0kg , 1  0.40 rev / s , I 0  0.40 kgm , L  1.8m , R  0.25m .
2

Since the angular momentum is conserved, I 11  I 2 2 ,

Therefore
1
(I 0     mL2 )1
I                12            [0.40  8.0  1.8 2 / 12]
2  1 1                                                     0.40  3.94(rev / s)
I2            1                0.5  8.0  0.252  0.40
mR  I 0
2

2
10-38 L  1.0m , m  0.50kg , M  40.0kg , v1  12 .0m / s

L     L          1          2v
mv1      mv  I , I  ML2 ,  
2     2          3          L
L
mv1                                                                          L
2           0.5  12  0.5                                    v1
v              
L 2                      2
m  ML 0.5  0.5   40.0  1
2 3                      3
3
               0.111(m / s )
0.25  26.67
2v
L
10-55                                                                  
1      1                                                           T                        
TR  MR2  MRa                                                                                 n
2      2

mg sin   T  f  ma                                                          f
mg cos  n

T
f  k n                                                                              
mg

mg sin    k mg cos
a                           , T  mg sin    k mg cos   ma .
mM /2
1
10-75 I A      I B , I A 0  ( I A  I B )
3
I A 0

IA  IB
, H  I A0  I A  I B  .
1
2
2
              2

1          4H
I A 0 
2
 3200( J )
2           3

22-60

26
          q              2q         q                         
E  j 4 ( y  a) 2  4 y 2  4 ( y  a) 2


     0                  0      0                           


q
j 2 2

  2a 4  6a 2 y 2          
4 0     y ( y  a2 )2

      6a 2 qj
For y  a , we get E 
4 0 y 4
22-69
        dy                                            x             y       
a) dE                  er , er  cosi  sin j                         i             j
4 0 r 2                                         x2  y2        x2  y2

        dy  a                dy            
E             e                        ( xi  yj )
4 0 r 2 r
0 4 0 ( x  y )
2   2 3/ 2

                     

4 0 x  (cosi  sin j )d
                   
         (sin i  cosj )
4 0 x
a
         y          x                        
                    i                            j
4 0 x  x 2  y 2    x2  y2                     
                                          0
        a         x      
            2       i         j  j
4 0 x  x  a 2    x2  a2      
a           a              Q 
When x  a ,                                  . Hence E            i
x2  a2           x             4 0 x 2

22-77
       ad                                                           
E             (cosi  sin j )          (sin i  cosj ) 2         ( i  j )
4 0 a 2
4 0 a                   0   4 0 a 
Q                                                                                a
                     (i  j )
2  0 a
2        2

 Q1  Q2  Q3
22-81 E P   E1  E 2  E 3 
2 0 A

Q1  Q2  Q3
E R  E1  E 2  E3 
2 0 A

Q1  Q2  Q3
E S  E1  E 2  E 3 
2 0 A

27
Q1  Q2  Q3
ET  E1  E 2  E 3 
2 0 A
 
23-3  S1     
S1
E  dA  CL2
 
 S 2   E  dA   DL2
S2

 S 3  CL2

 S 4  DL2

 S 5   BL2

 S 6  BL2

 t    Si  0

q1             q2                 q1  q 2                    q1  q3              q1  q 2  q3
23-6  S 1         , S2         ,  S3                   , S4                    ,  S5                    .
0             0                   0                          0                      0

23-25 a) 0  r  R , E1  0

Q                       Q
R  r  2R , E 2 4r 2               ,  E2                      .
0                   4 0 r 2

2Q                         Q
r  2R , E3 4r 2               ,  E3                     .
0                    2 0 r 2

Q                        Q
23-26 a) 0  r  a , E1 4r                ,  E1 
2
.
0                   4 0 r 2

a  r  b , E2  0

 2Q
r  b , E3 
4 0 r 2
Q
b)   
4a 2
4Q
c)   
4b 2
L             
23-33 a) a  r  b , E1 2rL             ,  E1 
0           0 2r

28
L             
b) r  c , E 2 2rL                 ,  E2 
0           0 2r

d)     b    , c  

    
23-37 a) E  E1  E 2  2                 ,
2 0  0

b) E  0

V
24-31 a) E x            ( Ay  2 Bx) ,
x
V
Ey           ( Ax  C )
y

Ez  0

b) ( Ay  2 Bx )  ( Ax  C )  0
2                    2

1                   Q
a
24-66 V                 dq 
4 0             4         0   a

24-69 r  R , E1  0 .

Q
R  r  2R , E 2 
4 0 r 2

2Q
r  2R , E 3 
4 0 r 2

  2 R Qdr              Q
V  V ( R)  V (2 R)   E  dl             
R 4 0 r
2
8 0 R

dq                    Q
24-73 V P     4    0 ( x0  x)
, dq  dx  dx .
a

0
dx                        x a
VP 
a
 4   0   ( x0  x)
        ln  0
4 0  x0 




 a        Q
When x0  a , VP                     
 x  4 x
4 0  0     0 0

29
0
dx                     x                 y0
VR        4
a             x y
2             2
,
y0
 tg , dx 
cos2 
d
0                   0

0
dx                                 d            1        
VR        4
a             x y
2             2
                         ln 
4 0 cos 4 0  cos
 tg 

0                   0
0
       y2  x2  x 
      ln  0        
4 0      y0     y0 
                                        a

                        y0  a 2
2
a 
      ln 1                                            
4 0                          y0                   y0 
                                          


25-50 a) V  Ex  Ea  E (d  a  x)  E (d  a )                             (d  a)
0

Q   A 0
C         
V d a

1   1   1        A 0          A 0
b)           , C1       , C2         ,
C C1 C 2          x          d ax

1   x    d ax d a       A 0
                  C
C A 0     A 0   A 0    d a

A 0
c) When a  0 , C                                   . When a  d , C  
d

Qr
25-53 a) r  R , E 
4 0 R 3
2
1      1  Qr 
u  0E2  0          
2      2  4 0 R 3 
          
2
1     1  Q
Q                                       
b) r  R , E            , u  0E  0                                       
2

4 0 r 2
2     2  4 0 r 2




                   2                     2
1  Qr                     1  Q                        
R

 4 R 3    4r dr 2  0  4 r 2
U   u 4r dr   4r dr  0 
2

22
                      

0
2        0       R                  0                 
c)
3Q 2

20 0 R

30
E0             E    
25-55 E1                    , E2  0 
K1  0 K1       K2 0K2

 d        d
V         (          )
 0 2 K1 2 K 2

Q A 2 0 A K 1 K 2
C               (          )
V   V   d    K1  K 2

m RB 2
28-23     
q   E
                                        
28-4 F  qv  B  ma a  qv  B / m
                                             
28-5 Fa  qva  B  qvj  Bi  qvBk
                             
Fb  qvb  B  qv(k )  Bi  qvBj
                      
Fc  qvc  B  qv(i )  Bi  0

                  2              2    
Fd  qv d  B  qv(i  k )    Bi      qvBj
2          2

                  2            2       
Fe  qve  B  qv( j  k )    Bi     qvB( j  k )
2        2
                               
28-51 a) Fab  Il ab  B  ILBk
                  
Fbc  Ilbc  B  ILBj
                    
Fcd  Ilcd  B  ILB( j  k )
                  
Fde  Ilde  B  ILBj
           
Fef  Ilef  B  0

                             
     0 qv  r
ˆ      0 qv        0 qv  r   0 qv
ˆ
29-3 B1              k        , B2             k
4 r   2
4 r 2
4 r   2
4 r 2

         0 I         0 I 
29-9 a) B  k                           
 2a        4a 

31
         0 I       0 I 
b) B  k                      
 2a         4a 

29-55

                                    0 Id l  r
ˆ
B  B1  B2  B3  B2  0  0   dB  
4 r     2

  0 Idl   0 I 2R   0 I
 k         k         k
4R 2       4R 2       8R

2R 3
R
3I
29-57 J  r , I            JdA   2rdrr 
0
3

2R 3
r                   r
 0 2r 3          0r 2 I

i) r  R , 2Br   0 JdA   0 2r dr                                          , B
2

0                   0
3              2R 3

0 I
ii) r  R , 2Br   0 I , B 
2r

                                                            0 In
29-65 a)  B  dl   0       I 2 BL                 0   InL ,  B 
2

d B
30-34 a)  B  A  B ,                     0
dt
 
b)  B  A  B  AB cos(t )

d B
             AB sin(t ) , E m ax  AB
dt

c)  m ax  AB

0 I
30-36 a) B           .
2r

0 I
b) d B  BdA  Ldr
2r

 0 IL b
c)  B           ln( )
2      a

d B    L dI b
d)            0     ln( ) .
dt     2 dt    a

32

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