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PHYS 195
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and 1
Physics Measurement
CHAPTER OUTLINE
1.1 1.2 1.3 1.4 1.5 1.6 1.7 Standards of Length, Mass, and Time Matter and ModelBuilding Density and Atomic Mass Dimensional Analysis Conversion of Units Estimates and OrderofMagnitude Calculations Significant Figures
ANSWERS TO QUESTIONS
Q1.1 Q1.2 Atomic clocks are based on electromagnetic waves which atoms emit. Also, pulsars are highly regular astronomical clocks. Density varies with temperature and pressure. It would be necessary to measure both mass and volume very accurately in order to use the density of water as a standard. People have different size hands. Defining the unit precisely would be cumbersome. (a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms (b) and (d). You cannot add or subtract quantities of different dimension. A dimensionally correct equation need not be true. Example: 1 chimpanzee = 2 chimpanzee is dimensionally correct. If an equation is not dimensionally correct, it cannot be correct.
Q1.3 Q1.4 Q1.5 Q1.6
Q1.7
If I were a runner, I might walk or run 10 1 miles per day. Since I am a college professor, I walk about 10 0 miles per day. I drive about 40 miles per day on workdays and up to 200 miles per day on vacation. On February 7, 2001, I am 55 years and 39 days old. 55 yr
Q1.8
F 365.25 d I + 39 d = 20 128 dFG 86 400 s IJ = 1.74 10 GH 1 yr JK H 1d K
9
s ~ 10 9 s .
Many college students are just approaching 1 Gs. Q1.9 Q1.10 Q1.11 Zero digits. An orderofmagnitude calculation is accurate only within a factor of 10. The mass of the fortysix chapter textbook is on the order of 10 0 kg . With one datum known to one significant digit, we have 80 million yr + 24 yr = 80 million yr.
1
2
Physics and Measurement
SOLUTIONS TO PROBLEMS
Section 1.1 Standards of Length, Mass, and Time
No problems in this section
Section 1.2 P1.1
Matter and ModelBuilding
From the figure, we may see that the spacing between diagonal planes is half the distance between diagonally adjacent atoms on a flat plane. This diagonal distance may be obtained from the Pythagorean theorem, Ldiag = L2 + L2 . Thus, since the atoms are separated by a distance 1 2 L + L2 = 0.141 nm . 2
L = 0.200 nm , the diagonal planes are separated by
Section 1.3 *P1.2
Density and Atomic Mass 4 3 4 r = 6.37 10 6 m 3 3
Modeling the Earth as a sphere, we find its volume as density is then =
e
j
3
= 1.08 10 21 m 3 . Its
m 5.98 10 24 kg = = 5.52 10 3 kg m3 . This value is intermediate between the V 1.08 10 21 m 3 tabulated densities of aluminum and iron. Typical rocks have densities around 2 000 to 3 000 kg m3 . The average density of the Earth is significantly higher, so higherdensity material must be down below the surface. P1.3 With V = base area height V = r 2 h and =
a
fb
g
e j
m , we have V
=
1 kg m = 2 r h 19.5 mm 2 39.0 mm
a
fa
.
F 10 mm I f GH 1 m JK
9 3 3
= 2.15 10 kg m
*P1.4
4
3
Let V represent the volume of the model, the same in =
gold =
m gold V
. Next,
gold iron
=
m gold 9.35 kg
and m gold
m for both. Then iron = 9.35 kg V and V 19.3 10 3 kg / m3 = 23.0 kg . = 9.35 kg 7.86 10 3 kg / m3
F GH
I JK
P1.5
V = Vo  Vi =
4 3 r2  r13 3
e
j
j e j
.
=
3 4 r2  r13 m 4 3 , so m = V = r2  r13 = V 3 3
FG IJ e H K
Chapter 1
3
P1.6
For either sphere the volume is V =
4 4 3 r and the mass is m = V = r 3 . We divide this equation 3 3 for the larger sphere by the same equation for the smaller:
4 r 3 3 r 3 m = = = 5. m s 4 rs3 3 rs3
Then r = rs 3 5 = 4.50 cm 1.71 = 7.69 cm . P1.7 Use 1 u = 1.66 10 24 g . (a) For He, m 0 = 4.00 u
a f
(b)
For Fe, m 0
(c) *P1.8 (a)
For Pb, m 0
F 1.66 10 GH 1 u F 1.66 10 = 55.9 uG H 1u F 1.66 10 = 207 uG H 1u
24
24
24
I = 6.64 10 JK gI JK = 9.29 10 gI JK = 3.44 10
g
24
g .
23
g .
22
g .
The mass of any sample is the number of atoms in the sample times the mass m 0 of one atom: m = Nm 0 . The first assertion is that the mass of one aluminum atom is m 0 = 27.0 u = 27.0 u 1.66 10 27 kg 1 u = 4.48 10 26 kg . Then the mass of 6.02 10 23 atoms is m = Nm 0 = 6.02 10 23 4.48 10 26 kg = 0.027 0 kg = 27.0 g . Thus the first assertion implies the second. Reasoning in reverse, the second assertion can be written m = Nm 0 . 0.027 0 kg = 6.02 10 23 m 0 , so m 0 = in agreement with the first assertion. 0.027 kg 6.02 10 23 = 4.48 10 26 kg ,
(b)
The general equation m = Nm 0 applied to one mole of any substance gives M g = NM u , where M is the numerical value of the atomic mass. It divides out exactly for all substances, giving 1.000 000 0 10 3 kg = N 1.660 540 2 10 27 kg . With eightdigit data, we can be quite sure of the result to seven digits. For one mole the number of atoms is N=
F 1 I 10 GH 1.660 540 2 JK
3 + 27
= 6.022 137 10 23 .
(c)
The atomic mass of hydrogen is 1.008 0 u and that of oxygen is 15.999 u. The mass of one molecule of H 2 O is 2 1.008 0 + 15.999 u = 18.0 u. Then the molar mass is 18.0 g .
b
g
(d)
For CO 2 we have 12.011 g + 2 15.999 g = 44.0 g as the mass of one mole.
b
g
4 P1.9
Physics and Measurement
Mass of gold abraded: m = 3.80 g  3.35 g = 0.45 g = 0.45 g Each atom has mass m 0 = 197 u = 197 u
F 1.66 10 GH 1 u
27
1 b gFGH 10kgg IJK = 4.5 10 kg I JK = 3.27 10 kg .
3 25
4
kg .
Now, m = N m 0 , and the number of atoms missing is N = The rate of loss is N t N t P1.10 (a) = 1 yr 1.38 10 21 atoms 365.25 d 50 yr m m0 = 4.5 10 4 kg 3.27 10 25 kg = 1.38 10 21 atoms .
FG H
IJ FG 1 d IJ FG 1 h IJ FG 1 min IJ K H 24 h K H 60 min K H 60 s K
= 8.72 10 11 atoms s .
m = L3 = 7.86 g cm 3 5.00 10 6 cm N=
e
je
j
3
= 9.83 10 16 g = 9.83 10 19 kg
(b)
9.83 10 19 kg m = = 1.06 10 7 atoms m 0 55.9 u 1.66 10 27 kg 1 u
e
j
P1.11
(a)
The crosssectional area is A = 2 0.150 m 0.010 m + 0.340 m 0.010 m = 6.40 10
3
a
fa
f a
fa
m .
2
f.
The volume of the beam is V = AL = 6.40 10 3 m 2 1.50 m = 9.60 10 3 m3 . Thus, its mass is FIG. P1.11
3 3 3 3 27
e
ja
f
(b)
je9.60 10 m j = 72.6 kg . F 1.66 10 kg I = 9.28 10 The mass of one typical atom is m = a55.9 ufG H 1 u JK
m = V = 7.56 10 kg / m
0
e
26
kg . Now
m = Nm 0 and the number of atoms is N =
72.6 kg m = = 7.82 10 26 atoms . 26 m 0 9.28 10 kg
Chapter 1
5
P1.12
(a)
The mass of one molecule is m 0 = 18.0 u molecules in the pail is N pail =
F 1.66 10 GH 1 u
27
kg
I = 2.99 10 JK
26
kg . The number of
1.20 kg m = = 4.02 10 25 molecules . m 0 2.99 10 26 kg
(b)
Suppose that enough time has elapsed for thorough mixing of the hydrosphere. N both = N pail or
F m I = (4.02 10 GH M JK
pail total
25
molecules)
F 1.20 kg I , GH 1.32 10 kg JK
21
N both = 3.65 10 4 molecules .
Section 1.4 P1.13
Dimensional Analysis
The term x has dimensions of L, a has dimensions of LT 2 , and t has dimensions of T. Therefore, the equation x = ka m t n has dimensions of L = LT 2
e
j aTf
m
n
or L1 T 0 = Lm T n  2 m .
The powers of L and T must be the same on each side of the equation. Therefore, L1 = Lm and m = 1 . Likewise, equating terms in T, we see that n  2m must equal 0. Thus, n = 2 . The value of k, a dimensionless constant, cannot be obtained by dimensional analysis . *P1.14 (a) (b) (c) Circumference has dimensions of L. Volume has dimensions of L3 . Area has dimensions of L2 .
Expression (i) has dimension L L2
e j
1/2
= L2 , so this must be area (c).
Expression (ii) has dimension L, so it is (a). Expression (iii) has dimension L L2 = L3 , so it is (b). Thus, (a) = ii; (b) = iii, (c) = i .
e j
6
Physics and Measurement
P1.15
(a) (b)
This is incorrect since the units of ax are m 2 s 2 , while the units of v are m s . This is correct since the units of y are m, and cos kx is dimensionless if k is in m 1 . a
a f
*P1.16
(a)
represents the proportionality of acceleration to resultant force and m m the inverse proportionality of acceleration to mass. If k has no dimensions, we have a = k M L T
2
F
or a = k
F
F F L M L , F = , 2 =1 . m T M T2
(b) P1.17
In units,
=
kg m s2
, so 1 newton = 1 kg m s 2 .
Inserting the proper units for everything except G,
LM kg m OP = G kg Ns Q m
2
2
2
.
Multiply both sides by m
2
and divide by kg ; the units of G are
2
m3 kg s 2
.
Section 1.5 *P1.18
Conversion of Units m jFGH 3.128 ft IJK
Each of the four walls has area 8.00 ft 12.0 ft = 96.0 ft 2 . Together, they have area 4 96.0 ft 2
e
2
a
fa
f
= 35.7 m 2 .
P1.19
Apply the following conversion factors: 1 in = 2.54 cm , 1 d = 86 400 s , 100 cm = 1 m , and 10 9 nm = 1 m
FG 1 H 32
*P1.20 8.50 in 3 = 8.50 in 3
in day
IJ b2.54 cm inge10 m cmje10 K 86 400 s day
2 3
9
nm m
j=
9.19 nm s .
This means the proteins are assembled at a rate of many layers of atoms each second!
FG 0.025 4 m IJ H 1 in K
= 1.39 10 4 m 3
Chapter 1
7
P1.21
Conceptualize: We must calculate the area and convert units. Since a meter is about 3 feet, we should expect the area to be about A 30 m 50 m = 1 500 m 2 .
a
fa
f
Categorize: We model the lot as a perfect rectangle to use Area = Length Width. Use the conversion: 1 m = 3.281 ft . Analyze: A = LW = 100 ft
a
1m 1m f FGH 3.281 ft IJK a150 ftfFGH 3.281 ft IJK = 1 390 m
2
= 1.39 10 3 m 2 .
Finalize: Our calculated result agrees reasonably well with our initial estimate and has the proper units of m 2 . Unit conversion is a common technique that is applied to many problems. P1.22 (a) V = 40.0 m 20.0 m 12.0 m = 9.60 10 3 m 3
a
fa
V = 9.60 10 m 3.28 ft 1 m
(b) The mass of the air is
3
3
b
fa
f
g
3
= 3.39 10 5 ft 3
m = air V = 1.20 kg m3 9.60 10 3 m3 = 1.15 10 4 kg . The student must look up weight in the index to find Fg = mg = 1.15 10 4 kg 9.80 m s 2 = 1.13 10 5 N . Converting to pounds,
e
je
j
e
je
j
Fg = 1.13 10 5 N 1 lb 4.45 N = 2.54 10 4 lb .
P1.23 (a) Seven minutes is 420 seconds, so the rate is r= (b) 30.0 gal = 7.14 10 2 gal s . 420 s
e
jb
g
Converting gallons first to liters, then to m3 , r = 7.14 10 2 gal s
e
jFGH 3.1786 L IJK FGH 10 1 Lm IJK gal
3 3
r = 2.70 10 4 m3 s . (c) At that rate, to fill a 1m3 tank would take t=
F 1m GH 2.70 10
3
4
m
3
IF 1 h I = s J H 3 600 K KG J
1.03 h .
8
Physics and Measurement
*P1.24
(a)
Length of Mammoth Cave = 348 mi
(b)
(c)
(d) P1.25
FG 1.609 km IJ = 560 km = 5.60 10 m = 5.60 10 cm . H 1 mi K F 0.304 8 m IJ = 491 m = 0.491 km = 4.91 10 cm . Height of Ribbon Falls = 1 612 ftG H 1 ft K F 0.304 8 m IJ = 6.19 km = 6.19 10 m = 6.19 10 cm . Height of Denali = 20 320 ftG H 1 ft K F 0.304 8 m IJ = 2.50 km = 2.50 10 m = 2.50 10 cm . Depth of King's Canyon = 8 200 ftG H 1 ft K
5 7 4 3 5 3 5
From Table 1.5, the density of lead is 1.13 10 4 kg m 3 , so we should expect our calculated value to be close to this number. This density value tells us that lead is about 11 times denser than water, which agrees with our experience that lead sinks. Density is defined as mass per volume, in = m . We must convert to SI units in the calculation. V
=
23.94 g 2.10 cm 3
F 1 kg I FG 100 cm IJ GH 1 000 g JK H 1 m K
3
= 1.14 10 4 kg m3
At one step in the calculation, we note that one million cubic centimeters make one cubic meter. Our result is indeed close to the expected value. Since the last reported significant digit is not certain, the difference in the two values is probably due to measurement uncertainty and should not be a concern. One important commonsense check on density values is that objects which sink in water must have a density greater than 1 g cm 3 , and objects that float must be less dense than water. P1.26 It is often useful to remember that the 1 600m race at track and field events is approximately 1 mile in length. To be precise, there are 1 609 meters in a mile. Thus, 1 acre is equal in area to 1 a1 acrefFGH 640mi IJK FGH 1 609 m IJK acres mi
2 2
= 4.05 10 3 m 2 .
*P1.27 P1.28
The weight flow rate is 1 200
ton 2 000 lb h ton
FG H
IJ FG 1 h IJ FG 1 min IJ = K H 60 min K H 60 s K
667 lb s .
1 mi = 1 609 m = 1.609 km ; thus, to go from mph to km h , multiply by 1.609. (a) (b) (c) 1 mi h = 1.609 km h 55 mi h = 88.5 km h 65 mi h = 104.6 km h . Thus, v = 16.1 km h .
P1.29
(a)
F 6 10 $ I F 1 h I FG 1 day IJ F 1 yr I = GH 1 000 $ s JK GH 3 600 s JK H 24 h K GH 365 days JK
12
Chapter 1
9
190 years
(b)
The circumference of the Earth at the equator is 2 6.378 10 3 m = 4.01 10 7 m . The length of one dollar bill is 0.155 m so that the length of 6 trillion bills is 9.30 10 11 m. Thus, the 6 trillion dollars would encircle the Earth 9.30 10 11 m = 2.32 10 4 times . 4.01 0 7 m
e
j
P1.30
N atoms =
1.99 10 30 kg mSun = = 1.19 10 57 atoms m atom 1.67 10 27 kg V 3.78 10 3 m 3 = = 1.51 10 4 m or 151 m 2 A 25.0 m
P1.31
V = At so t =
b
g
P1.32
13.0 acres 43 560 ft 2 acre 1 V = Bh = 3 3 7 3 = 9.08 10 ft , or V = 9.08 10 7 ft 3
a
fe
j a481 ftf
h
e
jFGH 2.83 110 ft jb
2 3
m3
I JK g
B
= 2.57 10 6 m3 P1.33 *P1.34
FIG. P1.32
Fg = 2.50 tons block 2.00 10 6 blocks 2 000 lb ton = 1.00 10 10 lbs
The area covered by water is
2 A w = 0.70 AEarth = 0.70 4 REarth = 0.70 4 6.37 10 6 m
b
ge
a fe
a
j a fa fe
g
j
2
= 3.6 10 14 m 2 .
The average depth of the water is d = 2.3 miles 1 609 m l mile = 3.7 10 3 m . The volume of the water is V = A w d = 3.6 10 14 m 2 3.7 10 3 m = 1.3 10 18 m 3 and the mass is
fb
e
je
j
m = V = 1 000 kg m3 1.3 10 18 m3 = 1.3 10 21 kg .
e
je
j
10 P1.35
Physics and Measurement
(a)
d nucleus, scale = d nucleus, real
d nucleus, scale = 6.79 10 3 Vatom = Vnucleus
3 4 ratom 3 3 4 rnucleus 3
e
Fd I = 2.40 10 m FG 300 ft IJ = 6.79 10 jH 1.06 10 m K GH d JK e ft jb304.8 mm 1 ft g = 2.07 mm
atom, scale atom, real 15 10
3
ft , or
(b)
=
FG r Hr
atom
nucleus
IJ = FG d K Hd
3
atom
nucleus
IJ = F 1.06 10 K GH 2.40 10
3 3 9
10 15
m m
I JK
3
= 8.62 10 13 times as large *P1.36 scale distance between P1.37
=
FG real IJ FG scale IJ = e4.0 10 H distanceK H factorK
S= 0.25 m
13
km
m jFGH 71..0410 m IJK = 10
200 km
The scale factor used in the "dinner plate" model is = 2.5 10 6 m lightyears .
1.0 10 lightyears
5
The distance to Andromeda in the scale model will be Dscale = Dactual S = 2.0 10 6 lightyears 2.5 10 6 m lightyears = 5.0 m .
2 AEarth 4 rEarth r = = Earth 2 A Moon 4 rMoon rMoon
e
je
j
P1.38
(a)
FG H
(b)
VEarth = VMoon
3 4 rEarth 3 3 4 rMoon 3
Fr =G Hr
Earth
Moon
F e6.37 10 mjb100 cm mg I =G GH 1.74 10 cm JJK = 13.4 IJ = FG e6.37 10 mjb100 cm mg IJ = 49.1 K GH 1.74 10 cm JK IJ K
2 6 2 8 3 6 3 8
P1.39
To balance, m Fe = m Al or FeVFe = Al VAl
Fe
FG 4IJ r H 3K
Fe
3
= Al
rAl = rFe
FG 4 IJ r H 3K FG IJ = a2.00 cmfFG 7.86 IJ H 2.70 K H K
Al 3 1/3 Fe Al
1/3
= 2.86 cm .
Chapter 1
11
P1.40
The mass of each sphere is m Al = Al VAl = and m Fe = FeVFe = Setting these masses equal, 4 Al rAl 3 4 Fe rFe 3 = and rAl = rFe 3 Fe . 3 3 Al 4 Fe rFe 3 . 3 4 Al rAl 3 3
Section 1.6 P1.41
Estimates and OrderofMagnitude Calculations
Model the room as a rectangular solid with dimensions 4 m by 4 m by 3 m, and each pingpong ball as a sphere of diameter 0.038 m. The volume of the room is 4 4 3 = 48 m3 , while the volume of one ball is 4 0.038 m 3 2 Therefore, one can fit about
FG H
IJ K
3
= 2.87 10 5 m3 .
48 ~ 10 6 pingpong balls in the room. 2.87 10 5 As an aside, the actual number is smaller than this because there will be a lot of space in the room that cannot be covered by balls. In fact, even in the best arrangement, the socalled "best 1 packing fraction" is 2 = 0.74 so that at least 26% of the space will be empty. Therefore, the 6 above estimate reduces to 1.67 10 6 0.740 ~ 10 6 .
P1.42
A reasonable guess for the diameter of a tire might be 2.5 ft, with a circumference of about 8 ft. Thus, the tire would make 50 000 mi 5 280 ft mi 1 rev 8 ft = 3 10 7 rev ~ 10 7 rev .
b
gb
gb
g
P1.43
In order to reasonably carry on photosynthesis, we might expect a blade of grass to require at least 1 in 2 = 43 10 5 ft 2 . Since 1 acre = 43 560 ft 2 , the number of blades of grass to be expected on a 16 quarteracre plot of land is about n= 0.25 acre 43 560 ft 2 acre total area = = 2.5 10 7 blades ~ 10 7 blades . area per blade 43 10 5 ft 2 blade
a
fe
j
12 P1.44
Physics and Measurement
A typical raindrop is spherical and might have a radius of about 0.1 inch. Its volume is then approximately 4 10 3 in 3 . Since 1 acre = 43 560 ft 2 , the volume of water required to cover it to a depth of 1 inch is
a1 acrefa1 inchf = a1 acre infFGH 431560 ft acre
The number of raindrops required is n= *P1.45
2
I F 144 in I 6.3 10 JK GH 1 ft JK
2 2
6
in 3 .
volume of water required 6.3 10 6 in 3 = = 1.6 10 9 ~ 10 9 . volume of a single drop 4 10 3 in 3
Assume the tub measures 1.3 m by 0.5 m by 0.3 m. Onehalf of its volume is then
V = 0.5 1.3 m 0.5 m 0.3 m = 0.10 m3 .
The mass of this volume of water is
a fa e
fa
fa
f
m water = water V = 1 000 kg m3 0.10 m3 = 100 kg ~ 10 2 kg .
Pennies are now mostly zinc, but consider copper pennies filling 50% of the volume of the tub. The mass of copper required is
je
j
mcopper = copper V = 8 920 kg m3 0.10 m3 = 892 kg ~ 10 3 kg .
P1.46 The typical person probably drinks 2 to 3 soft drinks daily. Perhaps half of these were in aluminum cans. Thus, we will estimate 1 aluminum can disposal per person per day. In the U.S. there are ~250 million people, and 365 days in a year, so
e
je
j
e250 10
6
cans day 365 days year 10 11 cans
jb
g
are thrown away or recycled each year. Guessing that each can weighs around 1 10 of an ounce, we estimate this represents
e10
P1.47
11
cans 0.1 oz can 1 lb 16 oz 1 ton 2 000 lb 3.1 10 5 tons year . ~ 10 5 tons
jb
gb
gb
g
Assume: Total population = 10 7 ; one out of every 100 people has a piano; one tuner can serve about 1 000 pianos (about 4 per day for 250 weekdays, assuming each piano is tuned once per year). Therefore, # tuners ~
F 1 tuner I F 1 piano I (10 GH 1 000 pianos JK GH 100 people JK
7
people) = 100 .
Chapter 1
13
Section 1.7 *P1.48
Significant Figures
METHOD ONE We treat the best value with its uncertainty as a binomial 21.3 0.2 cm 9.8 0.1 cm ,
a f a A = 21.3a9.8f 21.3a0.1f 0.2a9.8 f a0.2 fa0.1f cm
f
2
.
The first term gives the best value of the area. The cross terms add together to give the uncertainty and the fourth term is negligible. A = 209 cm 2 4 cm 2 . METHOD TWO We add the fractional uncertainties in the data. A = 21.3 cm 9.8 cm P1.49 (a)
a
fa
0 0 f FGH 21..23 + 9..1 IJK = 209 cm 8
2
2% = 209 cm 2 4 cm 2
r 2 = 10.5 m 0.2 m
= 346 m 2 13 m 2
a
f
2
= (10.5 m) 2 2(10.5 m)(0.2 m) + ( 0.2 m) 2
(b) P1.50 P1.51 (a)
2 r = 2 10.5 m 0.2 m = 66.0 m 1.3 m 3
(b)
a
f
4
(c)
3
(d)
2
a f a m = a1.85 0.02f kg
=
r = 6.50 0. 20 cm = 6.50 0.20 10 2 m m
f
c h r
4 3
3
also,
m 3 r = + . m r
In other words, the percentages of uncertainty are cumulative. Therefore,
0.02 3 0.20 = + = 0.103 , 6.50 1.85 =
a f
c h e6.5 10
4 3
1.85
2
m
j
3
= 1.61 10 3 kg m 3
and
= 1.61 0.17 10 3 kg m3 = 1.6 0.2 10 3 kg m3 .
a
f
a
f
14 P1.52
Physics and Measurement
(a)
756.?? 37.2? 0.83 + 2.5? 796.53 = 797 // 0.003 2 2 s.f. 356.3 4 s.f. = 1.140 16 = 2 s.f.
(b) (c) *P1.53
a
f
a
f
a
f
1.1
5.620 4 s.f. > 4 s.f. = 17.656 = 4 s.f.
a
f a
f
a
f
17.66
We work to nine significant digits: 1 yr = 1 yr
F 365.242 199 d I FG 24 h IJ FG 60 min IJ FG 60 s IJ = GH 1 yr JK H 1 d K H 1 h K H 1 min K
31 556 926.0 s .
P1.54
The distance around is 38.44 m + 19.5 m + 38.44 m + 19.5 m = 115.88 m , but this answer must be rounded to 115.9 m because the distance 19.5 m carries information to only one place past the decimal. 115.9 m V = 2V1 + 2V2 = 2 V1 + V2
P1.55
a fa fa f V = a10.0 mfa1.0 mfa0.090 mf = 0.900 m V = 2e1.70 m + 0.900 m j = 5.2 m U 0.12 m = = 0.0063  19.0 m  V w 0.01 m  = 0.006 + 0.010 + 0.011 = 0.027 = = = 0.010 V w 1.0 m V t 0.1 cm  = = 0.011  t 9.0 cm W
V1 = 17.0 m + 1.0 m + 1.0 m 1.0 m 0.09 m = 1.70 m 3
2 3 3 3 3 1 1 1 1 1 1
b
g
FIG. P1.55 3%
Additional Problems P1.56 It is desired to find the distance x such that 1 000 m x = 100 m x (i.e., such that x is the same multiple of 100 m as the multiple that 1 000 m is of x). Thus, it is seen that x 2 = 100 m 1 000 m = 1.00 10 5 m 2 and therefore x = 1.00 10 5 m 2 = 316 m .
a
fb
g
Chapter 1
15
*P1.57
Consider one cubic meter of gold. Its mass from Table 1.5 is 19 300 kg. One atom of gold has mass m 0 = 197 u So, the number of atoms in the cube is N= 19 300 kg 3.27 10 25 kg = 5.90 10 28 .
a
fFGH 1.66 110u
27
kg
I = 3.27 10 JK
25
kg .
The imagined cubical volume of each atom is d3 = So d = 2.57 10 10 m . 1 m3 5.90 10 28 = 1.69 10 29 m 3 .
P1.58
Atotal = N A drop = A total
total
F I V a fe j FG V IJ e A j = GG V JJ e4 r j H K H K F 3V IJ = 3FG 30.0 10 m IJ = 4.50 m =G H r K H 2.00 10 m K
total drop drop total 4 r 3 3 2 6 3 5 2
P1.59
One month is 1 mo = 30 day 24 h day 3 600 s h = 2.592 10 6 s . Applying units to the equation, V = 1.50 Mft 3 mo t + 0.008 00 Mft 3 mo 2 t 2 . Since 1 Mft 3 = 10 6 ft 3 , V = 1.50 10 6 ft 3 mo t + 0.008 00 10 6 ft 3 mo 2 t 2 . Converting months to seconds, V= 1.50 10 6 ft 3 mo 2.592 10 6 s mo t+ 0.008 00 10 6 ft 3 mo 2 t2.
b
gb
gb
g
e
j e j e
j
e
j
e2.592 10
6
s mo
j
2
Thus, V [ft 3 ] = 0.579 ft 3 s t + 1.19 10 9 ft 3 s 2 t 2 .
e
j e
j
16 P1.60
Physics and Measurement
(deg)
15.0 20.0 25.0 24.0 24.4 24.5 24.6 24.7
(rad)
0.262 0.349 0.436 0.419 0.426 0.428 0.429 0.431
tan
af
sin
af
difference 3.47% 6.43% 10.2% 9.34% 9.81% 9.87% 9.98% 10.1%
0.268 0.364 0.466 0.445 0.454 0.456 0.458 0.460
0.259 0.342 0.423 0.407 0.413 0.415 0.416 0.418
24.6
P1.61
2 r = 15.0 m r = 2.39 m h = tan 55.0 r h = 2.39 m tan( 55.0 ) = 3.41 m
a
f
h
55 r
FIG. P1.61 *P1.62 Let d represent the diameter of the coin and h its thickness. The mass of the gold is m = V = At = where t is the thickness of the plating. m = 19.3 2
F 2 d GH 4
2
+ dh t
I JK
= 0.003 64 grams
LM a2.41f NM 4
2
+ 2.41 0.178
a fa
fOPPe0.18 10 j Q
4
cost = 0.003 64 grams $10 gram = $0.036 4 = 3.64 cents This is negligible compared to $4.98. P1.63 The actual number of seconds in a year is
b86 400 s daygb365.25 day yrg = 31 557 600 s yr .
The percent error in the approximation is
e 10
7
s yr  31 557 600 s yr 31 557 600 s yr
j b
g 100% =
0.449% .
Chapter 1
17
P1.64
(a)
V = L3 , A = L2 , h = L V = A h L3 = L2 L = L3 . Thus, the equation is dimensionally correct.
(b)
Vcylinder = R 2 h = R 2 h = Ah , where A = R 2 Vrectangular object =
e
j wh = a w fh = Ah , where
v=
A= w
P1.65
(a)
The speed of rise may be found from
aVol rate of flowf = 16.5 cm
(Area:
D2 4 )
3
6 .30 cm 4
a
f
s
2
= 0.529 cm s .
(b)
Likewise, at a 1.35 cm diameter, v= 16.5 cm 3 s
1.35 cm 4
a
f
2
= 11.5 cm s .
P1.66
(a)
1 cubic meter of water has a mass m = V = 1.00 10 3 kg cm3 1.00 m 3 10 2 cm m
e
je
je
j
3
= 1 000 kg
(b)
As a rough calculation, we treat each item as if it were 100% water. cell: m = V =
FG 4 R IJ = FG 1 D IJ = e1 000 kg m jFG 1 IJ e1.0 10 H3 K H6 K H6 K
3 3 3
6
m
j
3
= 5.2 10 16 kg kidney: m = V =
FG 4 R IJ = e1.00 10 H3 K
3 2 3
3
kg cm3
4 jFGH 3 IJK ( 4.0 cm)
3
= 0.27 kg fly: m=
FG D hIJ = e1 10 H4 K
kg cm 3
jFGH IJK a2.0 mmf a4.0 mmfe10 4
2
1
cm mm
j
3
= 1.3 10 5 kg (10 8 cars)(10 4 mi yr ) = 5.0 10 10 gal yr 20 mi gal (10 8 cars)(10 4 mi yr ) = 4.0 10 10 gal yr 25 mi gal
P1.67
V20 mpg = V25 mpg =
Fuel saved = V25 mpg  V20 mpg = 1.0 10 10 gal yr
18 P1.68
Physics and Measurement
furlongs 220 yd 0.914 4 m 1 fortnight 1 day 1 hr = 8.32 10 4 m s fortnight 1 furlong 1 yd 14 days 24 hrs 3 600 s This speed is almost 1 mm/s; so we might guess the creature was a snail, or perhaps a sloth. v = 5.00 The volume of the galaxy is
F GH
IF JK GH
IF JK GH
IF JK GH
I FG JK H
IJ FG KH
I JK
P1.69
r 2 t = 10 21 m
e
j e10
2
19
m ~ 10 61 m3 .
j
If the distance between stars is 4 10 16 m , then there is one star in a volume on the order of
e4 10
The number of stars is about 10 61 m 3 10
50
16
m ~ 10 50 m 3 .
j
3
m star
3
~ 10 11 stars .
P1.70
The density of each material is =
4m m m . = = V r 2h D2h
g cm g cm3 g cm3 g cm3 g cm
3 3
Al:
=
Cu:
Brass:
Sn:
Fe:
f a3.75 cmf 4b56.3 g g = = a1.23 cmf a5.06 cmf 4b94.4 g g = = a1.54 cmf a5.69 cmf 4b69.1 g g = = a1.75 cmf a3.74 cmf 4b 216.1 g g = = a1.89 cmf a9.77 cmf
2.52 cm
2 2 2 2 2
a
4 51.5 g
b
g
= 2.75
9.36
FG g IJ is H cm K F g IJ is The tabulated value G 8.92 H cm K
The tabulated value 2.70
3 3
2% smaller.
5% smaller.
8.91
7.68
7.88
The tabulated value 7.86
FG H
g cm3
IJ is K
0.3% smaller.
P1.71
(a) (b)
b3 600 s hrgb24 hr daygb365.25 days yrg =
Vmm = Vcube Vmm
3.16 10 7 s yr
3 4 3 4 r = 5.00 10 7 m = 5.24 10 19 m3 3 3 1 m3 = = 1.91 10 18 micrometeorites 5.24 10 19 m3
e
j
This would take
1.91 10 18 micrometeorites 3.16 10 7 micrometeorites yr
= 6.05 10 10 yr .
Chapter 1
19
ANSWERS TO EVEN PROBLEMS
P1.2
5.52 10 3 kg m3 , between the densities of aluminum and iron, and greater than the densities of surface rocks.
23.0 kg 7.69 cm (a) and (b) see the solution, N A = 6.022 137 10 23 ; (c) 18.0 g; (d) 44.0 g (a) 9.83 10 16 g ; (b) 1.06 10 7 atoms (a) 4.02 10 25 molecules; (b) 3.65 10 4 molecules (a) ii; (b) iii; (c) i (a) M L ; (b) 1 newton = 1 kg m s 2 T2
P1.34 P1.36 P1.38 P1.40
1.3 10 21 kg 200 km (a) 13.4; (b) 49.1 rAl = rFe
P1.4 P1.6 P1.8
FG IJ H K
Fe Al
13
P1.42 P1.44 P1.46 P1.48 P1.50 P1.52 P1.54 P1.56 P1.58 P1.60
7
~ 10 7 rev ~ 10 9 raindrops
~ 10 11 cans; ~ 10 5 tons
P1.10 P1.12
a209 4f cm
2
P1.14 P1.16 P1.18 P1.20 P1.22 P1.24
(a) 3; (b) 4; (c) 3; (d) 2 (a) 797; (b) 1.1; (c) 17.66 115.9 m 316 m
35.7 m 2 1.39 10 4 m3
(a) 3.39 10 ft ; (b) 2.54 10 lb (a) 560 km = 5.60 10 m = 5.60 10 cm ; (b) 491 m = 0.491 km = 4.91 10 4 cm ; (c) 6.19 km = 6.19 10 3 m = 6.19 10 5 cm ; (d) 2.50 km = 2.50 10 3 m = 2.50 10 5 cm
5 5 3 4
4.50 m 2
see the solution; 24.6
P1.62 P1.64 P1.66
3.64 cents ; no
see the solution (a) 1 000 kg; (b) 5.2 10 16 kg ; 0. 27 kg ;
P1.26 P1.28
4.05 10 3 m 2
(a) 1 mi h = 1.609 km h ; (b) 88.5 km h ; (c) 16.1 km h
1.3 10 5 kg
P1.68 P1.70
8.32 10 4 m s ; a snail
see the solution
P1.30 P1.32
1.19 10 57 atoms 2.57 10 6 m3
2
Motion in One Dimension
CHAPTER OUTLINE
2.1 2.2 2.3 2.4 2.5 2.6 2.7 Position, Velocity, and Speed Instantaneous Velocity and Speed Acceleration Motion Diagrams OneDimensional Motion with Constant Acceleration Freely Falling Objects Kinematic Equations Derived from Calculus
ANSWERS TO QUESTIONS
Q2.1 If I count 5.0 s between lightning and thunder, the sound has traveled 331 m s 5.0 s = 1.7 km . The transit time for the light is smaller by
b
ga f
3.00 10 8 m s = 9.06 10 5 times, 331 m s so it is negligible in comparison. Q2.2 Q2.3 Q2.4 Yes. Yes, if the particle winds up in the +x region at the end. Zero. Yes. Yes.
Q2.5
No. Consider a sprinter running a straightline race. His average velocity would simply be the length of the race divided by the time it took for him to complete the race. If he stops along the way to tie his shoe, then his instantaneous velocity at that point would be zero. We assume the object moves along a straight line. If its average x velocity is zero, then the displacement must be zero over the time interval, according to Equation 2.2. The object might be stationary throughout the interval. If it is moving to the right at first, it must later move to the left to return to its starting point. Its velocity must be zero as it turns around. The graph of the motion shown to the right represents such motion, as the initial and final positions are the same. In an x vs. t graph, the instantaneous velocity at any time t is the slope of the curve at that point. At t 0 in the graph, the slope of the curve is zero, and thus the instantaneous velocity at that time is also zero.
Q2.6
t0
t
FIG. Q2.6 Q2.7 Yes. If the velocity of the particle is nonzero, the particle is in motion. If the acceleration is zero, the velocity of the particle is unchanging, or is a constant.
21
22 Q2.8
Motion in One Dimension
Yes. If you drop a doughnut from rest v = 0 , then its acceleration is not zero. A common misconception is that immediately after the doughnut is released, both the velocity and acceleration are zero. If the acceleration were zero, then the velocity would not change, leaving the doughnut floating at rest in midair. No: Car A might have greater acceleration than B, but they might both have zero acceleration, or otherwise equal accelerations; or the driver of B might have tramped hard on the gas pedal in the recent past. Yes. Consider throwing a ball straight up. As the ball goes up, its v velocity is upward v > 0 , and its acceleration is directed down v0 a < 0 . A graph of v vs. t for this situation would look like the figure to the right. The acceleration is the slope of a v vs. t graph, and is always negative in this case, even when the velocity is positive.
a
f
Q2.9
Q2.10
a f
a
f
t FIG. Q2.10 Q2.11 (a) (d) (g) (h) Q2.12 Q2.13 Accelerating East Braking West (b) (e) Braking East Accelerating West (c) (f) Cruising East Cruising West
Stopped but starting to move East Stopped but starting to move West
No. Constant acceleration only. Yes. Zero is a constant. The position does depend on the origin of the coordinate system. Assume that the cliff is 20 m tall, and that the stone reaches a maximum height of 10 m above the top of the cliff. If the origin is taken as the top of the cliff, then the maximum height reached by the stone would be 10 m. If the origin is taken as the bottom of the cliff, then the maximum height would be 30 m. The velocity is independent of the origin. Since the change in position is used to calculate the instantaneous velocity in Equation 2.5, the choice of origin is arbitrary. Once the objects leave the hand, both are in free fall, and both experience the same downward acceleration equal to the freefall acceleration, g. They are the same. After the first ball reaches its apex and falls back downward past the student, it will have a downward velocity equal to vi . This velocity is the same as the velocity of the second ball, so after they fall through equal heights their impact speeds will also be the same. With h = (a) (b) 1 2 gt , 2 1 2 g 0.707t . The time is later than 0.5t. 2 1 2 g 0.5t . The elevation is 0.75h, greater than 0.5h. 2
Q2.14 Q2.15
Q2.16
0.5 h =
a
f
The distance fallen is 0.25 h =
a f
Chapter 2
23
Q2.17
Above. Your ball has zero initial speed and smaller average speed during the time of flight to the passing point.
SOLUTIONS TO PROBLEMS
Section 2.1 P2.1 (a) (b) (c) *P2.2 (a) Position, Velocity, and Speed v = 2.30 m s v= v= v= x 57.5 m  9.20 m = = 16.1 m s t 3.00 s x 57.5 m  0 m = = 11.5 m s t 5.00 s x 20 ft 1m = t 1 yr 3.281 ft
FG H
IJ FG 1 yr IJ = 2 10 m s or in particularly windy times K H 3.156 10 s K x 100 ft F 1 m I F = v= GH 3.281 ft JK GH 3.1561 yr s IJK = 1 10 m s . t 1 yr 10
7 7 7 6
(b)
The time required must have been t = 3 000 mi 1 609 m x = v 10 mm yr 1 mi
FG H
IJ FG 10 mm IJ = KH 1 m K
3
5 10 8 yr .
P2.3
(a) (b) (c)
v= v= v=
x 10 m = = 5 ms t 2s 5m = 1.2 m s 4s x 2  x1 5 m  10 m = = 2.5 m s 4 s2 s t 2  t1 x 2  x 1 5 m  5 m = = 3.3 m s 7 s4 s t 2  t1 x 2  x1 0  0 = = 0 ms 80 t 2  t1
(d)
v=
(e)
v=
P2.4
x = 10t 2 : For
af xamf
ts
= 2.0 = 40
2.1 44.1
3.0 90
(a) (b)
v= v=
x 50 m = = 50.0 m s t 1.0 s x 4.1 m = = 41.0 m s t 0.1 s
24 P2.5
Motion in One Dimension
(a)
Let d represent the distance between A and B. Let t1 be the time for which the walker has d the higher speed in 5.00 m s = . Let t 2 represent the longer time for the return trip in t1 d d d 3.00 m s =  . Then the times are t1 = and t 2 = . The average speed t2 5.00 m s 3.00 m s is:
b
g
b
g
v=
Total distance = Total time 2 15.0 m 2 s 2 8.00 m s
b
d 5.00 m s
d+d = + 3 .00d m s
g b
b8.00 m sgd g e15.0 m s j
2 2
2d
v= (b)
e
j=
3.75 m s
She starts and finishes at the same point A. With total displacement = 0, average velocity = 0 .
Section 2.2 P2.6 (a)
Instantaneous Velocity and Speed At any time, t, the position is given by x = 3.00 m s 2 t 2 .
i i 2 2
e Thus, at t = 3.00 s: x = e3.00 m s ja3.00 sf = e ja b
j
27.0 m .
(b)
At t f = 3.00 s + t : x f = 3.00 m s 2 3.00 s + t , or x f = 27.0 m + 18.0 m s t + 3.00 m s 2 t
f
2
g e
ja f
2
.
(c)
The instantaneous velocity at t = 3.00 s is: v = lim
t 0
F x  x I = lim e18.0 m s + e3.00 m s jtj = GH t JK
f i 2 t 0
18.0 m s .
P2.7
(a)
at ti = 1.5 s , x i = 8.0 m (Point A) at t f = 4.0 s , x f = 2.0 m (Point B) v= x f  xi t f  ti =
a2.0  8.0f m =  6.0 m = a4  1.5f s 2.5 s
2.4 m s
(b)
The slope of the tangent line is found from points C and D. tC = 1.0 s, x C = 9.5 m and t D = 3.5 s, x D = 0 ,
b
g
b
g
v 3.8 m s . (c) The velocity is zero when x is a minimum. This is at t 4 s .
FIG. P2.7
Chapter 2
25
P2.8
(a)
(b)
At t = 5.0 s, the slope is v
58 m 2.5 s 54 m At t = 4.0 s, the slope is v 3s 49 m At t = 3.0 s, the slope is v 3.4 s 36 m At t = 2.0 s , the slope is v 4.0 s a= v 23 m s 4.6 m s 2 t 5.0 s
23 m s . 18 m s . 14 m s . 9.0 m s .
(c) (d) P2.9 (a)
Initial velocity of the car was zero . v= (5  0 ) m (1  0) s = 5 ms
(b)
v=
(5  10) m (4  2) s
= 2.5 m s
(c)
v=
(5 m  5 m) (5 s  4 s) 0  (5 m) (8 s  7 s )
= 0
(d) *P2.10
v=
= +5 m s
FIG. P2.9
Once it resumes the race, the hare will run for a time of t= x f  xi vx = 1 000 m  800 m = 25 s . 8 ms
In this time, the tortoise can crawl a distance x f  xi = 0.2 m s ( 25 s)= 5.00 m .
a
f
26
Motion in One Dimension
Section 2.3 P2.11
Acceleration
Choose the positive direction to be the outward direction, perpendicular to the wall. v f = vi + at : a = v 22.0 m s  25.0 m s = = 1.3410 4 m s 2 . t 3.50 103 s
a
f
P2.12
(a)
Acceleration is constant over the first ten seconds, so at the end, v f = vi + at = 0 + 2.00 m s 2 (10.0 s)= 20.0 m s . Then a = 0 so v is constant from t = 10.0 s to t = 15.0 s. And over the last five seconds the velocity changes to v f = vi + at = 20.0 m s + 3.00 m s 2 (5.00 s)= 5.00 m s .
c
h
c
h
(b)
In the first ten seconds, x f = x i + vi t + 1 2 1 2 at = 0 + 0 + 2.00 m s 2 (10.0 s) = 100 m . 2 2
c
h
Over the next five seconds the position changes to x f = xi + vi t + And at t = 20.0 s , x f = x i + vi t + *P2.13 (a) 1 2 1 2 at = 200 m + 20.0 m s (5.00 s)+ 3.00 m s 2 (5.00 s) = 262 m . 2 2 1 2 at = 100 m + 20.0 m s (5.00 s)+ 0 = 200 m . 2
a
f
a
f
c
h
The average speed during a time interval t is v =
distance traveled . During the first t quarter mile segment, Secretariat's average speed was v1 = 0.250 mi 1 320 ft = = 52.4 ft s 25.2 s 25.2 s
b35.6 mi hg .
During the second quarter mile segment, v2 = 1 320 ft = 55.0 ft s 24.0 s
b37.4 mi hg . b37.7 mi hg , b39.0 mi hg .
For the third quarter mile of the race, v3 = 1 320 ft = 55.5 ft s 23.8 s
and during the final quarter mile, v4 = continued on next page 1 320 ft = 57.4 ft s 23.0 s
Chapter 2
27
(b)
Assuming that v f = v 4 and recognizing that vi = 0 , the average acceleration during the race was a= v f  vi total elapsed time = 57.4 ft s  0 = 0.598 ft s 2 . ( 25. 2 + 24.0 + 23.8 + 23.0) s
a (m/s2) 2.0 1.6 1.0
P2.14
(a)
Acceleration is the slope of the graph of v vs t. For 0 < t < 5.00 s, a = 0 . For 15.0 s < t < 20.0 s , a = 0 . For 5.0 s < t < 15.0 s , a = v f  vi t f  ti .
a=
8.00  (8.00) 15.0  5.00
0.0
t (s) 0 5 10 15 20
= 1.60 m s 2
We can plot a(t ) as shown. (b) a= (i) v f  vi t f  ti For 5.00 s < t < 15.0 s , ti = 5.00 s , vi = 8.00 m s , t f = 15.0 s v f = 8.00 m s a= (ii) v f  vi t f  ti =
FIG. P2.14
8.00  8.00 = 1.60 m s 2 . 15.0  5.00
a
f
ti = 0 , vi = 8.00 m s , t f = 20.0 s , v f = 8.00 m s a= v f  vi t f  ti = 8.00  (8.00) 20.0  0 = 0.800 m s 2
P2.15
x = 2.00 + 3.00t  t 2 , v = At t = 3.00 s : (a) (b) (c)
dx dv = 3.00  2.00t , a = = 2.00 dt dt
x = ( 2.00 + 9.00  9.00) m = 2.00 m v = (3.00  6.00) m s = 3.00 m s
a = 2.00 m s 2
28 P2.16
Motion in One Dimension
(a)
At t = 2.00 s , x = 3.00( 2.00)  2.00( 2.00)+ 3.00 m = 11.0 m. At t = 3.00 s , x = 3.00 9.00 so v= x 24.0 m  11.0 m = 13.0 m s . = 3.00 s  2.00 s t
2
a f
2
 2.00 3.00 + 3.00 m = 24.0 m
a f
(b)
At all times the instantaneous velocity is v= d 3.00t 2  2.00t + 3.00 = (6.00t  2.00) m s dt
c
h
At t = 2.00 s , v = 6.00( 2.00) 2.00 m s = 10.0 m s . At t = 3.00 s , v = 6.00(3.00) 2.00 m s = 16.0 m s . (c) (d) P2.17 (a) (b) (c) (d) a= v 16.0 m s  10.0 m s = = 6.00 m s 2 t 3.00 s  2.00 s d (6.00  2.00)= 6.00 m s 2 . (This includes both t = 2.00 s and t = 3.00 s ). dt
At all times a = a=
v 8.00 m s = = 1.3 m s 2 t 6.00 s
Maximum positive acceleration is at t = 3 s, and is approximately 2 m s 2 . a = 0 , at t = 6 s , and also for t > 10 s . Maximum negative acceleration is at t = 8 s, and is approximately 1.5 m s 2 .
Section 2.4 P2.18 (a) (b) (c) (d) (e)
Motion Diagrams
continued on next page
Chapter 2
29
(f)
One way of phrasing the answer: The spacing of the successive positions would change with less regularity. Another way: The object would move with some combination of the kinds of motion shown in (a) through (e). Within one drawing, the accelerations vectors would vary in magnitude and direction.
Section 2.5 P2.19
OneDimensional Motion with Constant Acceleration
From v 2 = vi2 + 2 ax , we have 10.97 10 3 m s f which is a = 2.79 10 4 times g .
c
h
2
= 0 + 2 a( 220 m) , so that a = 2.7410 5 m s 2
P2.20
(a) (b)
x f  xi = a=
1 1 vi + v f t becomes 40 m = vi + 2.80 m s (8.50 s) which yields vi = 6.61 m s . 2 2
c
h
a
f
v f  vi t
=
2.80 m s  6.61 m s = 0.448 m s 2 8.50 s
P2.21
Given vi = 12.0 cm s when x i = 3.00 cm(t = 0) , and at t = 2.00 s , x f = 5.00 cm , x f  x i = vi t + 1 1 2 2 at : 5.00  3.00 = 12.0( 2.00)+ a( 2.00) 2 2 32.0 a = = 16.0 cm s 2 . 8.00 = 24.0 + 2 a 2
*P2.22
(a)
Let i be the state of moving at 60 mi h and f be at rest
2 2 v xf = v xi + 2 a x x f  xi 2
d 0 = b60 mi hg
2
ax
mi fFGH 5 1280 ft IJK 3 600 mi F 5 280 ft I F 1 h I = G JG J = 21.8 mi h s 242 h H 1 mi K H 3 600 s K F 1 609 m IJ FG 1 h IJ = 9.75 m s = 21.8 mi h s G H 1 mi K H 3 600 s K + 2 a x 121 ft  0
i
a
2
.
(b)
Similarly,
0 = 80 mi h ax =  (c)
b
6 400 5 280 422 3 600
b
b
g
2
+ 2 a x 211 ft  0
g
g
a
f
mi h s = 22.2 mi h s = 9.94 m s 2 .
Let i be moving at 80 mi h and f be moving at 60 mi h .
d i b60 mi hg = b80 mi hg + 2a a211 ft  121 ftf 2 800b5 280 g a = mi h s = 22.8 mi h s = 10.2 m s 2a90fb3 600g
2 2 v xf = v xi + 2 a x x f  x i 2 2 x x
2
.
30 *P2.23
Motion in One Dimension
(a)
Choose the initial point where the pilot reduces the throttle and the final point where the boat passes the buoy: x i = 0 , x f = 100 m , v xi = 30 m s , v xf = ?, a x = 3.5 m s 2 , t = ? x f = xi + v xi t + 1 axt 2 : 2 100 m = 0 + 30 m s t +
2 2
a
f
1 3.5 m s 2 t 2 2
c
h
c1.75 m s ht  a30 m sft + 100 m = 0 .
We use the quadratic formula: t= b b 2  4ac 2a
t=
30 m s 900 m 2 s 2  4 1.75 m s 2 (100 m) 2 1.75 m s
c
c
2
h
j
h
=
30 m s 14.1 m s 3.5 m s 2
= 12.6 s or 4.53 s .
The smaller value is the physical answer. If the boat kept moving with the same acceleration, it would stop and move backward, then gain speed, and pass the buoy again at 12.6 s. (b) P2.24 (a) v xf = v xi + a x t = 30 m s  3.5 m s 2 4.53 s = 14.1 m s Total displacement = area under the v , t curve from t = 0 to 50 s. 1 50 m s 15 s + 50 m s 40  15 s 2 1 + 50 m s 10 s 2 x = 1 875 m x = (b) From t = 10 s to t = 40 s , displacement is 1 50 m s + 33 m s 5 s + 50 m s 25 s = 1 457 m . 2 v (50  0) m s 0 t 15 s : a1 = = = 3.3 m s 2 t 15 s  0 15 s < t < 40 s : a 2 = 0 x =
e
a f ga
b b
ga f b ga f
f
b
ga f b
ga f
FIG. P2.24
(c)
40 s t 50 s : a 3 =
continued on next page
v (0  50) m s = = 5.0 m s 2 t 50 s  40 s
Chapter 2
31
(d)
(i) (ii) (iii)
x1 = 0 +
1 1 a1 t 2 = 3.3 m s 2 t 2 or x1 = 1.67 m s 2 t 2 2 2
c
h
c
h
1 x 2 = (15 s) 50 m s  0 + 50 m s (t  15 s) or x 2 = 50 m s t  375 m 2 For 40 s t 50 s , x3 = or x 3 = 375 m + 1 250 m + which reduces to x 3 = 250 m s t  2.5 m s 2 t 2  4 375 m . 1 5.0 m s 2 t  40 s 2
a
f
a
f
FG area under v vs t IJ + 1 a (t  40 s) + a50 m sf(t  40 s) H from t = 0 to 40 sK 2
3 2
e
ja
f + b50 m sgat  40 sf
2
b
g e
j
(e) P2.25 (a)
v=
total displacement 1 875 m = = 37.5 m s total elapsed time 50 s
Compare the position equation x = 2.00 + 3.00t  4.00t 2 to the general form x f = xi + vi t + 1 2 at 2
to recognize that x i = 2.00 m, vi = 3.00 m s, and a = 8.00 m s 2 . The velocity equation, v f = vi + at , is then v f = 3.00 m s  8.00 m s 2 t . The particle changes direction when v f = 0 , which occurs at t = time is: x = 2.00 m + 3.00 m s 3 s . The position at this 8
c
h
a
fFGH 3 sIJK  c4.00 m s hFGH 3 sIJK 8 8
2
2
= 2.56 m .
(b)
From x f = xi + vi t +
2v 1 2 at , observe that when x f = xi , the time is given by t =  i . Thus, a 2 when the particle returns to its initial position, the time is t= 2 3.00 m s 8.00 m s
2
a
f=3 s
4 3.00 m s .
and the velocity is v f = 3.00 m s  8.00 m s 2
c
hFGH 3 sIJK = 4
32 *P2.26
Motion in One Dimension
The time for the Ford to slow down we find from x f = xi + 1 v xi + v xf t 2 2 250 m 2 x t= = = 6.99 s . v xi + v xf 71.5 m s + 0
d
a
i
f
Its time to speed up is similarly t= 2(350 m) 0 + 71.5 m s = 9.79 s .
The whole time it is moving at less than maximum speed is 6.99 s + 5.00 s + 9.79 s = 21.8 s . The Mercedes travels x f = xi + 1 1 v xi + v xf t = 0 + 71.5 + 71.5 m s 21.8 s 2 2 = 1 558 m
d
i
a
fb ga
f
while the Ford travels 250 + 350 m = 600 m, to fall behind by 1 558 m  600 m = 958 m . P2.27 (a) vi = 100 m s , a = 5.00 m s 2 , v f = vi + at so 0 = 100  5t , v 2 = vi2 + 2 a x f  xi so f 0 = (100 )  2(5.00) x f  0 . Thus x f = 1 000 m and t = 20.0 s . (b) P2.28 (a) At this acceleration the plane would overshoot the runway: No . Take ti = 0 at the bottom of the hill where x i = 0 , vi = 30.0 m s, a = 2.00 m s 2 . Use these values in the general equation x f = xi + vi t + to find 1 2 at 2
2 2 2
c
h
c
h
x f = 0 + 30.0t m s + when t is in seconds
a
f 1 c2.00 m s ht 2 h
x f = 30.0t  t 2 m . To find an equation for the velocity, use v f = vi + at = 30.0 m s + 2.00 m s 2 t , v f = (30.0  2.00t ) m s . (b) The distance of travel x f becomes a maximum, x max , when v f = 0 (turning point in the motion). Use the expressions found in part (a) for v f to find the value of t when x f has its maximum value: From v f = (3.00  2.00t ) m s , v f = 0 when t = 15.0 s. Then x max = 30.0t  t 2 m = (30.0)(15.0)(15.0) = 225 m .
c
e
j
c
h
2
Chapter 2
33
P2.29
In the simultaneous equations:
R  Sx  T
v xf = v xi + a x t
f
 xi =
1 v xi + v xf 2
c
U   we have Rv = v  c5.60 m s h(4.20 s)U . S 62.4 m = 1 v + v (4.20 s)  V V t hW   c h 2 T W
xf xi 2 xi xf
So substituting for v xi gives 62.4 m =
1 v xf + 56.0 m s 2 ( 4.20 s)+ v xf ( 4.20 s) 2
c
h
14.9 m s = v xf + Thus
1 5.60 m s 2 ( 4.20 s). 2
c
h
v xf = 3.10 m s .
P2.30
Take any two of the standard four equations, such as substitute into the other: v xi = v xf  a x t x f  xi = Thus
R  Sx  T
v xf = v xi + a x t
f
 xi =
1 v xi + v xf 2
c
U . Solve one for v ht V  W
xi ,
and
1 v xf  a x t + v xf t . 2
c
h
1 x f  xi = v xf t  a x t 2 . 2 Back in problem 29, 62.4 m = v xf ( 4.20 s) v xf = v f  vi t 632 1 2 5.60 m s 2 ( 4. 20 s) 2
c
h
62.4 m  49.4 m = 3.10 m s . 4.20 s
P2.31
(a)
a=
=
1.40
e j=
5 280 3 600
662 ft s 2 = 202 m s 2
(b)
x f = vi t +
5 280 1 2 1 1.40  662 1.40 at = 632 2 3 600 2
a
fFGH
I a f a fa f JK
2
= 649 ft = 198 m
34 P2.32
Motion in One Dimension
(a)
The time it takes the truck to reach 20.0 m s is found from v f = vi + at . Solving for t yields t= The total time is thus 10.0 s + 20.0 s + 5.00 s = 35.0 s . v f  vi a = 20.0 m s  0 m s 2.00 m s 2 = 10.0 s .
(b)
The average velocity is the total distance traveled divided by the total time taken. The distance traveled during the first 10.0 s is x 1 = vt =
FG 0 + 20.0 IJ(10.0)= 100 m . H 2 K
With a being 0 for this interval, the distance traveled during the next 20.0 s is x 2 = vi t + 1 2 at = ( 20.0)( 20.0)+ 0 = 400 m. 2
The distance traveled in the last 5.00 s is x 3 = vt =
FG 20.0 + 0 IJ(5.00)= 50.0 m. H 2 K
The total distance x = x1 + x 2 + x 3 = 100 + 400 + 50 = 550 m , and the average velocity is x 550 = 15.7 m s . given by v = = t 35.0 P2.33 We have vi = 2.00 10 4 m s, v f = 6.00 10 6 m s , x f  xi = 1.50 102 m . 2 1.50 102 m 2 x f  xi 1 = = 4.98 109 s vi + v f t : t = 4 6 2 vi + v f 2.00 10 m s + 6.00 10 m s
(a)
x f  xi =
c
h
c
h
c
h
(b)
v 2 = vi2 + 2 a x x f  xi : f v 2  vi2 f 2( x f  xi )
d
i
ax =
e6.00 10 =
6
ms
j  e2.00 10
2
4
ms
j
2
2(1.50 10 2 m)
= 1.20 10 15 m s 2
Chapter 2
35
*P2.34
(a)
2 2 v xf = v xi + 2 a x x f  x i : 0.01 3 10 8 m s
c
h
c
h
2
= 0 + 2 a x ( 40 m)
ax (b)
c310 = i
6
ms
h
2
80 m
= 1.12 10 11 m s 2
We must find separately the time t1 for speeding up and the time t 2 for coasting: x f  xi = 1 1 v xf + v xi t1 : 40 m = 3 10 6 m s + 0 t1 2 2 t1 = 2.67 10 5 s
d
e
j
x f  xi =
1 1 v xf + v xi t 2 : 60 m = 3 10 6 m s + 3 10 6 m s t 2 2 2 t 2 = 2.00 10 5 s
d
i
e
j
total time = 4.67 105 s . *P2.35 (a) Along the time axis of the graph shown, let i = 0 and f = t m . Then v xf = v xi + a x t gives v c = 0 + am tm am = (b) The displacement between 0 and t m is x f  xi = v xi t + The displacement between t m and t 0 is x f  xi = v xi t + The total displacement is x = (c) 1 1 v c t m + v c t 0  v c t m = v c t 0  tm 2 2 1 a x t 2 = v c t0  tm + 0 . 2 1 vc 2 1 1 axt 2 = 0 + t m = v c tm . 2 tm 2 2 vc . tm
a
f
FG H
IJ K
.
For constant v c and t 0 , x is minimized by maximizing t m to t m = t 0 . Then v t 1 x min = v c t 0  t 0 = c 0 . 2 2
FG H
IJ K
(e) (d) (e)
This is realized by having the servo motor on all the time. We maximize x by letting t m approach zero. In the limit x = v c t 0  0 = v c t 0 . This cannot be attained because the acceleration must be finite.
a
f
36 *P2.36
Motion in One Dimension
Let the glider enter the photogate with velocity vi and move with constant acceleration a. For its motion from entry to exit, x f = xi + v xi t + 1 axt 2 2 1 2 = 0 + vi t d + at d = v d t d 2 1 v d = vi + at d 2
(a)
The speed halfway through the photogate in space is given by
2 v hs = vi2 + 2 a
FG IJ = v H 2K
2 i
+ av d t d .
v hs = vi2 + av d t d and this is not equal to v d unless a = 0 . (b) The speed halfway through the photogate in time is given by v ht = vi + a equal to v d as determined above. P2.37 (a) Take initial and final points at top and bottom of the incline. If the ball starts from rest, vi = 0 , a = 0.500 m s 2 , x f  xi = 9.00 m . Then v 2 = vi2 + 2 a x f  xi = 0 2 + 2 0.500 m s 2 9.00 m f v f = 3.00 m s . (b) x f  x i = vi t + 1 2 at 2
FG t IJ and this is H 2K
d
d
i
e
ja
f
9.00 = 0 +
1 0.500 m s 2 t 2 2 t = 6.00 s
e
j
(c)
Take initial and final points at the bottom of the planes and the top of the second plane, respectively: vi = 3.00 m s, v f = 0 , x f  xi = 15.00 m. v 2 = vi2 + 2 a x f  xi gives f a= v 2  vi2 f 2 x f  xi
c
h
c
h
=
0  3.00 m s 2(15.0 m)
a
f
2
= 0.300 m s 2 .
(d)
Take the initial point at the bottom of the planes and the final point 8.00 m along the second: vi = 3.00 m s, x f  xi = 8.00 m , a = 0.300 m s 2 v 2 = vi2 + 2 a x f  xi = 3.00 m s f v f = 2.05 m s .
d
i b
g + 2e0.300 m s ja8.00 mf = 4.20 m
2 2
2
s2
Chapter 2
37
P2.38
Take the original point to be when Sue notices the van. Choose the origin of the xaxis at Sue's car. For her we have x is = 0 , vis = 30.0 m s , a s = 2.00 m s 2 so her position is given by x s (t )= x is + vis t + 1 1 a s t 2 = 30.0 m s t + 2.00 m s 2 t 2 . 2 2
a
f
c
h
For the van, x iv = 155 m, viv = 5.00 m s , a v = 0 and x v (t )= xiv + viv t + 1 a v t 2 = 155 + 5.00 m s t + 0 . 2
a
f
To test for a collision, we look for an instant t c when both are at the same place:
2 30.0t c  t c = 155 + 5.00t c 2 0 = t c  25.0t c + 155 .
From the quadratic formula 25.0 ( 25.0)  4(155) 2
2
tc =
= 13.6 s or 11.4 s .
The smaller value is the collision time. (The larger value tells when the van would pull ahead again if the vehicles could move through each other). The wreck happens at position 155 m + 5.00 m s (11.4 s)= 212 m . *P2.39 As in the algebraic solution to Example 2.8, we let t represent the time the trooper has been moving. We graph x car = 45 + 45t and x (km) 1.5 car 1 police officer t (s)
a
f
x trooper = 1.5t .
They intersect at t = 31 s .
2
0.5
10
20
30
40
FIG. P2.39
38
Motion in One Dimension
Section 2.6 P2.40
Freely Falling Objects
Choose the origin y = 0 , t = 0 at the starting point of the ball and take upward as positive. Then yi = 0 , vi = 0 , and a = g = 9.80 m s 2 . The position and the velocity at time t become: y f  yi = vi t + and v f = vi + at : v f = gt =  9.80 m s 2 t . (a) at t = 1.00 s : y f =  1 2 9.80 m s 2 (1.00 s) = 4.90 m 2 1 2 at t = 2.00 s : y f =  9.80 m s 2 ( 2.00 s) = 19.6 m 2 1 2 at t = 3.00 s : y f =  9.80 m s 2 (3.00 s) = 44.1 m 2 1 1 1 2 at : y f =  gt 2 =  9.80 m s 2 t 2 2 2 2
a
f
e
j
c
h
c c c
h h h
(b)
at t = 1.00 s : v f =  9.80 m s 2 (1.00 s)= 9.80 m s at t = 2.00 s : v f at t = 3.00 s : v f
2 2
c h = c9.80 m s h( 2.00 s)= 19.6 m s = c9.80 m s h(3.00 s)= 29.4 m s
P2.41
Assume that air resistance may be neglected. Then, the acceleration at all times during the flight is that due to gravity, a = g = 9.80 m s 2 . During the flight, Goff went 1 mile (1 609 m) up and then 1 mile back down. Determine his speed just after launch by considering his upward flight: v 2 = vi2 + 2 a y f  yi : f
d
i
0 = vi2  2 9.80 m s 2 1 609 m vi = 178 m s .
e
jb
g
His time in the air may be found by considering his motion from just after launch to just before impact: y f  yi = vi t + 1 1 2 at : 0 = 178 m s t  9.80 m s 2 t 2 . 2 2
a
f
c
h
The root t = 0 describes launch; the other root, t = 36.2 s , describes his flight time. His rate of pay may then be found from pay rate = $1.00 = 0.027 6 $ s 3 600 s h = $99.3 h . 36.2 s
b
gb
g
We have assumed that the workman's flight time, "a mile", and "a dollar", were measured to threedigit precision. We have interpreted "up in the sky" as referring to the free fall time, not to the launch and landing times. Both the takeoff and landing times must be several seconds away from the job, in order for Goff to survive to resume work.
Chapter 2
39
P2.42
1 We have y f =  gt 2 + vi t + yi 2 0 =  4.90 m s 2 t 2  8.00 m s t + 30.0 m . Solving for t, t= 8.00 64.0 + 588 . 9.80
c
h a
f
Using only the positive value for t, we find that t = 1.79 s . P2.43 (a) (b) y f  yi = vi t + 1 2 2 at : 4.00 = (1.50)vi (4.90)(1.50) and vi = 10.0 m s upward . 2
v f = vi + at = 10.0 (9.80)(1.50) = 4.68 m s v f = 4.68 m s downward
P2.44
The bill starts from rest vi = 0 and falls with a downward acceleration of 9.80 m s 2 (due to gravity). Thus, in 0.20 s it will fall a distance of y = vi t  1 2 2 gt = 0  4.90 m s 2 (0. 20 s) = 0.20 m . 2
c
h
This distance is about twice the distance between the center of the bill and its top edge 8 cm . Thus, David will be unsuccessful . *P2.45 (a) From y = vi t + 1 2 at with vi = 0 , we have 2 t= 2 y a
a
f
a f=
h
2(23 m) 9.80 m s 2
= 2.17 s .
(b) (c)
The final velocity is v f = 0 + 9.80 m s 2 ( 2.17 s)= 21.2 m s . The time take for the sound of the impact to reach the spectator is t sound = y v sound = 23 m = 6.76 102 s , 340 m s
c
so the total elapsed time is t total = 2.17 s + 6.76 10 2 s 2.23 s .
40 P2.46
Motion in One Dimension
At any time t, the position of the ball released from rest is given by y1 = h  position of the ball thrown vertically upward is described by y 2 = vi t 
1 2 gt . The time at which the 2 h 1 h first ball has a position of y1 = is found from the first equation as = h  gt 2 , which yields 2 2 2 h h . To require that the second ball have a position of y 2 = at this time, use the second t= g 2 equation to obtain ball as vi = gh . P2.47 (a) v f = vi  gt : v f = 0 when t = 3.00 s , g = 9.80 m s 2 . Therefore, vi = gt = 9.80 m s 2 (3.00 s)= 29.4 m s . (b) y f  yi = 1 v f + vi t 2 h h 1 h = vi  g . This gives the required initial upward velocity of the second 2 g 2 g
1 2 gt . At time t, the 2
F I GH JK
c
h
c
h
y f  yi = *P2.48 (a)
1 29.4 m s 3.00 s = 44.1 m 2
b
ga
f
Consider the upward flight of the arrow.
2 2 v yf = v yi + 2 a y y f  yi 2
d i 0 = b100 m sg + 2e 9.8 m s jy
2
y = (b)
10 000 m 2 s 2 19.6 m s 2
= 510 m
Consider the whole flight of the arrow. y f = yi + v yi t + 1 ayt 2 2
0 = 0 + 100 m s t +
b
g
1 9.8 m s 2 t 2 2
e
j
The root t = 0 refers to the starting point. The time of flight is given by t= 100 m s 4.9 m s 2 = 20.4 s . 1 9.80 m s 2 t 2 , t = 0.782 s. 2
P2.49
Time to fall 3.00 m is found from Eq. 2.12 with vi = 0 , 3.00 m = (a) (b)
c
h
With the horse galloping at 10.0 m s, the horizontal distance is vt = 7.82 m . t = 0.782 s
Chapter 2
41
P2.50
Take downward as the positive y direction. (a) While the woman was in free fall, y = 144 ft , vi = 0 , and a = g = 32.0 ft s 2 . Thus, y = vi t + before impact is: 1 2 at 144 ft = 0 + 16.0 ft s 2 t 2 giving t fall = 3.00 s . Her velocity just 2
c
h
v f = vi + gt = 0 + 32.0 ft s 2 (3.00 s)= 96.0 ft s . (b) While crushing the box, vi = 96.0 ft s , v f = 0 , and y = 18.0 in. = 1.50 ft . Therefore, a= v 2  vi2 f 2 y
c
h
a f
=
0  96.0 ft s 2(1.50 ft )
a
f
2
= 3.07 10 3 ft s 2 , or a = 3.07 10 3 ft s 2 upward .
(c) P2.51
Time to crush box: t =
2(1.50 ft) y y = v +v = or t = 3.13 102 s . f i v 0 + 96.0 ft s
2 3
y = 3.00t 3 : At t = 2.00 s , y = 3.00 2.00
a f
= 24.0 m and dy = 9.00t 2 = 36.0 m s . dt 1 1 2 gt = 24.0 + 36.0t  (9.80)t 2 . 2 2
vy =
A
If the helicopter releases a small mailbag at this time, the equation of motion of the mailbag is y b = y bi + vi t  Setting y b = 0 , 0 = 24.0 + 36.0t  4.90t 2 . Solving for t, (only positive values of t count), t = 7.96 s . *P2.52 Consider the last 30 m of fall. We find its speed 30 m above the ground: y f = yi + v yi t + 1 ayt 2 2
0 = 30 m + v yi 1.5 s + v yi =
30 m + 11.0 m = 12.6 m s . 1.5 s
a f 1 e9.8 m s ja1.5 sf 2
2
2
Now consider the portion of its fall above the 30 m point. We assume it starts from rest
i b12.6 m sg = 0 + 2e9.8 m s jy
2 2 v yf = v yi + 2 a y y f  yi 2 2
d
y =
160 m 2 s 2 19.6 m s 2
= 8.16 m .
Its original height was then 30 m + 8.16 m = 38.2 m .
42
Motion in One Dimension
Section 2.7 P2.53 (a)
Kinematic Equations Derived from Calculus J= da = constant dt da = Jdt
a = J dt = Jt + c 1
but a = ai when t = 0 so c 1 = ai . Therefore, a = Jt + ai dv dt dv = adt a=
z
v = adt =
z zb
v=
Jt + ai dt =
g
1 2 Jt + ai t + c 2 2
but v = vi when t = 0, so c 2 = vi and v = dx dt dx = vdt
1 2 Jt + ai t + vi 2
x = vdt = x=
z z FGH
1 2 Jt + ai t + vi dt 2
IJ K
1 3 1 2 Jt + ai t + vi t + c 3 6 2 x = xi when t = 0, so c 3 = xi . Therefore, x = (b) 1 3 1 2 Jt + ai t + vi t + xi . 6 2
a 2 = Jt + ai
a
a 2 = ai2 + J 2 t 2 + 2 Jai t
c
f
2
= J 2 t 2 + ai2 + 2 Jai t
1 2 a 2 = ai2 + 2 J Jt + ai t 2
FG H
h
IJ K
1 2 1 Jt + ai t + vi . So v  vi = Jt 2 + ai t . Therefore, 2 2
Recall the expression for v: v =
a
f
a 2 = ai2 + 2 J v  vi
a
f
.
Chapter 2
43
P2.54
(a)
See the graphs at the right. Choose x = 0 at t = 0. At t = 3 s, x = 1 8 m s (3 s)= 12 m . 2
a
f
At t = 5 s, x = 12 m + 8 m s ( 2 s)= 28 m . At t = 7 s, x = 28 m + (b) For 0 < t < 3 s, a = 1 8 m s ( 2 s)= 36 m . 2
a
f
a
f
8 ms = 2.67 m s 2 . 3s For 3 < t < 5 s, a = 0 . 16 m s = 4 m s 2 . 4s
(c) (d) (e)
For 5 s < t < 9 s , a = 
At t = 6 s, x = 28 m + 6 m s (1 s)= 34 m . At t = 9 s, x = 36 m + 1 8 m s ( 2 s)= 28 m . 2
a
f
a
f
FIG. P2.54
P2.55
(a)
a=
dv d = 5.00 10 7 t 2 + 3.00 10 5 t dt dt a =  10.0 10 7 m s 3 t + 3.00 10 5 m s 2
c
h
Take x i = 0 at t = 0. Then v =
t
dx dt
t
x  0 = vdt =
0
z ze
0
5.00 10 7 t 2 + 3.00 10 5 t dt
j
x = 5.00 10 7
t3 t2 + 3.00 10 5 3 2 3 3 7 x =  1.67 10 m s t + 1.50 10 5 m s 2 t 2 .
e
j e
j
(b)
The bullet escapes when a = 0 , at  10.0 10 7 m s 3 t + 3.00 10 5 m s 2 = 0 t= 3.00 10 5 s = 3.00 103 s . 10.0 10 7
2 5 3
c
h
(c)
New v = 5.00 10 7 3.00 103
c
hc
h + c3.0010 hc3.0010 h
5 3 2
v = 450 m s + 900 m s = 450 m s . (d) x =  1.67 10 7 3.00 103
c
hc
h + c1.5010 hc3.0010 h
3
x = 0.450 m + 1.35 m = 0.900 m
44 P2.56
Motion in One Dimension
a=
dv = 3.00 v 2 , vi = 1.50 m s dt dv = 3.00 v 2 dt
Solving for v,
v = vi
z
v
v 2 dv = 3.00 dt
t =0
z
t
 When v =
1 1 1 1 + = 3.00t or 3.00t =  . v vi v vi
vi 1 , t= = 0.222 s . 2 3.00 vi
Additional Problems *P2.57 The distance the car travels at constant velocity, v 0 , during the reaction time is x 1 = v 0 t r . The time for the car to come to rest, from initial velocity v 0 , after the brakes are applied is t2 = v f  vi a = 0  v0 v = 0 a a
a f
and the distance traveled during this braking period is
axf
2
= vt 2 =
Fv GH
f
+ vi 2
I t = FG 0 + v IJ FG  v IJ =  v . JK H 2 K H a K 2 a
2 0 0 2 0 2
Thus, the total distance traveled before coming to a stop is sstop = x
a f + a x f
1
= v 0 t r 
2 v0 . 2a
*P2.58
(a)
If a car is a distance sstop = v 0 t r 
2 v0 (See the solution to Problem 2.57) from the 2a intersection of length s i when the light turns yellow, the distance the car must travel before the light turns red is v2 x = sstop + si = v 0 t r  0 + si . 2a
Assume the driver does not accelerate in an attempt to "beat the light" (an extremely dangerous practice!). The time the light should remain yellow is then the time required for the car to travel distance x at constant velocity v 0 . This is t light (b) v s x v 0 t r  2 0 + si a = = = t r  0 + i . v0 v0 2 a v0
v2
With si = 16 m, v = 60 km h , a = 2.0 m s 2 , and t r = 1.1 s , t light = 1.1 s 
F 0.278 m s I + 16 m F 1 km h I = G J G J 2e 2.0 m s j H 1 km h K 60 km h H 0.278 m s K
60 km h
2
6. 23 s .
Chapter 2
45
*P2.59
(a)
As we see from the graph, from about 50 s to 50 s Acela is cruising at a constant positive velocity in the +x direction. From 50 s to 200 s, Acela accelerates in the +x direction reaching a top speed of about 170 mi/h. Around 200 s, the engineer applies the brakes, and the train, still traveling in the +x direction, slows down and then stops at 350 s. Just after 350 s, Acela reverses direction (v becomes negative) and steadily gains speed in the x direction.
200 100 0 50 0 100
v t 100 200 300 400
t (s)
FIG. P2.59(a)
(b)
The peak acceleration between 45 and 170 mi/h is given by the slope of the steepest tangent to the v versus t curve in this interval. From the tangent line shown, we find a = slope = v (155  45) mi h = = 2. 2 mi h s = 0.98 m s 2 . (100  50) s t
a
f
(c)
Let us use the fact that the area under the v versus t curve equals the displacement. The train's displacement between 0 and 200 s is equal to the area of the gray shaded region, which we have approximated with a series of triangles and rectangles. x 0 200 s = area 1 + area 2 + area 3 + area 4 + area 5
200 100 0 3
5 4 1 2 0
100 200 300 400
t (s)
b ga f b ga f + b160 mi hga100 sf 1 + a50 sfb100 mi hg 2 1 + a100 sfb170 mi h  160 mi hg 2 = 24 000bmi hgasf
50 mi h 50 s + 50 mi h 50 s
FIG. P2.59(c)
Now, at the end of our calculation, we can find the displacement in miles by converting hours to seconds. As 1 h = 3 600 s , x 0 200 s
F 24 000 mi I asf = GH 3 600 s JK
6.7 mi .
46 *P2.60
Motion in One Dimension
Average speed of every point on the train as the first car passes Liz: x 8.60 m = = 5.73 m s. 1.50 s t The train has this as its instantaneous speed halfway through the 1.50 s time. Similarly, halfway 8.60 m through the next 1.10 s, the speed of the train is = 7.82 m s . The time required for the speed 1.10 s to change from 5.73 m/s to 7.82 m/s is 1 1 (1.50 s)+ (1.10 s)= 1.30 s 2 2 so the acceleration is: a x = v x 7.82 m s  5.73 m s = = 1.60 m s 2 . t 1.30 s
P2.61
The rate of hair growth is a velocity and the rate of its increase is an acceleration. Then mm d v xi = 1.04 mm d and a x = 0.132 . The increase in the length of the hair (i.e., displacement) w during a time of t = 5.00 w = 35.0 d is
FG H
IJ K
x = v xi t +
x = 1.04 mm d 35.0 d + or x = 48.0 mm . P2.62
b
1 axt 2 2
ga
f 1 b0.132 mm d wga35.0 dfa5.00 wf 2
Let point 0 be at ground level and point 1 be at the end of the engine burn. Let point 2 be the highest point the rocket reaches and point 3 be just before impact. The data in the table are found for each phase of the rocket's motion. (0 to 1) v 2  80.0 f
a f
2
2
= 2 4.00 1 000
a fb
g
h
so giving
v f = 120 m s t = 10.0 s x f  xi = 735 m t = 12.2 s FIG. P2.62
120 = 80.0 +( 4.00)t (1 to 2) 0 (120) = 2(9.80) x f  xi
c
giving
0  120 = 9.80t giving This is the time of maximum height of the rocket. (2 to 3)
v 2  0 = 2 9.80 1 735 f
v f = 184 = (9.80)t
a
fb
g
giving t = 18.8 s
(a) (b)
t total = 10 + 12.2 + 18.8 = 41.0 s
cx
f
 xi
h
continued on next page
total
= 1.73 km
Chapter 2
47
(c)
v final = 184 m s t 0.0 10.0 22.2 41.0 x 0 1 000 1 735 0 v 80 120 0 184 a +4.00 +4.00 9.80 9.80
0 #1 #2 #3 P2.63
Launch End Thrust Rise Upwards Fall to Earth
Distance traveled by motorist = 15.0 m s t 1 Distance traveled by policeman = 2.00 m s 2 t 2 2
a
c
f
h
(a) (b) (c) P2.64
intercept occurs when 15.0t = t 2 , or t = 15.0 s v(officer)= 2.00 m s 2 t = 30.0 m s x(officer )= 1 2.00 m s 2 t 2 = 225 m 2
c
h
c
h
Area A1 is a rectangle. Thus, A1 = hw = v xi t . 1 1 Area A 2 is triangular. Therefore A 2 = bh = t v x  v xi . 2 2 The total area under the curve is
b
g
vx vx vxi A1 0 FIG. P2.64 t t A2
A = A1 + A 2 = v xi t + and since v x  v xi = a x t A = v xi t +
bv
x
 v xi t 2
g
1 axt 2 . 2 1 a x t 2 , the 2
The displacement given by the equation is: x = v xi t + same result as above for the total area.
48 P2.65
Motion in One Dimension
(a)
Let x be the distance traveled at acceleration a until maximum speed v is reached. If this is achieved in time t1 we can use the following three equations: x= The first two give 100 = 10.2  1 v + vi t1 , 100  x = v 10.2  t1 and v = vi + at1 . 2
a
f
a
f
FG 1 t IJ v = FG10.2  1 t IJ at H 2 K H 2 K 200 a= b20.4  t gt .
1 1 1 1
1
For Maggie: a =
a18.4fa2.00f = 200 For Judy: a = a17.4fa3.00f = a fa f Judy: v = a3.83fa3.00f =
x=
200
5.43 m s 2 3.83 m s 2
(b)
v = a1 t Maggie: v = 5.43 2.00 = 10.9 m s 11.5 m s
(c)
At the sixsecond mark 1 2 at1 + v 6.00  t1 2
2
a
f
Maggie: x =
1 5.43 2.00 2 1 Judy: x = 3.83 3.00 2
a fa f + a10.9fa4.00f = 54.3 m a fa f + a11.5fa3.00f = 51.7 m
2
Maggie is ahead by 2.62 m . P2.66 a1 = 0.100 m s 2 1 1 2 2 x = 1 000 m = a1 t1 + v1 t 2 + a 2 t 2 2 2 at at 1 1 2 1 000 = a1 t1 + a1 t1  1 1 + a 2 1 1 2 a2 2 a2 a 2 = 0.500 m s 2 t = t1 + t 2 and v1 = a1 t1 = a 2 t 2
2
FG H
IJ K
FG IJ H K
1 000 = t1 =
a 1 2 a1 1  1 t1 2 a2
FG H
IJ K
t2 =
a1 t1 12.9 = 26 s a 2 0.500
20 000 = 129 s 1.20
Total time = t = 155 s
Chapter 2
49
P2.67
Let the ball fall 1.50 m. It strikes at speed given by
2 2 v xf = v xi + 2 a x f  x i : 2 v xf = 0 + 2 9.80 m s 2 (1.50 m)
c
h
c
h
v xf = 5.42 m s and its stopping is described by
2 2 v xf = v xi + 2 a x x f  x i
0 = 5.42 m s ax = 2.00 10
b
d
g
i e j
2
+ 2 a x 10 2 m
29.4 m 2 s 2
2
m
= +1.47 10 3 m s 2 .
Its maximum acceleration will be larger than the average acceleration we estimate by imagining constant acceleration, but will still be of order of magnitude ~ 10 3 m s 2 . *P2.68 (a) x f = xi + v xi t + 1 a x t 2 . We assume the package starts from rest. 2 145 m = 0 + 0 + 1 9.80 m s 2 t 2 2
c
h
t=
2(145 m) 9.80 m s 2
= 5. 44 s
(b)
x f = xi + v xi t +
1 1 2 a x t 2 = 0 + 0 + 9.80 m s 2 (5.18 s) = 131 m 2 2
c
h
distance fallen = x f = 131 m (c) (d) speed = v xf = v xi + a x t = 0 + 9.8 m s 2 5.18 s = 50.8 m s The remaining distance is 145 m  131.5 m = 13.5 m . During deceleration, v xi = 50.8 m s, v xf = 0, x f  xi = 13.5 m
2 2 v xf = v xi + 2 a x x f  x i :
e
j
c
h
0 = 50.8 m s ax =
a
f
2
+ 2 a x (13.5 m)
2 580 m 2 s 2 = +95.3 m s 2 = 95.3 m s 2 upward . 2 13.5 m
a
f
50 P2.69
Motion in One Dimension
(a)
1 1 2 at = 50.0 = 2.00t + (9.80)t 2 , 2 2 4.90t 2 + 2.00t  50.0 = 0 y f = v i1 t + t= 2.00 + 2.00 2  4( 4.90)(50.0) 2( 4.90)
Only the positive root is physically meaningful: t = 3.00 s after the first stone is thrown. (b) 1 2 at and t = 3.00  1.00 = 2.00 s 2 1 2 substitute 50.0 = vi 2 ( 2.00)+ (9.80)( 2.00) : 2 y f = vi 2 t + vi2 = 15.3 m s downward (c) v1 f = vi1 + at = 2.00 +(9.80)(3.00)= 31.4 m s downward v 2 f = vi 2 + at = 15.3 +(9.80)( 2.00)= 34.8 m s downward P2.70 (a) 1 2 d = (9.80)t1 2 t1 + t 2 = 2.40
2 4.90t 2
d = 336 t 2
336t 2 = 4.90 2.40  t 2
t2 = so
a
f
2
 359.5t 2 + 28.22 = 0
359.5 359.5 2  4( 4.90)( 28.22) 9.80
359.5 358.75 = 0.076 5 s t2 = 9.80 (b) P2.71 (a)
d = 336 t 2 = 26.4 m
1 2 Ignoring the sound travel time, d = (9.80)( 2.40) = 28.2 m , an error of 6.82% . 2 In walking a distance x , in a time t , the length of rope is only increased by x sin . x sin . The pack lifts at a rate t v= x x sin = v boy = v boy t x x + h2
2
(b)
d 1 dv v boy dx = + v boy x dt dt dt v boy v boy x d x d = v = v boy a = v boy  2 , but dt dt 2 2 2 v boy v boy h 2 h 2 v boy x2 a= = 1 2 = 2 3 2 x 2 + h2 a=
FG IJ HK
F GH
I JK
c
h
(c) (d)
2 v boy
h
,0
FIG. P2.71
v boy , 0
Chapter 2
51
P2.72
h = 6.00 m, v boy = 2.00 m s v = However, x = v boy t : v = (a) ts 0 1 1.5 2 2.5 3 3.5 4 4.5 5 (b)
v boy x x x sin = v boy = . 12 t x 2 + h2
c
h
2 v boy t
c
2 v boy t 2
+h
2 12
h
=
c 4t
4t
2
+ 36
h
12
.
a f vb m s g
0 0.32 0.63 0.89 1.11 1.28 1.41 1.52 1.60 1.66 1.71
2 h 2 v boy 2 h 2 v boy
0.5
FIG. P2.72(a) 144
2
From problem 2.71 above, a = ts 0 1 1.5 2 2.5 3. 3.5 4. 4.5 5
a f aem s j
2
cx
2
+h
2 3 2
h
=
c
2 v boy t 2
+h
2 3 2
h
=
c4t
+ 36
h
32
.
0.67 0.64 0.57 0.48 0.38 0.30 0.24 0.18 0.14 0.11 0.09
0.5
FIG. P2.72(b)
P2.73
(a)
We require x s = x k when t s = t k + 1.00 1 3.50 m s 2 t k + 1.00 2 t k + 1.00 = 1.183t k xs =
e
jb
g
2
=
1 4.90 m s 2 t k 2
e
jb g
2
= xk
t k = 5.46 s . (b) (c) xk = 1 4.90 m s 2 5.46 s 2
e
ja
f
2
= 73.0 m
v k = 4.90 m s 2 5.46 s = 26.7 m s vs
2
e ja f = e3.50 m s ja6.46 sf =
22.6 m s
52 P2.74
Motion in One Dimension
Time t (s) 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00 3.25 3.50 3.75 4.00 4.25 4.50 4.75 5.00
Height h (m) 5.00 5.75
h (m) 0.75 0.65
t (s) 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25
v (m/s) 3.00 2.60 2.16 1.76 1.36 0.96 0.56 0.12 0.24 0.68 1.12 1.48 1.92 2.28 2.72 3.16 3.52 3.96 4.36 4.76
midpt time t (s) 0.13 0.38 0.63 0.88 1.13 1.38 1.63 1.88 2.13 2.38 2.63 2.88 3.13 3.38 3.63 3.88 4.13 4.38 4.63 4.88 FIG. P2.74
6.40 0.54 6.94 0.44 7.38 0.34 7.72 0.24 7.96 0.14 8.10 0.03 8.13 0.06 8.07 0.17 7.90 0.28 7.62 0.37 7.25 0.48 6.77 0.57 6.20 0.68 5.52 0.79 4.73 0.88 3.85 0.99 2.86 1.09 1.77 1.19 0.58 TABLE P2.74
acceleration = slope of line is constant.
a =1.63 m s 2 = 1.63 m s 2 downward
Chapter 2
53
P2.75
The distance x and y are always related by x 2 + y 2 = L2 . Differentiating this equation with respect to time, we have dy dx 2x + 2y =0 dt dt Now dy dx = v . is v B , the unknown velocity of B; and dt dt From the equation resulting from differentiation, we have dy x dx x =  (v). = dt y dt y But
y
y B x L v A
O x
FG IJ H K
FIG. P2.75
y v v 3 1 = = 0.577 v . = tan so v B = v . When = 60.0 , v B = tan 60.0 3 x tan
FG H
IJ K
ANSWERS TO EVEN PROBLEMS
P2.2 (a) 2 10 7 m s ; 1 10 6 m s ; (b) 5 10 yr P2.4 P2.6 (a) 50.0 m s ; (b) 41.0 m s (a) 27.0 m ; 2 (b) 27.0 m + 18.0 m s t + 3.00 m s 2 t ;
8
P2.24
(a) 1.88 km; (b) 1.46 km; (c) see the solution; (d) (i) x 1 = 1.67 m s 2 t 2 ;
(ii) x 2 = 50 m s t  375 m ;
b
g e
ja f
(iii) x 3
b g = b 250 m sgt  e 2.5 m s jt
2
e
j
2
 4 375 m ;
(e) 37.5 m s P2.26 P2.28 958 m (a) x f = 30.0t  t 2 m; v f = 30.0  2t m s ; (b) 225 m P2.30 P2.32 P2.34 P2.36 P2.38 P2.40 x f  xi = v xf t  1 a x t 2 ; 3.10 m s 2
(c) 18.0 m s P2.8 P2.10 P2.12 P2.14 (a), (b), (c) see the solution; 4.6 m s 2 ; (d) 0 5.00 m (a) 20.0 m s ; 5.00 m s ; (b) 262 m (a) see the solution; (b) 1.60 m s 2 ; 0.800 m s 2 (a) 13.0 m s; (b) 10.0 m s; 16.0 m s; (c) 6.00 m s 2 ; (d) 6.00 m s 2 P2.18 P2.20 P2.22 see the solution (a) 6.61 m s; (b) 0. 448 m s 2 (a) 21.8 mi h s = 9.75 m s 2 ; (b) 22.2 mi h s = 9.94 m s 2 ; (c) 22.8 mi h s = 10.2 m s 2
e
j
a
f
(a) 35.0 s; (b) 15.7 m s (a) 1.12 10 11 m s 2 ; (b) 4.67 10 5 s (a) False unless the acceleration is zero; see the solution; (b) True Yes; 212 m; 11.4 s (a) 4.90 m ; 19.6 m; 44.1 m; (b) 9.80 m s; 19.6 m s; 29.4 m s 1.79 s
P2.16
P2.42
54 P2.44 P2.46
Motion in One Dimension
No; see the solution The second ball is thrown at speed vi = gh (a) 510 m; (b) 20.4 s (a) 96.0 ft s ; (b) a = 3.07 10 ft s upward ; (c) t = 3.13 10 2 s
3 2
P2.60 P2.62 P2.64 P2.66 P2.68
1.60 m s 2 (a) 41.0 s; (b) 1.73 km; (c) 184 m s v xi t + 1 a x t 2 ; displacements agree 2
P2.48 P2.50
155 s; 129 s (a) 5.44 s; (b) 131 m; (c) 50.8 m s ; (d) 95.3 m s 2 upward
P2.52 P2.54
38.2 m (a) and (b) see the solution; (c) 4 m s ; (d) 34 m; (e) 28 m 0.222 s (a) see the solution; (b) 6.23 s
2
P2.70 P2.72 P2.74
(a) 26.4 m; (b) 6.82% see the solution see the solution; a x = 1.63 m s 2
P2.56 P2.58
3
Vectors
CHAPTER OUTLINE
3.1 3.2 3.3 3.4 Coordinate Systems Vector and Scalar Quantities Some Properties of Vectors Components of a Vector and Unit Vectors
ANSWERS TO QUESTIONS
Q3.1 No. The sum of two vectors can only be zero if they are in opposite directions and have the same magnitude. If you walk 10 meters north and then 6 meters south, you won't end up where you started. No, the magnitude of the displacement is always less than or equal to the distance traveled. If two displacements in the same direction are added, then the magnitude of their sum will be equal to the distance traveled. Two vectors in any other orientation will give a displacement less than the distance traveled. If you first walk 3 meters east, and then 4 meters south, you will have walked a total distance of 7 meters, but you will only be 5 meters from your starting point.
Q3.2
Q3.3
The largest possible magnitude of R = A + B is 7 units, found when A and B point in the same direction. The smallest magnitude of R = A + B is 3 units, found when A and B have opposite directions. Only force and velocity are vectors. None of the other quantities requires a direction to be described. If the directionangle of A is between 180 degrees and 270 degrees, its components are both negative. If a vector is in the second quadrant or the fourth quadrant, its components have opposite signs. The book's displacement is zero, as it ends up at the point from which it started. The distance traveled is 6.0 meters. 85 miles. The magnitude of the displacement is the distance from the starting point, the 260mile mark, to the ending point, the 175mile mark. Vectors A and B are perpendicular to each other. No, the magnitude of a vector is always positive. A minus sign in a vector only indicates direction, not magnitude.
Q3.4 Q3.5
Q3.6 Q3.7 Q3.8 Q3.9
55
56
Vectors
Q3.10 Q3.11 Q3.12 Q3.13
Any vector that points along a line at 45 to the x and y axes has components equal in magnitude. A x = B x and A y = B y . Addition of a vector to a scalar is not defined. Think of apples and oranges. One difficulty arises in determining the individual components. The relationships between a vector and its components such as A x = A cos , are based on righttriangle trigonometry. Another problem would be in determining the magnitude or the direction of a vector from its components. Again,
2 2 A = A x + A y only holds true if the two component vectors, A x and A y , are perpendicular.
Q3.14
If the direction of a vector is specified by giving the angle of the vector measured clockwise from the positive yaxis, then the xcomponent of the vector is equal to the sine of the angle multiplied by the magnitude of the vector.
SOLUTIONS TO PROBLEMS
Section 3.1 P3.1 Coordinate Systems
a f a fa f y = r sin = a5.50 mf sin 240 = a5.50 mfa 0.866f =
(a)
x = r cos = 5.50 m cos 240 = 5.50 m 0.5 = 2.75 m 4.76 m
P3.2
x = r cos and y = r sin , therefore x1 = 2.50 m cos 30.0 , y1 = 2.50 m sin 30.0 , and
a
f
a
f
bx , y g = a2.17 , 1.25f m x = a3.80 mf cos 120 , y = a3.80 mf sin 120 , and bx , y g = a1.90, 3.29f m .
1 1
2
2
2
2
(b) P3.3
d = ( x) 2 + ( y) 2 = 16.6 + 4.16 = 4.55 m
The x distance out to the fly is 2.00 m and the y distance up to the fly is 1.00 m. (a) We can use the Pythagorean theorem to find the distance from the origin to the fly. distance = x 2 + y 2 = (b)
a2.00 mf + a1.00 mf
2
2
= 5.00 m 2 = 2.24 m
= tan 1
FG 1 IJ = 26.6 ; r = H 2K
2.24 m, 26.6
Chapter 3
57
P3.4
(a)
d=
bx
2
 x1
g + by
2
2
 y1
g
2
=
c2.00  3.00 h + a4.00  3.00f
2
2
d = 25.0 + 49.0 = 8.60 m (b) r1 =
1
a2.00f + a4.00f = 20.0 = F 4.00 IJ = 63.4 = tan G  H 2.00 K
2 2 1
4.47 m
r2 =
a3.00f + a3.00f
2
2
= 18.0 = 4.24 m
2 = 135 measured from the +x axis.
P3.5 We have 2.00 = r cos 30.0 r= and y = r sin 30.0 = 2.31 sin 30.0 = 1.15 . P3.6 We have r = x 2 + y 2 and = tan 1 (a) 2.00 = 2.31 cos 30.0
FG y IJ . H xK
The radius for this new point is
a x f
and its angle is
2
+ y2 = x2 + y2 = r
tan 1
FG y IJ = H x K
180 .
(b)
( 2 x) 2 + ( 2 y) 2 = 2r . This point is in the third quadrant if x , y is in the first quadrant or in the fourth quadrant if x , y is in the second quadrant. It is at an angle of 180+ .
b g
b g
(c)
( 3 x) 2 + ( 3 y) 2 = 3r . This point is in the fourth quadrant if x , y is in the first quadrant or in the third quadrant if x , y is in the second quadrant. It is at an angle of  .
b g
b g
58
Vectors
Section 3.2 Section 3.3 P3.7
Vector and Scalar Quantities Some Properties of Vectors x 100 m x = 100 m tan 35.0 = 70.0 m
tan 35.0 =
a
f
FIG. P3.7 P3.8 R = 14 km
= 65 N of E
R 13 km
1 km 6 km FIG. P3.8 P3.9  R = 310 km at 57 S of W (Scale: 1 unit = 20 km )
FIG. P3.9 P3.10 (a) Using graphical methods, place the tail of vector B at the head of vector A. The new vector A + B has a magnitude of 6.1 at 112 from the xaxis.
A+B y B B
A AB x O
(b)
The vector difference A  B is found by placing the negative of vector B at the head of vector A. The resultant vector A  B has magnitude 14.8 units at an angle of 22 from the + xaxis.
FIG. P3.10
Chapter 3
59
P3.11
(a)
d =  10.0 i = 10.0 m since the displacement is in a straight line from point A to point B. B
C 5.00 m d FIG. P3.11 A
(b)
The actual distance skated is not equal to the straightline displacement. The distance follows the curved path of the semicircle (ACB). s= 1 2 r = 5 = 15.7 m 2
b g
(c) P3.12
If the circle is complete, d begins and ends at point A. Hence, d = 0 .
Find the resultant F1 + F2 graphically by placing the tail of F2 at the head of F1 . The resultant force vector F1 + F2 is of magnitude 9.5 N and at an angle of 57 above the x axis . y
F1 + F2
F2
F1 0 1 2 3 N FIG. P3.12 P3.13 (a)
x
The large majority of people are standing or sitting at this hour. Their instantaneous foottohead vectors have upward vertical components on the order of 1 m and randomly oriented horizontal components. The citywide sum will be ~ 10 5 m upward .
(b)
Most people are lying in bed early Saturday morning. We suppose their beds are oriented north, south, east, west quite at random. Then the horizontal component of their total vector height is very nearly zero. If their compressed pillows give their height vectors vertical components averaging 3 cm, and if onetenth of one percent of the population are onduty nurses or police officers, we estimate the total vector height as ~ 10 5 0.03 m + 10 2 1 m
a
f
a f
~ 10 m upward .
3
60 P3.14
Vectors
Your sketch should be drawn to scale, and should look somewhat like that pictured to the right. The angle from the westward direction, , can be measured to be 4 N of W , and the distance R from the sketch can be converted according to the scale to be 7.9 m .
N 1m W 15.0 meters E
R
3.50 meters
30.0
8.20 meters
S
FIG. P3.14 P3.15 To find these vector expressions graphically, we draw each set of vectors. Measurements of the results are taken using a ruler and protractor. (Scale: 1 unit = 0.5 m ) (a) (b) (c) (d) A + B = 5.2 m at 60 A B = 3.0 m at 330 B A = 3.0 m at 150 A 2B = 5.2 m at 300.
FIG. P3.15 *P3.16 The three diagrams shown below represent the graphical solutions for the three vector sums: R 1 = A + B + C , R 2 = B + C + A , and R 3 = C + B + A . You should observe that R 1 = R 2 = R 3 , illustrating that the sum of a set of vectors is not affected by the order in which the vectors are added.
100 m
C B A B R1 C B A R2 A R3 C
FIG. P3.16
Chapter 3
61
P3.17
The scale drawing for the graphical solution should be similar to the figure to the right. The magnitude and direction of the final displacement from the starting point are obtained by measuring d and on the drawing and applying the scale factor used in making the drawing. The results should be d = 420 ft and = 3
(Scale: 1 unit = 20 ft ) FIG. P3.17
Section 3.4 P3.18
Components of a Vector and Unit Vectors
Coordinates of the superhero are:
a f a f y = a100 mf sina 30.0f =
P3.19 A x = 25.0 A y = 40.0
2 2 A = Ax + Ay =
x = 100 m cos 30.0 = 86.6 m 50.0 m FIG. P3.18
a25.0f + a40.0f
2
2
= 47.2 units
We observe that tan = Ay Ax . FIG. P3.19
So
= tan 1
F A I = tan 40.0 = tan a1.60f = 58.0 . GH A JK 25.0
y x 1
The diagram shows that the angle from the +x axis can be found by subtracting from 180:
= 180  58 = 122 .
P3.20 The person would have to walk 3.10 sin 25.0 = 1.31 km north , and 3.10 cos 25.0 = 2.81 km east .
a
f
a
f
62 P3.21
Vectors
x = r cos and y = r sin , therefore: (a) (b) (c) x = 12.8 cos 150 , y = 12.8 sin 150 , and x , y = 11.1i + 6.40 j m x = 3.30 cos 60.0 , y = 3.30 sin 60.0 , and x , y = 1.65 i + 2.86 j cm x = 22.0 cos 215 , y = 22.0 sin 215 , and x , y = 18.0 i  12.6 j in
b g e
j
b g e
j
b g e
j
P3.22
x = d cos = 50.0 m cos 120 = 25.0 m y = d sin = 50.0 m sin 120 = 43.3 m
d=
f a f f a f a25.0 mfi + a43.3 mf j
a a
*P3.23
(a)
Her net x (eastwest) displacement is 3.00 + 0 + 6.00 = +3.00 blocks, while her net y (northsouth) displacement is 0 + 4.00 + 0 = +4.00 blocks. The magnitude of the resultant displacement is R=
b x g + by g
net 2 net
2
=
a3.00f + a4.00f
2 1
2
= 5.00 blocks
and the angle the resultant makes with the xaxis (eastward direction) is
= tan 1
FG 4.00 IJ = tan a1.33f = 53.1 . H 3.00 K
The resultant displacement is then 5.00 blocks at 53.1 N of E . (b) *P3.24 The total distance traveled is 3.00 + 4.00 + 6.00 = 13.0 blocks . N
Let i = east and j = north. The unicyclist's displacement is, in meters 280 j + 220 i + 360 j  300 i  120 j + 60 i  40 j  90 i + 70 j . R = 110 i + 550 j = R
2
a110 mf + a550 mf
2
at tan 1
= 561 m at 11.3 west of north . The crow's velocity is v=
110 m west of north 550 m FIG. P3.24
E
x 561 m at 11.3 W of N = t 40 s
= 14.0 m s at 11.3 west of north .
Chapter 3
63
P3.25
+x East, +y North
x = 250 +125 cos 30 = 358 m y = 75 +125 sin 30150 = 12.5 m
c xh + c yh = a358f + a12.5f c yh =  12.5 = 0.0349 tan = c xh 358
d=
2 2 2
2
= 358 m
= 2.00 d = 358 m at 2.00 S of E
P3.26 The east and north components of the displacement from Dallas (D) to Chicago (C) are the sums of the east and north components of the displacements from Dallas to Atlanta (A) and from Atlanta to Chicago. In equation form: d DC east = d DA east + d AC east = 730 cos 5.00560 sin 21.0 = 527 miles. d DC north = d DA north + d AC north = 730 sin 5.00+560 cos 21.0 = 586 miles. By the Pythagorean theorem, d = ( d DC east ) 2 + ( d DC north ) 2 = 788 mi . Then tan = d DC north = 1.11 and = 48.0 . d DC east
Thus, Chicago is 788 miles at 48.0 northeast of Dallas . P3.27 (a) (b) See figure to the right. C = A + B = 2.00 i + 6.00 j + 3.00 i  2.00 j = 5.00 i + 4.00 j C = 25.0 + 16.0 at tan 1
FG 4 IJ = H 5K
1
6.40 at 38.7
D = A  B = 2.00 i + 6.00 j  3.00 i + 2.00 j = 1.00 i + 8.00 j
a1.00f + a8.00f at tan FGH 81.00 IJK .00 D = 8.06 at b180  82.9g = 8.06 at 97.2
D=
2 2
FIG. P3.27
P3.28
bx + x + x g + by + y + y g = a3.00  5.00 + 6.00f + a 2.00 + 3.00 + 1.00 f F 6.00 IJ = 56.3 = tan G H 4.00 K
d=
1 2 3 2 2 1 2 3 2 1
2
= 52.0 = 7.21 m
64 P3.29
Vectors
We have B = R  A : A x = 150 cos120 = 75.0 cm A y = 150 sin 120 = 130 cm R x = 140 cos 35.0 = 115 cm R y = 140 sin 35.0 = 80.3 cm Therefore, B = 115  75 i + 80.3  130 j = 190 i  49.7 j cm B = 190 2 + 49.7 2 = 196 cm FIG. P3.29
a f
e
j
= tan 1 
P3.30 A = 8.70 i + 15.0 j and B = 13. 2 i  6.60 j A  B + 3C = 0 :
FG H
49.7 = 14.7 . 190
IJ K
3C = B  A = 21.9 i  21.6 j C = 7.30 i  7.20 j or C x = 7.30 cm ; C y = 7.20 cm P3.31 (a) (b) (c) (d) (e)
aA + Bf = e3 i  2 jj + e i  4 jj = aA  Bf = e3i  2 jj  e i  4 jj =
A + B = 2 2 + 6 2 = 6.32 A  B = 4 2 + 2 2 = 4.47
2i  6 j 4i + 2 j
A+B = tan1  AB = tan1
FG 6 IJ = 71.6= H 2K FG 2 IJ = 26.6 H 4K
288
P3.32
(a)
D = A + B + C = 2i + 4 j D = 2 2 + 4 2 = 4. 47 m at = 63.4
(b)
E =  A  B + C = 6 i + 6 j E = 6 2 + 6 2 = 8.49 m at = 135
Chapter 3
65
P3.33
d1 = 3.50 j m d 2 = 8.20 cos 45.0 i + 8.20 sin 45.0 j = 5.80 i + 5.80 j m d 3 = 15.0 i m R = d1 + d 2 + d 3 = 15.0 + 5.80 i + 5.80  3.50 j = (or 9.20 m west and 2.30 m north) The magnitude of the resultant displacement is
2 2 R = Rx + R y =
e
j
e
j
e
j
a
f a
f e9.20i + 2.30 jj m a9.20f + a2.30f
2
2
= 9.48 m .
The direction is = arctan P3.34 Refer to the sketch
FG 2.30 IJ = H 9.20 K
166 .
A = 10.0 B = 15.0 C = 50.0 R
R = A + B + C = 10.0 i  15.0 j + 50.0 i = 40.0 i  15.0 j R = 40.0
a f + a15.0f
2
2 12
= 42.7 yards
FIG. P3.34 P3.35 (a) F = F1 + F2
a f F = 60.0 i + 104 j  20.7 i + 77.3 j = e39.3 i + 181 jj N f f
F = 39.3 2 + 181 2 = 185 N
F = 120 cos 60.0 i + 120 sin 60.0 j  80.0 cos 75.0 i + 80.0 sin 75.0 j
a
a
a
f
= tan 1
(b) P3.36 F3 =  F =
FG 181 IJ = H 39.3 K
77.8
e39.3 i  181 jj N
East x 0m 1.41 0.500 +0.914 R=
2
West y 4.00 m 1.41 0.866 4.55
2
x + y = 4.64 m at 78.6 N of E
66 P3.37
Vectors
A = 3.00 m, A = 30.0 A x = A cos A = 3.00 cos 30.0 = 2.60 m
B = 3.00 m , B = 90.0 A y = A sin A = 3.00 sin 30.0= 1.50 m A = A x i + A y j = 2.60 i + 1.50 j m
e
j
Bx = 0 , By = 3.00 m A + B = 2.60 i + 1.50 j + 3.00 j = P3.38
so
B = 3.00 j m
e
j
e2.60i + 4.50 jj m
Let the positive xdirection be eastward, the positive ydirection be vertically upward, and the positive zdirection be southward. The total displacement is then d = 4.80 i + 4.80 j cm + 3.70 j  3.70k cm = 4.80 i + 8.50 j  3.70k cm . (a) (b) The magnitude is d = ( 4.80) +(8.50) + (3.70) cm = 10.4 cm . Its angle with the yaxis follows from cos = 8.50 , giving = 35.5 . 10. 4
2 2 2
e
j
e
j
e
j
P3.39
B = Bx i + By j + Bz k = 4.00 i + 6.00 j + 3.00k B = 4.00 2 + 6.00 2 + 3.00 2 = 7.81
= cos 1 = cos 1 = cos 1
P3.40
FG 4.00 IJ = H 7.81 K FG 6.00 IJ = H 7.81 K FG 3.00 IJ = H 7.81 K
59.2 39.8 67.4
The y coordinate of the airplane is constant and equal to 7.60 10 3 m whereas the x coordinate is given by x = vi t where vi is the constant speed in the horizontal direction. is
At t = 30.0 s we have x = 8.0410 3 , so vi = 268 m s. The position vector as a function of time
P = 268 m s t i + 7.60 10 3 m j . At t = 45.0 s , P = 1. 21 10 4 i + 7.60 10 3 j m. The magnitude is
b
g e
j
P= and the direction is
c1.2110 h + c7.6010 h
4 2
3 2
m = 1.43 10 4 m
= arctan
F 7.6010 I = GH 1.2110 JK
3 4
32.2 above the horizontal .
Chapter 3
67
P3.41
(a) (b) (c)
A = 8.00 i + 12.0 j  4.00k B= A = 2.00 i + 3.00 j  1.00k 4
C = 3A = 24.0 i  36.0 j + 12.0k
P3.42
R = 75.0 cos 240 i + 75.0 sin 240 j + 125 cos 135 i + 125 sin 135 j + 100 cos 160 i + 100 sin 160 j R = 37.5 i  65.0 j  88.4i + 88.4 j  94.0 i + 34.2 j R = 220 i + 57.6 j R = (220 ) + 57.6 2 at arctan R = 227 paces at 165
2
FG 57.6 IJ above the xaxis H 220 K
P3.43
(a)
C=A+B=
2
e5.00 i  1.00 j  3.00kj m
2 2
C = (5.00) +(1.00) +(3.00) m = 5.92 m (b) D = 2A  B =
2
e4.00i  11.0 j + 15.0kj m
2 2
D = ( 4.00) +(11.0) +(15.0) m = 19.0 m P3.44 The position vector from radar station to ship is S = 17.3 sin 136 i + 17.3 cos 136 j km = 12.0 i  12.4 j km. From station to plane, the position vector is P = 19.6 sin 153 i + 19.6 cos 153 j + 2.20k km, or P = 8.90 i  17.5 j + 2.20k km. (a) To fly to the ship, the plane must undergo displacement D = S P = (b) The distance the plane must travel is D = D = (3.12) +(5.02) +( 2.20) km = 6.31 km .
2 2 2
e
j
e
j
e
j
e
j
e3.12 i + 5.02 j  2.20kj km .
68 P3.45
Vectors
The hurricane's first displacement is is
FG 25.0 km IJ(1.50 h) due North. With i H h K
FG 41.0 km IJ(3.00 h) at 60.0 N of W, and its second displacement H h K
representing east and j representing north, its total
displacement is:
FG 41.0 km cos 60.0IJ a3.00 hfe ij + FG 41.0 km sin 60.0IJ a3.00 hf j + FG 25.0 km IJ a1.50 hf j = 61.5 kme ij H h K H h K H hK
+144 km j with magnitude (61.5 km) +(144 km) = 157 km . P3.46 (a) E = 17.0 cm cos 27.0 i + 17.0 cm sin 27.0 j E= (b)
2 2
a
f
a
f
y
e15.1i + 7.72 jj cm
a a f e7.72i + 15.1 jj cm f
27.0 27.0 F G E 27.0 x
F =  17.0 cm sin 27.0 i + 17.0 cm cos 27.0 j F=
(c)
G = + 17.0 cm sin 27.0 i + 17.0 cm cos 27.0 j G=
a a f e+7.72 i + 15.1 jj cm
f
FIG. P3.46
P3.47
A x = 3.00 , A y = 2.00 (a) (b) A = A x i + A y j = 3.00 i + 2.00 j
2 2 A = A x + A y = (3.00) +( 2.00) = 3.61 2 2
tan =
Ay Ax
=
2.00 = 0.667 , tan1 (0.667)= 33.7 (3.00)
is in the 2 nd quadrant, so = 180+ 33.7 = 146 .
(c) R x = 0 , R y = 4.00 , R = A + B thus B = R  A and Bx = R x  A x = 0  (3.00)= 3.00 , By = R y  A y = 4.00  2.00 = 6.00 . Therefore, B = 3.00 i  6.00 j .
a
f
Chapter 3
69
P3.48
Let +x = East, +y = North, x 300 175 0 125 (a) (b) y 0 303 150 453 y = 74.6 N of E x
= tan1
R = x 2 + y 2 = 470 km R x = 40.0 cos 45.0+30.0 cos 45.0 = 49.5 R y = 40.0 sin 45.030.0 sin 45.0+20.0 = 27.1 R = 49.5 i + 27.1 j
y A B
P3.49
(a)
(b)
R=
a49.5f + a27.1f = 56.4 F 27.1 IJ = 28.7 = tan G H 49.5 K
2 2 1
O
45 45 C
x
FIG. P3.49 P3.50 Taking components along i and j , we get two equations:
6.00 a  8.00b + 26.0 = 0
and
8.00 a + 3.00b + 19.0 = 0 .
Solving simultaneously, a = 5.00 , b = 7.00 . Therefore, 5.00A + 7.00B + C = 0 .
70
Vectors
Additional Problems P3.51 Let represent the angle between the directions of A and B. Since A and B have the same magnitudes, A, B, and R = A + B form an isosceles triangle in which the angles are 180 , magnitude of R is then R = 2 A cos
/2
A
R
FG IJ . [Hint: apply the law of H 2K
, and . The 2 2
B
A D B FIG. P3.51
cosines to the isosceles triangle and use the fact that B = A .] Again, A, B, and D = A  B form an isosceles triangle with apex angle . Applying the law of cosines and the identity
a1  cos f = 2 sin FGH IJK 2
2
gives the magnitude of D as D = 2 A sin The problem requires that R = 100D . Thus, 2 A cos
FG IJ . H 2K
FG IJ = 200 A sinFG IJ . This gives tanFG IJ = 0.010 and H 2K H 2K H 2K
= 1.15 .
P3.52 Let represent the angle between the directions of A and B. Since A and B have the same magnitudes, A, B, and R = A + B form an isosceles triangle in which the angles are 180 , magnitude of R is then R = 2 A cos
FG IJ . [Hint: apply the law of H 2K
, and . The 2 2
cosines to the isosceles triangle and use the fact that B = A . ] Again, A, B, and D = A  B form an isosceles triangle with apex angle . Applying the law of cosines and the identity
FIG. P3.52
a1  cos f = 2 sin FGH IJK 2
2
gives the magnitude of D as D = 2 A sin
FG IJ . H 2K F I F I The problem requires that R = nD or cosG J = n sinG J giving H 2K H 2K F 1I = 2 tan G J . H nK
1
Chapter 3
71
P3.53
(a) (b)
R x = 2.00 , R y = 1.00 , R z = 3.00
2 2 2 R = R x + R y + R z = 4.00 + 1.00 + 9.00 = 14.0 = 3.74
(c)
cos x = cos y = cos z =
Rx R x = cos1 x = 57.7 from + x R R Ry R y = cos1
y
Rz z = cos1 R
F I GH JK F R I = 74.5 from + y GH R JK F R I = 36.7 from + z GH R JK
z
*P3.54
Take the xaxis along the tail section of the snake. The displacement from tail to head is 240 m i + 420  240 m cos 180105 i  180 m sin 75 j = 287 m i  174 m j . Its magnitude is (287) +(174) m = 335 m . From v =
2 2
a
f
a
f
distance , the time for each child's run is t
Inge: t = Olaf: t = Inge wins by 126  101 = 25.4 s . *P3.55
distance 335 m h 1 km 3 600 s = = 101 s v 12 km 1 000 m 1 h
a fa a fb
fb
ga f
g
420 m s = 126 s . 3.33 m
The position vector from the ground under the controller of the first airplane is r1 = 19.2 km cos 25 i + 19.2 km sin 25 j + 0.8 km k
a fa f a = e17.4i + 8.11 j + 0.8k j km .
fa
f a
f
The second is at r2 = 17.6 km cos 20 i + 17.6 km sin 20 j + 1.1 km k
a fa f a = e16.5 i + 6.02 j + 1.1k j km . e
fa
f a
f
Now the displacement from the first plane to the second is r2  r1 = 0.863 i  2.09 j + 0.3k km with magnitude (0.863 ) +( 2.09) +(0.3) = 2.29 km .
2 2 2
j
72 *P3.56
Vectors
Let A represent the distance from island 2 to island 3. The displacement is A = A at 159 . Represent the displacement from 3 to 1 as B = B at 298 . We have 4.76 km at 37 +A + B = 0 . For xcomponents
3 A 28 B 1 37 69 C 2 N E
a4.76 kmf cos 37+ A cos 159+B cos 298 = 0
3.80 km  0.934 A + 0.469B = 0 B = 8.10 km + 1.99 A For ycomponents,
FIG. P3.56
a4.76 kmf sin 37+ A sin 159+B sin 298 = 0
2.86 km + 0.358 A  0.883B = 0 (a) We solve by eliminating B by substitution: 2.86 km + 0.358 A  0.883 8.10 km + 1.99 A = 0 2.86 km + 0.358 A + 7.15 km  1.76 A = 0 10.0 km = 1.40 A A = 7.17 km (b) *P3.57 (a) B = 8.10 km + 1.99(7.17 km)= 6.15 km We first express the corner's position vectors as sets of components A = 10 m cos 50 i + 10 m sin 50 j = 6.43 m i +7.66 m j
a
f
a f a f B = a12 mf cos 30 i + a12 mf sin 30 j = 10.4 m i +6.00 mj .
10.4 m  6.43 m = 3.96 m .
The horizontal width of the rectangle is
Its vertical height is 7.66 m  6.00 m = 1.66 m . Its perimeter is 2(3.96 + 1.66) m = 11.2 m . (b) The position vector of the distant corner is Bx i + A y j = 10.4 mi +7.66 mj = 10.4 2 + 7.66 2 m at 7.66 m = 12.9 m at 36.4 . tan1 10.4 m
Chapter 3
73
P3.58
Choose the +xaxis in the direction of the first force. The total force, y in newtons, is then 31 N 12.0 i + 31.0 j  8.40 i  24.0 j = The magnitude of the total force is (3.60) +(7.00) N = 7.87 N and the angle it makes with our +xaxis is given by tan = (7.00) FIG. P3.58 ,
2 2
x R 12 N 35.0 horizontal 24 N
e3.60 ij + e7.00 jj N
.
8.4 N
(3.60) = 62.8 . Thus, its angle counterclockwise from the horizontal is 35.0+62.8 = 97.8 . d 1 = 100 i d 2 = 300 j
P3.59
a f a f d = 200 cosa60.0fi + 200 sina60.0f j = 100 i + 173 j R = d + d + d + d = e 130 i  202 jj m R = a 130f + a202f = 240 m F 202 IJ = 57.2 = tan G H 130 K
4 1 2 3 4 2 2 1
d 3 = 150 cos 30.0 i  150 sin 30.0 j = 130 i  75.0 j
FIG. P3.59
= 180 + = 237
dr d 4 i + 3 j  2 t j = = 0 + 0  2 j =  2.00 m s j dt dt
P3.60
e
j
b
g
The position vector at t = 0 is 4i + 3 j . At t = 1 s , the position is 4i + 1 j , and so on. The object is moving straight downward at 2 m/s, so dr represents its velocity vector . dt P3.61 v = v x i + v y j = 300 + 100 cos 30.0 i + 100 sin 30.0 j v = 387 i + 50.0 j mi h v = 390 mi h at 7.37 N of E
a
f a
f
e
j
74 P3.62
Vectors
(a)
You start at point A: r1 = rA = 30.0 i  20.0 j m. The displacement to B is rB  rA = 60.0 i + 80.0 j  30.0 i + 20.0 j = 30.0 i + 100 j . You cover half of this, 15.0 i + 50.0 j to move to r2 = 30.0 i  20.0 j + 15.0 i + 50.0 j = 45.0 i + 30.0 j . Now the displacement from your current position to C is rC  r2 = 10.0 i  10.0 j  45.0 i  30.0 j = 55.0 i  40.0 j . You cover onethird, moving to r3 = r2 + r23 = 45.0 i + 30.0 j + 1 55.0 i  40.0 j = 26.7 i + 16.7 j . 3
e
j
e
j
e
j
The displacement from where you are to D is rD  r3 = 40.0 i  30.0 j  26.7 i  16.7 j = 13.3 i  46.7 j . You traverse onequarter of it, moving to r4 = r3 + 1 1 rD  r3 = 26.7 i + 16.7 j + 13.3 i  46.7 j = 30.0 i + 5.00 j . 4 4
b
g
e
j
The displacement from your new location to E is rE  r4 = 70.0 i + 60.0 j  30.0 i  5.00 j = 100 i + 55.0 j of which you cover onefifth the distance, 20.0 i + 11.0 j, moving to r4 + r45 = 30.0 i + 5.00 j  20.0 i + 11.0 j = 10.0 i + 16.0 j . The treasure is at (10.0 m, 16.0 m) . (b) Following the directions brings you to the average position of the trees. The steps we took numerically in part (a) bring you to rA + r + rB 1 rB  rA = A 2 2
a
ar then to ar then to
A
+ rB
r + rB + rC + rD = A 3 4 4 arA +rB +rC +rD f r + r + r + r + r rA + rB + rC + rD r  B C D E 4 + E = A and last to . 4 5 5
D 3
2 + rB + rC A
f+r
C

ar
3
A +rB
2
f
=
f FGH
IJ K
f+r

ar
A + rB + rC
rA + rB + rC 3
f
a
f
This center of mass of the tree distribution is the same location whatever order we take the trees in.
Chapter 3
75
*P3.63
(a)
Let T represent the force exerted by each child. The xcomponent of the resultant force is
T cos 0 + T cos 120+T cos 240 = T 1 + T 0.5 + T 0.5 = 0 .
The ycomponent is T sin 0 + T sin 120 + T sin 240 = 0 + 0.866T  0.866T = 0 . Thus, FIG. P3.63
af a f a f
F = 0.
(b) If the total force is not zero, it must point in some direction. When each child moves one 360 space clockwise, the total must turn clockwise by that angle, . Since each child exerts N the same force, the new situation is identical to the old and the net force on the tire must still point in the original direction. The contradiction indicates that we were wrong in supposing that the total force is not zero. The total force must be zero. From the picture, R 1 = a i + b j and R 1 = a 2 + b 2 . R 2 = ai + b j + ck ; its magnitude is R1 + c 2 = a 2 + b 2 + c 2 .
2
P3.64
(a) (b)
FIG. P3.64
76 P3.65
Vectors
Since A + B = 6.00 j , we have
bA
giving
x
+ Bx i + A y + B y j = 0 i + 6.00 j FIG. P3.65 A x + B x = 0 or A x = Bx [1]
g e
j
and A y + B y = 6.00 . Since both vectors have a magnitude of 5.00, we also have
2 2 2 2 A x + A y = Bx + By = 5.00 2 .
[2]
From A x = Bx , it is seen that
2 2 A x = Bx . 2 2 2 2 Therefore, A x + A y = Bx + By gives 2 2 A y = By .
Then, A y = By and Eq. [2] gives A y = By = 3.00 . Defining as the angle between either A or B and the y axis, it is seen that cos = Ay A = By B = 3.00 = 0.600 and = 53.1 . 5.00
The angle between A and B is then = 2 = 106 .
Chapter 3
77
*P3.66
Let represent the angle the xaxis makes with the horizontal. Since angles are equal if their sides are perpendicular right side to right side and left side to left side, is also the angle between the weight and our y axis. The xcomponents of the forces must add to zero: 0.150 N sin + 0.127 N = 0 . (b) (a)
x y Ty 0.127 N
0.150 N
= 57.9
The ycomponents for the forces must add to zero: +Ty  0.150 N cos 57.9 = 0 , Ty = 0.079 8 N .
FIG. P3.66
a
f
(c) P3.67
The angle between the y axis and the horizontal is 90.057.9= 32.1 .
y y P
The displacement of point P is invariant under rotation of 2 the coordinates. Therefore, r = r and r 2 = r or, x 2 + y2 = x
b g + by g . Also, from the figure, =  F y I F yI tan G J = tan G J  H xK H x K y e j  tan = x 1 + e j tan
2 2 1 1 y x y x
bg
r
x
t
O
x
FIG. P3.67
Which we simplify by multiplying top and bottom by x cos . Then,
x = x cos + y sin , y = x sin + y cos .
ANSWERS TO EVEN PROBLEMS P3.2 (a) 2.17 m, 1.25 m ; 1.90 m, 3.29 m ; (b) 4.55 m (a) 8.60 m; (b) 4.47 m at 63.4 ; 4.24 m at 135 (a) r at 180 ; (b) 2r at 180+ ; (c) 3r at 14 km at 65 north of east (a) 6.1 at 112; (b) 14.8 at 22 9.5 N at 57 7.9 m at 4 north of west
a
fa
f
P3.16 P3.18 P3.20 P3.22 P3.24 P3.26 P3.28 P3.30
see the solution 86.6 m and 50.0 m 1.31 km north; 2.81 km east 25.0 m i + 43.3 m j 14.0 m s at 11.3 west of north 788 mi at 48.0 north of east 7.21 m at 56.3 C = 7.30 cm i  7.20 cm j
P3.4
P3.6 P3.8 P3.10 P3.12 P3.14
78 P3.32 P3.34 P3.36 P3.38 P3.40 P3.42 P3.44 P3.46
Vectors
(a) 4.47 m at 63.4; (b) 8.49 m at 135 42.7 yards 4.64 m at 78.6 (a) 10.4 cm; (b) 35.5 1.43 10 4 m at 32.2 above the horizontal 220 i + 57.6 j = 227 paces at 165 (a) 3.12 i + 5.02 j  2.20k km; (b) 6.31 km
P3.50 P3.52 P3.54 P3.56 P3.58
a = 5.00 , b = 7.00 2 tan 1 25.4 s (a) 7.17 km; (b) 6.15 km 7.87 N at 97.8 counterclockwise from a horizontal line to the right
FG 1 IJ H nK
e
j
P3.60 P3.62 P3.64
e j (b) e 7.72 i + 15.1 jj cm; (c) e +7.72 i + 15.1 jj cm
(a) 15.1i + 7.72 j cm; (a) 74.6 north of east; (b) 470 km
b2.00 m sg j ; its velocity vector (a) a10.0 m, 16.0 mf ; (b) see the solution
(a) R 1 = a i + b j ; R 1 = a 2 + b 2 ; (b) R 2 = ai + b j + ck ; R 2 = a 2 + b 2 + c 2
P3.48
P3.66
(a) 0.079 8N; (b) 57.9; (c) 32.1
4
Motion in Two Dimensions
CHAPTER OUTLINE
4.1 4.2 4.3 4.4 4.5 4.6 The Position, Velocity, and Acceleration Vectors TwoDimensional Motion with Constant Acceleration Projectile Motion Uniform Circular Motion Tangential and Radial Acceleration Relative Velocity and Relative Acceleration
ANSWERS TO QUESTIONS
Q4.1 Yes. An object moving in uniform circular motion moves at a constant speed, but changes its direction of motion. An object cannot accelerate if its velocity is constant. No, you cannot determine the instantaneous velocity. Yes, you can determine the average velocity. The points could be widely separated. In this case, you can only determine the average velocity, which is v= x . t
Q4.2
Q4.3
(a)
a
v a
v a v a v
(b)
v a
a
v a
v a v a v
Q4.4 Q4.5
(a)
10 i m s
(b)
9.80 j m s 2
The easiest way to approach this problem is to determine acceleration first, velocity second and finally position. Vertical: In free flight, a y =  g . At the top of a projectile's trajectory, v y = 0. Using this, the
2 maximum height can be found using v 2 = viy + 2 a y y f  yi . fy
d
i
Horizontal: a x = 0 , so v x is always the same. To find the horizontal position at maximum height, one needs the flight time, t. Using the vertical information found previously, the flight time can be found using v fy = viy + a y t . The horizontal position is x f = vix t . If air resistance is taken into account, then the acceleration in both the x and ydirections would have an additional term due to the drag. Q4.6 A parabola.
79
80 Q4.7
Motion in Two Dimensions
The balls will be closest together as the second ball is thrown. Yes, the first ball will always be moving faster, since its flight time is larger, and thus the vertical component of the velocity is larger. The time interval will be one second. No, since the vertical component of the motion determines the flight time. The ball will have the greater speed. Both the rock and the ball will have the same vertical component of the velocity, but the ball will have the additional horizontal component. (a) yes (b) no (c) no (d) yes (e) no
Q4.8 Q4.9 Q4.10 Q4.11 Q4.12 Q4.13 Q4.14
Straight up. Throwing the ball any other direction than straight up will give a nonzero speed at the top of the trajectory. No. The projectile with the larger vertical component of the initial velocity will be in the air longer. The projectile is in free fall. Its vertical component of acceleration is the downward acceleration of gravity. Its horizontal component of acceleration is zero. (a) no (b) yes (c) yes (d) no
60. The projection angle appears in the expression for horizontal range in the function sin 2 . This function is the same for 30 and 60. The optimal angle would be less than 45. The longer the projectile is in the air, the more that air resistance will change the components of the velocity. Since the vertical component of the motion determines the flight time, an angle less than 45 would increase range. The projectile on the moon would have both the larger range and the greater altitude. Apollo astronauts performed the experiment with golf balls. Gravity only changes the vertical component of motion. Since both the coin and the ball are falling from the same height with the same vertical component of the initial velocity, they must hit the floor at the same time. (a) no (b) yes
Q4.15
Q4.16 Q4.17
Q4.18
In the second case, the particle is continuously changing the direction of its velocity vector. Q4.19 Q4.20 Q4.21 The racing car rounds the turn at a constant speed of 90 miles per hour. The acceleration cannot be zero because the pendulum does not remain at rest at the end of the arc. (a) (b) Q4.22 (a) The velocity is not constant because the object is constantly changing the direction of its motion. The acceleration is not constant because the acceleration always points towards the center of the circle. The magnitude of the acceleration is constant, but not the direction. straight ahead (b) in a circle or straight ahead
Chapter 4
81
Q4.23
v v a a v a a v a v
v
a a a v
Q4.24
v
a
r
r a aa v v
r
r
r v a
Q4.25
The unit vectors r and are in different directions at different points in the xy plane. At a location along the xaxis, for example, r = i and = j, but at a point on the yaxis, r = j and =  i . The unit vector i is equal everywhere, and j is also uniform.
Q4.26
The wrench will hit at the base of the mast. If air resistance is a factor, it will hit slightly leeward of the base of the mast, displaced in the direction in which air is moving relative to the deck. If the boat is scudding before the wind, for example, the wrench's impact point can be in front of the mast. (a) (b) The ball would move straight up and down as observed by the passenger. The ball would move in a parabolic trajectory as seen by the ground observer. Both the passenger and the ground observer would see the ball move in a parabolic trajectory, although the two observed paths would not be the same. g downward (b) g downward
Q4.27
Q4.28
(a)
The horizontal component of the motion does not affect the vertical acceleration.
SOLUTIONS TO PROBLEMS
Section 4.1 P4.1 The Position, Velocity, and Acceleration Vectors xm 0 3 000 1 270 4 270 m (a)
a f
ym 3 600 0
a f
1 270 2 330 m
Net displacement = x 2 + y 2 = 4.87 km at 28.6 S of W FIG. P4.1
(b)
Average speed =
b20.0 m sga180 sf + b25.0 m sga120 sf + b30.0 m sga60.0 sf =
180 s + 120 s + 60.0 s 4.87 10 3 m = 13.5 m s along R 360 s
23.3 m s
(c)
Average velocity =
82 P4.2
Motion in Two Dimensions
(a) (b) (c) (d) (e) (f)
r = 18.0t i + 4.00t  4.90t 2 j v= a=
e
j
b18.0 m sgi + 4.00 m s  e9.80 m s jt j
2 2
e9.80 m s j j ra3.00 sf = a54.0 mf i  a32.1 mf j
v 3.00 s = 18.0 m s i  25.4 m s j
a 3.00 s =
a
f b
g b
2
g
a
f e9.80 m s j j
5.00 m s cos 60.0 = 2.50 m s .
*P4.3
The sun projects onto the ground the xcomponent of her velocity:
a
f
P4.4
(a)
From x = 5.00 sin t , the xcomponent of velocity is vx = dx d = 5.00 sin t = 5.00 cos t dt dt dv x = +5.00 2 sin t dt
FG IJ b H K
g
and a x =
similarly, v y = and a y =
FG d IJ b4.00  5.00 cos tg = 0 + 5.00 sin t H dt K
2
FG d IJ b5.00 sin tg = 5.00 H dt K
cos t .
At t = 0 , v = 5.00 cos 0 i + 5.00 sin 0 j = and a = 5.00 2 sin 0 i + 5.00 2 cos 0 j = (b) r = xi + yj = v= a= (c)
e5.00 i + 0 jj m s e0 i + 5.00 jj m s .
2 2
a4.00 mf j + a5.00 mfe sin t i  cos t jj a5.00 mf  cos t i + sin t j a5.00 mf sin t i + cos t j
2
The object moves in a circle of radius 5.00 m centered at 0 , 4.00 m .
a
f
Chapter 4
83
Section 4.2 P4.5 (a)
TwoDimensional Motion with Constant Acceleration v f = vi + a t a= v f  vi t =
e9.00i + 7.00 jj  e3.00i  2.00 jj =
3.00
e2.00i + 3.00 jj m s j
2
(b)
r f = ri + v i t +
1 2 1 a t = 3.00 i  2.00 j t + 2.00 i + 3.00 j t 2 2 2
e
j
e
x = 3.00t + t 2 m and y = 1.50t 2  2.00t m P4.6 (a)
e
j
e
j
FG IJ e j H K dv F d I a= = G J e 12.0t jj = 12.0 j m s dt H dt K
v= r = 3.00 i  6.00 j m; v = 12.0 j m s
dr d 3.00 i  6.00t 2 j = 12.0t j m s = dt dt
2
(b) P4.7
e
j
v i = 4.00 i + 1.00 j m s and v 20.0 = 20.0 i  5.00 j m s (a) v x 20.0  4.00 = m s 2 = 0.800 m s 2 t 20.0 v y 5.00  1.00 ay = = m s 2 = 0.300 m s 2 t 20.0 ax =
e
j
a f e
j
(b) (c)
= tan 1
FG 0.300 IJ = 20.6 = H 0.800 K
x f = xi + v xi t +
339 from + x axis
At t = 25.0 s 1 1 2 a x t 2 = 10.0 + 4.00 25.0 + 0.800 25.0 = 360 m 2 2 1 1 2 y f = yi + v yi t + a y t 2 = 4.00 + 1.00 25.0 + 0.300 25.0 = 72.7 m 2 2 v xf = v xi + a x t = 4 + 0.8 25 = 24 m s
yf yi y
a f a fa f a f a fa f
a f v = v + a t = 1  0.3a 25f = 6.5 m s Fv I F 6.50 IJ = 15.2 = tan G J = tan G H 24.0 K Hv K
1 y x 1
84 P4.8
Motion in Two Dimensions
a = 3.00 j m s 2 ; v i = 5.00 i m s ; ri = 0 i + 0 j (a) r f = ri + v i t + 1 2 at = 2
v f = vi + a t = (b)
LM5.00ti + 1 3.00t jOP m 2 N Q e5.00i + 3.00tjj m s
2
1 3.00 2.00 2 so x f = 10.0 m , y f = 6.00 m t = 2.00 s , r f = 5.00 2.00 i +
a f
a fa f j = e10.0 i + 6.00 jj m
2
v f = 5.00 i + 3.00 2.00 j = 5.00 i + 6.00 j m s
2 v f = v f = v xf 2 yf 2 2
a f e j + v = a5.00f + a6.00f j e
= 7.81 m s 1 axt 2 . 2
*P4.9
(a)
For the xcomponent of the motion we have x f = xi + v xi t + 0.01 m = 0 + 1.80 10 7 m s t +
e
e4 10
t= =
14
m s 2 t 2 + 1.80 10 7 m s t  10 2 m = 0
j e
1 8 10 14 m s 2 t 2 2
j
j
1.80 10 7 m s
e1.8 10 m sj  4e4 10 2e 4 10 m s j
7 2 14 2
14
m s 2 10 2 m
je
j
1.8 10 7 1.84 10 7 m s 8 10 14 m s 2
We choose the + sign to represent the physical situation t= Here y f = yi + v yi t + 4.39 10 5 m s 8 10
14
ms
2
= 5.49 10 10 s .
1 1 a y t 2 = 0 + 0 + 1.6 10 15 m s 2 5.49 10 10 s 2 2
e
je
j
2
= 2.41 10 4 m .
So, r f = 10.0 i + 0.241 j mm . (b) v f = v i + at = 1.80 10 7 m s i + 8 10 14 m s 2 i + 1.6 10 15 m s 2 j 5.49 10 10 s = 1.80 10 7 m s i + 4.39 10 5 m s i + 8.78 10 5 m s j = (c) (d) vf =
e
j
e
je
j
e
e1.84 10
7
j e j e m sj i + e8.78 10 m sj j
5 2 5 2 5
j
e1.84 10 m sj + e8.78 10 m sj = F 8.78 10 I = 2.73 Fv I = tan G J = tan G v K H H 1.84 10 JK
7 1 y x 1 7
1.85 10 7 m s
Chapter 4
85
Section 4.3 P4.10
Projectile Motion
x = v xi t = vi cos i t
x = 300 m s cos 55.0 42.0 s x = 7. 23 10 m y = v yi t 
3
b b
ga ga
fa
f
1 2 1 gt = vi sin i t  gt 2 2 2
y = 300 m s sin 55.0 42.0 s  P4.11 (a)
fa
f 1 e9.80 m s ja42.0 sf 2
2
2
= 1.68 10 3 m
The mug leaves the counter horizontally with a velocity v xi (say). If time t elapses before it hits the ground, then since there is no horizontal acceleration, x f = v xi t , i.e., t= xf v xi =
a1.40 mf
v xi FIG. P4.11
In the same time it falls a distance of 0.860 m with acceleration downward of 9.80 m s 2 . Then y f = yi + v yi t + Thus, 1 1 a y t 2 : 0 = 0.860 m + 9.80 m s 2 2 2
e
jFGH 1.40 m IJK v
xi
2
.
v xi = (b)
e4.90 m s je1.96 m j =
2 2
0.860 m
3.34 m s .
The vertical velocity component with which it hits the floor is v yf = v yi + a y t = 0 + 9.80 m s 2
e
1.40 m jFGH 3.34 m s IJK = 4.11 m s .
Hence, the angle at which the mug strikes the floor is given by
= tan 1
F v I = tan FG 4.11 IJ = GH v JK H 3.34 K
yf xf 1
50.9 .
86 P4.12
Motion in Two Dimensions
The mug is a projectile from just after leaving the counter until just before it reaches the floor. Taking the origin at the point where the mug leaves the bar, the coordinates of the mug at any time are x f = v xi t + 1 1 1 a x t 2 = v xi t + 0 and y f = v yi t + a y t 2 = 0  g t 2 . 2 2 2
When the mug reaches the floor, y f =  h so h =  which gives the time of impact as t= 2h . g 2h giving the g 1 2 gt 2
(a)
Since x f = d when the mug reaches the floor, x f = v xi t becomes d = v xi initial velocity as v xi = d g . 2h
(b)
Just before impact, the xcomponent of velocity is still v xf = v xi while the ycomponent is v yf = v yi + a y t = 0  g 2h . g
Then the direction of motion just before impact is below the horizontal at an angle of
= tan 1
Fv GG v H
yf xf
I JJ = tan K
1
Fg GG Hd
2h g g 2h
I JJ = K
tan 1
FG 2 h IJ HdK
.
Chapter 4
87
P4.13
(a)
The time of flight of the first snowball is the nonzero root of y f = yi + v yi t1 + 0 = 0 + 25.0 m s sin 70.0 t1  t1 = 2( 25.0 m s) sin 70.0 9.80 m s 2
1 2 a y t1 2
b
gb
g
1 2 9.80 m s 2 t1 2
e
j
= 4.79 s .
The distance to your target is x f  xi = v xi t1 = 25.0 m s cos 70.0 4.79 s = 41.0 m . Now the second snowball we describe by y f = yi + v yi t 2 + 1 ayt2 2 2
b
g
a
f
2 0 = 25.0 m s sin 2 t 2  4.90 m s 2 t 2
b
g
e
j
t 2 = 5.10 s sin 2 x f  xi = v xi t 2
41.0 m = 25.0 m s cos 2 5.10 s sin 2 = 128 m sin 2 cos 2 0.321 = sin 2 cos 2 Using sin 2 = 2 sin cos we can solve 0.321 = 1 sin 2 2 2
a
f
b
g
a
f
a
f
2 2 = sin 1 0.643 and 2 = 20.0 . (b) The second snowball is in the air for time t 2 = 5.10 s sin 2 = 5.10 s sin 20 = 1.75 s , so you throw it after the first by t1  t 2 = 4.79 s  1.75 s = 3.05 s . P4.14 From Equation 4.14 with R = 15.0 m , vi = 3.00 m s , max = 45.0 g = vi2 9.00 = = 0.600 m s 2 R 15.0
a
f
a
f
88 P4.15
Motion in Two Dimensions
h= so or
vi2 sin 2 i vi2 sin 2 i ; R= ; 3h = R , g 2g
2 3 vi2 sin 2 i vi sin 2 i = g 2g
b
g
b
g
2 sin 2 i tan i = = 3 sin 2 i 2 4 = 53.1 . thus i = tan 1 3
FG IJ HK
*P4.16
(a)
To identify the maximum height we let i be the launch point and f be the highest point:
2 2 v yf = v yi + 2 a y y f  yi
d
0= y max =
vi2 vi2
sin i
2
2
i + 2b g gb y
max
0
g
sin i . 2g
To identify the range we let i be the launch and f be the impact point; where t is not zero: y f = yi + v yi t + 1 ay t 2 2 1 0 = 0 + vi sin i t +  g t 2 2 2 vi sin i t= g
b g
x f = xi + v xi t +
1 axt 2 2 2 v sin i + 0. d = 0 + vi cos i i g
For this rock, d = y max vi2 sin 2 i 2 vi2 sin i cos i = g 2g sin i = tan i = 4 cos i
i = 76.0
(b) (c) Since g divides out, the answer is the same on every planet. The maximum range is attained for i = 45 : d max vi cos 45 2 vi sin 45 g = = 2.125 . d gvi cos 76 2 vi sin 76 So d max = 17d . 8
Chapter 4
89
P4.17
(a) (b)
x f = v xi t = 8.00 cos 20.0 3.00 = 22.6 m Taking y positive downwards, y f = v yi t + 1 2 gt 2
a f
y f = 8.00 sin 20.0 3.00 + (c) 10.0 = 8.00 sin 20.0 t +
a f 1 a9.80fa3.00f 2
2
= 52.3 m .
a
f
1 9.80 t 2 2
a f
4.90t 2 + 2.74t  10.0 = 0 t= *P4.18 2.74
a2.74f
9.80
2
+ 196
= 1.18 s
We interpret the problem to mean that the displacement from fish to bug is 2.00 m at 30 = 2.00 m cos30 i + 2.00 m sin30 j = 1.73 m i + 1.00 m j. If the water should drop 0.03 m during its flight, then the fish must aim at a point 0.03 m above the bug. The initial velocity of the water then is directed through the point with displacement
a
f
a
f
a
f a
f
a1.73 mfi + a1.03 mf j = 2.015 m at 30.7.
For the time of flight of a water drop we have x f = xi + v xi t + 1.73 m = 0 + vi cos 30.7 t + 0 so t= The vertical motion is described by y f = yi + v yi t + The "drop on its path" is 1 3.00 cm = 9.80 m s 2 2 1 ayt 2 . 2 1.73 m . vi cos 30.7 1 axt 2 2
b
g
e
.73 m jFGH v 1cos 30.7 IJK
i
2
.
Thus, vi = 1.73 m 9.80 m s 2 = 2.015 m 12.8 s 1 = 25.8 m s . cos30.7 2 0.03 m
e
j
90 P4.19
Motion in Two Dimensions
(a)
We use the trajectory equation: y f = x f tan i  With x f = 36.0 m, vi = 20.0 m s, and = 53.0 we find yf gx 2 f 2 vi2 cos 2 i .
e9.80 m s ja36.0 mf = 3.94 m. = a36.0 mf tan 53.0 2b 20.0 m sg cos a53.0f
2 2 2 2
The ball clears the bar by
a3.94  3.05f m =
(b) t1 =
0.889 m .
The time the ball takes to reach the maximum height is 20.0 m s sin53.0 vi sin i = = 1.63 s . g 9.80 m s 2 xf vix
b
ga
f
The time to travel 36.0 m horizontally is t 2 =
t2 =
36.0 m = 2.99 s . ( 20.0 m s) cos 53.0
a
f
Since t 2 > t1 the ball clears the goal on its way down . P4.20 The horizontal component of displacement is x f = v xi t = vi cos i t . Therefore, the time required to d reach the building a distance d away is t = . At this time, the altitude of the water is vi cos i y f = v yi t + g d d 1 a y t 2 = vi sin i  vi cos i 2 2 vi cos i
b
g
FG H
IJ FG K H
IJ K
2
.
Therefore the water strikes the building at a height h above ground level of h = y f = d tan i  gd 2 2 vi2 cos 2 i .
Chapter 4
91
*P4.21
(a)
For the horizontal motion, we have 1 axt 2 2 24 m = 0 + vi cos 53 2.2 s + 0 x f = xi + v xi t +
a
fa
f
vi = 18.1 m s . (b) As it passes over the wall, the ball is above the street by y f = yi + v yi t + y f = 0 + 18.1 m s sin 53 2.2 s + 1 ayt 2 2
b
ga
fa
f 1 e9.8 m s ja2.2 sf 2
2
2
= 8.13 m .
So it clears the parapet by 8.13 m  7 m = 1.13 m . (c) Note that the highest point of the ball's trajectory is not directly above the wall. For the whole flight, we have from the trajectory equation y f = tan i x f  or 6 m = tan 53 x f  Solving,
b
g FGH 2v
g
2 i
cos
2
Ix J K
i 2
2 f
a
F .8 f GG 2 18.19m sm scos g Hb
2 1 2 f
2
I Jx . 53 J K
2 f
e0.041 2 m jx
and xf = This yields two results:
 1.33 x f + 6 m = 0
1.33 1.33 2  4 0.0412 6 2 0.0412 m
e
a
1
j
fa f .
x f = 26.8 m or 5.44 m The ball passes twice through the level of the roof. It hits the roof at distance from the wall 26.8 m  24 m = 2.79 m .
92 *P4.22
Motion in Two Dimensions
When the bomb has fallen a vertical distance 2.15 km, it has traveled a horizontal distance x f given by xf =
a3.25 kmf  a2.15 kmf
2
2
= 2.437 km
y f = x f tan 
gx 2 f 2 vi2 cos 2 i
i 2 2
e9.8 m s jb2 437 mg 2 150 m = b 2 437 mg tan  2b 280 m sg cos 2 150 m = b 2 437 mg tan  a371.19 mfe1 + tan j
2 2 i i 2 i
tan  6.565 tan i  4.792 = 0 tan i = 1 6.565 2
2
F H
a6.565f  4a1fa4.792f IK = 3.283 3.945 .
2
Select the negative solution, since i is below the horizontal. tan i = 0.662 , i = 33.5 P4.23 The horizontal kick gives zero vertical velocity to the rock. Then its time of flight follows from y f = yi + v yi t + 40.0 m = 0 + 0 + 1 ayt 2 2
1 9.80 m s 2 t 2 2 t = 2.86 s .
e
j
The extra time 3.00 s  2.86 s = 0.143 s is the time required for the sound she hears to travel straight back to the player. It covers distance
b343 m sg0.143 s = 49.0 m =
x = 28.3 m = v xi t + 0t 2 v xi =
x 2 + 40.0 m
a
f
2
where x represents the horizontal distance the rock travels.
28.3 m = 9.91 m s 2.86 s
Chapter 4
93
P4.24
From the instant he leaves the floor until just before he lands, the basketball star is a projectile. His 2 2 vertical velocity and vertical displacement are related by the equation v yf = v yi + 2 a y y f  yi . Applying this to the upward part of his flight gives 0 =
2 v yi 2
d i + 2e 9.80 m s ja1.85  1.02f m . From this, e ja f
v yi = 4.03 m s . [Note that this is the answer to part (c) of this problem.] Thus the vertical velocity just before he lands is v yf = 4.32 m s. (a) His hang time may then be found from v yf = v yi + a y t :
2 For the downward part of the flight, the equation gives v yf = 0 + 2 9.80 m s 2 0.900  1.85 m .
4.32 m s = 4.03 m s + 9.80 m s 2 t or t = 0.852 s . (b) Looking at the total horizontal displacement during the leap, x = v xi t becomes
e
j
2.80 m = v xi 0.852 s
which yields v xi = 3.29 m s . (c) v yi = 4.03 m s . See above for proof. The takeoff angle is: = tan 1
a
f
(d) (e)
F v I = tan F 4.03 m s I = GH v JK GH 3.29 m s JK
yi xi 1
50.8 .
Similarly for the deer, the upward part of the flight gives 2 2 v yf = v yi + 2 a y y f  yi :
d
i
2 0 = v yi + 2 9.80 m s 2 2.50  1.20 m
e
ja
f
so v yi = 5.04 m s .
2 2 2 For the downward part, v yf = v yi + 2 a y y f  yi yields v yf = 0 + 2 9.80 m s 2 0.700  2.50 m
d
i
e
ja
f
and v yf = 5.94 m s. The hang time is then found as v yf = v yi + a y t : 5.94 m s = 5.04 m s + 9.80 m s 2 t and t = 1.12 s .
e
j
94 *P4.25
Motion in Two Dimensions
The arrow's flight time to the collision point is t= x f  xi v xi =
b
150 m = 5.19 s . 45 m s cos 50
g
The arrow's altitude at the collision is y f = yi + v yi t + 1 ayt 2 2
= 0 + 45 m s sin 50 5.19 s + (a)
b
ga
f
1 9.8 m s 2 5.19 s 2
e
ja
f
2
= 47.0 m .
The required launch speed for the apple is given by
2 2 v yf = v yi + 2 a y y f  yi
d
i
2 0 = v yi + 2 9.8 m s 2 47 m  0
e
ja
f
v yi = 30.3 m s . (b) The time of flight of the apple is given by v yf = v yi + a y t 0 = 30.3 m s  9.8 m s 2 t t = 3.10 s . So the apple should be launched after the arrow by 5.19 s  3.10 s = 2.09 s . *P4.26 For the smallest impact angle
= tan 1
Fv I, GH v JK
yf xf
we want to minimize v yf and maximize v xf = v xi . The final ycomponent
2 2 of velocity is related to v yi by v yf = v yi + 2 gh , so we want to minimize v yi
and maximize v xi . Both are accomplished by making the initial velocity horizontal. Then v xi = v , v yi = 0 , and v yf = 2 gh . At last, the impact angle is
FIG. P4.26
= tan 1
Fv I = GH v JK
yf xf
tan 1
F GH
2 gh v
I JK
.
Chapter 4
95
Section 4.4 P4.27
Uniform Circular Motion
20.0 m s v2 ac = = = 377 m s 2 r 1.06 m The mass is unnecessary information. v2 , T = 24 h 3 600 s h = 86 400 s R 2 R 2 ( 6.37 10 6 m) v= = = 463 m s T 86 400 s a=
b
g
2
P4.28
b
g
b463 m sg a=
6
2
6.37 10 m
= 0.033 7 m s 2 directed toward the center of Earth .
P4.29
r = 0.500 m; vt = a= 2 r 2 0.500 m = = 10. 47 m s = 10.5 m s 60 .0 s T 200 rev 10. 47 v2 = R 0.5 v2 r
a
f
a
f
2
= 219 m s 2 inward
P4.30
ac =
v = a c r = 3 9.8 m s 2 9.45 m = 16.7 m s Each revolution carries the astronaut over a distance of 2 r = 2 9.45 m = 59.4 m. Then the rotation rate is 16.7 m s P4.31 (a)
e
ja
f
a
f
FG 1 rev IJ = H 59.4 m K gb gb
0.281 rev s .
At 8.00 rev s , v = 0.600 m 8.00 rev s 2 rad rev = 30.2 m s = 9.60 m s . At 6.00 rev s , v = 0.900 m 6.00 rev s 2 rad rev = 33.9 m s = 10.8 m s . 6.00 rev s gives the larger linear speed. 9.60 m s v2 = Acceleration = r 0.600 m At 6.00 rev s , acceleration =
v = r
a a
fb fb
g g
(b)
b
g
2
= 1.52 10 3 m s 2 .
2
(c)
b10.8 m sg
0.900 m
= 1.28 10 3 m s 2 .
96 P4.32
Motion in Two Dimensions
The satellite is in free fall. Its acceleration is due to gravity and is by effect a centripetal acceleration. ac = g so v2 = g. r Solving for the velocity, v = rg =
a6 ,400 + 600fe10 mje8.21 m s j =
3 2
7.58 10 3 m s
v= and T=
2r T
3 2 r 2 7 ,000 10 m = = 5.80 10 3 s v 7.58 10 3 m s
e
j
T = 5.80 10 3 s
FG 1 min IJ = 96.7 min . H 60 s K
Section 4.5 P4.33
Tangential and Radial Acceleration
We assume the train is still slowing down at the instant in question. ac = at = v2 = 1.29 m s 2 r
40.0 km h 10 3 m km v = 15.0 s t
2 2
b
ge
je
1h 3 600 s
j = 0.741 m s
2 2
2
2 a = a c + a t2 =
at an angle of tan 1
e1.29 m s j + e0.741 m s j F a I = tan FG 0.741 IJ GH a JK H 1.29 K
t 1 c
FIG. P4.33
a = 1.48 m s 2 inward and 29.9 o backward
P4.34 (a)
at = 0.600 m s 2
4.00 m s v2 ar = = r 20.0 m
(b) (c)
b
g
2
= 0.800 m s 2
2 a = a t2 + a r = 1.00 m s 2
= tan 1
ar = 53.1 inward from path at
Chapter 4
97
P4.35
r = 2.50 m , a = 15.0 m s 2 (a) (b)
a c = a cos 30.0 o = 15.0 m s 2 cos 30 = 13.0 m s 2
v2 r so v 2 = ra c = 2.50 m 13.0 m s 2 = 32.5 m 2 s 2 ac =
e
ja
f
e
j
v = 32.5 m s = 5.70 m s
(c)
2 a 2 = a t2 + a r 2 so a t = a 2  a r =
FIG. P4.35
e15.0 m s j  e13.0 m s j =
2 2 2
7.50 m s 2
P4.36
(a) (b)
See figure to the right. The components of the 20.2 and the 22.5 m s 2 along the rope together constitute the centripetal acceleration:
a c = 22.5 m s 2 cos 90.036.9 + 20.2 m s 2 cos 36.9 = 29.7 m s 2
(c) v2 so v = a c r = 29.7 m s 2 1.50 m = 6.67 m s tangent to circle r v = 6.67 m s at 36.9 above the horizontal ac =
e
j a
f e
a
j
f
FIG. P4.36
*P4.37
Let i be the starting point and f be one revolution later. The curvilinear motion with constant tangential acceleration is described by x = v xi t + 2 r = 0 + at = t 1 axt 2 2 a
at
ar
1 2 at t 2 4 r
2
FIG. P4.37
and v xf = v xi + a x t , v f = 0 + a t t = Then tan = 4 r t 2 at 1 = 2 = a r t 16 2 r 4
v 2 16 2 r 2 4 r f = . The magnitude of the radial acceleration is a r = . r t t 2r
= 4.55 .
98
Motion in Two Dimensions
Section 4.6 P4.38 (a)
Relative Velocity and Relative Acceleration v H = 0 + a H t = 3.00 i  2.00 j m s 2 5.00 s v H = 15.0 i  10.0 j m s v J = 0 + a j t = 1.00 i + 3.00 j m s 2 5.00 s v J = 5.00 i +15.0 j m s v HJ = v H  v J = 15.0 i  10.0 j  5.00 i  15.0 j m s v HJ = 10.0 i  25.0 j m s v HJ = (10.0) 2 + ( 25.0) 2 m s = 26.9 m s
e
j
a
f
e e e
j
e
j
a
f j
j j
e
(b)
rH = 0 + 0 +
rH = 37.5 i  25.0 j m rJ = rHJ rHJ rHJ (c)
e
1 1 a H t 2 = 3.00 i  2.00 j m s 2 5.00 s 2 2
e
j
a
f
2
j
f = e12.5 i + 37.5 jj m = r  r = e37.5 i  25.0 j  12.5 i  37.5 jj m = e 25.0 i  62.5 jj m = a 25.0 f + a62.5 f m = 67.3 m
1 1.00 i + 3.00 j m s 2 5.00s 2
e
j
a
2
H
J
2
2
a HJ = a H  a J = 3.00 i  2.00 j  1.00 i  3.00 j m s 2 a HJ =
e
j
e2.00i  5.00 jj m s
2
*P4.39
v ce = the velocity of the car relative to the earth. v wc = the velocity of the water relative to the car. v we = the velocity of the water relative to the earth. These velocities are related as shown in the diagram at the right. (a) Since v we is vertical, v wc sin 60.0 = v ce = 50.0 km h or v wc = 57.7 km h at 60.0 west of vertical . Since v ce has zero vertical component,
vwe 60
vce
vwc
v we = v ce + v wc FIG. P4.39
(b)
v we = v wc cos 60.0 = 57.7 km h cos 60.0 = 28.9 km h downward .
b
g
Chapter 4
99
P4.40
The bumpers are initially 100 m = 0.100 km apart. After time t the bumper of the leading car travels 40.0 t, while the bumper of the chasing car travels 60.0t. Since the cars are side by side at time t, we have 0.100 + 40.0t = 60.0t , yielding t = 5.00 10 3 h = 18.0 s .
P4.41
Total time in still water t =
d 2 000 = = 1.67 10 3 s . v 1. 20
Total time = time upstream plus time downstream: t up = t down 1 000 = 1.43 10 3 s (1.20  0.500 ) 1 000 = = 588 s . 1.20 + 0.500
Therefore, ttotal = 1.43 10 3 + 588 = 2.02 10 3 s . P4.42 v = 150 2 + 30.0 2 = 153 km h
= tan 1
P4.43
FG 30.0 IJ = H 150 K
11.3 north of west
For Alan, his speed downstream is c + v, while his speed upstream is c  v . Therefore, the total time for Alan is t1 =
2L L L c + = 2 c+v cv 1  v2 c
.
For Beth, her crossstream speed (both ways) is c2  v2 . Thus, the total time for Beth is t 2 = 2L c2  v2 =
2L c
1
v2 c2
.
Since 1 
v2 < 1 , t1 > t 2 , or Beth, who swims crossstream, returns first. c2
100 P4.44
Motion in Two Dimensions
(a)
To an observer at rest in the train car, the bolt accelerates downward and toward the rear of the train. a= tan =
b2.50 m sg + b9.80 m sg
2
2
= 10.1 m s 2
2.50 m s 2 9.80 m s 2
= 0.255
= 14.3 to the south from the vertical
(b) P4.45
a = 9.80 m s 2 vertically downward
Identify the student as the S' observer and the professor as the S observer. For the initial motion in S', we have v y v x = tan 60.0 = 3 .
Let u represent the speed of S' relative to S. Then because there is no xmotion in S, we can write v x = v + u = 0 so x that v x = u = 10.0 m s . Hence the ball is thrown backwards in S'. Then,
v y = v = 3 v = 10.0 3 m s . y x
2 Using v y = 2 gh we find
e10.0 3 m sj = h= 2e9.80 m s j
2 2
FIG. P4.45 15.3 m .
The motion of the ball as seen by the student in S' is shown in diagram (b). The view of the professor in S is shown in diagram (c). *P4.46 Choose the xaxis along the 20km distance. The ycomponents of the displacements of the ship and the speedboat must agree:
N 40 15 x 25
b26 km hgt sina4015f = b50 km hgt sin
11.0 = 12.7 . = sin 1 50 The speedboat should head 15+12.7 = 27.7 east of north .
E y
FIG. P4.46
Chapter 4
101
Additional Problems *P4.47 (a) (b) The speed at the top is v x = vi cos i = 143 m s cos 45 = 101 m s . In free fall the plane reaches altitude given by
2 2 v yf = v yi + 2 a y y f  yi
b
g
i 0 = b143 m s sin 45g + 2e 9.8 m s jd y  31 000 ft i F 3.28 ft IJ = 3.27 10 ft . y = 31 000 ft + 522 mG H 1m K
2 2 f f 3
d
(c)
For the whole free fall motion v yf = v yi + a y t 101 m s = +101 m s  9.8 m s 2 t t = 20.6 s
e
j
(d)
ac =
v2 r
v = a c r = 0.8 9.8 m s 2 4,130 m = 180 m s P4.48 At any time t, the two drops have identical ycoordinates. The distance between the two drops is then just twice the magnitude of the horizontal displacement either drop has undergone. Therefore, d = 2 x t = 2 v xi t = 2 vi cos i t = 2 vi t cos i . P4.49 After the string breaks the ball is a projectile, and reaches the ground at time t: y f = v yi t + 1.20 m = 0 + so t = 0.495 s. Its constant horizontal speed is v x = x 2.00 m = = 4.04 m s t 0.495 s 1 9.80 m s 2 t 2 2 1 ayt 2 2
e
j
af b g b
g
e
j
4.04 m s v2 so before the string breaks a c = x = r 0.300 m
b
g
2
= 54.4 m s 2 .
102 P4.50
Motion in Two Dimensions
(a)
y f = tan i x f 
b
gd i b
g 2 vi2 cos 2 i
x2 f
Setting x f = d cos , and y f = d sin , we have d sin = tan i d cos 
gb
g
g 2 vi2 cos i
2
bd cos g .
2
FIG. P4.50
Solving for d yields, d =
2 vi2 cos i sin i cos  sin cos i g cos 2
or d =
2 vi2 cos i sin i  g cos
2
b
g
.
(b) P4.51
Setting
v 2 1  sin dd = 0 leads to i = 45+ and d max = i d i 2 g cos 2
b
g
.
Refer to the sketch: (b) x = v xi t ; substitution yields 130 = vi cos 35.0 t . 1 y = v yi t + at 2 ; substitution yields 2 20.0 = vi sin 35.0 t +
b
g
b
g
1 9.80 t 2 . 2 FIG. P4.51
a
f
Solving the above gives t = 3.81 s . (a) (c) vi = 41.7 m s v yf = vi sin i  gt , v x = vi cos i At t = 3.81 s , v yf = 41.7 sin 35.0 9.80 3.81 = 13.4 m s v x = 41.7 cos 35.0 = 34.1 m s
2 2 v f = v x + v yf = 36.7 m s .
a fa f a
f
Chapter 4
103
P4.52
(a)
The moon's gravitational acceleration is the probe's centripetal acceleration: (For the moon's radius, see end papers of text.) a= v2 r
v2 1 9.80 m s 2 = 6 1.74 10 6 m
e
j
v = 2.84 10 6 m 2 s 2 = 1.69 km s (b) 2 r T 2 r 2 (1.74 10 6 m) T= = = 6.47 10 3 s = 1.80 h v 1.69 10 3 m s v= 5.00 m s v2 ac = = r 1.00 m
P4.53
(a)
b
g
2
= 25.0 m s 2
at = g = 9.80 m s 2
(b) (c) See figure to the right.
2 a = a c + a t2 =
e25.0 m s j + e9.80 m s j
2 2 1
2 2
= 26.8 m s 2
= tan 1
P4.54
FG a IJ = tan Ha K
t c
9.80 m s 2 25.0 m s 2
= 21.4 FIG. P4.53
x f = vix t = vi t cos 40.0 Thus, when x f = 10.0 m , t = 10.0 m . vi cos 40.0 At this time, y f should be 3.05 m  2.00 m = 1.05 m .
i 2
Thus, 1.05 m =
From this, vi = 10.7 m s .
bv sin 40.0g10.0 m + 1 e9.80 m s jLM 10.0 m OP . v cos 40.0 2 N v cos 40.0 Q
2 i i
104 P4.55
Motion in Two Dimensions
The special conditions allowing use of the horizontal range equation applies. For the ball thrown at 45, D = R 45 = For the bouncing ball, v 2 sin 2 D = R1 + R 2 = i + g vi2 sin 90 . g
e j
vi 2 2
sin 2 g
where is the angle it makes with the ground when thrown and when bouncing. (a) We require: vi2 vi2 sin 2 vi2 sin 2 = + g g 4g sin 2 = 4 5 = 26.6
FIG. P4.55
(b)
The time for any symmetric parabolic flight is given by y f = v yi t  1 2 gt 2 1 0 = vi sin i t  gt 2 . 2 2 vi sin i is the time at landing. g
If t = 0 is the time the ball is thrown, then t = So for the ball thrown at 45.0 t 45 = For the bouncing ball,
2 vi sin 45.0 . g
2 v sin 26.6 2 t = t1 + t 2 = i + g The ratio of this time to that for no bounce is
3 vi sin 26.6 g 2 vi sin 45 .0 g
e j sin 26.6 = 3v sin 26.6 .
vi 2 i
g
g
=
1.34 = 0.949 . 1.41
Chapter 4
105
P4.56
Using the range equation (Equation 4.14) R= vi2 sin( 2 i ) g
the maximum range occurs when i = 45 , and has a value R =
vi2 . Given R, this yields vi = gR . g If the boy uses the same speed to throw the ball vertically upward, then
v y = gR  gt and y = gR t 
at any time, t. At the maximum height, v y = 0, giving t = R g = gR  g 2
gt 2 2
R , and so the maximum height reached is g
y max P4.57
F RI GH g JK
2
= R
R R = . 2 2
Choose upward as the positive ydirection and leftward as the positive xdirection. The vertical height of the stone when released from A or B is
vi
yi = 1.50 + 1.20 sin 30.0 m = 2.10 m
(a) The equations of motion after release at A are v y = vi sin 60.0 gt = 1.30  9.80t m s
a
f
B 30 1.20 m vi 30
A
a
f
e x = a0.750t f m 1.30 a1.30f When y = 0 , t =
A
v x = vi cos 60.0 = 0.750 m s y = 2.10 +1.30t  4.90t 2 m FIG. P4.57
2
j
+ 41.2
9.80
= 0.800 s. Then, x A = 0.750 0.800 m = 0.600 m .
a
fa
f
(b)
The equations of motion after release at point B are v y = vi  sin 60.0  gt = 1.30  9.80t m s v x = vi cos 60.0 = 0.750 m s yi = 2.10  1.30t  4.90t 2 m . When y = 0 , t = +1.30
a
f
a j
f
e
a1.30f
9.80
2
+ 41.2
= 0.536 s. Then, x B = 0.750 0.536 m = 0.402 m .
a
fa
f
(c) (d)
ar =
1.50 m s v2 = r 1.20 m
b
g
2
= 1.87 m s 2 toward the center
After release, a =  g j = 9.80 m s 2 downward
106 P4.58
Motion in Two Dimensions
The football travels a horizontal distance R= vi2 sin 2 i g
b g = a20.0f sina60.0f = 35.3 m.
2
9.80
Time of flight of ball is 2 v sin i 2( 20.0) sin 30.0 t= i = = 2.04 s . g 9.80 FIG. P4.58
The receiver is x away from where the ball lands and x = 35.3  20.0 = 15.3 m. To cover this distance in 2.04 s, he travels with a velocity v= P4.59 (a) y=  15.3 = 7.50 m s in the direction the ball was thrown . 2.04
1 2 g t ; x = vi t 2 Combine the equations eliminating t: 1 x y=  g vi 2 From this, x
FG IJ H K
2
.
b g = FGH 2g y IJK v
2
2 i
FIG. P4.59 thus x = vi (b) 2 y 2( 300 ) = 275 = 6.80 10 3 = 6.80 km . g 9.80
The plane has the same velocity as the bomb in the x direction. Therefore, the plane will be 3 000 m directly above the bomb when it hits the ground. When is measured from the vertical, tan = x y
(c)
therefore, = tan 1
F x I = tan F 6 800 I = GH y JK GH 3 000 JK
1
66. 2 .
Chapter 4
107
*P4.60
(a)
We use the approximation mentioned in the problem. The time to travel 200 m horizontally is x 200 m t= = = 0.200 s . The bullet falls by v x 1,000 m s y = v yi t + 1 1 a y t 2 = 0 + 9.8 m s 2 0.2 s 2 2
e
ja f
2
= 0.196 m . barrel axis bullet path scope axis 50 150 200 250 FIG. P4.60(b)
(b)
The telescope axis must point below the barrel axis 0.196 m = 0.056 1 . by = tan 1 200 m t= 50.0 m = 0.050 0 s . The bullet falls by only 1 000 m s y= 1 9.8 m s 2 0.05 s 2
(c)
e
ja
f
2
= 0.0122 m .
At range 50 m =
1 1 200 m , the scope axis points to a location 19.6 cm = 4.90 cm above the 4 4 barrel axis, so the sharpshooter must aim low by 4.90 cm  1.22 cm = 3.68 cm . t= 150 m = 0.150 s 1 000 m s
a
f
a
f
(d)
y=
1 2 9.8 m s 2 0.15 s = 0.110 m 2 150 19.6 cm  11.0 cm = 3.68 cm . Aim low by 200
e
ja a
f
f
(e)
t=
250 m = 0.250 s 1 000 m s 1 9.8 m s 2 0.25 s 2
y=
e
ja
f
2
= 0.306 m
Aim high by 30.6 cm 
250 19.6 cm = 6.12 cm . 200
a
f
(f), (g) Many marksmen have a hard time believing it, but they should aim low in both cases. As in case (a) above, the time of flight is very nearly 0.200 s and the bullet falls below the barrel axis by 19.6 cm on its way. The 0.0561 angle would cut off a 19.6cm distance on a vertical wall at a horizontal distance of 200 m, but on a vertical wall up at 30 it cuts off distance h as shown, where cos 30 = 19.6 cm h , h = 22.6 cm. The marksman must aim low by 22.6 cm  19.6 cm = 3.03 cm . The answer can be obtained by considering limiting cases. Suppose the target is nearly straight above or below you. Then gravity will not cause deviation of the path of the bullet, and one must aim low as in part (c) to cancel out the sightingin of the telescope.
barrel axis scope 30 19.6 cm 30 19.6 cm scope axis bullet hits here h
FIG. P4.60(fg)
108 P4.61
Motion in Two Dimensions
(a)
From Part (c), the raptor dives for 6.34  2.00 = 4.34 s undergoing displacement 197 m downward and 10.0 4.34 = 43.4 m forward.
a fa f
d v= t (b) (c) P4.62
a197f + a43.4f
2
2
4.34 77.6
= 46.5 m s
= tan 1
197 =
FG 197 IJ = H 43.4 K
1 2 gt , t = 6.34 s 2
FIG. P4.61
Measure heights above the level ground. The elevation y b of the ball follows yb = R + 0  with x = vi t so y b = R  (a) gx 2 2 vi2 . 1 2 gt 2
The elevation yr of points on the rock is described by
2 yr + x 2 = R 2 .
2 surface as in y b > y r . Then y b + x 2 > R 2
We will have y b = y r at x = 0 , but for all other x we require the ball to be above the rock
F R  gx I GH 2 v JK
2 2 i
2
+ x2 > R2 + x2 > R2 + x2 > gx 2 R vi2 .
R2 
gx 2 R vi2
+
g 2x4 4vi4 g 2x4 4vi4
If this inequality is satisfied for x approaching zero, it will be true for all x. If the ball's parabolic trajectory has large enough radius of curvature at the start, the ball will clear the gR whole rock: 1 > 2 vi
vi > gR .
(b) With vi = gR and y b = 0 , we have 0 = R  or x = R 2 . The distance from the rock's base is xR= gx 2 2 gR
e
2 1 R .
j
Chapter 4
109
P4.63
(a)
While on the incline v 2  vi2 = 2 ax f v f  vi = at v 2  0 = 2 4.00 50.0 f 20.0  0 = 4.00t v f = 20.0 m s t = 5.00 s FIG. P4.63
a fa f
(b)
Initial freeflight conditions give us v xi = 20.0 cos 37.0 = 16.0 m s and v yi = 20.0 sin 37.0 = 12.0 m s v xf = v xi since a x = 0 v yf =  2 a y y + v yi 2 =  2 9.80 30.0 + 12.0 v f = v xf 2 + v yf 2 =
a
fa
f a
f
2
= 27.1 m s
a16.0f + a27.1f
2
2
= 31.5 m s at 59.4 below the horizontal
(c)
t1 = 5 s ; t 2 =
v yf  v yi ay
=
27.1 + 12.0 = 1.53 s 9.80
t = t1 + t 2 = 6.53 s
(d) P4.64
x = v xi t1 = 16.0 1.53 = 24.5 m
y 2 = 16 x x = vi t 1 y =  g t2 2
a f
Equation of bank: Equations of motion:
a1f a 2f a3 f F I GH JK
2
Substitute for t from (2) into (3) y = 
1 x2 g . Equate y 2 vi2 from the bank equation to y from the equations of motion:
FIG. P4.64
2 3 4 i
From this, x = 0 or x
3
I OP g x  16x = xF g x  16I = 0 . JK PQ 4v GH 4v JK F 10 I = 18.8 m . Also, 64v = and x = 4G g H 9.80 JK 1 Fx I 1 a9.80fa18.8f y =  gG J =  2 Hv K 2 a10.0f = 17.3 m .
2 2 4 4 i 2 i 2 4 i 4 1/3 2 2 2 2 i 2
L 1 Fx 16 x = M g G MN 2 H v
110 P4.65
Motion in Two Dimensions
(a)
Coyote:
1 2 1 at ; 70.0 = 15.0 t 2 2 2 70.0 = vi t Roadrunner: x = vi t ; x =
a f
Solving the above, we get vi = 22.9 m s and t = 3.06 s. (b) At the edge of the cliff,
v xi = at = 15.0 3.06 = 45.8 m s .
Substituting into y = 1 a y t 2 , we find 2 1 9.80 t 2 2 t = 4.52 s 1 1 2 a x t 2 = 45.8 4.52 s + 15.0 4.52 s . 2 2
a fa f
100 =
a
f
x = v xi t + Solving,
a fa
f a fa
f
x = 360 m . (c) For the Coyote's motion through the air v xf = v xi + a x t = 45.8 + 15 4.52 = 114 m s v yf = v yi + a y P4.66 44.3 m s .
a f t = 0  9.80a 4.52f =
Think of shaking down the mercury in an old fever thermometer. Swing your hand through a circular arc, quickly reversing direction at the bottom end. Suppose your hand moves through onequarter of a circle of radius 60 cm in 0.1 s. Its speed is
1 4
a2 fa0.6 mf 9 m s
0.1 s
and its centripetal acceleration is
v 2 ( 9 m s) 2 ~ 10 2 m s 2 . r 0.6 m
The tangential acceleration of stopping and reversing the motion will make the total acceleration somewhat larger, but will not affect its order of magnitude.
Chapter 4
111
P4.67
(a)
x = v xi t , y = v yi t +
1 2 gt 2
d cos 50.0 = 10.0 cos 15.0 t
and  d sin 50.0 = 10.0 sin 15.0 t +
a
f
a
f
1 9.80 t 2 . 2
a
f
Solving, d = 43.2 m and t = 2.88 s . (b) Since a x = 0 , v xf = v xi = 10.0 cos 15.0 = 9.66 m s FIG. P4.67
v yf = v yi + a y t = 10.0 sin 15.09.80 2.88 = 25.6 m s . Air resistance would decrease the values of the range and maximum height. As an airfoil, he can get some lift and increase his distance. *P4.68 For one electron, we have y = viy t , D = vix t + 1 1 a x t 2 a x t 2 , v yf = v yi , and v xf = v xi + a x t a x t . 2 2 v yf v xf v yi axt v yi t axt
2
a f
The angle its direction makes with the xaxis is given by
= tan 1
= tan 1
= tan 1
= tan 1
y . 2D
FIG. P4.68
Thus the horizontal distance from the aperture to the virtual source is 2D. The source is at coordinate x =  D . *P4.69 (a) The ice chest floats downstream 2 km in time t, so that 2 km = v w t . The upstream motion of the boat is described by d = ( v  v w )15 min. The downstream motion is described by 2 km d + 2 km = ( v + v w )( t  15 min) . We eliminate t = and d by substitution: vw
bv  v g15 min + 2 km = bv + v gFGH 2vkm  15 minIJK v va15 minf  v a15 minf + 2 km = 2 km + 2 km  va15 minf  v a15 minf v v va30 minf = 2 km v
w w w w w w w
2 km vw = = 4.00 km h . 30 min (b) In the reference frame of the water, the chest is motionless. The boat travels upstream for 15 min at speed v, and then downstream at the same speed, to return to the same point. Thus it travels for 30 min. During this time, the falls approach the chest at speed v w , traveling 2 km. Thus vw = x 2 km = = 4.00 km h . t 30 min
112 *P4.70
Motion in Two Dimensions
Let the river flow in the x direction. (a) To minimize time, swim perpendicular to the banks in the y direction. You are in the water for time t in y = v y t , t = (b) (c)
vw vs vs + vw vs vw vs vs + vw vs + vw
80 m = 53.3 s . 1.5 m s
The water carries you downstream by x = v x t = 2.50 m s 53.3 s = 133 m .
vw
b
g
To minimize downstream drift, you should swim so that your resultant velocity v s + v w is perpendicular to your swimming velocity v s relative to the water. This condition is shown in the middle picture. It maximizes the angle between the resultant velocity and the shore. The angle 1.5 m s between v s and the shore is given by cos = , 2.5 m s
vs + vw v w = 2.5 m/s i
vs
= 53.1 .
(d) Now v y = v s sin = 1.5 m s sin 53.1 = 1.20 m s t= y 80 m = = 66.7 s v y 1.2 m s
x = v x t = 2.5 m s  1.5 m s cos 53.1 66.7 s = 107 m .
b
g
Chapter 4
113
*P4.71
Find the highest firing angle H for which the projectile will clear the mountain peak; this will yield the range of the closest point of bombardment. Next find the lowest firing angle; this will yield the maximum range under these conditions if both H and L are > 45 ; x = 2500 m, y = 1800 m , vi = 250 m s . 1 2 1 gt = vi sin t  gt 2 2 2 x f = v xi t = vi cos t y f = v yi t 
a f
a
f
Thus t= Substitute into the expression for y f y f = vi sin x x a f v cos  1 gFGH v cos IJK 2
f f i i 2
xf vi cos
.
= x f tan 
gx 2 f 2 vi2 cos 2
gx 2 1 f 2 = tan + 1 so y f = x f tan  2 tan 2 + 1 and but 2 cos 2 vi
e
j
0=
gx 2 f 2 vi2
tan 2  x f tan +
gx 2 f 2 vi2
+ yf .
Substitute values, use the quadratic formula and find tan = 3.905 or 1.197 , which gives H = 75.6 and L = 50.1 . Range at H =
b b
g
vi2 sin 2 H = 3.07 10 3 m from enemy ship g 3.07 10 3  2 500  300 = 270 m from shore.
Range at L =
g
vi2 sin 2 L = 6.28 10 3 m from enemy ship g 6.28 10 3  2 500  300 = 3.48 10 3 from shore.
Therefore, safe distance is < 270 m or > 3.48 10 3 m from the shore.
FIG. P4.71
114 *P4.72
Motion in Two Dimensions
We follow the steps outlined in Example 4.7, eliminating t =
d cos to find vi cos
vi sin d cos gd 2 cos 2  2 =  d sin . vi cos 2 vi cos 2 Clearing of fractions, 2 vi2 cos sin cos  gd cos 2 = 2 vi2 cos 2 sin . To maximize d as a function of , we differentiate through with respect to and set 2 vi2 cos cos cos + 2 vi2 sin  sin cos  g dd = 0: d
a
f
dd cos 2 = 2 vi2 2 cos  sin sin . d
a
f
We use the trigonometric identities from Appendix B4 cos 2 = cos 2  sin 2 and sin 1 sin 2 = 2 sin cos to find cos cos 2 = sin 2 sin . Next, = tan and cot 2 = give cos tan 2
cot 2 = tan = tan 902 so = 902 and = 45
a
f
. 2
ANSWERS TO EVEN PROBLEMS
P4.2
e j (b) v = 18.0 i + a 4.00  9.80t f j ; (c) a = e 9.80 m s j j ; (d) a54.0 mf i  a32.1 mf j ; (e) b18.0 m sg i  b 25.4 m sg j ; (f) e 9.80 m s j j
2 2
(a) r = 18.0t i + 4.00t  4.90t 2 j ;
P4.8
(a) r = 5.00t i + 1.50t 2 j m ; v = 5.00 i + 3.00t j m s ; (b) r = 10.0 i + 6.00 j m; 7.81 m s
e
j
e
j
e
j
P4.10
e7.23 10
(a) d
3
m, 1.68 10 3 m
j
P4.4
(a) v = 5.00 i + 0 j m s ; a = 0 i + 5.00 2 j m s 2 ; (b) r = 4.00 m j +5.00 m  sin t i  cos t j ; v = 5.00 m  cos t i + sin t j ; a = 5.00 m 2
e
j
P4.12
e
j
g horizontally; 2h 2h below the horizontal (b) tan 1 d
FG IJ H K
e
j
P4.14 P4.16 P4.18 P4.20
0.600 m s 2 (a) 76.0; (b) the same; (c) 25.8 m s d tan i  gd 2
2 i
e
j esin t i + cos t jj ;
17d 8
(c) a circle of radius 5.00 m centered at 0 , 4.00 m
a
f
P4.6
(a) v = 12.0t j m s ; a = 12.0 j m s ; (b) r = 3.00 i  6.00 j m; v = 12.0 j m s
2
e
j
e2 v
cos 2 i
j
Chapter 4
115
P4.22 P4.24
33.5 below the horizontal (a) 0.852 s; (b) 3.29 m s ; (c) 4.03 m s; (d) 50.8; (e) 1.12 s tan
1
P4.48 P4.50
2vi t cos i (a) see the solution; (b) i = 45+ ; d max = 2 vi2 1  sin g cos
2
b
g
P4.26
F GH
2gh v
I JK
P4.52 P4.54 P4.56 P4.58
(a) 1.69 km s ; (b) 6.47 10 3 s 10.7 m s R 2 7.50 m s in the direction the ball was thrown (a) 19.6 cm; (b) 0.0561; (c) aim low 3.68 cm; (d) aim low 3.68 cm; (e) aim high 6.12 cm; (f) aim low; (g) aim low (a) gR ; (b)
P4.28
0.033 7 m s 2 toward the center of the Earth 0.281 rev s 7.58 10 3 m s; 5.80 10 3 s (a) 0.600 m s 2 forward; (b) 0.800 m s 2 inward; (c) 1.00 m s 2 forward and 53.1 inward
P4.30 P4.32 P4.34
P4.60
P4.36
(a) see the solution; (b) 29.7 m s 2 ; (c) 6.67 m s at 36.9 above the horizontal (a) 26.9 m s ; (b) 67.3 m; (c) 2.00 i  5.00 j m s
P4.62 P4.64 P4.66 P4.68 P4.70
e
2 1 R
j
P4.38
e
j
a18.8 m;  17.3 mf
see the solution; ~ 10 2 m s 2 x = D (a) at 90 to the bank; (b) 133 m; (c) upstream at 53.1 to the bank; (d) 107 m see the solution
2
P4.40 P4.42 P4.44
18.0 s 153 km h at 11.3 north of west (a) 10.1 m s at 14.3 south from the vertical; (b) 9.80 m s 2 vertically downward
2
P4.72
P4.46
27.7 east of north
5
The Laws of Motion
CHAPTER OUTLINE
5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 The Concept of Force Newton's First Law and Inertial Frames Mass Newton's Second Law The Gravitational Force and Weight Newton's Third Law Some Applications of Newton's Laws Forces of Friction
ANSWERS TO QUESTIONS
Q5.1 (a) The force due to gravity of the earth pulling down on the ballthe reaction force is the force due to gravity of the ball pulling up on the earth. The force of the hand pushing up on the ballreaction force is ball pushing down on the hand. The only force acting on the ball in freefall is the gravity due to the earth the reaction force is the gravity due to the ball pulling on the earth.
(b)
Q5.2 Q5.3
The resultant force is zero, as the acceleration is zero. Mistake one: The car might be momentarily at rest, in the process of (suddenly) reversing forward into backward motion. In this case, the forces on it add to a (large) backward resultant.
Mistake two: There are no cars in interstellar space. If the car is remaining at rest, there are some large forces on it, including its weight and some force or forces of support. Mistake three: The statement reverses cause and effect, like a politician who thinks that his getting elected was the reason for people to vote for him. Q5.4 When the bus starts moving, the mass of Claudette is accelerated by the force of the back of the seat on her body. Clark is standing, however, and the only force on him is the friction between his shoes and the floor of the bus. Thus, when the bus starts moving, his feet start accelerating forward, but the rest of his body experiences almost no accelerating force (only that due to his being attached to his accelerating feet!). As a consequence, his body tends to stay almost at rest, according to Newton's first law, relative to the ground. Relative to Claudette, however, he is moving toward her and falls into her lap. (Both performers won Academy Awards.) First ask, "Was the bus moving forward or backing up?" If it was moving forward, the passenger is lying. A fast stop would make the suitcase fly toward the front of the bus, not toward the rear. If the bus was backing up at any reasonable speed, a sudden stop could not make a suitcase fly far. Fine her for malicious litigiousness. It would be smart for the explorer to gently push the rock back into the storage compartment. Newton's 3rd law states that the rock will apply the same size force on her that she applies on it. The harder she pushes on the rock, the larger her resulting acceleration.
Q5.5
Q5.6
117
118 Q5.7
The Laws of Motion
The molecules of the floor resist the ball on impact and push the ball back, upward. The actual force acting is due to the forces between molecules that allow the floor to keep its integrity and to prevent the ball from passing through. Notice that for a ball passing through a window, the molecular forces weren't strong enough. While a football is in flight, the force of gravity and air resistance act on it. When a football is in the process of being kicked, the foot pushes forward on the ball and the ball pushes backward on the foot. At this time and while the ball is in flight, the Earth pulls down on the ball (gravity) and the ball pulls up on the Earth. The moving ball pushes forward on the air and the air backward on the ball. It is impossible to string a horizontal cable without its sagging a bit. Since the cable has a mass, gravity pulls it downward. A vertical component of the tension must balance the weight for the cable to be in equilibrium. If the cable were completely horizontal, then there would be no vertical component of the tension to balance the weight. Some physics teachers demonstrate this by asking a beefy student to pull on the ends of a cord supporting a can of soup at its center. Some get two burly young men to pull on opposite ends of a strong rope, while the smallest person in class gleefully mashes the center of the rope down to the table. Point out the beauty of sagging suspensionbridge cables. With a laser and an optical lever, demonstrate that the mayor makes the courtroom table sag when he sits on it, and the judge bends the bench. Give them "I make the floor sag" buttons, available to instructors using this manual. Estimate the cost of an infinitely strong cable, and the truth will always win. As the barbell goes through the bottom of a cycle, the lifter exerts an upward force on it, and the scale reads the larger upward force that the floor exerts on them together. Around the top of the weight's motion, the scale reads less than average. If the iron is moving upward, the lifter can declare that she has thrown it, just by letting go of it for a moment, so our answer applies also to this case. As the sand leaks out, the acceleration increases. With the same driving force, a decrease in the mass causes an increase in the acceleration. As the rocket takes off, it burns fuel, pushing the gases from the combustion out the back of the rocket. Since the gases have mass, the total remaining mass of the rocket, fuel, and oxidizer decreases. With a constant thrust, a decrease in the mass results in an increasing acceleration. The friction of the road pushing on the tires of a car causes an automobile to move. The push of the air on the propeller moves the airplane. The push of the water on the oars causes the rowboat to move. As a man takes a step, the action is the force his foot exerts on the Earth; the reaction is the force of the Earth on his foot. In the second case, the action is the force exerted on the girl's back by the snowball; the reaction is the force exerted on the snowball by the girl's back. The third action is the force of the glove on the ball; the reaction is the force of the ball on the glove. The fourth action is the force exerted on the window by the air molecules; the reaction is the force on the air molecules exerted by the window. We could in each case interchange the terms `action' and `reaction.' The tension in the rope must be 9 200 N. Since the rope is moving at a constant speed, then the resultant force on it must be zero. The 49ers are pulling with a force of 9 200 N. If the 49ers were winning with the rope steadily moving in their direction or if the contest was even, then the tension would still be 9 200 N. In all of these case, the acceleration is zero, and so must be the resultant force on the rope. To win the tugofwar, a team must exert a larger force on the ground than their opponents do.
Q5.8
Q5.9
Q5.10
Q5.11 Q5.12
Q5.13
Q5.14
Q5.15
Chapter 5
119
Q5.16 Q5.17
The tension in the rope when pulling the car is twice that in the tugofwar. One could consider the car as behaving like another team of twenty more people. This statement contradicts Newton's 3rd law. The force that the locomotive exerted on the wall is the same as that exerted by the wall on the locomotive. The wall temporarily exerted on the locomotive a force greater than the force that the wall could exert without breaking. The sack of sand moves up with the athlete, regardless of how quickly the athlete climbs. Since the athlete and the sack of sand have the same weight, the acceleration of the system must be zero. The resultant force doesn't always add to zero. If it did, nothing could ever accelerate. If we choose a single object as our system, action and reaction forces can never add to zero, as they act on different objects. An object cannot exert a force on itself. If it could, then objects would be able to accelerate themselves, without interacting with the environment. You cannot lift yourself by tugging on your bootstraps. To get the box to slide, you must push harder than the maximum static frictional force. Once the box is moving, you need to push with a force equal to the kinetic frictional force to maintain the box's motion. The stopping distance will be the same if the mass of the truck is doubled. The stopping distance will decrease by a factor of four if the initial speed is cut in half. If you slam on the brakes, your tires will skid on the road. The force of kinetic friction between the tires and the road is less than the maximum static friction force. Antilock brakes work by "pumping" the brakes (much more rapidly that you can) to minimize skidding of the tires on the road. With friction, it takes longer to come down than to go up. On the way up, the frictional force and the component of the weight down the plane are in the same direction, giving a large acceleration. On the way down, the forces are in opposite directions, giving a relatively smaller acceleration. If the incline is frictionless, it takes the same amount of time to go up as it does to come down. (a) The force of static friction between the crate and the bed of the truck causes the crate to accelerate. Note that the friction force on the crate is in the direction of its motion relative to the ground (but opposite to the direction of possible sliding motion of the crate relative to the truck bed). It is most likely that the crate would slide forward relative to the bed of the truck.
Q5.18 Q5.19
Q5.20
Q5.21
Q5.22 Q5.23
Q5.24
Q5.25
(b) Q5.26
In Question 25, part (a) is an example of such a situation. Any situation in which friction is the force that accelerates an object from rest is an example. As you pull away from a stop light, friction is the force that accelerates forward a box of tissues on the level floor of the car. At the same time, friction of the ground on the tires of the car accelerates the car forward.
120
The Laws of Motion
SOLUTIONS TO PROBLEMS
The following problems cover Sections 5.15.6. Section 5.1 Section 5.2 Section 5.3 Section 5.4 Section 5.5 Section 5.6 P5.1 The Concept of Force Newton's First Law and Inertial Frames Mass Newton's Second Law The Gravitational Force and Weight Newton's Third Law
For the same force F, acting on different masses F = m 1 a1 and F = m2 a2 (a) (b) m1 a 1 = 2 = 3 m2 a1 F = m1 + m 2 a = 4m1 a = m1 3.00 m s 2
a
f
c
h
a = 0.750 m s
*P5.2
2
v f = 880 m s, m = 25.8 kg , x f = 6 m v 2 = 2 ax f = 2 x f f F= mv 2 f 2x f
FG F IJ H mK
= 1.66 10 6 N forward
P5.3
m = 3.00 kg a = 2.00 i + 5.00 j m s 2
e
j
F = ma = F =
e6.00i + 15.0 jj N a6.00f + a15.0f N =
2 2
16.2 N
Chapter 5
121
P5.4
Fg = weight of ball = mg v release = v and time to accelerate = t : a= (a) Distance x = vt : x= Fg v gt v v v = = i t t t
FG v IJ t = H 2K
vt 2
(b)
Fp  Fg j =
i Fg v gt
Fp =
i + Fg j
P5.5
m = 4.00 kg , v i = 3.00 i m s , v 8 = 8.00 i + 10.0 j m s , t = 8.00 s a= v 5.00 i + 10.0 j = m s2 t 8.00
e
j
F = ma =
e2.50i + 5.00 jj N
2 2
F = ( 2.50) +(5.00) = 5.59 N P5.6 (a) Let the xaxis be in the original direction of the molecule's motion. v f = vi + at: 670 m s = 670 m s + a 3.00 10 13 s a = 4. 47 10 15 m s 2 (b) For the molecule,
e
j
F = ma . Its weight is negligible.
Fwall on molecule = 4.68 10 26 kg 4.47 10 15 m s 2 = 2.09 10 10 N Fmolecule on wall = +2.09 10 10 N
e
j
122 P5.7
The Laws of Motion
(a)
F = ma and v 2 = vi2 + 2 ax f or a = f
Therefore,
v 2  vi2 f 2x f
.
F = m
ev
2 f
 vi2
j LMe7.00 10 kg N
c
5
2x f
F = 9.11 10 31
(b)
m s2
j  e3.00 10 2b0.050 0 mg
2
5
m s2
j OPQ
2
= 3.64 10 18 N .
The weight of the electron is Fg = mg = 9.111031 kg 9.80 m s 2 = 8.93 1030 N The accelerating force is 4.08 10 11 times the weight of the electron.
hc
h
P5.8
(a) (b)
Fg = mg = 120 lb = 4.448 N lb (120 lb)= 534 N m= Fg g = 534 N = 54.5 kg 9.80 m s 2 900 N = 91.8 kg 9.80 m s 2
a
f
P5.9
Fg = mg = 900 N , m =
cF h
g
on Jupiter
= 91.8 kg 25.9 m s 2 = 2.38 kN
c
h
P5.10
Imagine a quick trip by jet, on which you do not visit the rest room and your perspiration is just canceled out by a glass of tomato juice. By subtraction, Fg = mg p and Fg = mg C give
c h h
p
c h
C
Fg = m g p  g C . For a person whose mass is 88.7 kg, the change in weight is Fg = 88.7 kg 9.809 5  9.780 8 = 2.55 N . A precise balance scale, as in a doctor's office, reads the same in different locations because it compares you with the standard masses on its beams. A typical bathroom scale is not precise enough to reveal this difference.
c
b
g
Chapter 5
123
P5.11
(a)
F = F1 + F2 = e 20.0 i + 15.0 jj N F = ma:
or 20.0 i + 15.0 j = 5.00a a = 4.00 i + 3.00 j m s 2
e
j
a = 5.00 m s 2 at = 36.9
(b) F2 x = 15.0 cos 60.0 = 7.50 N F2 y = 15.0 sin 60.0 = 13.0 N F2 = 7.50 i + 13.0 j N
FIG. P5.11
e
j
F = F1 + F2 = e27.5 i + 13.0 jj N = ma = 5.00a
a= P5.12
e5.50i + 2.60 jj m s
2
= 6.08 m s 2 at 25.3
We find acceleration: r f  ri = v i t + 1 2 at 2
1 4.20 m i  3.30 mj = 0+ a 1.20 s 2
f = 0.720 s a a = e5.83 i  4.58 jj m s .
2 2 2
a
Now F = ma becomes Fg + F2 = ma F2 = 2.80 kg 5.83 i  4.58 j m s 2 + 2.80 kg 9.80 m s 2 j F2 = P5.13 (a)
e
j
b
ge
j
e16.3 i + 14.6 jj N
.
You and the earth exert equal forces on each other: m y g = M e a e . If your mass is 70.0 kg, ae
a70.0 kg fc9.80 m s h = =
2
5.98 10 24 kg
~ 1022 m s 2 . 1 2 at . If the seat is 2
(b)
You and the planet move for equal times intervals according to x = 50.0 cm high, 2xy ay xe = = 2xe ae
my 70.0 kg 0.500 m ae ~ 10 23 m . xy = xy = ay me 5.98 10 24 kg
a
f
124 P5.14
The Laws of Motion
F = ma reads
e2.00 i + 2.00 j + 5.00i  3.00 j  45.0ij N = me3.75 m s ja
2
where a represents the direction of a
e42.0 i  1.00 jj N = me3.75 m s ja
2
FG 1.00 IJ below the xaxis H 42.0 K F = 42.0 N at 181 = mc3.75 m s ha .
F =
(42.0) +(1.00) N at tan1
2 2 2
For the vectors to be equal, their magnitudes and their directions must be equal. (a) (b) (d) a is at 181 counterclockwise from the xaxis m= 42.0 N = 11.2 kg 3.75 m s 2
v f = v i + at = 0 + 3.75 m s 2 at 181 10.0 s so v f = 37.5 m s at 181 v f = 37.5 m s cos 181 i + 37.5 m s sin 181 j so v f =
e
j
e37.5 i  0.893 jj m s
(c) P5.15 (a) (b) (c)
v f = 37.5 2 + 0.893 2 m s = 37.5 m s 15.0 lb up 5.00 lb up 0
Section 5.7 P5.16 vx =
Some Applications of Newton's Laws
dy dx = 10t , v y = = 9t 2 dt dt dv y dv x = 10 , a y = = 18t ax = dt dt At t = 2.00 s , a x = 10.0 m s 2 , a y = 36.0 m s 2
Fx = ma x : 3.00 kg e10.0 Fy = ma y : 3.00 kg e36.0
m s 2 = 30.0 N
2
j m s j = 108 N
F=
Fx2 + Fy2 = 112 N
Chapter 5
125
P5.17
m = 1.00 kg mg = 9.80 N 0.200 m tan = 25.0 m = 0.458 Balance forces, 2T sin = mg T= 9.80 N = 613 N 2 sin (1) (2) (3)
50.0 m 0.200 m
T
T
mg
FIG. P5.17
P5.18
T3 = Fg T1 sin 1 + T2 sin 2 = Fg T1 cos 1 = T2 cos 2 Eliminate T2 and solve for T1 T1 =
1
2
Fg
bsin
Fg cos 2
1
cos 2 + cos 1 sin 2
g
=
Fg cos 2 sin 1 + 2
b
g
T1
T3 = Fg = 325 N T1 = Fg T2 P5.19
1
FG cos 25.0 IJ = 296 N H sin 85.0 K F cos IJ = 296 NFG cos 60.0 IJ = =T G H cos 25.0 K H cos K
1 2
1
T3 163 N
2
T2
FIG. P5.18
See the solution for T1 in Problem 5.18.
126 P5.20
The Laws of Motion
(a)
An explanation proceeding from fundamental physical principles will be best for the parents and for you. Consider forces on the bit of string touching the weight hanger as shown in the freebody diagram: Horizontal Forces: Vertical Forces:
Fy = ma y : Fg + T sin = 0
Fx = ma x : Tx + T cos = 0
FIG. P5.20
You need only the equation for the vertical forces to find that the tension in the string is Fg . The force the child feels gets smaller, changing from T to T cos , while given by T = sin the counterweight hangs on the string. On the other hand, the kite does not notice what you are doing and the tension in the main part of the string stays constant. You do not need a level, since you learned in physics lab to sight to a horizontal line in a building. Share with the parents your estimate of the experimental uncertainty, which you make by thinking critically about the measurement, by repeating trials, practicing in advance and looking for variations and improvements in technique, including using other observers. You will then be glad to have the parents themselves repeat your measurements. (b) P5.21 (a) T= Fg sin = 0.132 kg 9.80 m s 2 sin 46.3
e
j=
1.79 N
Isolate either mass T + mg = ma = 0 T = mg . The scale reads the tension T, so T = mg = 5.00 kg 9.80 m s 2 = 49.0 N . FIG. P5.21(a)
e
j
(b)
Isolate the pulley T2 + 2T1 = 0 T2 = 2 T1 = 2mg = 98.0 N .
(c)
F = n + T + mg = 0
Take the component along the incline n x + Tx + mg x = 0 or 0 + T  mg sin 30.0 = 0 T = mg sin 30.0 = = 24.5 N . mg 5.00 9.80 = 2 2 FIG. P5.21(b)
a f
FIG. P5.21(c)
Chapter 5
127
P5.22
The two forces acting on the block are the normal force, n, and the weight, mg. If the block is considered to be a point mass and the xaxis is chosen to be parallel to the plane, then the free body diagram will be as shown in the figure to the right. The angle is the angle of inclination of the plane. Applying Newton's second law for the accelerating system (and taking the direction up the plane as the positive x direction) we have
Fy = n  mg cos = 0: n = mg cos Fx = mg sin = ma : a = g sin
(a) When = 15.0
FIG. P5.22
a = 2.54 m s 2
(b) Starting from rest v 2 = vi2 + 2 a x f  xi = 2 ax f f v f = 2 ax f = 2 2.54 m s 2 2.00 m = 3.18 m s P5.23 Choose a coordinate system with i East and j North.
d
i
e
ja
f
a5.00 Nf j + F = a10.0 Nf30.0 = a5.00 Nf j + a8.66 Nfi
1
F = ma = 1.00 kg e10.0
m s 2 at 30.0
j
F1 = 8.66 N (East ) *P5.24 First, consider the block moving along the horizontal. The only force in the direction of movement is T. Thus, Fx = ma T = 5 kg a
FIG. P5.23
n +x T T 9 kg Fg = 88.2 N +y
a f
(1)
5 kg
49 N
Next consider the block that moves vertically. The forces on it are the tension T and its weight, 88.2 N. We have
FIG. P5.24
Fy = ma
88.2 N  T = 9 kg a
a f
(2)
Note that both blocks must have the same magnitude of acceleration. Equations (1) and (2) can be added to give 88.2 N = 14 kg a. Then
b
g
a = 6.30 m s 2 and T = 31.5 N .
128 P5.25
The Laws of Motion
After it leaves your hand, the block's speed changes only because of one component of its weight:
Fx = ma x
Taking v f
i = 0 , v = 5.00 m s, and a = g sina 20.0f gives
v 2 = vi2 + 2 a x f  xi . f
i
 mg sin 20.0 = ma
d
0 = (5.00)  2(9.80) sin 20.0 x f  0 or xf = P5.26 25.0 = 3.73 m . 2(9.80) sin 20.0
2
a
fc
h
FIG. P5.25
a
f
m1 = 2.00 kg , m 2 = 6.00 kg , = 55.0 (a)
Fx = m 2 g sin  T = m 2 a
and T  m1 g = m1 a a= m 2 g sin  m1 g = 3.57 m s 2 m1 + m 2 FIG. P5.26
(b) (c) *P5.27
T = m1 a + g = 26.7 N Since vi = 0 , v f = at = 3.57 m s 2 ( 2.00 s)= 7.14 m s .
a
f
c
h
We assume the vertical bar is in compression, pushing up on the pin with force A, and the tilted bar is in tension, exerting force B on the pin at 50 .
30 2 500 N A
50 B B cos50 A
Fx = 0: Fy = 0:
2 500 N cos 30+ B cos 50 = 0 B = 3.37 10 3 N 2 500 N sin 30+ A  3.37 10 3 N sin 50 = 0 A = 3.83 10 3 N
2 500 N cos30
2 500 N sin30 FIG. P5.27
B sin50
Positive answers confirm that B is in tension and A is in compression.
Chapter 5
129
P5.28
First, consider the 3.00 kg rising mass. The forces on it are the tension, T, and its weight, 29.4 N. With the upward direction as positive, the second law becomes
Fy = ma y : T  29.4 N = a3.00 kg fa
(1)
The forces on the falling 5.00 kg mass are its weight and T, and its acceleration is the same as that of the rising mass. Calling the positive direction down for this mass, we have
Fy = ma y : 49 N  T = a5.00 kg fa
(2)
FIG. P5.28
Equations (1) and (2) can be solved simultaneously by adding them: T  29.4 N + 49.0 N  T = 3.00 kg a + 5.00 kg a (b) This gives the acceleration as a= (a) Then T  29.4 N = 3.00 kg 2.45 m s 2 = 7.35 N . The tension is T = 36.8 N . (c) Consider either mass. We have y = vi t + *P5.29 1 1 2 2 at = 0 + 2.45 m s 2 (1.00 s) = 1.23 m . 2 2 19.6 N = 2.45 m s 2 . 8.00 kg
a
f a
f
a
fc
h
c
h
As the man rises steadily the pulley turns steadily and the tension in the rope is the same on both sides of the pulley. Choose manpulleyandplatform as the system:
T
Fy = ma y
+T  950 N = 0 T = 950 N . The worker must pull on the rope with force 950 N . 950 N FIG. P5.29
130 *P5.30
The Laws of Motion
Both blocks move with acceleration a =
FG m Hm
 m1 g: 2 + m1
2 2
IJ K
a= (a)
F 7 kg  2 kg I 9.8 m s GH 7 kg + 2 kg JK
= 5.44 m s 2 .
Take the upward direction as positive for m1 .
2 2 v xf = v xi + 2 a x x f  x i :
d
i
0 = 2.4 m s xf = 
b
g + 2e5.44 m s jdx  0i
2 2 f
5.76 m 2 s 2 2 5.44 m s 2
e
j
= 0.529 m
x f = 0.529 m below its initial level (b) v xf = v xi + a x t: v xf = 2.40 m s + 5.44 m s 2 1.80 s v xf = 7. 40 m s upward P5.31 Forces acting on 2.00 kg block: T  m1 g = m 1 a Forces acting on 8.00 kg block: Fx  T = m 2 a (a) Eliminate T and solve for a: a= Fx  m1 g m1 + m 2 (2) (1)
e
ja
f
a > 0 for Fx > m1 g = 19.6 N . (b) Eliminate a and solve for T: T= m1 Fx + m 2 g m1 + m 2
a
f
FIG. P5.31
T = 0 for Fx m 2 g = 78.4 N . (c) Fx , N ax , m s 2 100 12.5 78.4 9.80 50.0 6.96 0 1.96 50.0 3.04 100 8.04
Chapter 5
131
*P5.32
(a)
For force components along the incline, with the upward direction taken as positive,
Fx = ma x :
For the upward motion,
 mg sin = ma x
a x =  g sin =  9.8 m s 2 sin 35 = 5.62 m s 2 .
e
j
2 2 v xf = v xi + 2 a x x f  xi
d
i
2 f
0= 5 m s xf = (b)
b
g + 2e5.62 m s jdx  0i
2
2 5.62 m s 2
e
25 m 2 s 2
j
= 2.22 m .
The time to slide down is given by x f = xi + v xi t + 1 axt 2 2 1 5.62 m s 2 t 2 2
0 = 2.22 m + 0 + t= For the second particle, x f = xi + v xi t + 1 axt 2 2 2 2.22 m 5.62 m s
e
j
a
2
f = 0.890 s .
0 = 10 m + v xi 0.890 s + 5.62 m s 2 0.890 s v xi = 10 m + 2.22 m = 8.74 m s 0.890 s
a
f e
ja
f
2
speed = 8.74 m s .
132 P5.33
The Laws of Motion
First, we will compute the needed accelerations:
a1f a 2f a3 f a4f
Before it starts to move: During the first 0.800 s:
ay = 0 v yf  v yi 1. 20 m s  0 = ay = t 0.800 s = 1.50 m s 2
While moving at constant velocity: a y = 0 v yf  v yi 0  1.20 m s = During the last 1.50 s: ay = 1.50 s t = 0.800 m s 2
Newton's second law is:
Fy = ma y
+S  72.0 kg 9.80 m s 2 = 72.0 kg a y
FIG. P5.33
b
ge
g S = 706 N + b72.0 kg ga
j b
y
.
(a) (b) (c) (d) P5.34 (a)
When a y = 0 , S = 706 N . When a y = 1.50 m s 2 , S = 814 N . When a y = 0 , S = 706 N . When a y =0.800 m s 2 , S = 648 N . Pulley P1 has acceleration a 2 . Since m1 moves twice the distance P1 moves in the same time, m1 has twice the acceleration of P1 , i.e., a1 = 2 a 2 . From the figure, and using
(b)
F = ma:
m 2 g  T2 = m 2 a 2 T1 = m1 a1 = 2m1 a 2 T2  2T1 = 0
a1f a 2f a3 f
IJ K
FIG. P5.34
Equation (1) becomes m 2 g  2T1 = m 2 a 2 . This equation combined with Equation (2) yields T1 m 2m1 + 2 = m 2 g m1 2 T1 = (c) m1 m 2 m1 m 2 g and T2 = g . 2m1 + 1 m 2 m1 + 1 m 2 2 4
FG H
From the values of T1 and T2 we find that a1 = m2 g T1 = 2m1 + 1 m 2 m1 2 and a 2 = m2 g 1 a1 = . 2 4m 1 + m 2
Chapter 5
133
Section 5.8 *P5.35
Forces of Friction +y n ground = Fg /2 = 85.0 lb +y n tip f +x F = 45.8 lb 22.0
22.0 22.0 F2 F1 +x
Fg = 170 lb FreeBody Diagram of Person
FreeBody Diagram of Crutch Tip
FIG. P5.35 From the freebody diagram of the person,
Fx = F1 sina22.0f  F2 sina22.0f = 0 ,
which gives F1 = F2 = F . Then, Fy = 2 F cos 22.0+85.0 lbs  170 lbs = 0 yields F = 45.8 lb. (a) Now consider the freebody diagram of a crutch tip.
Fx = f ( 45.8 lb) sin 22.0= 0 ,
or f = 17. 2 lb .
Fy = n tip ( 45.8 lb) cos 22.0= 0 ,
which gives n tip = 42.5 lb . For minimum coefficient of friction, the crutch tip will be on the verge of slipping, so f 17.2 lb f = f s max = s n tip and s = = = 0.404 . n tip 42.5 lb
a f
(b)
As found above, the compression force in each crutch is F1 = F2 = F = 45.8 lb .
134 P5.36
The Laws of Motion
For equilibrium: f = F and n = Fg . Also, f = n i.e.,
= s =
and
f F = n Fg 75.0 N = 0.306 25.0 9.80 N
a f
FIG. P5.36 60.0 N = 0.245 . 25.0(9.80) N
k =
P5.37
Fy = ma y :
+n  mg = 0 fs sn = s mg
This maximum magnitude of static friction acts so long as the tires roll without skidding.
Fx = ma x :
The maximum acceleration is
 f s = ma
a =  s g . The initial and final conditions are: x i = 0 , vi = 50.0 mi h = 22.4 m s , v f = 0 v 2 = vi2 + 2 a x f  xi :  vi2 = 2 s gx f f (a) xf = xf = vi2 2 g
d
i
a22.4 m sf = 2(0.100 )c9.80 m s h
2 2
256 m
(b)
xf = xf =
vi2 2 g
a22.4 m sf = 2(0.600)c9.80 m s h
2 2
42.7 m
Chapter 5
135
P5.38
If all the weight is on the rear wheels, (a) F = ma: s mg = ma But x = so s = 2 x : gt 2 at 2 s gt 2 = 2 2
s =
(b) *P5.39 (a)
2 0.250 mi 1 609 m mi
a
e9.80 m s ja4.96 sf
2
fb
2
g=
3.34 .
Time would increase, as the wheels would skid and only kinetic friction would act; or perhaps the car would flip over. The person pushes backward on the floor. The floor pushes forward on the person with a force of friction. This is the only horizontal force on the person. If the person's shoe is on the point of slipping the static friction force has its maximum value.
Fx = ma x : Fy = ma y :
ma x = s mg x f = xi + v xi t + 1 ax t 2 2
f = sn = ma x n  mg = 0 a x = s g = 0.5 9.8 m s 2 = 4.9 m s 2 1 3 m = 0 + 0 + 4.9 m s 2 t 2 2 t = 1.11 s
e e
j j
FIG. P5.39
(b)
xf =
2x f 2(3 m) 1 s gt 2 , t = = = 0.875 s 2 s g (0.8) 9.8 m s 2
c
h
P5.40
m suitcase = 20.0 kg , F = 35.0 N
Fx = ma x : Fy = ma y :
(a) F cos = 20.0 N cos =
20.0 N + F cos = 0 +n + F sin  Fg = 0
20.0 N = 0.571 35.0 N
FIG. P5.40
= 55.2
(b)
n = Fg  F sin = 196  35.0(0.821) N
n = 167 N
136 P5.41
The Laws of Motion
m = 3.00 kg , = 30.0 , x = 2.00 m, t = 1.50 s (a) x= 1 2 at : 2 2.00 m = 1 2 a 1.50 s 2 4.00 a= = 1.78 m s 2 2 1.50
a
f
FIG. P5.41
a f
F = n + f + mg = m a :
Along x: 0  f + mg sin 30.0 = ma f = m g sin 30.0 a
b
g
Along y: n + 0  mg cos 30.0 = 0 n = mg cos 30.0 (b) (c) (d)
k =
f m g sin 30.0a a , k = tan 30.0 = = 0.368 n mg cos 30.0 g cos 30.0
a
f
f = m g sin 30.0a , f = 3.00 9.80 sin 30.01.78 = 9.37 N v 2 = vi2 + 2 a x f  xi f where x f  xi = 2.00 m v 2 = 0 + 2 1.78 2.00 = 7.11 m 2 s 2 f v f = 7.11 m 2 s 2 = 2.67 m s
a
f
a
f
c
h
a fa f
Chapter 5
137
*P5.42
First we find the coefficient of friction:
n
Fy = 0: Fx = ma x :
 s mg =  mvi2 2 x
+n  mg = 0 f = sn = s mg v2 f = vi2 + 2 a x x = 0 f mg n mg sin10 mg cos10 FIG. P5.42 f
88 ft s v2 s = i = = 0.981 2 gx 2 32.1 ft s 2 123 ft
e
b
g ja
2
f
Now on the slope
Fy = 0: Fx = ma x :
+n  mg cos 10 = 0 f s = sn = s mg cos 10  s mg cos 10+ mg sin 10 =  x = = mvi2 2 x
vi2 2 g s cos 10 sin 10
b88 ft sg = 2e32.1 ft s ja0.981 cos 10 sin 10f
2 2
b
g
152 ft .
P5.43
T  f k = 5.00 a (for 5.00 kg mass) 9.00 g  T = 9.00 a (for 9.00 kg mass) Adding these two equations gives: 9.00 9.80  0.200 5.00 9.80 = 14.0 a a = 5.60 m s 2 T = 5.00 5.60 + 0.200 5.00 9.80
a f
a fa f
a f
a fa f
FIG. P5.43
= 37.8 N
138 P5.44
The Laws of Motion
Let a represent the positive magnitude of the acceleration a j of m1 , of the acceleration a i of m 2 , and of the acceleration +aj of m 3 . Call T12 the tension in the left rope and T23 the tension in the cord on the right. For m1 , For m 2 , and for m 3 ,
n
T12
T23 f = kn m2 g
Fy = ma y
+T12  m1 g = m1 a T12 + k n + T23 = m 2 a n  m2 g = 0 T23  m 3 g = +m 3 a T12
Fx = ma x
Fy = ma y Fy = ma y
T23
we have three simultaneous equations
+T12
b g  0.350a9.80 N f  T = b1.00 kg ga +T  19.6 N = b 2.00 kg ga .
T12 + 39.2 N = 4.00 kg a
23 23
m1 g FIG. P5.44
m3 g
(a)
Add them up: +39.2 N  3. 43 N  19.6 N = 7.00 kg a
a
f
a = 2.31 m s 2 , down for m1 , left for m 2 , and up for m 3 .
(b) Now T12 + 39. 2 N = 4.00 kg 2.31 m s 2
a
fc
h
T12 = 30.0 N and T23  19.6 N = 2.00 kg 2.31 m s 2
a
fc
h
T23 = 24.2 N .
T m2 n2 T T f2 = k n 2 m2 g = 176 N m2 F
P5.45
(a) (b)
See Figure to the right 68.0  T  m 2 g = m 2 a (Block #2) T  m1 g = m1 a (Block #1) Adding, 68.0  m1 + m 2 g = m1 + m 2 a
1 2
m1 n1 m1 f1 = k n 1
F
b
g b g 68.0 a= bm + m g  g =
m1 g = 118 N
1.29 m s 2
FIG. P5.45
T = m1 a + m1 g = 27. 2 N
Chapter 5
139
P5.46
(Case 1, impending upward motion) Setting
Fx = 0:
P cos 50.0n = 0 fs , max = s P cos 50.0 = 0.250 0.643 P = 0.161 P
fs , max = sn:
a
f
Setting
Fy = 0:
P sin 50.00.161P  3.00 9.80 = 0 Pmax = 48.6 N
a f
(Case 2, impending downward motion) As in Case 1,
fs, max = 0.161P
Setting
FIG. P5.46
Fy = 0:
*P5.47
P sin 50.0+0.161P  3.00 9.80 = 0 Pmin = 31.7 N
a f
When the sled is sliding uphill
y n x mg sin mg cos FIG. P5.47 f
Fy = ma y : Fx = ma x :
+n  mg cos = 0 f = k n = k mg cos + mg sin + k mg cos = ma up v f = 0 = vi + a up t up vi =  a up t up
1 vi + v f t up 2 1 1 2 x = a up t up + 0 t up = a up t up 2 2 x =
d e
i
j
When the sled is sliding down, the direction of the friction force is reversed: mg sin  k mg cos = ma down x= Now t down = 2t up 1 1 2 a up tup = a down 2t up 2 2 a up = 4a down 1 2 a down t down . 2
e j
2
g sin + k g cos = 4 g sin  k g cos 5 k cos = 3 sin
b
g
k =
FG 3 IJ tan H 5K
140 *P5.48
The Laws of Motion
Since the board is in equilibrium, Fx = 0 and we see that the normal forces must be the same on both sides of the board. Also, if the minimum normal forces (compression forces) are being applied, the board is on the verge of slipping and the friction force on each side is f = fs
n
f
f
n
a f
max
= sn . Fg = 95.5 N FIG. P5.48
The board is also in equilibrium in the vertical direction, so
Fy = 2 f  Fg = 0 , or
f=
Fg 2
.
The minimum compression force needed is then n= *P5.49 (a) f = Fg 2 s = 95.5 N = 72.0 N . 2(0.663) n F 15 f s, max 25 For equilibrium: F cos 15+24.67  0.094F  75 sin 25= 0 . This gives F = 8.05 N . 75 N FIG. P5.49(a) (b) F cos 15( 24.67  0.094F ) 75 sin 25 = 0 . This gives F = 53.2 N . f s, max 25 75 N FIG. P5.49(b) (c) f k = k n = 10.6  0.040 F . Since the velocity is constant, the net force is zero: F cos 15(10.6  0.040 F ) 75 sin 25 = 0 . This gives F = 42.0 N . fk n F 15 25 75 N FIG. P5.49(c) n F 15
s
n + F sin 15 75 N cos 25 = 0 n = 67.97  0.259 F fs , max = s n = 24.67  0.094F
a
f
Chapter 5
141
*P5.50
We must consider separately the disk when it is in contact with the roof and when it has gone over the top into free fall. In the first case, we take x and y as parallel and perpendicular to the surface of the roof:
Fy = ma y :
+n  mg cos = 0 n = mg cos FIG. P5.50
then friction is f k = k n = k mg cos
Fx = ma x :
 f k  mg sin = ma x
a x =  k g cos  g sin = 0.4 cos 37 sin 37 9.8 m s 2 = 9.03 m s 2
a
f
The Frisbee goes ballistic with speed given by
2 2 v xf = v xi + 2 a x x f  xi = 15 m s
d
i b d
g + 2e9.03 m s ja10 m  0f = 44.4 m
2 2
2
s2
v xf = 6.67 m s For the free fall, we take x and y horizontal and vertical:
2 2 v yf = v yi + 2 a y y f  yi
0 = 6.67 m s sin 37 yf
g + 2e9.8 m s jdy b4.01 m sg = 6.84 m = 6.02 m +
2 2 2
b
i
f
 10 m sin 37
i
19.6 m s 2
Additional Problems P5.51 (a) (b) see figure to the right First consider Pat and the chair as the system. Note that two ropes support the system, and T = 250 N in each rope. Applying F = ma 2T  480 = ma , where m = Solving for a gives a= (c) 500  480 = 0.408 m s 2 . 49.0 480 = 49.0 kg . 9.80 FIG. P5.51
F = ma on Pat: F = n + T  320 = ma , where m = 9.80 = 32.7 kg
n = ma + 320  T = 32.7(0.408)+ 320  250 = 83.3 N . 320
142 P5.52
The Laws of Motion
F = ma gives the object's acceleration
a=
F = e8.00 i  4.00tjj N
m 2.00 kg
a = 4.00 m s 2 i  2.00 m s 3 t j = Its velocity is
e
j e
j
dv . dt
vi
z
v
dv = v  v i = v  0 = adt
0
z
t
j e v = e 4.00t m s ji  e1.00t
v=
0 2
ze
t
4.00 m s 2 i  2.00 m s 3 t j dt
2 3
j m s jj.
(a)
We require v = 15.0 m s , v = 225 m 2 s 2 16.0t 2 m 2 s 4 + 1.00t 4 m 2 s 6 = 225 m 2 s 2 1.00t 4 + 16.0 s 2 t 2  225 s 4 = 0 t2 = 16.0
2
a16.0f  4a225f = 9.00 s
2
2
2.00
t = 3.00 s . Take ri = 0 at t = 0. The position is r = vdt =
0
z ze
t t 0
4.00t m s 2 i  1.00t 2 m s 3 j dt
j e
j
r = 4.00 m s 2 at t = 3 s we evaluate. (c) (b) r=
e
j t2 i  e1.00 m s j t3 j
3
2
3
e18.0i  9.00 jj m
2 2
So r = (18.0) +(9.00) m = 20.1 m
Chapter 5
143
*P5.53
(a)
Situation A
y
Fx = ma x : Fy = ma y :
FA + sn  mg sin = 0 +n  mg cos = 0
n FA mg sin fs mg cos FIG. P5.53(a)
x
Eliminate n = mg cos to solve for FA = mg sin  s cos (b) Situation B
a
f.
FB mg sin
y FB cos + s n  mg sin = 0  FB sin + n  mg cos = 0 n fs mg cos FIG. P5.53(b) x
Fx = ma x : Fy = ma y :
Substitute n = mg cos + FB sin to find FB cos + s mg cos + s FB sin  mg sin = 0 FB = (c) mg sin  s cos cos + s sin
a
f
FA = 2 kg 9.8 m s 2 sin 250.16 cos 25 = 5.44 N FB = 19.6 N 0.278 = 5.59 N cos 25+0.16 sin 25
a
a
f
f
Student A need exert less force. (d) FB = FA F = A cos 25+0.38 sin 25 1.07
Student B need exert less force. P5.54 18 N  P = 2 kg a
b g P  Q = b3 kg ga Q = b 4 kg ga
Adding gives 18 N = 9 kg a so
a f
FIG. P5.54
a = 2.00 m s 2 .
(b) Q = 4 kg 2 m s 2 = 8.00 N net force on the 4 kg P  8 N = 3 kg 2 m s 2 = 6.00 N net force on the 3 kg and P = 14 N 18 N  14 N = 2 kg 2 m s 2 = 4.00 N net force on the 2 kg continued on next page
e
e
j
e
j
j
144
The Laws of Motion
(c) (d)
From above, Q = 8.00 N and P = 14.0 N . The 3kg block models the heavy block of wood. The contact force on your back is represented by Q, which is much less than the force F. The difference between F and Q is the net force causing acceleration of the 5kg pair of objects. The acceleration is real and nonzero, but lasts for so short a time that it never is associated with a large velocity. The frame of the building and your legs exert forces, small relative to the hammer blow, to bring the partition, block, and you to rest again over a time large relative to the hammer blow. This problem lends itself to interesting lecture demonstrations. One person can hold a lead brick in one hand while another hits the brick with a hammer. First, we note that F = T1 . Next, we focus on the mass M and write T5 = Mg . Next, we focus on the bottom pulley and write T5 = T2 + T3 . Finally, we focus on the top pulley and write T4 = T1 + T2 + T3 . Since the pulleys are not starting to rotate and are frictionless, T1 = T3 , and T2 = T3 . From this Mg information, we have T5 = 2T2 , soT2 = . 2 Then T1 = T2 = T3 = T5 = Mg . Mg 3 Mg , and T4 = , and 2 2
P5.55
(a)
(b)
Since F = T1 , we have F =
Mg . 2 FIG. P5.55
P5.56
We find the diver's impact speed by analyzing his freefall motion: v 2 = vi2 + 2 ax = 0 + 2 9.80 m s 2 (10.0 m) so v f = 14.0 m s. f Now for the 2.00 s of stopping, we have v f = vi + at : 0 = 14.0 m s + a 2.00 s a = +7.00 m s . Call the force exerted by the water on the diver R. Using
2
c
h
a
f
Fy = ma ,
+ R  70.0 kg 9.80 m s 2 = 70.0 kg 7.00 m s 2 R = 1.18 kN .
e
j
e
j
Chapter 5
145
P5.57
(a)
The crate is in equilibrium, just before it starts to move. Let the normal force acting on it be n and the friction force, fs . Resolving vertically: n = Fg + P sin Horizontally: P cos = fs But, fs sn i.e., P cos s Fg + P sin or P cos  s sin s Fg . Divide by cos : P 1 s tan s Fg sec . Then Pminimum = 0.400(100 N ) sec 1  0.400 tan 0.00 40.0 15.0 46.4 30.0 60.1 45.0 94.3 60.0 260 FIG. P5.57
c
h
a a
f
f
s Fg sec
1  s tan
.
(b)
P=
deg
b g Pa N f
If the angle were 68.2 or more, the expression for P would go to infinity and motion would become impossible.
146 P5.58
The Laws of Motion
(a)
Following the inchapter Example about a block on a frictionless incline, we have a = g sin = 9.80 m s 2 sin 30.0
c
h
a = 4.90 m s 2
(b) The block slides distance x on the incline, with sin 30.0 = 0.500 m x
x = 1.00 m: v 2 = vi2 + 2 a x f  xi = 0 + 2 4.90 m s 2 (1.00 m) f v f = 3.13 m s after time t s = 2x f vf = 2(1.00 m) 3.13 m s = 0.639 s .
c
h
c
h
(c)
Now in free fall y f  yi = v yi t +
1 ayt 2 : 2 2.00 = 3.13 m s sin 30.0 t 
b
g
e4.90 m s jt + b1.56 m sgt  2.00 m = 0
2 2
1 9.80 m s 2 t 2 2
e
j
t= Only one root is physical t = 0.499 s
1.56 m s
b1.56 m sg  4e4.90 m s ja2.00 mf
2 2
9.80 m s 2
x f = v x t = 3.13 m s cos 30.0 0.499 s = 1.35 m (d) (e) total time = t s + t = 0.639 s + 0.499 s = 1.14 s The mass of the block makes no difference.
b
g
a
f
Chapter 5
147
P5.59
With motion impending, n + T sin  mg = 0
f = s mg  T sin and T cos  s mg + sT sin = 0 so T= To minimize T, we maximize cos + s sin
b
g
FIG. P5.59
s mg . cos + s sin
d cos + s sin = 0 =  sin + s cos . d (a)
b
g
= tan1 s = tan1 0.350 = 19.3
0.350 1.30 kg 9.80 m s 2 cos 19.3+0.350 sin 19.3
(b) *P5.60 (a) (b) (c)
T=
a
fc
h=
4.21 N mg = 36.4 kg 9.8 m s 2 = 357 N
See Figure (a) to the right. See Figure (b) to the right. For the pin,
a
fc
h
Fy = ma y :
C cos  357 N = 0 357 N . C= cos FIG. P5.60(a) FIG. P5.60(b)
For the foot,
Fy = ma y :
(d) For the foot with motion impending,
+n B  C cos = 0 n B = 357 N .
Fx = ma x :
+ f s  C sin s = 0
sn B = C sin s 357 N cos s sin s C sin s s = = = tan s . nB 357 N
b
g
(e)
The maximum coefficient is
s = tan s = tan 50.2 = 1.20 .
148 P5.61
The Laws of Motion
F = ma
For m1 : For m 2 : Eliminating T, a= For all 3 blocks: m2 g m1 FIG. P5.61 F = M + m1 + m 2 a = T = m1 a T  m2 g = 0
a
f
aM + m + m fFGH m g IJK m
1 2 2 1
P5.62 0
ts 1.02 1.53 2.01 2.64 3.30 3.75
af
0
t2 s2 1.04 0 2.34 1 4.04 0 6.97 0 10.89 14.06
e j
xm 0 0.100 0.200 0.350 0.500 0.750 1.00
a f
FIG. P5.62 From x = 1 2 1 at the slope of a graph of x versus t 2 is a , and 2 2
a = 2 slope = 2 0.071 4 m s 2 = 0.143 m s 2 .
From a = g sin , a = 9.80 m s 2
e
j
FG 1.77 4 IJ = 0.137 m s H 127.1 K
2
, different by 4%.
The difference is accounted for by the uncertainty in the data, which we may estimate from the third point as 0.350  0.071 4 4.04 0.350
b
ga f = 18%.
Chapter 5
149
P5.63
(1)
m1 a  A = T a =
a
f
T +A m1 T M
(2) (3) (a)
MA = R x = T A =
m2 a = m2 g  T T = m2 g  a
b
g
T = m2 g 
FIG. P5.63
Substitute the value for a from (1) into (3) and solve for T:
LM F T + AI OP . N GH m JK Q
1
Substitute for A from (2): T = m2 g  (b)
LM F T + T I OP = N GH m M JK Q
1
m2 g
LM m M N m M + m am
1 1 2
1
OP + Mf Q
.
Solve (3) for a and substitute value of T: a= m 2 g m1 + M
a
m1 M + m 2 M + m1
a
f
f
.
(c)
From (2), A =
T , Substitute the value of T: M A= m1 m 2 g m1 M + m 2 m1 + M
a
f
.
(d)
a A =
Mm 2 g m1 M + m 2 m1 + M
a
f
150 P5.64
The Laws of Motion
(a), (b) Motion impending n = 49.0 N f s1 P f s1 f s2 Fg = 49.0 N fs1 = n = 14.7 N 196 N 147 N n = 49.0 N
5.00 kg
15.0 kg
fs2 = 0.500 196 N = 98.0 N
FIG. P5.64
a
f
P = f s1 + f s 2 = 14.7 N + 98.0 N = 113 N (c) Once motion starts, kinetic friction acts. 112.7 N  0.100 49.0 N  0.400 196 N = 15.0 kg a 2 0.100 49.0 N = 5.00 kg a1 a1 = 0.980 m s 2 *P5.65 (a)
2 Let x represent the position of the glider along the air track. Then z 2 = x 2 + h0 , dx 1 2 dz dz 2 12 2 1 2 is the rate at which string passes = z  h0 2z x = z 2  h0 , vx = . Now dt 2 dt dt over the pulley, so it is equal to v y of the counterweight.
a
f
a
f b
g
a
a 2 = 1.96 m s 2
f b
g
e
j
e
j a f c
2 v x = z z 2  h0
h
1 2
v y = uv y
(b) (c)
ax =
dv y dv x du d at release from rest, v y = 0 and a x = ua y . = uv y = u + vy dt dt dt dt
80.0 cm 2 , z = 1.60 m , u = z 2  h0 z For the counterweight sin 30.0 =
e
j
1 2
z = 1.6 2  0.8 2
e
j a1.6f = 1.15 .
1 2
Fy = ma y :
For the glider
T  0.5 kg 9.8 m s 2 = 0.5 kga y a y = 2T + 9.8
Fx = ma x :
T cos 30 = 1.00 kg a x = 1.15 a y = 1.15 2T + 9.8 = 2.31T + 11.3 N 3.18T = 11.3 N T = 3.56 N
a
f
Chapter 5
151
*P5.66
The upward acceleration of the rod is described by y f = yi + v yi t + 1 10 3 m = 0 + 0 + 1 ayt 2 2
1 a y 8 10 3 s 2 a y = 31.2 m s 2
e
j
2
The distance y moved by the rod and the distance x moved by the wedge in the same time are related y y . Then their speeds and by tan 15 = x = x tan 15 accelerations are related by dy dx 1 = dt tan 15 dt and d2x dt
2
FIG. P5.66
=
d2y 1 1 = 31.2 m s 2 = 117 m s 2 . 2 tan 15 dt tan 15
FG H
IJ K
The free body diagram for the rod is shown. Here H and H are forces exerted by the guide.
Fy = ma y :
n cos 15 mg = ma y
n cos 150.250 kg 9.8 m s 2 = 0.250 kg 31.2 m s 2 10.3 N = 10.6 N n= cos 15 For the wedge,
e
j
e
j
Fx = Ma x :
*P5.67 (a)
n sin 15+ F = 0.5 kg 117 m s 2
F = 10.6 N sin 15+58.3 N = 61.1 N Consider forces on the midpoint of the rope. It is nearly in equilibrium just before the car begins to move. Take the yaxis in the direction of the force you exert:
a
e
f
j
Fy = ma y :
T sin + f  T sin = 0 T= f . 2 sin FIG. P5.67
(b)
T=
100 N = 410 N 2 sin 7
152 P5.68
The Laws of Motion
Since it has a larger mass, we expect the 8.00kg block to move down the plane. The acceleration for both blocks should have the same magnitude since they are joined together by a nonstretching string. Define up the left hand plane as positive for the 3.50kg object and down the right hand plane as positive for the 8.00kg object.
F1 = m1 a1 : F2 = m 2 a 2 :
and
 m1 g sin 35.0+T = m1 a m 2 g sin 35.0T = m 2 a
FIG. P5.68
 3.50 9.80 sin 35.0+T = 3.50 a
a fa f a8.00fa9.80f sin 35.0T = 8.00a .
Adding, we obtain +45.0 N  19.7 N = 11.5 kg a . (b) Thus the acceleration is
a
f
a = 2.20 m s 2 .
By substitution, 19.7 N + T = 3.50 kg 2.20 m s 2 = 7.70 N . (a) The tension is T = 27.4 N . P5.69 Choose the xaxis pointing down the slope. v f = vi + at: 30.0 m s = 0 + a 6.00 s
2
a
fc
h
a = 5.00 m s 2
a
f
a = 5.00 m s . Consider forces on the toy.
Fx = ma x :
mg sin = m 5.00 m s 2
e
j fa f
FIG. P5.69
= 30.7 Fy = ma y :  mg cos + T = 0 T = mg cos = 0.100 9.80 cos 30.7 T = 0.843 N
a
Chapter 5
153
*P5.70
Throughout its up and down motion after release the block has
Fy = ma y :
+n  mg cos = 0 n = mg cos .
Let R = R x i + R y j represent the force of table on incline. We have
Fx = ma x :
+ R x  n sin = 0 R x = mg cos sin Fy = ma y :  Mg  n cos + Ry = 0
R y = Mg + mg cos 2 .
R = mg cos sin to the right + M + m cos 2 g upward *P5.71 Take +x in the direction of motion of the tablecloth. For the mug:
e
j
FIG. P5.70
Fx = ma x
0.1 N = 0.2 kg a x
a x = 0.5 m s 2 .
Relative to the tablecloth, the acceleration of the mug is 0.5 m s 2  3 m s 2 = 2.5 m s 2 . The mug reaches the edge of the tablecloth after time given by x = v xi t + 0.3 m = 0 + 1 axt 2 2
1 2.5 m s 2 t 2 2 t = 0.490 s .
e
j
The motion of the mug relative to tabletop is over distance 1 1 a x t 2 = 0.5 m s 2 0.490 s 2 2
e
ja
f
2
= 0.060 0 m .
The tablecloth slides 36 cm over the table in this process.
154 P5.72
The Laws of Motion
Fy = ma y : n  mg cos = 0
or
a f n = a82.3 N f cos
Fx = ma x : mg sin = ma
or a = g sin
n = 8.40 9.80 cos
a = 9.80 m s 2 sin
e
j
, deg n , N
0.00 5.00 10.0 15.0 20.0 25.0 30.0 35.0 40.0 45.0 50.0 55.0 60.0 65.0 70.0 75.0 80.0 85.0 90.0 82.3 82.0 81.1 79.5 77.4 74.6 71.3 67.4 63.1 58.2 52.9 47.2 41.2 34.8 28.2 21.3 14.3 7.17 0.00
a, m s 2 0.00 0.854 1.70 2.54 3.35 4.14 4.90 5.62 6.30 6.93 7.51 8.03 8.49 8.88 9.21 9.47 9.65 9.76 9.80 FIG. P5.72
At 0, the normal force is the full weight and the acceleration is zero. At 90, the mass is in free fall next to the vertical incline.
Chapter 5
155
P5.73
(a)
Apply Newton's second law to two points where butterflies are attached on either half of mobile (other half the same, by symmetry) (1) (2) (3) (4) T2 cos 2  T1 cos 1 = 0 T1 sin 1  T2 sin 2  mg = 0 T2 cos 2  T3 = 0 T2 sin 2  mg = 0
Substituting (4) into (2) for T2 sin 2 , T1 sin 1  mg  mg = 0 . Then T1 = Substitute (3) into (1) for T2 cos 2 : T3  T1 cos 1 = 0 , T3 = T1 cos 1 Substitute value of T1 : T3 = 2mg From Equation (4), T2 = (b) Divide (4) by (3): mg T2 sin 2 . = T2 cos 2 T3 Substitute value of T3 : tan 2 = mg tan 1 tan 1 , 2 = tan1 2 2mg mg . sin 2 2mg cos 1 = = T3 . sin 1 tan 1 2mg . sin 1 FIG. P5.69
FG H
IJ K
.
Then we can finish answering part (a): T2 = (c) mg sin tan1 1 2
b
tan 1
g
.
D is the horizontal distance between the points at which the two ends of the string are attached to the ceiling. D = 2 cos 1 + 2 cos 2 + and L = 5 D= 1 L 2 cos 1 + 2 cos tan1 tan 1 2 5
R S T
LM N
FG H
IJ OP + 1U KQ V W
156
The Laws of Motion
ANSWERS TO EVEN PROBLEMS
P5.2 P5.4 P5.6 1.66 10 6 N forward Fg v vt (a) ; (b) i + Fg j gt 2 (a) 4.47 10 15 m s 2 away from the wall; (b) 2.09 10 10 N toward the wall (a) 534 N down; (b) 54.5 kg 2.55 N for an 88.7 kg person P5.42 P5.44 152 ft (a) 2.31 m s 2 down for m1 , left for m 2 and up for m 3 ; (b) 30.0 N and 24.2 N Any value between 31.7 N and 48.6 N 72.0 N 6.84 m (a) 3.00 s; (b) 20.1 m; (c) 18.0 i  9.00 j m (a) 2.00 m s 2 to the right; (b) 8.00 N right on 4 kg; 6.00 N right on 3 kg; 4 N right on 2 kg; (c) 8.00 N between 4 kg and 3 kg; 14.0 N between 2 kg and 3 kg; (d) see the solution 1.18 kN (a) 4.90 m s 2 ; (b) 3.13 m s at 30.0 below the horizontal; (c) 1.35 m; (d) 1.14 s; (e) No (a) and (b) see the solution; (c) 357 N; (d) see the solution; (e) 1.20 see the solution; 0.143 m s 2 agrees with 0.137 m s 2 P5.64 (a) see the solution; (b) on block one: 49.0 N j  49.0 N j + 14.7 N i ; on block two: 49.0 N j  14.7 N i  147 N j +196 N j  98.0 N i + 113 N i ; (c) for block one: 0.980 i m s 2 ; for block two: 1.96 m s 2 i P5.66 P5.68 P5.70 61.1 N (a) 2.20 m s 2 ; (b) 27.4 N mg cos sin to the right + M + m cos 2 g upward P5.72 see the solution
F I GH JK
P5.46 P5.48 P5.50 P5.52 P5.54
P5.8 P5.10 P5.12 P5.14
e16.3i + 14.6 jj N
(a) 181; (b) 11.2 kg; (c) 37.5 m s ; (d) 37.5 i  0.893 j m s
e
j
e
j
P5.16 P5.18 P5.20 P5.22
112 N T1 = 296 N ; T2 = 163 N ; T3 = 325 N (a) see the solution; (b) 1.79 N (a) 2.54 m s 2 down the incline; (b) 3.18 m s see the solution; 6.30 m s 2 ; 31.5 N (a) 3.57 m s 2 ; (b) 26.7 N; (c) 7.14 m s (a) 36.8 N; (b) 2.45 m s 2 ; (c) 1.23 m (a) 0.529 m; (b) 7.40 m s upward (a) 2.22 m; (b) 8.74 m s (a) a1 = 2 a 2 ; m1 m 2 g m m g ; T2 = 1 2 ; (b) T1 = m2 m 2m1 + 2 m1 + 42 m2 g m2 g ; a2 = (c) a1 = m2 4m1 + m 2 2m1 + 2 P5.56 P5.58 P5.60 P5.62
P5.24 P5.26 P5.28 P5.30 P5.32 P5.34
P5.36 P5.38 P5.40
s = 0.306 ; k = 0.245
(a) 3.34; (b) Time would increase (a) 55.2; (b) 167 N
e
j
6
Circular Motion and Other Applications of Newton's Laws
CHAPTER OUTLINE
6.1 Newton's Second Law Applied to Uniform Circular Motion Nonuniform Circular Motion Motion in Accelerated Frames Motion in the Presence of Resistive Forces Numerical Modeling in Particle Dynamics
ANSWERS TO QUESTIONS
Q6.1 Mud flies off a rapidly spinning tire because the resultant force is not sufficient to keep it moving in a circular path. In this case, the force that plays a major role is the adhesion between the mud and the tire. The spring will stretch. In order for the object to move in a circle, the force exerted on the object by the spring must have a mv 2 . Newton's third law says that the force exerted on size of r the object by the spring has the same size as the force exerted by the object on the spring. It is the force exerted on the spring that causes the spring to stretch. Driving in a circle at a constant speed requires a centripetal acceleration but no tangential acceleration.
6.2 6.3 6.4 6.5
Q6.2
Q6.3 Q6.4 (a) (b) Q6.5 Q6.6
The object will move in a circle at a constant speed. The object will move in a straight line at a changing speed.
The speed changes. The tangential force component causes tangential acceleration. Consider the force required to keep a rock in the Earth's crust moving in a circle. The size of the force is proportional to the radius of the circle. If that rock is at the Equator, the radius of the circle through which it moves is about 6400 km. If the rock is at the north pole, the radius of the circle through which it moves is zero! Consider standing on a bathroom scale. The resultant force on you is your actual weight minus the normal force. The scale reading shows the size of the normal force, and is your `apparent weight.' If you are at the North or South Pole, it can be precisely equal to your actual weight. If you are at the equator, your apparent weight must be less, so that the resultant force on you can be a downward force large enough to cause your centripetal acceleration as the Earth rotates. A torque is exerted by the thrust force of the water times the distance between the nozzles.
Q6.7
Q6.8
157
158 Q6.9
Circular Motion and Other Applications of Newton's Laws
I would not accept that statement for two reasons. First, to be "beyond the pull of gravity," one would have to be infinitely 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. This is the same principle as the centrifuge. All the material inside the cylinder tends to move along a straightline path, but the walls of the cylinder exert an inward force to keep everything moving around in a circular path. The ball would not behave as it would when dropped on the Earth. As the astronaut holds the ball, she and the ball are moving with the same angular velocity. The ball, however, being closer to the center of rotation, is moving with a slower tangential velocity. Once the ball is released, it acts according to Newton's first law, and simply drifts with constant velocity in the original direction of its velocity when releasedit is no longer "attached" to the rotating space station. Since the ball follows a straight line and the astronaut follows a circular path, it will appear to the astronaut that the ball will "fall to the floor". But other dramatic effects will occur. Imagine that the ball is held so high that it is just slightly away from the center of rotation. Then, as the ball is released, it will move very slowly along a straight line. Thus, the astronaut may make several full rotations around the circular path before the ball strikes the floor. This will result in three obvious variations with the Earth drop. First, the time to fall will be much larger than that on the Earth, even though the feet of the astronaut are pressed into the floor with a force that suggests the same force of gravity as on Earth. Second, the ball may actually appear to bob up and down if several rotations are made while it "falls". As the ball moves in a straight line while the astronaut rotates, sometimes she is on the side of the circle on which the ball is moving toward her and other times she is on the other side, where the ball is moving away from her. The third effect is that the ball will not drop straight down to her feet. In the extreme case we have been imagining, it may actually strike the surface while she is on the opposite side, so it looks like it ended up "falling up". In the least extreme case, in which only a portion of a rotation is made before the ball strikes the surface, the ball will appear to move backward relative to the astronaut as it falls. The water has inertia. The water tends to move along a straight line, but the bucket pulls it in and around in a circle. There is no such force. If the passenger slides outward across the slippery car seat, it is because the passenger is moving forward in a straight line while the car is turning under him. If the passenger pushes hard against the outside door, the door is exerting an inward force on him. No object is exerting an outward force on him, but he should still buckle his seatbelt. Blood pressure cannot supply the force necessary both to balance the gravitational force and to provide the centripetal acceleration, to keep blood flowing up to the pilot's brain. 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 When you are not accelerating, the normal force and your weight are equal in size. Your body interprets the force of the floor pushing up on you as your weight. When you accelerate in an elevator, this normal force changes so that you accelerate with the elevator. In free fall, you are never weightless since the Earth's gravity and your mass do not change. It is the normal forceyour apparent weightthat is zero.
Q6.10
Q6.11
Q6.12 Q6.13
Q6.14 Q6.15
Q6.16
Chapter 6
159
Q6.17
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: m dv y dt = mg  D A 2 vy 2
where D is the coefficient of drag of the parachutist, and A is the projected area of the parachutist's body. At terminal speed, ay = dv y dt
F 2mg I = 0 and V G H D A JK
T
12
.
When the parachute opens, the coefficient 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 m dv y dt = mg  D A 2 L A 2 vy  vx 2 2
where v y is the vertical velocity, and v x 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. Q6.18 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. Lower air density reduces air resistance, so a tanktruckload of fuel takes you farther. Suppose the rock is moving rapidly when it enters the water. The speed of the rock decreases until it reaches terminal velocity. The acceleration, which is upward, decreases to zero as the rock approaches terminal velocity. 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.
Q6.19 Q6.20
Q6.21
160
Circular Motion and Other Applications of Newton's Laws
SOLUTIONS TO PROBLEMS
Section 6.1 P6.1 Newton's Second Law Applied to Uniform Circular Motion
m = 3.00 kg , r = 0.800 m. The string will break if the tension exceeds the weight corresponding to 25.0 kg, so
Tmax = Mg = 25.0 9.80 = 245 N .
When the 3.00 kg mass rotates in a horizontal circle, the tension causes the centripetal acceleration, so T= 3.00 v 2 mv 2 = . r 0.800 0.800 T 0.800 Tmax 0.800 245 rT = = = 65.3 m 2 s 2 m 3.00 3.00 3.00
a f
a f
Then and or P6.2 In
v2 =
a
f a
f
a f
0 v 65.3 0 v 8.08 m s . FIG. P6.1
v2 , both m and r are unknown but remain constant. Therefore, F is proportional to v 2 r 2 18.0 and increases by a factor of as v increases from 14.0 m/s to 18.0 m/s. The total force at the 14.0 higher speed is then
F = m
FG IJ H K
Ffast = G 14.0 J H K
Symbolically, write
F 18.0 I a130 Nf = 215 N .
2
Fslow = G r J b14.0 H K
=
F mI
2
ms
g
2
and
Ffast = G r J b18.0 H K
F mI
ms .
g
2
Dividing gives
Ffast Fslow
FG 18.0 IJ H 14.0 K
, or
Ffast = G 14.0 J Fslow = G 14.0 J H K H K
F 18.0 I
2
F 18.0 I a130 Nf =
2
215 N .
This force must be horizontally inward to produce the driver's centripetal acceleration.
Chapter 6
161
P6.3
(a)
9.11 0 31 kg 2.20 10 6 m s mv 2 = F= r 0.530 10 10 m
e
je
2
j
2
= 8.32 10 8 N inward
(b)
2.20 10 6 m s v2 = = 9.13 10 22 m s 2 inward a= r 0.530 10 10 m mv 2 r
e
j
P6.4
Neglecting relativistic effects. F = ma c =
F = 2 1.661 10 P6.5 (a) (b) static friction ma i = f i + nj + mg  j
e
27
e2.998 10 m sj kg j a0.480 mf
7
2
= 6.22 10 12 N
e j
b g
Fy = 0 = n  mg
v2 = f = n = mg . r 2 50.0 cm s v2 = = 0.085 0 . Then = rg 30.0 cm 980 cm s 2 thus n = mg and
Fr = m
a
fe
j
P6.6
(a)
Fy = ma y , mg moon down =
v = g moon r =
2
e1.52 m s je1.7 10
e j
mv 2 down r
6
m + 100 10 3 m = 1.65 10 3 m s
j
(b) P6.7
2 1.8 10 6 m 2r v= ,T= = 6.84 10 3 s = 1.90 h T 1.65 10 3 m s
n = mg since a y = 0 The force causing the centripetal acceleration is the frictional force f. From Newton's second law f = ma c = But the friction condition is f sn i.e., mv 2 s mg r mv 2 . r FIG. P6.7
v s rg = 0.600 35.0 m 9.80 m s 2
a
fe
j
v 14.3 m s
162 P6.8 P6.9
Circular Motion and Other Applications of Newton's Laws
a=
v = r
2
b86.5 km hge
b
1h 3 600 s
je
1 000 m 1 km
61.0 m
j F 1g I= GH 9.80 m s JK
2
2
0.966 g
T cos 5.00 = mg = 80.0 kg 9.80 m s 2 (a) (b) T = 787 N : T =
ge
j
a68.6 Nfi + a784 Nf j
T sin 5.00 = ma c : a c = 0.857 m s 2 toward the center of the circle. The length of the wire is unnecessary information. We could, on the other hand, use it to find the radius of the circle, the speed of the bob, and the period of the motion.
FIG. P6.9
P6.10
(b)
v=
235 m = 6.53 m s 36.0 s 1 2r = 235 m 4 r = 150 m
The radius is given by
(a)
ar =
F v I toward center GH r JK b6.53 m sg at 35.0 north of west =
2 2
= 0.285 m s 2 cos 35.0  i + sin 35.0 j = 0.233 m s 2 i + 0.163 m s 2 j
e
150 m
je
e j
j
(c)
a= =
dv
f
 vi t
i
36.0 s
e6.53 m s j  6.53 m s ij
= 0.181 m s 2 i + 0.181 m s 2 j
Chapter 6
163
*P6.11
Fg = mg = 4 kg 9.8 m s 2 = 39.2 N 1.5 m 2m = 48.6 r = 2 m cos 48.6 = 1.32 m sin =
b ge
j
Ta
39.2 N Tb ac
2
a f
mv 2 Fx = ma x = r Ta cos 48.6+Tb cos 48.6 = Ta + Tb =
forces v motion FIG. P6.11
b gb
4 kg 6 m s 1.32 m
g
109 N = 165 N cos 48.6
Fy = ma y
+Ta sin 48.6Tb sin 48.639.2 N = 0 Ta  Tb = (a) 39. 2 N = 52.3 N sin 48.6 To solve simultaneously, we add the equations in Ta and Tb : Ta + Tb + Ta  Tb = 165 N + 52.3 N Ta = (b) *P6.12 ac = 217 N = 108 N 2
Tb = 165 N  Ta = 165 N  108 N = 56.2 N
v2 . Let f represent the rotation rate. Each revolution carries each bit of metal through distance r 2r , so v = 2rf and ac = v2 = 4 2 rf 2 = 100 g . r
A smaller radius implies smaller acceleration. To meet the criterion for each bit of metal we consider the minimum radius:
F 100 g IJ f =G H 4 r K
2
12
F 100 9.8 m s I =G H 4 a0.021 mf JK
2 2
12
= 34.4
1 60 s = 2.06 10 3 rev min . s 1 min
FG H
IJ K
164
Circular Motion and Other Applications of Newton's Laws
Section 6.2 P6.13
Nonuniform Circular Motion T T Mg
M = 40.0 kg , R = 3.00 m, T = 350 N (a)
F = 2T  Mg =
v 2 = 2T  Mg v2
gFGH IJK F 3.00 IJ = 23.1 em s j = 700  a 40.0fa9.80f G H 40.0 K b
2 2
Mv 2 R R M
v = 4.81 m s (b) n  Mg = F = n = Mg + P6.14 (a) Mv 2 R
Mg child + seat FIG. P6.13(a)
n child alone FIG. P6.13(b)
Mv 2 23.1 = 40.0 9.80 + = 700 N R 3.00
FG H
IJ K
Consider the forces acting on the system consisting of the child and the seat:
Fy = ma y 2T  mg = m
v2 = R v= (b)
FG 2T  gIJ Hm K F 2T  gIJ RG Hm K
F
v2 R
Consider the forces acting on the child alone:
Fy = ma y n = mG g + H
and from above, v 2 = R
v2 R
I JK
FG 2T  gIJ , so Hm K
n=m g+
FG H
2T  g = 2T . m
IJ K
P6.15
Let the tension at the lowest point be T.
F = ma:
I JK L b8.00 m sg T = b85.0 kg gM9.80 m s + 10.0 m MN
T=m g+ v2 r
2
T  mg = ma c =
F GH
mv 2 r
2
OP PQ = 1.38 kN > 1 000 N
FIG. P6.15
He doesn't make it across the river because the vine breaks.
Chapter 6
165
P6.16
(a) (b)
4.00 m s v2 = ac = r 12.0 m
2 a = a c + a t2
b
g
2
= 1.33 m s 2
a=
a1.33f + a1.20f
2
2
= 1.79 m s 2
at an angle = tan 1 mv 2 = mg + n r
FG a IJ = Ha K
c t
48.0 inward
FIG. P6.16
P6.17
Fy =
But n = 0 at this minimum speed condition, so mv 2 = mg v = gr = r P6.18
e9.80 m s ja1.00 mf =
2
3.13 m s .
FIG. P6.17
At the top of the vertical circle, T=m v2  mg R
or T = 0.400 P6.19 (a)
a
f a4..00f  a0.400fa9.80f = 0 500
2
8.88 N C 10 m A FIG. P6.19 B 15 m
v = 20.0 m s, n = force of track on roller coaster, and R = 10.0 m . Mv 2 F = R = n  Mg From this we find n = Mg +
500 kg 20.0 m s 2 Mv 2 = 500 kg 9.80 m s 2 + R 10.0 m
b
ge
j
b
ge
j
n = 4 900 N + 20 000 N = 2.49 10 4 N (b) At B, n  Mg =  Mv 2 R n=0  Mg = 
2 Mv max v max = Rg = 15.0 9.80 = 12.1 m s R
The max speed at B corresponds to
a f
166 P6.20
Circular Motion and Other Applications of Newton's Laws
(a)
v2 ac = r
13.0 m s v2 = = 8.62 m r= a c 2 9.80 m s 2
b e
g
2
j
(b)
Let n be the force exerted by the rail. Newton's law gives Mv 2 Mg + n = r v2 n=M  g = M 2 g  g = Mg , downward r FIG. P6.20
F GH
I JK
b
g
(c)
ac =
v2 r
ac =
b13.0 m sg
20.0 m
2
= 8.45 m s 2
If the force exerted by the rail is n1 then Mv 2 = Ma c r n1 = M a c  g which is < 0 since a c = 8.45 m s 2 n1 + Mg =
b
g
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 c > g . We need v2 > g or v > rg = r
a20.0 mfe9.80 m s j , v > 14.0 m s .
2
Section 6.3 P6.21 (a)
Motion in Accelerated Frames
Fx = Ma , a = M = 5.00 kg =
to the right.
T
18.0 N
3.60 m s 2
(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 fictitious force (Ma). Someone at rest outside the car (inertial observer) claims that T is the only force on M in the xdirection.
5.00 kg
FIG. P6.21
Chapter 6
167
*P6.22
We adopt the view of an inertial observer. If it is on the verge of sliding, the cup is moving on a circle with its centripetal acceleration caused by friction.
n f mg FIG. P6.22
Fy = ma y : Fx = ma x :
+n  mg = 0 f= mv 2 = sn = s mg r
v = s gr = 0.8 9.8 m s 2 30 m = 15.3 m s
e
ja
f
If you go too fast the cup will begin sliding straight across the dashboard to the left. P6.23 The only forces acting on the suspended object are the force of gravity mg and the force of tension T, as shown in the freebody diagram. Applying Newton's second law in the x and y directions,
T cos T sin mg FIG. P6.23
Fx = T sin = ma Fy = T cos  mg = 0
or (a) T cos = mg
(1)
(2)
Dividing equation (1) by (2) gives tan = Solving for , = 17.0 a 3.00 m s 2 = = 0.306 . g 9.80 m s 2
(b)
From Equation (1), T= 0.500 kg 3.00 m s 2 ma = = 5.12 N . sin sin 17.0
a
fc a f f
h
*P6.24
The water moves at speed v= 2r 2 0.12 m = = 0.104 m s . T 7.25 s
a
The top layer of water feels a downward force of gravity mg and an outward fictitious force in the turntable frame of reference, mv 2 m 0.104 m s = r 0.12 m
b
g
2
= m9.01 10 2 m s 2 .
It behaves as if it were stationary in a gravity field pointing downward and outward at tan 1 0.090 1 m s 2 9.8 m s 2 = 0.527 .
Its surface slopes upward toward the outside, making this angle with the horizontal.
168 P6.25
Circular Motion and Other Applications of Newton's Laws
Fmax = Fg + ma = 591 N Fmin = Fg  ma = 391 N (a) (b) (c) Adding, 2 Fg = 982 N, Fg = 491 N Since Fg = mg , m = 491 N = 50.1 kg 9.80 m s 2
Subtracting the above equations, 2ma = 200 N
a = 2.00 m s 2
P6.26
(a)
Fr = mar
mg = g= T= 4 2 R T2
mv 2 m 2R = R R T
FG H
IJ K
2
4 2 R 6.37 10 6 m = 2 = 5.07 10 3 s = 1.41 h g 9.80 m s 2 v new T 2R Tcurrent 24.0 h = = current = = 17.1 2R 1.41 h v current Tnew Tnew
(b) *P6.27
speed increase factor =
FG H
IJ K
The car moves to the right with acceleration a. We find the acceleration of a b of the block relative to the Earth. The block moves to the right also.
Fy = ma y : Fx = ma x :
+n  mg = 0 , n = mg , f = k mg + k mg = ma b , a b = k g
The acceleration of the block relative to the car is a b  a = k g  a . In this frame the block starts from rest and undergoes displacement  and gains speed according to
2 2 v xf = v xi + 2 a x x f  xi 2 v xf
= 0 + 2 kg  a   0 = 2 a  kg .
b
d
ga
i
f b
g
(a)
v = 2 a  kg
d b
gi
12
to the left
continued on next page
Chapter 6
169
(b)
The time for which the box slides is given by x = 1 v xi + v xf t 2 1  = 0  2 a  kg 2
12
d i LM d b gi OQPt N F 2 I . t=G H a  g JK
12 k
The car in the Earth frame acquires finals speed v xf = v xi of the box in the Earth frame is then v be = v bc + v ce =  2 a  k g
12 k
F 2 I + at = 0 + aG H a  g JK
k
12
. The speed
g a 2 f b a  g g + a 2 f = ba  g g
12 k
b
12
+a =
12
a
F 2 I GH a  g JK ga2 f ba  g g
k k k
12
12 12
=
2 a  kg
b
k g2
g
g
12
=
2 k g . v
*P6.28
Consider forces on the backpack as it slides in the Earth frame of reference.
Fy = ma y : Fx = ma x :
+n  mg = ma , n = m g + a , f k = k m g + a  k m g + a = ma x
b
b
g
b
g
The motion across the floor is described by L = vt + We solve for k : vt  L = P6.29
2 vt  L 1 = k . k g + a t2 , 2 g + a t2
b
g
b
a
g
f
1 1 a x t 2 = vt  k g + a t 2 . 2 2
b
g
In an inertial reference frame, the girl is accelerating horizontally inward at 5.70 m s v2 = r 2.40 m
b
g
2
= 13.5 m s 2
In her own noninertial frame, her head feels a horizontally outward fictitious force equal to its mass times this acceleration. Together this force and the weight of her head add to have a magnitude equal to the mass of her head times an acceleration of g2 +
Fv I GH r JK
2
2
=
a9.80f + a13.5f
2
2
m s 2 = 16.7 m s 2
This is larger than g by a factor of
16.7 = 1.71 . 9.80 Thus, the force required to lift her head is larger by this factor, or the required force is F = 1.71 55.0 N = 93.8 N .
a
f
170 *P6.30
Circular Motion and Other Applications of Newton's Laws
(a)
The chunk is at radius r =
0.137 m + 0.080 m = 0.054 2 m . Its speed is 4 v= 20 000 2r = 2 0.054 2 m = 114 m s 60 s T
b
g
and its acceleration ac = 114 m s v2 = = 2.38 10 5 m s 2 horizontally inward r 0.054 2 m
b
g
2
= 2.38 10 5 m s 2 (b)
F g I= GH 9.8 m s JK
2
2.43 10 4 g . n mv 2 r FIG. P6.30(b) a
In the frame of the turning cone, the chunk feels a mv 2 . In this frame its horizontally outward force of r 3.3 cm acceleration is up along the cone, at tan 1 a13 .7  8 f cm = 49.2 . Take the y axis perpendicular to the cone:
2
f
49.2
Fy = ma y :
+n 
n = 2 10 3 kg 2.38 10 5 m s 2 sin 49.2 = 360 N (c)
e
je
mv 2 sin 49.2 = 0 r
j
f = k n = 0.6 360 N = 216 N
a
Fx = ma x :
e2 10
P6.31 ar =
2 e
3
kg 2.38 10 5 m s 2 cos 49.2216 N = 2 10 3 kg a x
je
mv cos 49. 2 f = ma x r
2
f
j
e
j
a x = 47.5 10 4 m s 2 radially up the wall of the cone
F 4 R I cos 35.0 = 0.027 6 m s GH T JK We take the y axis along the local vertical. ba g = 9.80  ba g = 9.78 m s ba g = 0.015 8 m s
2 2 net y net x r y 2 2
N
(exaggerated size)
35.0 a r
= arctan
ax = 0.092 8 ay
g0 a net 35.0
Equator
FIG. P6.31
Chapter 6
171
Section 6.4 P6.32
Motion in the Presence of Resistive Forces
2 DAvT DA mg = 2 = 0.314 kg m 2 2 vT
m = 80.0 kg , vT = 50.0 m s , mg = (a) At v = 30.0 m s a=g (b)
DAv 2 2
m
= 9.80 
a0.314fa30.0f
80.0
2
= 6. 27 m s 2 downward
At v = 50.0 m s , terminal velocity has been reached. Fy = 0 = mg  R R = mg = 80.0 kg 9.80 m s 2 = 784 N directed up
b
ge
j
(c)
At
v = 30.0 m s DAv 2 = 0.314 30.0 2
a
fa f
2
= 283 N upward
P6.33
(a)
a = g  bv When v = vT , a = 0 and g = bvT b= g vT
The Styrofoam falls 1.50 m at constant speed vT in 5.00 s. Thus, Then (b) (c) P6.34 (a) At t = 0 , v = 0 and vT = b= y 1.50 m = = 0.300 m s 5.00 s t
9.80 m s 2 = 32.7 s 1 0.300 m s
a = g = 9.80 m s 2 down
When v = 0.150 m s, a = g  bv = 9.80 m s 2  32.7 s 1 0.150 m s = 4.90 m s 2
e
jb
g
down
=
m 1 2 , A = 0.020 1 m 2 , R = air ADvT = mg V 2 m = beadV = 0.830 g cm 3
LM 4 a8.00 cmf OP = 1.78 kg N3 Q
3
Assuming a drag coefficient of D = 0.500 for this spherical object, and taking the density of air at 20C from the endpapers, we have
vT =
b ge j = 0.500e1.20 kg m je0.020 1 m j
2 1.78 kg 9.80 m s 2
3 2
53.8 m s
(b)
v2 f
=
vi2
b53.8 m sg = = + 2 gh = 0 + 2 gh : h = 2 g 2e9.80 m s j
v2 f
2 2
148 m
172 P6.35
Circular Motion and Other Applications of Newton's Laws
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 + bv . The mass of the copper ball is m= 4r 3 4 = 8.92 10 3 kg m3 2.00 10 2 m 3 3
FG IJ e HK a
je
j
3
= 0.299 kg .
The applied force is then F = mg + bv = 0.299 9.80 + 0.950 9.00 10 2 = 3.01 N . P6.36
fa f a
fe
j
Fy = ma y
+T cos 40.0 mg = 0 T= cos 40.0 Fx = ma x
b620 kgge9.80 m s j = 7.93 10
2
3
N
 R + T sin 40.0 = 0 R = 7.93 10 3 N sin 40.0 = 5.10 10 3 N = 2 5.10 10 3 N
e
j
D=
2R = Av 2 1. 20 kg m 2 3.80 m 2 40.0 m s
e
e
je
jFH
kg m s 2 N
jb
IK
1 DAv 2 2
FIG. P6.36
g
2
= 1. 40
P6.37
(a)
At terminal velocity,
R = vT b = mg b = 3.00 10 3 kg 9.80 m s 2 mg = = 1.47 N s m vT 2.00 10 2 m s
e
je
j
(b)
In the equation describing the time variation of the velocity, we have v = vT 1  e  bt m or at time
e
j
v = 0.632 vT when e  bt m = 0.368 t=
FG m IJ lna0.368f = HbK
2.04 10 3 s
(c) P6.38
At terminal velocity,
R = vT b = mg = 2.94 10 2 N
The resistive force is 1 1 DAv 2 = 0.250 1.20 kg m3 2.20 m 2 27.8 m s 2 2 R = 255 N R=
a
fe
je
jb
g
2
a=
255 N R = = 0.212 m s 2 m 1200 kg
Chapter 6
173
P6.39
(a)
v t = vi e  ct
af
v 20.0 s = 5.00 = vi e 20 .0 c , vi = 10.0 m s .
So 5.00 = 10.0 e 20 .0 c and 20.0 c = ln (b) (c) P6.40 At t = 40.0 s v = vi e  ct
FG 1 IJ H 2K v = b10.0 m sge
a
f
c=
ln
20.0
c h=
1 2
3.47 10 2 s 1
40 .0 c
= 10.0 m s 0.250 = 2.50 m s
b
ga
f
s=
dv =  cvi e  ct =  cv dt
F = ma
 kmv 2 = m  kdt =
t
 k dt = v 2 dv
0 v0
z z
v
dv v2
dv dt
k t  0 =
a f
v 1 1
v
=
v0
1 1 + v v0
1 + v 0 kt 1 1 = + kt = v v0 v0 v0 v= 1 + v 0 kt *P6.41 (a) From Problem 40, v=
x
z z
t 0 0
v0 dx = dt 1 + v 0 kt
dx = v 0
x
dt 1 v 0 kdt = 1 + v 0 kt k 0 1 + v 0 kt
z
t
t 1 ln 1 + v 0 kt 0 k 1 x  0 = ln 1 + v 0 kt  ln 1 k 1 x = ln 1 + v 0 kt k
x0 =
b b
g g
b
g
(b)
We have ln 1 + v 0 kt = kx v0 v0 = kx = v 0 e  kx = v 1 + v 0 kt = e kx so v = 1 + v 0 kt e 1 DAv 2 so 2 k=
3 3 2 DA 0.305 1.20 kg m 4.2 10 m = = 5.3 10 3 m 2m 2 0.145 kg
b
g
*P6.42
We write  kmv 2 = 
e
b
je
g
j
v = v 0 e  kx = 40.2 m s e
b
g
 5.3 10
e
3
m 18 .3 m
ja
f=
36.5 m s
174 P6.43
Circular Motion and Other Applications of Newton's Laws
In R =
1 DAv 2 , we estimate that D = 1.00 , = 1.20 kg m3 , A = 0.100 m 0.160 m = 1.60 10 2 m 2 2 and v = 27.0 m s. The resistance force is then R= or 1 1.00 1.20 kg m3 1.60 10 2 m 2 27.0 m s 2
a
fa
f
a fe
je
jb
g
2
= 7.00 N
R ~ 10 1 N
Section 6.5
Numerical Modeling in Particle Dynamics
Note: In some problems we compute each new position as x t + t = x t + v t + t t , rather than x t + t = x t + v t t as quoted in the text. This method has the same theoretical validity as that presented in the text, and in practice can give quicker convergence.
a
f af af
(a)
a
f af a
e
f
P6.44
At v = vT , a = 0,  mg + bvT = 0
vT =
3.00 10 3 kg 9.80 m s 2 mg = = 0.980 m s b 3.00 10 2 kg s
je
j
(b)
ts
0 0.005 0.01 0.015 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65
af
xm
a f
vms
b g
F mN
29.4 27.93 26.534 25.2 17.6 10.5 6.31 3.78 2.26 1.35 0.811 0.486 0.291 0.174 0.110 0.062 4 0.037 4
a f
a m s2 9.8 9.31 8.844 5 8.40 5.87 3.51 2.10 1.26 0.754 0.451 0.270 0.162 0.096 9 0.058 0 0.034 7 0.020 8 0.012 5
e
j
2 2 1.999 755 1.999 3 1.990 1.965 1.930 1.889 1.845 1.799 1.752 1.704 1.65 1.61 1.56 1.51 1.46
0 0.049 0.095 55 0.139 77 0.393 0.629 0.770 0.854 0.904 0.935 0.953 0.964 0.970 0.974 0.977 0.978 0.979
. . . we list the result after each tenth iteration
Terminal velocity is never reached. The leaf is at 99.9% of vT after 0.67 s. The fall to the ground takes about 2.14 s. Repeating with t = 0.001 s , we find the fall takes 2.14 s.
Chapter 6
175
P6.45
(a)
When v = vT , a = 0,
2 F =  mg + CvT = 0
mg = vT =  C (b)
e4.80 10
vms
4
kg 9.80 m s 2
5
je
2.50 10
kg m
j=
13.7 m s
ts
0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2 3 4 5
af
xm
0 0 0.392 1.168 2.30 3.77 5.51 7.48 9.65 11.96 14.4 27.4 41.0 54.7
a f
b g
F mN
4.704 4.608 4.327 6 3.896 5 3.369 3 2.807 1 2.263 5 1.775 3 1.361 6 1.03 0.762
a f
a m s2 9.8 9.599 9 9.015 9 8.117 8 7.019 3 5.848 1 4.715 6 3.698 6 2.836 6 2.14 1.59 0.321 0.060 6 0.011 3
e
j
0 1.96 3.88 5.683 2 7.306 8 8.710 7 9.880 3 10.823 11.563 12.13 12.56 13.49 13.67 13.71
. . . listing results after each fifth step 0.154 0.029 1 0.005 42
The hailstone reaches 99% of vT after 3.3 s, 99.95% of vT after 5.0 s, 99.99% of vT after 6.0 s, 99.999% of vT after 7.4 s. P6.46 (a) At terminal velocity,
2 F = 0 =  mg + CvT
C=
mg
2 vT
=
b0.142 kgge9.80 m s j = b42.5 m sg
2 2
7.70 10 4 kg m
(b) (c)
Cv 2 = 7.70 10 4 kg m 36.0 m s
Elapsed Time (s) 0.000 00 0.050 00 ... 2.950 00 3.000 00 3.050 00 ... 6.250 00 6.300 00 Altitude (m) 0.000 00 1.757 92 48.623 27 48.640 00 48.632 24 1.250 85 0.106 52
e
jb
g
2
= 0.998 N
Resistance Force (N) 0.998 49 0.952 35 0.000 52 0.000 09 0.000 02 0.555 55 0.567 80 Net Force (N) 2.390 09 2.343 95 1.392 12 1.391 69 1.391 58 0.836 05 0.823 80 Acceleration m s2
Speed (m/s) 36.000 00 35.158 42 0.824 94 0.334 76 0.155 27 26.852 97 27.147 36
e
j
16.831 58 16.506 67 9.803 69 9.800 61 9.799 87 5.887 69 5.801 44
Maximum height is about 49 m . It returns to the ground after about 6.3 s with a speed of approximately 27 m s .
176 P6.47
Circular Motion and Other Applications of Newton's Laws
(a)
At constant velocity mg = vT =  C vT = 
2 F = 0 =  mg + CvT 2
b50.0 kg ge9.80 m s j =
0.200 kg m
49.5 m s with chute closed and
b50.0 kg gb9.80 m sg =
20.0 kg m
4.95 m s with chute open.
(b)
We use time increments of 0.1 s for 0 < t < 10 s , then 0.01 s for 10 s < t < 12 s , and then 0.1 s again. time(s) 0 1 2 4 7 10 10.1 10.3 11 12 50 100 145 height(m) 1000 995 980 929 812 674 671 669 665 659 471 224 0 velocity(m/s) 0 9.7 18.6 32.7 43.7 47.7 16.7 8.02 5.09 4.95 4.95 4.95 4.95
6.48
(a)
We use a time increment of 0.01 s. time(s) 0 0.100 0.200 0.400 1.00 1.92 2.00 4.00 5.00 6.85 x(m) 0 7.81 14.9 27.1 51.9 70.0 70.9 80.4 81.4 81.8 y(m) 0 5.43 10.2 18.3 32.7 38.5 38.5 26.7 17.7 0 with 30.0 35.0 25.0 20.0 15.0 10.0 17.0 16.0 15.5 15.8 16.1 15.9 we find range 86.410 m 81.8 m 90.181 m 92.874 m 93.812 m 90.965 m 93.732 m 93.839 8 m 93.829 m 93.839 m 93.838 m 93.840 2 m
(b) (c)
range = 81.8 m So we have maximum range at = 15.9
Chapter 6
177
P6.49
(a)
At terminal speed,
F =  mg + Cv 2 = 0 . Thus,
C= mg v2 =
b0.046 0 kgge9.80 m s j = b44.0 m sg
2 2
2.33 10 4 kg m
(b)
We set up a spreadsheet to calculate the motion, try different initial speeds, and home in on 53 m s as that required for horizontal range of 155 m, thus: tan 1
Time t (s) 0.000 0 0.002 7 ... 2.501 6 2.504 3 2.506 9 ... 3.423 8 3.426 5 3.429 1 ... 5.151 6 5.154 3 (c)
x (m) 0.000 0 0.121 1
vx (m/s)
e
ax m s2
j
y (m)
vy (m/s)
em s j
2
ay
v=
2 vx
+
2 vy
(m/s) 53.300 0 53.257 4 28.937 5 28.926 3 28.915 1 26.998 4 26.998 4 26.998 4 31.269 2 31.279 2
Fv I GH v JK
y x
(deg) 31.000 0 30.982 2 0.046 6 0.004 8 0.056 3 19.226 2 19.282 2 19.338 2 48.195 4 48.226 2
45.687 0 10.565 9 0.000 0 27.451 5 13.614 6 45.659 0 10.552 9 0.072 7 27.415 5 13.604 6
90.194 6 28.937 5 4.238 8 32.502 4 0.023 5 9.800 0 90.271 3 28.926 3 4.235 5 32.502 4 0.002 4 9.800 0 90.348 0 28.915 0 4.232 2 32.502 4 0.028 4 9.800 0 115.229 8 25.492 6 3.289 6 28.397 2 8.890 5 9.399 9 115.297 4 25.483 9 3.287 4 28.373 6 8.915 4 9.397 7 115.364 9 25.475 1 3.285 1 28.350 0 8.940 3 9.395 4 154.996 8 20.843 8 2.199 2 0.005 9 23.308 7 7.049 8 155.052 0 20.838 0 2.198 0 0.055 9 23.327 4 7.045 4
Similarly, the initial speed is 42 m s . The motion proceeds thus: tan 1
Time t (s)
x (m)
vx (m/s)
e
ax m s2
j
y (m)
vy (m/s)
em s j
2
ay
v=
2 vx
+
2 vy
(m/s) 42.150 0 42.102 6 20.548 5 20.541 0 20.533 5 20.105 8 20.105 8 20.105 8 29.762 3 29.779 5
Fv I GH v JK
y x
(deg) 47.000 0 46.967 1 0.072 5 0.023 1 0.118 8 11.307 7 11.406 7 11.505 6 58.073 1 58.103 7
0.000 0 0.000 0 0.003 5 0.100 6 ... 2.740 5 66.307 8 2.744 0 66.379 7 2.747 5 66.451 6 ... 3.146 5 74.480 5 3.150 0 74.549 5 3.153 5 74.618 5 ... 5.677 0 118.969 7 5.680 5 119.024 8
28.746 2 4.182 9 28.731 6 4.178 7
0.000 0 30.826 6 14.610 3 0.107 9 30.775 4 14.594 3
20.548 4 2.137 4 39.485 4 0.026 0 9.800 0 20.541 0 2.135 8 39.485 5 0.008 3 9.800 0 20.533 5 2.134 3 39.485 5 0.042 6 9.800 0 19.715 6 1.967 6 38.696 3 3.942 3 9.721 3 19.708 7 1.966 2 38.682 5 3.976 4 9.720 0 19.701 8 1.964 9 38.668 6 4.010 4 9.718 6 15.739 4 1.254 0 0.046 5 25.260 0 6.570 1 15.735 0 1.253 3 0.041 9 25.283 0 6.564 2
The trajectory in (c) reaches maximum height 39 m, as opposed to 33 m in (b). In both, the ball reaches maximum height when it has covered about 57% of its range. Its speed is a minimum somewhat later. The impact speeds are both about 30 m/s.
178
Circular Motion and Other Applications of Newton's Laws
Additional Problems *P6.50 When the cloth is at a lower angle , the radial component of F = ma reads mv 2 n + mg sin = . r At = 68.0 , the normal force drops to zero and v2 . g sin 68 = r v = rg sin 68 = The rate of revolution is angular speed = 1.73 m s
2
p 68 R mg
p
mg sin68
mg cos68
FIG. P6.50
a0.33 mfe9.8 m s j sin 68 = 1.73 m s
0.835 rev s = 50.1 rev min .
b
gFGH 12rev IJK FGH 2 a02.r mf IJK = r 33
*P6.51
(a)
v = 30 km h
b
1h gFGH 3 600 s IJK FGH 1 1000 m IJK = 8.33 m s km mv 2 r
n
L F vI b8.33 m sg n = mG g  J = 1 800 kg M9.8 m s  20. 4 m MN H rK
2 2
Fy = ma y : +n  mg = 
2
OP PQ
mg FIG. P6.51
= 1.15 10 4 N up (b) Take n = 0 . Then mg = mv 2 . r
v = gr =
e9.8 m s ja20.4 mf =
2
14.1 m s = 50.9 km h
P6.52
(a)
Fy = ma y =
mg  n = mv 2 R
mv 2 R n = mg  mv 2 R mv 2 R
(b)
When n = 0 , Then,
mg =
v=
gR .
Chapter 6
179
*P6.53
(a)
slope =
0.160 N  0 = 0.016 2 kg m 9.9 m 2 s 2 R v
2
(b)
slope =
=
1 2
DAv 2 v
2
=
1 D A 2
(c)
1 DA = 0.016 2 kg m 2 2 0.016 2 kg m D= 1.20 kg m3 0.105 m
e
b
ja
g
f
2
= 0.778
(d)
From the table, the eighth point is at force mg = 8 1.64 10 3 kg 9.8 m s 2 = 0.129 N and horizontal coordinate 2.80 m s . The vertical coordinate of the line is here 0.129 N  0.127 N 2 = 1.5%. 0.016 2 kg m 2.8 m s = 0.127 N . The scatter percentage is 0.127 N
b
gb
b
g
g
e
je
j
2
(e)
The interpretation of the graph can be stated thus: For stacked coffee filters 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 1 R = DAv 2 . The value of the constant slope of the graph implies that the drag coefficient 2 for coffee filters is D = 0.78 2% . While the car negotiates the curve, the accelerometer is at the angle . Horizontally: Vertically: T sin = mv 2 r
P6.54
(a)
T cos = mg
where r is the radius of the curve, and v is the speed of the car. By division, v2 = g tan : Then a c = r tan = v2 rg
2
a c = 9.80 m s
e
j tan 15.0
FIG. P6.54
a c = 2.63 m s 2
v2 r= ac v 2 = rg tan = 201 m 9.80 m s 2 tan 9.00
(b)
b23.0 m sg r= a fe j
v = 17.7 m s
2
2.63 m s 2
= 201 m
(c)
180 P6.55
Circular Motion and Other Applications of Newton's Laws
Take xaxis up the hill
Fx = ma x :
+T sin  mg sin = ma T a = sin  g sin m Fy = ma y : +T cos  mg cos = 0 mg cos T= cos g cos sin  g sin a= cos a = g cos tan  sin
b
g
n ac x y
*P6.56
(a)
The speed of the bag is
2 7.46 m = 1.23 m s . The 38 s total force on it must add to ma c =
a
f
fs
b30 kg gb1.23 m sg
7.46 m
2
= 6.12 N
mg FIG. P6.56
Fx = ma x : Fy = ma y :
fs cos 20  n sin 20 = 6.12 N
fs sin 20 + n cos 20  30 kg 9.8 m s 2 = 0
b
ge
j
f cos 20  6.12 N n= s sin 20 Substitute: fs sin 20 + fs cos 2 20 cos 20  6.12 N = 294 N sin 20 sin 20 f s 2.92 = 294 N + 16.8 N
a
f
a f
fs = 106 N (b) 2 7.94 m = 1.47 m s 34 s 2 30 kg 1.47 m s = 8.13 N ma c = 7.94 m fs cos 20  n sin 20 = 8.13 N v=
a b
f gb
g
fs sin 20 + n cos 20 = 294 N n= fs cos 20  8.13 N sin 20 cos 2 20 cos 20  8.13 N = 294 N fs sin 20 + fs sin 20 sin 20 fs 2.92 = 294 N + 22.4 N
a
a f
f
fs = 108 N n=
sin 20 fs 108 N = 0.396 s = = n 273 N
a108 Nf cos 20  8.13 N = 273 N
Chapter 6
181
P6.57
(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, Fg = Fg  mv 2 or Fg > Fg r
(b)
At the poles v = 0 and Fg = Fg = mg = 75.0 9.80 = 735 N down.
a f b
At the equator, Fg = Fg  ma c = 735 N  75.0 0.033 7 N = 732 N down. P6.58 (a) Since the object of mass m 2 is in equilibrium, or (b)
g
FIG. P6.57
Fy = T  m 2 g = 0
T = m2 g .
The tension in the string provides the required centripetal acceleration of the puck. Thus, Fc = T = m 2 g . Fc = v= 88 ft gFGH 60.0.0 mi sh IJK = 440 ft s m1 v 2 R RFc = m1
(c)
From we have
FG m IJ gR Hm K
2 1
.
P6.59
(a)
v = 300 mi h
b
At the lowest point, his seat exerts an upward force; therefore, his weight seems to increase. His apparent weight is Fg = mg + m (b) v2 160 = 160 + r 32.0
FG H
IJ a440f K 1 200
2
= 967 lb .
At the highest point, the force of the seat on the pilot is directed down and Fg = mg  m v2 = 647 lb . r
Since the plane is upside down, the seat exerts this downward force. (c) When Fg = 0 , then mg = mv 2 . If we vary the aircraft's R and v such that the above is true, R then the pilot feels weightless.
182 P6.60
Circular Motion and Other Applications of Newton's Laws
For the block to remain stationary,
Fy = 0 and Fx = mar .
n1 = m p + m b g so f s1n1 = s1 m p + m b g . At the point of slipping, the required centripetal force equals the maximum friction force: mp + mb or v max = s1 rg =
mb g n1 fp mp g
e
j
e
j
e
j r = em + m j g a0.750fa0.120fa9.80f = 0.939 m s .
s1 p b
2 v max
f mb g mp g n2
For the penny to remain stationary on the block:
Fy = 0 n 2  m p g = 0 or n 2 = m p g
and
Fx = ma r f p = m p
2 v max r
v . r
2
When the penny is about to slip on the block, f p = f p , max = s 2 n 2 or s 2 m p g = m p
fp
mp g
v max = s 2 rg =
a0.520fa0.120fa9.80f = 0.782 m s
v max 1 rev = 0.782 m s 2r 2 0.120 m
FIG. P6.60
This is less than the maximum speed for the block, so the penny slips before the block starts to slip. The maximum rotation frequency is Max rpm =
b
gLM a N
OPFG 60 s IJ = f QH 1 min K
62.2 rev min .
P6.61
v= (a) (b) (c) (d)
2r 2 9.00 m = = 3.77 m s T 15.0 s
a
a
f
f
ar =
v2 = 1.58 m s 2 r
Flow = m g + a r = 455 N Fhigh
r
b g = mb g  a g =
328 N ar 1.58 = tan 1 = 9.15 inward . g 9.8
2 Fmid = m g 2 + a r = 397 N upward and at = tan 1
P6.62
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 s 2 centripetal acceleration: ac = v2 : r
The period of rotation comes from v = so the frequency of rotation is
2r : T
e3.00 m s ja60.0 mf = 13.4 m s 2r 2 a60.0 mf = = 28.1 s T= v 13. 4 m s 1 1 1 F 60 s I = f= = G J = 2.14 rev min T 28.1 s 28.1 s H 1 min K
v = ac r =
2
.
Chapter 6
183
P6.63
(a)
The mass at the end of the chain is in vertical equilibrium. T Thus T cos = mg . l = 2.50 m mv 2 Horizontally T sin = ma r = r
R = 4.00 m
a f r = a 2.50 sin 28.0+4.00f m = 5.17 m
r = 2.50 sin + 4.00 m v2 . Then a r = 5.17 m By division tan =
mg
r
ar v2 = g 5.17 g
FIG. P6.63
v 2 = 5.17 g tan = 5.17 9.80 tan 28.0 m 2 s 2 v = 5.19 m s (b) T cos = mg
a fa fa j
f
50.0 kg 9.80 m s 2 mg T= = = 555 N cos cos 28.0 P6.64 (a) The putty, when dislodged, rises and returns to the original level in time t. To find t, we use 2v where v is the speed of a point on the rim of the wheel. v f = vi + at : i.e.,  v = + v  gt or t = g 2R 2 v 2R If R is the radius of the wheel, v = = , so t = . t g v Thus, v 2 = Rg and v = Rg . (b) The putty is dislodged when F, the force holding it to the wheel is F= mv 2 R mv 2 = m g . R
f
b
ge
P6.65
(a)
n=
f  mg = 0 v= 2R T
f = sn T= 4 2 R s g
(b)
T = 2.54 s # rev 1 rev 60 s rev = = 23.6 min 2.54 s min min
n
FG H
IJ K
mg
FIG. P6.65
184 P6.66
Circular Motion and Other Applications of Newton's Laws
Let the xaxis point eastward, the yaxis upward, and the zaxis point southward. (a) The range is Z = vi2 sin 2 i g The initial speed of the ball is therefore vi = gZ = sin 2 i
a9.80fa285f = 53.0 m s
sin 96.0 1 a y t 2 as 2
The time the ball is in the air is found from y = viy t +
0 = 53.0 m s sin 48.0 t  4.90 m s 2 t 2 giving t = 8.04 s .
6 2R e cos i 2 6.37 10 m cos 35.0 = = 379 m s 86 400 s 86 400 s
b
ga
f e
j
(b) (c)
vix =
e
j
360 of latitude corresponds to a distance of 2R e , so 285 m is a change in latitude of =
FG S IJ a360f = FG 285 m IJ a360f = 2.56 10 GH 2 e6.37 10 mj JK H 2R K
e 6
3
degrees
The final latitude is then f = i  = 35.00.002 56 = 34.997 4 . The cup is moving eastward at a speed v fx = velocity of the tee by v x = v fx  v fi = = 2R e 2R e cos f  cos i = cos i   cos i 86 400 s 86 400 s 2R e cos f 86 400 s , which is larger than the eastward
b
g
2R e cos i cos + sin i sin  cos i 86 400 s 2R e sin i sin . 86 400 s 1.19 10 2 m s
Since is such a small angle, cos 1 and v x 2 6.37 10 6 m 86 400 s
v x (d)
e
j sin 35.0 sin 0.002 56 =
x = v x t = 1.19 10 2 m s 8.04 s = 0.095 5 m = 9.55 cm
b g e
ja
f
Chapter 6
185
P6.67
(a)
If the car is about to slip down the incline, f is directed up the incline.
t f
n
Fy = n cos + f sin  mg = 0 where
n=
f = sn gives
mg s mg and f = . cos 1 + s tan cos 1 + s tan
b
g
b
g
t
mg
Then,
Fx = n sin  f cos = m
v min =
2 v min yields R
Rg tan  s 1 + s tan
b
g
n cos
.
f sin f cos n sin
When the car is about to slip up the incline, f is directed down the incline. Then, Fy = n cos  f sin  mg = 0 with f = sn yields
mg
s mg mg n= and f = . cos 1  s tan cos 1  s tan
b
g
b
g
t
n
In this case,
Fx = n sin + f cos = m
v max =
2 v max , which gives R
Rg tan + s 1  s tan
b
g
f
.
mg
t
(b)
If v min =
Rg tan  s 1 + s tan
b
g = 0 , then
2
s = tan .
n cos
(c)
v min =
v max =
a100 mfe9.80 m s jatan 10.00.100f = 1 + a0.100 f tan 10.0 a100 mfe9.80 m s jatan 10.0+0.100f = 1  a0.100f tan 10.0
2
8.57 m s
n sin f cos
16.6 m s
mg
f sin
FIG. P6.67
186 P6.68
Circular Motion and Other Applications of Newton's Laws
(a)
The bead moves in a circle with radius v = R sin at a speed of v= 2r 2R sin = T T
The normal force has an inward radial component of n sin and an upward component of n cos
Fy = ma y :
or n=
n cos  mg = 0
mg cos
FIG. P6.68(a)
Then
Fx = n sin = m
v2 becomes r
FG mg IJ sin = m FG 2R sin IJ K H cos K R sin H T
g sin 4 2 R sin = cos T2 sin = 0 = 0 cos = gT 2 4 2 R
2
which reduces to This has two solutions: and
(1) (2)
If R = 15.0 cm and T = 0.450 s, the second solution yields
e9.80 m s ja0.450 sf cos = 4 a0.150 mf
2 2
2
= 0.335 and = 70.4
Thus, in this case, the bead can ride at two positions = 70. 4 and = 0 . (b) At this slower rotation, solution (2) above becomes
e9.80 m s ja0.850 sf cos = 4 a0.150 mf
2 2
2
= 1.20 , which is impossible.
In this case, the bead can ride only at the bottom of the loop, = 0 . The loop's rotation must be faster than a certain threshold value in order for the bead to move away from the lowest position.
Chapter 6
187
P6.69
At terminal velocity, the accelerating force of gravity is balanced by frictional drag: mg = arv + br 2 v 2 (a) mg = 3.10 10 9 v + 0.870 10 10 v 2 For water, m = V = 1 000 kg m3 4.11 10 11
e
j e
LM 4 e10 mj OP N3 Q = e3.10 10 jv + e0.870 10 jv
5 3 9 10
j
2
Assuming v is small, ignore the second term on the right hand side: v = 0.013 2 m s . (b) mg = 3.10 10 8 v + 0.870 10 8 v 2 Here we cannot ignore the second term because the coefficients are of nearly equal magnitude. 4.11 10 8 = 3.10 10 8 v + 0.870 10 8 v 2 v= (c) 3.10
2
e
j e
j
e j e j a3.10f + 4a0.870fa4.11f = 1.03 m s 2a0.870f
v = 6.87 m s
mg = 3.10 10 7 v + 0.870 10 6 v 2 Assuming v > 1 m s , and ignoring the first term: 4.11 10 5 = 0.870 10 6 v 2
e
j e
j
e
j
P6.70
v=
FG mg IJ LM1  expFG bt IJ OP where expaxf = e H b KN H m KQ
x
is the exponential function. v vT = mg b
At t , At t = 5.54 s
0.500 vT = vT 1  exp exp
F ba5.54 sf I = 0.500 ; GH 9.00 kg JK  b a5.54 sf = ln 0.500 = 0.693 ;
9.00 kg 5.54 s
LM NM
F ba5.54 sf I OP GH 9.00 kg JK QP
b= mg b
b9.00 kg ga0.693f = 1.13 m s b9.00 kg ge9.80 m s j =
2
(a)
vT =
vT =
1.13 kg s
78.3 m s
(b)
0.750 vT = vT 1  exp
LM N
FG 1.13t IJ OP H 9.00 s K Q
exp
FG 1.13t IJ = 0.250 H 9.00 s K 9.00aln 0.250 f t= s=
1.13
11.1 s
continued on next page
188
Circular Motion and Other Applications of Newton's Laws
(c)
mg dx = dt b
FG IJ LM1  expFG  bt IJ OP ; H K N H mKQ
x0
z z FGH
x
dx =
t
0
mg b
IJ LM1  expFG bt IJ OPdt KN H m KQ F GH
t 2 0 2
At t = 5.54 s ,
I FG IJ = mgt + F m g I LexpFG bt IJ  1O JK H K b GH b JK MN H m K PQ F b9.00 kg g e9.80 m s j I 5.54 s J expa0.693f  1 x = 9.00 kg e9.80 m s j +G 1.13 kg s G H b1.13 m sg JK x = 434 m + 626 ma 0.500f = 121 m
mgt m2 g  bt + x  x0 = exp 2 b m b
2 2 2 2
P6.71
Fy = L y  Ty  mg = L cos 20.0T sin 20.07.35 N = ma y = 0 Fx = L x + Tx = L sin 20.0+T cos 20.0 = m
35.0 m s v2 = 0.750 kg = 16.3 N m r 60.0 m cos 20.0
a
b
f
g
v2 r
2
L sin 20.0+T cos 20.0 = 16.3 N L cos 20.0T sin 20.0 = 7.35 N cos 20.0 16.3 N = sin 20.0 sin 20.0 sin 20.0 7.35 N = LT cos 20.0 cos20.0 16.3 N 7.35 N  T cot 20.0+ tan 20.0 = sin 20.0 cos 20.0 T 3.11 = 39.8 N L+T FIG. P6.71
a a f
f
T = 12.8 N
Chapter 6
189
P6.72
(a)
ts dm 4.88 1.00 2.00 3.00 4.00 18.9 42.1 73.8
af
a f
(b)
d (m) 900 800 700 600 500 400 300 200 100 0 0 2 4 6 8 10 12 14 16 18 20 t (s)
5.00 112 6.00 154 7.00 199 8.00 246 9.00 296 10.0 11.0 12.0 13.0 14.0 15.0 16.0 17.0 18.0 19.0 20.0 (c) 347 399 452 505 558 611 664 717 770 823 876
A straight line fits the points from t = 11.0 s to 20.0 s quite precisely. Its slope is the terminal speed. vT = slope = 876 m  399 m = 53.0 m s 20.0 s  11.0 s a= dv dx =0k =  kv dt dt
*P6.73
v = vi  kx implies the acceleration is Then the total force is The resistive force is opposite to the velocity:
F = ma = ma kvf
F = kmv
.
ANSWERS TO EVEN PROBLEMS
P6.2 P6.4 P6.6 P6.8 P6.10 215 N horizontally inward 6.22 10 12 N (a) 1.65 km s ; (b) 6.84 10 3 s 0.966 g P6.16 P6.12 P6.14 2.06 10 3 rev min (a) R
FG 2T  gIJ ; (b) 2T upward Hm K
(a) 1.33 m s 2 ; (b) 1.79 m s 2 forward and 48.0 inward 8.88 N
e j (c) e 0.181 i + 0.181 jj m s
(a) 0.233 i + 0.163 j m s 2 ; (b) 6.53 m s ;
2
P6.18
190 P6.20
Circular Motion and Other Applications of Newton's Laws
(a) 8.62 m; (b) Mg downward; (c) 8.45 m s 2 , Unless they are belted in, the riders will fall from the cars. 15.3 m s Straight across the dashboard to the left 0.527 (a) 1.41 h; (b) 17.1
P6.46
(a) 7.70 10 4 kg m; (b) 0.998 N; (c) The ball reaches maximum height 49 m. Its flight lasts 6.3 s and its impact speed is 27 m s . (a) see the solution; (b) 81.8 m; (c) 15.9 0.835 rev s (a) mg  mv 2 ; (b) v = gR R
P6.22
P6.48 P6.50 P6.52 P6.54 P6.56 P6.58 P6.60 P6.62 P6.64
P6.24 P6.26 P6.28
k =
b g + agt
2 vt  L
a
f
2
(a) 2.63 m s 2 ; (b) 201 m; (c) 17.7 m s (a) 106 N; (b) 0.396 (a) m 2 g ; (b) m 2 g ; (c) 62.2 rev min 2.14 rev min (a) v = Rg ; (b) m g (a) 8.04 s; (b) 379 m s; (c) 1.19 cm s ; (d) 9.55 cm (a) either 70.4 or 0; (b) 0 (a) 78.3 m s ; (b) 11.1 s; (c) 121 m (a) and (b) see the solution; (c) 53.0 m s
P6.30
(a) 2.38 10 5 m s 2 horizontally inward = 2.43 10 4 g ; (b) 360 N inward perpendicular to the cone; (c) 47.5 10 4 m s 2 (a) 6.27 m s 2 downward ; (b) 784 N up; (c) 283 N up (a) 53.8 m s ; (b) 148 m 1.40 0. 212 m s 2 see the solution 36.5 m s (a) 0.980 m s ; (b) see the solution
FG m IJ gR Hm K
2 1
P6.32
P6.34 P6.36 P6.38 P6.40 P6.42 P6.44
P6.66
P6.68 P6.70 P6.72
7
Energy and Energy Transfer
CHAPTER OUTLINE
7.1 7.2 7.3 7.4 7.5 Systems and Environments Work Done by a Constant Force The Scalar Product of Two Vectors Work Done by a Varying Force Kinetic Energy and the WorkKinetic Energy Theorem The NonIsolated SystemConservation of Energy Situations Involving Kinetic Friction Power Energy and the Automobile
ANSWERS TO QUESTIONS
Q7.1 Q7.2 The force is perpendicular to every increment of displacement. Therefore, F r = 0 . (a) (b) (c) (d) (e) Q7.3 Positive work is done by the chicken on the dirt. No work is done, although it may seem like there is. Positive work is done on the bucket. Negative work is done on the bucket. Negative work is done on the person's torso.
7.6
7.7 7.8 7.9
Yes. Force times distance over which the toe is in contact with the ball. No, he is no longer applying a force. Yes, both air friction and gravity do work.
Q7.4 Q7.5
Force of tension on a ball rotating on the end of a string. Normal force and gravitational force on an object at rest or moving across a level floor. (a) (c) Tension (b) Air resistance
Positive in increasing velocity on the downswing. Negative in decreasing velocity on the upswing.
Q7.6 Q7.7 Q7.8
No. The vectors might be in the third and fourth quadrants, but if the angle between them is less than 90 their dot product is positive. The scalar product of two vectors is positive if the angle between them is between 0 and 90. The scalar product is negative when 90 < < 180 . If the coils of the spring are initially in contact with one another, as the load increases from zero, the graph would be an upwardly curved arc. After the load increases sufficiently, the graph will be linear, described by Hooke's Law. This linear region will be quite large compared to the first region. The graph will then be a downward curved arc as the coiled spring becomes a completely straight wire. As the load increases with a straight wire, the graph will become a straight line again, with a significantly smaller slope. Eventually, the wire would break. k = 2 k . To stretch the smaller piece one meter, each coil would have to stretch twice as much as one coil in the original long spring, since there would be half as many coils. Assuming that the spring is ideal, twice the stretch requires twice the force. 191
Q7.9
192 Q7.10
Energy and Energy Transfer
Kinetic energy is always positive. Mass and squared speed are both positive. A moving object can always do positive work in striking another object and causing it to move along the same direction of motion. Work is only done in accelerating the ball from rest. The work is done over the effective length of the pitcher's armthe distance his hand moves through windup and until release. Kinetic energy is proportional to mass. The first bullet has twice as much kinetic energy. The longer barrel will have the higher muzzle speed. Since the accelerating force acts over a longer distance, the change in kinetic energy will be larger. (a) (b) Kinetic energy is proportional to squared speed. Doubling the speed makes an object's kinetic energy four times larger. If the total work on an object is zero in some process, its speed must be the same at the final point as it was at the initial point.
Q7.11 Q7.12 Q7.13 Q7.14
Q7.15
The larger engine is unnecessary. Consider a 30 minute commute. If you travel the same speed in each car, it will take the same amount of time, expending the same amount of energy. The extra power available from the larger engine isn't used. If the instantaneous power output by some agent changes continuously, its average power in a process must be equal to its instantaneous power at least one instant. If its power output is constant, its instantaneous power is always equal to its average power. It decreases, as the force required to lift the car decreases. As you ride an express subway train, a backpack at your feet has no kinetic energy as measured by you since, according to you, the backpack is not moving. In the frame of reference of someone on the side of the tracks as the train rolls by, the backpack is moving and has mass, and thus has kinetic energy. The rock increases in speed. The farther it has fallen, the more force it might exert on the sand at the bottom; but it might instead make a deeper crater with an equalsize average force. The farther it falls, the more work it will do in stopping. Its kinetic energy is increasing due to the work that the gravitational force does on it. The normal force does no work because the angle between the normal force and the direction of motion is usually 90. Static friction usually does no work because there is no distance through which the force is applied. An argument for: As a glider moves along an airtrack, the only force that the track applies on the glider is the normal force. Since the angle between the direction of motion and the normal force is 90, the work done must be zero, even if the track is not level. Against: An airtrack has bumpers. When a glider bounces from the bumper at the end of the airtrack, it loses a bit of energy, as evidenced by a decreased speed. The airtrack does negative work. Gaspard de Coriolis first stated the workkinetic energy theorem. Jean Victor Poncelet, an engineer who invaded Russia with Napoleon, is most responsible for demonstrating its wide practical applicability, in his 1829 book Industrial Mechanics. Their work came remarkably late compared to the elucidation of momentum conservation in collisions by Descartes and to Newton's Mathematical Principles of the Philosophy of Nature, both in the 1600's.
Q7.16
Q7.17 Q7.18
Q7.19
Q7.20
Q7.21
Q7.22
Chapter 7
193
SOLUTIONS TO PROBLEMS
Section 7.1 Section 7.2 P7.1 (a) Systems and Environments Work Done by a Constant Force
W = Fr cos = 16.0 N 2.20 m cos 25.0 = 31.9 J
a
fa
f
(b), (c) The normal force and the weight are both at 90 to the displacement in any time interval. Both do 0 work. (d) P7.2
W = 31.9 J + 0 + 0 =
31.9 J
The component of force along the direction of motion is F cos = 35.0 N cos 25.0 = 31.7 N . The work done by this force is W = F cos r = 31.7 N 50.0 m = 1.59 10 3 J .
a
f
a
f
a
fa
f
P7.3
Method One. Let represent the instantaneous angle the rope makes with the vertical as it is swinging up from i = 0 to f = 60 . In an incremental bit of motion from angle to + d , the definition of radian measure implies that r = 12 m d . The angle between the incremental displacement and the force of gravity is = 90+ . Then cos = cos 90+ =  sin . The work done by the gravitational force on Batman is W = F cos dr =
i
a
f
b
g
FIG. P7.3
z
f
= 60 =0
z
mg  sin 12 m d
2 60 0
b
ga
f
f z sin d = b80 kg ge9.8 m s ja12 mfb cos g = a 784 N fa12 mfa cos 60+1f = 4.70 10 J
=  mg 12 m
a
60 0
3
Method Two. The force of gravity on Batman is mg = 80 kg 9.8 m s 2 = 784 N down. Only his vertical displacement contributes to the work gravity does. His original ycoordinate below the tree limb is 12 m. His final ycoordinate is 12 m cos 60 = 6 m . His change in elevation is 6 m  12 m = 6 m . The work done by gravity is
b
ge
j
a
f
a
f
W = Fr cos = 784 N 6 m cos 180 = 4.70 kJ .
a
fa f
194 P7.4
Energy and Energy Transfer
(a) (b)
W = mgh = 3.35 10 5 9.80 100 J = 3.28 10 2 J Since R = mg , Wair resistance = 3.28 10 2 J
e
ja fa f
Section 7.3 P7.5
The Scalar Product of Two Vectors
A = 5.00 ; B = 9.00 ; = 50.0 A B = AB cos = 5.00 9.00 cos 50.0 = 28.9
a fa f
x y
P7.6
A B = A x i + A y j + A z k B x i + B y j + Bz k
x x x z y
je j A B = A B e i i j + A B e i jj + A B e i k j + A B e j i j + A B e j jj + A B e j k j + A B ek i j + A B ek jj + A B ek k j
y x z x y y z z y z z
e
A B = A x Bx + A y B y + A z Bz P7.7 (a) (b)
W = F r = Fx x + Fy y = 6.00 3.00 N m + 2.00 1.00 N m = 16.0 J
a fa f
2
a
fa f
2
= cos 1
16 FG F r IJ = cos = H Fr K a6.00f + a2.00f jea3.00f + a1.00f j e
1 2 2
36.9
P7.8
We must first find the angle between the two vectors. It is:
= 36011890.0132 = 20.0
Then F v = Fv cos = 32.8 N 0.173 m s cos 20.0 or F v = 5.33 N m J = 5.33 = 5.33 W s s
a
fb
g
FIG. P7.8 A B 12.0 + 8.00 = cos 1 = 11.3 AB 13.0 32.0
P7.9
(a)
A = 3.00 i  2.00 j B = 4.00 i  4.00 j
= cos 1
a fa f
(b)
B = 3.00 i  4.00 j + 2.00k A = 2.00 i + 4.00 j cos =
AB 6.00  16.0 = AB 20.0 29.0
a fa f
1
= 156
(c)
A = i  2.00 j + 2.00k B = 3.00 j + 4.00k
= cos 1
FG A B IJ = cos FG 6.00 + 8.00 IJ = H AB K H 9.00 25.0 K
82.3
Chapter 7
195
P7.10
A  B = 3.00 i + j  k   i + 2.00 j + 5.00k A  B = 4.00 i  j  6.00k
e
j e
j j a f a f
C A  B = 2.00 j  3.00k 4.00 i  j  6.00k = 0 + 2.00 + +18.0 = 16.0
a
f e
je
Section 7.4 P7.11
Work Done by a Varying Force
W = Fdx = area under curve from xi to x f
i
z
f
(a)
xi = 0
x f = 8.00 m
W = area of triangle ABC = W0 8 = (b)
FG 1 IJ 8.00 m 6.00 N = H 2K
FG 1 IJ AC altitude, H 2K
24.0 J FIG. P7.11
xi = 8.00 m W = area of CDE = W8 10
x f = 10.0 m
FG 1 IJ CE altitude, H 2K F 1I = G J a 2.00 mf a 3.00 N f = 3.00 J H 2K a f
(c) P7.12
W0 10 = W0 8 + W810 = 24.0 + 3.00 = 21.0 J
Fx = 8 x  16 N (a) (b) See figure to the right Wnet =  2.00 m 16.0 N 1.00 m 8.00 N + = 12.0 J 2 2
a
f
a
fa
f a
fa
f
FIG. P7.12
196 P7.13
Energy and Energy Transfer
W = Fx dx and W equals the area under the ForceDisplacement curve (a) For the region 0 x 5.00 m , W= (b)
z
a3.00 Nfa5.00 mf =
2
7.50 J FIG. P7.13
For the region 5.00 x 10.0 ,
W = 3.00 N 5.00 m = 15.0 J
(c) For the region 10.0 x 15.0 , W= (d)
a
fa
f
a3.00 Nfa5.00 mf =
2
7.50 J
For the region 0 x 15.0
W = 7.50 + 7.50 + 15.0 J = 30.0 J
a
f
P7.14
W = F dr =
i 5m 0
z
f
5m 0
ze
4x i + 3 y j N dx i
j
zb
x2 4 N m xdx + 0 = 4 N m 2
g
b
g
5m
= 50.0 J
0
P7.15
k=
4.00 9.80 N F Mg = = = 1.57 10 3 N m y y 2.50 10 2 m For 1.50 kg mass y = Work = 1.50 9.80 mg = = 0.938 cm k 1.57 10 3
a fa f
(a)
a fa f
(b)
1 2 ky 2 1 Work = 1.57 10 3 N m 4.00 10 2 m 2
e
je
j
2
= 1.25 J
P7.16
(a)
Spring constant is given by F = kx k= 230 N F = = 575 N m x 0.400 m
a
a
f
f
(b)
Work = Favg x =
1 230 N 0.400 m = 46.0 J 2
a
fa
f
Chapter 7
197
*P7.17
(a)
Fapplied = k leaf x + k helper x h = k x + k h x  y 0 5 10 5 N = 5.25 10 5
b
g g a f
2
N N x + 3.60 10 5 x  0.5 m m m 6.8 10 5 N x = = 0.768 m 8.85 10 5 N m
b
(b)
W=
N 1 1 1 2 k x 2 + khxh = 5.25 10 5 0.768 m m 2 2 2
FG H
IJ a K
f
2
+
N 1 3.60 10 5 0.268 m m 2
= 1.68 10 5 J P7.18 (a) W = F dr W=
i 0.600 m
z
f
ze
0
15 000 N + 10 000 x N m  25 000 x 2 N m 2 dx cos 0
0.600 m
j
10 000 x 2 25 000 x 3  W = 15 000 x + 2 3
0
W = 9.00 kJ + 1.80 kJ  1.80 kJ = 9.00 kJ (b) Similarly,
b10.0 kN mga1.00 mf  e25.0 kN m ja1.00 mf W = a15.0 kN fa1.00 mf + 2 3
2 2
3
W = 11.7 kJ , larger by 29.6% P7.19 1 2 k 0.100 m 2 k = 800 N m and to stretch the spring to 0.200 m requires 4.00 J = W = P7.20 (a) 1 800 0.200 2
a
f
a fa
f
2
 4.00 J = 12.0 J
The radius to the object makes angle with the horizontal, so its weight makes angle with the negative side of the xaxis, when we take the xaxis in the direction of motion tangent to the cylinder. Fx = ma x F  mg cos = 0 F = mg cos FIG. P7.20
(b)
W = F dr We use radian measure to express the next bit of displacement as dr = Rd in terms of the next bit of angle moved through:
2
i
z
f
W=
z
0
mg cos Rd = mgR sin
2 0
W = mgR 1  0 = mgR
a f
198 *P7.21
Energy and Energy Transfer
The same force makes both light springs stretch. (a) The hanging mass moves down by x = x1 + x 2 = mg mg 1 1 + = mg + k1 k2 k1 k 2
= 1.5 kg 9.8 m s 2 (b)
F 1m + 1m I= GH 1 200 N 1 800 N JK
FG H
IJ K
2.04 10 2 m
We define the effective spring constant as k=
FG b g H F 1m + 1m I = =G H 1 200 N 1 800 N JK
1
mg F 1 1 = = + k1 k 2 x mg 1 k1 + 1 k 2
IJ K
1
720 N m
*P7.22
See the solution to problem 7.21. (a)
P7.23
FG 1 + 1 IJ Hk k K F1 1I (b) k=G + J Hk k K L F O N kg m s = k =M P= = NxQ m m
x = mg
1 2 1 1 2 2
kg s2
Section 7.5 Section 7.6 P7.24 (a)
Kinetic Energy and the WorkKinetic Energy Theorem The NonIsolated SystemConservation of Energy KA = 1 0.600 kg 2.00 m s 2 2K B = m 1
b
gb
g
2
= 1.20 J
(b) (c) P7.25 (a) (b)
1 2 mv B = K B : v B = 2
a2fa7.50f =
0.600
5.00 m s 6.30 J
2 W = K = K B  K A = 2 mevB  v 2 j = 7.50 J  1.20 J = A
K= K=
1 1 mv 2 = 0.300 kg 15.0 m s 2 2 1 0.300 30.0 2
b
gb
g
2
= 33.8 J
2
a
fa f
2
=
1 0.300 15.0 2
a
fa f a4f = 4a33.8f =
135 J
Chapter 7
199
P7.26
v i = 6.00 i  2.00 j = m s (a)
2 2 vi = vix + viy = 40.0 m s
e
j
Ki = (b)
1 1 mvi2 = 3.00 kg 40.0 m 2 s 2 = 60.0 J 2 2
b
ge
j
v f = 8.00 i + 4.00 j
v 2 = v f v f = 64.0 + 16.0 = 80.0 m 2 s 2 f 1 3.00 K = K f  K i = m v 2  vi2 = 80.0  60.0 = 60.0 J f 2 2
e
j
a f
P7.27
Consider the work done on the pile driver from the time it starts from rest until it comes to rest at the end of the fall. Let d = 5.00 m represent the distance over which the driver falls freely, and h = 0.12 m the distance it moves the piling.
W = K :
so Thus,
Wgravity + Wbeam =
b ga f d ia f bmg gah + df = b2 100 kgge9.80 m s ja5.12 mf = F=
2
1 1 mv 2  mvi2 f 2 2 mg h + d cos 0+ F d cos 180 = 0  0 . d 0.120 m
8.78 10 5 N . The force on the pile
driver is upward . P7.28 (a) K = K f  K i = vf = (b)
1 mv 2  0 = W = (area under curve from x = 0 to x = 5.00 m) f 2 2 area 2 7.50 J = = 1.94 m s m 4.00 kg
a f
a
f
K = K f  K i = vf =
1 mv 2  0 = W = (area under curve from x = 0 to x = 10.0 m) f 2 2 area 2 22.5 J = = 3.35 m s m 4.00 kg
a f a f e
a
f
(c)
K = K f  K i = vf =
1 mv 2  0 = W = (area under curve from x = 0 to x = 15.0 m) f 2 2 area 2 30.0 J = = 3.87 m s m 4.00 kg
a
f
P7.29
(a)
Ki + W = K f = 0 + W =
1 15.0 10 3 kg 780 m s 2
1 mv 2 f 2
jb
g
2
= 4.56 kJ
(b)
(c) (d)
a f v v b780 m sg  0 = 422 km s = a= 2x 2a0.720 mf F = ma = e15 10 kg je422 10 m s j =
2 f 2 i 2 2 f 3 3 2
F=
4.56 10 3 J W = = 6.34 kN r cos 0.720 m cos 0
6.34 kN
200 P7.30
Energy and Energy Transfer
(a)
v f = 0.096 3 10 8 m s = 2.88 10 7 m s Kf = 1 1 mv 2 = 9.11 10 31 kg 2.88 10 7 m s f 2 2
e
j
e
je
j
2
= 3.78 10 16 J
(b)
Ki + W = K f :
0 + Fr cos = K f F 0.028 m cos 0 = 3.78 10 16 J
a
f
F = 1.35 10 14 N
(c)
F = ma ;
v xf = v xi + a x t
a=
F =
m
1.35 10 14 N = 1. 48 10 +16 m s 2 9.11 10 31 kg
(d)
2.88 10 7 m s = 0 + 1.48 10 16 m s 2 t
e
j
t = 1.94 10 9 s
Check: 1 v xi + v xf t 2 1 0.028 m = 0 + 0 + 2.88 10 7 m s t 2 x f = xi +
d
i
e
j
t = 1.94 10 9 s
Section 7.7 P7.31
Situations Involving Kinetic Friction n  392 N = 0 n = 392 N
Fy = ma y :
(a) (b) (c) (d) (e)
WF
a fa f = Fr cos = a130 fa5.00f cos 0 = a fa f a fa f
f k = k n = 0.300 392 N = 118 N 650 J
Eint = f k x = 118 5.00 = 588 J Wn = nr cos = 392 5.00 cos 90 = 0 W g = mgr cos = 392 5.00 cos 90 = 0 K = K f  K i = Wother  Eint 1 mv 2  0 = 650 J  588 J + 0 + 0 = 62.0 J f 2 vf = 2K f m = 2 62.0 J = 1.76 m s 40.0 kg
FIG. P7.31
a fa f a f
(f)
a
f
Chapter 7
201
P7.32
(a)
1 2 1 2 1 kx i  kx f = 500 5.00 10 2 2 2 2 1 1 1 Ws = mv 2  mvi2 = mv 2  0 f f 2 2 2 Ws = so vf = 2 m 2.00
a fe
j
2
 0 = 0.625 J
c W h = 2a0.625f m s =
0.791 m s
(b)
fa fa fb g 1 0.282 J = b 2.00 kg gv 2 2a0.282 f v = m s = 0.531 m s
2 f f
1 1 mvi2  f k x + Ws = mv 2 f 2 2
0  0.350 2.00 9.80 0.050 0 J + 0.625 J =
a
1 mv 2 f 2 FIG. P7.32
2.00
P7.33
(a)
W g = mg cos 90.0+ Wg
a f = b10.0 kg gd9.80 m s ia5.00 mf cos 110 =
2
168 J
(b)
f k = k n = k mg cos Eint = f k = k mg cos
Eint = 5.00 m 0.400 10.0 9.80 cos 20.0 = 184 J (c) (d) (e) WF = F = 100 5.00 = 500 J K = Wother  Eint = WF + W g  Eint = 148 J 1 1 mv 2  mvi2 f 2 2 2 K 2 148 + vi2 = + 1.50 vf = 10.0 m K = FIG. P7.33
a
fa
fa fa f
a fa f
a f a
a f a f f
2
= 5.65 m s
P7.34
Fy = ma y :
n + 70.0 N sin 20.0147 N = 0 n = 123 N f k = k n = 0.300 123 N = 36.9 N
f
a
(a) (b) (c) (d) (e)
W = Fr cos = 70.0 N 5.00 m cos 20.0 = 329 J W = Fr cos = 123 N 5.00 m cos 90.0 = 0 J W = Fr cos = 147 N 5.00 m cos 90.0 = 0 Eint = Fx = 36.9 N 5.00 m = 185 J K = K f  K i = W  Eint = 329 J  185 J = +144 J
a a a
fa
f
fa fa
f
f
FIG. P7.34
a
fa
f
202 P7.35
Energy and Energy Transfer
k = 0.100 1 K i  f k x + Wother = K f : mvi2  f k x = 0 2 2 2.00 m s v2 1 x = i = = 2.04 m mvi2 = k mgx 2 2 k g 2 0.100 9.80
vi = 2.00 m s
b a
fa f
g
Section 7.8 *P7.36
Power W K f mv 2 0.875 kg 0.620 m s = = = 2 t t t 2 21 10 3 s
Pav =
e
b
j
g
2
= 8.01 W
P7.37 P7.38
Power =
W t
P=
mgh 700 N 10.0 m = = 875 W t 8.00 s
a
fa
f
A 1 300kg car speeds up from rest to 55.0 mi/h = 24.6 m/s in 15.0 s. The output work of the engine is equal to its final kinetic energy, 1 1 300 kg 24.6 m s 2 with power P =
b
gb
g
2
= 390 kJ
390 000 J ~ 10 4 W around 30 horsepower. 15.0 s
P7.39
(a)
distance of 60.0 m sin 30.0 = 30.0 m . Thus,
W = K , but K = 0 because he moves at constant speed. The skier rises a vertical
a
f
Win = Wg = 70.0 kg 9.8 m s 2 30.0 m = 2.06 10 4 J = 20.6 kJ .
(b) The time to travel 60.0 m at a constant speed of 2.00 m/s is 30.0 s. Thus,
b
ge
ja
f
Pinput =
P7.40 (a)
W 2.06 10 4 J = = 686 W = 0.919 hp . t 30.0 s
The distance moved upward in the first 3.00 s is y = vt =
LM 0 + 1.75 m s OPa3.00 sf = 2.63 m . N 2 Q
The motor and the earth's gravity do work on the elevator car: 1 1 mvi2 + Wmotor + mgy cos 180 = mv 2 f 2 2 1 2 Wmotor = 650 kg 1.75 m s  0 + 650 kg g 2.63 m = 1.77 10 4 J 2
b
gb
g
b
ga
f
Also, W = P t so P = (b)
W 1.77 10 4 J = = 5.91 10 3 W = 7.92 hp. t 3.00 s
When moving upward at constant speed v = 1.75 m s the applied force equals the weight = 650 kg 9.80 m s
b
j = 6.37 10 N . Therefore, P = Fv = e6.37 10 N jb1.75 m sg = 1.11 10
2 3 3
ge
b
g
4
W = 14.9 hp .
Chapter 7
203
P7.41
energy = power time For the 28.0 W bulb: Energy used = 28.0 W 1.00 10 4 h = 280 kilowatt hrs total cost = $17.00 + 280 kWh $0.080 kWh = $39.40 For the 100 W bulb: Energy used = 100 W 1.00 10 4 h = 1.00 10 3 kilowatt hrs 1.00 10 4 h = 13.3 # bulb used = 750 h bulb
a
a
fe
fb
j
g
a
fe
j
total cost = 13.3 $0.420 + 1.00 10 3 kWh $0.080 kWh = $85.60 Savings with energyefficient bulb = $85.60  $39.40 = $46.20
b
g e
jb
g
*P7.42
(a)
Burning 1 lb of fat releases energy
1 lb
FG 454 g IJ FG 9 kcal IJ FG 4 186 J IJ = 1.71 10 H 1 lb K H 1 g K H 1 kcal K
7
7
J.
The mechanical energy output is Then
e1.71 10 Jja0.20f = nFr cos .
3.42 10 6 J = nmgy cos 0 3.42 10 6 J = n 50 kg 9.8 m s 2 80 steps 0.150 m 3.42 10 6
3
b ge jb J = ne5.88 10 Jj
ga
f
3.42 10 6 J = 582 . 5.88 10 3 J This method is impractical compared to limiting food intake. where the number of times she must climb the steps is n = (b) Her mechanical power output is
P=
1 hp W 5.88 10 3 J = = 90.5 W = 90.5 W = 0.121 hp . t 65 s 746 W 1h 3 mi 220 kcal h
FG H
IJ K
*P7.43
(a)
The fuel economy for walking is
(b)
IJ FG 1 kcal IJ FG 1.30 10 J IJ = K H 4 186 J K H 1 gal K 1 h F 10 mi I F 1 kcal I F 1.30 10 J I For bicycling G JH KH 400 kcal H h K G 4 186 J J G 1 gal J K = 776 mi gal . FG H
8 8
423 mi gal .
204
Energy and Energy Transfer
Section 7.9 P7.44
Energy and the Automobile
At a speed of 26.8 m/s (60.0 mph), the car described in Table 7.2 delivers a power of P1 = 18.3 kW to the wheels. If an additional load of 350 kg is added to the car, a larger output power of
P2 = P1 + (power input to move 350 kg at speed v)
will be required. The additional power output needed to move 350 kg at speed v is: Pout = f v = r mg v . Assuming a coefficient of rolling friction of r = 0.016 0 , the power output now needed from the engine is
b g b
jb
g
P2 = P1 + 0.016 0 350 kg 9.80 m s 2 26.8 m s = 18.3 kW + 1.47 kW .
With the assumption of constant efficiency of the engine, the input power must increase by the same factor as the output power. Thus, the fuel economy must decrease by this factor:
b
gb
ge
g
P 18 bfuel economyg = FGH P IJK bfuel economyg = FGH 18.3 +.31.47 IJK b6.40 km Lg or bfuel economy g = 5.92 km L .
1 2 2 1 2
P7.45
(a)
fuel needed = =
1 2
mv 2  1 mvi2 f 2
2
b900 kggb24.6 m sg = a0.150fe1.34 10 J galj
1 2 8
useful energy per gallon
=
eff. energy content of fuel 1.35 10 2 gal
b
1 2
mv 2  0 f
g
(b)
73.8 power =
(c)
FG 1 gal IJ FG 55.0 mi IJ FG 1.00 h IJ FG 1.34 10 H 38.0 mi K H 1.00 h K H 3 600 s K H 1 gal
8
J
I a0.150f = JK
8.08 kW
Additional Problems P7.46 At start, v = 40.0 m s cos 30.0 i + 40.0 m s sin 30.0 j
b g b g At apex, v = b 40.0 m sg cos 30.0 i + 0 j = b34.6 m sgi 1 1 And K = mv = b0.150 kg gb34.6 m sg = 90.0 J 2 2
2 2
Chapter 7
205
P7.47
Concentration of Energy output = 0.600 J kg step 60.0 kg
b
gb
1 step gFGH 1.50 m IJK = 24.0 J m
F = 24.0 J m 1 N m J = 24.0 N
b
gb
g
P = Fv
70.0 W = 24.0 N v v = 2.92 m s P7.48 (a) A i = A 1 cos . But also, A i = A x . Thus, Similarly, and where
a
f
a fa f
a Afa1f cos = A
cos = cos = Ay A Az A
x
or cos =
Ax . A
2 2 2 A = Ax + Ay + Az .
(b) P7.49 (a)
cos 2 + cos 2 + cos 2 = x = t + 2.00t 3 Therefore,
FG A IJ + FG A IJ + FG A IJ H AK H AK H AK
x 2 y 2 z
2
=
A2 =1 A2
dx = 1 + 6.00t 2 dt 1 1 K = mv 2 = 4.00 1 + 6.00t 2 2 2 v=
a fe
j = e2.00 + 24.0t
2
2
+ 72.0t 4 J
j
(b)
a12.0tf m s F = ma = 4.00a12.0t f = a 48.0t f N
a= dv = dt
2
(c)
P = Fv = 48.0t 1 + 6.00t 2 =
W=
2.00 0
a
fe
j e48.0t + 288t j W
3
(d)
z
Pdt =
2 .00 0
ze
48.0t + 288 t 3 dt = 1 250 J
j
206 *P7.50
Energy and Energy Transfer
(a)
We write F = ax b
1 000 N = a 0.129 m
b
a f 5 000 N = aa0.315 mf F 0.315 IJ = 2.44 5=G H 0.129 K
b
b b
ln 5 = b ln 2.44 b= ln 5 = 1.80 = b ln 2.44 1 000 N a= = 4.01 10 4 N m1.8 = a 1.80 0.129 m
a
f
(b)
W=
0. 25 m 0
z
Fdx =
4
0. 25 m 0
z
4.01 10 4
N 1.8 x dx m1.8 = 4.01 10
4
= 4.01 10 = 294 J *P7.51
N x 2 .8 m1.8 2.8
0. 25 m
0
N 0.25 m 2.8 m1.8
a
f
2.8
The work done by the applied force is W = Fapplied dx =
i x max
z
f
x max 0
z
  k1 x + k 2 x 2 dx
2 x max 0
e
j
x2 = k 1 x dx + k 2 x dx = k1 2 0 0 = k1 P7.52 (a)
2 x max
z
x max
z
x3 + k2 3
xmax 0
2
+ k2
3 x max
3
The work done by the traveler is mghs N where N is the number of steps he climbs during the ride. N = (time on escalator)(n) where and Then, h atime on escalatorf = vertical velocity of person vertical velocity of person = v + nhs N= nh v + nhs mgnhhs v + nhs
and the work done by the person becomes Wperson = continued on next page
Chapter 7
207
(b)
The work done by the escalator is
We = power time = force exerted speed time = mgvt
where Thus, t= h as above. v + nhs mgvh . v + nhs
b
ga f a
fb
ga f
We =
As a check, the total work done on the person's body must add up to mgh, the work an elevator would do in lifting him. It does add up as follows: 1 mv 2  0 = W , so 2 2W and v = m 2W m W d
W = Wperson + We =
mgnhhs mgvh mgh nhs + v + = = mgh v + nhs v + nhs v + nhs
b
g
P7.53
(a)
K = v2 =
(b) *P7.54
W = F d = Fx d Fx =
During its whole motion from y = 10.0 m to y = 3.20 mm, the force of gravity and the force of the plate do work on the ball. It starts and ends at rest Ki + W = K f 0 + Fg y cos 0+ Fp x cos 180 = 0 mg 10.003 2 m  Fp 0.003 20 m = 0
2
Fp
g b g 5 kg e9.8 m s ja10 mf = = 1.53 10
3.2 10
3
b
5
m
2
N upward
P7.55
(a)
P = Fv = F vi + at = F 0 +
b
g FGH
F t = m
IJ FG F IJ t K HmK
(b)
P=
LM a20.0 Nf MN 5.00 kg
2
OPa3.00 sf = PQ
240 W
208 *P7.56
Energy and Energy Transfer
(a)
W1 = F1 dx =
i  xi 2 + x a  xi 2
z
f
xi 1 + x a xi 1
z
k1 x dx =
1 k 1 x i1 + x a 2
b
g
2
 xi2 = 1
1 2 k 1 x a + 2 x a x i1 2
e
j
(b) (c)
W2 =
z
k 2 x dx =
1 k 2  xi 2 + x a 2
b
g
2
 xi22 =
1 2 k 2 x a  2 x a xi 2 2
e
j
Before the horizontal force is applied, the springs exert equal forces: k 1 xi1 = k 2 xi 2 xi 2 = 1 1 2 2 k1 x a + k1 x a xi1 + k 2 x a  k 2 x a x i 2 2 2 k x 1 1 2 2 = k 1 x a + k 2 x a + k1 x a xi1  k 2 x a 1 i1 2 2 k2 1 2 = k1 + k 2 x a 2 k1 xi1 k2
(d)
W1 + W2 =
b
g
*P7.57
(a)
v = a dt =
0
z ze
t t 0
1.16t  0.21t 2 + 0.24t 3 dt
t
j
t2 t3 t4 = 1.16  0.21 + 0. 24 2 3 4 At t = 0 , vi = 0. At t = 2.5 s ,
= 0.58t 2  0.07t 3 + 0.06t 4
0
v f = 0.58 m s 3 2.5 s Ki + W = K f 0+W = (b) At t = 2.5 s ,
e
ja f  e0.07 m s ja2.5 sf + e0.06 m s ja2.5 sf
2 4 3 5
4
= 4.88 m s
1 1 mv 2 = 1 160 kg 4.88 m s f 2 2
b
g
2
= 1.38 10 4 J
a = 1.16 m s 3 2.5 s  0.210 m s 4 2.5 s
e
j
e
ja f + e0.240 m s ja2.5 sf
2 5
3
= 5.34 m s 2 .
Through the axles the wheels exert on the chassis force
F = ma = 1 160 kg 5.34
and inject power
m s 2 = 6.19 10 3 N
P = Fv = 6.19 10 3 N 4.88 m s = 3.02 10 4 W .
b
g
Chapter 7
209
P7.58
(a)
The new length of each spring is
x 2 + L2 , so its extension is
x 2 + L2  L and the force it exerts is k
FH
x 2 + L2  L toward its
IK
fixed end. The y components of the two spring forces add to zero. Their x components add to F = 2 ik
FH
x 2 + L2  L
IK
x x 2 + L2
= 2 kx i 1 
F GH
L x 2 + L2
I JK F GH
.
FIG. P7.58
(b)
W = Fx dx
i
z
f
W = 2 kx 1 
A
z
0
W = 2 k x dx + kL
A
z
0
A
ze
0
x2 + L
2 1 2
j
2 x dx
x2 W = 2 k 2
0
A
I dx J x +L K ex + L j + kL b1 2g
L
2 2 2
0 2 12
A
W = 0 + kA 2 + 2 kL2  2 kL A 2 + L2 *P7.59 For the rocket falling at terminal speed we have
W = 2 kL2 + kA 2  2 kL A 2 + L2
F = ma
+ R  Mg = 0 Mg = (a) 1 2 DAvT 2
For the rocket with engine exerting thrust T and flying up at the same speed,
F = ma
+T  Mg  R = 0 T = 2 Mg The engine power is P = Fv = TvT = 2 MgvT . (b) For the rocket with engine exerting thrust Tb and flying down steadily at 3vT , 1 2 Rb = DA 3 vT = 9 Mg 2
b g
F = ma
Tb  Mg + 9 Mg = 0 Tb = 8 Mg The engine power is P = Tv = 8 Mg 3 vT = 24MgvT .
210 P7.60
Energy and Energy Transfer
(a)
F1 = 25.0 N cos 35.0 i + sin 35.0 j = F2
a fe j e20.5i + 14.3 jj N = a 42.0 N fecos 150 i + sin 150 jj = e36.4i + 21.0 jj N e15.9 i + 35.3 jj N
2
(b)
F = F1 + F2 =
a=
(c) (d)
F =
m
e3.18 i + 7.07 jj m s e j
v f = v i + at = 4.00 i + 2.50 j m s + 3.18 i + 7.07 j m s 2 3.00 s vf =
e
je
ja
f
e5.54i + 23.7 jj m s e jb ga e2.30i + 39.3 jj m b
1 2 at 2
(e)
r f = ri + v i t +
r f = 0 + 4.00 i + 2.50 j m s 3.00 s + r = r f = (f) (g) Kf = Kf =
f 1 e3.18i + 7.07 jjem s ja3.00 sf 2
2 2 2
2
1 1 mv 2 = 5.00 kg f 2 2
g a5.54f + a23.7f em s j =
2 2 2
1.48 kJ
1 mvi2 + F r 2 1 2 K f = 5.00 kg 4.00 + 2.50 2 K f = 55.6 J + 1 426 J = 1.48 kJ
b
g a f a f bm sg + a15.9 Nfa2.30 mf + a35.3 Nfa39.3 mf
Ws + W g = 0
P7.61
(a)
W = K :
1 2 kxi  0 + mgx cos 90+60 = 0 2 1 2 1.40 10 3 N m 0.100  0.200 9.80 sin 60.0 x = 0 2 x = 4.12 m
a
f
e
j a
f a
fa fa
f
(b)
W = K + Eint :
Ws + W g  Eint = 0
1 2 kxi + mgx cos 150 k mg cos 60 x = 0 2 1 2 1.40 10 3 N m 0.100  0.200 9.80 sin 60.0 x  0.200 9.80 0.400 cos 60.0 x = 0 2 x = 3.35 m
e
j a
f a
fa fa
f
a
fa fa
fa
f
Chapter 7
211
P7.62
(a)
FN 2.00 4.00 6.00 8.00 10.0 12.0
a f Lammf FaNf Lammf
15.0 32.0 49.0 64.0 79.0 98.0 FIG. P7.62 14.0 16.0 18.0 20.0 22.0 112 126 149 175 190
(b)
A straight line fits the first eight points, together with the origin. By leastsquare fitting, its slope is 0.125 N mm 2% = 125 N m 2% In F = kx , the spring constant is k = F , the same as the slope of the Fversusx graph. x
(c) P7.63
F = kx = 125 N m 0.105 m = 13.1 N
b
ga
f
K i + Ws + W g = K f 1 1 1 1 mvi2 + kx i2  kx 2 + mgx cos = mv 2 f f 2 2 2 2 1 1 FIG. P7.63 0 + kxi2  0 + mgxi cos 100 = mv 2 f 2 2 1 1 1.20 N cm 5.00 cm 0.050 0 m  0.100 kg 9.80 m s 2 0.050 0 m sin 10.0 = 0.100 kg v 2 2 2 3 2 0.150 J  8.51 10 J = 0.050 0 kg v
b
ga
fb b
g b g j
ge
jb
g
b
g
v=
0.141 = 1.68 m s 0.050 0 Eint =  K =  1 1 m v 2  vi2 : Eint =  0.400 kg f 2 2
P7.64
(a) (b)
e
b
gea6.00f  a8.00f jbm sg
2 2
2
= 5.60 J
Eint = fr = k mg 2r :
Thus,
a f
5.60 J = k 0.400 kg 9.80 m s 2 2 1.50 m
b
ge
j a
f
k = 0.152 .
(c)
After N revolutions, the object comes to rest and K f = 0 . Thus, or This gives Eint =  K = 0 + K i = 1 mvi2 2
k mg N 2r =
1 2
a f
mvi2
1 mvi2 . 2
1 2 2 2
b8.00 m sg = = N= mg a 2r f a0.152fe9.80 m s j2 a1.50 mf
k
2.28 rev .
212 P7.65
Energy and Energy Transfer
If positive F represents an outward force, (same as direction as r), then W = F dr =
i
z
f
rf ri
ze
6
2 F0 13 r 13  F0 7 r 7 dr
rf ri
j
2 F 13 r 12 F0 7 r 6  W= 0 12 6 W=  F0
13
e
r f12
77
 ri12 r f6
j + F er
0 7
6 f
 ri6
134
6
j= F
0
7
6
r f6  ri6 
F0 13 12 r f  ri12 6
W = 1.03 10
 ri6
 1.89 10
r f12
 ri12
W = 1.03 10 77 1.88 10 6  2.44 10 6 10 60  1.89 10 134 3.54 10 12  5.96 10 8 10 120 W = 2.49 10 21 J + 1.12 10 21 J = 1.37 10 21 J P7.66
Pt = W = K =
The density is
a m f v
2
2
=
m m = . vol Ax x = v, t FIG. P7.66
Substituting this into the first equation and solving for P , since for a constant speed, we get Also, since P = Fv,
P=
F=
Av 2
3
.
Av 2 . 2
Our model predicts the same proportionalities as the empirical equation, and gives D = 1 for the drag coefficient. Air actually slips around the moving object, instead of accumulating in front of it. For this reason, the drag coefficient is not necessarily unity. It is typically less than one for a streamlined object and can be greater than one if the airflow around the object is complicated. P7.67 We evaluate 375dx by calculating 12.8 x + 3.75 x
3 23 .7
z
f + 375a0.100f + ... 375a0.100f = 0.806 a12.8f + 3.75a12.8f a12.9f + 3.75a12.9f a23.6f + 3.75a23.6f
375 0.100
3 3 3
a
and
f + 375a0.100f + ... 375a0.100f = 0.791 . a12.9f + 3.75a12.9f a13.0f + 3.75a13.0f a23.7f + 3.75a23.7f
375 0.100
3 3 3
a
The answer must be between these two values. We may find it more precisely by using a value for x smaller than 0.100. Thus, we find the integral to be 0.799 N m .
Chapter 7
213
*P7.68
P=
(a)
1 Dr 2 v 3 2
Pa =
1 1 1.20 kg m3 1.5 m 2
e
ja
f b8 m sg
2
3
= 2.17 10 3 W
(b)
3 24 m s Pb v b = 3 = Pa v a 8 ms
Pb = 27 2.17 10 3 W = 5.86 10 4 W
P7.69 (a) The suggested equation Pt = bwd implies all of the following cases: (1) v = constant n fk = k n d F
e
F GH
I JK
3
= 3 3 = 27
j
(3)
FG w IJ a2df H 2K F t I F d I P G J = bwG J H 2 K H 2K
Pt = b
(2)
and
(4)
FG t IJ = bFG w IJ d H 2 K H 2K FG P IJ t = bFG w IJ d H 2K H 2K
P
These are all of the proportionalities Aristotle lists.
w FIG. P7.69
(b)
For one example, consider a horizontal force F pushing an object of weight w at constant velocity across a horizontal floor with which the object has coefficient of friction k .
F = ma implies that:
+n  w = 0 and F  k n = 0 so that F = k w As the object moves a distance d, the agent exerting the force does work W = Fd cos = Fd cos 0 = k wd and puts out power P = W t
This yields the equation Pt = k wd which represents Aristotle's theory with b = k . Our theory is more general than Aristotle's. Ours can also describe accelerated motion. *P7.70 (a) So long as the spring force is greater than the friction force, the block will be gaining speed. The block slows down when the friction force becomes the greater. It has maximum speed when kx a  f k = ma = 0.
0
e1.0 10
(b)
3
N m x a  4.0 N = 0
j j
x = 4.0 10 3 m
0
By the same logic,
e1.0 10
3
N m x b  10.0 N = 0
x = 1.0 10 2 m
FIG. P7.70
214
Energy and Energy Transfer
ANSWERS TO EVEN PROBLEMS
P7.2 P7.4 P7.6 P7.8 P7.10 P7.12 P7.14 P7.16 P7.18 P7.20 P7.22 P7.24 P7.26 P7.28 P7.30
1.59 10 3 J
(a) 3.28 10 2 J ; (b) 3.28 10 2 J see the solution 5.33 W 16.0 (a) see the solution; (b) 12.0 J 50.0 J (a) 575 N m ; (b) 46.0 J (a) 9.00 kJ; (b) 11.7 kJ, larger by 29.6% (a) see the solution; (b) mgR mg mg 1 1 + + ; (b) (a) k1 k 2 k1 k2
P7.44 P7.46 P7.48
5.92 km L 90.0 J Ay Ax A ; cos = ; cos = z ; A A A (b) see the solution (a) cos = (a) a = 40.1 kN ; b = 1.80 ; (b) 294 J m 1.8
P7.50
P7.52 P7.54 P7.56 P7.58
(a)
mgnhhs mgvh ; (b) v + nhs v + nhs
1.53 10 5 N upward see the solution (a) see the solution; (b) 2 kL2 + kA 2  2 kL A 2 + L2
FG H
IJ K
1
P7.60
(a) F1 = 20.5 i + 14.3 j N ; F2 = 36.4i + 21.0 j N ;
e
j
(a) 1.20 J; (b) 5.00 m s ; (c) 6.30 J (a) 60.0 J; (b) 60.0 J (a) 1.94 m s ; (b) 3.35 m s ; (c) 3.87 m s (a) 3.78 10 16 J ; (b) 1.35 10 14 N ; (c) 1.48 10 +16 m s 2 ; (d) 1.94 ns (a) 0.791 m s; (b) 0.531 m s (a) 329 J; (b) 0; (c) 0; (d) 185 J; (e) 144 J 8.01 W ~ 10 4 W (a) 5.91 kW; (b) 11.1 kW No. (a) 582; (b) 90.5 W = 0.121 hp P7.62
e
j
e j (c) e 3.18 i + 7.07 jj m s ; (d) e 5.54i + 23.7 jj m s ; (e) e 2.30 i + 39.3 jj m ; (f) 1.48 kJ; (g) 1.48 kJ
(b) 15.9 i + 35.3 j N ;
2
P7.32 P7.34 P7.36 P7.38 P7.40 P7.42
(a) see the solution; (b) 125 N m 2% ; (c) 13.1 N (a) 5.60 J; (b) 0.152; (c) 2.28 rev see the solution (a) 2.17 kW; (b) 58.6 kW (a) x = 4.0 mm ; (b) 1.0 cm
P7.64 P7.66 P7.68 P7.70
8
Potential Energy
CHAPTER OUTLINE
8.1 8.2 Potential Energy of a System The Isolated SystemConservation of Mechanical Energy Conservative and Nonconservative Forces Changes in Mechanical Energy for Nonconservative Forces Relationship Between Conservative Forces and Potential Energy Energy Diagrams and the Equilibrium of a System
ANSWERS TO QUESTIONS
Q8.1 The final speed of the children will not depend on the slide length or the presence of bumps if there is no friction. If there is friction, a longer slide will result in a lower final speed. Bumps will have the same effect as they effectively lengthen the distance over which friction can do work, to decrease the total mechanical energy of the children. Total energy is the sum of kinetic and potential energies. Potential energy can be negative, so the sum of kinetic plus potential can also be negative. Both agree on the change in potential energy, and the kinetic energy. They may disagree on the value of gravitational potential energy, depending on their choice of a zero point.
8.3 8.4
8.5
Q8.2
8.6
Q8.3
Q8.4
(a) (b)
mgh is provided by the muscles. No further energy is supplied to the objectEarth system, but some chemical energy must be supplied to the muscles as they keep the weight aloft. The object loses energy mgh, giving it back to the muscles, where most of it becomes internal energy.
(c)
Q8.5
Lift a book from a low shelf to place it on a high shelf. The net change in its kinetic energy is zero, but the bookEarth system increases in gravitational potential energy. Stretch a rubber band to encompass the ends of a ruler. It increases in elastic energy. Rub your hands together or let a pearl drift down at constant speed in a bottle of shampoo. Each system (two hands; pearl and shampoo) increases in internal energy. Three potential energy terms will appear in the expression of total mechanical energy, one for each conservative force. If you write an equation with initial energy on one side and final energy on the other, the equation contains six potentialenergy terms.
Q8.6
215
216 Q8.7
Potential Energy
(a)
It does if it makes the object's speed change, but not if it only makes the direction of the velocity change. Yes, according to Newton's second law.
(b) Q8.8
The original kinetic energy of the skidding can be degraded into kinetic energy of random molecular motion in the tires and the road: it is internal energy. If the brakes are used properly, the same energy appears as internal energy in the brake shoes and drums. All the energy is supplied by foodstuffs that gained their energy from the sun. Elastic potential energy of plates under stress plus gravitational energy is released when the plates "slip". It is carried away by mechanical waves. The total energy of the ballEarth system is conserved. Since the system initially has gravitational energy mgh and no kinetic energy, the ball will again have zero kinetic energy when it returns to its original position. Air resistance will cause the ball to come back to a point slightly below its initial position. On the other hand, if anyone gives a forward push to the ball anywhere along its path, the demonstrator will have to duck. Using switchbacks requires no less work, as it does not change the change in potential energy from top to bottom. It does, however, require less force (of static friction on the rolling drive wheels of a car) to propel the car up the gentler slope. Less power is required if the work can be done over a longer period of time. There is no work done since there is no change in kinetic energy. In this case, air resistance must be negligible since the acceleration is zero. There is no violation. Choose the book as the system. You did work and the earth did work on the book. The average force you exerted just counterbalanced the weight of the book. The total work on the book is zero, and is equal to its overall change in kinetic energy. Kinetic energy is greatest at the starting point. Gravitational energy is a maximum at the top of the flight of the ball. Gravitational energy is proportional to mass, so it doubles. In stirring cake batter and in weightlifting, your body returns to the same conformation after each stroke. During each stroke chemical energy is irreversibly converted into output work (and internal energy). This observation proves that muscular forces are nonconservative.
Q8.9 Q8.10
Q8.11
Q8.12
Q8.13
Q8.14
Q8.15
Q8.16 Q8.17
Chapter 8
217
Q8.18
Let the gravitational energy be zero at the lowest point in the motion. If you start the vibration by pushing down on the block (2), its kinetic energy becomes extra elastic potential energy in the spring ( Us ). After the block starts moving up at its lower turning point (3), this energy becomes both kinetic energy (K) and gravitational potential energy ( U g ), and then just gravitational energy when the block is at its greatest height (1). The energy then turns back into kinetic and elastic potential energy, and the cycle repeats.
FIG. Q8.18
Q8.19
(a)
Kinetic energy of the running athlete is transformed into elastic potential energy of the bent pole. This potential energy is transformed to a combination of kinetic energy and gravitational potential energy of the athlete and pole as the athlete approaches the bar. The energy is then all gravitational potential of the pole and the athlete as the athlete hopefully clears the bar. This potential energy then turns to kinetic energy as the athlete and pole fall to the ground. It immediately becomes internal energy as their macroscopic motion stops. Rotational kinetic energy of the athlete and shot is transformed into translational kinetic energy of the shot. As the shot goes through its trajectory as a projectile, the kinetic energy turns to a mix of kinetic and gravitational potential. The energy becomes internal energy as the shot comes to rest. Kinetic energy of the running athlete is transformed to a mix of kinetic and gravitational potential as the athlete becomes projectile going over a bar. This energy turns back into kinetic as the athlete falls down, and becomes internal energy as he stops on the ground.
(b)
(c)
The ultimate source of energy for all of these sports is the sun. See question 9. Q8.20 Chemical energy in the fuel turns into internal energy as the fuel burns. Most of this leaves the car by heat through the walls of the engine and by matter transfer in the exhaust gases. Some leaves the system of fuel by work done to push down the piston. Of this work, a little results in internal energy in the bearings and gears, but most becomes work done on the air to push it aside. The work on the air immediately turns into internal energy in the air. If you use the windshield wipers, you take energy from the crankshaft and turn it into extra internal energy in the glass and wiper blades and wipermotor coils. If you turn on the air conditioner, your end effect is to put extra energy out into the surroundings. You must apply the brakes at the end of your trip. As soon as the sound of the engine has died away, all you have to show for it is thermal pollution. A graph of potential energy versus position is a straight horizontal line for a particle in neutral equilibrium. The graph represents a constant function. The ball is in neutral equilibrium. The ball is in stable equilibrium when it is directly below the pivot point. The ball is in unstable equilibrium when it is vertically above the pivot.
Q8.21 Q8.22 Q8.23
218
Potential Energy
SOLUTIONS TO PROBLEMS
Section 8.1 P8.1 (a) Potential Energy of a System With our choice for the zero level for potential energy when the car is at point B, UB = 0 . When the car is at point A, the potential energy of the carEarth system is given by U A = mgy where y is the vertical height above zero level. With 135 ft = 41.1 m , this height is found as:
FIG. P8.1
y = 41.1 m sin 40.0 = 26.4 m .
Thus,
a
f
U A = 1 000 kg 9.80 m s 2 26.4 m = 2.59 10 5 J .
The change in potential energy as the car moves from A to B is
b
ge
ja
f
U B  U A = 0  2.59 10 5 J = 2.59 10 5 J .
(b) With our choice of the zero level when the car is at point A, we have U A = 0 . The potential energy when the car is at point B is given by U B = mgy where y is the vertical distance of point B below point A. In part (a), we found the magnitude of this distance to be 26.5 m. Because this distance is now below the zero reference level, it is a negative number. Thus,
U B = 1 000 kg 9.80 m s 2 26.5 m = 2.59 10 5 J .
The change in potential energy when the car moves from A to B is
b
ge
ja
f
U B  U A = 2.59 10 5 J  0 = 2.59 10 5 J .
Chapter 8
219
P8.2
(a)
We take the zero configuration of system potential energy with the child at the lowest point of the arc. When the string is held horizontal initially, the initial position is 2.00 m above the zero level. Thus, U g = mgy = 400 N 2.00 m = 800 J .
a
fa
f
(b)
From the sketch, we see that at an angle of 30.0 the child is at a vertical height of 2.00 m 1  cos 30.0 above the lowest point of the arc. Thus,
a
fa
f
FIG. P8.2
U g = mgy = 400 N 2.00 m 1  cos 30.0 = 107 J . (c) The zero level has been selected at the lowest point of the arc. Therefore, U g = 0 at this location. *P8.3 The volume flow rate is the volume of water going over the falls each second: 3 m 0.5 m 1.2 m s = 1.8 m3 s The mass flow rate is m V = = 1 000 kg m3 1.8 m3 s = 1 800 kg s t t
a
fa
fa
f
a
fb
g
e
je
j
If the stream has uniform width and depth, the speed of the water below the falls is the same as the speed above the falls. Then no kinetic energy, but only gravitational energy is available for conversion into internal and electric energy. energy mgy m = = gy = 1 800 kg s 9.8 m s 2 5 m = 8.82 10 4 J s t t t The output power is Puseful = efficiency Pin = 0.25 8.82 10 4 W = 2.20 10 4 W The input power is Pin =
b
g
b e
ge
ja f
j
The efficiency of electric generation at Hoover Dam is about 85%, with a head of water (vertical drop) of 174 m. Intensive research is underway to improve the efficiency of low head generators.
Section 8.2 *P8.4 (a)
The Isolated SystemConservation of Mechanical Energy One child in one jump converts chemical energy into mechanical energy in the amount that her body has as gravitational energy at the top of her jump: mgy = 36 kg 9.81 m s 2 0.25 m = 88.3 J . For all of the jumps of the children the energy is
e ja f 12e1.05 10 j88.3 J = 1.11 10 J .
6 9
(b)
The seismic energy is modeled as E =
0.01 1.11 10 9 J = 1.11 10 5 J , making the Richter 100 log E  4.8 log 1.11 10 5  4.8 5.05  4.8 magnitude = = = 0.2 . 1.5 1.5 1.5
220 P8.5
Potential Energy
Ui + K i = U f + K f :
mgh + 0 = mg 2 R +
a f 1 mv 2 1 g a3.50 Rf = 2 g a Rf + v 2
2
2
v = 3.00 gR
F = m
v : R
2
n + mg = m n=m
2
LM v  g OP = m L 3.00 gR  g O = 2.00mg N R Q MN R PQ n = 2.00e5.00 10 kg je9.80 m s j
3 2
v2 R
= 0.098 0 N downward P8.6 From leaving ground to the highest point,
FIG. P8.5
K i + Ui = K f + U f 1 2 m 6.00 m s + 0 = 0 + m 9.80 m s 2 y 2
b
g
e
j
The mass makes no difference:
b6.00 m sg = y = a2fe9.80 m s j
2 2
1.84 m
*P8.7
(a)
1 1 1 1 mvi2 + kx i2 = mv 2 + kx 2 f f 2 2 2 2 1 1 2 0 + 10 N m 0.18 m = 0.15 kg v 2 + 0 f 2 2 vf
f b ga g F 10 N I FG 1 kg m IJ = = a0.18 mf G H 0.15 kg m JK H 1 N s K b
2
1.47 m s
(b)
K i + U si = K f + U sf 1 0 + 10 N m 0.18 m 2
b
ga
f
2
0.162 J = vf =
2 0.138 J = 1.35 m s 0.15 kg
a
1 0.15 kg v 2 + 0.024 5 J f 2
b
g
1 0.15 kg v 2 f 2 1 + 10 N m 0. 25 m  0.18 m 2 =
b
b
g ga
f
2
FIG. P8.7
f
Chapter 8
221
*P8.8
The energy of the car is E = E=
1 mv 2 + mgd sin where d is the distance it has moved along the track. 2 dE dv P= = mv + mgv sin dt dt (a) When speed is constant,
1 mv 2 + mgy 2
P = mgv sin = 950 kg 9.80 m s 2 2.20 m s sin 30 = 1.02 10 4 W
(b) 2. 2 m s  0 dv =a= = 0.183 m s 2 dt 12 s Maximum power is injected just before maximum speed is attained:
e
jb
g
P = mva + mgv sin = 950 kg 2.2 m s 0.183 m s 2 + 1.02 10 4 W = 1.06 10 4 W
(c) At the top end, 1 1 2.20 m s mv 2 + mgd sin = 950 kg 2 2
b
ge
j
FG b H
g + e9.80 m s j1 250 m sin 30IJK =
2 2
5.82 10 6 J
*P8.9
(a)
Energy of the objectEarth system is conserved as the object moves between the release point and the lowest point. We choose to measure heights from y = 0 at the top end of the string.
eK + U j = eK + U j :
g i g f
0 + mgyi =
2
e9.8 m s ja2 m cos 30f = 1 v + e9.8 m s ja2 mf 2 v = 2e9.8 m s ja 2 mfa1  cos 30f = 2.29 m s
2 f 2 f 2
1 mv 2 + mgy f f 2
(b)
Choose the initial point at = 30 and the final point at = 15 : 0 + mg  L cos 30 = vf =
a
f 2 gLacos 15 cos 30f = 2e9.8 m s ja 2 mfacos 15 cos 30f =
1 mv 2 + mg  L cos 15 f 2
2
f
a
1.98 m s
P8.10
Choose the zero point of gravitational potential energy of the objectspringEarth system as the configuration in which the object comes to rest. Then because the incline is frictionless, we have EB = E A : K B + U gB + U sB = K A + U gA + U sA or 0 + mg d + x sin + 0 = 0 + 0 + kx 2 x . 2mg sin
a
f
1 2 kx . 2
Solving for d gives
d=
222 P8.11
Potential Energy
From conservation of energy for the blockspringEarth system, U gt = U si , or
b0.250 kg ge9.80 m s jh = FGH 1 IJK b5 000 N mga0.100 mf 2
2
2
This gives a maximum height h = 10.2 m . P8.12 (a) The force needed to hang on is equal to the force F the trapeze bar exerts on the performer. From the freebody diagram for the performer's body, as shown, F  mg cos = m or F = mg cos + m v2 v2
FIG. P8.11
FIG. P8.12
Apply conservation of mechanical energy of the performerEarth system as the performer moves between the starting point and any later point: mg mv 2
b
 cos i = mg
g
a
 cos +
f
1 mv 2 2
Solve for (b)
and substitute into the force equation to obtain F = mg 3 cos  2 cos i
b
g
.
At the bottom of the swing, = 0 so F = mg 3  2 cos i
b
F = 2mg = mg 3  2 cos i which gives
b
g
g
i = 60.0 .
Chapter 8
223
P8.13
Using conservation of energy for the system of the Earth and the two objects (a)
b5.00 kg gga4.00 mf = b3.00 kg gga4.00 mf + 1 a5.00 + 3.00fv 2
v = 19.6 = 4.43 m s
2
(b)
Now we apply conservation of energy for the system of the 3.00 kg object and the Earth during the time interval between the instant when the string goes slack and the instant at which the 3.00 kg object reaches its highest position in its free fall. 1 3.00 v 2 = mg y = 3.00 gy 2 y = 1.00 m y max = 4.00 m + y = 5.00 m
a f
FIG. P8.13
P8.14
m1 > m 2 (a) m1 gh = v= 1 m1 + m 2 v 2 + m 2 gh 2
b
g
2 m1  m 2 gh
1 + m2
b bm
g g
(b)
Since m 2 has kinetic energy
1 m 2 v 2 , it will rise an additional height h determined from 2 m 2 g h = 1 m2 v 2 2
or from (a), h = m1  m 2 h v2 = m1 + m 2 2g
b b
g g
The total height m 2 reaches is h + h = P8.15
2m1 h . m1 + m 2
The force of tension and subsequent force of compression in the rod do no work on the ball, since they are perpendicular to each step of displacement. Consider energy conservation of the ballEarth system between the instant just after you strike the ball and the instant when it reaches the top. The speed at the top is zero if you hit it just hard enough to get it there. K i + U gi = K f + U gf : 1 mvi2 + 0 = 0 + mg 2L 2 vi = 4 gL = 4 9.80 0.770 vi = 5.49 m s
initial L
final
L
a f a fa f
vi FIG. P8.15
224 *P8.16
Potential Energy
efficiency =
useful output energy useful output power = total input energy total input power = 2 water v water t gy
e=
b1 2gm ev tj
air 2
m water gy t
b
air r
2
g e v tj
2
=
2 w v w t gy
b
g
a r v
2 3
where is the length of a cylinder of air passing through the mill and v w is the volume of water pumped in time t. We need inject negligible kinetic energy into the water because it starts and ends at rest.
3 v w e a r 2 v 3 0.275 1.20 kg m 1.15 m 11 m s = = 2 w gy t 2 1 000 kg m 3 9.80 m s 2 35 m
= 2.66 10 3 m3 P8.17 (a) K i + U gi = K f + U gf
e ja fb g e je j F 1 000 L IJ FG 60 s IJ = 160 L min sG H 1 m K H 1 min K
2 3
3
1 1 mvi2 + 0 = mv 2 + mgy f f 2 2 1 1 1 2 2 2 mv xi + mv yi = mv xf + mgy f 2 2 2 But v xi = v xf , so for the first ball
2 v yi
yf = and for the second
2g
=
b1 000 sin 37.0g 2a9.80f b1 000g = = 2a9.80f
2
2
= 1.85 10 4 m
yf (b)
5.10 10 4 m
The total energy of each is constant with value 1 20.0 kg 1 000 m s 2
b
gb
g
2
= 1.00 10 7 J .
Chapter 8
225
P8.18
In the swing down to the breaking point, energy is conserved: mgr cos = at the breaking point consider radial forces 1 mv 2 2
Fr = mar
+Tmax  mg cos = m Eliminate v2 = 2 g cos r Tmax  mg cos = 2mg cos Tmax = 3mg cos v2 r
= cos 1
F T I = cos GH 3mg JK
max
1
F I 44 GG 3 2.00 kg .59.N m s JJ ge 80 j K Hb
2
= 40.8
*P8.19 (a) For a 5m cord the spring constant is described by F = kx , mg = k 1.5 m . For a longer cord of length L the stretch distance is longer so the spring constant is smaller in inverse proportion:
a
f
k=
eK + U d
5 m mg = 3.33 mg L L 1.5 m
g
+ Us
j = eK + U
i
g
+ Us
j
f
1 0 + mgyi + 0 = 0 + mgy f + kx 2 f 2 mg 2 1 1 mg yi  y f = kx 2 = 3.33 xf f 2 2 L
initial
final
FIG. P8.19(a)
i
here yi  y f = 55 m = L + x f 1 2 55.0 mL = 3.33 55.0 m  L 2 55.0 mL = 5.04 10 3 m 2  183 mL + 1.67 L2
a
f
0 = 1.67L2  238 L + 5.04 10 3 = 0 L= 238 238 2  4 1.67 5.04 10 3
a fe 2a1.67f
j = 238 152 =
3.33
25.8 m
only the value of L less than 55 m is physical. (b) mg 25.8 m F = ma k = 3.33 x max = x f = 55.0 m  25.8 m = 29.2 m + kx max  mg = ma mg 3.33 29. 2 m  mg = ma 25.8 m a = 2.77 g = 27.1 m s 2
226 *P8.20
Potential Energy
When block B moves up by 1 cm, block A moves down by 2 cm and the separation becomes 3 cm. v h We then choose the final point to be when B has moved up by and has speed A . Then A has 3 2 2h moved down and has speed v A : 3
eK
A
+ KB + Ug
j = eK
i
A
+ KB + Ug
j
0+0+0=
v 1 1 2 mv A + m A 2 2 2
FG IJ H K
f
2
+
mgh mg 2 h  3 3
mgh 5 2 = mv A 3 8 vA = 8 gh 15
Section 8.3 P8.21
Conservative and Nonconservative Forces
Fg = mg = 4.00 kg 9.80 m s 2 = 39.2 N (a) Work along OAC = work along OA + work along AC = Fg OA cos 90.0+ Fg AC cos 180
b
ge
j
y B C (5.00, 5.00) m
a f a f = a39.2 N fa5.00 mf + a39.2 N fa5.00 mfa 1f
= 196 J
(b)
W along OBC = W along OB + W along BC = 39.2 N 5.00 m cos 180+ 39.2 N 5.00 m cos 90.0
O
A
x
a
fa
f
a
fa
f
FIG. P8.21
= 196 J (c) Work along OC = Fg OC cos135 = 39.2 N 5.00
a
fe
a f F 1 IJ = 2 mjG  H 2K
196 J
The results should all be the same, since gravitational forces are conservative. P8.22 (a)
W = F dr and if the force is constant, this can be written as
W = F dr = F r f  ri , which depends only on end points, not path.
z
z
z
d
i
(b)
W = F dr =
ze
3 i + 4 j dx i + dyj = 3.00 N
5.00 m 0
W = 3.00 N x 0
a
f
5.00 m
je j a f z dx + a4.00 Nf z dy + a 4.00 N fy = 15.0 J + 20.0 J = 35.0 J
5.00 m 0 5.00 m 0
The same calculation applies for all paths.
Chapter 8
227
P8.23
(a) and since along this path, y = 0
WOA =
5.00 m 0
z z
0
dx i 2 y i + x 2 j =
e
j
5.00 m
z
0
2 ydx
WOA = 0 W AC =
5 .00 m
dyj 2 y i + x 2 j =
e
j
5 .00 m 2 0
z
x dy
For x = 5.00 m, and
W AC = 125 J WOAC = 0 + 125 = 125 J WOB =
5 .00 m 0
(b) since along this path, x = 0 ,
z
0
dyj 2 y i + x 2 j =
e
j
5.00 m 2 0
z
x dy
WOB = 0 WBC =
5.00 m
z
dx i 2 yi + x 2 j =
e
j
5.00 m
z
0
2 ydx
since y = 5.00 m,
WBC = 50.0 J WOBC = 0 + 50.0 = 50.0 J
(c)
WOC = Since x = y along OC, WOC =
5.00 m 0
ze ze
dx i + dyj 2 y i + x 2 j = 2 x + x 2 dx = 66.7 J
je
j
j z e2ydx + x dyj
2
(d) P8.24 (a)
F is nonconservative since the work done is path dependent.
a K f
= W = W g = mgh = mg 5.00  3.20 1 1 2 mv B  mv 2 = m 9.80 1.80 A 2 2 v B = 5.94 m s
AB
a
f
A B 5.00 m 3.20 m C 2.00 m
a fa f a
2 Similarly, vC = v A + 2 g 5.00  2.00 = 7.67 m s
f
FIG. P8.24
(b)
Wg
A C
= mg 3.00 m = 147 J
a
f
228 P8.25
Potential Energy
(a)
F = 3.00 i + 5.00 j N m = 4.00 kg r = 2.00 i  3.00 j m W = 3.00 2.00 + 5.00 3.00 = 9.00 J The result does not depend on the path since the force is conservative.
e
j
e
a f
j
a
f
(b)
W = K 4.00 4.00 v 2 9.00 =  4.00 2 2 so v = 32.0  9.00 = 3.39 m s 2.00
Fa f I GH JK
2
(c)
U = W = 9.00 J
Section 8.4 P8.26 (a)
Changes in Mechanical Energy for Nonconservative Forces U f = K i  K f + Ui U f = 30.0  18.0 + 10.0 = 22.0 J E = 40.0 J
(b) P8.27
Yes, Emech = K + U is not equal to zero. For conservative forces K + U = 0 .
The distance traveled by the ball from the top of the arc to the bottom is R . The work done by the nonconservative force, the force exerted by the pitcher, is
E = Fr cos 0 = F R .
a f
We shall assign the gravitational energy of the ballEarth system to be zero with the ball at the bottom of the arc. Then becomes or Emech = 1 1 mv 2  mvi2 + mgy f  mgyi f 2 2 1 1 mv 2 = mvi2 + mgyi + F R f 2 2
a f
v f = vi2 + 2 gyi + v f = 26.5 m s
2 F R = m
a f a15.0f + 2a9.80fa1.20f + 2a30.0f a0.600f 0.250
2
*P8.28
The useful output energy is 120 Wh 1  0.60 = mg y f  yi = Fg y y = 120 W 3 600 s 0.40 890 N
a
b
f
g FG J IJ FG N m IJ = H W sKH J K
d
i
194 m
Chapter 8
229
*P8.29
As the locomotive moves up the hill at constant speed, its output power goes into internal energy plus gravitational energy of the locomotiveEarth system:
Pt = mgy + fr = mgr sin + fr
As the locomotive moves on level track,
P = mgv f sin + fv f
F 746 W I = f b27 m sg f = 2.76 10 N GH 1 hp JK F 5 m IJ + e2.76 10 Njv Then also 746 000 W = b160 000 kg ge9.8 m s jv G H 100 m K
P = fvi
1 000 hp
4 2 f 4
f
vf = P8.30
746 000 W 1.06 10 5 N
= 7.04 m s
We shall take the zero level of gravitational potential energy to be at the lowest level reached by the diver under the water, and consider the energy change from when the diver started to fall until he came to rest. E = 1 1 mv 2  mvi2 + mgy f  mgyi = f k d cos 180 f 2 2 mg yi  y f d
0  0  mg yi  y f =  f k d fk = P8.31
d
d
i = b70.0 kg ge9.80 m s ja10.0 m + 5.00 mf =
2
i
5.00 m
2.06 kN
Ui + K i + Emech = U f + K f :
m 2 gh  fh =
1 1 m1 v 2 + m 2 v 2 2 2 1 m1 + m 2 v 2 2
f = n = m1 g m 2 gh  m1 gh = v2 =
b
g
2 m 2  m1 hg m1 + m 2
b
gb g
ja
FIG. P8.31
v= Emech = K f  K i + U gf  U gi
2 9.80 m s 2 1.50 m 5.00 kg  0.400 3.00 kg 8.00 kg
e
f
b
g
= 3.74 m s
P8.32
d
i e
j
But Emech = Wapp  fx , where Wapp is the work the boy did pushing forward on the wheels. Thus, or Wapp = K f  K i + U gf  U gi + fx Wapp Wapp
2 f 2 i
j 1 = me v  v j + mg a  hf + fx 2 1 = a 47.0 f a6. 20f  a1.40f  a 47.0fa9.80fa 2.60 f + a 41.0 fa12.4f 2
2 2
d
i e
FIG. P8.32
Wapp = 168 J
230 P8.33
Potential Energy
(a) (b) (c)
K =
1 1 m v 2  vi2 =  mvi2 = 160 J f 2 2
e
j
U = mg 3.00 m sin 30.0 = 73.5 J The mechanical energy converted due to friction is 86.5 J f= 86.5 J = 28.8 N 3.00 m FIG. P8.33
a
f
(d)
f = k n = k mg cos 30.0 = 28.8 N 28.8 N k = = 0.679 5.00 kg 9.80 m s 2 cos 30.0
b
ge
j
P8.34
Consider the whole motion: K i + U i + Emech = K f + U f (a) 0 + mgyi  f1 x1  f 2 x 2 =
2
b80.0 kg ge9.80 m s j1 000 m  a50.0 Nfa800 mf  b3 600 Nga200 mf = 1 b80.0 kg gv 2 1 784 000 J  40 000 J  720 000 J = b80.0 kg gv 2 2b 24 000 Jg = 24.5 m s v =
2 f f
1 mv 2 + 0 f 2
2 f
80.0 kg
(b) (c)
Yes this is too fast for safety. Now in the same energy equation as in part (a), x 2 is unknown, and x1 = 1 000 m  x 2 : 784 000 J  50.0 N 1 000 m  x 2  3 600 N x 2 = 784 000 J  50 000 J  3 550 N x 2 = 1 000 J x 2 = 733 000 J = 206 m 3 550 N
a
fb
g b
g
b
g
1 80.0 kg 5.00 m s 2
b
gb
g
2
(d)
Really the air drag will depend on the skydiver's speed. It will be larger than her 784 N weight only after the chute is opened. It will be nearly equal to 784 N before she opens the chute and again before she touches down, whenever she moves near terminal speed.
P8.35
(a)
aK + Uf + E
i
mech
= K +U f :
a
f
Chapter 8
231
0+
1 8.00 N m 5.00 10 2 m 2 v= (b) 2 5. 20 10 3 J 5.30 10
3
b
1 2 1 kx  fx = mv 2 + 0 2 2
ge
j  e3.20 10
2
2
N 0.150 m =
ja
f 1 e5.30 10 2
3
kg v 2
j
e
kg
j=
1.40 m s
When the spring force just equals the friction force, the ball will stop speeding up. Here Fs = kx ; the spring is compressed by 3. 20 10 2 N = 0.400 cm 8.00 N m and the ball has moved 5.00 cm  0.400 cm = 4.60 cm from the start.
(c)
Between start and maximum speed points, 1 2 1 1 kxi  fx = mv 2 + kx 2 f 2 2 2 2 1 1 1 8.00 5.00 10 2  3.20 10 2 4.60 10 2 = 5.30 10 3 v 2 + 8.00 4.00 10 3 2 2 2 v = 1.79 m s
e
j e
je
j e
j
e
j
2
P8.36
Fy = n  mg cos 37.0 = 0
n = mg cos 37.0 = 400 N f = n = 0.250 400 N = 100 N
a
f
f
a100fa20.0f = U + U + K + K U = m g d h  h i = a50.0fa9.80fa 20.0 sin 37.0f = 5.90 10 U = m g d h  h i = a100 fa9.80fa 20.0f = 1.96 10 1 K = m e v  v j 2 m 1 K = m e v  v j = K = 2 K 2 m
A i B A B A A B B f i 4 A A 2 f 2 i B B 2 f 2 i B A A A
 fx = Emech
3
Adding and solving, K A = 3.92 kJ . FIG. P8.36
232 P8.37
Potential Energy
(a)
The object moved down distance 1.20 m + x. Choose y = 0 at its lower point. K i + U gi + U si + Emech = K f + U gf + U sf 0 + mgyi + 0 + 0 = 0 + 0 +
2
b1.50 kg ge9.80 m s ja1.20 m + xf = 1 b320 N mgx 2 0 = b160 N mgx  a14.7 N fx  17.6 J 14.7 N a 14.7 N f  4b160 N mga17.6 N mf x= 2b160 N mg
2 2 2
1 2 kx 2
x=
14.7 N 107 N 320 N m
The negative root tells how high the object will rebound if it is instantly glued to the spring. We want x = 0.381 m (b) From the same equation,
b1.50 kg ge1.63 m s ja1.20 m + xf = 1 b320 N mgx 2
2
2
0 = 160 x 2  2.44x  2.93 The positive root is x = 0.143 m . (c) The equation expressing the energy version of the nonisolated system model has one more term: mgyi  fx = 1 2 kx 2
b1.50 kg ge9.80 m s ja1.20 m + xf  0.700 Na1.20 m + xf = 1 b320 N mgx 2
2
2
17.6 J + 14.7 Nx  0.840 J  0.700 Nx = 160 N m x 2 160 x 2  14.0 x  16.8 = 0 x= 14.0
a14.0f  4a160fa16.8f
2
320
x = 0.371 m
Chapter 8
233
P8.38
The total mechanical energy of the skysurferEarth system is Emech = K + U g = Since the skysurfer has constant speed, dEmech dv dh = mv + mg = 0 + mg  v =  mgv . dt dt dt The rate the system is losing mechanical energy is then dEmech = mgv = 75.0 kg 9.80 m s 2 60.0 m s = 44.1 kW . dt 1 mv 2 + mgh . 2
a f
b
ge
jb
g
*P8.39
(a)
Let m be the mass of the whole board. The portion on the rough surface has mass normal force supporting it is a=
k gx opposite to the motion. L
mgx mxg and the frictional force is k = ma . Then L L
mx . The L
(b)
In an incremental bit of forward motion dx, the kinetic energy converted into internal mgx energy is f k dx = k dx . The whole energy converted is L
L k mgx mg x 2 1 mv 2 = dx = k 2 2 L L 0
z
L
=
0
k mgL 2
v = k gL
Section 8.5 P8.40 (a)
Relationship Between Conservative Forces and Potential Energy U =   Ax + Bx 2 dx =
0
ze
x
j
Ax 2 Bx 3  2 3
(b)
U =  K =
3.00 m
FG  5.00 A + 19.0 BIJ H 2 3 K
5 .00 m 1
2.00 m
z
Fdx =
A 3.00 2  2.00 2
e
j a f
2

B 3.00
a f  a2.00f
3
3
3
=
5.00 19.0 A B 2 3
P8.41
(a) (b) (c)
W = Fx dx = K + U = 0 K = K f 
z
z a2x + 4fdx = FGH 2x2
2
+ 4x
I JK
5 .00 m
= 25.0 + 20.0  1.00  4.00 = 40.0 J
1
U =  K = W = 40.0 J K f = K +
2 mv1 = 62.5 J 2
2 mv1 2
234 P8.42
Potential Energy
3x 3 y  7x U = =  9x 2 y  7 = 7  9x 2 y Fx =  x x 3x 3 y  7x U = =  3 x 3  0 = 3 x 3 Fy =  y y
e e
j e j e
j
j
Thus, the force acting at the point x , y is F = Fx i + Fy j = P8.43 Ur =
b g
e7  9 x y ji  3 x j .
2 3
A r d A A U Fr =  = = 2 . The positive value indicates a force of repulsion. dr r r r
af
FG IJ H K
Section 8.6 P8.44
Energy Diagrams and the Equilibrium of a System
stable
unstable FIG. P8.44
neutral
P8.45
(a) (b) (c)
Fx is zero at points A, C and E; Fx is positive at point B and negative at point D. A and E are unstable, and C is stable. Fx B A C E D FIG. P8.45 x (m)
Chapter 8
235
P8.46
(a)
There is an equilibrium point wherever the graph of potential energy is horizontal: At r = 1.5 mm and 3.2 mm, the equilibrium is stable. At r = 2.3 mm , the equilibrium is unstable. A particle moving out toward r approaches neutral equilibrium.
(b) (c)
The system energy E cannot be less than 5.6 J. The particle is bound if 5.6 J E < 1 J . If the system energy is 3 J, its potential energy must be less than or equal to 3 J. Thus, the particle's position is limited to 0.6 mm r 3.6 mm . K + U = E . Thus, K max = E  U min = 3.0 J  5.6 J = 2.6 J . Kinetic energy is a maximum when the potential energy is a minimum, at r = 1.5 mm .
(d) (e) (f) P8.47 (a)
a
f
3 J + W = 1 J . Hence, the binding energy is W = 4 J .
When the mass moves distance x, the length of each spring changes from L to k
FH
x 2 + L2  L towards its fixed end. The ycomponents
IK
x 2 + L2 , so each exerts force
cancel out and the x components add to: Fx = 2 k
FH
x
x 2 + L2  L
IK FG H
x x +L
2 2
I = 2kx + JK
2 kLx x 2 + L2 FIG. P8.47(a)
Choose U = 0 at x = 0 . Then at any point the potential energy of the system is
(b)
z z FGH 2kx + x2kLxL IJK dx = 2kz xdx  2kLz + U a x f = kx + 2 kLF L  x + L I H K U a x f = 40.0 x + 96.0F 1.20  x + 1.44 I H K
U x =  Fx dx = 
0 2
af
x
x
x 2
x x + L2
0
2
2
dx
0
0
2
2
2
2
For negative x, U x has the same value as for positive x. The only equilibrium point (i.e., where Fx = 0) is x = 0 . (c) K i + U i + Emech = K f + U f 1 0 + 0.400 J + 0 = 1.18 kg v 2 + 0 f 2 v f = 0.823 m s
af
b
g
FIG. P8.47(b)
236
Potential Energy
Additional Problems P8.48 The potential energy of the blockEarth system is mgh. An amount of energy k mgd cos is converted into internal energy due to friction on the incline. Therefore the final height y max is found from mgy max = mgh  k mgd cos where y max sin mgy max = mgh  k mgy max cot d= Solving, y max = P8.49 h 1 + k cot . h y max
FIG. P8.48
At a pace I could keep up for a halfhour exercise period, I climb two stories up, traversing forty steps each 18 cm high, in 20 s. My output work becomes the final gravitational energy of the system of the Earth and me, mgy = 85 kg 9.80 m s 2 40 0.18 m = 6 000 J making my sustainable power 6 000 J = ~ 10 2 W . 20 s
b
ge
ja
f
P8.50
v = 100 km h = 27.8 m s The retarding force due to air resistance is R= 1 1 DAv 2 = 0.330 1.20 kg m 3 2.50 m 2 27.8 m s 2 2
a
fe
je
jb
g
2
= 382 N
Comparing the energy of the car at two points along the hill, K i + U gi + E = K f + U gf or K i + U gi + We  R s = K f + U gf
a f
where We is the work input from the engine. Thus, We = R s + K f  K i + U gf  U gi
a f d
i e
j
Recognizing that K f = K i and dividing by the travel time t gives the required power input from the engine as
FG W IJ = RFG s IJ + mgFG y IJ = Rv + mgv sin H t K H t K H t K P = a382 N fb 27.8 m sg + b1 500 kg ge9.80 m s jb 27.8 m sg sin 3.20
P=
e 2
P = 33.4 kW = 44.8 hp
Chapter 8
237
P8.51
m = mass of pumpkin R = radius of silo top
Fr = mar n  mg cos = m
v2 R
When the pumpkin first loses contact with the surface, n = 0 . Thus, at the point where it leaves the surface: v 2 = Rg cos .
FIG. P8.51
Choose U g = 0 in the = 90.0 plane. Then applying conservation of energy for the pumpkinEarth system between the starting point and the point where the pumpkin leaves the surface gives K f + U gf = K i + U gi 1 mv 2 + mgR cos = 0 + mgR 2 Using the result from the force analysis, this becomes 1 mRg cos + mgR cos = mgR , which reduces to 2 cos = 2 , and gives = cos 1 2 3 = 48.2 3
b g
ge
as the angle at which the pumpkin will lose contact with the surface. P8.52 (a) (b) U A = mgR = 0. 200 kg 9.80 m s 2 0.300 m = 0.588 J K A + U A = KB + UB K B = K A + U A  U B = mgR = 0.588 J (c) vB = 2K B = m 2 0.588 J = 2.42 m s 0.200 kg
b
ja
f
a
f
(d)
UC = mghC = 0.200 kg 9.80 m s 2 0.200 m = 0.392 J K C = K A + U A  U C = mg h A  hC K C = 0.200 kg 9.80 m s 2 0.300  0.200 m = 0.196 J
b
ge
b
ge
b
ja
g
ja
f
FIG. P8.52
f
P8.53
(a) (b)
KB =
1 1 2 mv B = 0.200 kg 1.50 m s 2 2
b
gb
g
2
= 0.225 J
Emech = K + U = K B  K A + U B  U A = K B + mg hB  hA
2
b g = 0.225 J + b0.200 kg ge9.80 m s ja0  0.300 mf
= 0.225 J  0.588 J = 0.363 J
(c)
It's possible to find an effective coefficient of friction, but not the actual value of since n and f vary with position.
238 P8.54
Potential Energy
The gain in internal energy due to friction represents a loss in mechanical energy that must be equal to the change in the kinetic energy plus the change in the potential energy. Therefore,  k mgx cos = K + and since vi = v f = 0 , K = 0. Thus,  k 2.00 9.80 cos 37.0 0.200 = 1 2 kx  mgx sin 2
a fa fa
fa
f a100fa02.200f  a2.00fa9.80fasin 37.0fa0.200f
2
and we find k = 0.115 . Note that in the above we had a gain in elastic potential energy for the spring and a loss in gravitational potential energy. P8.55 (a) Since no nonconservative work is done, E = 0 Also K = 0 therefore, Ui = U f where Ui = mg sin x and U f = 1 2 kx 2 FIG. P8.55 k = 100 N/m 2.00 kg
b
g
Substituting values yields 2.00 9.80 sin 37.0 = 100
a fa f
x a f 2 and solving we find
x = 0.236 m (b)
F = ma . Only gravity and the spring force act on the block, so
 kx + mg sin = ma For x = 0.236 m ,
a = 5.90 m s 2 . The negative sign indicates a is up the incline.
The acceleration depends on position . (c) U(gravity) decreases monotonically as the height decreases. U(spring) increases monotonically as the spring is stretched. K initially increases, but then goes back to zero.
Chapter 8
239
P8.56
k = 2.50 10 4 N m, x A = 0.100 m, (a) Emech = K A + U gA + U sA
m = 25.0 kg
Ug
x =0
= Us
x=0
=0
Emech = 0 + mgx A + Emech
1 2 kx A 2 = 25.0 kg 9.80 m s 2 0.100 m
b
ge
Emech (b)
1 2.50 10 4 2 = 24.5 J + 125 J = 100 J +
e
f ja N mja 0.100 mf
2
Since only conservative forces are involved, the total energy of the childpogostickEarth system at point C is the same as that at point A. K C + U gC + U sC = K A + U gA + U sA : 0 + 25.0 kg 9.80 m s 2 xC + 0 = 0  24.5 J + 125 J x C = 0.410 m
b
ge
j
(c)
K B + U gB + U sB = K A + U gA + U sA :
1 2 25.0 kg v B + 0 + 0 = 0 + 24.5 J + 125 J 2 v B = 2.84 m s
b
g
a
f
(d)
K and v are at a maximum when a = F m = 0 (i.e., when the magnitude of the upward spring force equals the magnitude of the downward gravitational force). This occurs at x < 0 where or Thus,
k x = mg
x=
b25.0 kg ge9.8 m s j = 9.80 10
2
2.50 10 4 N m
3
m
K = K max at x = 9.80 mm
(e)
K max = K A + U gA  U g or
e
j + eU  U j 1 b25.0 kg gv = b25.0 kgge9.80 m s j a0.100 mf  b0.009 8 mg 2 1 + e 2.50 10 N mj a 0.100 mf  b0.009 8 mg 2
x =9.80 mm sA s x =9.80 mm 2 max 2 4 2 2
yielding P8.57 Emech =  fx E f  Ei =  f d BC 1 2 kx  mgh =  mgd BC 2 mgh  1 kx 2 2 = 0.328 = mgd BC
v max = 2.85 m s
FIG. P8.57
240 P8.58
Potential Energy
(a) (b)
F= F=0
d  x3 + 2x 2 + 3x i = dx
e
j e3 x
2
 4x  3 i
j
when x = 1.87 and  0.535 (c) The stable point is at x = 0.535 point of minimum U x . The unstable point is at x = 1.87 maximum in U x . P8.59 FIG. P8.58
af
af
aK + U f = aK + U f 1 0 + b30.0 kg ge9.80 m s ja0.200 mf + b 250 N mga0.200 mf 2 1 = b50.0 kg gv + b 20.0 kg ge9.80 m s ja0.200 mf sin 40.0 2 58.8 J + 5.00 J = b 25.0 kg gv + 25.2 J
i f 2 2 2 2 2
v = 1.24 m s P8.60 (a) Between the second and the third picture, Emech = K + U  mgd =  1 1 mvi2 + kd 2 2 2
FIG. P8.59
1 1 50.0 N m d 2 + 0.250 1.00 kg 9.80 m s 2 d  1.00 kg 3.00 m s 2 = 0 2 2 2.45 21.25 N d= = 0.378 m 50.0 N m
b
g
b
ge
j
b
ge
j
(b)
Between picture two and picture four, Emech = K + U
a f 1 mv  1 mv 2 2 2 v = b3.00 m sg  b1.00 kg g a2.45 Nfa2fa0.378 mf
 f 2d =
2 2 i 2
= 2.30 m s (c) For the motion from picture two to picture five, Emech = K + U 1 2 1.00 kg 3.00 m s 2 9.00 J D=  2 0.378 m = 1.08 m 2 0.250 1.00 kg 9.80 m s 2  f D + 2d = 
a
f
b
gb
g
a
fb
ge
j
a
f
FIG. P8.60
Chapter 8
241
P8.61
(a)
Initial compression of spring: 1 450 N m x 2 x = 0. 400 m
1 2 1 kx = mv 2 2 2
b
ga f
2
=
1 0.500 kg 12.0 m s 2
b
gb
g
2
(b)
Speed of block at top of track: Emech =  fx
FIG. P8.61
FG mgh + 1 mv IJ  FG mgh + 1 mv IJ =  f aRf H K H K 2 2 b0.500 kgge9.80 m s ja2.00 mf + 1 b0.500 kg gv  1 b0.500 kggb12.0 m sg 2 2 = a7.00 N fa fa1.00 mf
T 2 T B 2 B 2 2 T 2 0.250 vT = 4.21
2
vT = 4.10 m s (c) Does block fall off at or before top of track? Block falls if a c < g ac =
2 4.10 vT = = 16.8 m s 2 R 1.00
a f
2
Therefore a c > g and the block stays on the track . P8.62 Let represent the mass of each one meter of the chain and T represent the tension in the chain at the table edge. We imagine the edge to act like a frictionless and massless pulley. (a) For the five meters on the table with motion impending,
Fy = 0 :
+n  5 g = 0 fs sn = 0.6 5g = 3g
n = 5 g
b g
Fx = 0 :
Fy = 0 :
+T  f s = 0
T = fs
T 3 g
FIG. P8.62
The maximum value is barely enough to support the hanging segment according to +T  3 g = 0 T = 3 g
so it is at this point that the chain starts to slide. continued on next page
242
Potential Energy
(b)
Let x represent the variable distance the chain has slipped since the start. Then length 5  x remains on the table, with now
Fy = 0 :
a f +n  a5  x fg = 0 n = a5  xfg f = n = 0.4a5  xfg = 2g  0.4xg
k k
Consider energies of the chainEarth system at the initial moment when the chain starts to slip, and a final moment when x = 5 , when the last link goes over the brink. Measure heights above the final position of the leading end of the chain. At the moment the final link slips off, the center of the chain is at y f = 4 meters. Originally, 5 meters of chain is at height 8 m and the middle of the dangling segment is at 3 height 8  = 6.5 m . 2 K i + U i + Emech = K f + U f :
FG 1 mv + mgyIJ H2 K b5g g8 + b3g g6.5  z b2g  0.4xg gdx = 1 b8 gv + b8g g4 2
0 + m1 gy1 + m 2 gy 2 i  f k dx =
i 5 0
b
g
z
f
2
f
2
40.0 g + 19.5 g  2.00 g dx + 0.400 g x dx = 4.00 v 2 + 32.0 g
0 0
z
5
z
5
27.5 g  2.00 gx 0 + 0.400 g
5
27.5 g  2.00 g 5.00 + 0.400 g 12.5 = 4.00 v 2 22.5 g = 4.00 v 2 v= P8.63 Launch speed is found from mg
a f
4.00
x 2
2 5 0
= 4.00 v 2
a f
a22.5 mfe9.80 m s j =
2
7.42 m s
FG 4 hIJ = 1 mv : H5 K 2
2
v = 2g
v y = v sin
FG 4 IJ h H 5K
FIG. P8.63
The height y above the water (by conservation of energy for the childEarth system) is found from mgy = 1 h 2 mv y + mg 2 5 (since 1 2 mv x is constant in projectile motion) 2 1 2 h 1 2 h y= vy + = v sin 2 + 2g 5 2g 5 1 4 2g h 2g 5
y=
LM FG IJ OP sin + h = 5 N H KQ
2
4 h h sin 2 + 5 5
Chapter 8
243
*P8.64
(a)
The length of string between glider and pulley is given by Now
2
2 = x 2 + h0 . Then 2
d d x = v y = v x = cos v x . is the rate at which string goes over the pulley: dt dt
a
d dx = 2x + 0. dt dt
f
(b)
eK
A
+ KB + U g
j = eK
i
A
+ KB + U g
j
f
0 + 0 + m B g y 30  y 45
b
g
1 1 2 2 = m A v x + mB v y 2 2
Now y 30  y 45 is the amount of string that has gone over the pulley, 30  45 . We have h h h0 h0 sin 30 = 0 and sin 45 = 0 , so 30  45 =  = 0. 40 m 2  2 = 0.234 m . sin 30 sin 45 30 45 From the energy equation
e
j
0.5 kg 9.8 m s 2 0.234 m = vx = (c) (d)
1 1 2 2 1.00 kg v x + 0.500 kg v x cos 2 45 2 2
1.15 J = 1.35 m s 0.625 kg
v y = v x cos = 1.35 m s cos 45 = 0.958 m s The acceleration of neither glider is constant, so knowing distance and acceleration at one point is not sufficient to find speed at another point.
b
g
P8.65
The geometry reveals D = L sin + L sin , 50.0 m = 40.0 m sin 50+ sin , = 28.9 (a) From takeoff to alighting for the JaneEarth system
b
g
eK + U j + W = eK + U j 1 mv + mg a  L cos f + FDa 1f = 0 + mg b L cos g 2 1 50 kg v + 50 kg e9.8 m s ja 40 m cos 50f  110 Na50 mf = 50 kg e9.8 m s ja 40 m cos 28.9f 2
g i wind g f 2 i 2 i 2 2
1 50 kg vi2  1.26 10 4 J  5.5 10 3 J = 1.72 10 4 J 2 2 947 J = 6.15 m s 50 kg
vi = (b)
a
f
For the swing back 1 mvi2 + mg  L cos + FD +1 = 0 + mg  L cos 2 1 130 kg vi2 + 130 kg 9.8 m s 2 40 m cos 28.9 + 110 N 50 m 2
b
g e
a f ja
a
ja
f f
a
f
= 130 kg 9.8 m s 2 40 m cos 50
e
f
1 130 kg vi2  4.46 10 4 J + 5 500 J = 3.28 10 4 J 2 vi = 2 6 340 J 130 kg
b
g=
9.87 m s
244 P8.66
Potential Energy
Case I: Surface is frictionless
1 1 mv 2 = kx 2 2 2 2 5.00 kg 1.20 m s mv k= 2 = x 10 2 m 2
b
gb
g
2
= 7. 20 10 2 N m
Case II: Surface is rough,
5.00 kg 2 1 v = 7.20 10 2 N m 10 1 m 2 2 v = 0.923 m s
e
k = 0.300 1 1 mv 2 = kx 2  k mgx 2 2
je
j  a0.300fb5.00 kg ge9.80 m s je10
2 2
1
m
j
*P8.67
(a)
eK + U j = eK + U j
g A g
1 2 0 + mgy A = mv B + 0 2 (b) (c) 11.1 m s v2 = ac = r 6.3 m
B
v B = 2 gy A = 2 9.8 m s 2 6.3 m = 11.1 m s = 19.6 m s 2 up
e
j
b
g
2
Fy = ma y
+n B  mg = ma c
n B = 76 kg 9.8 m s 2 + 19.6 m s 2 = 2.23 10 3 N up
(d) (e)
e
j
W = Fr cos = 2.23 10 3 N 0.450 m cos 0 = 1.01 10 3 J
a
f
eK + U j
g
1 1 2 2 mv B + 0 + 1.01 10 3 J = mv D + mg y D  y B 2 2 1 1 2 2 76 kg 11.1 m s + 1.01 10 3 J = 76 kg v D + 76 kg 9.8 m s 2 6.3 m 2 2
B
+ W = K +Ug
e
j
D
b
g
b
g
e
j
e5.70 10
(f)
g D
3
J  4.69 10 3 J 2 76 kg
j
= v D = 5.14 m s
eK + U j = eK + U j 1 mv + 0 = 0 + mg b y 2
g 2 D
E
where E is the apex of his motion
E  yD
g
y E  yD =
2 5.14 m s vD = = 1.35 m 2 g 2 9.8 m s 2
b
e
g
2
j
(g)
Consider the motion with constant acceleration between takeoff and touchdown. The time is the positive root of 1 y f = yi + v yi t + a y t 2 2 1 2.34 m = 0 + 5.14 m s t + 9.8 m s 2 t 2 2 2 4.9t  5.14t  2.34 = 0
e
j
t=
5.14 5.14 2  4 4.9 2.34 9.8
a fa
f=
1.39 s
Chapter 8
245
*P8.68
If the spring is just barely able to lift the lower block from the table, the spring lifts it through no noticeable distance, but exerts on the block a force equal to its weight Mg. The extension of the spring, from Fs = kx , must be Mg k . Between an initial point at release and a final point when the moving block first comes to rest, we have K i + U gi + U si = K f + U gf + U sf : 0 + mg  4mg Mg 1 Mg 1 4mg + k = 0 + mg + k 2 2 k k k k 2 2 2 2 2 2 2 mMg M g 4m g 8m g  + = + k k k 2k M2 4m 2 = mM + 2 M2 + mM  4m 2 = 0 2 m m 2  4
1 2 1 2 2
FG H
IJ K
FG H
IJ K
2
FG IJ H K
FG IJ H K
2
c he4m j = m M= 2c h Only a positive mass is physical, so we take M = ma3  1f = 2m .
P8.69 (a) Take the original point where the ball is released and the final point where its upward swing stops at height H and horizontal displacement x = L2  L  H
9m 2
a
f
2
= 2LH  H 2
Since the wind force is purely horizontal, it does work
Wwind = F ds = F dx = F 2LH  H 2
The workenergy theorem can be written: K i + U gi + Wwind = K f + U gf , or
z
z
FIG. P8.69
0 + 0 + F 2LH  H 2 = 0 + mgH giving F 2 2LH  F 2 H 2 = m 2 g 2 H 2 Here H = 0 represents the lower turning point of the ball's oscillation, and the upper limit is at F 2 2L = F 2 + m 2 g 2 H . Solving for H yields
a f e
j
H=
2LF 2 2L = F + m2 g 2 1 + mg F
2
b
g
2
As F 0 , H 0 as is reasonable. As F , H 2L , which would be hard to approach experimentally. (b) H= 2 2.00 m
1 + 2.00 kg 9.80 m s 2 14.7 N
b
ge
a
f j
2
= 1.44 m
continued on next page
246
Potential Energy
(c)
Call the equilibrium angle with the vertical.
Fx = 0 T sin = F , and Fy = 0 T cos = mg
Dividing: tan = F 14.7 N = = 0.750 , or = 36.9 mg 19.6 N
Therefore, H eq = L 1  cos = 2.00 m 1  cos 36.9 = 0.400 m (d) As F , tan , 90.0 and H eq L A very strong wind pulls the string out horizontal, parallel to the ground. Thus,
a
f a
fa
f
eH j
eq
max
=L .
vi = Rg
P8.70
Call = 180 the angle between the upward vertical and the radius to the release point. Call v r the speed here. By conservation of energy K i + U i + E = K r + U r 1 1 2 mvi2 + mgR + 0 = mv r + mgR cos 2 2 2 gR + 2 gR = v r + 2 gR cos v r = 3 gR  2 gR cos The components of velocity at release are v x = v r cos and v y = v r sin so for the projectile motion we have x = vxt R sin = v r cos t  R cos = v r sin t 
The path after string is cut
R
C
FIG. P8.70
1 y = v y t  gt 2 2
1 2 gt 2
By substitution  R cos = v r sin with sin 2 + cos 2 = 1 ,
2 gR sin 2 = 2 v r cos = 2 cos 3 gR  2 gR cos
R sin g R 2 sin 2  2 v r cos 2 v r cos 2
b
g
sin 2 = 6 cos  4 cos 2 = 1  cos 2 3 cos 2  6 cos + 1 = 0 cos = 6 36  12 6
Only the sign gives a value for cos that is less than one: cos = 0.183 5
= 79.43
so = 100.6
Chapter 8
247
P8.71
Applying Newton's second law at the bottom (b) and top (t) of the circle gives Tb  mg =
2 mv b
vt mg Tt
R
and Tt  mg =  m
mv t2 R v t2
Adding these gives
Tb = Tt + 2mg +
e
2 vb

j
Tb
R mg
Also, energy must be conserved and U + K = 0 So,
2 m v b  v t2
vb
e
2
j + b0  2mgRg = 0 and mev
2 b
 v t2
R
j = 4mg
FIG. P8.71
Substituting into the above equation gives Tb = Tt + 6mg . P8.72 (a) Energy is conserved in the swing of the pendulum, and the stationary peg does no work. So the ball's speed does not change when the string hits or leaves the peg, and the ball swings equally high on both sides. Relative to the point of suspension, Ui = 0, U f =  mg d  L  d From this we find that 1  mg 2d  L + mv 2 = 0 2 Also for centripetal motion, mg = mv 2 where R = L  d . R 3L . 5
L Peg
d
(b)
a
f
FIG. P8.72
a
f
Upon solving, we get d =
248 *P8.73
Potential Energy
(a)
At the top of the loop the car and riders are in free fall:
Fy = ma y :
mg down = v = Rg
mv 2 down R
Energy of the carridersEarth system is conserved between release and top of loop: K i + U gi = K f + U gf : 0 + mgh = gh = 1 mv 2 + mg 2 R 2
a f
1 Rg + g 2 R 2 h = 2.50 R
a f
(b)
Let h now represent the height 2.5 R of the release point. At the bottom of the loop we have mgh = 1 2 mv b 2 or
2 v b = 2 gh 2 mv b up R m 2 gh
Fy = ma y :
n b  mg = n b = mg +
b g b g
R
At the top of the loop, mgh =
1 mv t2 + mg 2 R 2 v t2 = 2 gh  4 gR mv t2
a f
FIG. P8.73
Fy = ma y :
R m 2 gh  4 gR n t =  mg + R m 2 gh  5mg nt = R
n t  mg = 
b
g
b g
Then the normal force at the bottom is larger by n b  n t = mg + m 2 gh R
b g  mb2 ghg + 5mg =
R
6mg .
Chapter 8
249
*P8.74
(a)
Conservation of energy for the sledriderEarth system, between A and C: K i + U gi = K f + U gf 1 m 2.5 m s 2 vC =
b
g
2
+ m 9.80 m s 2 9.76 m =
2 2
e
ja
f
b2.5 m sg + 2e9.80 m s ja9.76 mf = b gb
1 2 mvC + 0 2 14.1 m s FIG. P8.74(a)
(b)
Incorporating the loss of mechanical energy during the portion of the motion in the water, we have, for the entire motion between A and D (the rider's stopping point), K i + U gi  f k x = K f + U gf : 1 2 80 kg 2.5 m s + 80 kg 9.80 m s 2 9.76 m  f k x = 0 + 0 2  f k x = 7.90 10 3 J fk = 7.90 10 3 J 7.90 10 3 N m = = 158 N x 50 m
g b
ge
ja
f
(c)
The water exerts a frictional force and also a normal force of The magnitude of the water force is
n = mg = 80 kg 9.80 m s 2 = 784 N
b
ge
j
a158 Nf + a784 Nf
2
2
= 800 N
(d)
The angle of the slide is
= sin 1
9.76 m = 10.4 54.3 m
For forces perpendicular to the track at B,
Fy = ma y :
n B  mg cos = 0
FIG. P8.74(d)
n B = 80.0 kg 9.80 m s 2 cos 10.4 = 771 N (e)
b
ge
j
Fy = ma y :
2 mvC r nC = 80.0 kg 9.80 m s 2
+nC  mg =
b ge j b80.0 kggb14.1 m sg +
2
20 m
nC = 1.57 10 3 N up
FIG. P8.74(e)
The rider pays for the thrills of a giddy height at A, and a high speed and tremendous splash at C. As a bonus, he gets the quick change in direction and magnitude among the forces we found in parts (d), (e), and (c).
250
Potential Energy
ANSWERS TO EVEN PROBLEMS
P8.2 P8.4 P8.6 P8.8 P8.10 P8.12 P8.14 P8.16 P8.18 P8.20 P8.22 P8.24 P8.26 (a) 800 J; (b) 107 J; (c) 0 (a) 1.11 10 9 J ; (b) 0.2 1.84 m (a) 10.2 kW; (b) 10.6 kW; (c) 5.82 10 J d= kx 2 x 2mg sin
6
P8.42 P8.44 P8.46
e7  9 x y ji  3 x j
2 3
see the solution (a) r = 1.5 mm and 3.2 mm, stable; 2.3 mm and unstable; r neutral; (b) 5.6 J E < 1 J ; (c) 0.6 mm r 3.6 mm ; (d) 2.6 J; (e) 1.5 mm; (f) 4 J see the solution 33.4 kW (a) 0.588 J; (b) 0.588 J; (c) 2.42 m s; (d) 0.196 J; 0.392 J 0.115 (a) 100 J; (b) 0.410 m; (c) 2.84 m s ; (d) 9.80 mm ; (e) 2.85 m s (a) 3 x 2  4x  3 i ; (b) 1.87; 0.535; (c) see the solution
P8.48 P8.50 P8.52
(a) see the solution; (b) 60.0 (a) 2 m1  m 2 gh
1 2
b bm
g +m g
; (b)
2m1 h m1 + m 2
160 L min 40.8
P8.54 P8.56
FG 8 gh IJ H 15 K
12
P8.58
e
j
(a) see the solution; (b) 35.0 J (a) v B = 5.94 m s; vC = 7.67 m s ; (b) 147 J (a) U f = 22.0 J ; E = 40.0 J ; (b) Yes. The total mechanical energy changes. P8.60 P8.62 P8.64
(a) 0.378 m; (b) 2.30 m s ; (c) 1.08 m (a) see the solution; (b) 7.42 m s (a) see the solution; (b) 1.35 m s ; (c) 0.958 m s ; (d) see the solution 0.923 m s 2m 100.6 see the solution (a) 14.1 m s; (b) 7.90 J ; (c) 800 N; (d) 771 N; (e) 1.57 kN up
P8.28 P8.30 P8.32 P8.34 P8.36 P8.38 P8.40
194 m 2.06 kN up 168 J (a) 24.5 m s ; (b) yes; (c) 206 m; (d) Air drag depends strongly on speed. 3.92 kJ 44.1 kW (a) Ax 2 Bx 3  ; 2 3 5 A 19B 19B 5 A (b) U =  ; K =  2 3 3 2 P8.66 P8.68 P8.70 P8.72 P8.74
9
Linear Momentum and Collisions
CHAPTER OUTLINE
9.1 9.2 9.3 9.4 9.5 9.6 9.7 Linear Momentum and Its Conservation Impulse and Momentum Collisions in One Dimension TwoDimensional Collisions The Center of Mass Motion of a System of Particles Rocket Propulsion
ANSWERS TO QUESTIONS
Q9.1 Q9.2 No. Impulse, Ft , depends on the force and the time for which it is applied. The momentum doubles since it is proportional to the speed. The kinetic energy quadruples, since it is proportional to the speedsquared. The momenta of two particles will only be the same if the masses of the particles of the same. (a) (b) It does not carry force, for if it did, it could accelerate itself. It cannot deliver more kinetic energy than it possesses. This would violate the law of energy conservation.
Q9.3 Q9.4
(c) Q9.5
It can deliver more momentum in a collision than it possesses in its flight, by bouncing from the object it strikes.
Provided there is some form of potential energy in the system, the parts of an isolated system can move if the system is initially at rest. Consider two airtrack gliders on a horizontal track. If you compress a spring between them and then tie them together with a string, it is possible for the system to start out at rest. If you then burn the string, the potential energy stored in the spring will be converted into kinetic energy of the gliders. No. Only in a precise headon collision with momenta with equal magnitudes and opposite directions can both objects wind up at rest. Yes. Assume that ball 2, originally at rest, is struck squarely by an equalmass ball 1. Then ball 2 will take off with the velocity of ball 1, leaving ball 1 at rest. Interestingly, mutual gravitation brings the ball and the Earth together. As the ball moves downward, the Earth moves upward, although with an acceleration 10 25 times smaller than that of the ball. The two objects meet, rebound, and separate. Momentum of the ballEarth system is conserved. (a) (b) Linear momentum is conserved since there are no external forces acting on the system. Kinetic energy is not conserved because the chemical potential energy initially in the explosive is converted into kinetic energy of the pieces of the bomb. 251
Q9.6
Q9.7
Q9.8
252 Q9.9
Linear Momentum and Collisions
Momentum conservation is not violated if we make our system include the Earth along with the clay. When the clay receives an impulse backwards, the Earth receives the same size impulse forwards. The resulting acceleration of the Earth due to this impulse is significantly smaller than the acceleration of the clay, but the planet absorbs all of the momentum that the clay loses. Momentum conservation is not violated if we choose as our system the planet along with you. When you receive an impulse forward, the Earth receives the same size impulse backwards. The resulting acceleration of the Earth due to this impulse is significantly smaller than your acceleration forward, but the planet's backward momentum is equal in magnitude to your forward momentum. As a ball rolls down an incline, the Earth receives an impulse of the same size and in the opposite direction as that of the ball. If you consider the Earthball system, momentum conservation is not violated. Suppose car and truck move along the same line. If one vehicle overtakes the other, the fastermoving one loses more energy than the slower one gains. In a headon collision, if the speed of the m + 3m c times the speed of the car, the car will lose more energy. truck is less than T 3mT + m c The rifle has a much lower speed than the bullet and much less kinetic energy. The butt distributes the recoil force over an area much larger than that of the bullet. His impact speed is determined by the acceleration of gravity and the distance of fall, in v 2 = vi2  2 g 0  yi . The force exerted by the pad depends also on the unknown stiffness of the pad. f
Q9.10
Q9.11
Q9.12
Q9.13 Q9.14
b
g
Q9.15 Q9.16
The product of the mass flow rate and velocity of the water determines the force the firefighters must exert. The sheet stretches and pulls the two students toward each other. These effects are larger for a fastermoving egg. The time over which the egg stops is extended so that the force stopping it is never too large. (c) In this case, the impulse on the Frisbee is largest. According to Newton's third law, the impulse on the skater and thus the final speed of the skater will also be largest. Usually but not necessarily. In a onedimensional collision between two identical particles with the same initial speed, the kinetic energy of the particles will not change. g downward. As one finger slides towards the center, the normal force exerted by the sliding finger on the ruler increases. At some point, this normal force will increase enough so that static friction between the sliding finger and the ruler will stop their relative motion. At this moment the other finger starts sliding along the ruler towards the center. This process repeats until the fingers meet at the center of the ruler. The planet is in motion around the sun, and thus has momentum and kinetic energy of its own. The spacecraft is directed to cross the planet's orbit behind it, so that the planet's gravity has a component pulling forward on the spacecraft. Since this is an elastic collision, and the velocity of the planet remains nearly unchanged, the probe must both increase speed and change direction for both momentum and kinetic energy to be conserved.
Q9.17 Q9.18 Q9.19 Q9.20
Q9.21
Chapter 9
253
Q9.22 Q9.23 Q9.24 Q9.25
Noan external force of gravity acts on the moon. Yes, because its speed is constant. The impulse given to the egg is the same regardless of how it stops. If you increase the impact time by dropping the egg onto foam, you will decrease the impact force. Yes. A boomerang, a kitchen stool. The center of mass of the balls is in free fall, moving up and then down with the acceleration due to gravity, during the 40% of the time when the juggler's hands are empty. During the 60% of the time when the juggler is engaged in catching and tossing, the center of mass must accelerate up with a somewhat smaller average acceleration. The center of mass moves around in a little circle, making three revolutions for every one revolution that one ball makes. Letting T represent the time for one cycle and Fg the weight of one ball, we have FJ 0.60T = 3 Fg T and FJ = 5 Fg . The average force exerted by the juggler is five times the weight of one ball.
Q9.26
In empty space, the center of mass of a rocketplusfuel system does not accelerate during a burn, because no outside force acts on this system. According to the text's `basic expression for rocket propulsion,' the change in speed of the rocket body will be larger than the speed of the exhaust relative to the rocket, if the final mass is less than 37% of the original mass. The gun recoiled. Inflate a balloon and release it. The air escaping from the balloon gives the balloon an impulse. There was a time when the English favored position (a), the Germans position (b), and the French position (c). A Frenchman, Jean D'Alembert, is most responsible for showing that each theory is consistent with the others. All are equally correct. Each is useful for giving a mathematically simple solution for some problems.
Q9.27 Q9.28 Q9.29
SOLUTIONS TO PROBLEMS
Section 9.1 P9.1 Linear Momentum and Its Conservation v = 3.00 i  4.00 j m s
m = 3.00 kg , (a)
e
j
p = mv = 9.00 i  12.0 j kg m s Thus, and p x = 9.00 kg m s p y = 12.0 kg m s
e
j
(b)
2 2 p = px + py =
= tan 1
a9.00f + a12.0f = 15.0 kg m s F p I = tan a1.33f = 307 GH p JK
2 2 y x 1
254 P9.2
Linear Momentum and Collisions
(a) (b)
At maximum height v = 0 , so p = 0 . Its original kinetic energy is its constant total energy, Ki = 1 1 mvi2 = 0.100 kg 15.0 m s 2 2
a
f b
g
2
= 11.2 J .
At the top all of this energy is gravitational. Halfway up, onehalf of it is gravitational and the other half is kinetic: K = 5.62 J = v= Then p = mv = 0.100 kg 10.6 m s j p = 1.06 kg m s j . P9.3 I have mass 85.0 kg and can jump to raise my center of gravity 25.0 cm. I leave the ground with speed given by v 2  vi2 = 2 a x f  x i : f 1 0.100 kg v 2 2 2 5.62 J = 10.6 m s 0.100 kg
b
g
b
gb
g
d
i
0  vi2 = 2 9.80 m s 2 0.250 m vi = 2.20 m s
e
ja
f
Total momentum of the system of the Earth and me is conserved as I push the earth down and myself up: 0 = 5.98 10 24 kg v e + 85.0 kg 2.20 m s v e ~ 10 23 m s P9.4 (a) For the system of two blocks p = 0 , or Therefore, Solving gives left). (b) 1 2 1 1 2 2 kx = Mv M + 3 M v 3 M = 8. 40 J 2 2 2 pi = p f 0 = Mv m + 3 M 2.00 m s
e
j b
gb
g
a fb
g
v m = 6.00 m s (motion toward the
a f
FIG. P9.4
Chapter 9
255 p2 . 2m
P9.5
(a)
The momentum is p = mv , so v = 1 mv 2 implies v = 2
p p 1 1 and the kinetic energy is K = mv 2 = m m m 2 2 2mK .
FG IJ H K
2
=
(b)
K=
2K 2K , so p = mv = m = m m
Section 9.2 *P9.6
Impulse and Momentum
From the impulsemomentum theorem, F t = p = mv f  mvi , the average force required to hold onto the child is F= m v f  vi
a f
d
i = b12 kg gb0  60 mi hg F 1 m s I = 6.44 10 GH 2.237 mi h JK 0.050 s  0 a t f
3
N.
Therefore, the magnitude of the needed retarding force is 6.44 10 3 N , or 1 400 lb. A person cannot exert a force of this magnitude and a safety device should be used. P9.7 (a)
I = Fdt = area under curve
I= 1 1.50 10 3 s 18 000 N = 13.5 N s 2
z
e
jb
g
(b) (c) *P9.8
F=
13.5 N s = 9.00 kN 1.50 10 3 s FIG. P9.7
From the graph, we see that Fmax = 18.0 kN
The impact speed is given by
1 1 2 2 mv1 = mgy1 . The rebound speed is given by mgy 2 = mv 2 . The 2 2 impulse of the floor is the change in momentum, mv 2 up  mv1 down = m v 2 + v1 up
b = me
g
2 gh2 + 2 gh1 up
j
= 0.15 kg 2 9.8 m s 2
e
je
0.960 m + 1.25 m up
j
= 1.39 kg m s upward
256 P9.9
Linear Momentum and Collisions
p = Ft p x
p y = m v fy  viy = m v cos 60.0  mv cos 60.0 = 0
f e j a = ma  v sin 60.0 v sin 60.0f = 2mv sin 60.0 = 2b3.00 kg gb10.0 m sga0.866f
= 52.0 kg m s p x 52.0 kg m s = = 260 N 0.200 s t
Fave = P9.10
FIG. P9.9
Assume the initial direction of the ball in the x direction. (a) (b) Impulse, I = p = p f  pi = 0.060 0 40.0 i  0.060 0 50.0  i = 5.40 i N s Work = K f  K i = 1 0.060 0 2
b
ga f b
2
ga fe j
b
g a40.0f  a50.0f
2
= 27.0 J
P9.11
Take xaxis toward the pitcher (a) pix + I x = p fx : piy + I y = p fy :
b0.200 kg gb15.0 m sga cos 45.0f + I = b0.200 kggb40.0 m sg cos 30.0
x
b0.200 kg gb15.0 m sga sin 45.0f + I = b0.200 kg gb40.0 m sg sin 30.0 I = e9.05 i + 6.12 jj N s
y
I x = 9.05 N s
(b)
1 1 0 + Fm 4.00 ms + Fm 20.0 ms + Fm 4.00 ms 2 2 Fm 24.0 10 3 s = 9.05 i + 6.12 j N s I=
b
ga
f
a
f
a
f
e
j
Fm = P9.12
e377 i + 255 jj N
If the diver starts from rest and drops vertically into the water, the velocity just before impact is found from K f + U gf = K i + U gi 1 2 mv impact + 0 = 0 + mgh v impact = 2 gh 2 With the diver at rest after an impact time of t , the average force during impact is given by F= m 0  v impact t
e
j = m
2 gh t
or F =
m 2 gh t
(directed upward).
Assuming a mass of 55 kg and an impact time of 1.0 s , the magnitude of this average force is
b55 kg g 2e9.8 m s ja10 mf = 770 N , or F =
2
1.0 s
~ 10 3 N .
Chapter 9
257
P9.13
The force exerted on the water by the hose is F= 0.600 kg 25.0 m s  0 p water mv f  mvi = = = 15.0 N . 1.00 s t t
b
gb
g
According to Newton's third law, the water exerts a force of equal magnitude back on the hose. Thus, the gardener must apply a 15.0 N force (in the direction of the velocity of the exiting water stream) to hold the hose stationary. *P9.14 (a) Energy is conserved for the springmass system: K i + U si = K f + U sf : 0+ 1 2 1 kx = mv 2 + 0 2 2 k v=x m k larger. m
(b)
From the equation, a smaller value of m makes v = x k = x km m
(c) (d) (e)
I = p f  p i = mv f = 0 = mx
From the equation, a larger value of m makes I = x km larger. For the glider, W = K f  K i = 1 1 mv 2  0 = kx 2 2 2
The mass makes no difference to the work.
Section 9.3 P9.15
Collisions in One Dimension
b200 g gb55.0 m sg = b46.0 g gv + b200 g gb40.0 m sg
v = 65.2 m s
*P9.16
bm v
g = bm v + m v g 22.5 g b35 m sg + 300 g b2.5 m sg = 22.5 gv
1 1
+ m2 v2
i
1 1
2 2 f
1f
+0
v1 f =
37.5 g m s = 1.67 m s 22.5 g
FIG. P9.16
258 P9.17
Linear Momentum and Collisions
Momentum is conserved 10.0 10 3 kg v = 5.01 kg 0.600 m s
e
j b
gb
g
v = 301 m s P9.18 (a) mv1i + 3mv 2 i = 4mv f where m = 2.50 10 4 kg vf = (b) P9.19 (a) 4.00 + 3 2.00 = 2.50 m s 4 1 1 1 2 2 4m v 2  mv1i + 3m v 2i = 2.50 10 4 12.5  8.00  6.00 = 3.75 10 4 J f 2 2 2
a f
K f  Ki =
a f
LM N
a f OPQ e
ja
f
The internal forces exerted by the actor do not change the total momentum of the system of the four cars and the movie actor
a4mfv = a3mfb2.00 m sg + mb4.00 m sg
i
6.00 m s + 4.00 m s = 2.50 m s vi = 4 (b) Wactor = K f  K i = Wactor (c)
FIG. P9.19
1 1 2 2 3m 2.00 m s + m 4.00 m s  4 m 2.50 m s 2 2 2.50 10 4 kg 2 = 12.0 + 16.0  25.0 m s = 37.5 kJ 2
e
a fb ja
g
b g fb g
a fb
g
2
The event considered here is the time reversal of the perfectly inelastic collision in the previous problem. The same momentum conservation equation describes both processes.
P9.20
v1 , speed of m1 at B before collision. 1 2 m1 v1 = m1 gh 2 v1 = 2 9.80 5.00 = 9.90 m s v1 f , speed of m1 at B just after collision. m  m2 1 v1 f = 1 v1 =  9.90 m s = 3.30 m s m1 + m 2 3 At the highest point (after collision)
a fa f
a f
2
FIG. P9.20
m1 ghmax =
1 m1 3.30 2
a
f
hmax =
b3.30 m sg = 2e9.80 m s j
2 2
0.556 m
Chapter 9
259
P9.21
(a), (b) Let v g and v p be the velocity of the girl and the plank relative to the ice surface. Then we may say that v g  v p is the velocity of the girl relative to the plank, so that v g  v p = 1.50 But also we must have m g v g + m p v p = 0 , since total momentum of the girlplank system is zero relative to the ice surface. Therefore 45.0 v g + 150 v p = 0 , or v g = 3.33 v p Putting this into the equation (1) above gives 3.33 v p  v p = 1.50 or v p = 0.346 m s Then v g = 3.33 0.346 = 1.15 m s FIG. P9.21 (1)
a
f
*P9.22
For the cartruckdriverdriver system, momentum is conserved: p 1i + p 2 i = p 1 f + p 2 f :
b4 000 kg gb8 m sgi + b800 kg gb8 m sge ij = b4 800 kg gv i
f
vf =
25 600 kg m s = 5.33 m s 4 800 kg
For the driver of the truck, the impulsemomentum theorem is Ft = p f  pi :
F 0.120 s = 80 kg 5.33 m s i  80 kg 8 m s i
F = 1.78 10 3 N  i on the truck driver
a a
f b f b
gb gb
g b g b
gb gb
g
e j
For the driver of the car,
F 0.120 s = 80 kg 5.33 m s i  80 kg 8 m s  i
ge j
F = 8.89 10 3 Ni on the car driver , 5 times larger.
P9.23 (a) According to the Example in the chapter text, the fraction of total kinetic energy transferred to the moderator is f2 =
bm
4m1 m 2
1
+ m2
g
2
where m 2 is the moderator nucleus and in this case, m 2 = 12m1 f2 = 4m1 12m1
b13m g
1
b
2
g = 48 =
169
0. 284 or 28.4%
of the neutron energy is transferred to the carbon nucleus. (b)
K C = 0.284 1.6 10 13 J = 4.54 10 14 J Kn
13
a fe j Jj = = a0.716 fe1.6 10
1.15 10 13 J
260 P9.24
Linear Momentum and Collisions
Energy is conserved for the bobEarth system between bottom and top of swing. At the top the stiff rod is in compression and the bob nearly at rest. K i + Ui = K f + U f : 1 2 Mv b + 0 = 0 + Mg 2 2 2 v b = g 4 so v b = 2 g FIG. P9.24
Momentum of the bobbullet system is conserved in the collision: mv = m P9.25 v +M 2 g 2
e
j
v=
4M m
g
At impact, momentum of the clayblock system is conserved, so: mv1 = m1 + m 2 v 2 After impact, the change in kinetic energy of the clayblocksurface system is equal to the increase in internal energy: 1 2 m1 + m 2 v 2 = f f d = m1 + m 2 gd 2 1 2 0.112 kg v 2 = 0.650 0.112 kg 9.80 m s 2 7.50 m 2 2 v 2 = 95.6 m 2 s 2 v 2 = 9.77 m s
b
g
b b
g g
b
b
g ge
ja
f
FIG. P9.25
e12.0 10
P9.26
3
kg v1 = 0.112 kg 9.77 m s
j b
gb
g
v1 = 91.2 m s
We assume equal firing speeds v and equal forces F required for the two bullets to push wood fibers apart. These equal forces act backward on the two bullets. For the first, For the second, K i + Emech = K f pi = p f 1 7.00 10 3 kg v 2  F 8.00 10 2 m = 0 2 7.00 10 3 kg v = 1.014 kg v f
e
e
vf Again, Substituting for v f , K i + Emech = K f :
j b e7.00 10 jv =
3
j
e
g
j
1.014
1 1 7.00 10 3 kg v 2  Fd = 1.014 kg v 2 f 2 2
e
j
b
g
1 1 7.00 10 3 v 7.00 10 3 kg v 2  Fd = 1.014 kg 2 2 1.014
e
j
b
3 1 1 7.00 10 3 2 Fd = 7.00 10 v  2 2 1.014
e
j
e
j
gFGH
2
I JK
2
v2 d = 7.94 cm
Substituting for v,
Fd = F 8.00 10 2 m 1 
e
jFGH
7.00 10 3 1.014
I JK
Chapter 9
261
*P9.27
(a)
Using conservation of momentum,
c ph = c ph , gives a4.0 + 10 + 3.0f kg v = b4.0 kggb5.0 m sg + b10 kg gb3.0 m sg + b3.0 kggb4.0 m sg .
after before
Therefore, v = +2.24 m s , or 2. 24 m s toward the right . (b) No . For example, if the 10kg and 3.0kg mass were to stick together first, they would move with a speed given by solving
b13 kggv = b10 kg gb3.0 m sg + b3.0 kg gb4.0 m sg , or v
1
1
= +1.38 m s .
Then when this 13 kg combined mass collides with the 4.0 kg mass, we have
b17 kggv = b13 kggb1.38 m sg + b4.0 kg gb5.0 m sg , and v = +2.24 m s
just as in part (a). Coupling order makes no difference.
Section 9.4 P9.28 (a)
TwoDimensional Collisions First, we conserve momentum for the system of two football players in the x direction (the direction of travel of the fullback).
b90.0 kg gb5.00 m sg + 0 = b185 kg gV cos
where is the angle between the direction of the final velocity V and the x axis. We find V cos = 2.43 m s (1)
Now consider conservation of momentum of the system in the y direction (the direction of travel of the opponent).
b95.0 kg gb3.00 m sg + 0 = b185 kggaV sin f
which gives, Divide equation (2) by (1) From which Then, either (1) or (2) gives (b) 1 90.0 kg 5.00 m s 2 1 K f = 185 kg 2.88 m s 2 Ki = V sin = 1.54 m s tan = 1.54 = 0.633 2.43 (2)
= 32.3
V = 2.88 m s + 1 95.0 kg 3.00 m s 2
b b
gb gb
g g
2
b
gb
g
2
= 1.55 10 3 J
2
= 7.67 10 2 J
Thus, the kinetic energy lost is 783 J into internal energy.
262 P9.29
Linear Momentum and Collisions
p xf = p xi mvO cos 37.0+ mv Y cos 53.0 = m 5.00 m s 0.799 vO + 0.602 v Y = 5.00 m s p yf = p yi mvO sin 37.0 mv Y sin 53.0 = 0 0.602 vO = 0.799 v Y Solving (1) and (2) simultaneously, vO = 3.99 m s and v Y = 3.01 m s . FIG. P9.29 (2)
b
g
(1)
P9.30
p xf = p xi :
mvO cos + mv Y cos 90.0 = mvi
vO cos + v Y sin = vi (1)
a
f
p yf = p yi :
mvO sin  mv Y sin 90.0 = 0
vO sin = v Y cos (2)
a
f
From equation (2), vO = v Y
FG cos IJ H sin K
sin = vi
(3) FIG. P9.30
Substituting into equation (1), vY so
F cos I + v GH sin JK
2
Y
v Y cos 2 + sin 2 = vi sin , and v Y = vi sin .
e
j
Then, from equation (3), vO = vi cos . We did not need to write down an equation expressing conservation of mechanical energy. In the problem situation, the requirement of perpendicular final velocities is equivalent to the condition of elasticity.
Chapter 9
263
P9.31
The initial momentum of the system is 0. Thus,
a1.20mfv
and Ki =
Bi
= m 10.0 m s
b
g a fb g FG e H e j
v Bi = 8.33 m s 1 1 1 2 2 m 10.0 m s + 1. 20m 8.33 m s = m 183 m 2 s 2 2 2 2 1 1 1 1 2 2 K f = m vG + 1.20m v B = m 183 m 2 s 2 2 2 2 2
b b g
g a
fb g
jIJK
or
2 2 vG + 1.20 v B = 91.7 m 2 s 2
(1)
From conservation of momentum,
mvG = 1.20m v B
or vG = 1.20 v B (2)
a
f
Solving (1) and (2) simultaneously, we find vG = 7.07 m s (speed of green puck after collision) and P9.32 v B = 5.89 m s (speed of blue puck after collision)
We use conservation of momentum for the system of two vehicles for both northward and eastward components. For the eastward direction: M 13.0 m s = 2 MV f cos 55.0 For the northward direction: Mv 2i = 2 MV f sin 55.0 Divide the northward equation by the eastward equation to find: v 2 i = 13.0 m s tan 55.0 = 18.6 m s = 41.5 mi h Thus, the driver of the north bound car was untruthful. FIG. P9.32
b
g
b
g
264 P9.33
Linear Momentum and Collisions
By conservation of momentum for the system of the two billiard balls (with all masses equal), 5.00 m s + 0 = 4.33 m s cos 30.0+ v 2 fx v 2 fx = 1.25 m s 0 = 4.33 m s sin 30.0+ v 2 fy v 2 fy = 2.16 m s v 2 f = 2.50 m s at  60.0 Note that we did not need to use the fact that the collision is perfectly elastic. FIG. P9.33
b
g
b
g
P9.34
(a)
pi = p f
so and
p xi = p xf p yi = p yf mvi = mv cos + mv cos 0 = mv sin + mv sin (1) (2)
From (2), so
sin =  sin
=  Furthermore, energy conservation for the system of two protons requires 1 1 1 mvi2 = mv 2 + mv 2 2 2 2 vi v= so 2
2 vi cos 2 3.00 5.00 i  6.00 j = 5.00 v
FIG. P9.34
(b) P9.35
Hence, (1) gives vi =
= 45.0
= 45.0
m1 v 1i + m 2 v 2i = m1 + m 2 v f :
b
g
a f v = e3.00 i  1.20 jj m s
P9.36
xcomponent of momentum for the system of the two objects: p1ix + p 2ix = p1 fx + p 2 fx :  mvi + 3mvi = 0 + 3mv 2 x ycomponent of momentum of the system: by conservation of energy of the system: we have also So the energy equation becomes 0 + 0 =  mv1 y + 3 mv 2 y 1 1 1 1 2 2 2 mvi2 + 3mvi2 = mv1 y + 3m v 2 x + v 2 y 2 2 2 2 2v v2x = i 3 v1 y = 3 v 2 y +
e
j
2 4vi2 = 9 v 2 y +
or continued on next page
8 vi2 2 = 12 v 2 y 3 2 vi v2y = 3
4vi2 2 + 3v2 y 3
Chapter 9
265
(a)
The object of mass m has final speed and the object of mass 3 m moves at
v1 y = 3 v 2 y =
2 2 v2x + v2y =
2 vi
4vi2 2 vi2 + 9 9 2 vi 3
2 2 v2x + v2y =
(b) P9.37
= tan 1
Fv I GH v JK
2y 2x
= tan 1
F GH
2 vi 3 = 35.3 3 2 vi
I JK
m 0 = 17.0 10 27 kg m1 = 5.00 10 27 kg m 2 = 8.40 10 27 kg (a) m1 v 1 + m 2 v 2 + m 3 v 3 = 0
v i = 0 (the parent nucleus) v 1 = 6.00 10 6 j m s v 2 = 4.00 10 6 i m s
where m 3 = m 0  m1  m 2 = 3.60 10 27 kg
e
5.00 10 27 6.00 10 6 j + 8.40 10 27 4.00 10 6 i + 3.60 10 27 v 3 = 0
je
j e
je
j e
j
FIG. P9.37
v3 = (b) E=
e9.33 10 i  8.33 10 jj m s
6 6
1 1 1 2 2 2 m 1 v1 + m 2 v 2 + m 3 v 3 2 2 2 1 5.00 10 27 6.00 10 6 E= 2
LMe N
je
j + e8.40 10 je4.00 10 j + e3.60 10 je12.5 10 j OQP
2 27 6 2 27 6 2
E = 4.39 10 13 J
Section 9.5 P9.38
The Center of Mass
The xcoordinate of the center of mass is x CM =
m i xi mi
=
b
0+0+0+0 2.00 kg + 3.00 kg + 2.50 kg + 4.00 kg
g
xCM = 0 and the ycoordinate of the center of mass is yCM =
m i yi mi
=
b2.00 kgga3.00 mf + b3.00 kg ga2.50 mf + b2.50 kg ga0f + b4.00 kg ga0.500 mf
2.00 kg + 3.00 kg + 2.50 kg + 4.00 kg
yCM = 1.00 m
266 P9.39
Linear Momentum and Collisions
Take xaxis starting from the oxygen nucleus and pointing toward the middle of the V. Then yCM = 0 x CM =
and
mi x i mi
=
x CM =
0 + 1.008 u 0.100 nm cos 53.0+1.008 u 0.100 nm cos 53.0 15.999 u + 1.008 u + 1.008 u
a
f
a
f
FIG. P9.39
xCM = 0.006 73 nm from the oxygen nucleus *P9.40 Let the x axis start at the Earth's center and point toward the Moon. x CM =
24 22 8 m1 x1 + m 2 x 2 5.98 10 kg 0 + 7.36 10 kg 3.84 10 m = m1 + m 2 6.05 10 24 kg
e
j
= 4.67 10 6 m from the Earth's center The center of mass is within the Earth, which has radius 6.37 10 6 m. P9.41 Let A1 represent the area of the bottom row of squares, A 2 the middle square, and A3 the top pair. A = A1 + A 2 + A 3 M = M1 + M 2 + M 3 M1 M = A1 A A1 = 300 cm 2 , A 2 = 100 cm 2 , A3 = 200 cm 2 , A = 600 cm 2 A 300 cm 2 M M1 = M 1 = M= 2 A 2 600 cm M2 M3 x CM x CM
FG IJ H K F A IJ = 100 cm = MG H A K 600 cm F A IJ = 200 cm = MG H A K 600 cm
2 3
2 2 2
M= M=
M 6
FIG. P9.41
M 3 x M + x 2 M 2 + x 3 M 3 15.0 cm = 1 1 = M = 11.7 cm
2 1 2
c M h + 5.00 cmc Mh + 10.0 cmc Mh
1 2 1 6 1 3
M
yCM =
M 5.00 cm + 1 M 15.0 cm + 6 M
a
f
a
f c Mha25.0 cmf = 13.3 cm
1 3
yCM = 13.3 cm
Chapter 9
267
*P9.42
(a)
Represent the height of a particle of mass dm within the object as y. Its contribution to the gravitational energy of the objectEarth system is dm gy . The total gravitational energy is 1 Ug = gy dm = g y dm . For the center of mass we have yCM = y dm , so U g = gMyCM . M all mass
z
z
a f f
z
(b)
The volume of the ramp is
V = 3 800 kg m3
e
je
1 3.6 m 15.7 m 64.8 m = 1.83 10 3 m 3 . Its mass is 2 1.83 10 3 m 3 = 6.96 10 6 kg . Its center of mass is above its base by one
a
fa
fa
j
1 third of its height, yCM = 15.7 m = 5.23 m . Then 3 U g = MgyCM = 6.96 10 6 kg 9.8 m s 2 5.23 m = 3.57 10 8 J .
e
j
P9.43
(a)
M=
0.300 m 0
z
dx =
0 .300 m 0
z
50.0 g m + 20.0 x g m 2 dx
0.300 m 0
M = 50.0 x g m + 10.0 x 2 g m 2
= 15.9 g
(b)
x CM = x CM
all mass
z
xdm =
M
20 x g m 2 1 25.0 x 2 g m + = 15.9 g 3
LM NM
1 M
0.300 m 0
z
xdx =
1 M
3
0.300 m 0
z
50.0 x g m + 20.0 x 2 g m 2 dx
OP QP
0.300 m
= 0.153 m
0
*P9.44
Take the origin at the center of curvature. We have L = r= 2L
1 2r , 4
y
mass given by
dm M Mr = , dm = d where we have used the rd L L definition of radian measure. Now yCM = = 1 1 Mr r2 y dm = r sin d = M all mass M = 45 L L
2 135 45
. An incremental bit of the rod at angle from the x axis has
x FIG. P9.44
z
135
z
135 45
z
sin d
FG 2L IJ 1 a cos f HK L
= 2
=
4L
2
FG 1 + 1 IJ = 4 2L H 2 2K
2
The top of the bar is above the origin by r = by 2L
2L

4 2L
2
F1  2 2 I L = G H JK
, so the center of mass is below the middle of the bar
0.063 5L .
268
Linear Momentum and Collisions
Section 9.6 P9.45 (a)
Motion of a System of Particles v CM = = v CM =
mi v i
M
=
b2.00 kg ge2.00i m s  3.00 j m sj + b3.00 kg ge1.00i m s + 6.00 j m sj
5.00 kg
m1 v 1 + m 2 v 2 M
e1.40i + 2.40 jj m s b ge j e7.00i + 12.0 jj kg m s
(b) P9.46 (a) (b)
p = Mv CM = 5.00 kg 1. 40 i + 2.40 j m s = See figure to the right.
Using the definition of the position vector at the center of mass, rCM = rCM = rCM = m1 r1 + m 2 r2 m1 + m 2
b2.00 kg ga1.00 m, 2.00 mf + b3.00 kg ga4.00 m,  3.00 mf
2.00 kg + 3.00 kg
e2.00i  1.00 jj m b gb b g b gb g
FIG. P9.46
(c)
The velocity of the center of mass is v CM = v CM = 2.00 kg 3.00 m s , 0.50 m s + 3.00 kg 3.00 m s ,  2.00 m s P m1 v 1 + m 2 v 2 = = M m1 + m 2 2.00 kg + 3.00 kg
g
e3.00i  1.00 jj m s
P = m1 v 1 + m 2 v 2 P=
(d)
The total linear momentum of the system can be calculated as P = Mv CM or as Either gives
e15.0 i  5.00 jj kg m s
P9.47
Let x = distance from shore to center of boat = length of boat x = distance boat moves as Juliet moves toward Romeo The center of mass stays fixed. Before: After: x CM = Mb x + M J x 
B
h + M cx + h dM + M + M i M ax  x f + M cx +  x h + M c x +  x h = x dM + M + M i FG  55.0 + 77.0 IJ = x a80.0  55.0  77.0f + a55.0 + 77.0f H 2 2K 2 55.0a 2.70f 55.0 x = = = 0.700 m
2 R 2 J R 2 B J R 2 CM B J R
c
FIG. P9.47
212
212
Chapter 9
269
P9.48
(a)
Conservation of momentum for the twoball system gives us: 0.200 kg 1.50 m s + 0.300 kg 0.400 m s = 0.200 kg v1 f + 0.300 kg v 2 f Relative velocity equation: v 2 f  v1 f = 1.90 m s Then 0.300  0.120 = 0.200 v1 f + 0.300 1.90 + v1 f v1 f = 0.780 m s v 1 f = 0.780 i m s
b
g
b
g
d
i
v 2 f = 1.12 m s v 2 f = 1.12 i m s
(b)
Before,
v CM =
b0.200 kg gb1.50 m sgi + b0.300 kg gb0.400 m sgi b g
0.500 kg
v CM = 0.360 m s i Afterwards, the center of mass must move at the same velocity, as momentum of the system is conserved.
Section 9.7 P9.49 (a) (b)
Rocket Propulsion Thrust = v e dM dt Thrust = 2.60 10 3 m s 1.50 10 4 kg s = 3.90 10 7 N 3.90 10 7  3.00 10 6 9.80 = 3.00 10 6 a
e
je
j
Fy = Thrust  Mg = Ma :
e
ja f e
j
a = 3.20 m s 2
*P9.50 (a) The fuel burns at a rate Thrust = v e dM : dt dM 12.7 g = = 6.68 10 3 kg s dt 1.90 s 5.26 N = v e 6.68 10 3 kg s v e = 787 m s (b) v f  vi = v e ln
e
j
F M I: GH M JK
i f
v f  0 = 797 m s ln v f = 138 m s
b
g g g FGH 53.553.+5 25.+5 25.512.7 g IJK g g
P9.51
v = v e ln (a)
Mi Mf
Mi = e v v e M f
The mass of fuel and oxidizer is
Mi = e 5 3.00 10 3 kg = 4.45 10 5 kg M = Mi  M f = 445  3.00 10 3 kg = 442 metric tons
e
a
j
f
(b)
M = e 2 3.00 metric tons  3.00 metric tons = 19.2 metric tons Because of the exponential, a relatively small increase in fuel and/or engine efficiency causes a large change in the amount of fuel and oxidizer required.
a
f
270 P9.52
Linear Momentum and Collisions
(a)
From Equation 9.41, v  0 = v e ln Now, M f
M With the definition, Tp i , this becomes k
F M I =  v lnF M I GH M JK GH M JK F M  kt IJ =  v lnFG 1  k tIJ = M  kt , so v =  v lnG H M K H M K
i f e f i i e i i e i
(b)
With v e = 1 500 m s, and Tp ts 0 20 40 60 80 100 120 132
I JK F t IJ = 144 s , v = b1 500 m sg lnG 1  H 144 s K
v t =  v e ln 1 
af
F GH
t Tp
a f vbm sg
0 224 488 808 1220 1780 2690 3730 d  v e ln 1  Tt dt
v (m/s) 4000 3500 3000 2500 2000 1500 1000 500 0
t (s)
0
20
40
60
80
100
120
FIG. P9.52(b)
(c)
dv at = = dt at =
af af
LM N
FH
p
IK OP F 1 Q = v G GH 1 
e
t Tp
IF 1 I F v IF 1 JJ GH  T JK = GH T JK GG 1  K H
e p p
t Tp
I JJ , or K
ve Tp  t 1 500 m s 144 s  t
a (m/s ) 140 120 100 80 60 40 20 0
0 20 40 60 80 100 120
2
(d)
With v e = 1 500 m s, and Tp = 144 s , a = ts 0
a f aem s j
2
10.4 12.1 14.4
20 40 60 80 100 120 132
17.9 23.4 34.1 62.5 125
140
t (s)
FIG. P9.52(d)
continued on next page
140
Chapter 9
271
(e)
LM F t I OP L t OF dt I  v lnG 1  J dt = v T z ln M1  PG  J z z MN H T K PQ MN T PQH T K LF t I F t I F t I O xat f = v T MG 1  J lnG 1  J  G 1  J P MNH T K H T K H T K PQ F tI xat f = v eT  t j lnG 1  J + v t H TK
x t = 0 + vdt =
0
af
t
t
t
e
0
p
e p t
0
p
p
e p
p
p
p
0
e
p
p
e
(f)
With v e = 1 500 m s = 1.50 km s , and Tp = 144 s , x = 1.50 144  t ln 1  ts 0 20 40 60 80 100 120 132
a
f FGH
t + 1.50t 144
x (km) 160 140 120 100 80 60 40 20 0
IJ K
a f xakmf
0 2.19 9.23 22.1 42.2 71.7 115 153
t (s)
0
20
40
60
80
100
120
FIG. P9.52(f) *P9.53 The thrust acting on the spacecraft is
F = ma :
thrust =
F = b3 500 kg ge2.50 10 6 je9.80
e
m s 2 = 8.58 10 2 N
j
FG dM IJ v : H dt K
8.58 10 2 N = M = 4.41 kg
F M I b70 m sg GH 3 600 s JK
140
272
Linear Momentum and Collisions
Additional Problems P9.54 (a) When the spring is fully compressed, each cart moves with same velocity v. Apply conservation of momentum for the system of two gliders pi = p f : (b) m1 v 1 + m 2 v 2 = m1 + m 2 v
b
g
v=
m1 v 1 + m 2 v 2 m1 + m 2
Only conservative forces act, therefore E = 0 . Substitute for v from (a) and solve for x m .
2 xm
1 1 1 1 2 2 2 m1 v1 + m 2 v 2 = m1 + m 2 v 2 + kx m 2 2 2 2
b
g
bm =
1
2 2 + m 2 m 1 v1 + m 1 + m 2 m 2 v 2  m 1 v1
g
b
g
k m1 + m 2
b
b
g
g  bm v g
2 2 2 1 2 1 2
2
 2m 1 m 2 v1 v 2
xm = (c)
2 2 m 1 m 2 v1 + v 2  2 v 1 v 2
e
k m1 + m 2
b
g
j = bv
1
 v2
m g kbm mm g +
m1 v 1 + m 2 v 2 = m1 v 1 f + m 2 v 2 f Conservation of momentum: Conservation of energy: which simplifies to: Factoring gives m1 v 1  v 1 f = m 2 v 2 f  v 2
d
i
d
i
(1)
1 1 1 1 2 2 2 2 m 1 v1 + m 2 v 2 = m 1 v1 f + m 2 v 2 f 2 2 2 2
e m dv
1
2 2 2 2 m 1 v 1  v1 f = m 2 v 2 f  v 2
1
 v1 f
j e j i dv + v i = m dv
1 1f 2
2f
 v2 v2 f + v2
id
i
and with the use of the momentum equation (equation (1)), this reduces to or
dv
1
+ v1 f = v 2 f + v 2
i d
i
(2)
v1 f = v 2 f + v 2  v1
Substituting equation (2) into equation (1) and simplifying yields: v2 f =
FG 2m IJ v + FG m Hm +m K Hm
1 1 2 1
 m1 v2 + m2 1
2
IJ K
Upon substitution of this expression for v 2 f into equation 2, one finds v1 f =
FG m Hm
 m2 2m 2 v1 + v2 m1 + m 2 1 + m2
1
IJ K
FG H
IJ K
Observe that these results are the same as Equations 9.20 and 9.21, which should have been expected since this is a perfectly elastic collision in one dimension.
P9.55
(a)
b60.0 kg g4.00 m s = a120 + 60.0f kgv
v f = 1.33 m s i
Chapter 9
f
273
(b)
Fy = 0 :
n  60.0 kg 9.80 m s 2 = 0
b
g
f k = k n = 0.400 588 N = 235 N
fk = 235 N i (c) For the person, pi + I = p f
a
f
FIG. P9.55
b60.0 kg g4.00 m s  a235 Nft = b60.0 kg g1.33 m s
t = 0.680 s (d) person: cart: (e) (f) (g) (h) (i) x f  xi = x f  xi = mv f  mv i = 60.0 kg 1.33  4.00 m s = 160 N si 120 kg 1.33 m s  0 = +160 N si 1 1 4.00 + 1.33 m s 0.680 s = 1.81 m vi + v f t = 2 2
mvi + Ft = mv f
a
f
b
g
d d
i i
a
f
1 1 vi + v f t = 0 + 1.33 m s 0.680 s = 0.454 m 2 2
b
g
1 1 1 mv 2  mvi2 = 60.0 kg 1.33 m s f 2 2 2
b
g
2

2
1 60.0 kg 4.00 m s 2
b
g
2
= 427 J
1 1 1 mv 2  mvi2 = 120.0 kg 1.33 m s f 2 2 2
b
g
 0 = 107 J
The force exerted by the person on the cart must equal in magnitude and opposite in direction to the force exerted by the cart on the person. The changes in momentum of the two objects must be equal in magnitude and must add to zero. Their changes in kinetic energy are different in magnitude and do not add to zero. The following represent two ways of thinking about ' why. ' The distance the cart moves is different from the distance moved by the point of application of the friction force to the cart. The total change in mechanical energy for both objects together,  320 J, becomes +320 J of additional internal energy in this perfectly inelastic collision.
P9.56
The equation for the horizontal range of a projectile is R = velocity is vi = Rg =
vi2 sin 2 . Thus, with = 45.0 , the initial g
I = F t = p = mvi  0 Therefore, the magnitude of the average force acting on the ball during the impact is: F= 46.0 10 3 kg 44.3 m s mvi = = 291 N . t 7.00 10 3 s
a f
a200 mfe9.80 m s j = 44.3 m s
2
e
jb
g
274 P9.57
Linear Momentum and Collisions
We hope the momentum of the wrench provides enough recoil so that the astronaut can reach the ship before he loses life support! We might expect the elapsed time to be on the order of several minutes based on the description of the situation. No external force acts on the system (astronaut plus wrench), so the total momentum is constant. Since the final momentum (wrench plus astronaut) must be zero, we have final momentum = initial momentum = 0. m wrench v wrench + m astronaut v astronaut = 0 0.500 kg 20.0 m s m wrench v wrench = = 0.125 m s m astronaut 80.0 kg At this speed, the time to travel to the ship is Thus v astronaut =  t= 30.0 m = 240 s = 4.00 minutes 0.125 m s
b
gb
g
The astronaut is fortunate that the wrench gave him sufficient momentum to return to the ship in a reasonable amount of time! In this problem, we were told that the astronaut was not drifting away from the ship when he threw the wrench. However, this is not quite possible since he did not encounter an external force that would reduce his velocity away from the ship (there is no air friction beyond earth's atmosphere). If this were a reallife situation, the astronaut would have to throw the wrench hard enough to overcome his momentum caused by his original push away from the ship. P9.58 Using conservation of momentum from just before to just after the impact of the bullet with the block: mvi = M + m v f or vi =
vi m M
a
f
FG M + m IJ v H m K
f
(1)
h
The speed of the block and embedded bullet just after impact may be found using kinematic equations: d = v f t and h = Thus, t = 1 2 gt 2 gd 2 2h
d
FIG. P9.58
g 2h d = and v f = = d g t 2h
Substituting into (1) from above gives vi =
FG M + m IJ H m K
gd 2 . 2h
Chapter 9
275
*P9.59
(a)
Conservation of momentum: 0.5 kg 2 i  3 j + 1k m s + 1.5 kg 1i + 2 j  3k m s = 0.5 kg 1i + 3 j  8k m s + 1.5 kg v 2 f v2 f =
e
j
e
j
e
j
e0.5 i + 1.5 j  4kj kg m s + e0.5i  1.5 j + 4kj kg m s =
1.5 kg
0
The original kinetic energy is 1 1 0.5 kg 2 2 + 3 2 + 1 2 m 2 s 2 + 1.5 kg 1 2 + 2 2 + 3 2 m 2 s 2 = 14.0 J 2 2
e
j
e
j
The final kinetic energy is
1 0.5 kg 1 2 + 3 2 + 8 2 m 2 s 2 + 0 = 18.5 J different from the original 2 energy so the collision is inelastic .
e
j
(b)
We follow the same steps as in part (a):
e0.5 i + 1.5 j  4kj kg m s = 0.5 kge0.25i + 0.75 j  2kj m s + 1.5 kg v e0.5i + 1.5 j  4kj kg m s + e0.125 i  0.375 j + 1kj kg m s v =
2f 2f
1.5 kg
=
e0.250i + 0.750 j  2.00kj m s
We see v 2 f = v 1 f , so the collision is perfectly inelastic . (c) Conservation of momentum:
e0.5 i + 1.5 j  4kj kg m s = 0.5 kge1i + 3 j + akj m s + 1.5 kg v e0.5 i + 1.5 j  4kj kg m s + e0.5i  1.5 j  0.5akj kg m s v =
2f 2f
1.5 kg
= Conservation of energy: 14.0 J =
a2.67  0.333 afk m s a f
2
1 1 0.5 kg 1 2 + 3 2 + a 2 m 2 s 2 + 1.5 kg 2.67 + 0.333 a 2 2 2 = 2.5 J + 0. 25 a + 5.33 J + 1.33 a + 0.083 3 a 2
e
j
m2 s 2
0 = 0.333 a 2 + 1.33 a  6.167 a= 1.33 1.33 2  4 0.333 6.167
a
fa
f
0.667 a = 2.74 or  6.74. Either value is possible. a = 2.74 , v 2 f = 2.67  0.333 2.74 k m s = 3.58k m s 0.419 k m s
a = 6.74 , v 2 f
a fh c = c2.67  0.333a 6.74fhk m s =
276 P9.60
Linear Momentum and Collisions
(a)
The initial momentum of the system is zero, which remains constant throughout the motion. Therefore, when m1 leaves the wedge, we must have m 2 v wedge + m1 v block = 0 or so
b3.00 kg gv
wedge
+ 0.500 kg +4.00 m s = 0
b
gb
g
v wedge
v wedge = 0.667 m s
+x
v block = 4.00 m/s
(b)
Using conservation of energy for the blockwedgeEarth system as the block slides down the smooth (frictionless) wedge, we have K block + U system + K wedge = K block + U system
i i f
FIG. P9.60 + K wedge
f
or *P9.61 (a)
0 + m1 gh + 0 =
LM 1 m a4.00f + 0OP + 1 m a0.667f N2 Q 2
1 2 2
2
which gives h = 0.952 m .
Conservation of the x component of momentum for the cartbucketwater system: mvi + 0 = m + V v
b
g
vi =
m + V v m
(b)
Raindrops with zero xcomponent of momentum stop in the bucket and slow its horizontal motion. When they drip out, they carry with them horizontal momentum. Thus the cart slows with constant acceleration.
Chapter 9
277
P9.62
Consider the motion of the firefighter during the three intervals: (1) before, (2) during, and (3) after collision with the platform. (a) While falling a height of 4.00 m, his speed changes from vi = 0 to v1 as found from E = K f + U f  K i  U i , or K f = E  U f + K i + U i When the initial position of the platform is taken as the zero level of gravitational potential, we have 1 2 mv1 = fh cos 180  0 + 0 + mgh 2 Solving for v1 gives v1 = (b) 2  fh + mgh m
v1 v2
d
i b
g
a f
FIG. P9.62
b
g = 2c300a4.00f + 75.0a9.80f4.00h =
75.0
6.81 m s
During the inelastic collision, momentum is conserved; and if v 2 is the speed of the firefighter and platform just after collision, we have mv1 = m + M v 2 or
a
f
v2 =
75.0 6.81 m 1 v1 = = 5.38 m s m + M 75.0 + 20.0
a f
Following the collision and again solving for the work done by nonconservative forces, using the distances as labeled in the figure, we have (with the zero level of gravitational potential at the initial position of the platform): E = K f + U fg + U fs  K i  U ig  U is , or 1 1  fs = 0 + m + M g  s + ks 2  m + M v 2  0  0 2 2
a
fa f
a
f
This results in a quadratic equation in s: 2 000s 2  931 s + 300s  1 375 = 0 or s = 1.00 m
a f
278 *P9.63
Linear Momentum and Collisions
(a)
Each object swings down according to mgR = 1 2 mv1 2 MgR = 1 2 Mv1 2 v1 = 2 gR
The collision: mv1 + Mv1 = + m + M v 2 Mm v2 = v1 M+m Swinging up: 1 2 M + m v 2 = M + m gR 1  cos 35 2 v 2 = 2 gR 1  cos 35 0.425 M + 0.425m = M  m 1.425m = 0.575 M m = 0.403 M (b) No change is required if the force is different. The nature of the forces within the system of colliding objects does not affect the total momentum of the system. With strong magnetic attraction, the heavier object will be moving somewhat faster and the lighter object faster still. Their extra kinetic energy will all be immediately converted into extra internal energy when the objects latch together. Momentum conservation guarantees that none of the extra kinetic energy remains after the objects join to make them swing higher. Use conservation of the horizontal component of momentum for the system of the shell, the cannon, and the carriage, from just before to just after the cannon firing. p xf = p xi : or (b)
a
f
f a f a f 2 gRa1  cos 35fa M + mf = a M  m f 2 gR f
a
a
P9.64
(a)
a200fa125f cos 45.0+b5 000gv
v recoil = 3.54 m s
m shell v shell cos 45.0+ m cannon v recoil = 0
recoil
=0 FIG. P9.64
Use conservation of energy for the system of the cannon, the carriage, and the spring from right after the cannon is fired to the instant when the cannon comes to rest. K f + U gf + U sf = K i + U gi + U si : 0+0+ x max = 1 2 1 2 kx max = mv recoil + 0 + 0 2 2
2 mv recoil = k
b5 000ga3.54f
2.00 10 4
2
m = 1.77 m
(c) (d)
Fs, max = kx max
Fs, max = 2.00 10 4 N m 1.77 m = 3.54 10 4 N
e
ja
f
No. The rail exerts a vertical external force (the normal force) on the cannon and prevents it from recoiling vertically. Momentum is not conserved in the vertical direction. The spring does not have time to stretch during the cannon firing. Thus, no external horizontal force is exerted on the system (cannon, carriage, and shell) from just before to just after firing. Momentum of this system is conserved in the horizontal direction during this interval.
Chapter 9
279
P9.65
(a)
Utilizing conservation of momentum, m 1 v1 A = m 1 + m 2 v B v1 A m + m2 = 1 m1 2 gh
v1i
b
g
y
v1 A 6.29 m s (b) Utilizing the two equations, 1 2 gt = y and x = v1 A t 2 we combine them to find v1 A = x
2y g
x FIG. P9.65
From the data, v1 A = 6.16 m s Most of the 2% difference between the values for speed is accounted for by the uncertainty 0.01 0.1 1 1 0.1 in the data, estimated as + + + + = 1.1% . 8.68 68.8 263 257 85.3 *P9.66 The ice cubes leave the track with speed determined by mgyi = v = 2 9.8 m s 2 1.5 m = 5.42 m s . Its speed at the apex of its trajectory is 5.42 m s cos 40 = 4.15 m s . For its collision with the wall we have mvi + Ft = mv f 1 mv 2 ; 2
e
j
0.005 kg 4.15 m s + Ft = 0.005 kg  Ft = 3.12 10 2 kg m s
FG H
1 4.15 m s 2
IJ K
The impulse exerted by the cube on the wall is to the right, +3.12 10 2 kg m s. Here F could refer to a large force over a short contact time. It can also refer to the average force if we interpret t as 1 s, the time between one cube's tap and the next's. 10 Fav = 3.12 10 2 kg m s = 0.312 N to the right 0.1 s
280 P9.67
Linear Momentum and Collisions
(a)
Find the speed when the bullet emerges from the block by using momentum conservation: mvi = MVi + mv The block moves a distance of 5.00 cm. Assume for an approximation that the block quickly reaches its maximum velocity, Vi , and the bullet kept going with a constant velocity, v. The block then compresses the spring and stops.
400 m/s
5.00 cm
v
1 1 MVi2 = kx 2 2 2 Vi = v=
FIG. P9.67
b900 N mge5.00 10
1.00 kg
2
m
j
2
= 1.50 m s
5.00 10 3 kg 400 m s  1.00 kg 1.50 m s mvi  MVi = m 5.00 10 3 kg
e
jb
g b
gb
g
v = 100 m s (b) E = K + U =
e 1 + b900 N mge5.00 10 2
1 5.00 10 3 kg 100 m s 2
2 2
jb mj
g
2

1 5.00 10 3 kg 400 m s 2
e
jb
g
2
E = 374 J , or there is an energy loss of 374 J .
*P9.68 The orbital speed of the Earth is vE = 2r 2 1.496 10 11 m = = 2.98 10 4 m s T 3.156 10 7 s FIG. P9.68 S CM E
In six months the Earth reverses its direction, to undergo momentum change
m E v E = 2m E v E = 2 5.98 10 24 kg 2.98 10 4 m s = 3.56 10 25 kg m s . Relative to the center of mass, the sun always has momentum of the same magnitude in the opposite direction. Its 6month momentum change is the same size, mS v S = 3.56 10 25 kg m s . Then v S = 3.56 10 25 kg m s 1.991 10 30 kg = 0.179 m s .
e
je
j
Chapter 9
281
P9.69
(a)
p i + Ft = p f :
b3.00 kg gb7.00 m sg j + e12.0 Nija5.00 sf = b3.00 kg gv v = e 20.0 i + 7.00 jj m s
f
f
(b) (c) (d)
a= a=
v f  vi t
:
a= a=
e20.0 i + 7.00 j  7.00 jj m s =
5.00 s 12.0 N i = 4.00 i m s 2 3.00 kg
4.00 i m s 2
F :
m 1 2 at : 2
r = v i t +
r = 7.00 m s j 5.00 s + r =
e
ja
e50.0i + 35.0 jj m e je je b g
f 1 e4.00 m s ija5.00 sf 2
2
2
(e) (f)
W = F r :
W = 12.0 N i 50.0 m i + 35.0 m j = 600 J
j
1 1 mv 2 = 3.00 kg 20.0 i + 7.00 j 20.0 i + 7.00 j m 2 s 2 f 2 2 1 mv 2 = 1.50 kg 449 m 2 s 2 = 674 J f 2
b
ge
j
ge
j
(g) P9.70
1 1 mvi2 + W = 3.00 kg 7.00 m s 2 2
b
gb
2
+ 600 J = 674 J
We find the mass from We find the acceleration from
M = 360 kg  2.50 kg s t . a=
b
g
1 500 m s 2.50 kg s 3 750 N Thrust v e dM dt = = = M M M M We find the velocity and position according to Euler, v new = v old + a t from x new = x old + v t and If we take t = 0.132 s , a portion of the output looks like this:
b
gb
g
a f a f
Time t(s) 0.000 0.132 0.264 ... 65.868 66.000 66.132 ... 131.736 131.868 132.000 (a) (b)
Total mass (kg) 360.00 359.67 359.34 195.330 195.000 194.670 30.660 30.330 30.000
Acceleration a m s2
e
j
Speed, v (m/s) 0.0000 1.3750 2.7513 916.54 919.08 921.61 3687.3 3703.5 3719.8
Position x(m) 0.0000 0.1815 0.54467 27191 27312 27433 152382 152871 153362
10.4167 10.4262 10.4358 19.1983 19.2308 19.2634 122.3092 123.6400 125.0000 v f = 3.7 km s 153 km
The final speed is The rocket travels
282 P9.71
Linear Momentum and Collisions
The force exerted by the table is equal to the change in momentum of each of the links in the chain. By the calculus chain rule of derivatives, F1 = dp d mv dm dv = =v +m . dt dt dt dt FIG. P9.71
a f
We choose to account for the change in momentum of each link by having it pass from our area of interest just before it hits the table, so that v dm dv 0 and m = 0. dt dt
Since the mass per unit length is uniform, we can express each link of length dx as having a mass dm: dm = M dx . L
The magnitude of the force on the falling chain is the force that will be necessary to stop each of the elements dm. F1 = v dm M dx M 2 =v = v dt L dt L
FG IJ H K
FG IJ H K
After falling a distance x, the square of the velocity of each link v 2 = 2 gx (from kinematics), hence F1 = 2 Mgx . L
The links already on the table have a total length x, and their weight is supported by a force F2 : F2 = Hence, the total force on the chain is Ftotal = F1 + F2 = 3 Mgx . L Mgx . L
That is, the total force is three times the weight of the chain on the table at that instant.
Chapter 9
283
P9.72
A picture one second later differs by showing five extra kilograms of sand moving on the belt. (a) (b) 5.00 kg 0.750 m s p x = = 3.75 N t 1.00 s The only horizontal force on the sand is belt friction, so from (c) p xi + ft = p xf this is f= p x = 3.75 N t
b
gb
g
The belt is in equilibrium:
Fx = ma x :
(d) (e) (f)
+ Fext  f = 0
and
Fext = 3.75 N
W = Fr cos = 3.75 N 0.750 m cos 0 = 2.81 J 1 1 m v 2 = 5.00 kg 0.750 m s 2 2
a
f
a f
b
g
2
= 1.41 J
Friction between sand and belt converts half of the input work into extra internal energy.
*P9.73
x CM =
m i xi mi
=
m1 R +
c
2
m1 + m 2
h + m a0 f =
2
m1 R +
c
2
h
y x R
m1 + m 2
2 FIG. P9.73
ANSWERS TO EVEN PROBLEMS
P9.2 P9.4 P9.6 P9.8 P9.10 P9.12 P9.14 P9.16 P9.18 (a) 0; (b) 1.06 kg m s ; upward (a) 6.00 m s to the left; (b) 8.40 J The force is 6.44 kN 1.39 kg m s upward (a) 5.40 N s toward the net; (b) 27.0 J ~ 10 3 N upward (a) and (c) see the solution; (b) small; (d) large; (e) no difference 1.67 m s (a) 2.50 m s ; (b) 3.75 10 4 J P9.24 P9.26 P9.28 v= P9.20 P9.22 0.556 m 1.78 kN on the truck driver; 8.89 kN in the opposite direction on the car driver 4M m g
7.94 cm (a) 2.88 m s at 32.3; (b) 783 J becomes internal energy v Y = vi sin ; vO = vi cos No; his speed was 41.5 mi h (a) v = vi 2 ; (b) 45.0 and 45.0
P9.30 P9.32 P9.34
284 P9.36 P9.38 P9.40 P9.42 P9.44 P9.46
Linear Momentum and Collisions
(a)
2vi ;
2 vi ; (b) 35.3 3
(c) v 1 f = v2 f = P9.56 P9.58 291 N
a0, 1.00 mf
4.67 10 6 m from the Earth's center (a) see the solution; (b) 3.57 10 8 J 0.063 5L
FG 2m IJ v + FG m Hm +m K Hm
1 1 2 1
FG m Hm
 m2 2m 2 v1 + v2 ; m1 + m 2 1 + m2
1
IJ K
FG H
 m1 v2 1 + m2
2
IJ K
IJ K
FG M + m IJ H m K
gd 2 2h
a f (c) e3.00 i  1.00 jj m s ; (d) e15.0 i  5.00 jj kg m s
(a) 0.780 i m s ; 1.12 i m s; (b) 0.360 i m s (a) 787 m s; (b) 138 m s see the solution (a) m1 v 1 + m 2 v 2 ; m1 + m 2
(a) see the solution; (b) 2.00 m,  1.00 m ;
P9.60 P9.62 P9.64
(a) 0.667 m s; (b) 0.952 m (a) 6.81 m s; (b) 1.00 m (a) 3.54 m s ; (b) 1.77 m; (c) 35.4 kN; (d) No. The rails exert a vertical force to change the momentum 0.312 N to the right 0.179 m s (a) 3.7 km s ; (b) 153 km (a) 3.75 N to the right; (b) 3.75 N to the right; (c) 3.75 N; (d) 2.81 J; (e) 1.41 J; (f) Friction between sand and belt converts half of the input work into extra internal energy.
P9.48 P9.50 P9.52 P9.54
P9.66 P9.68 P9.70 P9.72
(b) v1  v 2
b
m g kbm mm g ; +
1 2 1 2
10
Rotation of a Rigid Object About a Fixed Axis
CHAPTER OUTLINE
10.1 10.2 Angular Position, Velocity, and Acceleration Rotational Kinematics: Rotational Motion with Constant Angular Acceleration Angular and Linear Quantities Rotational Energy Calculation of Moments of Inertia Torque Relationship Between Torque and Angular Acceleration Work, Power, and Energy in Rotational Motion Rolling Motion of a Rigid Object
ANSWERS TO QUESTIONS
Q10.1 1 rev/min, or
rad/s. Into the wall (clockwise rotation). = 0. 30
10.3 10.4 10.5 10.6 10.7
FIG. Q10.1 Q10.2 Q10.3 +k , k Yes, they are valid provided that is measured in degrees per second and is measured in degrees per secondsquared.
10.8 10.9
Q10.4 Q10.5 Q10.6
The speedometer will be inaccurate. The speedometer measures the number of revolutions per second of the tires. A larger tire will travel more distance in one full revolution as 2r . Smallest I is about x axis and largest I is about y axis. ML2 if the mass was nonuniformly distributed, nor 12 could it be calculated if the mass distribution was not known. The moment of inertia would no longer be The object will start to rotate if the two forces act along different lines. Then the torques of the forces will not be equal in magnitude and opposite in direction. No horizontal force acts on the pencil, so its center of mass moves straight down. You could measure the time that it takes the hanging object, m, to fall a measured distance after being released from rest. Using this information, the linear acceleration of the mass can be calculated, and then the torque on the rotating object and its angular acceleration. You could use = t and v = at . The equation v = R is valid in this situation since a = R . The angular speed would decrease. The center of mass is farther from the pivot, but the moment of inertia increases also. 285
Q10.7 Q10.8 Q10.9
Q10.10 Q10.11
286 Q10.12
Rotation of a Rigid Object About a Fixed Axis
The moment of inertia depends on the distribution of mass with respect to a given axis. If the axis is changed, then each bit of mass that makes up the object is a different distance from the axis. In example 10.6 in the text, the moment of inertia of a uniform rigid rod about an axis perpendicular to the rod and passing through the center of mass is derived. If you spin a pencil back and forth about this axis, you will get a feeling for its stubbornness against changing rotation. Now change the axis about which you rotate it by spinning it back and forth about the axis that goes down the middle of the graphite. Easier, isn't it? The moment of inertia about the graphite is much smaller, as the mass of the pencil is concentrated near this axis. Compared to an axis through the center of mass, any other parallel axis will have larger average squared distance from the axis to the particles of which the object is composed. A quick flip will set the hardboiled egg spinning faster and more smoothly. The raw egg loses mechanical energy to internal fluid friction. I CM = MR 2 , I CM = MR 2 , I CM = 1 1 MR 2 , I CM = MR 2 3 2
Q10.13 Q10.14
Q10.15 Q10.16 Q10.17 Q10.18 Q10.19 Q10.20
Yes. If you drop an object, it will gain translational kinetic energy from decreasing gravitational potential energy. No, just as an object need not be moving to have mass. No, only if its angular momentum changes. Yes. Consider a pendulum at its greatest excursion from equilibrium. It is momentarily at rest, but must have an angular acceleration or it would not oscillate. Since the source reel stops almost instantly when the tape stops playing, the friction on the source reel axle must be fairly large. Since the source reel appears to us to rotate at almost constant angular velocity, the angular acceleration must be very small. Therefore, the torque on the source reel due to the tension in the tape must almost exactly balance the frictional torque. In turn, the frictional torque is nearly constant because kinetic friction forces don't depend on velocity, and the radius of the axle where the friction is applied is constant. Thus we conclude that the torque exerted by the tape on the source reel is essentially constant in time as the tape plays. v must increase to keep the As the source reel radius R shrinks, the reel's angular speed = R tape speed v constant. But the biggest change is to the reel's moment of inertia. We model the reel as a roll of tape, ignoring any spool or platter carrying the tape. If we think of the roll of tape as a 1 uniform disk, then its moment of inertia is I = MR 2 . But the roll's mass is proportional to its base 2 area R 2 . Thus, on the whole the moment of inertia is proportional to R 4 . The moment of inertia decreases very rapidly as the reel shrinks! The tension in the tape coming into the readandwrite heads is normally dominated by balancing frictional torque on the source reel, according to TR friction . Therefore, as the tape plays the tension is largest when the reel is smallest. However, in the case of a sudden jerk on the tape, the rotational dynamics of the source reel becomes important. If the source reel is full, then the moment of inertia, proportional to R 4 , will be so large that higher tension in the tape will be required to give the source reel its angular acceleration. If the reel is nearly empty, then the same tape acceleration will require a smaller tension. Thus, the tape will be more likely to break when the source reel is nearly full. One sees the same effect in the case of paper towels; it is easier to snap a towel free when the roll is new than when it is nearly empty.
Chapter 10
287
Q10.21 Q10.22
The moment of inertia would decrease. This would result in a higher angular speed of the earth, shorter days, and more days in the year! There is very little resistance to motion that can reduce the kinetic energy of the rolling ball. Even though there is static friction between the ball and the floor (if there were none, then no rotation would occur and the ball would slide), there is no relative motion of the two surfacesby the definition of "rolling"and so no force of kinetic friction acts to reduce K. Air resistance and friction associated with deformation of the ball eventually stop the ball. In the frame of reference of the ground, no. Every point moves perpendicular to the line joining it to the instantaneous contact point. The contact point is not moving at all. The leading and trailing edges of the cylinder have velocities at 45 to the vertical as shown.
Q10.23
v CM vCM v P
FIG. Q10.23 Q10.24 The sphere would reach the bottom first; the hoop would reach the bottom last. If each object has the same mass and the same radius, they all have the same torque due to gravity acting on them. The one with the smallest moment of inertia will thus have the largest angular acceleration and reach the bottom of the plane first. To win the race, you want to decrease the moment of inertia of the wheels as much as possible. Small, light, solid disklike wheels would be best!
Q10.25
SOLUTIONS TO PROBLEMS
Section 10.1 P10.1 (a) Angular Position, Velocity, and Acceleration
t= 0 = 5.00 rad t =0 = t=0 =
d dt d dt
t=0
= 10.0 + 4.00t t = 0 = 10.0 rad s = 4.00 rad s 2
t=0
(b)
t= 3.00 s = 5.00 + 30.0 + 18.0 = 53.0 rad t = 3.00 s = t = 3.00 s =
d dt d dt
t = 3 .00 s
= 10.0 + 4.00t t = 3.00 s = 22.0 rad s = 4.00 rad s 2
t = 3 .00 s
288
Rotation of a Rigid Object About a Fixed Axis
Section 10.2 *P10.2
Rotational Kinematics: Rotational Motion with Constant Angular Acceleration
f = 2.51 10 4 rev min = 2.63 10 3 rad s
(a) (b)
=
f i
t
=
2.63 10 3 rad s  0 = 8.22 10 2 rad s 2 3.2 s
1 1 f = i t + t 2 = 0 + 8.22 10 2 rad s 2 3.2 s 2 2
e
ja f
2
= 4.21 10 3 rad
P10.3
(a) (b)
=
 i 12.0 rad s = = 4.00 rad s 2 t 3.00 s
1 1 = i t + t 2 = 4.00 rad s 2 3.00 s 2 2
e
ja
f
2
= 18.0 rad
P10.4
i = 2 000 rad s , = 80.0 rad s 2
(a) (b)
f = i + t = 2 000  80.0 10.0 = 1 200 rad s
0 = i + t t= 
a fa f
i
=
2 000 = 25.0 s 80.0
P10.5
i =
(a) (b)
100 rev 1 min 1.00 min 60.0 s t=
FG H
IJ FG 2 rad IJ = 10 K H 1.00 rev K 3
0  10 3 2.00 +i 2 s = 5.24 s
rad s , f = 0
f i
=
f
f = t =
FG H
IJ t = FG 10 K H6
rad s
IJ FG 10 sIJ = KH 6 K
27.4 rad
P10.6
i = 3 600 rev min = 3.77 10 2 rad s = 50.0 rev = 3.14 10 2 rad and f = 0 2 = i2 + 2 f
0 = 3.77 10 2 rad s
e
j
2
+ 2 3.14 10 2 rad
e
j
= 2.26 10 2 rad s 2
P10.7
= 5.00 rev s = 10.0 rad s . We will break the motion into two stages: (1) a period during which the tub speeds up and (2) a period during which it slows down.
While speeding up, While slowing down, So,
1 = t = 2
a f 10.0 rad s + 0 = t = a12.0 sf = 60.0 rad 2
0 + 10.0 rad s 8.00 s = 40.0 rad 2
total = 1 + 2 = 100 rad = 50.0 rev
Chapter 10
289
P10.8
1 f  i = i t + t 2 and f = i + t are two equations in two unknowns i and 2
i = f  t :
1 1 f  i = f  t t + t 2 = f t  t 2 2 2 2 rad 1 37.0 rev = 98.0 rad s 3.00 s  3.00 s 1 rev 2
d
i
FG H
IJ K
a
f
a
f
2
232 rad = 294 rad  4.50 s 2 :
e
j
=
61.5 rad = 13.7 rad s 2 2 4.50 s
P10.9
(a)
=
1 rev 2 rad = = = 7.27 10 5 rad s t 1 day 86 400 s = 107 2 rad = 2.57 10 4 s or 428 min 5 7.27 10 rad s 360
(b) *P10.10
t =
FG H
The location of the dog is described by d
IJ K = b0.750 rad sgt . For the bone,
b =
We look for a solution to 0.75t =
1 1 2 rad + 0.015 rad s 2 t 2 . 3 2
2 + 0.007 5t 2 3 0 = 0.007 5t 2  0.75t + 2.09 = 0 t= 0.75 0.75 2  4 0.007 5 2.09 0.015
b
g
= 2.88 s or 97.1 s
The dog and bone will also pass if 0.75t =
2 2  2 + 0.007 5t 2 or if 0.75t = + 2 + 0.007 5t 2 that is, if 3 3 either the dog or the turntable gains a lap on the other. The first equation has t= 0.75 0.75 2  4 0.007 5 4.19 0.015
b
ga
f = 105 s or  5.30 s
only one positive root representing a physical answer. The second equation has t= 0.75 0.75 2  4 0.007 5 8.38 0.015
b
g
= 12.8 s or 87.2 s .
In order, the dog passes the bone at 2.88 s after the merrygoround starts to turn, and again at 12.8 s and 26.6 s, after gaining laps on the bone. The bone passes the dog at 73.4 s, 87.2 s, 97.1 s, 105 s, and so on, after the start.
290
Rotation of a Rigid Object About a Fixed Axis
Section 10.3 P10.11
Angular and Linear Quantities
Estimate the tire's radius at 0.250 m and miles driven as 10 000 per year.
=
s 1.00 10 4 mi 1 609 m = = 6.44 10 7 rad yr 0.250 m 1 mi r
7
= 6.44 10 7
FG IJ H K F 1 rev IJ = 1.02 10 rad yr G H 2 rad K
rev yr or ~ 10 7 rev yr
P10.12
(a)
v = r ; =
v 45.0 m s = = 0.180 rad s r 250 m
(b) P10.13
45.0 m s v2 = ar = r 250 m
b
g
2
= 8.10 m s 2 toward the center of track
Given r = 1.00 m, = 4.00 rad s 2 , i = 0 and i = 57.3 = 1.00 rad (a)
f = i + t = 0 + t
At t = 2.00 s , f = 4.00 rad s 2 2.00 s = 8.00 rad s
a
f
(b)
v = r = 1.00 m 8.00 rad s = 8.00 m s a r = a c = r 2 = 1.00 m 8.00 rad s
b
g
b
g
2
= 64.0 m s 2
a t = r = 1.00 m 4.00 rad s 2 = 4.00 m s 2 The magnitude of the total acceleration is:
2 a = a r + a t2 =
e
j
e64.0 m s j + e4.00 m s j
2 2
2 2
= 64.1 m s 2
The direction of the total acceleration vector makes an angle with respect to the radius to point P:
= tan 1
FG a IJ = tan FG 4.00 IJ = H 64.0 K Ha K
t 1 c
3.58
(c)
1 1 f = i + i t + t 2 = 1.00 rad + 4.00 rad s 2 2.00 s 2 2
a
f e
ja
f
2
= 9.00 rad
Chapter 10
291
*P10.14
(a)
Consider a tooth on the front sprocket. It gives this speed, relative to the frame, to the link of the chain it engages: v = r =
FG 0.152 m IJ 76 rev min FG 2 rad IJ FG 1 min IJ = H 2 K H 1 rev K H 60 s K
=
v 0.605 m s = 0.07 m = 17.3 rad s r 2
0.605 m s
(b)
Consider the chain link engaging a tooth on the rear sprocket:
c
h
(c)
Consider the wheel tread and the road. A thread could be unwinding from the tire with this speed relative to the frame: v = r =
FG 0.673 m IJ 17.3 rad s = H 2 K
5.82 m s
(d)
We did not need to know the length of the pedal cranks, but we could use that information to find the linear speed of the pedals: v = r = 0.175 m 7.96 rad s v 25.0 m s = = 25.0 rad s r 1.00 m
FG 1 IJ = 1.39 m s H 1 rad K
P10.15
(a) (b)
=
2 = i2 + 2 f
= 2  i2 f
a f
2
b25.0 rad sg  0 = = 2a f 2 a1.25 revfb 2 rad rev g
= 25.0 rad s 39.8 rad s 2 = 0.628 s
39.8 rad s 2
(c) P10.16 (a)
t =
s = vt = 11.0 m s 9.00 s = 99.0 m
b
ga
f
=
s 99.0 m = = 341 rad = 54.3 rev r 0.290 m vf r = 22.0 m s = 75.9 rad s = 12.1 rev s 0.290 m
(b)
f =
292 P10.17
Rotation of a Rigid Object About a Fixed Axis
(a) (b) (c) (d)
= 2f =
2 rad 1 200 rev = 126 rad s 1 rev 60.0 s
FG H
IJ K
v = r = 126 rad s 3.00 10 2 m = 3.77 m s a c = 2 r = 126
b
ge
j
a f e8.00 10 j = 1 260 m s
2 2
2
so a r = 1.26 km s 2 toward the center
s = r = rt = 126 rad s 8.00 10 2 m 2.00 s = 20.1 m
b
ge
ja
f
P10.18
The force of static friction must act forward and then more and more inward on the tires, to produce both tangential and centripetal acceleration. Its tangential component is m 1.70 m s 2 . Its radially
e
j
inward component is
mv . This takes the maximum value r
2
m 2 r = mr i2 + 2 = mr 0 + 2 f With skidding impending we have
e
j
FG H
= mr = ma t = m 1.70 m s 2 . 2
IJ K
e
j
Fy = ma y , + n  mg = 0, n = mg
fs = sn = s mg = m 2 1.70 m s 2
e
j
2
+ m 2 2 1.70 m s 2
e
j
2
s =
*P10.19 (a)
1.70 m s 2 g
1 + 2 = 0.572
Let RE represent the radius of the Earth. The base of the building moves east at v1 = RE where is one revolution per day. The top of the building moves east at v 2 = RE + h . Its eastward speed relative to the ground is v 2  v1 = h . The object's time of fall is given by
b
g
y = 0 +
1 2 gt , t = 2
2h . During its fall the object's eastward motion is unimpeded so its g
deflection distance is x = v 2  v1 t = h
b
g
2h 2 = h3 2 g g
F I GH JK
12
.
(b) (c)
2 rad 50 m 86 400 s
a
f FGH 92.8sm IJK
32 2
12
= 1.16 cm
The deflection is only 0.02% of the original height, so it is negligible in many practical cases.
Chapter 10
293
Section 10.4 P10.20
Rotational Energy
m1 = 4.00 kg , r1 = y1 = 3.00 m ; m 2 = 2.00 kg , r2 = y 2 = 2.00 m; m 3 = 3.00 kg , r3 = y 3 = 4.00 m ;
= 2.00 rad s about the xaxis
(a) I x = m1 r12 + m 2 r22 + m 3 r32 I x = 4.00 3.00 KR = 1 I x 2 2
a f
a f + 3.00a4.00f 1 = a92.0fa 2.00f = 184 J 2
2
+ 2.00 2.00
2
2
= 92.0 kg m 2
2
FIG. P10.20 1 1 2 2 m1 v1 = 4.00 6.00 = 72.0 J 2 2 1 1 2 2 K 2 = m 2 v 2 = 2.00 4.00 = 16.0 J 2 2 1 1 2 2 K 3 = m 3 v 3 = 3.00 8.00 = 96.0 J 2 2 K1 =
(b)
v1 = r1 = 3.00 2.00 = 6.00 m s v2 v3
2
a f = r = 2.00a 2.00f = = r = 4.00a 2.00f =
3
4.00 m s 8.00 m s
a fa f a fa f a fa f
y (m) 4
K = K 1 + K 2 + K 3 = 72.0 + 16.0 + 96.0 = 184 J = P10.21 (a) I = m j r j2
j
1 I x 2 2
In this case, r1 = r2 = r3 = r4 r= I=
3.00 kg
3 2
2.00 kg
a3.00 mf + a2.00 mf
2
2
= 13.0 m
1 x (m) 0 1 2 3
13.0 m
2
3.00 + 2.00 + 2.00 + 4.00 kg
= 143 kg m 2 (b) KR = 1 2 1 I = 143 kg m 2 6.00 rad s 2 2
e
jb
g
2
2.00 kg
4.00 kg
= 2.57 10 3 J
FIG. P10.21
294 P10.22
Rotation of a Rigid Object About a Fixed Axis
I = Mx 2 + m L  x
a
f
2
x
dI = 2 Mx  2m L  x = 0 (for an extremum) dx mL x = M+m d2I = 2m + 2 M ; therefore I is minimum when the axis of dx 2 mL which is also the center rotation passes through x = M+m of mass of the system. The moment of inertia about an axis passing through x is I CM = M
a
f
L M x Lx m
LM mL OP NM + mQ
2
+m 1
LM N
m M+m
OP L Q
2
2
=
Mm 2 L = L2 M+m
FIG. P10.22
Mm . where = M+m Section 10.5 P10.23 Calculation of Moments of Inertia y
We assume the rods are thin, with radius much less than L. Call the junction of the rods the origin of coordinates, and the axis of rotation the zaxis. For the rod along the yaxis, I = 1 mL2 from the table. 3
x
For the rod parallel to the zaxis, the parallelaxis theorem gives I= 1 L mr 2 + m 2 2
axis of rotation z FIG. P10.23
FG IJ H K
2
1 mL2 4
In the rod along the xaxis, the bit of material between x and x + dx has mass distance r = x 2 +
FG L IJ H 2K
FG m IJ dx and is at H LK
2
from the axis of rotation. The total rotational inertia is: 1 1 L2 mL2 + mL2 + x2 + 3 4 4 L 2
L2
I total =
z FGH
I FG m IJ dx JK H L K
L2
7 m x3 = mL2 + 12 L 3 =
2
FG IJ H K
L2
+
L 2
mL x 4 L 2
7 11mL2 mL mL2 + = mL2 + 12 12 4 12
Note: The moment of inertia of the rod along the x axis can also be calculated from the parallelaxis 2 1 L theorem as . mL2 + m 12 2
FG IJ H K
Chapter 10
295
P10.24
Treat the tire as consisting of three parts. The two sidewalls are each treated as a hollow cylinder of inner radius 16.5 cm, outer radius 30.5 cm, and height 0.635 cm. The tread region is treated as a hollow cylinder of inner radius 30.5 cm, outer radius 33.0 cm, and height 20.0 cm. Use I = 1 2 2 m R1 + R 2 for the moment of inertia of a hollow cylinder. 2
e
j
Sidewall: m = 0.305 m I side Tread: m = 0.330 m I tread Entire Tire:
f  a0.165 mf e6.35 10 mje1.10 10 1 = b1.44 kg g a0.165 mf + a0.305 mf = 8.68 10 2
2 2 3 2 2
a
3
kg m 3 = 1.44 kg kg m 2
j
2
f  a0.305 mf a0.200 mfe1.10 10 kg m j = 11.0 kg 1 = b11.0 kg g a0.330 mf + a0.305 mf = 1.11 kg m 2
2 2 3 3 2 2 2
a
I total = 2 I side + I tread = 2 8.68 10 2 kg m 2 + 1.11 kg m 2 = 1.28 kg m 2
P10.25 Every particle in the door could be slid straight down into a highdensity rod across its bottom, without changing the particle's distance from the rotation axis of the door. Thus, a rod 0.870 m long with mass 23.0 kg, pivoted about one end, has the same rotational inertia as the door: I= 1 1 ML2 = 23.0 kg 0.870 m 3 3
e
j
b
ga
f
2
= 5.80 kg m 2 .
The height of the door is unnecessary data. P10.26 Model your body as a cylinder of mass 60.0 kg and circumference 75.0 cm. Then its radius is 0.750 m = 0.120 m 2 and its moment of inertia is 1 1 MR 2 = 60.0 kg 0.120 m 2 2
b
ga
f
2
= 0.432 kg m 2 ~ 10 0 kg m 2 = 1 kg m 2 .
296 P10.27
Rotation of a Rigid Object About a Fixed Axis
For a spherical shell dI =
2 2 4r 2 dr r 2 dmr 2 = 3 3
e
j
e j af 2 rI F 4r jG 14. 2  11.6 J e10 kg m jdr I= H 3 RK R R F 2I F 2I = G J 4 e14. 2 10 j  G J 4 e11.6 10 j H 3K H 3K 5 6 8 14.2 11.6 I I= e10 jR FGH 5  6 JK 3 rI F M = z dm = z 4r G 14.2  11.6 J 10 dr H RK F 14.2  11.6 IJ R = 4 10 G H3 4K b8 3ge10 jR b14.2 5  11.6 6g = 2 FG .907 IJ = 0.330 I = 4 10 R R b14.2 3  11.6 4g 3 H 1.83 K MR
I = dI =
R 0
z z ze
2 4r 2 r 2 r dr 3
4
3
3
3
5
3
5
3
5
R 0
2
3
3
3
3
5
2
3
3
2
I = 0.330 MR 2 *P10.28 (a) By similar triangles, y h hx . The area of the front face = , y= L x L 1 1 is hL. The volume of the plate is hLw . Its density is 2 2 2M M M . The mass of the ribbon is = = 1 = V hLw 2 hLw dm = dV = ywdx = The moment of inertia is I= 2 Mywdx 2 Mhx 2 Mxdx . = dx = hLw hLL L2
y x L FIG. P10.28
h
all mass
z
r 2 dm =
x=0
z
L
x2
2 Mxdx 2 M 3 2 M L4 ML2 = 2 x dx = 2 = . 2 2 L L 0 L 4
z
L
(b)
From the parallel axis theorem I = I CM + M I h = I CM + M
FG L IJ H 3K
2
FG 2L IJ H3K
2
= I CM +
4ML2 and 9
= I CM +
inertia I CM +
4ML2 + I CM 9
ML2 . The two triangles constitute a rectangle with moment of 9 1 ML2 1 + = 2 M L2 . Then 2 I CM = ML2 9 3 9
a f
I = I CM +
4ML2 1 8 1 = ML2 + ML2 = ML2 . 9 18 18 2
Chapter 10
297
*P10.29
We consider the cam as the superposition of the original solid disk and a disk of negative mass cut from it. With half the radius, the cutaway part has onequarter the face area and onequarter the volume and onequarter the mass M 0 of the original solid cylinder: M0  1 M0 = M 4 M0 = 4 M. 3
By the parallelaxis theorem, the original cylinder had moment of inertia
FG R IJ = 1 M R + M R = 3 M R . H 2K 2 4 4 1F 1 I F R I M R . The whole cam has The negativemass portion has I = G  M J G J =  H 4 KH 2 K 2 32
I CM + M 0
2 0 2 2 0 0 2 2 0 0 2
I=
M R 2 23 3 23 4 23 1 1 23 23 M0 R 2  0 = M0 R 2 = MR 2 = MR 2 and K = I 2 = MR 2 2 = MR 2 2 . 4 32 32 32 3 24 2 2 24 48
Section 10.6 P10.30
Torque
Resolve the 100 N force into components perpendicular to and parallel to the rod, as Fpar = 100 N cos 57.0 = 54.5 N and Fperp = 100 N sin 57.0 = 83.9 N
a
f
a
f
The torque of Fpar is zero since its line of action passes through the pivot point. The torque of Fperp is = 83.9 N 2.00 m = 168 N m (clockwise) P10.31
a
f
FIG. P10.30
= 0.100 ma12.0 N f  0.250 ma9.00 N f  0.250 ma10.0 N f =
The thirtydegree angle is unnecessary information.
3.55 N m
FIG. P10.31 P10.32 The normal force exerted by the ground on each wheel is 1 500 kg 9.80 m s 2 mg n= = = 3 680 N 4 4 The torque of friction can be as large as max = f max r = sn r = 0.800 3 680 N 0.300 m = 882 N m
b
ge
j
b g a
fb
ga
f
The torque of the axle on the wheel can be equally as large as the light wheel starts to turn without slipping.
298 P10.33
Rotation of a Rigid Object About a Fixed Axis
In the previous problem we calculated the maximum torque that can be applied without skidding to be 882 N m. This same torque is to be applied by the frictional force, f, between the brake pad and the rotor for this wheel. Since the wheel is slipping against the brake pad, we use the coefficient of kinetic friction to calculate the normal force.
= fr = k n r , so n =
b g
882 N m = = 8.02 10 3 N = 8.02 kN 0.500 0.220 m kr
a
fa
f
Section 10.7 P10.34 (a)
Relationship Between Torque and Angular Acceleration 1 1 MR 2 = 2.00 kg 7.00 10 2 m 2 2 0.600 = = = 122 rad s 2 I 4.90 10 3 = t 2 1 200 60 t = = = 1.03 s 122 I=
b
ge
j
2
= 4.90 10 3 kg m 2
c h
(b) P10.35
=
1 2 1 t = 122 rad s 1.03 s 2 2
b
ga
f
2
= 64.7 rad = 10.3 rev
m = 0.750 kg , F = 0.800 N (a) (b) (c)
= rF = 30.0 m 0.800 N = 24.0 N m =
at
a
f
a f = r = 0.035 6a30.0 f = 1.07 m s
2
24.0 rF = = I mr 2 0.750 30.0
= 0.035 6 rad s 2
2
FIG. P10.35
P10.36
f = i + t :
10.0 rad s = 0 + 6.00 s 10.00 rad s 2 = 1.67 rad s 2 = 6.00
I=
a
f
(a) (b)
= 36.0 N m = I :
f = i + t :
=
36.0 N m = 21.6 kg m 2 2 1.67 rad s
0 = 10.0 + 60.0
a f
je j
1 2 t 2
= 0.167 rad s 2 = I = 21.6 kg m 2 0.167 rad s 2 = 3.60 N m
(c) Number of revolutions f = i + i t + During first 6.00 s During next 60.0 s
e
1 2 1.67 6.00 = 30.1 rad 2 1 2 f = 10.0 60.0  0.167 60.0 = 299 rad 2 1 rev total = 329 rad = 52.4 rev 2 rad
f =
a fa f a f a fa f FG IJ H K
Chapter 10
299
P10.37
For m1 , Fy = ma y :
+n  m 1 g = 0 n1 = m1 g = 19.6 N f k 1 = k n1 = 7.06 N
Fx = ma x : = I :
7.06 N + T1 = 2.00 kg a
b
g
(1)
For the pulley,
T1 R + T2 R = T1 + T2 T1 + T2
FG IJ H K 1 = b10.0 kg ga 2 = b5.00 kg ga
1 a MR 2 2 R
(2)
For m 2 ,
+n 2  m 2 g cos = 0
n 2 = 6.00 kg 9.80 m s 2 cos 30.0 = 50.9 N
e
ja
f
FIG. P10.37
fk 2 = kn 2 = 18.3 N : 18.3 N  T2 + m 2 sin = m 2 a 18.3 N  T2 + 29.4 N = 6.00 kg a (3)
b
g
(a)
Add equations (1), (2), and (3): 7.06 N  18.3 N + 29.4 N = 13.0 kg a a= 4.01 N = 0.309 m s 2 13.0 kg
b
g
(b)
T1 = 2.00 kg 0.309 m s 2 + 7.06 N = 7.67 N T2 = 7.67 N + 5.00 kg 0.309 m s 2 = 9.22 N
e
e
j
j
P10.38
1 1 mR 2 = 100 kg 0.500 m 2 2 i = 50.0 rev min = 5.24 rad s I=
b
ga
f
2
= 12.5 kg m 2
=
f i
t
=
0  5.24 rad s = 0.873 rad s 2 6.00 s
= I = 12.5 kg m 2 0.873 rad s 2 = 10.9 N m
The magnitude of the torque is given by fR = 10.9 N m, where f is the force of friction. Therefore, yields f= 10.9 N m 0.500 m f 21.8 N = = 0.312 n 70.0 N and f = kn FIG. P10.38
e
j
k =
300 *P10.39
Rotation of a Rigid Object About a Fixed Axis
= I = 2 MR 2
1
135 N 0.230 m + T 0.230 m = T = 21.5 N
a
f a
25 f 1 b80 kggFGH 1.2 mIJK e1.67 rad s j 2
2 2
Section 10.8 P10.40
Work, Power, and Energy in Rotational Motion 1 ML2 . The total rotational 3
The moment of inertia of a thin rod about an axis through one end is I = kinetic energy is given as KR = with and In addition, while Therefore, Ih = m h L2 60.0 kg 2.70 m h = 3 3 1 1 2 I h 2 + I m m h 2 2
a
f
2
= 146 kg m 2
100 kg 4.50 m m L2 = 675 kg m 2 Im = m m = 3 3 2 rad 1h h = = 1.45 10 4 rad s 12 h 3 600 s
m
KR
a f F I GH JK 2 rad F 1 h I = 1 h G 3 600 s J H K = 1.75 10 rad s 1 1 = a146 fe1.45 10 j + a675 fe1.75 10 j 2 2
2 3 4 2
3 2
= 1.04 10 3 J
*P10.41
The power output of the bus is P =
1 11 E where E = I 2 = MR 2 2 is the stored energy and 2 22 t 1 Px x is the time it can roll. Then MR 2 2 = Pt = and t = 4 v v
2 MR 2 2 v 1 600 kg 0.65 m 4 000 60 s 11.1 m s x = = = 24.5 km . 4P 4 18 746 W
a
a
fc
2
f
h
2
P10.42
Work done = Fr = 5.57 N 0.800 m = 4.46 J 1 1 and Work = K = I 2  I i2 f 2 2 (The last term is zero because the top starts from rest.) Thus, 4.46 J = 1 4.00 10 4 kg m 2 2 f 2
F
a
fa
f
A
e
j
A
and from this, f = 149 rad s . FIG. P10.42
Chapter 10
301
*P10.43
(a)
g a f a f = 2.28 10 kg m j b b g e j . 1 b0.850 kg gb0.82 m sg + 1 b0.42 kggb0.82 m sg + 1 e2.28 10 kg m jFGH 0082 m s IJK 2 2 2 .03 m +0.42 kg e9.8 m s ja0.7 mf  0.25b0.85 kg ge9.8 m s ja0.7 mf F v IJ 1 1 1 = b0.85 kg gv + b0.42 kg gv + e 2.28 10 kg m jG 2 2 2 H 0.03 m K 0.512 J + 2.88 J  1.46 J = b0.761 kg gv
I= 1 1 2 2 2 M R1 + R 2 = 0.35 kg 0.02 m + 0.03 m 2 2 K 1 + K 2 + K rot + U g 2  f k x = K 1 + K 2 + K rot f
e
2
4
2
i
2
2
4
2
2
2
2
2 f
2 f
4
2
f
2
2 f
vf =
1.94 J = 1.59 m s 0.761 kg
(b) P10.44
=
v 1.59 m s = = 53.1 rad s 0.03 m r
We assume the rod is thin. For the compound object 1 2 M rod L2 + m ball R 2 + M ball D 2 3 5 1 2 2 I = 1.20 kg 0.240 m + 2.00 kg 4.00 10 2 m 3 5 I = 0.181 kg m 2 I=
a
LM N
f
OP Q
e
j
2
+ 2.00 kg 0.280 m
a
f
2
(a)
K f + U f = K i + U i + E 1 2 L + M ball g L + R + 0 I + 0 = 0 + M rod g 2 2 1 0.181 kg m 2 2 = 1.20 kg 9.80 m s 2 0.120 m + 2.00 kg 9.80 m s 2 0.280 m 2 1 0.181 kg m 2 2 = 6.90 J 2
FG IJ H K
a
e e
j j
e
f ja
f
e
ja
f
(b) (c) (d)
= 8.73 rad s
v = r = 0.280 m 8.73 rad s = 2.44 m s v 2 = vi2 + 2 a y f  yi f
a
f
d
i
v f = 0 + 2 9.80 m s 2 0.280 m = 2.34 m s The speed it attains in swinging is greater by 2.44 = 1.043 2 times 2.34
e
ja
f
302 P10.45
Rotation of a Rigid Object About a Fixed Axis
(a)
For the counterweight,
Fy = ma y becomes:
For the reel where
50.0  T =
FG 50.0 IJ a H 9.80 K
a R
= I
reads TR = I = I I=
1 MR 2 = 0.093 8 kg m 2 2
We substitute to eliminate the acceleration: 50.0  T = 5.10 T = 11.4 N a= v 2 = vi2 + 2 a x f  xi : f (b)
F TR I GH I JK
2
and
d
i
50.0  11.4 = 7.57 m s 2 5.10
FIG. P10.45
v f = 2 7.57 6.00 = 9.53 m s
a f
Use conservation of energy for the system of the object, the reel, and the Earth:
aK + U f = aK + U f :
i f
mgh =
1 1 mv 2 + I 2 2 2 v2 I 2mgh = mv 2 + I 2 = v 2 m + 2 R R 2mgh m + I2
R
v=
F I FG IJ GH JK H K 2a50.0 N fa6.00 mf = =
5.10 kg +
a0. 250 f
0.093 8
9.53 m s
2
P10.46
Choose the zero gravitational potential energy at the level where the masses pass. K f + U gf = K i + U gi + E 1 1 1 m1 v 2 + m 2 v 2 + I 2 = 0 + m1 gh1i + m 2 gh2 i + 0 2 2 2 1 1 1 v 15.0 + 10.0 v 2 + 3.00 R 2 = 15.0 9.80 1.50 + 10.0 9.80 1.50 2 2 2 R 1 26.5 kg v 2 = 73.5 J v = 2.36 m s 2
a b
f
g
LM a f OPFG IJ N QH K
2
a fa f
a fa
f
P10.47
From conservation of energy for the objectturntablecylinderEarth system, 1 v I 2 r I
FG IJ H K
2
+
1 mv 2 = mgh 2
v2 = 2mgh  mv 2 r2 2 gh I = mr 2 1 v2
FG H
IJ K
FIG. P10.47
Chapter 10
303
P10.48
The moment of inertia of the cylinder is I= 1 1 mr 2 = 81.6 kg 1.50 m 2 2
b
ga
f
2
= 91.8 kg m 2
and the angular acceleration of the merrygoround is found as
=
50.0 N 1.50 m Fr = = = 0.817 rad s 2 . 2 I I 91.8 kg m
a f a
e
fa
j
f
At t = 3.00 s, we find the angular velocity
= i + t
= 0 + 0.817 rad s 2 3.00 s = 2.45 rad s
and K = P10.49 (a) 1 2 1 I = 91.8 kg m 2 2. 45 rad s 2 2
e
ja
f
e
jb
g
2
= 276 J .
Find the velocity of the CM
i
aK + U f = aK + U f
f
Pivot
R g
1 0 + mgR = I 2 2 2mgR 2mgR = 3 = 2 I 2 mR vCM = R Rg 3 Rg 4g Rg = 2 3R 3 FIG. P10.49
(b)
v L = 2 vCM = 4 2mgR = 2m
(c) *P10.50 (a)
vCM =
The moment of inertia of the cord on the spool is 1 1 2 2 M R1 + R 2 = 0.1 kg 0.015 m 2 2
e
j
ea
f + a0.09 mf j = 4.16 10
2 2
4
kg m 2 .
The protruding strand has mass 10 2 kg m 0.16 m = 1.6 10 3 kg and I = I CM + Md 2 = 1 1 0.16 m ML2 + Md 2 = 1.6 10 3 kg 12 12
e
j
FG a H
FG H
f + a0.09 m + 0.08 mf IJK
2 2 2
= 4.97 10 5 kg m 2 For the whole cord, I = 4.66 10 4 kg m 2 . In speeding up, the average power is
P=
E = t
1 2
I 2 t
=
4.66 10 4 kg m 2 2 500 2 2 0.215 s 60 s
a
f
IJ K
= 74.3 W
(b)
P = = 7.65 N 0.16 m + 0.09 m
a
fa
fFGH 2 000 s2 IJK = 60
401 W
304
Rotation of a Rigid Object About a Fixed Axis
Section 10.9 P10.51 (a) (b) (c) P10.52
Rolling Motion of a Rigid Object K trans = K rot = 1 1 mv 2 = 10.0 kg 10.0 m s 2 2
b
gb
g
2
= 500 J
2
1 2 1 1 I = mr 2 2 2 2
FG H
IJ FG v IJ = 1 b10.0 kg gb10.0 m sg KH r K 4
2 2 trans
= 250 J
K total = K trans + K rot = 750 J
W = K f  K i = K trans + K rot W= or
b
g  bK
f
+ K rot
1 1 1 1 2 Mv 2 + I 2  0  0 = Mv 2 + MR 2 2 2 5 2 2 7 W= Mv 2 10
FG IJ H K
FG H
g
i
IJ FG v IJ KH RK
2
P10.53
(a)
= I mgR sin = I CM + mR 2
e
j
a=
mgR 2 sin I CM + mR
2
R f
a hoop =
mgR 2 sin 2mR 2 mgR 2 sin
1 = g sin 2
n
2 = a disk = 3 g sin 3 mR 2 2 4 The disk moves with the acceleration of the hoop. 3 (b) Rf = I f = n = mg cos
I f R = = = mg cos mg cos
mg
FIG. P10.53
c
2 3
g sin
2
he
1 2
mR 2
R mg cos where and
j=
1 tan 3 since no slipping.
P10.54
1 1 1 I mv 2 + I 2 = m + 2 v2 2 2 2 R Also, Ui = mgh , U f = 0 , K= Therefore, Thus, For a disk, So For a ring, Since v disk v2 = 2 gh 1+ 1 2
LM N
OP Q
=
v R vi = 0
1 I m + 2 v 2 = mgh 2 R 2 gh v2 = 1+ I 2
LM N
OP Q
e j
mR
1 I = mR 2 2 or v disk = 4 gh 3
2 gh or v ring = gh 2 > v ring , the disk reaches the bottom first. I = mR 2 so v 2 =
Chapter 10
305
P10.55
1 x 3.00 m = = 2.00 m s = 0 + v f 1.50 s 2 t vf 4.00 m s 8.00 rad s v f = 4.00 m s and f = = = 2 r 6.38 10 2 6.38 10 m 2 v=
d
i
e
j
We ignore internal friction and suppose the can rolls without slipping.
e K + K + U j + E b0 + 0 + mgy g + 0 = FGH 1 mv 2
trans rot g i i
mech 2 f
= K trans + K rot + U g 1 2 I f + 0 2
e
+
IJ K
j
f
0.215 kg 9.80 m s 2
e j a3.00 mf sin 25.0 = 1 b0.215 kg gb4.00 m sg 2 2.67 J = 1.72 J + e7 860 s jt
2
2
+
1 8.00 rad s I 2 6.38 10 2
FG H
IJ K
2
I= P10.56 (a)
0.951 kg m 2 s 2 7 860 s 2
= 1.21 10 4 kg m 2
The height of the can is unnecessary data.
Energy conservation for the system of the ball and the Earth between the horizontal section and top of loop: 1 1 1 1 2 2 2 mv 2 + I 2 + mgy 2 = mv1 + I 1 2 2 2 2 2 1 1 2 2 mv 2 + mr 2 2 2 3 1 2 mv1 2 5 2 5 2 v 2 + gy 2 = v1 6 6 =
2 v 2 = v1 
FG H
IJ FG v IJ + mgy KH r K 1F2 IF v I + G mr J G J KH r K 2H3
2 2 2 2 1
2
FIG. P10.56
6 gy 2 = 5
b4.03 m sg
2

v2 2 The centripetal acceleration is 2 >g 0.450 m r Thus, the ball must be in contact with the track, with the track pushing downward on it. (b) 1 1 2 2 mv 3 + mr 2 2 2 3
2 v 3 = v1 
f e ja b2.38 m sg = 12.6 m s =
2 2 1 2 1 2
6 9.80 m s 2 0.900 m = 2.38 m s 5
FG H
6 gy 3 5
IJ FG v IJ + mgy = 1 mv + 1 FG 2 mr IJ FG v IJ KH r K KH r K 2 2H3 6 = b 4.03 m sg  e9.80 m s ja 0.200 mf = 5
3 2 3 2 2
4.31 m s
(c)
1 1 2 2 mv 2 + mgy 2 = mv1 2 2
2 v 2 = v1  2 gy 2 =
b4.03 m sg  2e9.80 m s ja0.900 mf =
2 2
1.40 m 2 s 2
This result is imaginary. In the case where the ball does not roll, the ball starts with less energy than in part (a) and never makes it to the top of the loop.
306
Rotation of a Rigid Object About a Fixed Axis
Additional Problems P10.57 2 3 g sin = 2 3 g sin r at = 2 3 g Then r > g sin 2 2 for r > 3 1 About the length of the chimney will have a 3 tangential acceleration greater than g sin . mg sin = 1 m 2 3
FG H
FG IJ H K
IJ K
t
at
t
g sin
g
FIG. P10.57
P10.58
The resistive force on each ball is R = DAv 2 . Here v = r , where r is the radius of each ball's path. The resistive torque on each ball is = rR , so the total resistive torque on the three ball system is total = 3rR . The power required to maintain a constant rotation rate is P = total = 3rR . This required power may be written as
With
or (a)
a f = e3r DA j 2 rad F 10 rev I F 1 min I 1 000 = 1 rev G 1 min J H 60.0 s K K G J = 30.0 rad s H F 1 000 IJ P = 3a0.100 mf a0.600 fe 4.00 10 m jG H 30.0 s K P = e0.827 m s j , where is the density of the resisting medium.
P = total = 3r DA r
3 2 3 3 3 4 2 3 5 3
In air, = 1.20 kg m3 , and P = 0.827 m 5 s 3 1. 20 kg m 3 = 0.992 N m s = 0.992 W
e
j
(b) P10.59 (a)
In water, = 1 000 kg m3 and P = 827 W .
(b) (c)
e j F 1IF I = G JG Jb f b ga g H 2KH K  0 r b8.00 rad sga0.500 mf = = = 1.60 s t=
W = K =
f
1 2 1 2 1 1 I f  I i = I 2  i2 where I = mR 2 f 2 2 2 2 1 2 2 1.00 kg 0.500 m 8.00 rad s  0 = 4.00 J 2 2.50 m s 2
a
1 f = i + i t + t 2 ; i = 0 ; i = 0 2 1 2 1 2.50 m s 2 2 1.60 s = 6.40 rad f = t = 2 2 0.500 m
Ia JK s = r = a0.500 mfa6. 40 radf =
F GH
f
3.20 m < 4.00 m Yes
Chapter 10
307
*P10.60
The quantity of tape is constant. Then the area of the rings you see it fill is constant. This is expressed by
Start rt rs v
rt2  rs2 = r 2  rs2 + r22  rs2 or r2 = rt2 + rs2  r 2 is the outer radius of spool 2.
(a) v Where the tape comes off spool 1, 1 = . Where the r 1 2 v 2 2 . tape joins spool 2, 2 = = v rs + rt  r 2 r2
e
j
Later r r2
(b)
At the start, r = rt and r2 = rs so 1 =
v v and 2 = . The rt rs takeup reel must spin at maximum speed. At the end, v v and 1 = . The angular r = rs and r2 = rt so 2 = rt rs speeds are just reversed. Since only conservative forces act within the system of the rod and the Earth, E = 0 so K f + U f = K i + Ui 1 2 L I + 0 = 0 + Mg 2 2
v FIG. P10.60
P10.61
(a)
where I = Therefore, (b)
1 ML2 3
FG IJ H K
FIG. P10.61
=
3g L
= I , so that in the horizontal orientation,
Mg
FG L IJ = ML H 2K 3
3g 2L
2
=
(c) (d)
a x = a r =  r 2 = 
FG L IJ H 2K
2
= 
3g 2
a y =  a t =  r = 
FG L IJ = H 2K

3g 4
Using Newton's second law, we have R x = Ma x =  3 Mg 2 3 Mg 4 Ry = Mg 4
R y  Mg = Ma y = 
308 P10.62
Rotation of a Rigid Object About a Fixed Axis
= 10.0 rad s 2  5.00 rad s 3 t =
e
j
d = = 65.0 rad s  10.0 rad s 2 t  2.50 rad s 3 t 2 dt
65.0
z z
d =
t 0
d dt
10.0  5.00t dt = 10.0t  2.50t 2 =  65.0 rad s
e
j e
j
(a)
At t = 3.00 s,
= 65.0 rad s  10.0 rad s 2 3.00 s  2.50 rad s 3 9.00 s 2 = 12.5 rad s
e
ja
f e
je
j
(b)
z z
t 0
d = dt =
= 65.0 rad s t  5.00 rad s 2 t 2  0.833 rad s 3 t 3
At t = 3.00 s,
b
0
z
t 0
65.0 rad s  10.0 rad s 2 t  2.50 rad s 3 t 2 dt
e
j e
j
g e
j e
j
= 65.0 rad s 3.00 s  5.00 rad s 2 9.00 s 2  0.833 rad s 3 27.0 s 3 = 128 rad
P10.63 The first drop has a velocity leaving the wheel given by 1 mvi2 = mgh1 , so 2
b
ga
f e
j
e
j
v1 = 2 gh1 = 2 9.80 m s 2 0.540 m = 3.25 m s The second drop has a velocity given by v 2 = 2 gh2 = 2 9.80 m s 2 0.510 m = 3.16 m s From = v , we find r
e
ja
f
e
ja
f
1 =
or
3.16 m s v1 3.25 m s v = = 8.53 rad s and 2 = 2 = = 8.29 rad s 0.381 m 0.381 m r r
=
2 8.29 rad s  8.53 rad s 2 1 2 = 2 4
b
g b
2
g
2
= 0.322 rad s 2
Chapter 10
309
P10.64
At the instant it comes off the wheel, the first drop has a velocity v1 , directed upward. The magnitude of this velocity is found from K i + U gi = K f + U gf 1 2 mv1 + 0 = 0 + mgh1 or v1 = 2 gh1 2 and the angular velocity of the wheel at the instant the first drop leaves is
1 =
v1 = R
2 gh1 R2
. 2 gh2 R2
Similarly for the second drop: v 2 = 2 gh2 and 2 = The angular acceleration of the wheel is then
2 2 1 = a= 2 2 2 gh2 R2
v2 = R
.
2 2
a f

2 gh1 R2
=
g h2  h1 2R
2
b
g
.
P10.65
1 1 1 1 Mv 2 + I 2 : U f = Mgh f = 0 ; K i = Mvi2 + I i2 = 0 f f 2 2 2 2 1 v Ui = Mgh i : f = N = Mg cos ; = ; h = d sin and I = mr 2 2 r Kf =
b g
(a)
E = E f  Ei or  fd = K f + U f  K i  U i 1 1  fd = Mv 2 + I 2  Mgh f f 2 2 1 mr 2  Mg cos d = Mv 2 + 2 2
2
I  Mgd sin b g JK 2 1L m MNM + 2 OPQv = Mgd sin  bMg cos gd or 2 bsin  cos g v = 2 Mgd
v2 r2 2
F GH
vd (b)
L M = M4 gd bsin  cos gOP N am + 2 M f Q
FG H IJ b K
m 2
+M
12
2 v 2 = vi2 + 2 ax , v d = 2 ad f
a=
2 vd M sin  cos = 2g 2d m + 2M
g
310 P10.66
Rotation of a Rigid Object About a Fixed Axis
(a)
E=
1 2 MR 2 2 2 5
FG H
IJ e j K je
j FGH 862400 IJK
2 2
1 2 E = 5.98 10 24 6.37 10 6 2 5
e
= 2.57 10 29 J
(b)
dE d 1 2 MR 2 = dt dt 2 5
2 2
IJ FG 2 IJ OP K H T K PQ 1 dT = MR a 2 f e 2T j 5 dt 1 F 2 I F 2 I dT = MR G J G J H T K H T K dt 5 F 2 I F 10 10 = e 2.57 10 JjG H 86 400 s JK GH 3.16 10
2 3 2 2 29
LM FG MN H
6 7
s
I b86 400 s dayg J sK
dE = 1.63 10 17 J day dt *P10.67 (a)
f = i + t = f i
t =
2 Tf
 t
2 Ti
=
2 Ti  T f Ti T f t
2
d
i
7
j F 1 d I FG 1 yr IJ = ~ 1 d 1 d 100 yr G 86 400 s J H 3.156 10 s K H K
2 10 3 s (b) I=
e
10 22 s 2
The Earth, assumed uniform, has moment of inertia 2 2 MR 2 = 5.98 10 24 kg 6.37 10 6 m 5 5
e
= I ~ 9.71 10 37
je j = 9.71 10 kg m e2.67 10 s j = 10
2 2 22 2
37
kg m 2
16
Nm
The negative sign indicates clockwise, to slow the planet's counterclockwise rotation. (c)
= Fd . Suppose the person can exert a 900N force.
d=
2.59 10 16 N m ~ 10 13 m = 900 N F
This is the order of magnitude of the size of the planetary system.
Chapter 10
311
P10.68
= t t= =
0.800 m = 139 m s v= 0.005 74 s
c h rev = 0.005 74 s
31.0 360 900 rev 60 s
= 31 v
d
FIG. P10.68 P10.69
f will oppose the torque due to the hanging object:
= I = TR  f : Fy = T  mg =  ma :
also y = vi t + and at 2 2
f = TR  I
(1)
Now find T, I and in given or known terms and substitute into equation (1). T=m ga a= 2y t2 2y a = 2: R Rt
b
g
(2) (3) (4) FIG. P10.69
=
I=
1 R M R2 + 2 2
LM NM
FG IJ OP = 5 MR H K QP 8
2 2 2
2
(5)
Substituting (2), (3), (4), and (5) into (1), we find P10.70 (a) W = K + U W = K f  K i + U f  Ui 0=
f =m g
FG H
2y t2
IJ R  5 MR b2yg = RLMmFG g  2 y IJ  5 My OP K 8 Rt N H t K 4t Q
2 2
1 1 1 mv 2 + I 2  mgd sin  kd 2 2 2 2 1 2 1 2 I + mR 2 = mgd sin + kd 2 2
e
j
=
2mgd sin + kd 2 I + mR 2 FIG. P10.70 2 0.500 kg 9.80 m s 2 0.200 m sin 37.0 + 50.0 N m 0.200 m 1.00 kg m 2 + 0.500 kg 0.300 m 1.18 + 2.00 = 3.04 = 1.74 rad s 1.05
(b)
= =
b
ge
ja
fa
a
f
f
a
f
2
2
312 P10.71
Rotation of a Rigid Object About a Fixed Axis
(a)
m 2 g  T2 = m 2 a
T2 = m 2 g  a = 20.0 kg 9.80 m s 2  2.00 m s 2 = 156 N T1  m1 g sin 37.0 = m1 a T1 = 15.0 kg 9.80 sin 37.0+2.00 m s 2 = 118 N (b)
b
g
e
j
b
ga
f
bT
FG a IJ H RK bT  T gR = a156 N  118 Nfa0.250 mf I=
2
 T1 R = I = I
2 1 2
g
2
a
2.00 m s 2
= 1.17 kg m 2
FIG. P10.71
P10.72
For the board just starting to move,
= I :
FG IJ cos = FG 1 m IJ H 2K H3 K 3 F gI = G J cos 2H K
mg
2
3 g cos 2 3 The vertical component is a y = a t cos = g cos 2 2 If this is greater than g, the board will pull ahead of the ball falling: The tangential acceleration of the end is at = = (a) (b) 3 2 g cos 2 g gives cos 2 so 2 3 cos 2 3 and
FIG. P10.72
35.3
rc = cos
When = 35.3 , the cup will land underneath the releasepoint of the ball if When = 1.00 m, and = 35.3 rc = 1.00 m 2 = 0.816 m 3
so the cup should be 1.00 m  0.816 m = 0.184 m from the moving end P10.73 At t = 0 , = 3.50 rad s = 0 e 0 . Thus, 0 = 3.50 rad s
a
f
At t = 9.30 s, = 2.00 rad s = 0 e  a 9.30 s f , yielding = 6.02 10 2 s 1 d d 0 e = dt dt At t = 3.00 s,
(a)
=
e
 t
j = a fe
0
t
= 3.50 rad s 6.02 10 2 s 1 e
(b)
b
ge
j
3.00 6 .02 10 2
e
j=
0.176 rad s 2
= 0 e t dt =
0
z
t
0  t e  1 = 0 1  e t 
At t = 2.50 s , 3.50 rad s  6 .02 10 2 ja 2.50 f 1e e = = 8.12 rad = 1.29 rev 2 6.02 10 1 s
e
LM j N e
OP Q
(c)
As t ,
3.50 rad s 0 1  e  = = 58.2 rad = 9.26 rev 6.02 10 2 s 1
j
Chapter 10
313
P10.74
Consider the total weight of each hand to act at the center of gravity (midpoint) of that hand. Then the total torque (taking CCW as positive) of these hands about the center of the clock is given by
= m h g
FG L IJ sin H 2K
h
h
 mm g
FG L IJ sin H2K
m
m
=
g m h L h sin h + m m Lm sin m 2
b
g
If we take t = 0 at 12 o'clock, then the angular positions of the hands at time t are
h = ht ,
where and where Therefore, or (a) (i)
h =
rad h 6
m = mt , m = 2 rad h
f FGH 6t IJK + 100 kga4.50 mf sin 2tOPQ L F t I O = 794 N mMsinG J + 2.78 sin 2t P , where t is in hours. H6K N Q
= 4.90 m s 2 60.0 kg 2.70 m sin
LM N
a
At 3:00, t = 3.00 h , so
= 794 N m sin
LM FG IJ + 2.78 sin 6 OP = N H 2K Q
794 N m
(ii)
At 5:15, t = 5 h +
15 h = 5.25 h , and substitution gives: 60
= 2 510 N m
(iii) (iv) (v) (b) At 6:00, At 8:20, At 9:45,
= 0 N m = 1 160 N m = 2 940 N m
The total torque is zero at those times when sin
FG t IJ + 2.78 sin 2t = 0 H6K
12:58:19 3:33:22 6:00:00 8:26:38 11:01:41 1:32:31 3:56:55 6:29:08 9:03:31 11:29:05 1:57:01 4:32:24 7:01:46 9:26:35
We proceed numerically, to find 0, 0.515 295 5, ..., corresponding to the times 12:00:00 2:33:25 4:58:14 7:27:36 10:02:59 12:30:55 2:56:29 5:30:52 8:03:05 10:27:29
314 *P10.75
Rotation of a Rigid Object About a Fixed Axis
(a)
As the bicycle frame moves forward at speed v, the center of each wheel moves forward at v the same speed and the wheels turn at angular speed = . The total kinetic energy of the R bicycle is K = K trans + K rot or K= 1 1 1 1 m frame + 2m wheel v 2 + 2 I wheel 2 = m frame + 2m wheel v 2 + m wheel R 2 2 2 2 2
b
g
FG H
IJ b K
g FGH
IJ FG v IJ . KH R K
2 2
This yields K= (b) 1 1 m frame + 3m wheel v 2 = 8.44 kg + 3 0.820 kg 2 2
b
g
b
g b3.35 m sg
2
= 61.2 J .
As the block moves forward with speed v, the top of each trunk moves forward at the same v speed and the center of each trunk moves forward at speed . The angular speed of each 2 v roller is = . As in part (a), we have one object undergoing pure translation and two 2R identical objects rolling without slipping. The total kinetic energy of the system of the stone and the trees is K = K trans + K rot or K= 1 1 v m stone v 2 + 2 m tree 2 2 2
FG IJ H K
2
+2
FG 1 I H2
tree
2
IJ = 1 FG m K 2H
stone
+
1 1 m tree v 2 + m tree R 2 2 2
IJ K
FG H
IJ FG v IJ . K H 4R K
2 2
This gives K= P10.76 1 3 1 m stone + m tree v 2 = 844 kg + 0.75 82.0 kg 2 4 2
FG H
IJ K
b
g b0.335 m sg
2
= 50.8 J .
Energy is conserved so U + K rot + K trans = 0 mg R  r cos  1 + Since r = v , this gives
a
fa
f LMN 1 mv 2 f
2
0 +
OP 1 LM 2 mr OP Q 2 N5 Q
2
2
=0
R
=
or
10 R  r 1  cos g 7 r2 10 Rg 1  cos 7r
2
a
fa
=
a
f
since R >> r .
FIG. P10.76
Chapter 10
315
P10.77
F = T  Mg =  Ma: = TR = I = 2 MR 2 G R J H K
(a) Combining the above two equations we find T =M ga and a= 2T M T= Mg 3 FIG. P10.77
1
F aI g
b
thus 2T 2 Mg 2 = = g 3 M M 3
(b)
a=
FG IJ H K
(c)
v 2 = vi2 + 2 a x f  xi f
d
i
v2 = 0 + 2 f vf =
FG 2 gIJ ah  0f H3 K
4 gh 3
For comparison, from conservation of energy for the system of the disk and the Earth we have U gi + K rot i + K trans i = U gf + K rot f + K trans f : Mgh + 0 + 0 = 0 + vf = P10.78 (a) 4 gh 3 1 1 MR 2 2 2
FG H
IJ FG v IJ KH R K
f
2
+
1 Mv f 2 2
Fx = F  f = Ma: = fR = I
Using I = 1 a 2F MR 2 and = , we find a = 2 3M R
(b)
When there is no slipping, f = Mg . Substituting this into the torque equation of part (a), we have
MgR =
1 F . MRa and = 2 3 Mg
316 P10.79
Rotation of a Rigid Object About a Fixed Axis
(a)
K rot + K trans + U = 0 Note that initially the center of mass of the sphere is a distance h + r above the bottom of the loop; and as the mass reaches the top of the loop, this distance above the reference level is 2R  r . The conservation of energy requirement gives 1 1 mg h + r = mg 2 R  r + mv 2 + I 2 2 2 For the sphere I = gh + 2 gr = 2 gR +
m
r
h
R
P
a f
a
f
FIG. P10.79
2 mr 2 and v = r so that the expression becomes 5 (1)
7 2 v 10
Note that h = hmin when the speed of the sphere at the top of the loop satisfies the condition
F = mg = aR  r f
mv 2
or v 2 = g R  r
a
f a f
Substituting this into Equation (1) gives
hmin = 2 R  r + 0.700 R  r or hmin = 2.70 R  r = 2.70 R
(b) When the sphere is initially at h = 3 R and finally at point P, the conservation of energy equation gives 1 1 mg 3 R + r = mgR + mv 2 + mv 2 , or 2 5 10 2 2R + r g v = 7
a
f
a
f
a
f
a
f
Turning clockwise as it rolls without slipping past point P, the sphere is slowing down with counterclockwise angular acceleration caused by the torque of an upward force f of static 2 friction. We have Fy = ma y and = I becoming f  mg =  m r and fr = mr 2 . 5
FG IJ H K
Eliminating f by substitution yields = mv 2
5g so that 7r 20mg 7
Fy =
5  mg 7
Fx = n =  R  r = 
c h( 2R + r ) mg =
10 7
Rr
(since R >> r )
Chapter 10
317
P10.80
Consider the freebody diagram shown. The sum of torques about the chosen pivot is
pivot
Hy Hx
= I F
(a)
=
FG 1 ml IJ FG a IJ = FG 2 mlIJ a H 3 KH K H 3 K
2 CM l 2
CM
(1) CM mg l
= l = 1.24 m : In this case, Equation (1) becomes 3 14.7 N 3F = = 35.0 m s 2 aCM = 2m 2 0.630 kg
b
a
f
g
Fx = maCM F + H x = maCM or H x = maCM  F
Thus, H x = 0.630 kg 35.0 m s 2  14.7 N = +7.35 N or
F = 14.7 N FIG. P10.80
b
ge
j
H x = 7.35 i N .
(b) = 1 = 0.620 m : For this situation, Equation (1) yields 2 aCM = Again, 3 14.7 N 3F = = 17.5 m s 2 . 4m 4 0.630 kg
b
a
f
g
Fx = maCM H x = maCM  F , so
H x = 0.630 kg 17.5 m s 2  14.7 N = 3.68 N or H x = 3.68 i N .
b
ge
j
(c)
If H x = 0, then
Fx = maCM F = maCM , or aCM = m .
F
Thus, Equation (1) becomes F =
FG 2 mlIJ FG F IJ so H 3 KH mK
2 2 = l = 1.24 m = 0.827 m from the top . 3 3
a
f
b
g
P10.81
2 Let the ball have mass m and radius r. Then I = mr 2 . If the ball takes four seconds to go down 5 twentymeter alley, then v = 5 m s . The translational speed of the ball will decrease somewhat as the ball loses energy to sliding friction and some translational kinetic energy is converted to rotational kinetic energy; but its speed will always be on the order of 5.00 m s , including at the starting point. As the ball slides, the kinetic friction force exerts a torque on the ball to increase the angular speed. v When = , the ball has achieved pure rolling motion, and kinetic friction ceases. To determine the r elapsed time before pure rolling motion is achieved, consider:
= I b k mg gr = G 5 mr 2 J M H K
t= 2( 5.00 m s) 2.00 m s = 5 k g kg
F 2 I L b5.00 m sg r OP which gives MN t PQ
Note that the mass and radius of the ball have canceled. If k = 0.100 for the polished alley, the sliding distance will be given by x = vt = 5.00 m s
b
L 2 ms O gMM a0.100f.00.80 m s PP = 10.2 m or x ~ jQ N e9
2
10 1 m .
318 P10.82
Rotation of a Rigid Object About a Fixed Axis
Conservation of energy between apex and the point where the grape leaves the surface: mgy = 1 1 mv 2 + I 2 f f 2 2
i y = RR cos
1 1 2 mgR 1  cos = mv 2 + mR 2 f 2 2 5
2 7 vf which gives g 1  cos = 10 R
a
f
FG H
IJ FG v IJ KH R K
f
2
R
f
a
f
F I GH JK
mv 2 f R
n (1) mg cos FIG. P10.82 . mg sin
Consider the radial forces acting on the grape: mg cos  n =
At the point where the grape leaves the surface, n 0 . Thus, mg cos = mv 2 f R or v2 f R = g cos .
Substituting this into Equation (1) gives g  g cos = P10.83 (a) 10 7 and = 54.0 . g cos or cos = 17 10
There are not any horizontal forces acting on the rod, so the center of mass will not move horizontally. Rather, the center of mass drops straight downward (distance h/2) with the rod rotating about the center of mass as it falls. From conservation of energy: K f + U gf = K i + U gi
1 1 h 2 or MvCM + I 2 + 0 = 0 + Mg 2 2 2 1 1 1 2 MvCM + Mh 2 2 2 12 vCM = (b) 3 gh 4
FG H
IJ FG v KH
CM h 2
FG IJ H K I = MgFG h IJ which reduces to JK H 2K
2
In this case, the motion is a pure rotation about a fixed pivot point (the lower end of the rod) with the center of mass moving in a circular path of radius h/2. From conservation of energy: K f + U gf = K i + U gi
1 2 h or I + 0 = 0 + Mg 2 2 1 1 Mh 2 2 3 vCM =
FG H
IJ FG v IJ KH K
CM h 2
FG IJ H K
2
= Mg
FG h IJ which reduces to H 2K
3 gh 4
Chapter 10
319
P10.84
(a)
The mass of the roll decreases as it unrolls. We have m = the roll. Since E = 0 , we then have U g + K trans
2
Mr 2 where M is the initial mass of R2 mr 2 , + K rot = 0 . Thus, when I = 2
2
bmgr  MgRg + mv + LM mr 2 N2
Since r = v , this becomes v = 4g R3  r 3 3r
2
2 =0 2
OP Q
e
j
(b) (c)
Using the given data, we find v = 5.31 10 4 m s We have assumed that E = 0 . When the roll gets to the end, we will have an inelastic collision with the surface. The energy goes into internal energy . With the assumption we made, there are problems with this question. It would take an infinite time to unwrap the tissue since dr 0 . Also, as r approaches zero, the velocity of the center of mass approaches infinity, which is physically impossible.
P10.85
(a)
Fx = F + f = MaCM = FR  fR = I
FR  MaCM  F R =
F Mg IaCM R aCM = 1 F 3 4F 3M n f
b
g
(b) (c)
f = MaCM  F = M
FG 4F IJ  F = H 3M K i
FIG. P10.85
v 2 = vi2 + 2 a x f  xi f vf = 8 Fd 3M
d
320 P10.86
Rotation of a Rigid Object About a Fixed Axis
Call ft the frictional force exerted by each roller backward on the plank. Name as fb the rolling resistance exerted backward by the ground on each roller. Suppose the rollers are equally far from the ends of the plank. For the plank,
M m R m R
F
Fx = ma x
6.00 N  2 f t = 6.00 kg a p ap 2
b
g
FIG. P10.86
The center of each roller moves forward only half as far as the plank. Each roller has acceleration and angular acceleration ap 2 ap
a5.00 cmf a0.100 mf
Then for each,
=
So
b g a2 1 f a5.00 cmf + f a5.00 cmf = b 2.00 kg ga5.00 cmf = I 2 F1 I f + f = G kg J a H2 K
Fx = ma x
+ f t  f b = 2.00 kg
t b p t b p
2
ap 10.0 cm
Add to eliminate fb : 2 f t = 1.50 kg a p (a) And 6.00 N  1.50 kg a p = 6.00 kg a p ap = For each roller, a = (b) ap 2 = 0.400 m s 2
b
g
b
g b
g
a6.00 Nf = b7.50 kg g
0.800 m s 2
Substituting back, 2 f t = 1.50 kg 0.800 m s 2 ft = 0.600 N 0.600 N + f b = fb = 0.200 N 1 kg 0.800 m s 2 2
b
g
Mg 6.00 N
e
j
ft
nt nt ft mg
ft
nt nt ft mg
The negative sign means that the horizontal force of ground on each roller is 0.200 N forward rather than backward as we assumed.
fb
nb
fb
nb
FIG. P10.86(b)
Chapter 10
321
P10.87
Rolling is instantaneous rotation about the contact point P. The weight and normal force produce no torque about this point. Now F1 produces a clockwise torque about P and makes the spool roll forward. Counterclockwise torques result from F3 and F4 , making the spool roll to the left. The force F2 produces zero torque about point P and does not cause the spool to roll. If F2 were strong enough, it would cause the spool to slide to the right, but not roll. P FIG. P10.87 F4
F3 F2
c
F1
P10.88
The force applied at the critical angle exerts zero torque about the spool's contact point with the ground and so will not make the spool roll. From the right triangle shown in the sketch, observe that c = 90 = 90 90 = .
F2
b
g
R
r
r Thus, cos c = cos = . R
P
c
FIG. P10.88 P10.89 (a) Consider motion starting from rest over distance x along the incline:
bK
trans
+ K rot + U i + E = K trans + K rot + U 1 1 Mv 2 + 2 mR 2 2 2
g
b
g
0 + 0 + Mgx sin + 0 =
2 Mgx sin = M + 2m v 2 Since acceleration is constant, v 2 = vi2 + 2 ax = 0 + 2 ax , so
y x
a
f
FG H
IJ FG v IJ KH RK
f
2
+0
2 Mgx sin = M + 2m 2 ax
a=
a
f
a
Mg sin M + 2m
f
x
FIG. P10.88 continued on next page
322
Rotation of a Rigid Object About a Fixed Axis
(c)
Suppose the ball is fired from a cart at rest. It moves with acceleration g sin = a x down the incline and a y =  g cos perpendicular to the incline. For its range along the ramp, we have y  yi = v yi t  t= 2 v yi g cos 1 axt 2 2 1 g cos t 2 = 0  0 2
x  xi = v xi t + d=0+
2 4v yi 1 g sin 2 2 g cos 2
F GH
I JK
d= (b) In the same time the cart moves
2 2 v yi sin
g cos 2
x  xi = v xi t + dc = 0 + dc = So the ball overshoots the cart by x = d  d c = x = x =
2 2 v yi
1 2
F g sinM I FG 4v IJ GH aM + 2mf JK H g cos K
2 yi 2 2
1 axt 2 2
g M + 2m cos 2
a
2 2 v yi sin M
f
2 2 v yi sin
g cos 2
2 4v yi 2

2 2 v yi sin M
g cos 2 M + 2m
2 2 v yi
a
f
sin M +
2 4mv yi
sin m 
g cos M + 2m
a
f
sin M
aM + 2mfg cos
sin
2
Chapter 10
323
P10.90
Fx = ma x reads  f + T = ma . If we take torques around the center of mass, we can use = I , which reads + fR 2  TR1 = I . For rolling without
slipping, = a . By substitution, R2 f Ia I = T f R2 R 2 m
mg T n FIG. P10.90
fR 2  TR1 =
b
g
2 fR 2 m  TR1 R 2 m = IT  If 2 f I + mR 2 = T I + mR1 R 2
e
f=
F I + mR R I T GH I + mR JK
1 2 2 2
j b
g
Since the answer is positive, the friction force is confirmed to be to the left.
ANSWERS TO EVEN PROBLEMS
P10.2 P10.4 P10.6 P10.8 P10.10 P10.12 (a) 822 rad s 2 ; (b) 4.21 10 3 rad (a) 1.20 10 2 rad s ; (b) 25.0 s 226 rad s 2 13.7 rad s 2 (a) 2.88 s; (b) 12.8 s (a) 0.180 rad s; (b) 8.10 m s 2 toward the center of the track (a) 0.605 m s ; (b) 17.3 rad s ; (c) 5.82 m s ; (d) The crank length is unnecessary (a) 54.3 rev; (b) 12.1 rev s 0.572 (a) 92.0 kg m 2 ; 184 J ; (b) 6.00 m s ; 4.00 m s ; 8.00 m s ; 184 J see the solution 1.28 kg m
2
P10.28 P10.30 P10.32 P10.34 P10.36 P10.38 P10.40 P10.42 P10.44
1 ML2 2 168 N m clockwise 882 N m (a) 1.03 s; (b) 10.3 rev (a) 21.6 kg m 2 ; (b) 3.60 N m ; (c) 52.4 rev 0.312
1.04 10 3 J
149 rad s (a) 6.90 J; (b) 8.73 rad s ; (c) 2.44 m s ; (d) 1.043 2 times larger 2.36 m s 276 J (a) 74.3 W; (b) 401 W 7 Mv 2 10 The disk; 4 gh versus 3 gh
P10.14
P10.16 P10.18 P10.20
P10.46 P10.48 P10.50
P10.22 P10.24 P10.26
P10.52
~ 10 0 kg m 2
P10.54
324 P10.56
Rotation of a Rigid Object About a Fixed Axis
(a) 2.38 m s ; (b) 4.31 m s; (c) It will not reach the top of the loop. (a) 0.992 W; (b) 827 W see the solution (a) 12.5 rad s ; (b) 128 rad g h2  h1 2R
2
P10.76 P10.78 P10.80
10 Rg 1  cos 7r 2
a
f
P10.58 P10.60 P10.62 P10.64 P10.66 P10.68 P10.70 P10.72 P10.74
see the solution (a) 35.0 m s 2 ; 7.35 i N ; (b) 17.5 m s 2 ; 3.68 i N ; (c) At 0.827 m from the top.
b
g
P10.82
54.0 4g R3  r 3
2
(a) 2.57 10 29 J ; (b) 1.63 10 17 J day 139 m s (a) 2mgd sin + kd 2 I + mR 2 ; (b) 1.74 rad s
P10.84
(a)
e
3r (c) It becomes internal energy.
j ; (b) 5.31 10
4
m s;
P10.86
see the solution (a) 794 N m ; 2 510 N m; 0; 1 160 N m; 2 940 N m; (b) see the solution P10.88 P10.90
(a) 0.800 m s 2 ; 0.400 m s 2 ; (b) 0.600 N between each cylinder and the plank; 0.200 N forward on each cylinder by the ground see the solution see the solution; to the left
11
Angular Momentum
CHAPTER OUTLINE
11.1 11.2 11.3 11.4 11.5 11.6 The Vector Product and Torque Angular Momentum Angular Momentum of a Rotating Rigid Object Conservation of Angular Momentum The Motion of Gyroscopes and Tops Angular Momentum as a Fundamental Quantity
ANSWERS TO QUESTIONS
Q11.1 No to both questions. An axis of rotation must be defined to calculate the torque acting on an object. The moment arm of each force is measured from the axis. A B C is a scalar quantity, since B C is a vector. Since A B is a scalar, and the cross product between a scalar and a vector is not defined, A B C is undefined.
Q11.2
a
f
a
f
a f
Q11.3
(a) (b)
Downcrossleft is away from you:  j  i =  k Leftcrossdown is toward you:  i  j = k
e j
e j
FIG. Q11.3 Q11.4 Q11.5 The torque about the point of application of the force is zero. You cannot conclude anything about the magnitude of the angular momentum vector without first defining your axis of rotation. Its direction will be perpendicular to its velocity, but you cannot tell its direction in threedimensional space until an axis is specified. Yes. If the particles are moving in a straight line, then the angular momentum of the particles about any point on the path is zero. Its angular momentum about that axis is constant in time. You cannot conclude anything about the magnitude of the angular momentum. No. The angular momentum about any axis that does not lie along the instantaneous line of motion of the ball is nonzero.
Q11.6 Q11.7 Q11.8
325
326 Q11.9
Angular Momentum
There must be two rotors to balance the torques on the body of the helicopter. If it had only one rotor, the engine would cause the body of the helicopter to swing around rapidly with angular momentum opposite to the rotor. The angular momentum of the particle about the center of rotation is constant. The angular momentum about any point that does not lie along the axis through the center of rotation and perpendicular to the plane of motion of the particle is not constant in time. The long pole has a large moment of inertia about an axis along the rope. An unbalanced torque will then produce only a small angular acceleration of the performerpole system, to extend the time available for getting back in balance. To keep the center of mass above the rope, the performer can shift the pole left or right, instead of having to bend his body around. The pole sags down at the ends to lower the system center of gravity. The diver leaves the platform with some angular momentum about a horizontal axis through her center of mass. When she draws up her legs, her moment of inertia decreases and her angular speed increases for conservation of angular momentum. Straightening out again slows her rotation. Suppose we look at the motorcycle moving to the right. Its drive wheel is turning clockwise. The wheel speeds up when it leaves the ground. No outside torque about its center of mass acts on the airborne cycle, so its angular momentum is conserved. As the drive wheel's clockwise angular momentum increases, the frame of the cycle acquires counterclockwise angular momentum. The cycle's front end moves up and its back end moves down. The angular speed must increase. Since gravity does not exert a torque on the system, its angular momentum remains constant as the gas contracts. Mass moves away from axis of rotation, so moment of inertia increases, angular speed decreases, and period increases. The turntable will rotate counterclockwise. Since the angular momentum of the mouseturntable system is initially zero, as both are at rest, the turntable must rotate in the direction opposite to the motion of the mouse, for the angular momentum of the system to remain zero. Since the cat cannot apply an external torque to itself while falling, its angular momentum cannot change. Twisting in this manner changes the orientation of the cat to feetdown without changing the total angular momentum of the cat. Unfortunately, humans aren't flexible enough to accomplish this feat. The angular speed of the ball must increase. Since the angular momentum of the ball is constant, as the radius decreases, the angular speed must increase. Rotating the book about the axis that runs across the middle pages perpendicular to the bindingmost likely where you put the rubber bandis the one that has the intermediate moment of inertia and gives unstable rotation. The suitcase might contain a spinning gyroscope. If the gyroscope is spinning about an axis that is oriented horizontally passing through the bellhop, the force he applies to turn the corner results in a torque that could make the suitcase swing away. If the bellhop turns quickly enough, anything at all could be in the suitcase and need not be rotating. Since the suitcase is massive, it will want to follow an inertial path. This could be perceived as the suitcase swinging away by the bellhop.
Q11.10
Q11.11
Q11.12
Q11.13
Q11.14 Q11.15 Q11.16
Q11.17
Q11.18 Q11.19
Q11.20
Chapter 11
327
SOLUTIONS TO PROBLEMS
Section 11.1 The Vector Product and Torque i P11.1 MN= 6 j 2 k 1 = 7.00 i + 16.0 j  10.0k
2 1 3 P11.2 (a) (b) area = A B = AB sin = 42.0 cm 23.0 cm sin 65.015.0 = 740 cm 2
a
fa
f a
f
a f a f A + B = a50.3 cmfi + a31.7 cmf j length = A + B = a50.3 cmf + a31.7 cmf
2
A + B = 42.0 cm cos 15.0+ 23.0 cm cos 65.0 i + 42.0 cm sin 15.0+ 23.0 cm sin 65.0 j
2
a
f
a
f
= 59.5 cm
i P11.3 (a) 2 (b)
j k 3 0
A B = 3 4 0 = 17.0k
A B = A B sin 17 = 5 13 sin
P11.4
FG 17 IJ = 70.6 H 5 13 K A B = 3.00a6.00f + 7.00a 10.0f + a 4.00fa9.00 f = 124 AB = a3.00f + a7.00f + a 4.00f a6.00f + a 10.0 f + a9.00 f F A B IJ = cos a0.979f = 168 (a) cos G H AB K
= arcsin
2 2 2 2 2 1 1
2
= 127
i (b) A B = 3.00
j
k 9.00
2 2
7.00 4.00 = 23.0 i + 3.00 j  12.0k
6.00 10.0 AB = sin 1
2
a23.0f + a3.00f + a12.0f = 26.1 FG A B IJ = sin a0.206f = 11.9 or 168 H AB K
1
(c)
Only the first method gives the angle between the vectors unambiguously.
328 *P11.5
Angular Momentum
= r F = 0.450 m 0.785 N sin 9014 up east
= 0.343 N m north
a
f a
f
FIG. P11.5 P11.6 The crossproduct vector must be perpendicular to both of the factors, so its dot product with either factor must be zero: Does 2 i  3 j + 4k 4i + 3 j  k = 0 ? 8  9  4 = 5 0
e
je
j
No . The cross product could not work out that way.
P11.7
A B = A B AB sin = AB cos tan = 1 or
= 45.0
i P11.8 (a) j k
=rF= 1 3 0 = i 00  j 00 +k 29 = 3 2 0
a f a f a f a7.00 N mfk
(b)
The particle's position vector relative to the new axis is 1 i + 3 j  6 j = 1 i  3 j . i j k
= 1 3 0 = 11.0 N m k 3 2 0
P11.9 F3 = F1 + F2 The torque produced by F3 depends on the perpendicular distance OD, therefore translating the point of application of F3 to any other point along BC will not change the net torque .
A C F2 B
a
f
F3 D O
F1
FIG. P11.9
Chapter 11
329
*P11.10
i i = 1 1 sin 0 = 0 j j and k k are zero similarly since the vectors being multiplied are parallel. i j = 1 1 sin 90 = 1
j i k
i j=k jk = i ki= j FIG. P11.10
j i = k k j = i i k = j
Section 11.2 P11.11
Angular Momentum
y
L = mi vi ri
= 4.00 kg 5.00 m s 0.500 m + 3.00 kg 5.00 m s 0.500 m
b
gb
ga
f b
gb
ga
f
1.00 m 4.00 kg
3.00 kg
L = 17.5 kg m s , and
L = 17.5 kg m 2 s k
2
e
j
x
FIG. P11.11 P11.12
L=rp
ge e j b j L = e 8.10k  13.9k j kg m s = e 22.0 kg m sjk
L = 1.50 i + 2.20 j m 1.50 kg 4.20 i  3.60 j m s
2 2
P11.13
r = 6.00 i + 5.00t j m so
e
j e
v=
dr = 5.00 j m s dt
p = mv = 2.00 kg 5.00 j m s = 10.0 j kg m s i j 5.00t 10.0 k 0 = 0
j
and
L = r p = 6.00 0
e60.0 kg m sjk
2
330 P11.14
Angular Momentum
Fx = ma x Fy = ma y
So sin v = cos rg sin cos
2
T sin =
mv 2 r
T cos = mg
v = rg sin cos
l
L = rmv sin 90.0 L = rm rg FIG. P11.14
m
sin cos r = sin , so L = m 2 gr 3 L= m2 g
3
sin 4 cos vt . R
y v m
P11.15
The angular displacement of the particle around the circle is = t = The vector from the center of the circle to the mass is then R cos i + R sin j . The vector from point P to the mass is r = R i + R cos i + R sin j r = R 1 + cos The velocity is v= So dr vt vt =  v sin i + v cos j dt R R
R P
Q x
LMFG NH
FG vt IJ IJ i + sinFG vt IJ jOP H R KK H R K Q FG IJ H K FG IJ H K
FIG. P11.15
L = r mv
L = mvR 1 + cos t i + sin t j  sin t i + cos t j L=
f L F vt I O mvRk McosG J + 1P N H RK Q
a
P11.16
(a)
The net torque on the counterweightcordspool system is:
= r F = 8.00 10 2 m 4.00 kg 9.80 m s 2 = 3.14 N m .
(b) L = r mv + I dL = 0.400 kg m a dt L = Rmv + 1 v M MR 2 =R m+ v= 2 R 2
b
ge
j
FG IJ FG H K H
IJ K
b0.400 kg mgv
(c)
=
b
g
a=
3.14 N m = 7.85 m s 2 0.400 kg m
Chapter 11
331
P11.17
(a) (b)
zero
At the highest point of the trajectory, v 2 sin 2 1 x= R= i and 2 2g y = hmax = vi O
vi = vxi i
R FIG. P11.17
v2
b
vi sin 2g
g
2
L 1 = r1 mv 1
Lv =M MN
2 i
vi sin sin 2 i+ 2g 2g
b
=
 m vi sin
b
2g
g v cos k
2 i
g jOP mv PQ
2
xi i
(c)
L 2 = R i mv 2 , where R =
vi2 sin 2 g
= mR i vi cos i  vi sin j =  mRvi sin k = (d) P11.18
e
j
 mvi3 sin 2 sin k g
The downward force of gravity exerts a torque in the z direction.
Whether we think of the Earth's surface as curved or flat, we interpret the problem to mean that the plane's line of flight extended is precisely tangent to the mountain at its peak, and nearly parallel to the wheat field. Let the positive x direction be eastward, positive y be northward, and positive z be vertically upward. (a)
j p = mv = 12 000 kg e 175 i m sj = 2.10 10 i kg m s L = r p = e 4.30 10 k mj e 2.10 10 i kg m sj = e 9.03 10
r = 4.30 km k = 4.30 10 3 m k
6 3 6
a
f e
9
kg m 2 s j
j
(b)
No . L = r p sin = mv r sin , and r sin is the altitude of the plane. Therefore, L =
constant as the plane moves in level flight with constant velocity.
a
f
(c)
Zero . The position vector from Pike's Peak to the plane is antiparallel to the velocity of
the plane. That is, it is directed along the same line and opposite in direction. Thus, L = mvr sin180 = 0 .
332 P11.19
Angular Momentum
The vector from P to the falling ball is 1 r = ri + v i t + at 2 2 r=
m
e cos i +
sin j + 0 
j
FG 1 gt IJ j H2 K
2
l
The velocity of the ball is v = v i + at = 0  gt j So
P
L = r mv
L=m
LMe cos i + N
sin j + 0 
j
FG 1 gt IJ jOP e gtjj H2 K Q
2
FIG. P11.19
L =  m gt cos k P11.20 In the vertical section of the hose, the water has zero angular momentum about our origin (point O between the fireman's feet). As it leaves the nozzle, a parcel of mass m has angular momentum: L = r mv = mrv sin 90.0 = m 1.30 m 12.5 m s L = 16.3 m s m The torque on the hose is the rate of change in angular momentum. Thus,
vf
a
fb
e
2
j
g
1.30 m
O vi
=
dL dm = 16.3 m 2 s = 16.3 m 2 s 6.31 kg s = 103 N m dt dt
e
j
e
jb
g
FIG. P11.20
Section 11.3 *P11.21 P11.22 K=
Angular Momentum of a Rotating Rigid Object 1 2 1 I 2 2 L2 I = = 2 2 I 2I
The moment of inertia of the sphere about an axis through its center is I= 2 2 MR 2 = 15.0 kg 0.500 m 5 5
b
ga
f
2
= 1.50 kg m 2
Therefore, the magnitude of the angular momentum is L = I = 1.50 kg m 2 3.00 rad s = 4.50 kg m 2 s Since the sphere rotates counterclockwise about the vertical axis, the angular momentum vector is directed upward in the +z direction. Thus, L = 4.50 kg m 2 s k .
e
jb
g
e
j
Chapter 11
333
P11.23
(a)
(b)
FG 1 MR IJ = 1 b3.00 kg ga0.200 mf b6.00 rad sg = H2 K 2 L1 F RI O L = I = M MR + M G J P H 2 K PQ MN 2 3 = b3.00 kg ga0.200 mf b6.00 rad sg = 0.540 kg m s 4
L = I =
2 2 2 2 2 2
0.360 kg m 2 s
P11.24
The total angular momentum about the center point is given by L = I h h + I m m with and In addition, Ih = m h L2 60.0 kg 2.70 m h = 3 3
a
f
2
= 146 kg m 2
2
I m3 =
while Thus, or P11.25 (a) I=
F I GH JK 2 rad F 1 h I = 1 h G 3 600 s J H K = 1.75 10 rad s L = 146 kg m e1.45 10 rad sj + 675 kg m e1.75 10
h =
2 rad 1h = 1.45 10 4 rad s 12 h 3 600 s
3 m 2 4 2
100 kg 4.50 m m m L2 m = 3 3
a
f
= 675 kg m 2
3
rad s
j
L = 1.20 kg m 2 s 1 m1 L2 + m 2 0.500 12
a
f a
2
=
L = I = 0.108 3 4.00 = 0.433 kg m 2 s (b) I= 1 1 m1 L2 + m 2 R 2 = 0.100 1.00 3 3
a f
1 0.100 1.00 12
a
fa f
2
+ 0.400 0.500
a
f
2
= 0.108 3 kg m 2
fa f
2
+ 0.400 1.00
a f
2
= 0.433
L = I = 0.433 4.00 = 1.73 kg m 2 s *P11.26
a f
Fx = ma x :
+ fs = ma x 88 cm n
Fg fs
We must use the center of mass as the axis in
= I : Fy = ma y :
Fg 0  n 77.5 cm + fs 88 cm = 0
af a
f a
f
+n  Fg = 0
155 cm 2 FIG. P11.26
We combine the equations by substitution:  mg 77.5 cm + ma x 88 cm = 0
2
ax
a f a f e9.80 m s j77.5 cm = 8.63 m s =
88 cm
2
334 *P11.27
Angular Momentum
We require a c = g =
v2 = 2r r g = r
2
=
e9.80 m s j = 0.313 rad s
2
100 m
I = Mr = 5 10 4 kg 100 m (a)
a
f
2
= 5 10 8 kg m 2
L = I = 5 10 8 kg m 2 0.313 s = 1.57 10 8 kg m 2 s
I f i t
(c)
= I =
d
i
t = I f  I i = L f  Li
This is the angular impulseangular momentum theorem. (b) t = Lf 0 = 1.57 10 8 kg m 2 s = 6.26 10 3 s = 1.74 h 2 125 N 100 m
a
fa
f
Section 11.4 P11.28 (a)
Conservation of Angular Momentum From conservation of angular momentum for the system of two cylinders:
bI
(b)
1
+ I 2 f = I 1 i 1 I1 + I 2 2 f 2 Kf Ki =
1 2
g
or
f =
I1 i I1 + I 2 1 I 1 i2 2
Kf =
b
g
and
Ki =
so
bI
1
+ I2
2 1 2 I 1 i 2
g FG I IJ HI +I K
1 1 2 i
2
=
I1 which is less than 1 . I1 + I 2
2
P11.29
Ii i = I f f :
e250 kg m jb10.0 rev ming = 250 kg m
+ 25.0 kg 2.00 m
a
f
2
2
2 = 7.14 rev min
Chapter 11
335
P11.30
(a)
The total angular momentum of the system of the student, the stool, and the weights about the axis of rotation is given by I total = I weights + I student = 2 mr 2 + 3.00 kg m 2 Before: Thus, After: Thus,
e j
r = 1.00 m .
I i = 2 3.00 kg 1.00 m
b
ga
f
2
+ 3.00 kg m 2 = 9.00 kg m 2
r = 0.300 m I f = 2 3.00 kg 0.300 m
b
ga
f
2
+ 3.00 kg m 2 = 3.54 kg m 2
We now use conservation of angular momentum.
I f f = I i i
or
f =
F I I = FG 9.00 IJ b0.750 rad sg = GH I JK H 3.54 K
i f i
1.91 rad s
(b)
Ki = Kf =
1 1 I i i2 = 9.00 kg m 2 0.750 rad s 2 2
e
jb
g
2
= 2.53 J = 6.44 J
1 1 I f 2 = 3.54 kg m 2 1.91 rad s f 2 2
e
jb
g
2
P11.31
(a)
Let M = mass of rod and m = mass of each bead. From Ii i = I f f , we have
LM 1 M N 12
2
+ 2mr12 i =
OP Q
LM 1 M N 12
2
+ 2mr22 f
OP Q
When = 0.500 m , r1 = 0.100 m , r2 = 0.250 m , and with other values as stated in the problem, we find
f = 9.20 rad s .
(b) Since there is no external torque on the rod,
L = constant and is unchanged .
*P11.32 Let M represent the mass of all the ribs together and L the length of each. The original moment of 1 inertia is ML2 . The final effective length of each rib is L sin 22.5 and the final moment of inertia is 3 1 2 M L sin 22.5 angular momentum of the umbrella is conserved: 3
a
f
1 1 ML2 i = ML2 sin 2 22.5 f 3 3 1.25 rad s f = = 8.54 rad s sin 2 22.5
336 P11.33
Angular Momentum
(a)
The table turns opposite to the way the woman walks, so its angular momentum cancels that of the woman. From conservation of angular momentum for the system of the woman and the turntable, we have L f = Li = 0 so, and
L f = I woman woman + I table table = 0
table =  table
F I FG v IJ =  m I JK GH JK H r K 60.0 kg a 2.00 mfb1.50 m sg = = 0.360 rad s F GH
I woman m r2 woman =  woman I table I table 500 kg m 2
woman
woman rv woman
I table
or (b)
table = 0.360 rad s counterclockwise
a
f g
f
work done = K = K f  0 = W= 1 60 kg 1.50 m s 2
1 1 2 2 m woman v woman + I table 2 2 1 500 kg m 2 0.360 rad s 2
b
gb
g
2
+
e
jb
2
= 99.9 J
P11.34
When they touch, the center of mass is distant from the center of the larger puck by yCM = 0 + 80.0 g 4.00 cm + 6.00 cm = 4.00 cm 120 g + 80.0 g
a
(a) (b)
L = r1 m1 v1 + r2 m 2 v 2 = 0 + 6.00 10 2 m 80.0 10 3 kg 1.50 m s = 7.20 10 3 kg m 2 s The moment of inertia about the CM is
e
je
jb
g
FG 1 m r + m d IJ + FG 1 m r + m d IJ H2 K H2 K 1 I = b0.120 kg ge6.00 10 mj + b0.120 kg ge 4.00 10 j 2 1 + e80.0 10 kg je 4.00 10 mj + e80.0 10 kg je6.00 10 2
I=
2 1 1 2 1 1 2 2 2 2 2 2 2 2 2 2 3 2 2 3
2
m
j
2
I = 7.60 10 4 kg m 2 Angular momentum of the twopuck system is conserved: L = I
=
L 7.20 10 3 kg m 2 s = = 9. 47 rad s I 7.60 10 4 kg m 2
Chapter 11
337
P11.35
(a)
Li = mv
Lf = m + M vf vf
a f F m IJ v =G H m + MK a
ext = 0 , so L f
= Li = mv
l M v
(b)
1 mv 2 2 1 K f = M + m v2 f 2 m vf = v velocity of the bullet M+m and block Ki =
f
FG H
IJ K
FIG. P11.35 M M+m
Fraction of K lost = P11.36 For one of the crew,
1 2
mv 2  1 2
1 2
mv
m2 M +m 2
v2
=
Fr = mar :
We require Now,
n=
mv 2 = m i2 r r g r
n = mg , so i =
Ii i = I f f
5.00 10 8 kg m 2 + 150 65.0 kg 100 m
a
f
2
F 5.98 10 I GH 5.32 10 JK
8 8
g = 5.00 10 8 kg m 2 + 50 65.0 kg 100 m r
a
f
2
f
g g = f = 1.12 r r a r = 2 r = 1.26 g = 12.3 m s 2 f
Now, P11.37 (a)
Consider the system to consist of the wad of clay and the cylinder. No external forces acting on this system have a torque about the center of the cylinder. Thus, angular momentum of the system is conserved about the axis of the cylinder.
L f = Li :
or Thus,
I = mvi d
LM 1 MR N2
=
2
+ mR 2 = mvi d 2mvi d .
OP Q
FIG. P11.37
a M + 2 m fR
2
(b)
No . Some mechanical energy changes to internal energy in this perfectly inelastic collision.
338 *P11.38
Angular Momentum
(a)
Let be the angular speed of the signboard when it is vertical. 1 2 I = Mgh 2 1 1 1 ML2 2 = Mg L 1  cos 2 3 2
1 L 2
FG H
= =
3 g 1  cos L
a
IJ K
a
f
m Mg v FIG. P11.38
f
3 9.80 m s 2 1  cos 25.0 0.50 m
e
ja
f
= 2.35 rad s (b)
I f f = Ii i  mvL represents angular momentum conservation
FG 1 ML H3
1 3
2
+ mL2 f =
1 3
IJ K
1 ML2 i  mvL 3
f = =
1 3
b2.40 kg ga0.5 mfb2.347 rad sg  b0.4 kg gb1.6 m sg = b2.40 kgg + 0.4 kg a0.5 mf
1 3
c
ML i  mv M+m L
h
0. 498 rad s
(c)
Let hCM = distance of center of mass from the axis of rotation. hCM =
b2.40 kg ga0.25 mf + b0.4 kgga0.50 mf = 0.285 7 m .
2.40 kg + 0.4 kg
Apply conservation of mechanical energy:
aM + mfgh a1  cos f = 1 FGH 1 ML + mL IJK 2 3 L c M + mhL OP = cos M1  MN 2aM + mfgh PQ R b2.40 kg g + 0.4 kg a0.50 mf b0.498 rad sg  = cos S1   2b2.40 kg + 0.4 kg ge9.80 m s jb0.285 7 mg T
CM 2 2 2 1 1 3 2 2 CM 1 1 3 2 2
2
U  V  W
= 5.58
Chapter 11
339
P11.39
The meteor will slow the rotation of the Earth by the largest amount if its line of motion passes farthest from the Earth's axis. The meteor should be headed west and strike a point on the equator tangentially. Let the z axis coincide with the axis of the Earth with +z pointing northward. Then, conserving angular momentum about this axis, L f = Li I f = I i + mv r or Thus, 2 2 MR 2 f k = MR 2 i k  mvRk 5 5 mvR 5mv or = i  f = 2 2 2 MR 5 MR
i  f =
5 3.00 10 13 kg 30.0 10 3 m s 2 5.98 10
e
e
24
je j = 5.91 10 kg je6.37 10 mj
6
14
rad s
max ~ 10 13 rad s
Section 11.5 *P11.40
The Motion of Gyroscopes and Tops
Angular momentum of the system of the spacecraft and the gyroscope is conserved. The gyroscope and spacecraft turn in opposite directions. 0 = I 1 1 + I 2 2 :  I 1 1 = I 2
t
20 kg m 2 100 rad s = 5 10 5 kg m 2 t= 2.62 10 5 s = 131 s 2 000
b
g
FG 30 IJ FG rad IJ H t K H 180 K
*P11.41
I=
2 2 MR 2 = 5.98 10 24 kg 6.37 10 6 m 5 5
e
L = I = 9.71 10 37 kg m 2
= L p = 7.06 10 33
e
F 2 rad I = 7.06 10 kg m s GH 86 400 s JK F 2 rad I FG 1 yr IJ F 1 d I = kg m sjG H 2.58 10 yr JK H 365.25 d K GH 86 400 s JK
33 2 2 2 4
je
j
2
= 9.71 10 37 kg m 2
5.45 10 22 N m
Section 11.6 P11.42 (a)
Angular Momentum as a Fundamental Quantity L= h h = mvr so v = 2mr 2 v=
2 9.11 10 31 kg 0.529 10 10 m
e
6.626 1 10 34 J s
je
j
= 2.19 10 6 m s
(b) (c)
K=
1 1 mv 2 = 9.11 10 31 kg 2.19 10 6 m s 2 2
e
je
j
2
= 2.18 10 18 J = 4.13 10 16 rad s
=
L 1.055 10 34 J s = = I mr 2 9.11 10 31 kg 0.529 10 10 m
e
je
j
2
340
Angular Momentum
Additional Problems *P11.43 First, we define the following symbols: I P = moment of inertia due to mass of people on the equator I E = moment of inertia of the Earth alone (without people) = angular velocity of the Earth (due to rotation on its axis) 2 T= = rotational period of the Earth (length of the day) R = radius of the Earth The initial angular momentum of the system (before people start running) is Li = I P i + I E i = I P + I E i When the Earth has angular speed , the tangential speed of a point on the equator is v t = R . Thus, when the people run eastward along the equator at speed v relative to the surface of the Earth, vp v = + . their tangential speed is v p = v t + v = R + v and their angular speed is P = R R The angular momentum of the system after the people begin to run is L f = I P p + I E = I P +
b
g
FG H
I v v + I E = I P + I E + P . R R
IJ K
b
g
Since no external torques have acted on the system, angular momentum is conserved L f = Li , giving I P + I E + IPv = I P + I E i . Thus, the final angular velocity of the Earth is R IPv IPv = i  = i 1  x = , where x . I P + I E R i IP + IE R
d
i
b
g
b
g
b
g
a f LM MN bI
b
g
The new length of the day is T = day is T = T  Ti Ti x = Ti
2
=
P
IPv Ti2 I P v 2 , this may be written as T . Since i = . Ti + I E R i 2 I P + I E R
OP g PQ
T 2 = i Ti 1 + x , so the increase in the length of the i 1 x 1 x
a f
a f
b
g
To obtain a numeric answer, we compute I P = m p R 2 = 5.5 10 9 70 kg and IE = 2 2 m E R 2 = 5.98 10 24 kg 6.37 10 6 m 5 5
e
jb
g e6.37 10 mj
6
2
= 1.56 10 25 kg m 2
e
je
j
2
= 9.71 10 37 kg m 2 .
e8.64 10 sj e1.56 10 kg m jb2.5 m sg = Thus, T 2 e1.56 10 + 9.71 10 j kg m e6.37 10 mj
4 2 25 2 25 37 2 6
7.50 10 11 s .
Chapter 11
341
*P11.44
(a)
bK + U g = bK + U g
s A
s B
1 2 0 + mgy A = mv B + 0 2 v B = 2 gy A = 2 9.8 m s 2 6.30 m = 11.1 m s (b) L = mvr = 76 kg 11.1 m s 6.3 m = 5.32 10 3 kg m 2 s toward you along the axis of the channel. (c) The wheels on his skateboard prevent any tangential force from acting on him. Then no torque about the axis of the channel acts on him and his angular momentum is constant. His legs convert chemical into mechanical energy. They do work to increase his kinetic energy. The normal force acts forward on his body on its rising trajectory, to increase his linear momentum.
e
j
(d)
L = mvr
v=
5.32 10 3 kg m 2 s = 12.0 m s 76 kg 5.85 m
g
(e)
eK + U j + W = e K + U j 1 1 76 kg b11.1 m sg + 0 + W = 76 kg b12.0 m sg 2 2
g B C 2
2
+ 76 kg 9.8 m s 2 0.45 m
W = 5.44 kJ  4.69 kJ + 335 J = 1.08 kJ (f)
eK + U j = e K + U j 1 1 76 kg b12.0 m sg + 0 = 76 kgv 2 2
g C g D 2
2 D
+ 76 kg 9.8 m s 2 5.85 m
v D = 5.34 m s (g) Let point E be the apex of his flight:
g D g E
eK + U j = e K + U j 1 76 kg b5.34 m sg + 0 = 0 + 76 kg e9.8 m s jb y 2 by  y g = 1.46 m
2 2 E D
E
 yD
g
(h)
For the motion between takeoff and touchdown 1 y f = yi + v yi t + a y t 2 2 2.34 m = 0 + 5.34 m s t  4.9 m s 2 t 2 t= 5.34 5.34 2 + 4 4.9 2.34 9.8
a fa f =
1. 43 s
(i)
This solution is more accurate. In chapter 8 we modeled the normal force as constant while the skateboarder stands up. Really it increases as the process goes on.
342 P11.45
Angular Momentum
(a)
I = mi ri2 =m
FG 4d IJ H3K
2
+m
FG d IJ H 3K
2
+m
FG 2d IJ H3K
2
m 1 d
m 2
2d 3 P 3 d m
d2 = 7m 3
FIG. P11.45 (b) Think of the whole weight, 3mg, acting at the center of gravity.
=rF=
FG d IJ e ij 3mge jj = bmgdgk H 3K
(c)
=
3g 3mgd = = counterclockwise 2 I 7md 7d
(d)
a = r =
FG 3 g IJ FG 2d IJ = H 7d K H 3 K
2g up 7
The angular acceleration is not constant, but energy is.
a K + U f + E = a K + U f F dI 1 0 + a3m f g G J + 0 = I H 3K 2
i f
2 f
+0
(e)
maximum kinetic energy = mgd 6g 7d 7md 2 3 6g = 7d
(f)
f =
(g)
L f = I f =
FG 14 g IJ H3K
12
md 3 2
(h)
vf = fr =
6g d = 7d 3
2 gd 21
Chapter 11
343
P11.46
(a)
The radial coordinate of the sliding mass is r t = 0.012 5 m s t . Its angular momentum is
af b
g
L = mr 2 = 1.20 kg 2.50 rev s 2 rad rev 0.012 5 m s t 2
or L = 2.95 10 3 kg m 2 s 3 t 2
b
gb
gb
gb
g
2
e
j
The drive motor must supply torque equal to the rate of change of this angular momentum:
=
(b) (c) (d) (e)
dL = 2.95 10 3 kg m 2 s 3 2t = dt
e
ja f b0.005 89 Wgt
f = 0.005 89 W 440 s = 2.59 N m
P = = 0.005 89 W t 5 rad s =
b
ga
f
b
gb
g b0.092 5 W sgt
gb g b3.70 N sgt
P f = 0.092 5 W s 440 s = 40.7 W
T =m v2 = mr 2 = 1.20 kg 0.012 5 m s t 5 rad s r
b
ga
f
b
gb
2
=
(f) (g)
W=
440 s 0
z
Pdt =
440 s 0
zb
0.092 5 W s tdt =
g
1 0.092 5 J s 2 440 s 2
e
ja
f
2
= 8.96 kJ
The power the brake injects into the sliding block through the string is
Pb = F v = Tv cos 180 =  3.70 N s t 0.012 5 m s =  0.046 3 W s t =
Wb =
440 s 0
b
gb
g b
g
z
Pb dt = 
440 s 0
zb
dWb dt
0.046 3 W s tdt
g
= (h)
1 0.046 3 W s 440 s 2
b
ga
f
2
= 4.48 kJ
W = W + Wb = 8.96 kJ  4.48 kJ =
4.48 kJ
Just half of the work required to increase the angular momentum goes into rotational kinetic energy. The other half becomes internal energy in the brake. P11.47 Using conservation of angular momentum, we have
Laphelion = Lperihelion or mra2 a = mrp2 p .
Thus, mra2
e j
e j
e j v = emr j r r
a a 2 p
vp
p
giving
ra v a = rp v p or v a =
rp ra
vp =
0.590 AU 54.0 km s = 0.910 km s . 35.0 AU
b
g
344 P11.48
Angular Momentum
(a) (b)
= MgR  MgR = =
dL , and since dt
0
= 0 , L =
constant.
Since the total angular momentum of the system is zero, the monkey and bananas move upward with the same speed at any instant, and he will not reach the bananas (until they get tangled in the pulley). Also, since the tension in the rope is the same on both sides, Newton's second law applied to the monkey and bananas give the same acceleration upwards.
FIG. P11.48 P11.49 (a)
= r F = r F sin180 = 0
Angular momentum is conserved. L f = Li mrv = mri vi v= m ri vi mv 2 = r r3 ri vi r
(b) (c)
T=
b g
2
The work is done by the centripetal force in the negative direction. Method 1:
FIG. P11.49
b g W = z F d =  z Tdr =  z dr = 2a r f I F I mbr v g F 1 = GH r  r1 JK = 1 mv GH rr  1JK 2 2
r
m ri vi
ri
b g ar f
3
2
m ri vi
2 r
2
ri
i i
2
2
2 i
2 i
2 i 2
Method 2: W = K = (d)
r2 1 1 1 mv 2  mvi2 = mvi2 i2  1 2 2 2 r
F GH
I JK
W = 0.450 J
Using the data given, we find
v = 4.50 m s
T = 10.1 N
Chapter 11
345
P11.50
(a)
Angular momentum is conserved: mvi d 1 d = Md 2 + m 2 12 2
F GH
FG H
IJ IJ KK
2
m vi
=
6mvi Md + 3md
(a)
O
d
O
(b)
1 The original energy is mvi2 . 2 The final energy is 1 2 1 1 md 2 I = Md 2 + 2 2 12 4 The loss of energy is
(b)
FIG. P11.50
F GH
I 36m v JK aMd + 3mdf
2 2 i
2
=
3m 2 vi2 d . 2 Md + 3md
a
f
3m 2 vi2 d mMvi2 d 1 = mvi2  2 2 Md + 3md 2 Md + 3md
a
f a
=
f
and the fractional loss of energy is
2 Md + 3md P11.51 (a)
a
mMvi2 d 2
f
mvi2
M . M + 3m
FG d IJ H 2K L = 2b75.0 kg gb5.00 m sga5.00 mf
Li = m1 v1i r1i + m 2 v 2i r2i = 2mv
i
Li = 3 750 kg m 2 s (b) 1 1 2 2 m1 v1i + m 2 v 2i 2 2 1 75.0 kg 5.00 m s Ki = 2 2 Ki =
FG IJ b H K
Lf
gb
g
2
= 1.88 kJ
FIG. P11.51
(c)
Angular momentum is conserved: L f = Li = 3 750 kg m 2 s 3 750 kg m 2 s = 10.0 m s 2 75.0 kg 2.50 m
(d)
vf =
2 mr f
d i b
=
ga
f
(e) (f)
Kf = 2
FG 1 IJ b75.0 kggb10.0 m sg H 2K
2
= 7.50 kJ
W = K f  K i = 5.62 kJ
346 P11.52
Angular Momentum
(a)
(b) (c)
LM FG d IJ OP = Mvd N H 2K Q F1 I K = 2G Mv J = Mv H2 K
Li = 2 Mv
2 2
L f = Li = Mvd
Mvd = = 2v vf = 2 Mr f 2 M d 4 Lf FIG. P11.52
(d)
ch
(e)
Kf = 2
FG 1 Mv IJ = Ma2 vf H2 K
2 f
2
= 4Mv 2
(f) *P11.53
W = K f  K i = 3 Mv 2
The moment of inertia of the rest of the Earth is I= 2 2 MR 2 = 5.98 10 24 kg 6.37 10 6 m 5 5
e
j
2
= 9.71 10 37 kg m 2 .
For the original ice disks, I= 1 1 Mr 2 = 2.30 10 19 kg 6 10 5 m 2 2
e
j
2
= 4.14 10 30 kg m 2 .
For the final thin shell of water, I= 2 2 Mr 2 = 2.30 10 19 kg 6.37 10 6 m 3 3
e
j
2
= 6.22 10 32 kg m 2 .
Conservation of angular momentum for the spinning planet is expressed by Ii i = I f f 2 e4.14 10 + 9.71 10 j 86 2 s = e6.22 10 + 9.71 10 j b86 400 s + g 400 F1 + I F 1 + 4.14 10 I = F 1 + 6.22 10 I GH 86 400 s JK GH 9.71 10 JK GH 9.71 10 JK
30 37 32 37 30 32 37 37
6.22 10 32 4.14 10 30 =  86 400 s 9.71 10 37 9.71 10 37 = 0.550 s
Chapter 11
347
P11.54
For the cube to tip over, the center of mass (CM) must rise so that it is over the axis of rotation AB. To do this, the CM must be raised a distance of a
e
2 1 . Mga
j
CM
D
e
2 1 =
j
1 I cube 2 2
A
From conservation of angular momentum, 4a 8 Ma mv = 3 3 mv = 2 Ma
F GH
2
I JK
4a/3 A
C D B
1 8 Ma 2 m 2 v 2 = Mga 2 3 4M 2 a 2 v= P11.55 M 3 ga m
F GH
I JK
e
2 1
j
FIG. P11.54
e
2 1
j
Angular momentum is conserved during the inelastic collision. Mva = I Mva 3 v = I 8a
=
The condition, that the box falls off the table, is that the center of mass must reach its maximum height as the box rotates, hmax = a 2 . Using conservation of energy: 1 2 I = Mg a 2  a 2 1 8 Ma 2 2 3
2
I FG 3v IJ = Mgea JK H 8 a K 16 v = gae 2  1j 3 O L ga v = 4M e 2  1jP 3 Q N
2 12
F GH
e
j
2 a
j
FIG. P11.55
P11.56
(a)
The net torque is zero at the point of contact, so the angular momentum before and after the collision must be equal.
FG 1 MR IJ = FG 1 MR IJ + eMR j H2 K H2 K
2 i 2 2
=
i 3
(b)
E = E
1 1 2 2
e
MR 2
je j
i 2 3
+1M 2
1 1 2 2
e j e MR j
2
R i 2 3 2 i

1 1 2 2
e
MR 2 i2
j
= 
2 3
348 P11.57
Angular Momentum
(a)
t =
p R i Mv MR = = = Mg Mg 3 g f 1 2 1 I = MR 2 i2 2 18 x= R 2 i2 18 g (See Problem 11.56)
(b)
W = K =
Mgx =
1 MR 2 i2 18
ANSWERS TO EVEN PROBLEMS
P11.2 P11.4 P11.6 P11.8 P11.10 P11.12 P11.14 P11.16 (a) 740 cm 2 ; (b) 59.5 cm (a) 168; (b) 11.9 principal value; (c) Only the first is unambiguous. No; see the solution (a) 7.00 N m k ; (b) 11.0 N m k see the solution P11.32 P11.34 P11.36 8.54 rad s (a) 7.20 10 3 kg m 2 s ; (b) 9.47 rad s 12.3 m s 2 (a) 2.35 rad s; (b) 0.498 rad s ; (c) 5.58 131 s (a) 2.19 10 6 m s ; (b) 2.18 10 18 J ; (c) 4.13 10 16 rad s P11.44 (a) 11.1 m s; (b) 5.32 10 3 kg m 2 s ; (c) see the solution; (d) 12.0 m s; (e) 1.08 kJ ; (f) 5.34 m s; (g) 1.46 m; (h) 1.43 s; (i) see the solution
a
f
a
f
P11.38 P11.40 P11.42
e22.0 kg m sjk
2
see the solution (a) 3.14 N m ; (b) 0. 400 kg m v ; (c) 7.85 m s
2
b
g
P11.18
(a) +9.03 10 9 kg m 2 s south; (b) No; (c) 0
e
j
P11.46
b g b g (e) b3.70 N sgt ; (f) 8.96 kJ; (g) 4.48 kJ
(a) 0.005 89 W t ; (b) 2.59 N m ; (c) 0.092 5 W s t ; (d) 40.7 W ; (h) +4.48 kJ (a) 0; (b) 0; no (a) 6mvi M ; (b) M + 3m Md + 3md
P11.20 P11.22 P11.24
103 N m
e4.50 kg m sj up
2
P11.48 P11.50 P11.52
1.20 kg m s perpendicularly into the clock face 8.63 m s 2 Kf I1 I = (a) 1 i ; (b) Ki I1 + I 2 I1 + I 2 (a) 1.91 rad s ; (b) 2.53 J; 6.44 J
2
P11.26 P11.28 P11.30
(a) Mvd ; (b) Mv 2 ; (c) Mvd ; (d) 2v; (e) 4Mv 2 ; (f) 3 Mv 2 M 3 ga m (a)
P11.54 P11.56
e
2 1
j
i 2 E = ; (b) E 3 3
12
Static Equilibrium and Elasticity
CHAPTER OUTLINE
12.1 12.2 12.3 12.4 The Conditions for Equilibrium More on the Center of Gravity Examples of Rigid Objects in Static Equilibrium Elastic Properties of Solids
ANSWERS TO QUESTIONS
Q12.1 When you bend over, your center of gravity shifts forward. Once your CG is no longer over your feet, gravity contributes to a nonzero net torque on your body and you begin to rotate. Yes, it can. Consider an object on a spring oscillating back and forth. In the center of the motion both the sum of the torques and the sum of the forces acting on the object are (separately) zero. Again, a meteoroid flying freely through interstellar space feels essentially no forces and keeps moving with constant velocity. Noone condition for equilibrium is that
Q12.2
Q12.3
F = 0 . For this to
be true with only a single force acting on an object, that force would have to be of zero magnitude; so really no forces act on that object. Q12.4 (a) Consider pushing up with one hand on one side of a steering wheel and pulling down equally hard with the other hand on the other side. A pair of equalmagnitude oppositelydirected forces applied at different points is called a couple. An object in free fall has a nonzero net force acting on it, but a net torque of zero about its center of mass.
(b) Q12.5 Q12.6
No. If the torques are all in the same direction, then the net torque cannot be zero. (a) (b) Yes, provided that its angular momentum is constant. Yes, provided that its linear momentum is constant.
Q12.7 Q12.8
A Vshaped boomerang, a barstool, an empty coffee cup, a satellite dish, and a curving plastic slide at the edge of a swimming pool each have a center of mass that is not within the bulk of the object. Suspend the plywood from the nail, and hang the plumb bob from the nail. Trace on the plywood along the string of the plumb bob. Now suspend the plywood with the nail through a different point on the plywood, not along the first line you drew. Again hang the plumb bob from the nail and trace along the string. The center of gravity is located halfway through the thickness of the plywood under the intersection of the two lines you drew.
349
350 Q12.9 Q12.10
Static Equilibrium and Elasticity
The center of gravity must be directly over the point where the chair leg contacts the floor. That way, no torque is applied to the chair by gravity. The equilibrium is unstable. She can be correct. If the dog stands on a relatively thick scale, the dog's legs on the ground might support more of its weight than its legs on the scale. She can check for and if necessary correct for this error by having the dog stand like a bridge with two legs on the scale and two on a book of equal thicknessa physics textbook is a good choice. If their base areas are equal, the tall crate will topple first. Its center of gravity is higher off the incline than that of the shorter crate. The taller crate can be rotated only through a smaller angle before its center of gravity is no longer over its base. The free body diagram demonstrates that it is necessary to have friction on the ground to counterbalance the normal force of the wall and to keep the base of the ladder from sliding. Interestingly enough, if there is friction on the floor and on the wall, it is not possible to determine whether the ladder will slip from the equilibrium conditions alone.
Q12.11
Q12.12
FIG. Q12.12 Q12.13 When you lift a load with your back, your back muscles must supply the torque not only to rotate your upper body to a vertical position, but also to lift the load. Since the distance from the pivotyour hipsto the loadessentially your shouldersis great, the force required to supply the lifting torque is very large. When lifting from your knees, your back muscles need only keep your back straight. The force required to do that is much smaller than when lifting with your back, as the torque required is small, because the moment arm of the load is smallthe line of action of the load passes close to your hips. When you lift from your knees, your much stronger leg and hip muscles do the work. Shear deformation. The vertical columns experience simple compression due to gravity acting upon their mass. The horizontal slabs, however, suffer significant shear stress due to gravity. The bottom surface of a sagging lintel is under tension. Stone is much stronger under compression than under tension, so horizontal slabs are more likely to fail.
Q12.14 Q12.15
Chapter 12
351
SOLUTIONS TO PROBLEMS
Section 12.1 P12.1 The Conditions for Equilibrium
F 0.600 m
To hold the bat in equilibrium, the player must exert both a force and a torque on the bat to make
Fx = Fy = 0
upward force of F = 10.0 N
and
= 0
O
Fy = 0 F  10.0 N = 0 , or the player must exert a net
To satisfy the second condition of equilibrium, the player must exert an applied torque a to make
10.0 N
= a  a0.600 mfa10.0 N f = 0 . Thus, the required torque is
a = +6.00 N m or 6.00 N m counterclockwise
FIG. P12.1
P12.2
Use distances, angles, and forces as shown. The conditions of equilibrium are:
Fy
Fy = 0 Fx = 0 = 0
Fy + R y  Fg = 0 Fx  R x = 0 Fy cos  Fg
l
Fx
FG IJ cos  F H 2K
Ry
x
sin = 0
Rx O
Fg
FIG. P12.2 P12.3 Take torques about P. m1 g mb g d m1 2 CG x nO nP P m2 m2 g
p = n0 M 2 + d P + m1 g M 2 + d P + m b gd  m 2 gx = 0 N Q N Q
We want to find x for which n 0 = 0 . x=
L
O
L
O
bm g + m ggd + m g
1 b 1
2
m2 g
=
bm
1
+ m b d + m1 m2
g
O
2
FIG. P12.3
352
Static Equilibrium and Elasticity
Section 12.2 P12.4
More on the Center of Gravity
The hole we can count as negative mass xCG = Call the mass of each unit of pizza area. m 1 x1  m 2 x 2 m1  m 2
xCG = xCG = P12.5
R 2 0  R 2
R 8 3 4
c h c h  c h
R 2 2 R 2 R 2 2
=
R 6
4.00 cm
The coordinates of the center of gravity of piece 1 are x1 = 2.00 cm and y1 = 9.00 cm . The coordinates for piece 2 are x 2 = 8.00 cm and y 2 = 2.00 cm . The area of each piece is
18.0 cm
1
2
12.0 cm
4.00 cm
A1 = 72.0 cm 2 and A 2 = 32.0 cm 2 .
And the mass of each piece is proportional to the area. Thus,
FIG. P12.5
2
m i xi xCG = mi
and yCG =
e72.0 cm ja2.00 cmf + e32.0 cm ja8.00 cmf = =
2
72.0 cm 2 + 32.0 cm 2
3.85 cm
m i yi mi
=
e72.0 cm ja9.00 cmf + e32.0 cm ja2.00 cmf =
2 2
104 cm 2
6.85 cm .
Chapter 12
353
P12.6
Let represent the massperface area. A vertical strip at position x, with width dx and height
y 1.00 m
ax  3.00f
9
2
has mass
x  3.00 dx dm = . 9
The total mass is
x
3.00
a
f
2
y = (x  3.00)2/9
x  3 dx M = dm = 9 x =0
z
z
a f
2
x 0 dx 3.00 m
FG IJ z ex  6 x + 9jdx H 9K F I L x 6x + 9xOP M = G JM  H 9 KN 3 2 Q
M=
3.00 0 2 3 2
FIG. P12.6 =
3.00
0
The xcoordinate of the center of gravity is
xCG =
z
xdm M
=
1 9
3.00 0
z
x x  3 dx =
a f
2
9
3.00 0
ze
x 3  6 x 2 + 9 x dx =
j
1 x 4 6x3 9x 2  + 9 4 3 2
LM N
OP Q
3.00
=
0
6.75 m = 0.750 m 9.00
P12.7
Let the fourth mass (8.00 kg) be placed at (x, y), then xCG = 0 = x= Similarly, yCG = 0 =
a3.00fa4.00f + m axf
4
12.0 + m 4
a3.00fa4.00f + 8.00byg
12.0 + 8.00
12.0 = 1.50 m 8.00
y = 1.50 m
P12.8 In a uniform gravitational field, the center of mass and center of gravity of an object coincide. Thus, the center of gravity of the triangle is located at x = 6.67 m , y = 2.33 m (see the Example on the center of mass of a triangle in Chapter 9). The coordinates of the center of gravity of the threeobject system are then: xCG = xCG = yCG = yCG =
m i xi mi m i yi mi
=
b6.00 kg ga5.50 mf + b3.00 kgga6.67 mf + b5.00 kg ga3.50 mf a6.00 + 3.00 + 5.00f kg
14.0 kg
35.5 kg m = 2.54 m and 14.0 kg =
b6.00 kg ga7.00 mf + b3.00 kgga2.33 mf + b5.00 kgga+3.50 mf
66.5 kg m = 4.75 m 14.0 kg
354
Static Equilibrium and Elasticity
Section 12.3 P12.9
Examples of Rigid Objects in Static Equilibrium 3r
= 0 = mga3r f  Tr
2T  Mg sin 45.0 = 0 T= Mg sin 45.0 1 500 kg g sin 45.0 = 2 2 = 530 9.80 N
bg
a fa f
m=
T 530 g = = 177 kg 3g 3g
1 500 kg
m
= 45
FIG. P12.9 *P12.10 (a) For rotational equilibrium of the lowest rod about its point of support, +12.0 g g 3 cm  m1 g 4 cm (b)
= 0 .
m1 = 9.00 g
For the middle rod, + m 2 2 cm  12.0 g + 9.0 g 5 cm = 0
b
g
m 2 = 52.5 g
(c)
For the top rod, 52.5 g + 12.0 g + 9.0 g 4 cm  m3 6 cm = 0
b
g
m3 = 49.0 g
24.0 cm 26.0 cm
P12.11
Fg standard weight
f a f F 13 I F = F G J H 12 K F F  F I 100 = F 13  1I 100 = GH F JK GH 12 JK
Fg 0.240 = Fg 0.260
g g g g g
Fg weight of goods sold
a
Fg
Fg
FIG. P12.11 8.33%
T H V d 30.0 196 N
*P12.12
(a)
Consider the torques about an axis perpendicular to the page and through the left end of the horizontal beam.
= +aT sin 30.0fd  a196 N fd = 0 ,
giving T = 392 N . (b) From From
FIG. P12.12
Fx = 0 ,
H  T cos 30.0 = 0 , or H = 392 N cos 30.0 = 339 N to the right . 0 .
a
f
Fy = 0 , V + T sin 30.0200 N = 0 , or V = 196 N  a392 N f sin 30.0 =
Chapter 12
355
P12.13
(a)
Fx = f  n w = 0 Fy = n g  800 N  500 N = 0
Taking torques about an axis at the foot of the ladder,
nw
a800 Nfa4.00 mf sin 30.0+a500 Nfa7.50 mf sin 30.0 n a15.0 cmf cos 30.0 = 0
w
500 N ng
Solving the torque equation, nw
a4.00 mfa800 Nf + a7.50 mfa500 Nf tan 30.0 = 268 N . =
15.0 m
f A
800 N
Next substitute this value into the Fx equation to find
FIG. P12.13
f = n w = 268 N
Solving the equation
in the positive x direction.
Fy = 0 ,
n g = 1 300 N in the positive y direction.
(b)
In this case, the torque equation
A = 0
gives:
w
a9.00 mfa800 Nf sin 30.0+a7.50 mfa500 Nf sin 30.0a15.0 mfbn g sin 60.0 = 0
or n w = 421 N . Since f = n w = 421 N and f = fmax = n g , we find
=
fmax 421 N = = 0.324 . 1 300 N ng (1) (2)
nw m2 g
P12.14
(a)
Fx = f  n w = 0 Fy = n g  m1 g  m 2 g = 0
A = m1 g G 2 J cos  m 2 gx cos + n w L sin = 0 H K
From the torque equation, nw =
1 2
F LI
LM 1 m g + FG x IJ m g OP cot N2 H LK Q
m1 g
Then, from equation (1): and from equation (2): (b)
L1 F xI O f = n = M m g + G J m g P cot N2 H LK Q n = bm + m g g
w 1 2 g 1 2
f A
ng
FIG. P12.14
If the ladder is on the verge of slipping when x = d , then
=
f
x=d
ng
=
e
m1 2
+
m2d L
j cot
m1 + m 2
.
356 P12.15
Static Equilibrium and Elasticity
(a)
Taking moments about P,
aR sin 30.0f0 + aR cos 30.0fa5.00 cmf  a150 Nfa30.0 cmf = 0
R = 1 039.2 N = 1.04 kN The force exerted by the hammer on the nail is equal in magnitude and opposite in direction:
1.04 kN at 60 upward and to the right.
(b) f = R sin 30.0150 N = 370 N n = R cos 30.0 = 900 N Fsurface = 370 N i + 900 N j P12.16 See the freebody diagram at the right. When the plank is on the verge of tipping about point P, the normal force n1 goes to zero. Then, summing torques about point P gives
Mg 3.00 m x P n1 d n2
FIG. P12.15
a
f a
f
p = mgd + Mgx = 0
or
F mI x = G Jd . H MK
mg 6.00 m
1.50 m
From the dimensions given on the freebody diagram, observe that d = 1.50 m Thus, when the plank is about to tip, x= P12.17
FIG. P12.16
F 30.0 kg I a1.50 mf = GH 70.0 kg JK a fb g a
0.643 m .
Torque about the front wheel is zero.
0 = 1.20 m mg  3.00 m 2 Fr
Thus, the force at each rear wheel is
fb g
Fr = 0.200mg = 2.94 kN .
The force at each front wheel is then Ff = mg  2 Fr = 4. 41 kN . 2 FIG. P12.17
Chapter 12
357
P12.18
Fx = Fb  Ft + 5.50 N = 0 Fy = n  mg = 0
Summing torques about point O,
(1)
5.50 N
O = Ft a1.50 mf  a5.50 mfa10.0 mf = 0
which yields Ft = 36.7 N to the left Then, from Equation (1),
10.0 m
mg
Ft 1.50 m Fb O n
Fb = 36.7 N  5.50 N = 31.2 N to the right
FIG. P12.18 P12.19 (a) (b) P12.20 Te sin 42.0 = 20.0 N Te cos 42.0 = Tm
Te = 29.9 N Tm = 22.2 N
Ry T x Rx 20.0 y 4.00 m 5.00 m 7.00 m 19.6 kN 9.80 kN
a f vertically down a distance y = a5.00 mf sin 20.0 = 1.71 m . The
cable then makes the following angle with the horizontal:
Relative to the hinge end of the bridge, the cable is attached horizontally out a distance x = 5.00 m cos 20.0 = 4.70 m and
= tan 1
(a)
LM a12.0 + 1.71f m OP = 71.1 . N 4.70 m Q a a f f a f
Take torques about the hinge end of the bridge: R x 0 + R y 0  19.6 kN 4.00 m cos 20.0 9.80 kN 7.00 m cos 20.0 = 0 which yields T = 35.5 kN
af
af
T cos 71.1 1.71 m + T sin 71.1 4.70 m
a
f
FIG. P12.20
(b)
Fx = 0 Rx  T cos 71.1 = 0
or R x = 35.5 kN cos 71.1 = 11.5 kN right
a
f
b
g
(c)
Fy = 0 R y  19.6 kN + T sin 71.19.80 kN = 0
Thus, R y = 29. 4 kN  35.5 kN sin 71.1 = 4.19 kN = 4.19 kN down
a
f
358 *P12.21
Static Equilibrium and Elasticity
(a)
We model the horse as a particle. The drawbridge will fall out from under the horse.
Ry
0
Rx
= mg
=
1 2 1 3
cos 0 m
2
=
3g cos 0 2
3 9.80 m s 2 cos 20.0 2 8.00 m
e
a
j
f
= 1.73 rad s
2
mg FIG. P12.21(a)
(b)
1 2 I = mgh 2 1 1 1 1  sin 0 m 2 2 = mg 2 3 2
b
g
1.56 rad s
= (c)
3g
b1  sin g = e 8.00 m j a1  sin 20f =
3 9.80 m s 2
0
The linear acceleration of the bridge is: 1 1 a= = 8.0 m 1.73 rad s 2 = 6.907 m s 2 2 2
Ry
a
fe
j
0
Rx
The force at the hinge + the force of gravity produce the acceleration a = 6.907 m s 2 at right angles to the bridge. R x = ma x = 2 000 kg 6.907 m s 2 cos 250 = 4.72 kN
a
mg
b
ge
j
FIG. P12.21(c)
Ry  mg = ma y
R y = m g + a y = 2 000 kg 9.80 m s 2 + 6.907 m s 2 sin 250 = 6.62 kN Thus: R = 4.72 i + 6.62 j kN . (d) Rx = 0 a=
2
e
j b
g
e
j
e
j
FG 1 IJ = b1.56 rad sg a4.0 mf = 9.67 m s H2 K
2
Ry
2
a mg FIG. P12.21(d)
Rx
R y  mg = ma
R y = 2 000 kg 9.8 m s 2 + 9.67 m s 2 = 38.9 kN Thus: R y = 38.9 j kN
b
ge
j
Chapter 12
359
P12.22
Call the required force F, with components Fx = F cos 15.0 and
Fy
400 N
Fy =  F sin 15.0 , transmitted to the
center of the wheel by the handles. Just as the wheel leaves the ground, the ground exerts no force on it.
R b 8.00 cm a distances
Fx b nx ny a forces
Fx = 0 : Fy = 0 :
F cos 15.0n x
(1)
 F sin 15.0400 N + n y = 0 (2)
FIG. P12.22
Take torques about its contact point with the brick. The needed distances are seen to be: b = R  8.00 cm = 20.0  8.00 cm = 12.0 cm a = R  b = 16.0 cm (a)
2 2
a
f
= 0 :
F  12.0 cm cos 15.0+ 16.0 cm sin 15.0 + 400 N 16.0 cm = 0 so (b) F= 6 400 N cm = 859 N 7.45 cm
a
 Fx b + Fy a + 400 N a = 0 , or
a
f
f
a
f
a
fa
f
Then, using Equations (1) and (2), n x = 859 N cos 15.0 = 830 N and n y = 400 N + 859 N sin 15.0 = 622 N
2 2 n = n x + n y = 1.04 kN
a
f
a
f
= tan 1
*P12.23
F n I = tan a0.749f = GH n JK
y x 1
36.9 to the left and upward
When x = x min , the rod is on the verge of slipping, so f = fs From
2.0 m n f x Fg Fg
b g
max
= sn = 0.50n .
37 2.0 m
Fx = 0 , n  T cos 37 = 0 , or n = 0.799T .
Thus, f = 0.50 0.799T = 0.399T From Using
a
f
FIG. P12.23
Fy = 0 , = 0
f + T sin 372 Fg = 0 , or 0.399T  0.602T  2 Fg = 0 , giving T = 2.00 Fg .
for an axis perpendicular to the page and through the left end of the beam gives
 Fg x min  Fg 2.0 m + 2 Fg sin 37 4.0 m = 0 , which reduces to x min = 2.82 m .
a
f e j
a
f
360 P12.24
Static Equilibrium and Elasticity
x=
3L 4
L
If the CM of the two bricks does not lie over the edge, then the bricks balance. If the lower brick is placed L over the edge, then the 4
x
second brick may be placed so that its end protrudes over the edge. P12.25
3L 4
FIG. P12.24
To find U, measure distances and forces from point A. Then, balancing torques,
a0.750fU = 29.4a2.25f a0.750fD = a1.50fa29.4f
Also, notice that U = D + Fg , so *P12.26
U = 88.2 N
To find D, measure distances and forces from point B. Then, balancing torques,
D = 58.8 N
Fy = 0 .
Consider forces and torques on the beam.
Fx = 0 : Fy = 0 : = 0 :
(a)
R cos  T cos 53 = 0
aT sin 53f8 m  a600 Nfx  a200 Nf4 m = 0
R sin + T sin 53800 N = 0
Then T =
600 Nx + 800 N m = 93.9 N m x + 125 N . As x increases from 2 m, this expression 8 m sin 53 grows larger.
b
g
(b)
From substituting back, R cos = 93.9 x + 125 cos 53 R sin = 800 N  93.9 x + 125 sin 53 Dividing, tan = 800 N R sin =  tan 53+ 93.9 x +125 cos 53 R cos
f F 32  1IJ tan = tan 53 G H 3x + 4 K
a
As x increases the fraction decreases and decreases . continued on next page
Chapter 12
361
(c)
To find R we can work out R 2 cos 2 + R 2 sin 2 = R 2 . From the expressions above for R cos and R sin , R 2 = T 2 cos 2 53+T 2 sin 2 531 600 NT sin 53+ 800 N R 2 = T 2  1 600T sin 53+640 000 R 2 = 93.9 x + 125
a
f
2
a
f
2
 1 278 93.9 x + 125 + 640 000
a
f
R = 8 819 x 2  96 482 x + 495 678
e
j
12
At x = 0 this gives R = 704 N . At x = 2 m , R = 581 N . At x = 8 m , R = 537 N . Over the range of possible values for x, the negative term 96 482x dominates the positive term
8 819 x 2 , and R decreases as x increases.
Section 12.4 P12.27
Elastic Properties of Solids
L F =Y A Li L =
200 9.80 4.00 FLi = = 4.90 mm AY 0.200 10 4 8.00 10 10 stress = F F = 2 A r
a fa fa f e je j a e f FGH d IJK 2
8 2
P12.28
(a)
F = stress
F = 1.50 10 F = 73.6 kN (b)
F 2.50 10 m I N m j G H 2 JK
2 2
2
stress = Y strain =
i
a
f
YL Li
8
L =
astressfL = e1.50 10
Y
N m 2 0.250 m
10
ja
1.50 10
N m
2
f=
2.50 mm
*P12.29
The definition of Y =
stress means that Y is the slope of the graph: strain Y= 300 10 6 N m 2 = 1.0 10 11 N m 2 . 0.003
362 P12.30
Static Equilibrium and Elasticity
Count the wires. If they are wrapped together so that all support nearly equal stress, the number should be 20.0 kN = 100 . 0.200 kN Since crosssectional area is proportional to diameter squared, the diameter of the cable will be
a1 mmf
P12.31
100 ~ 1 cm .
From the defining equation for the shear modulus, we find x as 5.00 10 3 m 20.0 N hf x = = = 2.38 10 5 m 4 6 2 2 SA 3.0 10 N m 14.0 10 m
e
e
je
ja
f
j
or x = 2.38 10 2 mm . P12.32 The force acting on the hammer changes its momentum according to
mvi + F t = mv f so F =
Hence, F = 30.0 kg 10.0 m s  20.0 m s 0.110 s = 8.18 10 3 N .
a f
m v f  vi t
.
By Newton's third law, this is also the magnitude of the average force exerted on the spike by the hammer during the blow. Thus, the stress in the spike is: stress = F 8.18 10 3 N = = 1.97 10 7 N m 2 b0.023 0 mg2 A
4
and the strain is: strain = P12.33 (a) F = A stress
stress 1.97 10 7 N m 2 = = 9.85 10 5 . Y 20.0 10 10 N m 2
F
a fa f = e5.00 10
3
m
j e4.00 10
2
8
N m2
j
3.0 ft
= 3.14 10 4 N (b) The area over which the shear occurs is equal to the circumference of the hole times its thickness. Thus, A = 2r t = 2 5.00 10 3 m 5.00 10 3 m = 1.57 10
4
t A
a f
e
je
j
j
FIG. P12.33
m
2
So, F = A Stress = 1.57 10 4 m 2 4.00 10 8 N m 2 = 6.28 10 4 N .
af
e
je
Chapter 12
363
P12.34
Let the 3.00 kg mass be mass #1, with the 5.00 kg mass, mass # 2. Applying Newton's second law to each mass gives: m1 a = T  m1 g where T is the tension in the wire. Solving equation (1) for the acceleration gives: a = and substituting this into equation (2) yields: Solving for the tension T gives T=
2 2m1 m 2 g 2 3.00 kg 5.00 kg 9.80 m s = = 36.8 N . 8.00 kg m 2 + m1
(1)
and
m2 a = m2 g  T
(2)
T  g, m1
m2 T  m2 g = m2 g  T . m1
b
gb
ge
j
From the definition of Young's modulus, Y =
i
a f a36.8 Nfa2.00 mf TL L = = YA e2.00 10 N m j e2.00 10
11 2 9 2
FLi , the elongation of the wire is: A L
3
m
j
2
= 0.029 3 mm .
P12.35
Consider recompressing the ice, which has a volume 1.09V0 .
F V IJ = e2.00 10 N m ja0.090f = P =  BG 1.09 HV K
i
1.65 10 8 N m 2
*P12.36
B=
P
V Vi
=
PVi V 1.13 10 8 N m 2 1 m3 PVi = = 0.053 8 m 3 B 0.21 10 10 N m 2 1.03 10 3 kg 0.946 m
3
(a) (b)
V = 
e
j
The quantity of water with mass 1.03 10 3 kg occupies volume at the bottom 1 m 3  0.053 8 m 3 = 0.946 m 3 . So its density is = 1.09 10 3 kg m3 .
(c) *P12.37
With only a 5% volume change in this extreme case, liquid water is indeed nearly incompressible.
Part of the load force extends the cable and part compresses the column by the same distance : F= = YA A A
A
+
Ys As
s
F
YA A A
A
+
Ys As
s
=
7 10 10 0.162 4 2  0 .161 4 2 4 3. 25
e
8 500 N
a f
j + 20 10 b0.012 7 g 4a 5.75 f
10
2
= 8.60 10
4
m
364
Static Equilibrium and Elasticity
Additional Problems *P12.38 (a) (b) The beam is perpendicular to the wall, since 3 2 + 4 2 = 5 2 . Then sin = 4m ; = 53.1 . 5m
hinge = 0 :
+T sin 3 m  250 N 10 m = 0
T= 2 500 Nm = 1.04 10 3 N 3 m sin 53.1
a f
a
f
(c)
1.04 10 3 N T = = 0.126 m k 8.25 10 3 N m The cable is 5.126 m long. From the law of cosines, x= 4 2 = 5.126 2 + 3 2  2 3 5.126 cos
4m
5.126 m
a fa
2
f
3m
= cos 1
(d)
3 + 5.126  4 = 51.2 2 3 5.126
2
2
a fa
f
FIG. P12.38 sin sin 51.2 = 5.126 m 4m
From the law of sines, the angle the hinge makes with the wall satisfies sin = 0.998 58
hinge = 0
+T 3 m sin 51.2250 N 10 m 0.998 58 = 0 T = 1.07 10 3 N (e) x= 1.07 10 3 N = 0.129 m 8.25 10 3 N m 3 2 + 5.129 2  4 2 = 51.1 2 3 5.129
a f
a
fa
f
= cos 1
(f)
a fa
f
Now the answers are selfconsistent: sin 51.1 = 0.998 51 4m T 3 m sin 51.1250 N 10 m 0.998 51 = 0 sin = 5.129 m
a f
a
fa
f
T = 1.07 10 N x = 0.129 5 m
3
= 51.1
P12.39 Let n A and n B be the normal forces at the points of support. Choosing the origin at point A with we find: n A + n B  8.00 10 4 g  3.00 10 4 g = 0 and
4 4 B
Fy = 0 and = 0,
A
B
j e j e3.00 10 jb g g15.0  e8.00 10 jb g g25.0 + n a50.0 f = 0
e
15.0 m 50.0 m
FIG. P12.39
The equations combine to give n A = 5.98 10 5 N and b B = 4.80 10 5 N .
Chapter 12
365
P12.40
When the concrete has cured and the prestressing tension has been released, the rod presses in on the concrete and with equal force, T2 , the concrete produces tension in the rod. (a) In the concrete: stress = 8.00 10 6 N m 2 = Y strain = Y Thus, L = or (b)
a
astressfL = e
i
8.00 10 6 N m 2 1.50 m 30.0 10 N m
9 2
ja
f
f FGH L IJK L
i
Y
L = 4.00 10 4 m = 0.400 mm . T2 = 8.00 10 6 N m 2 , so Ac
In the concrete: stress =
T2 = 8.00 10 6 N m 2 50.0 10 4 m 2 = 40.0 kN (c) For the rod: T2 Li T2 L = Ysteel so L = AR Li A R Ysteel
e
je
j
FG IJ H K
e4.00 10 Nja1.50 mf L = = 2.00 10 e1.50 10 m je20.0 10 N m j
4 4 2 10 2
3
m = 2.00 mm
(d)
The rod in the finished concrete is 2.00 mm longer than its unstretched length. To remove stress from the concrete, one must stretch the rod 0.400 mm farther, by a total of 2. 40 mm . For the stretched rod around which the concrete is poured:
(e)
FG L IJ A Y H L K F 2.40 10 m I e1.50 10 m je20.0 10 T =G H 1.50 m JK
Ltotal T1 = Ysteel AR Li
3 1
FG H
IJ K
or T1 =
total i
R steel
4
2
10
N m 2 = 48.0 kN
j
*P12.41
With as large as possible, n1 and n 2 will both be large. The equality sign in f 2 sn 2 will be true, but the lessthan sign in f1 < sn1 . Take torques about the lower end of the pole. n 2 cos + Fg
n2
f2 d
FG 1 IJ cos  f H2 K f
2
sin = 0
Fg
n1
f1 FIG. P12.41
Setting f 2 = 0.576n 2 , the torque equation becomes n 2 1  0.576 tan + Since n 2 > 0 , it is necessary that 1  0.576 tan < 0 1 = 1.736 0.576 > 60.1 tan > = 7.80 ft d < = 9.00 ft sin sin 60.1
a
1 Fg = 0 2
366 P12.42
Static Equilibrium and Elasticity
Call the normal forces A and B. They make angles and with the vertical.
Mg
Fx = 0: Fy = 0:
Substitute B = A sin sin
A sin  B sin = 0 A cos  Mg + B cos = 0
A
B
A cos + A cos
A cos sin + sin cos = Mg sin sin A = Mg sin +
A sin A cos
b
sin = Mg sin
g
Mg B sin B cos
b
g
sin B = Mg sin +
b
g
Ry Rx O x 700 N 3.00 m
FIG. P12.42
T 60.0
P12.43
(a) (b)
See the diagram. If x = 1.00 m , then
O = a700 N fa1.00 mf  a200 N fa3.00 mf
a fa f +aT sin 60.0fa6.00 mf = 0
 80.0 N 6.00 m
200 N 3.00 m
80.0 N
FIG. P12.43
Solving for the tension gives: T = 343 N . From From (c)
Fx = 0 , R x = T cos 60.0 =
171 N . 683 N .
Fy = 0 , R y = 980 N  T sin 60.0 =
If T = 900 N :
O = a700 N fx  a 200 N fa3.00 mf  a80.0 N fa6.00 mf +
Solving for x gives: x = 5.13 m .
a900 Nf sin 60.0 a6.00 mf = 0 .
Chapter 12
367
P12.44
(a)
Sum the torques about top hinge:
D C
T sin 30.0 T cos 30.0
= 0:
C 0 + D 0 + 200 N cos 30.0 0
af af
+200 N sin 30.0 3.00 m
Giving A = (b)
af 160 N bright g
+B 0 = 0
392 N 1.50 m + A 1.80 m
a
f a
a
af
f
1.80 m
392 N
f
A 1.50 m 1.50 m
.
B
Fx = 0 :
C  200 N cos 30.0+ A = 0 C = 160 N  173 N = 13.2 N In our diagram, this means 13.2 N to the right .
FIG. P12.44
(c)
Fy = 0 : +B + D  392 N + 200 N sin 30.0 = 0
B + D = 392 N  100 N = 292 N up
b g
(d)
Given C = 0: Take torques about bottom hinge to obtain
A 0 + B 0 + 0 1.80 m + D 0  392 N 1.50 m + T sin 30.0 3.00 m + T cos 30.0 1.80 m = 0
so T = P12.45 Using
af af a a
f af
a
f
a
f
a
f
588 N m = 192 N . 1.50 m + 1.56 m
f
Fx = Fy = = 0, choosing the origin at the left end
of the beam, we have (neglecting the weight of the beam)
Fx = Rx  T cos = 0 , Fy = Ry + T sin  Fg = 0 ,
and
=  Fg aL + d f + T sin a2L + d f = 0.
Solving these equations, we find: (a) T=
(b)
Rx
a f sin a 2L + d f F aL + d f cot =
Fg L + d
g
2L + d
Ry =
Fg L 2L + d
FIG. P12.45
368 P12.46
Static Equilibrium and Elasticity
point 0 = 0 gives
aT cos 25.0fFGH 34 sin 65.0IJK + aT sin 25.0fFGH 34 cos 65.0IJK F I = b 2 000 N ga cos 65.0f + b1 200 N gG cos 65.0J H2 K
From which, T = 1 465 N = 1.46 kN From
T sin 25.0 l T cos 25.0 3l 4 1 200 N H 65.0 V FIG. P12.46 2 000 N
Fx = 0 ,
H = T cos 25.0 = 1 328 N toward right = 1.33 kN From
b
g
Fy = 0 ,
V = 3 200 N  T sin 25.0 = 2 581 N upward = 2.58 kN P12.47 We interpret the problem to mean that the support at point B is frictionless. Then the support exerts a force in the x direction and FBy = 0
b
g
g and = b3 000 g ga 2.00f  b10 000 g ga6.00f + F a1.00 f = 0 .
FAy  3 000 + 10 000 g = 0
Bx
Fx = FBx  FAx = 0
b
These equations combine to give FAx = FBx = 6.47 10 N FAy = 1.27 10 5 N P12.48
5
FIG. P12.47
n= M+m g H = f
a
f
H
H max = f max = s m + M g
a
f
A = 0 =
mgL cos 60.0+ Mgx cos 60.0 HL sin 60.0 2 m + M tan 60.0 m x H tan 60.0 m =  = s  2M 2M L Mg M
x Mg mg 60.0 A f
a
f
3 1 = s tan 60.0 = 0.789 2 4
n
FIG. P12.48
Chapter 12
369
P12.49
From the freebody diagram, the angle T makes with the rod is
T 20
= 60.0+20.0 = 80.0
and the perpendicular component of T is T sin 80.0. Summing torques around the base of the rod,
= 0: Fx = 0 :
Fy = 0 :
 4.00 m 10 000 N cos 60+T 4.00 m sin 80 = 0 T=
a
b10 000 Ng cos 60.0 =
sin 80.0
fb
g
a
f
10 000 N FV 60 FH
5.08 10 3 N
FH  T cos 20.0 = 0
FH = T cos 20.0 = 4.77 10 3 N
FV + T sin 20.010 000 N = 0
FIG. P12.49
and FV = 10 000 N  T sin 20.0 = 8.26 10 3 N P12.50 Choosing the origin at R, (1) (2) (3) T 90
b
g
Fx = + R sin 15.0T sin = 0 Fy = 700  R cos 15.0+T cos = 0 = 700 cos a0.180f + T b0.070 0g = 0
1 800 sin cos sin 15.0
R 15.0
Solve the equations for from (3), T = 1 800 cos from (1), R =
n
18.0 cm 25.0 cm
1 800 sin cos cos 15.0 Then (2) gives 700  + 1 800 cos 2 = 0 sin 15.0 or Let cos 2 + 0.388 9  3.732 sin cos = 0 u = cos 2 then using the quadratic equation, u = 0.011 65 or 0.869 3 Squaring, cos 4  0.880 9 cos 2 + 0.010 13 = 0
FIG. P12.50
Only the second root is physically possible, = cos 1 0.869 3 = 21.2 T = 1.68 10 3 N P12.51 and R = 2.34 10 3 N
Choosing torques about R, with 
= 0
2L L 350 N + T sin 12.0  200 N L = 0 . 2 3
a
f a
fFGH IJK a
f
From which, T = 2.71 kN . Let R x = compression force along spine, and from R x = Tx = T cos 12.0 = 2.65 kN .
Fx = 0
FIG. P12.51
370 P12.52
Static Equilibrium and Elasticity
(a)
Just three forces act on the rod: forces perpendicular to the sides of the trough at A and B, and its weight. The lines of action of A and B will intersect at a point above the rod. They will have no torque about this point. The rod's weight will cause a torque about the point of intersection as in Figure 12.52(a), and the rod will not be in equilibrium unless the center of the rod lies vertically below the intersection point, as in Figure 12.52(b). All three forces must be concurrent. Then the line of action of the weight is a diagonal of the rectangle formed by the trough and the normal forces, and the rod's center of gravity is vertically above the bottom of the trough. In Figure (b), AO cos 30.0 = BO cos 60.0 and L2 = AO + BO = AO + AO AO = L 1+
cos 2 30.0 cos 2 60.0 2 2 2 2
B A Fg
O
FIG. P12.52(a)
(b)
F cos GH cos
2
30.0 2 60.0
I JK
A 30.0
O
B Fg 60.0
L = 2
So cos = P12.53 (a)
AO 1 = and = 60.0 . 2 L
FIG. P12.52(b)
Locate the origin at the bottom left corner of the cabinet and let x = distance between the resultant normal force and the front of the cabinet. Then we have
Fx = 200 cos 37.0 n = 0 Fy = 200 sin 37.0+n  400 = 0 = na0.600  xf  400a0.300f + 200 sin 37.0 a0.600 f
200 cos 37.0 0.400 = 0
From (2), From (3), n = 400  200 sin 37.0 = 280 N 72.2  120 + 280 0.600  64.0 280 x = 20.1 cm to the left of the front edge x=
(1) (2)
a
f
(3)
a
f
From (1), (b)
k =
200 cos 37.0 = 0.571 280 FIG. P12.53
In this case, locate the origin x = 0 at the bottom right corner of the cabinet. Since the cabinet is about to tip, we can use = 0 to find h:
= 400a0.300f  a300 cos 37.0fh = 0
h=
120 = 0.501 m 300 cos 37.0
Chapter 12
371
P12.54
(a), (b) Use the first diagram and sum the torques about the lower front corner of the cabinet. = 0  F 1.00 m + 400 N 0.300 m = 0
0.300 m F 400 N 1.00 m
f a fa a400 Nfa0.300 mf = 120 N yielding F =
1.00 m Fx = 0  f + 120 N = 0 , or so
a
f
Fy = 0 400 N + n = 0 ,
f 120 N Thus, s = = = 0.300 . n 400 N (c)
f = 120 N n = 400 N
n
f
Apply F at the upper rear corner and directed so + = 90.0 to obtain the largest possible lever arm.
F'
= tan 1
FG 1.00 m IJ = 59.0 H 0.600 m K
2
Thus, = 90.059.0 = 31.0 . Sum the torques about the lower front corner of the cabinet: F
2
1.00 m 400 N
f n 0.600 m
a1.00 mf + a0.600 mf + a400 Nfa0.300 mf = 0
120 N m = 103 N . so F = 1.17 m Therefore, the minimum force required to tip the cabinet is 103 N applied at 31.0 above the horizontal at the upper left corner . P12.55 (a) We can use Then
FIG. P12.54
Fx = Fy = 0 and = 0 with pivot point at
P T L/2
the contact on the floor.
Fx = T  sn = 0 ,
Fy = n  Mg  mg = 0, and
= MgaL cos f + mg G 2 cos J  T aL sin f = 0 H K
Solving the above equations gives M= m 2 s sin  cos 2 cos  s sin
FL
I
Mg L/2 mg n
FG H
IJ K
f
FIG. P12.55
This answer is the maximum vaue for M if s < cot . If s cot , the mass M can increase without limit. It has no maximum value, and part (b) cannot be answered as stated either. In the case s < cot , we proceed. (b) At the floor, we have the normal force in the ydirection and frictional force in the xdirection. The reaction force then is R = n 2 + sn
b g
2
2
=
a M + mf g
a
2 1 + s .
At point P, the force of the beam on the rope is F = T 2 + Mg
b g
2 = g M2 + s M + m
f
2
.
372 P12.56
Static Equilibrium and Elasticity
(a)
The height of pin B is
1000 N B 10.0 m nA 30.0 45.0 C A nC
a10.0 mf sin 30.0 = 5.00 m .
The length of bar BC is then BC = 5.00 m = 7.07 m. sin 45.0
Consider the entire truss:
FIG. P12.56(a)
Fy = n A  1 000 N + nC = 0 A = b1 000 N g10.0 cos 30.0+nC 10.0 cos 30.0+7.07 cos 45.0
Which gives nC = 634 N . Then, n A = 1 000 N  nC = 366 N . (b) Suppose that a bar exerts on a pin a force not along the length of the bar. Then, the pin exerts on the bar a force with a component perpendicular to the bar. The only other force on the bar is the pin force on the other end. For F = 0 , this force must also have a component perpendicular to the bar. Then, the total torque on the bar is not zero. The contradiction proves that the bar can only exert forces along its length. Joint A:
=0
FIG. P12.56(b)
(c)
CAB A TAC nA = 366 N
Fy = 0 : C AB sin 30.0+366 N = 0 ,
so C AB = 732 N
Fx = 0 : C AB cos 30.0+TAC = 0
TAC = 732 N cos 30.0 = 634 N Joint B:
30.0
a
f
1000 N B 45.0 CBC
Fx = 0 : 732 N cos 30.0CBC cos 45.0 = 0
C BC =
a732 Nf cos 30.0 =
cos 45.0
a
f
CAB = 732 N
897 N
FIG. P12.56(c)
Chapter 12
373
P12.57
From geometry, observe that cos = 1 4 and
= 75.5
For the left half of the ladder, we have
Fx = T  R x = 0 Fy = R y + n A  686 N = 0 top = 686 Na1.00 cos 75.5f + T a2.00 sin 75.5f
n A 4.00 cos 75.5 = 0
For the right half of the ladder we have
(1) (2) (3) FIG. P12.57
a
f
Fx = Rx  T = 0 Fy = nB  Ry = 0 top = nB a 4.00 cos 75.5f  T a 2.00 sin 75.5f = 0
Solving equations 1 through 5 simultaneously yields: (a) (b) (c) T = 133 N n A = 429 N R x = 133 N and and n B = 257 N R y = 257 N
(4) (5)
The force exerted by the left half of the ladder on the right half is to the right and downward. P12.58 (a) x CG = = yCG =
b1 000 kg g10.0 m + b125 kgg0 + b125 kg g0 + b125 kg g20.0 m =
1 375 kg 1 375 kg
m i xi mi
9.09 m
b1 000 kg g10.0 m + b125 kgg20.0 m + b125 kgg20.0 m + b125 kg g0
= 10.9 m (b) By symmetry, x CG = 10.0 m There is no change in yCG = 10.9 m (c) P12.59 vCG =
FG 10.0 m  9.09 m IJ = H 8.00 s K
0.114 m s
Considering the torques about the point at the bottom of the bracket yields:
b0.050 0 mga80.0 Nf  Fb0.060 0 mg = 0 so
F = 66.7 N .
374 P12.60
Static Equilibrium and Elasticity
When it is on the verge of slipping, the cylinder is in equilibrium. and f1 = n 2 = sn1 f 2 = sn 2 Fx = 0 :
Fy = 0 : = 0:
P + n1 + f 2 = Fg P = f1 + f 2 and and becomes n 2 n1 = 2 4 n n 3 P = 1 + 1 = n1 2 4 4 5 4 P+ P = Fg 4 3 f2 =
As P grows so do f1 and f 2 n 1 Therefore, since s = , f1 = 1 2 2 n then (1) P + n1 + 1 = Fg 4 5 So P + n1 = Fg 4 3 Therefore, P = Fg 8 P12.61 (a)
FIG. P12.60 (2) or 8 P = Fg 3
FG IJ H K
F = k L , Young's modulus is Y =
Thus, Y =
L 0
a f
F A L Li
=
FLi A L
a f
a L f
2 Li
2
kLi YA and k = A Li
L 0
(b) P12.62 (a)
W =  Fdx = 
z
za
 kx dx =
f
YA Li
L 0
z
xdx = YA
Take both balls together. Their weight is 3.33 N and their CG is at their contact point.
Fx = 0 : + P3  P1 = 0 P2 = 3.33 N Fy = 0 : + P2  3.33 N = 0 A = 0:  P3 R + P2 R  3.33 NaR + R cos 45.0f
+ P1 R + 2 R cos 45.0 = 0
P1
a
f
3.33 N P3 Fg
Substituting,
 P1 R + 3.33 N R  3.33 N R 1 + cos 45.0
a
a3.33 Nf cos 45.0 = 2 P cos 45.0
1
fa + P Ra1 + 2 cos 45.0f = 0
1
f a
f
P2
P1 = 1.67 N so P3 = 1.67 N (b) Take the upper ball. The lines of action of its weight, of P1 , and of the normal force n exerted by the lower ball all go through its center, so for rotational equilibrium there can be no frictional force.
FIG. P12.62(a)
1.67 N
P1
Fx = 0 : n cos 45.0 P1 = 0
n=
n cos 45.0
1.67 N = 2.36 N cos 45.0 Fy = 0 : n sin 45.01.67 N = 0 gives the same result
n sin 45.0
FIG. P12.62(b)
Chapter 12
375
P12.63
Fy = 0 :
+380 N  Fg + 320 N = 0 Fg = 700 N
Take torques about her feet:
= 0:
380 N 2.00 m + 700 N x + 320 N 0 = 0
x = 1.09 m FIG. P12.63
a
f a
f a
f
P12.64
The tension in this cable is not uniform, so this becomes a fairly difficult problem. dL F = L YA At any point in the cable, F is the weight of cable below that point. Thus, F = gy where is the mass per unit length of the cable. Then, y =
Li 0
z FGH
2 g i 1 gLi dL dy = ydy = 2 YA L YA 0 2
IJ K
L
z
y =
P12.65
(a)
a fa fa f e je j a10.0  1.00f m s = F v I F = mG J = b1.00 kg g H t K 0.002 s
a fa f
2. 40 9.80 500 1 = 0.049 0 m = 4.90 cm 2 2.00 10 11 3.00 10 4 4 500 N
(b) (c)
stress =
4 500 N F = = 4.50 10 6 N m 2 0.010 m 0.100 m A
Yes . This is more than sufficient to break the board.
376 P12.66
Static Equilibrium and Elasticity
The CG lies above the center of the bottom. Consider a disk of water at height y above the bottom. Its radius is y fFGH 30.0y cm IJK = 25.0 cm + 3 yI yI yI F F F Its area is G 25.0 cm + J . Its volume is G 25.0 cm + J dy and its mass is G 25.0 cm + J H K H K H 3 3 3K 25.0 cm + 35.0  25.0 cm
a
2
2
2
dy . The
whole mass of the water is
M=
30 .0 cm y =0
z
dm =
30 .0 cm 0
z
625 +
2 3
L 50.0 y y O + P M = M625 y + 6 27 P MN Q L 50.0a30.0f a30.0f OP + M = M625a30.0f + 6 27 P MN Q M = e10 kg cm je 27 250 cm j = 85.6 kg
30.0 0 2 3 3 3 3
F GH
50.0 y y 2 + dy 3 9
I JK
The height of the center of gravity is yCG =
30 .0 cm y =0
z
ydm M
=
30 .0 cm 0
z
LM OP MN PQ .0 L 625a30.0f = MM 2 + 50.0a30.0f + a3036 f 9 MN e10 kg cm j 453 750 cm =
625 y 2 50.0 y 3 y 4 = + + 2 9 36 M
2 30 .0 cm 0 3 3 3 4
F 625y + 50.0 y GH 3
2
+
y 3 dy 9 M
I JK
4
OP PQ
yCG
M 1.43 10 3 kg cm = = 16.7 cm 85.6 kg
Chapter 12
377
P12.67
Let represent the angle of the wire with the vertical. The radius of the circle of motion is r = 0.850 m sin . For the mass:
a
f
r
T
mg
v2 = mr 2 r T sin = m 0.850 m sin 2
Fr = mar = m
a
f
T Further, = Y strain or T = AY strain A Thus, AY strain = m 0.850 m 2 , giving
4
a f f a f a f e3.90 10 mj e7.00 10 N m je1.00 10 j AY astrainf = = ma0.850 mf b1.20 kg ga0.850 mf a
2 10 2 3
FIG. P12.67
or P12.68
= 5.73 rad s .
B D
For the bridge as a whole:
A = n A a0f  a13.3 kNfa100 mf + nE a 200 mf = 0
so
nE =
a13.3 kNfa100 mf =
200 m
6.66 kN
A C
nA
E
100 m 100 m 13.3 kN
nE
Fy = n A  13.3 kN + n E = 0 gives
n A = 13.3 kN  n E = 6.66 kN At Pin A:
FAB =
6.66 kN = 10.4 kN compression sin 40.0 Fx = FAC  10.4 kN cos 40.0 = 0 so
Fy =  FAB sin 40.0+6.66 kN = 0 or
FAB
FAC = 10.4 kN cos 40.0 = 7.94 kN tension At Pin B:
a
f
a
f
b
g f
40.0
40.0 FAC nA = 6.66 kN
a
Fy = a10.4 kN f sin 40.0 FBC sin 40.0 = 0
FBD 40.0 FBC FAB = 10.4 kN
Thus, FBC = 10. 4 kN tension
a
f b g
Fx = FAB cos 40.0+ FBC cos 40.0 FBD = 0
FBD = 2 10.4 kN cos 40.0 = 15.9 kN compression By symmetry: FDE = FAB = 10.4 kN compression FDC = FBC and FEC = FAC
a
f
b g = 10.4 kN atensionf = 7.94 kN atensionf
10.4 kN 40.0 7.94 kN
10.4 kN 40.0 7.94 kN
We can check by analyzing Pin C:
13.3 kN
Fx = +7.94 kN  7.94 kN = 0 or 0 = 0 Fy = 2a10.4 kNf sin 40.013.3 kN = 0
which yields 0 = 0 .
FIG. P12.68
378 P12.69
Static Equilibrium and Elasticity
Member AC is not in pure compression or tension. It also has shear forces present. It exerts a downward force S AC and a tension force FAC on Pin A and on Pin C. Still, this member is in equilibrium. Fx = FAC  FAC = 0 FAC = FAC A = 0: a14.7 kNfa25.0 mf + S AC a50.0 mf = 0 or
SAC FAC A
25.0 m
25.0 m
SAC
FAC C 14.7 kN
S = 7.35 kN AC Fy = S AC  14.7 kN + 7.35 kN = 0 S AC = 7.35 kN
B
D
Then S AC = S and we have proved that the loading by the car A AC is equivalent to onehalf the weight of the car pulling down on n A each of pins A and C, so far as the rest of the truss is concerned. For the Bridge as a whole:
E C
75.0 m 25.0 m 14.7 kN nE
A = 0:
 14.7 kN 25.0 m + n E 100 m = 0 n E = 3.67 kN
a
fa
f a
f
7.35 kN
FAB 30.0 FAC nA = 11.0 kN
Fy = n A  14.7 kN + 3.67 kN = 0
n A = 11.0 kN At Pin A:
Fy = 7.35 kN + 11.0 kN  FAB sin 30.0 = 0 Fx = FAC  a7.35 kN f cos 30.0 = 0
FAC = 6.37 kN tension FAB = 7.35 kN compression
FBD 30.0 7.35 kN 60.0 FBC
b
g
a
f
At Pin B:
4.24 kN
Fy = a7.35 kN f sin 30.0 FBC sin 60.0 = 0
FBC = 4.24 kN tension
x
FCD 60.0 FCE
60.0 6.37 kN
a f F = a7.35 kNf cos 30.0+a 4.24 kN f cos 60.0 F F = 8.49 kN bcompressiong
BD
BD
=0
7.35 kN
At Pin C:
Fy = a 4.24 kNf sin 60.0+ FCD sin 60.07.35 kN = 0 Fx = 6.37 kN  a4.24 kNf cos 60.0+a 4.24 kNf cos 60.0+ FCE = 0
FCE = 6.37 kN tension FCD = 4.24 kN tension
FDE 30.0 6.37 kN 3.67 kN
a
f
a
f
At Pin E:
FIG. P12.69
Fy =  FDE sin 30.0+3.67 kN = 0
FDE = 7.35 kN compression
b
g
or Fx = 6.37 kN  FDE cos 30.0 = 0 which gives FDE = 7.35 kN as before.
Chapter 12
379
P12.70
(1) (2)
ph = I p = MvCM 2 I I I = = = R 5 p MvCM MR If the acceleration is a, we have Px = ma and Py + n  Fg = 0 . Taking the origin at the center of gravity, the torque equation gives Py L  d + Px h  nd = 0 . Solving these equations, we find
d
p h vCM
If the ball rolls without slipping, R = vCM So, h = P12.71 (a) FIG. P12.70
H L
a
f
CG P h
(b) (c)
F d  ah I . L G H g JK ah e 2.00 m s ja1.50 mf = = If P = 0 , then d =
Py = Fg
2 y
n
Fy g
FIG. P12.71 0.306 m .
g
9.80 m s 2
Using the given data, Px = 306 N and Py = 553 N . Thus, P = 306 i + 553 j N .
e
j
*P12.72
When the cyclist is on the point of tipping over forward, the normal force on the rear wheel is zero. Parallel to the plane we have f1  mg sin = ma . Perpendicular to the plane, n1  mg cos = 0 . Torque about the center of mass:
mg
mg 0  f1 1.05 m + n1 0.65 m = 0 .
Combining by substitution, ma = f1  mg sin = a = g cos 20
af a
f a
f
f1
n1
FIG. P12.72 n1 0.65 m 0.65 m  mg sin = mg cos  mg sin 1.05 m 1.05 m
*P12.73
0.65  sin 20 = 2.35 m s 2 1.05 When the car is on the point of rolling over, the normal force on its inside wheels is zero.
FG H
IJ K
Fy = ma y :
n  mg = 0 f= mv 2 R d = 0. 2 v max = gdR 2h
mg h f d FIG. P12.73 mg
Fx = ma x :
Take torque about the center of mass: fh  n Then by substitution
2 mv max
R
h
mgd =0 2
A wider wheelbase (larger d) and a lower center of mass (smaller h) will reduce the risk of rollover.
380
Static Equilibrium and Elasticity
ANSWERS TO EVEN PROBLEMS
P12.2 Fy + R y  Fg = 0 ; Fx  R x = 0 ; Fy cos  Fg P12.4 P12.6 P12.8 P12.10 P12.12 P12.14
FG IJ cos  F H 2K
P12.40
x
sin = 0 P12.42 P12.44
(a) 0.400 mm; (b) 40.0 kN; (c) 2.00 mm; (d) 2.40 mm; (e) 48.0 kN at A: Mg sin sin ; at B: Mg sin + sin +
see the solution 0.750 m
b
g
b
g
a2.54 m, 4.75 mf
(a) 9.00 g; (b) 52.5 g; (c) 49.0 g (a) 392 N; (b) 339 i + 0 j N (a) f = ng P12.46 P12.48 P12.50 P12.52 P12.54 P12.56
(a) 160 N to the right; (b) 13.2 N to the right; (c) 292 N up; (d) 192 N 1.46 kN ; 1.33 i + 2.58 j kN 0.789 T = 1.68 kN ; R = 2.34 kN; = 21.2 (a) see the solution; (b) 60.0 (a) 120 N; (b) 0.300; (c) 103 N at 31.0 above the horizontal to the right (a), (b) see the solution; (c) C AB = 732 N ; TAC = 634 N ; C BC = 897 N (a) 9.09 m, 10.9 m ; (b) 10.0 m, 10.9 m ; (c) 0.114 m s to the right 3 Fg 8 (a) P1 = 1.67 N ; P2 = 3.33 N ; P3 = 1.67 N ; (b) 2.36 N 4.90 cm 16.7 cm above the center of the bottom C AB = 10.4 kN ; TAC = 7.94 kN ; TBC = 10.4 kN ; C BD = 15.9 kN ; C DE = 10.4 kN ; TDC = 10. 4 kN ; TEC = 7.94 kN 2 R 5 2.35 m s 2
e
j
e
j
LM m g + m gx OP cot ; N2 L Q e + j cot = bm + m g g ; (b) = m +m
1 2 m1 2 m2d L 1 2 1 2
P12.16 P12.18 P12.20 P12.22
see the solution; 0.643 m 36.7 N to the left ; 31.2 N to the right (a) 35.5 kN; (b) 11.5 kN to the right; (c) 4.19 kN down (a) 859 N; (b) 104 kN at 36.9 above the horizontal to the left 3L 4 (a) see the solution; (b) decreases ; (c) R decreases (a) 73.6 kN; (b) 2.50 mm ~ 1 cm 9.85 10 5 0.029 3 mm (a) 0.053 8 m3 ; (b) 1.09 10 3 kg m3 ; (c) Yes, in most practical circumstances (a) 53.1; (b) 1.04 kN; (c) 0.126 m, 51.2; (d) 1.07 kN; (e) 0.129 m, 51.1; (f) 51.1 P12.70 P12.72
P12.58
a
f
a
f
P12.60 P12.62
P12.24 P12.26 P12.28 P12.30 P12.32 P12.34 P12.36
P12.64 P12.66 P12.68
P12.38
13
Universal Gravitation
CHAPTER OUTLINE
13.1 13.2 13.3 13.4 13.5 13.6 13.7 Newton's Law of Universal Gravitation Measuring the Gravitational Constant FreeFall Acceleration and the Gravitational Force Kepler's Laws and the Motion of Planets The Gravitational Field Gravitational Potential Energy Energy Considerations in Planetary and Satellite Motion
ANSWERS TO QUESTIONS
Q13.1 Because g is the same for all objects near the Earth's surface. The larger mass needs a larger force to give it just the same acceleration. To a good first approximation, your bathroom scale reading is unaffected because you, the 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 weigh slightly less at noon and at midnight than you do at sunrise or sunset. The Sun's gravitational field is a little weaker at the center of the Earth than at the surface subsolar 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.
Q13.2
Q13.3
Kepler's second law states that the angular momentum of the Earth is constant as the Earth orbits the sun. Since L = mr , as the orbital radius decreases from June to December, then the orbital speed must increase accordingly. Because both the Earth and Moon are moving in orbit about the Sun. As described by Fgravitational = ma centripetal , the gravitational force of the Sun merely keeps the Moon (and Earth) in a nearly circular orbit of radius 150 million kilometers. Because of its velocity, the Moon is kept in its orbit about the Earth by the gravitational force of the Earth. There is no imbalance of these forces, at new moon or full moon.
Q13.4
Q13.5
Air resistance causes a decrease in the energy of the satelliteEarth system. This reduces the diameter of the orbit, bringing the satellite closer to the surface of the Earth. A satellite in a smaller orbit, however, must travel faster. Thus, the effect of air resistance is to speed up the satellite! Kepler's third law, which applies to all planets, tells us that the period of a planet is proportional to r 3 2 . Because Saturn and Jupiter are farther from the Sun than Earth, they have longer periods. The Sun's gravitational field is much weaker at a distant Jovian planet. Thus, an outer planet experiences much smaller centripetal acceleration than Earth and has a correspondingly longer period.
Q13.6
381
382 Q13.7
Universal Gravitation
Ten terms are needed in the potential energy: U = U 12 + U 13 + U 14 + U 15 + U 23 + U 24 + U 25 + U 34 + U 35 + U 45 . With N particles, you need
ai  1f =
i =1
N
N2  N terms. 2
Q13.8
No, the escape speed does not depend on the mass of the rocket. If a rocket is launched at escape speed, then the total energy of the rocketEarth system will be zero. When the separation distance GM E m 1 = 0 , the mass becomes infinite U = 0 the rocket will stop K = 0 . In the expression mv 2  2 r m of the rocket divides out.
a
f
a
f
Q13.9
It takes 100 times more energy for the 10 5 kg spacecraft to reach the moon than the 10 3 kg spacecraft. Ideally, each spacecraft can reach the moon with zero velocity, so the only term that need be analyzed is the change in gravitational potential energy. U is proportional to the mass of the spacecraft. The escape speed from the Earth is 11.2 km/s and that from the Moon is 2.3 km/s, smaller by a factor of 5. The energy requiredand fuelwould be proportional to v 2 , or 25 times more fuel is required to leave the Earth versus leaving the Moon. The satellites used for TV broadcast are in geosynchronous orbits. The centers of their orbits are the center of the Earth, and their orbital planes are the Earth's equatorial plane extended. This is the plane of the celestial equator. The communication satellites are so far away that they appear quite close to the celestial equator, from any location on the Earth's surface. For a satellite in orbit, one focus of an elliptical orbit, or the center of a circular orbit, must be located at the center of the Earth. If the satellite is over the northern hemisphere for half of its orbit, it must be over the southern hemisphere for the other half. We could share with Easter Island a satellite that would look straight down on Arizona each morning and vertically down on Easter Island each evening. The absolute value of the gravitational potential energy of the EarthMoon system is twice the kinetic energy of the moon relative to the Earth. In a circular orbit each increment of displacement is perpendicular to the force applied. The dot product of force and displacement is zero. The work done by the gravitational force on a planet in an elliptical orbit speeds up the planet at closest approach, but negative work is done by gravity and the planet slows as it sweeps out to its farthest distance from the Sun. Therefore, net work in one complete orbit is zero. Every point q on the sphere that does not lie along the axis connecting the center of the sphere and the particle will have companion point q' for which the components of the gravitational force perpendicular to the axis will cancel. Point q' can be found by rotating the sphere through 180 about the axis. The forces will not necessarily cancel if the mass is not uniformly distributed, unless the center of mass of the nonuniform sphere still lies along the axis.
q
Q13.10
Q13.11
Q13.12
Q13.13 Q13.14
Q13.15
Fpq Fpq
q' (behind the sphere)
p
FIG. Q13.15
Chapter 13
383
Q13.16 Q13.17
Speed is maximum at closest approach. Speed is minimum at farthest distance. Set the universal description of the gravitational force, Fg = Fg = ma gravitational , where M X and R X mass of a "test particle." Divide both sides by m. , equal to the local description, 2 RX are the mass and radius of planet X, respectively, and m is the GM X m
Q13.18
The gravitational force of the Earth on an extra particle at its center must be zero, not infinite as one interpretation of Equation 13.1 would suggest. All the bits of matter that make up the Earth will pull in different outward directions on the extra particle. Cavendish determined G. Then from g = GM , one may determine the mass of the Earth. R2
Q13.19 Q13.20
The gravitational force is conservative. An encounter with a stationary mass cannot permanently speed up a spacecraft. Jupiter is moving. A spacecraft flying across its orbit just behind the planet will gain kinetic energy as the planet's gravity does net positive work on it. Method one: Take measurements from an old kinescope of Apollo astronauts on the moon. From the motion of a freely falling object or from the period of a swinging pendulum you can find the acceleration of gravity on the moon's surface and calculate its mass. Method two: One could determine the approximate mass of the moon using an object hanging from an extremely sensitive balance, with knowledge of the position and distance of the moon and the radius of the Earth. First weigh the object when the moon is directly overhead. Then weigh of the object when the moon is just rising or setting. The slight difference between the measured weights reveals the cause of tides in the Earth's oceans, which is a difference in the strength of the moon's gravity between different points on the Earth. Method three: Much more precisely, from the motion of a spacecraft in orbit around the moon, its mass can be determined from Kepler's third law. The spacecraft did not have enough fuel to stop dead in its highspeed course for the Moon.
Q13.21
Q13.22
SOLUTIONS TO PROBLEMS
Section 13.1 P13.1 Newton's Law of Universal Gravitation
For two 70kg persons, modeled as spheres, Fg = Gm1 m 2 r2 Gm1 m 2 r
2
e6.67 10 =
11
N m 2 kg 2 70 kg 70 kg
a2 mf
jb
gb
g
2
~ 10 7 N .
P13.2
F = m1 g = Gm 2 r
2
g=
=
e6.67 10
11
N m 2 kg 2 4.00 10 4 10 3 kg
2
je a100 mf
j=
2.67 10 7 m s 2
384 P13.3
Universal Gravitation
(a)
At the midpoint between the two objects, the forces exerted by the 200kg and 500kg objects are oppositely directed, and from we have Fg = Gm1 m 2 r2 G 50.0 kg 500 kg  200 kg
F =
b
a0.200 mf
gb
2
g=
2.50 10 5 N toward the 500kg object.
(b)
At a point between the two objects at a distance d from the 500kg objects, the net force on the 50.0kg object will be zero when G 50.0 kg 200 kg
2
b gb g = Gb50.0 kggb500 kg g d a0.400 m  df
2
or P13.4 m1 + m 2 = 5.00 kg F =G
d = 0.245 m m 2 = 5.00 kg  m1 m1 m 2 r
2
1.00 10 8 N = 6.67 10 11 N m 2 kg 2
e
b5.00 kg gm
or
2 1  m1 =
e1.00 10
8
N 0.040 0 m 2 N m kg
2 2
je
6.67 10
11
j = 6.00 kg
j m ba5..00 kg  m g 0 200 mf
1 1 2 2
2 Thus, m1  5.00 kg m1 + 6.00 kg = 0 1 1
g bm  3.00 kg gbm
b
 2.00 kg = 0
g
giving m1 = 3.00 kg, so m 2 = 2.00 kg . The answer m1 = 2.00 kg and m 2 = 3.00 kg is physically equivalent. P13.5 The force exerted on the 4.00kg mass by the 2.00kg mass is directed upward and given by F24 = G m4m2
2 r24
j = 6.67 10 11 N m 2 kg 2
e
2 j b4.00a3kg gbm.00 kg g j .00 f
2
= 5.93 10 11 j N The force exerted on the 4.00kg mass by the 6.00kg mass is directed to the left F64 = G m 4 m6
2 r64
e ij = e6.67 10
11
N m 2 kg 2
6 j b4.00a4kg gbm.00 kg g i .00 f
2
FIG. P13.5
= 10.0 10 11 iN Therefore, the resultant force on the 4.00kg mass is F4 = F24 + F64 =
e10.0i + 5.93 jj 10
11
N .
Chapter 13
385
P13.6
(a)
The SunEarth distance is 1.496 10 11 m and the EarthMoon distance is 3.84 10 8 m , so the distance from the Sun to the Moon during a solar eclipse is 1.496 10 11 m  3.84 10 8 m = 1.492 10 11 m The mass of the Sun, Earth, and Moon are M S = 1.99 10 30 kg M E = 5.98 10 24 kg and Gm1 m 2 r2
11 30
M M = 7.36 10 22 kg
We have FSM =
e6.67 10 je1.99 10 je7.36 10 j = = e1.492 10 j
22 11 2
4.39 10 20 N
(b)
FEM =
e6.67 10
11
N m 2 kg 2 5.98 10 24 7.36 10 22
je
je
e3.84 10 j
11
8 2
j=
1.99 10 20 N
(c)
FSE
e6.67 10 =
N m 2 kg 2 1.99 10 30 5.98 10 24
je
je
e1.496 10 j
11 2
j=
3.55 10 22 N
Note that the force exerted by the Sun on the Moon is much stronger than the force of the Earth on the Moon. In a sense, the Moon orbits the Sun more than it orbits the Earth. The Moon's path is everywhere concave toward the Sun. Only by subtracting out the solar orbital motion of the EarthMoon system do we see the Moon orbiting the center of mass of this system.
Section 13.2
Measuring the Gravitational Constant
P13.7
1.50 kg 15.0 10 3 kg GMm 11 2 2 F= = 6.67 10 N m kg = 7.41 10 10 N 2 2 r2 4.50 10 m
e
j
b
ge
j
e
j
386 P13.8
Universal Gravitation
Let represent the angle each cable makes with the vertical, L the cable length, x the distance each ball scrunches in, and d = 1 m the original distance between them. Then r = d  2 x is the separation of the balls. We have
Fy = 0 :
T cos  mg = 0 T sin  tan = Gmm =0 r2 x L x
2 2
Fx = 0 :
Then
FIG. P13.8 =
Gmm r 2 mg
g d  2x
a
Gm
f
2
x d  2x
a
f
2
=
Gm 2 L  x2 . g
The factor small.
Gm is numerically small. There are two possibilities: either x is small or else d  2 x is g
Possibility one: We can ignore x in comparison to d and L, obtaining
e6.67 10 xa1 mf =
2
11
N m 2 kg 2 100 kg
2
e9.8 m s j e
jb
g 45 m
x = 3.06 10 8 m.
The separation distance is r = 1 m  2 3.06 10 8 m = 1.000 m  61.3 nm . Possibility two: If d  2 x is small, x 0.5 m and the equation becomes m a0.5 mfr = e6.67 10 b9N N kgkg jb100 kgg a45 mf  a0.5 mf .8 g
11 2 2 2 2 2
j
r = 2.74 10 4 m .
For this answer to apply, the spheres would have to be compressed to a density like that of the nucleus of atom.
Section 13.3 P13.9 a=
FreeFall Acceleration and the Gravitational Force
b 4R g
E
MG
2
=
9.80 m s 2 = 0.613 m s 2 16.0
toward the Earth.
P13.10 If then
GM G 3 g= 2 = R R2 gM 1 = = gE 6 g M = M gE E
e j = 4 GR
4 R 3
3
F I F R I = FG 1 IJ a4f = GH JK GH R JK H 6 K
E M
4G M R M 3 4G E RE 3
2 . 3
Chapter 13
387
P13.11
(a)
At the zerototal field point, so rM = rE
GmM E
2 rE
=
GmM M
2 rM
MM r 7.36 10 22 = rE = E 24 ME 9.01 5.98 10 r 8 rE + rM = 3.84 10 m = rE + E 9.01 3.84 10 8 m rE = = 3.46 10 8 m 1.11
(b)
At this distance the acceleration due to the Earth's gravity is gE = GM E
2 rE
e6.67 10 =
11
N m 2 kg 2 5.98 10 24 kg
e3.46 10
8
je mj
j
2
g E = 3.34 10 3 m s 2 directed toward the Earth
Section 13.4 P13.12 (a) (b)
Kepler's Laws and the Motion of Planets v=
3 2r 2 384 400 10 m = = 1.02 10 3 m s . T 27.3 86 400 s
b
b
g
g
In one second, the Moon falls a distance
3 1 1 v 2 2 1 1.02 10 x = at 2 = t = 1.00 2 2 r 2 3.844 10 8
e e
j a f j
2
2
= 1.35 10 3 m = 1.35 mm .
The Moon only moves inward 1.35 mm for every 1020 meters it moves along a straightline path. P13.13 Applying Newton's 2nd Law, GMM =
F = ma yields Fg = ma c for each star:
or M= 4v 2 r . G
a 2r f
2
Mv 2 r
We can write r in terms of the period, T, by considering the time and distance of one complete cycle. The distance traveled in one orbit is the circumference of the stars' common orbit, so 2r = vT . Therefore M= 4v 2 r 4v 2 vT = G G 2
FG IJ H K
FIG. P13.13
3 2 v 3 T 2 220 10 m s 14.4 d 86 400 s d = = 1.26 10 32 kg = 63.3 solar masses so, M = 11 2 2 G 6.67 10 N m kg
e
e
ja
3
fb
j
g
388 P13.14
Universal Gravitation
Since speed is constant, the distance traveled between t1 and t 2 is equal to the distance traveled between t 3 and t 4 . The area of a triangle is equal to onehalf its (base) width across one side times its (height) dimension perpendicular to that side. So 1 1 bv t 2  t1 = bv t 4  t 3 2 2
b
g
b
g
states that the particle's radius vector sweeps out equal areas in equal times. P13.15 T2 = 4 2 a 3 GM 4 2 a 3 GT 2 = (Kepler's third law with m << M ) 4 2 4.22 10 8 m
M=
e
e
6.67 10 11 N m 2 kg 2 1.77 86 400 s
jb
j
3
g
2
= 1.90 10 27 kg
(Approximately 316 Earth masses) P13.16 By conservation of angular momentum for the satellite, rp v p = ra v a vp va = ra 2 289 km + 6.37 10 3 km 8 659 km = = = 1.27 . 6 829 km rp 459 km + 6.37 10 3 km
We do not need to know the period. P13.17 By Kepler's Third Law, T 2 = ka 3 (a = semimajor axis)
For any object orbiting the Sun, with T in years and a in A.U., k = 1.00 . Therefore, for Comet Halley
a75.6f = a1.00fFGH 0.570 + y IJK 2
2
3
The farthest distance the comet gets from the Sun is y = 2 75.6
a f
FIG. P13.17
23
 0.570 = 35.2 A. U. (out around the orbit of Pluto) Gm planet M star r2 m planet v 2 r
P13.18
F = ma :
=
GM star = v 2 = r 2 2 r 3 3 GM star = r 3 3 = rx 2 = ry 2 x y
y =x
Fr I GH r JK
x y
3 2
y =
F 90.0 I 3 GH 5.00 yr JK
3 2
=
468 5.00 yr
So planet Y has turned through 1.30 revolutions .
FIG. P13.18
Chapter 13
389
P13.19
dR + di
J
GM J
2
=
4 2 R J + d T
2
d
i
GM J T 2 = 4 2 R J + d
d
i
3
e6.67 10
P13.20
11
N m 2 kg 2 1.90 10 27 kg 9.84 3 600
je
jb
g
2
= 4 2 6.99 10 7 + d
e
j
3
d = 8.92 10 7 m = 89 200 km above the planet The gravitational force on a small parcel of material at the star's equator supplies the necessary centripetal force: GM s m Rs2 = mv 2 = mRs 2 Rs
so
=
GM s
3 Rs
=
e6.67 10
11
N m 2 kg 2 2 1.99 10 30 kg
e10.0 10
3
j e mj
j
3
= 1.63 10 4 rad s
*P13.21 The speed of a planet in a circular orbit is given by
F = ma :
GM sun m r2
=
mv 2 r
v=
GM sun . r
For Mercury the speed is
vM =
e6.67 10 je1.99 10 j m e5.79 10 j s
11 30 10 2
2
= 4.79 10 4 m s
and for Pluto,
vP =
e6.67 10 je1.99 10 j m e5.91 10 j s
11 30 12 2
2
= 4.74 10 3 m s .
With greater speed, Mercury will eventually move farther from the Sun than Pluto. With original distances rP and rM perpendicular to their lines of motion, they will be equally far from the Sun after time t where
2 2 2 2 rP + v P t 2 = rM + v M t 2 2 2 2 2 rP  rM = v M  v P t 2
e
j
t=
e5.91 10 e4.79 10
4
12
j  e5.79 10 m sj  e 4.74 10
m
2 2
10 3
m
j
2
ms
j
2
=
3.49 10 25 m 2 = 1.24 10 8 s = 393 yr . 2.27 10 9 m 2 s 2
390 *P13.22
Universal Gravitation
For the Earth, Then
F = ma :
GM s m r2
=
mv 2 m 2r = r r T
FG IJ H K
2
.
Also the angular momentum We eliminate GM s T 2 = 4 2
FG LT IJ H 2m K
GM s T 2 = 4 2 r 3 . 2r L = mvr = m r is a constant for the Earth. T LT r= between the equations: 2m GM sT 1 2 = 4 2
32
FG L IJ H 2m K
s
3 2
.
Now the rate of change is described by GM s
FG 1 T H2
1 2
dM s 1 2 dT +G 1 =0 T dt dt
T  t
dM s dt
dM F T I T dT IJ IJ FG = G2 J K H K dt dt H M K T FG 2 T IJ = 5 000 yrF 3.16 10 s I e3.64 10 kg sjF 2 1 yr GH 1 yr JK GH 1.991 10 H MK
s 7 9 s
30
I J kg K
T = 1.82 10 2 s
Section 13.5 P13.23 g=
The Gravitational Field Gm Gm Gm i + 2 j + 2 cos 45.0 i + sin 45.0 j 2l l2 l
e
j
y m l m
so
g= Gm l2
GM 1 1+ 2 l 2 2 2+
FG H
IJ ei + jj or K
l
g=
FG H
1 toward the opposite corner 2
IJ K
O
m
x
FIG. P13.23 P13.24 (a) 6.67 10 11 N m 2 kg 2 100 1.99 10 30 kg 10 3 kg GMm F= = = 1.31 10 17 N 2 4 r2 1.00 10 m + 50.0 m
e
j e
je
j
e
j
(b)
F =
GMm GMm  2 2 rfront rback
2 2 F GM rback  rfront = 2 2 m rfront rback 11
g =
e
j
30 4
e6.67 10 j 100e1.99 10 g = e1.00 10
g = 2.62 10 12 N kg
j LNMe1.01 10 mj  e1.00 10 mj OQP mj e1.01 10 mj
4 2 4 2 2 4 2
FIG. P13.24
Chapter 13
391
P13.25
g1 = g 2 =
g1y =  g 2y
MG r 2 + a2
g 1 x = g 2 x = g 2 cos g = 2 g2x i or g=
g y = g1y + g 2 y r cos = 2 a + r2
e j
j
3 2
e
j
12
2 MGr
e
r + a2
2
toward the center of mass FIG. P13.25
Section 13.6 P13.26 (a)
Gravitational Potential Energy 6.67 10 11 N m 2 kg 2 5.98 10 24 kg 100 kg GM E m = = 4.77 10 9 J . U= 6 r 6.37 + 2.00 10 m
e
a
je f
jb
g
(b), (c) Planet and satellite exert forces of equal magnitude on each other, directed downward on the satellite and upward on the planet. F= GM E m r
2
=
e6.67 10
GM E
2 RE
11
N m 2 kg 2 5.98 10 24 kg 100 kg
je
jb
e8.37 10 mj
6
2
g=
569 N
P13.27
U = G so that
Mm r
and
g=
FG 1  1 IJ = 2 mgR H 3R R K 3 2 U = b1 000 kg ge9.80 m s je6.37 10 mj = 3
U = GMm
E E E 2 6
4.17 10 10 J .
P13.28
The height attained is not small compared to the radius of the Earth, so U = mgy does not apply; GM 1 M 2 U= does. From launch to apogee at height h, r GM E M p GM E M p 1 +0=0 K i + Ui + Emch = K f + U f : M p vi2  2 RE RE + h 1 10.0 10 3 m s 2
e
j e
2
 6.67 10 11 N m 2 kg 2
24
=  6.67 10 11 N m 2 kg 2
e
.98 jFGH 56.3710 10 5.98 10 jFGH 6.37 10 mkgh IJK +
6
24 6
I J mK
kg
e5.00 10
7
m 2 s 2  6.26 10 7 m 2 s 2 = 3.99 10 14 m3 s 2 1. 26 10 7 m 2 s 2
j e
j
3.99 10 14 m3 s 2 6.37 10 6 m + h
6.37 10 6 m + h = h = 2.52 10 7 m
= 3.16 10 7 m
392 P13.29
Universal Gravitation
(a)
=
2 4 3 rE
MS
=
3 1.99 10 30 kg 4 6.37 10
e
e
6
j mj
3
= 1.84 10 9 kg m 3
(b)
g=
GM S
2 rE
=
e6.67 10 e
11
N m 2 kg 2 1.99 10 30 kg
e6.37 10 IJ K
6
je mj
2
j=
3.27 10 6 m s 2
(c)
Ug = 
6.67 10 11 N m 2 kg 2 1.99 10 30 kg 1.00 kg GM S m = = 2.08 10 13 J rE 6.37 10 6 m
1 2
je
jb
g
P13.30
W =  U = 
e+6.67 10 W=
P13.31 (a)
FG Gm m H r
11
0
N m 2 kg 2 7.36 10 22 kg 1.00 10 3 kg 1.74 10 m
6
je
je
j=
2.82 10 9 J
U Tot = U 12 + U 13 + U 23 = 3U 12 = 3  U Tot = 
FG H
Gm 1 m 2 r12
IJ K
3 6.67 10 11 N m 2 kg 2 5.00 10 3 kg 0.300 m
e
je
j
2
= 1.67 10 14 J
(b) *P13.32 (a)
At the center of the equilateral triangle Energy conservation of the objectEarth system from release to radius r:
eK + U j
g
altitude h
= K +Ug
e
j
radius r
GM E m GM E m 1 = mv 2  0 RE + h 2 r v = 2GM E
f f
F GH
FG 1  1 IJ I H r R + h K JK
E
12
=
dr dt
(b)
We can enter this expression directly into a mathematical calculation program. r Alternatively, to save typing we can change variables to u = 6 . Then 10 1 2 6. 87 1 2 10 6 6.87 1 1 1 1 10 6 du = 3.541 10 8 t = 7.977 10 14   1 2 6 6 u 6.87 6.87 10 6. 37 10 u 6. 37 10 6
z z z z FGH FGH z LMN
dt = 
i i
dr i dr = . The time of fall is v f v 2GM E 1 1  r RE + h
t = t =
RE + h RE
IJ I K JK
1 2
dr
6.87 10 6 m
2 6.67 10 11 5.98 10 24
6.37 10 6 m
FG 1  1 IJ OP H r 6.87 10 m K Q
6
1 2
dr
e
j
z FGH
IJ K
e j
z FGH
IJ K
1 2
du
A mathematics program returns the value 9.596 for this integral, giving for the time of fall t = 3.541 10 8 10 9 9.596 = 339.8 = 340 s .
Chapter 13
393
Section 13.7 P13.33
Energy Considerations in Planetary and Satellite Motion
1 1 1 1  = mv 2 mvi2 + GM E m f 2 2 r f ri or and
F GH
I JK
1 2 1 1 = v2 vi + GM E 0  f 2 2 RE
2 v 2 = v1  f 2 1
FG H
IJ K
vf vf
F 2GM IJ = Gv  H R K = Le 2.00 10 j  1.25 10 O MN PQ
12 E E 4 2 8
2GM E RE
12
= 1.66 10 4 m s
P13.34
(a) (b)
v solar escape =
2 M SunG = 42.1 km s RESun
Let r = RES x represent variable distance from the Sun, with x in astronomical units. v= If v = 2 M SunG 42.1 = RES x x 125 000 km , then x = 1.47 A.U. = 2.20 10 11 m 3 600 s
(at or beyond the orbit of Mars, 125 000 km/h is sufficient for escape). P13.35 To obtain the orbital velocity, we use or We can obtain the escape velocity from or vi2 GM E = RE + h RE + h
F =
v=
mMG mv 2 = R R2
MG R
1 mMG 2 mv esc = 2 R 2 MG = v esc = R
2v
P13.36
b
g
2
1 1 GM E m 1 = K i = mvi2 = +h 2 2 RE 2
FG H
IJ K
LM e6.67 10 N m kg je5.98 10 kg jb500 kgg OP MN PQ = 1.45 10 e6.37 10 mj + e0.500 10 mj
11 2 2 24 6 6
10
J
The change in gravitational potential energy of the satelliteEarth system is U =
Also,
Kf
F I GH JK = e6.67 10 N m kg je5.98 10 kg jb500 kg ge 1.14 10 1 1 = mv = b500 kg ge 2.00 10 m sj = 1.00 10 J . 2 2
GM E m GM E m 1 1  = GM E m  Ri Rf Ri R f
11 2 2 24 2 2 f 3 9
8
m 1 = 2.27 10 9 J
j
The energy transformed due to friction is
Eint = K i  K f  U = 14.5  1.00 + 2.27 10 9 J = 1.58 10 10 J .
a
f
394 P13.37
Universal Gravitation
Fc = FG gives
mv 2 GmM E = r r2 GM E r
which reduces to v =
and period = (a)
2r r = 2r . v GM E
r = RE + 200 km = 6 370 km + 200 km = 6 570 km Thus,
period = 2 6.57 10 m
3
e
6
j e6.67 10 je
11
e6.57 10 mj N m kg je5.98 10
6 2 2
24
kg
j
T = 5.30 10 s = 88.3 min = 1.47 h GM E = r
(b)
v=
e6.67 10
11
N m 2 kg 2 5.98 10 24 kg
e6.57 10 mj
6
j=
7.79 km s
(c)
K f + U f = K i + Ui + energy input, gives input = GM E m GM E m 1 1  mv 2  mvi2 + f 2 2 rf ri
F GH
I F JK GH
IJ K
(1)
ri = RE = 6.37 10 6 m vi = 2RE = 4.63 10 2 m s 86 400 s
Substituting the appropriate values into (1) yields the minimum energy input = 6.43 10 9 J
Chapter 13
395
P13.38
The gravitational force supplies the needed centripetal acceleration. Thus,
b R + hg b
E 2
GM E m
=
mv 2 RE + h
g
or
v2 =
GM E RE + h
(a)
2r 2 RE + h = T= GM E v
b
g
T = 2
bR + hg
E
3
b R + hg
E
GM E
(b)
v=
GM E RE + h Emin = K f + U gf  K i  U gi .
(c)
Minimum energy input is
e
j e
j
It is simplest to launch the satellite from a location on the equator, and launch it toward the east. This choice has the object starting with energy with vi = 2RE 2RE = 1.00 day 86 400 s and 1 mvi2 2 GM E m . U gi =  RE Ki =
2 E
Thus,
Emin =
2 GM E GM E m 1 4 2 RE 1 m   m 2 RE + h RE + h 2 86 400 s
FG H
IJ K
or
Emin = GM E m
LM R + 2 h OP  2 R m NM 2 R bR + hg QP b86 400 sg
E 2 2 E E E
LM MN b
OP GM m + g PQ R
E
2
P13.39
Etot =  E =
GMm 2r
6.67 10 11 5.98 10 24 10 3 kg 1 1 GMm 1 1  =  ri r f 2 2 10 3 m 6 370 + 100 6 370 + 200
F GH
I e JK
je
j
F GH
I JK
E = 4.69 10 8 J = 469 MJ P13.40 gE = Gm E
2 rE
gU =
GmU
2 rU
(a)
2 gU mU rE 1 = = 14.0 2 3.70 g E m E rU
FG IJ H K
2
= 1.02 2GmU : rU
gU = 1.02 9.80 m s 2 = 10.0 m s 2
v esc ,E v esc ,U mU rE 14.0 = = 1.95 3.70 m E rU
a fe
=
j
(b)
v esc ,E =
2Gm E ; v esc ,U = rE
For the Earth, from the text's table of escape speeds, v esc ,E = 11.2 km s v esc ,U = 1.95 11.2 km s = 21.8 km s
a fb
g
396 P13.41
Universal Gravitation
The rocket is in a potential well at Ganymede's surface with energy 6.67 10 11 N m 2 m 2 1.495 10 23 kg Gm 1 m 2 = U1 =  r kg 2 2.64 10 6 m
e
e
j
j
U1 = 3.78 10 m 2 m s
6
2
2
The potential well from Jupiter at the distance of Ganymede is U2 =  6.67 10 11 N m 2 m 2 1.90 10 27 kg Gm1 m 2 = r kg 2 1.071 10 9 m
e
e
j
j
U 2 = 1.18 10 m 2 m s To escape from both requires
8
2
2
1 2 m 2 v esc = + 3.78 10 6 + 1.18 10 8 m 2 m 2 s 2 2
e
j
v esc = 2 1.22 10 8 m 2 s 2 = 15.6 km s P13.42 We interpret "lunar escape speed" to be the escape speed from the surface of a stationary moon alone in the Universe: GM m m 1 2 mv esc = 2 Rm v esc = 2GM m Rm 2GM m Rm
v launch = 2 Now for the flight from moon to Earth
aK + U f = a K + U f
i
f
GmM m GmM E 1 GmM m GmM E 1 2 2  = mv impact   mv launch  2 2 Rm rel rm 2 RE 4GM m GM m GM E 1 2 GM m GM E   = v impact   2 Rm Rm rel rm 2 RE
v impact
L F 3 M + M + M  M I OP = M 2G G MN H R r R r JK PQ L F 3 7.36 10 kg + 7.36 10 = M 2G G NM H 1.74 10 m 3.84 10
m m E E m m2 E el 22 6
12
22 8
kg m
+
5.98 10 24 kg 6.37 10 6 m

5.98 10 24 kg 3.84 10 8
12
I OP m JQ KP
12
= 2G 1.27 10 17 + 1.92 10 14 + 9.39 10 17  1.56 10 16 kg m = 2 6.67 10 11 N m 2 kg 2 10.5 10 17 kg m
e
j
e
j
12
= 11.8 km s
Chapter 13
397
*P13.43
(a)
Energy conservation for the objectEarth system from firing to apex:
eK + U j = eK + U j
g i g
f
GmM E GmM E 1 =0 mvi2  RE RE + h 2 where GmM E 1 2 . Then mv esc = RE 2 RE 1 2 1 2 1 2 vi  v esc =  v esc 2 2 2 RE + h
2 v esc  vi2 = 2 v esc RE RE + h RE + h 2 v esc RE 2 2 v esc RE  v esc RE + vi2 RE 2 v esc  vi2
1
2 v esc
 vi2
=
h= h= 6.37 10 6 m 8.76
2 2
2 v esc  vi2 2 v esc  vi2
2 v esc RE
 RE =
RE vi2
(b) (c)
h=
a f a11.2f  a8.76f FG H
2
= 1.00 10 7 m
The fall of the meteorite is the timereversal of the upward flight of the projectile, so it is described by the same energy equation
2 vi2 = v esc 1 
RE h 2 = v esc = 11.2 10 3 m s RE + h RE + h
IJ K
FG H
IJ e K
2 m j FGH 6.37 10.51 +10.51 10 m IJK m 2
2 7 6 7
= 1.00 10 8 m 2 s 2 vi = 1.00 10 4 m s RE vi2
2 v esc
(d)
With vi << v esc , h
=
0 2 = vi2 + 2  g h  0 . P13.44
b ga f
RE vi2 RE GM E v2 . But g = , so h = i , in agreement with 2 2GM E 2g RE
For a satellite in an orbit of radius r around the Earth, the total energy of the satelliteEarth system is GM E E= . Thus, in changing from a circular orbit of radius r = 2 RE to one of radius r = 3 RE , the 2r required work is W = E =  GM E m GM E m GM E m 1 1 + = GM E m  = . 2r f 2ri 4 RE 6 R E 12 RE
LM N
OP Q
398 *P13.45
Universal Gravitation
(a)
The major axis of the orbit is Further, in Figure 13.5, Then
2 a = 50.5 AU so a + c = 50 AU so c 24.75 = 0.980 e= = a 25.25
a = 25.25 AU c = 24.75 AU
(b)
In T 2 = K s a 3 for objects in solar orbit, the Earth gives us
b1 yrg
Then 6.67 10 11 GMm U= = r For the satellite v0 = (b)
2
2
= K s 1 AU
2
a
f
3
Ks
3
(c)
e
b1 yrg a25.25 AU f T = 127 yr T = a1 AU f N m kg je1.991 10 kg je1.2 10 kg j = 2.13 10 50e1.496 10 mj
3 2 2 30 10 11
b1 yrg = a1 AU f
2 3
17
J
*P13.46
(a)
F = ma
GmM E r2
=
FG GM IJ H r K
E
2 mv 0 r
12
Conservation of momentum in the forward direction for the exploding satellite:
c mvh = c mvh
i
f
5mv 0 = 4mvi + m0 vi = (c) 5 5 GM E v0 = r 4 4
FG H
IJ K
12
With velocity perpendicular to radius, the orbiting fragment is at perigee. Its apogee distance and speed are related to r and vi by 4mrvi = 4mr f v f and GM E 4m 1 vr GM E 4m 1 = 4mv 2  . Substituting v f = i we have 4mvi2  f 2 2 r rf rf 1 2 GM E 1 vi2 r 2 GM E 25 GM E gives =  . Further, substituting vi2 = vi  2 2 2 rf 16 r r rf 25 GM E GM E 25 GM E r GM E  =  32 r 32 r f2 r rf 7 25r 1 =  32r 32r f2 r f Clearing of fractions, 7r f2 = 25r 2  32rr f or 7 rf r = +32 32 2  4 7 25 14 rf
14 14 25 25r = ; rf = . end of the orbit has 7 7 r
a fa f = 50 or 14 . The latter root describes the starting point. The outer
FG r IJ HrK
f
2
 32
FG r IJ + 25 = 0 giving HrK
f
Chapter 13
399
Additional Problems P13.47 Let m represent the mass of the spacecraft, rE the radius of the Earth's orbit, and x the distance from Earth to the spacecraft. The Sun exerts on the spacecraft a radial inward force of Fs = FE =
br  x g
E
GM s m
2
while the Earth exerts on it a radial outward force of
GM E m x2
The net force on the spacecraft must produce the correct centripetal acceleration for it to have an orbital period of 1.000 year. Thus, FS  FE = GM S
E
br  x g
E
GM S m
2

GM E m x2
which reduces to
br  x g
2

GM E x2
=
4 2
b br  x g .
E
mv 2 m = = rE  x rE  x
g b
LM 2 br  xg OP g NM T QP
E
2
T2
(1)
Cleared of fractions, this equation would contain powers of x ranging from the fifth to the zeroth. We do not solve it algebraically. We may test the assertion that x is between 1.47 10 9 m and 1.48 10 9 m by substituting both of these as trial solutions, along with the following data: M S = 1.991 10 30 kg , M E = 5.983 10 24 kg , rE = 1.496 10 11 m, and T = 1.000 yr = 3.156 10 7 s . With x = 1. 47 10 9 m substituted into equation (1), we obtain
6.052 10 3 m s 2  1.85 10 3 m s 2 5.871 10 3 m s 2
or
5.868 10 3 m s 2 5.871 10 3 m s 2
With x = 1. 48 10 9 m substituted into the same equation, the result is
6.053 10 3 m s 2  1.82 10 3 m s 2 5.870 8 10 3 m s 2
or
5.870 9 10 3 m s 2 5.870 8 10 3 m s 2 .
Since the first trial solution makes the lefthand side of equation (1) slightly less than the right hand side, and the second trial solution does the opposite, the true solution is determined as between the trial values. To threedigit precision, it is 1.48 10 9 m . As an equation of fifth degree, equation (1) has five roots. The SunEarth system has five Lagrange points, all revolving around the Sun synchronously with the Earth. The SOHO and ACE satellites are at one. Another is beyond the far side of the Sun. Another is beyond the night side of the Earth. Two more are on the Earth's orbit, ahead of the planet and behind it by 60. Plans are under way to gain perspective on the Sun by placing a spacecraft at one of these two coorbital Lagrange points. The Greek and Trojan asteroids are at the coorbital Lagrange points of the JupiterSun system.
400 P13.48
Universal Gravitation
The acceleration of an object at the center of the Earth due to the gravitational force of the Moon is given by M a = G Moon d2 MM At the point A nearest the Moon, a+ = G 2 dr
a f ad + r f
MM
At the point B farthest from the Moon, a  = G
2
FIG. P13.48
a = a +  a = GM M For d >> r , Across the planet, *P13.49 a = 2GM M r d
3
LM 1 MN ad  r f
2

1 d2
OP PQ
= 1.11 10 6 m s 2
g 2 a 2. 22 10 6 m s 2 = = = 2. 26 10 7 g g 9.80 m s 2
Energy conservation for the twosphere system from release to contact:  Gmm Gmm 1 1 = + mv 2 + mv 2 R 2r 2 2
Gm (a)
FG 1  1 IJ = v H 2r R K
12
2
v = Gm
FG L 1  1 OIJ H MN 2r R PQK
12
The injected impulse is the final momentum of each sphere, mv = m 2 2 Gm
FG L 1  1 OIJ H MN 2r R PQK
= Gm 3
LM N
FG 1  1 IJ OP H 2r R K Q
12
12
.
(b)
If they now collide elastically each sphere reverses its velocity to receive impulse mv   mv = 2mv = 2 Gm 3
a f
LM N
FG 1  1 IJ OP H 2r R K Q
P13.50
Momentum is conserved: m1 v 1i + m 2 v 2i = m1 v 1 f + m 2 v 2 f 0 = Mv 1 f + 2 Mv 2 f v2 f =  Energy is conserved: 1 v1 f 2
aK + U f + E = aK + U f
i
f
0 
Gm 1 m 2 Gm1 m 2 1 1 2 2 + 0 = m 1 v1 f + m 2 v 2 f  rf ri 2 2
GM 2 M 1 1 1 2 v1 f = Mv1 f + 2 M 12 R 2 2 2 2 GM R 3 v2 f =
a f
a fFGH
IJ K
2

GM 2 M 4R
a f
v1 f =
1 1 GM v1 f = R 2 3
Chapter 13
401
P13.51
(a)
v2 ac = r
ac
e1.25 10 =
GM r2
2
6
ms
11
j
2
1.53 10
m
= 10.2 m s 2
(b)
diff = 10.2  9.90 = 0.312 m s 2 =
e0.312 m s je1.53 10 mj M=
2 11
6.67 10
11
N m kg
2
2
= 1.10 10 32 kg GM E r2
FIG. P13.51 = GM E r 2 (directed downward)
P13.52
(a)
The freefall acceleration produced by the Earth is g = Its rate of change is
dg = GM E 2 r 3 = 2GM E r 3 . dr dg 2GM E = . 3 dr RE
a f
The minus sign indicates that g decreases with increasing height. At the Earth's surface, (b) For small differences, g r = g h = 2GM E
3 RE
Thus,
g =
2GM E h
3 RE
(c)
g =
2 6.67 10 11 N m 2 kg 2 5.98 10 24 kg 6.00 m
e
je
ja
e6.37 10 mj
6
3
f=
1.85 10 5 m s 2
*P13.53
(a)
Each bit of mass dm in the ring is at the same distance from the object at A. The separate GmM ring Gmdm to the system energy add up to  . When the object is at A, contributions  r r this is 6.67 10 11 N m 2 1 000 kg 2.36 10 20 kg kg 2 = 7.04 10 4 J .
e1 10 mj + e2 10 mj
8 2 8
2
(b)
When the object is at the center of the ring, the potential energy is  6.67 10 11 N m 2 1 000 kg 2.36 10 20 kg kg 2 1 10 8 m = 1.57 10 5 J .
(c)
Total energy of the objectring system is conserved:
e K + U j = eK + U j
g A g 4
B
1 2 0  7.04 10 J = 1 000 kgv B  1.57 10 5 J 2 vB =
F 2 8.70 10 J I GH 1 000 kg JK
4
12
= 13.2 m s
402 P13.54
Universal Gravitation
To approximate the height of the sulfur, set mv 2 = mg Io h 2 v = 2 g Io h A more precise answer is given by 1 GMm GMm = mv 2  2 r1 r2 1 2 v = 6.67 10 11 8.90 10 22 2 h = 70 000 m g Io = GM = 1.79 m s 2 r2
v = 2 1.79 70 000 500 m s over 1 000 mi h
a fb
g
b
g
e
je
jFGH 1.82 1 10 a
6

1 1.89 10
6
IJ K
v = 492 m s 25 000 m = 3.98 10 3 m 2
P13.55
From the walk, 2r = 25 000 m. Thus, the radius of the planet is r = From the drop: so, y = g= 1 2 1 gt = g 29. 2 s 2 2
f
2
= 1.40 m
3
2 1.40 m
a
a
29.2 s
f
2
f = 3.28 10
m s2 =
MG r2
M = 7.79 10 14 kg
*P13.56
The distance between the orbiting stars is d = 2r cos 30 = 3 r since 3 . The net inward force on one orbiting star is cos 30 = 2 Gmm GMm Gmm mv 2 cos 30+ 2 + 2 cos 30 = r d2 r d Gm 2 cos 30 GM 4 2 r 2 + 2 = r rT 2 3r 2 m 4 2 r 3 +M = G T2 3
30 r d r 60 30
FG H
IJ K
F F F
e j F r T = 2 G GH GeM +
G M+
m 3 3
T2 =
4 2 r 3
m 3
I JJ jK
12
FIG. P13.56
P13.57
For a 6.00 km diameter cylinder, r = 3 000 m and to simulate 1 g = 9.80 m s 2 v2 = 2r r g = 0.057 2 rad s = r g= The required rotation rate of the cylinder is 1 rev 110 s
(For a description of proposed cities in space, see Gerard K. O'Neill in Physics Today, Sept. 1974.)
Chapter 13
403
P13.58
(a)
G has units
N m 2 kg m m 2 m3 = 2 = 2 kg 2 s kg 2 s kg L3 . T2 M L , and Planck's constant has the same dimensions T
and dimensions G =
The speed of light has dimensions of c = as angular momentum or h = M L2 . T
We require G p c q h r = L , or L3 p T 2 p M  p Lq T  q M r L2 r T  r = L1 M 0 T 0 . Thus, 3 p + q + 2r = 1 2 p  q  r = 0 p + r = 0 which reduces (using r = p ) to 3p + q + 2p = 1 2 p  q  p = 0 These equations simplify to Then, 5 p  3 p = 1 , yielding p = 5 p + q = 1 and q = 3 p . 1 3 1 , q =  , and r = . 2 2 2
Therefore, Planck length = G 1 2 c  3 2 h1 2 . (b) P13.59
e6.67 10 j e3 10 j e6.63 10 j
11 1 2 8 3 2
34 1 2
= 1.64 10 69
e
j
12
= 4.05 10 35 m ~ 10 34 m
Gm p m 0 1 2 m 0 v esc = 2 R v esc = 2Gm p R
4 With m p = R 3 , we have 3 v esc = = So, v esc R .
4 2G 3 R 3
R 8G R 3
404 *P13.60
Universal Gravitation
For both circular orbits,
F = ma :
GM E m r2
=
mv 2 r
v=
GM E r
FIG. P13.60
11 2 2 24
(a)
The original speed is vi =
e6.67 10 N m kg je5.98 10 e6.37 10 m + 2 10 mj
6 5
kg
j= j=
7.79 10 3 m s .
(b)
The final speed is
vi =
e6.67 10
11
N m 2 kg 2 5.98 10 24 kg
e
je
6.47 10 m
6
j
7.85 10 3 m s .
The energy of the satelliteEarth system is K +Ug = GM E m GM E m 1 GM E GM E 1 = m  = mv 2  2 2 2r r r r
11
(c)
Originally
e6.67 10 E =
i
N m 2 kg 2 5.98 10 24 kg 100 kg 2 6.57 10 m
e
je
6
j j
jb
g= g=
3.04 10 9 J .
(d)
Finally
Ef = 
e6.67 10
11
N m 2 kg 2 5.98 10 24 kg 100 kg 2 6. 47 10 m
e
je
jb
6
3.08 10 9 J .
(e)
Thus the object speeds up as it spirals down to the planet. The loss of gravitational energy is so large that the total energy decreases by
Ei  E f = 3.04 10 9 J  3.08 10 9 J = 4.69 10 7 J .
(f) The only forces on the object are the backward force of air resistance R, comparatively very small in magnitude, and the force of gravity. Because the spiral path of the satellite is not perpendicular to the gravitational force, one component of the gravitational force pulls forward on the satellite to do positive work and make its speed increase.
e
j
Chapter 13
405
P13.61
(a)
At infinite separation U = 0 and at rest K = 0 . Since energy of the twoplanet system is conserved we have, 0= Gm1 m 2 1 1 2 2 m 1 v1 + m 2 v 2  d 2 2 (1)
The initial momentum of the system is zero and momentum is conserved. Therefore, Combine equations (1) and (2): 0 = m1 v1  m 2 v 2 v1 = m 2 2G d m1 + m 2 (2)
b
g b
d
and
v 2 = m1
2G d m1 + m 2
b
g
Relative velocity (b)
v r = v1   v 2 =
b g
2G m 1 + m 2
g
Substitute given numerical values into the equation found for v1 and v 2 in part (a) to find v1 = 1.03 10 4 m s Therefore, K1 = 1 2 m1 v1 = 1.07 10 32 J 2 and and v 2 = 2.58 10 3 m s K2 = 1 2 m 2 v 2 = 2.67 10 31 J 2
P13.62
(a)
The net torque exerted on the Earth is zero. Therefore, the angular momentum of the Earth is conserved; mra v a = mrp v p and v a = v p
F r I = e3.027 10 GH r JK
p a
4
ms
471 jFGH 1..521 IJK = 1
2.93 10 4 m s
(b)
Kp =
1 1 2 mv p = 5.98 10 24 3.027 10 4 2 2
e
je
j
2
= 2.74 10 33 J
6.673 10 11 5.98 10 24 1.99 10 30 GmM = = 5.40 10 33 J Up =  11 rp 1.471 10 (c) Using the same form as in part (b), K a = 2.57 10 33 J and U a = 5.22 10 33 J . Compare to find that K p + U p = 2.66 10 33 J and K a + U a = 2.65 10 33 J . They agree.
e
je
je
j
406 P13.63
Universal Gravitation
(a)
The work must provide the increase in gravitational energy W = U g = U gf  U gi = = GM E M p rf GM E M p RE + y + + GM E M p ri GM E M p RE  1 RE + y
= GM E M p =
F 6.67 10 GH kg
F1 GH R
E
11 2
N m2
I 5.98 10 JK e
I JK
24
kg 100 kg
jb
gFGH 6.37 110
6
m

1 7.37 10 6 m
IJ K
W = 850 MJ (b) In a circular orbit, gravity supplies the centripetal force:
bR
Then, 1 1 GM E M p Mp v2 = 2 2 RE + y
GM E M p
E
+y
g bR
2
=
Mpv2
E
+y
g
b
g
11 N m 2 5.98 10 24 kg 100 kg 1 6.67 10 2 kg 2 7.37 10 6 m
So, additional work = kinetic energy required =
e
e je
je
j
jb
g
W = 2.71 10 9 J P13.64 Centripetal acceleration comes from gravitational acceleration. v 2 M c G 4 2 r 2 = 2 = r r T 2r GM c T 2 = 4 2 r 3
11
e6.67 10 ja20fe1.99 10 je5.00 10 j
30
3 2
= 4 2 r 3
rorbit = 119 km P13.65 (a) T=
15 2r 2 30 000 9.46 10 m = = 7 10 15 s = 2 10 8 yr v 2.50 10 5 m s
e
j
(b)
4 2 30 000 9.46 10 15 m 4 2 a 3 M= = GT 2 6.67 10 11 N m 2 kg 2 7.13 10 15 s
e
j
3
e
je
j
2
= 2.66 10 41 kg
M = 1.34 10 11 solar masses ~ 10 11 solar masses
The number of stars is on the order of 10 11 .
Chapter 13
407
P13.66
(a)
From the data about perigee, the energy of the satelliteEarth system is E= or GM E m 1 1 2 = 1.60 8.23 10 3 mv p  rp 2 2
a fe b
6.67 10 je5.98 10 ja1.60f j e 7.02 10
2 11 24

6
E = 3.67 10 7 J
(b)
L = mvr sin = mv p rp sin 90.0 = 1.60 kg 8.23 10 3 m s 7.02 10 6 m = 9.24 10 10 kg m 2 s
ge
je
j
(c)
Since both the energy of the satelliteEarth system and the angular momentum of the Earth are conserved, at apogee we must have and Thus, and 1 GMm 2 =E mv a  2 ra mv a ra sin 90.0 = L . 6.67 10 11 5.98 10 24 1.60 1 2 = 3.67 10 7 J 1.60 v a  2 rs
a f
e
je
ja f
b1.60 kg gv r a f
a a
= 9.24 10 10 kg m 2 s .
6.67 10 11 5.98 10 24 1.60 1.60 v a 1 2 Solving simultaneously, = 3.67 10 7 1.60 v a  2 9.24 10 10 which reduces to
2 0.800 v a  11 046 v a + 3.672 3 10 7 = 0
e
je
ja fa f
7
so
va =
11 046
b11 046g  4a0.800fe3.672 3 10 j . 2a0.800 f
2
This gives v a = 8 230 m s or 5 580 m s . The smaller answer refers to the velocity at the apogee while the larger refers to perigee. Thus, ra = 9.24 10 10 kg m 2 s L = = 1.04 10 7 m . mv a 1.60 kg 5.58 10 3 m s
b
ge
j
(d)
The major axis is 2a = rp + ra , so the semimajor axis is a= 1 7.02 10 6 m + 1.04 10 7 m = 8.69 10 6 m 2
e
j
(e)
T=
4 2 a 3 = GM E
4 2 8.69 10 6 m
e6.67 10
e
11
N m 2 kg 2 5.98 10 24 kg
je
j
3
j
T = 8 060 s = 134 min
408 *P13.67
Universal Gravitation
Let m represent the mass of the meteoroid and vi its speed when far away. No torque acts on the meteoroid, so its angular momentum is conserved as it moves between the distant point and the point where it grazes the Earth, moving perpendicular to the radius: Li = L f : mri v i = mr f v f m 3 RE vi = mRE v f v f = 3 vi Now energy of the meteoroidEarth system is also conserved:
FIG. P13.67
b
g
eK + U j = eK + U j :
g i g f
GM E m 1 1 mvi2 + 0 = mv 2  f 2 2 RE GM E 1 2 1 vi = 9 vi2  2 2 RE
e j
GM E = 4vi2 : RE *P13.68
vi =
GM E 4 RE
From Kepler's third law, minimum period means minimum orbit size. The "treetop satellite" in Figure P13.35 has minimum period. The radius of the satellite's circular orbit is essentially equal to the radius R of the planet.
F = ma :
FG H R R e 4 R j GV =
GMm
2
=
mv 2 m 2R = R R T
2 2 2
IJ K
2
RT 2 4 4 2 R 3 G R 3 = 3 T2
FG H
IJ K
The radius divides out: T 2 G = 3 P13.69
T=
3 G
If we choose the coordinate of the center of mass at the origin, then 0=
bMr
2
 mr1
g
M+m
and
Mr2 = mr1
(Note: this is equivalent to saying that the net torque must be zero and the two experience no angular acceleration.) For each mass F = ma so
2 mr1 1 =
MGm d2
and
2 Mr2 2 =
MGm d2
Combining these two equations and using d = r1 + r2 gives r1 + r2 2 = with 1 = 2 = 2 and T =
b
g
aM + mfG
d2
FIG. P13.69
we find T 2 =
4 2 d 3 G M+m
a
f
.
Chapter 13
409
P13.70
(a)
The gravitational force exerted on m 2 by the Earth (mass m1 ) accelerates m 2 according to: Gm1 m 2 . The equal magnitude force exerted on the Earth by m 2 produces negligible m2 g 2 = r2 acceleration of the Earth. The acceleration of relative approach is then g2 = Gm1 r
2
=
e6.67 10
11
N m 2 kg 2 5.98 10 24 kg
e1.20 10
7
je mj
2
j=
2.77 m s 2 .
(b)
Again, m 2 accelerates toward the center of mass with g 2 = 2.77 m s 2 . Now the Earth accelerates toward m 2 with an acceleration given as m1 g 1 = g1 = Gm1 m 2 r2
Gm 2 r
2
e6.67 10 =
11
N m 2 kg 2 2.00 10 24 kg
e1.20 10
7
je mj
2
j = 0.926 m s
2
The distance between the masses closes with relative acceleration of
g rel = g 1 + g 2 = 0.926 m s 2 + 2.77 m s 2 = 3.70 m s 2 .
P13.71 Initial Conditions and Constants: Mass of planet: Radius of planet: Initial x: Initial y: Initial v x : Initial 5.98 10 24 kg 6.37 10 6 m 0.0 planet radii 2.0 planet radii +5 000 m/s 0.0 m/s 10.9 s FIG. P13.71 vx (m/s) 5 000.0 4 999.9 4 999.7 4 999.3 7 523.0 7 523.2 7 522.8 7 521.9 4 999.9 5 000.0 5 000.0 vy (m/s) 0.0 26.7 53.4 80.1 39.9 20.5 80.9 141.4 53.3 26.6 0.1
vy :
Time interval:
t (s) 0.0 10.9 21.7 32.6 ... 5 431.6 5 442.4 5 453.3 5 464.1 ... 10 841.3 10 852.2 10 863.1
x (m) 0.0 54 315.3 108 629.4 162 941.1 112 843.8 31 121.4 50 603.4 132 324.3 108 629.0 54 314.9 0.4
y (m) 12 740 000.0 12 740 000.0 12 739 710.0 12 739 130.0 8 466 816.0 8 467 249.7 8 467 026.9 8 466 147.7 12 739 134.4 12 739 713.4 12 740 002.4
r (m) 12 740 000.0 12 740 115.8 12 740 173.1 12 740 172.1 8 467 567.9 8 467 306.9 8 467 178.2 8 467 181.7 12 739 597.5 12 739 829.2 12 740 002.4
em s j
2
ax
em s j
2
ay
0.000 0 0.010 0 0.021 0 0.031 0 0.074 0 0.020 0 0.033 0 0.087 0 0.021 0 0.010 0 0.000 0
2.457 5 2.457 4 2.457 3 2.457 2 5.562 5 5.563 3 5.563 4 5.562 8 2.457 5 2.457 5 2.457 5
The object does not hit the Earth ; its minimum radius is 1.33 RE . Its period is 1.09 10 4 s . A circular orbit would require a speed of 5.60 km s .
410
Universal Gravitation
ANSWERS TO EVEN PROBLEMS
P13.2 P13.4 P13.6 2.67 10 7 m s 2 3.00 kg and 2.00 kg (a) 4.39 10 20 N toward the Sun; (b) 1.99 10 20 N toward the Earth; (c) 3.55 10 22 N toward the Sun see the solution; either 1 m  61.3 nm or 2.74 10 4 m 2 3 (a) 1.02 km s ; (b) 1.35 mm see the solution 1.27 Planet Y has turned through 1.30 revolutions 1.63 10 4 rad s 18.2 ms (a) 1.31 10 17 N toward the center; (b) 2.62 10 12 N kg (a) 4.77 10 9 J ; (b) 569 N down; (c) 569 N up 2.52 10 7 m P13.48 P13.50 P13.52 P13.40 P13.42 P13.44 (a) 10.0 m s 2 ; (b) 21.8 km s 11.8 km s GM E m 12 RE
P13.8
P13.46
(a) v 0
F GM IJ =G H r K
E
12
; (b) vi =
5
e j
4
GM E 1 2 r
;
P13.10 P13.12 P13.14 P13.16 P13.18 P13.20 P13.22 P13.24
(c) r f =
25r 7
2.26 10 7 2 GM 1 GM ; 3 R 3 R (a), (b) see the solution; (c) 1.85 10 5 m s 2 492 m s see the solution (a) G 1 2 c  3 2 h1 2 ; (b) ~ 10 34 m (a) 7.79 km s; (b) 7.85 km s;(c) 3.04 GJ ; (d) 3.08 GJ ; (e) loss = 46.9 MJ ; (f) A component of the Earth's gravity pulls forward on the satellite in its downward banking trajectory. (a) 29.3 km s ; (b) K p = 2.74 10 33 J ;
P13.54 P13.56 P13.58 P13.60
P13.26 P13.28 P13.30 P13.32 P13.34 P13.36 P13.38
2.82 10 9 J
(a) see the solution; (b) 340 s (a) 42.1 km s ; (b) 2.20 10 11 m
P13.62
U p = 5.40 10 33 J ;(c) K a = 2.57 10 33 J;
U a = 5.22 10 33 J; yes P13.64 P13.66 119 km (a) 36.7 MJ ; (b) 9.24 10 10 kg m 2 s ; (c) 5.58 km s; 10.4 Mm; (d) 8.69 Mm; (e) 134 min see the solution (a) 2.77 m s 2 ; (b) 3.70 m s 2
1.58 10 10 J
g bGM g ; (b) bGM g b R + hg ; L R + 2 h OP  2 R m (c) GM m M MN 2 R bR + hg PQ b86 400 sg
(a) 2 RE + h
E 12 3 2 E 1 2 E 1 2 E E 2 2 E E E
b
P13.68
2
The satellite should be launched from the Earth's equator toward the east.
P13.70
14
Fluid Mechanics
CHAPTER OUTLINE
14.1 14.2 14.3 14.4 14.5 14.6 14.7 Pressure Variation of Pressure with Depth Pressure Measurements Buoyant Forces and Archimede's Principle Fluid Dynamics Bernoulli's Equation Other Applications of Fluid Dynamics
ANSWERS TO QUESTIONS
Q14.1 Q14.2 The weight depends upon the total volume of glass. The pressure depends only on the depth. Both must be built the same. The force on the back of each dam is the average pressure of the water times the area of the dam. If both reservoirs are equally deep, the force is the same.
FIG. Q14.2 Q14.3 If the tube were to fill up to the height of several stories of the building, the pressure at the bottom of the depth of the tube of fluid would be very large according to Equation 14.4. This pressure is much larger than that originally exerted by inward elastic forces of the rubber on the water. As a result, water is pushed into the bottle from the tube. As more water is added to the tube, more water continues to enter the bottle, stretching it thin. For a typical bottle, the pressure at the bottom of the tube can become greater than the pressure at which the rubber material will rupture, so the bottle will simply fill with water and expand until it bursts. Blaise Pascal splintered strong barrels by this method. About 1 000 N: that's about 250 pounds. The submarine would stop if the density of the surrounding water became the same as the average density of the submarine. Unfortunately, because the water is almost incompressible, this will be much deeper than the crush depth of the submarine. Yes. The propulsive force of the fish on the water causes the scale reading to fluctuate. Its average value will still be equal to the total weight of bucket, water, and fish. The boat floats higher in the ocean than in the inland lake. According to Archimedes's principle, the magnitude of buoyant force on the ship is equal to the weight of the water displaced by the ship. Because the density of salty ocean water is greater than fresh lake water, less ocean water needs to be displaced to enable the ship to float. 411
Q14.4 Q14.5
Q14.6 Q14.7
412 Q14.8
Fluid Mechanics
In the ocean, the ship floats due to the buoyant force from salt water. Salt water is denser than fresh water. As the ship is pulled up the river, the buoyant force from the fresh water in the river is not sufficient to support the weight of the ship, and it sinks. Exactly the same. Buoyancy equals density of water times volume displaced. At lower elevation the water pressure is greater because pressure increases with increasing depth below the water surface in the reservoir (or water tower). The penthouse apartment is not so far below the water surface. The pressure behind a closed faucet is weaker there and the flow weaker from an open faucet. Your fire department likely has a record of the precise elevation of every fire hydrant. As the wind blows over the chimney, it creates a lower pressure at the top of the chimney. The smoke flows from the relatively higher pressure in front of the fireplace to the low pressure outside. Science doesn't suck; the smoke is pushed from below. The rapidly moving air above the ball exerts less pressure than the atmospheric pressure below the ball. This can give substantial lift to balance the weight of the ball. The skijumper gives her body the shape of an airfoil. She deflects downward the air stream as it rushes past and it deflects her upward by Newton's third law. The air exerts on her a lift force, giving her a higher and longer trajectory. To say it in different words, the pressure on her back is less than the pressure on her front. FIG. Q14.13
Q14.9 Q14.10
Q14.11
Q14.12 Q14.13
Q14.14
The horizontal force exerted by the outside fluid, on an area element of the object's side wall, has equal magnitude and opposite direction to the horizontal force the fluid exerts on another element diametrically opposite the first. The glass may have higher density than the liquid, but the air inside has lower density. The total weight of the bottle can be less than the weight of an equal volume of the liquid. Breathing in makes your volume greater and increases the buoyant force on you. You instinctively take a deep breath if you fall into the lake. No. The somewhat lighter barge will float higher in the water. The level of the pond falls. This is because the anchor displaces more water while in the boat. A floating object displaces a volume of water whose weight is equal to the weight of the object. A submerged object displaces a volume of water equal to the volume of the object. Because the density of the anchor is greater than that of water, a volume of water that weighs the same as the anchor will be greater than the volume of the anchor. The metal is more dense than water. If the metal is sufficiently thin, it can float like a ship, with the lip of the dish above the water line. Most of the volume below the water line is filled with air. The mass of the dish divided by the volume of the part below the water line is just equal to the density of water. Placing a bar of soap into this space to replace the air raises the average density of the compound object and the density can become greater than that of water. The dish sinks with its cargo.
Q14.15 Q14.16 Q14.17 Q14.18
Q14.19
Chapter 14
413
Q14.20
The excess pressure is transmitted undiminished throughout the container. It will compress air inside the wood. The water driven into the wood raises its average density and makes if float lower in the water. Add some thumbtacks to reach neutral buoyancy and you can make the wood sink or rise at will by subtly squeezing a large clearplastic softdrink bottle. Bored with graph paper and proving his own existence, Ren Descartes invented this toy or trick. The plate must be horizontal. Since the pressure of a fluid increases with increasing depth, other orientations of the plate will give a nonuniform pressure on the flat faces. The air in your lungs, the blood in your arteries and veins, and the protoplasm in each cell exert nearly the same pressure, so that the wall of your chest can be in equilibrium. Use a balance to determine its mass. Then partially fill a graduated cylinder with water. Immerse the rock in the water and determine the volume of water displaced. Divide the mass by the volume and you have the density. When taking off into the wind, the increased airspeed over the wings gives a larger lifting force, enabling the pilot to take off in a shorter length of runway. Like the ball, the balloon will remain in front of you. It will not bob up to the ceiling. Air pressure will be no higher at the floor of the sealed car than at the ceiling. The balloon will experience no buoyant force. You might equally well switch off gravity. Styrofoam is a little more dense than air, so the first ship floats lower in the water. We suppose the compound object floats. In both orientations it displaces its own weight of water, so it displaces equal volumes of water. The water level in the tub will be unchanged when the object is turned over. Now the steel is underwater and the water exerts on the steel a buoyant force that was not present when the steel was on top surrounded by air. Thus, slightly less wood will be below the water line on the block. It will appear to float higher. A breeze from any direction speeds up to go over the mound and the air pressure drops. Air then flows through the burrow from the lower entrance to the upper entrance. Regular cola contains a considerable mass of dissolved sugar. Its density is higher than that of water. Diet cola contains a very small mass of artificial sweetener and has nearly the same density as water. The lowdensity air in the can has a bigger effect than the thin aluminum shell, so the can of diet cola floats. (a) (b) Lowest density: oil; highest density: mercury The density must increase from top to bottom. Since the velocity of the air in the righthand section of the pipe is lower than that in the middle, the pressure is higher. The equation that predicts the same pressure in the far right and lefthand sections of the tube assumes laminar flow without viscosity. Internal friction will cause some loss of mechanical energy and turbulence will also progressively reduce the pressure. If the pressure at the left were not higher than at the right, the flow would stop.
Q14.21 Q14.22 Q14.23
Q14.24 Q14.25
Q14.26 Q14.27
Q14.28 Q14.29
Q14.30
Q14.31
(a) (b)
414 Q14.32
Fluid Mechanics
Clap your shoe or wallet over the hole, or a seat cushion, or your hand. Anything that can sustain a force on the order of 100 N is strong enough to cover the hole and greatly slow down the escape of the cabin air. You need not worry about the air rushing out instantly, or about your body being "sucked" through the hole, or about your blood boiling or your body exploding. If the cabin pressure drops a lot, your ears will pop and the saliva in your mouth may boilat body temperaturebut you will still have a couple of minutes to plug the hole and put on your emergency oxygen mask. Passengers who have been drinking carbonated beverages may find that the carbon dioxide suddenly comes out of solution in their stomachs, distending their vests, making them belch, and all but frothing from their ears; so you might warn them of this effect.
SOLUTIONS TO PROBLEMS
Section 14.1 P14.1 Pressure 4 jLMN 3 b0.015 0 mg OPQ
3
M = ironV = 7 860 kg m3 M = 0.111 kg
e
P14.2
The density of the nucleus is of the same order of magnitude as that of one proton, according to the assumption of close packing:
=
m 1.67 10 27 kg ~ ~ 10 18 kg m3 . 3 15 V 4 10 m
3
e
j
With vastly smaller average density, a macroscopic chunk of matter or an atom must be mostly empty space. P14.3 P= 50.0 9.80 F = A 0.500 10 2
e
a f
j
2
= 6.24 10 6 N m 2
P14.4
Let Fg be its weight. Then each tire supports so yielding P=
Fg 4
,
F Fg = A 4A Fg = 4 AP = 4 0.024 0 m 2 200 10 3 N m 2 = 1.92 10 4 N
e
je
j
P14.5
The Earth's surface area is 4R 2 . The force pushing inward over this area amounts to F = P0 A = P0 4R 2 . This force is the weight of the air: Fg = mg = P0 4R 2 so the mass of the air is P0 4R 2 g
e
j
e
j
m=
e
j = e1.013 10
5
N m 2 4 6.37 10 6 m 9.80 m s 2
jLMN e
j OPQ
2
= 5.27 10 18 kg .
Chapter 14
415
Section 14.2 P14.6 (a)
Variation of Pressure with Depth P = P0 + gh = 1.013 10 5 Pa + 1 024 kg m3 9.80 m s 2 1 000 m
e
je
jb
g
P = 1.01 10 7 Pa
(b) The gauge pressure is the difference in pressure between the water outside and the air inside the submarine, which we suppose is at 1.00 atmosphere.
Pgauge = P  P0 = gh = 1.00 10 7 Pa
The resultant inward force on the porthole is then F = Pgauge A = 1.00 10 7 Pa 0.150 m P14.7 Fel = Ffluid and h= kx gA
2 3
a
f
2
= 7.09 10 5 N .
or
kx = ghA
e1 000 N m je5.00 10 mj h= e10 kg m je9.80 m s jLMN e1.00 10
3 3 2
2
m
OP = jQ
2
1.62 m FIG. P14.7
P14.8
Since the pressure is the same on both sides, In this case, 15 000 F = 2 200 3.00 or
F1 F = 2 A1 A 2 F2 = 225 N
P14.9
Fg = 80.0 kg 9.80 m s 2 = 784 N When the cup barely supports the student, the normal force of the ceiling is zero and the cup is in equilibrium. Fg = F = PA = 1.013 10 5 Pa A A= Fg P = 784 = 7.74 10 3 m 2 5 1.013 10
e
j
e
j
FIG. P14.9 P14.10 (a) Suppose the "vacuum cleaner" functions as a highvacuum pump. The air below the brick will exert on it a lifting force F = PA = 1.013 10 5 Pa 1.43 10 2 m (b)
LM e N
j OPQ =
2
65.1 N .
The octopus can pull the bottom away from the top shell with a force that could be no larger than F = PA = P0 + gh A = 1.013 10 5 Pa + 1 030 kg m3 9.80 m s 2 32.3 m F = 275 N
b
g
e
je
ja
f LNM e1.43 10
2
m
j OQP
2
416 P14.11
Fluid Mechanics
The excess water pressure (over air pressure) halfway down is Pgauge = gh = 1 000 kg m3 9.80 m s 2 1.20 m = 1.18 10 4 Pa . The force on the wall due to the water is
e
je
ja
f
F = Pgauge A = 1.18 10 4 Pa 2.40 m 9.60 m = 2.71 10 5 N
horizontally toward the back of the hole. P14.12 The pressure on the bottom due to the water is Pb = gz = 1.96 10 4 Pa So, On each end, On the side, P14.13
e
ja
fa
f
Fb = Pb A = 5.88 10 6 N
F = PA = 9.80 10 3 Pa 20.0 m 2 = 196 kN F = PA = 9.80 10 3
2
e j Pae60.0 m j =
588 kN
In the reference frame of the fluid, the cart's acceleration causes a fictitious force to act backward, as if a the acceleration of gravity were g 2 + a 2 directed downward and backward at = tan 1 from the g d vertical. The center of the spherical shell is at depth below the air bubble and the pressure there is 2
F I GH JK
P = P0 + g eff h = P0 + P14.14
1 d g 2 + a 2 . 2
The air outside and water inside both exert atmospheric pressure, so only the excess water pressure gh counts for the net force. Take a strip of hatch between depth h and h + dh . It feels force
dF = PdA = gh 2.00 m dh .
(a) The total force is F = dF =
a
f
2.00 m
1.00 m 2.00 m
z
2.00 m
(b)
a2.00 mf a2.00 mf  a1.00 mf = e1 000 kg m je9.80 m s j f 2 F = 29.4 kN bto the right g The lever arm of dF is the distance a h  1.00 mf from hinge to strip: = z d = z gha 2.00 mfa h  1.00 mfdh Lh h O = g a 2.00 mfM  a1.00 mf P 2 Q N3 F 7.00 m  3.00 m I = e1 000 kg m je9.80 m s ja 2.00 mfG JK 2 H 3
h2 F = g 2.00 m 2
h = 1.00 m
z
gh 2.00 m dh
2.00 m 3 2 2
a
f
FIG. P14.14
2
a
1.00 m
2.00 m
h = 1.00 m
3
2
2.00 m
1.00 m
3
2
3
3
= 16.3 kN m counterclockwise
Chapter 14
417
P14.15
The bell is uniformly compressed, so we can model it with any shape. We choose a sphere of diameter 3.00 m. The pressure on the ball is given by: P = Patm + w gh so the change in pressure on the ball from when it is on the surface of the ocean to when it is at the bottom of the ocean is P = w gh . In addition: V =
ghV 4 w ghr 3 VP = w = , where B is the Bulk Modulus. 3B B B
4 1 030 kg m3 9.80 m s 2 10 000 m 1.50 m
10
V = 
e
je jb a3fe14.0 10 Paj
ga
f
3
= 0.010 2 m3
Therefore, the volume of the ball at the bottom of the ocean is V  V = 4 1.50 m 3
a
f
3
 0.010 2 m 3 = 14.137 m3  0.010 2 m 3 = 14.127 m 3 .
This gives a radius of 1.499 64 m and a new diameter of 2.999 3 m. Therefore the diameter decreases by 0.722 mm .
Section 14.3 P14.16 (a)
Pressure Measurements We imagine the superhero to produce a perfect vacuum in the straw. Take point 1 at the water surface in the basin and point 2 at the water surface in the straw: P1 + gy1 = P2 + gy 2 1.013 10 5 N m 2 + 0 = 0 + 1 000 kg m 3 9.80 m s 2 y 2
e
je
j
y 2 = 10.3 m
(b) P14.17
No atmosphere can lift the water in the straw through zero height difference.
P0 = gh h= P0 10.13 10 5 Pa = = 10.5 m g 0.984 10 3 kg m3 9.80 m s 2
e
je
j
No. Some alcohol and water will evaporate. The equilibrium
vapor pressures of alcohol and water are higher than the vapor pressure of mercury.
FIG. P14.17
418 P14.18
Fluid Mechanics
(a)
Using the definition of density, we have hw = 100 g m water = = 20.0 cm 2 A 2 water 5.00 cm 1.00 g cm 3
e
j
(b)
Sketch (b) at the right represents the situation after the water is added. A volume A 2 h2 of mercury has been displaced by water in the right tube. The additional volume of mercury now in the left tube is A1 h . Since the total volume of mercury has not changed,
b
g
FIG. P14.18 A1 h A2 (1)
A 2 h2 = A1 h
or
h2 =
At the level of the mercurywater interface in the right tube, we may write the absolute pressure as: P = P0 + water ghw The pressure at this same level in the left tube is given by P = P0 + Hg g h + h2 = P0 + water ghw which, using equation (1) above, reduces to
b
g
Hg h 1 +
or h =
LM N
A1 = water hw A2
OP Q
water hw Hg 1 +
e
A1 A2
j
.
3
e1.00 g cm ja20.0 cmf = Thus, the level of mercury has risen a distance of h = e13.6 g cm jc1 + h
3 10 .0 5.00
0.490 cm
above the original level. P14.19 P14.20 P0 = gh = 2.66 10 3 Pa :
P = P0 + P0 = 1.013  0.026 6 10 5 Pa = 0.986 10 5 Pa
b
g
Let h be the height of the water column added to the right side of the Utube. Then when equilibrium is reached, the situation is as shown in the sketch at right. Now consider two points, A and B shown in the sketch, at the level of the watermercury interface. By Pascal's Principle, the absolute pressure at B is the same as that at A. But, PA = P0 + w gh + Hg gh2 and PB = P0 + w g h1 + h + h2 . Thus, from PA = PB , w h1 + w h + w h2 = w h + Hg h2 , or h1 = B
h1 h h2 water Mercury A
b
g
FIG. P14.20
LM N
Hg w
 1 h2 = 13.6  1 1.00 cm = 12.6 cm .
OP Q
a
fa
f
Chapter 14
419
*P14.21
(a)
P = P0 + gh The gauge pressure is
P  P0 = gh = 1 000 kg 9.8 m s 2 0.160 m = 1.57 kPa = 1.57 10 3 Pa = 0.015 5 atm . It would lift a mercury column to height 1 568 Pa P  P0 h= = = 11.8 mm . g 13 600 kg m 3 9.8 m s 2
e
ja
f
FG 1 atm IJ H 1.013 10 Pa K
5
e
je
j
(b) (c)
Increased pressure of the cerebrospinal fluid will raise the level of the fluid in the spinal tap. Blockage of the fluid within the spinal column or between the skull and the spinal column would prevent the fluid level from rising.
Section 14.4 P14.22 (a)
Buoyant Forces and Archimede's Principle The balloon is nearly in equilibrium:  Fg Fy = ma y B  Fg
e j
helium
e j
payload
=0
or
air gV  helium gV  m payload g = 0 This reduces to m payload = air  helium V = 1.29 kg m3  0.179 kg m 3 400 m 3
b e
g e
je
j j
m payload = 444 kg (b) Similarly, m payload = air  hydrogen V = 1.29 kg m3  0.089 9 kg m3 400 m 3
j e
je
m payload = 480 kg The air does the lifting, nearly the same for the two balloons. P14.23 At equilibrium
F = 0 or
Fapp + mg = B B is the buoyant force. Fapp = B  mg B = Vol water g
where The applied force, where and
P14.24
a f So, F = aVolf g b g b 4 F = e1.90 10 mj e9.80 m s je10 kg m 3 F = bm + V g g must be equal to F = Vg
app water app 2 3 2 3 g s b w
b
g
m = Vol ball . 4  ball = r 3 g water  ball 3
3
g j
FIG. P14.23
 84.0 kg m 3 = 0.258 N
Since V = Ah , m + s Ah = w Ah and A =
b
m w  s h
g
FIG. P14.24
420 P14.25
Fluid Mechanics
(a)
Before the metal is immersed:
Fy = T1  Mg = 0 or
T1 = Mg = 1.00 kg 9.80 m s 2 = 9.80 N
T1
b
ge
j
scale
B
T2
(b)
After the metal is immersed:
Fy = T2 + B  Mg = 0
or
Mg Mg
T2 = Mg  B = Mg  w V g V= M
b
g
=
1.00 kg 2 700 kg m3
a b
Thus, T2 = Mg  B = 9.80 N  1 000 kg m3 *P14.26 (a) Fg (b)
FIG. P14.25
e
F 1. kg jGH 2 70000kg m
3
I 9.80 m s = j JK e
2
6.17 N .
Fy = 0 :
15 N  10 N + B = 0
B = 25.0 N
T B (c) (d) The oil pushes horizontally inward on each side of the block. String tension increases . The oil causes the water below to be under greater pressure, and the water pushes up more strongly on the bottom of the block. (e) Consider the equilibrium just before the string breaks: 15 N  60 N + 25 N+ Boil = 0 Boil = 50 N For the buoyant force of the water we have B = Vg 25 N = 1 000 kg m3 0.25Vblock 9.8 m s 2 60 N Boil 15 N
FIG. P14.26(a)
e
jb
g
25 N
Vblock = 1.02 10 2 m3 For the buoyant force of the oil 50 N = 800 kg m3 f e 1.02 10 2 m 3 9.8 m s 2 f e = 0.625 = 62.5% (f) 15 N + 800 kg m3 f f 1.02 10 2 m 3 9.8 m s 2 = 0 f f = 0.187 = 18.7%
FIG. P14.26(e)
e
j e
j
e
j e
j
15 N
Boil FIG. P14.26(f)
Chapter 14
421
P14.27
(a)
P = P0 + gh Taking P0 = 1.013 10 5 N m 2 and h = 5.00 cm we find For h = 17.0 cm, we get Since the areas of the top and bottom are we find and
Ptop = 1.017 9 10 5 N m 2
Pbot = 1.029 7 10 5 N m 2 A = 0.100 m
a
f
2
= 10 2 m 2
Ftop = Ptop A = 1.017 9 10 3 N Fbot = 1.029 7 10 3 N
FIG. P14.27
(b)
T + B  Mg = 0 where and Therefore,
B = w Vg = 10 3 kg m3 1.20 10 3 m3 9.80 m s 2 = 11.8 N
Mg = 10.0 9.80 = 98.0 N
T = Mg  B = 98.0  11.8 = 86.2 N
a f g
e
je
je
j
(c)
Fbot  Ftop = 1.029 7  1.017 9 10 3 N = 11.8 N which is equal to B found in part (b).
b
P14.28
Consider spherical balloons of radius 12.5 cm containing helium at STP and immersed in air at 0C and 1 atm. If the rubber envelope has mass 5.00 g, the upward force on each is B  Fg ,He  Fg , env = air Vg  HeVg  m env g Fup = air  He Fup
3 env
4 b gFGH 3 r IJK g  m g L4 O = a1. 29  0.179 f kg m M a0.125 mf Pe9.80 m s j  5.00 10 3 N Q
3 3 2
3
kg 9.80 m s 2 = 0.040 1 N
e
j
If your weight (including harness, strings, and submarine sandwich) is 70.0 kg 9.80 m s 2 = 686 N you need this many balloons: P14.29 (a) 686 N = 17 000 ~ 10 4 . 0.040 1 N
e
j
According to Archimedes, B = water Vwater g = 1.00 g cm 3 20.0 20.0 20.0  h g But B = Weight of block = mg = woodVwood g = 0.650 g cm 3 20.0 cm g
e
j
a
f
a f a fa fa f 20.0  h = 20.0a0.650 f so h = 20.0a1  0.650f = 7.00 cm
0.650 20.0 g = 1.00 20.0 20.0 20.0  h g
3
e
ja
f
3
(b)
B = Fg + Mg where M = mass of lead 1.00 20.0 g = 0.650 20.0 g + Mg
3 3 3
a f a f M = a1.00  0.650fa 20.0f = 0.350a 20.0 f
3
= 2 800 g = 2.80 kg
422 *P14.30
Fluid Mechanics
(a)
The weight of the ball must be equal to the buoyant force of the water: 1.26 kgg = water router 4 3 router g 3
13 3
F 3 1.26 kg I =G H 4 1 000 kg m JK FG 4 r H3 j
3 0
= 6.70 cm
(b)
The mass of the ball is determined by the density of aluminum: 4  ri3 3 4 1.26 kg = 2 700 kg m3 0.067 m 3 m = Al V = Al 1.11 10 4 m 3 = 3.01 10 4 m 3  ri3 ri = 1.89 10 4 m 3
IJ K FG IJ ea H K
f
3
 ri3
j
e
13
= 5.74 cm
*P14.31
Let A represent the horizontal crosssectional area of the rod, which we presume to be constant. The rod is in equilibrium:
Fy = 0 :
The density of the liquid is *P14.32
 mg + B = 0 =  0 Vwhole rod g + fluidVimmersed g
0 ALg = A L  h g
= 0L . Lh
a
f
We use the result of Problem 14.31. For the rod floating in a liquid of density 0.98 g cm 3 ,
= 0
L Lh
0.98 g cm3 =
3
a
0L L  0.2 cm
0.98 g cm L  0.98 g cm 3 0.2 cm = 0 L For floating in the dense liquid, 1.14 g cm 3 = 1.14 g cm 3 (a) By substitution,
e
f
j
aL  1.8 cmf  e1.14 g cm j1.8 cm = L
3 0
0L
1.14L  1.14 1.8 cm = 0.98L  0.2 0.98 0.16L = 1.856 cm L = 11.6 cm
a
f
a f
(b)
Substituting back,
0.98 g cm3 11.6 cm  0.2 cm = 0 11.6 cm
a
f
0 = 0.963 g cm
(c)
3
The marks are not equally spaced. Because =
0L is not of the form = a + bh , equalsize Lh steps of do not correspond to equalsize steps of h.
Chapter 14
423
P14.33
The balloon stops rising when Therefore,
b
air
 He gV = Mg
g
and
b
air
 He V = M ,
g
V=
400 M = air  He 1.25 e 1  0.180
V = 1 430 m3
P14.34
Since the frog floats, the buoyant force = the weight of the frog. Also, the weight of the displaced water = weight of the frog, so
oozeVg = m frog g
or m frog = oozeV = ooze 1 4 3 2 r = 1.35 10 3 kg m3 6.00 10 2 m 2 3 3
FG H
IJ e K
j e
j
3
Hence, m frog = 0.611 kg . P14.35 B = Fg V = sphere gV 2 1 sphere = H 2O = 500 kg m3 2 4 glycerin g V  sphere gV = 0 10 10 glycerin = 500 kg m3 = 1 250 kg m3 4
H 2O g
FG H
IJ K
FIG. P14.35
e
j
P14.36
Constant velocity implies zero acceleration, which means that the submersible is in equilibrium under the gravitational force, the upward buoyant force, and the upward resistance force:
Fy = ma y = 0
 1. 20 10 4 kg + m g + w gV + 1 100 N = 0
e
j
where m is the mass of the added water and V is the sphere's volume. 1.20 10 4 kg + m = 1.03 10 3 so P14.37
3
LM 4 a1.50f OP + 1 100 N N3 Q 9.8 m s
2
m = 2.67 10 3 kg
By Archimedes's principle, the weight of the fifty planes is equal to the weight of a horizontal slice of water 11.0 cm thick and circumscribed by the water line:
a f 50e 2.90 10 kg j g = e1 030 kg m j g a0.110 mf A
B = water g V
4 3
giving A = 1.28 10 4 m 2 . The acceleration of gravity does not affect the answer.
424
Fluid Mechanics
Section 14.5 Section 14.6 P14.38
Fluid Dynamics Bernoulli's Equation 1 1 1 000 v 2 = 6.00 10 4 N m 2 + 1 000 16 v 2 2 2 1 2.00 10 4 N m 2 = 1 000 15 v 2 2 v = 1.63 m s dm = Av = 1 000 5.00 10 2 dt
By Bernoulli's equation,
8.00 10 4 N m 2 +
b b
g g
b
g
e
jb
2
1.63 m s = 12.8 kg s
g
FIG. P14.38
P14.39
Assuming the top is open to the atmosphere, then P1 = P0 . Note P2 = P0 . Flow rate = 2.50 10 3 m3 min = 4.17 10 5 m3 s . (a) so A1 >> A 2 Assuming v1 = 0 , v1 << v 2
2 v 1 v 2 + gy1 = P2 + 2 + gy 2 2 2
P1 +
v 2 = 2 gy1 (b) Flow rate = A 2 v 2 =
b
g
12
= 2 9.80 16.0
5
a fa f
m3 s
12
= 17.7 m s
F d GH 4
2
I a17.7f = 4.17 10 JK
d = 1.73 10 3 m = 1.73 mm
*P14.40 Take point 1 at the free surface of the water in the tank and 2 inside the nozzle. F Fair water 1 2 1 2 (a) With the cork in place P1 + gy1 + v1 = P2 + gy 2 + v 2 becomes 2 2 f P0 + 1 000 kg m3 9.8 m s 2 7.5 m + 0 = P2 + 0 + 0 ; P2  P0 = 7.35 10 4 Pa . For the stopper Fx = 0 FIG. P14.40 Fwater  Fair  f = 0 P2 A  P0 A = f f = 7.35 10 4 Pa 0.011 m (b) Now Bernoulli's equation gives P0 + 7.35 10 4 Pa + 0 = P0 + 0 + v 2 = 12.1 m s The quantity leaving the nozzle in 2 h is 1 2 1 000 kg m 3 v 2 2
a
f
2
= 27.9 N
e
j
V = Av 2 t = 1 000 kg m3 0.011 m
continued on next page
e
ja
f b12.1 m sg7 200 s =
2
3.32 10 4 kg .
Chapter 14
425
(c)
Take point 1 in the wide hose and 2 just outside the nozzle. Continuity: A1 v 1 = A 2 v 2
FG 6.6 cm IJ H 2 K
2
v1 =
FG 2.2 cm IJ H 2 K
2
12.1 m s
12.1 m s = 1.35 m s 9 1 2 1 2 P1 + gy1 + v1 = P2 + gy 2 + v 2 2 2 1 1 2 P1 + 0 + 1 000 kg m3 1.35 m s = P0 + 0 + 1 000 kg m 3 12.1 m s 2 2 v1 =
e
jb
g
e
jb
g
2
P1  P0 = 7.35 10 4 Pa  9.07 10 2 Pa = 7.26 10 4 Pa P14.41 Flow rate Q = 0.012 0 m3 s = v1 A1 = v 2 A 2 v2 = *P14.42 (a) (b) *P14.43 Q 0.012 0 = = 31.6 m s A2 A2
P= PEL
FG IJ H K = 0.85e8.5 10 ja9.8 fa87f =
5
E mgh m = = gh = Rgh t t t 616 MW
The volume flow rate is 125 cm3 0.96 cm = Av1 = 16.3 s 2 The speed at the top of the falling column is v1 = Take point 2 at 13 cm below: P1 + gy1 + 1 2 1 2 v1 = P2 + gy 2 + v 2 2 2 7.67 cm 3 s 0.724 cm 2 = 10.6 cm s .
FG H
IJ K
2
v1 .
P0 + 1 000 kg m 3 9.8 m s 2 0.13 m + = P0 + 0 + 1 2 1 000 kg m3 v 2 2
e
je
j
1 1 000 kg m 3 0.106 m s 2
e
jb
g
2
e
j
v 2 = 2 9.8 m s 2 0.13 m + 0.106 m s The volume flow rate is constant: 7.67 cm 3 s = d = 0. 247 cm
e
j
b
g
2
= 1.60 m s
FG d IJ H 2K
2
160 cm s
426 *P14.44
Fluid Mechanics
(a)
Between sea surface and clogged hole: 1 atm + 0 + 1 030 kg m3 9.8 m s 2 2 m = P2 + 0 + 0
P1 +
1 2 1 2 v1 + gy1 = P2 + v 2 + gy 2 2 2
e
je
ja f
2
P2 = 1 atm + 20.2 kPa
The air on the back of his hand pushes opposite the water, so the net force on his hand is F = PA = 20. 2 10 3 N m 2 (b)
e
jFGH IJK e1.2 10 4 e
m
j
2
F = 2.28 N
Now, Bernoulli's theorem is 1 atm + 0 + 20. 2 kPa = 1 atm + The volume rate of flow is One acrefoot is Requiring 1 2 1 030 kg m 3 v 2 + 0 2
j
v 2 = 6. 26 m s
A2 v 2 =
1.2 10 2 m 4
e
j b6.26 m sg = 7.08 10
2
4
m3 s
4 047 m 2 0.304 8 m = 1 234 m3 1 234 m 3 7.08 10
4 3
m s
= 1.74 10 6 s = 20.2 days
P14.45
(a)
Suppose the flow is very slow:
FG P + 1 v + gyIJ = FG P + 1 v + gyIJ K H 2 K H 2 P + 0 + g a564 mf = 1 atm + 0 + g b 2 096 mg P = 1 atm + e1 000 kg m je9.8 m s jb1 532 mg = 1 atm + 15.0 MPa
2 2 river 3 2
rim
(b)
The volume flow rate is
4 500 m3 d = Av =
d 2 v 4
2.95 m s
v = 4 500 m3 d (c)
e
F 4 I d jFGH 861400 s IJK GH a0.150 mf JK =
2 2
Imagine the pressure as applied to stationary water at the bottom of the pipe:
FG P + 1 v H 2
IJ = FG P + 1 v + gyIJ K H 2 K 1 P + 0 = 1 atm + e1 000 kg m jb 2.95 m sg + 1 000 kg e9.8 m s jb1 532 mg 2
2
+ gy
bottom
top
3
2
2
P = 1 atm + 15.0 MPa + 4.34 kPa The additional pressure is 4.34 kPa .
Chapter 14
427
P14.46
(a)
For upward flight of a waterdrop projectile from geyser vent to fountaintop, 2 2 v yf = v yi + 2 a y y Then 0 = vi2 + 2 9.80 m s 2 +40.0 m and
e
ja
f
vi = 28.0 m s P1 + 1 2 1 2 v1 + gy1 = P2 + v 2 + gy 2 2 2
(b)
Between geyser vent and fountaintop: Air is so low in density that very nearly Then,
P1 = P2 = 1 atm 1 2 vi + 0 = 0 + 9.80 m s 2 40.0 m 2
e
ja
f
v1 = 28.0 m s (c) Between the chamber and the fountaintop: P1 + 1 2 1 2 v1 + gy1 = P2 + v 2 + gy 2 2 2
f je ja P  P = e1 000 kg m je9.80 m s ja 215 mf = e
1 0 3 2
P1 + 0 + 1 000 kg m 3 9.80 m s 2 175 m = P0 + 0 + 1 000 kg m 3 9.80 m s 2 +40.0 m 2.11 MPa
e
je
ja
f
P14.47
P1 +
2 v 1 2 A = P2 + 2 (Bernoulli equation), v1 A1 = v 2 A 2 where 1 = 4 2 2 A2 2 2 2 A1 v 2 2  1 and P = 1 15 = 21 000 Pa v 2  v1 = v 1 2 2 2 2 A2
P = P1  P2 =
e
j
F GH
I JK
v1 = 2.00 m s ; v 2 = 4v1 = 8.00 m s : The volume flow rate is
v1 A1 = 2.51 10 3 m3 s
Section 14.7 P14.48
Other Applications of Fluid Dynamics
Mg = P1  P2 A
where
b
g
for a balanced condition A = 80.0 m 2
16 000 9.80 = 7.00 10 4  P2 A
a f
P2 = 7.0 10 4  0.196 10 4 = 6.80 10 4 Pa
P14.49
air
v=
v2 = P = Hg gh 2 2 Hg gh
v air A h Mercury
air
= 103 m s
FIG. P14.49
428 P14.50
Fluid Mechanics
The assumption of incompressibility is surely unrealistic, but allows an estimate of the speed: 1 2 1 2 v1 = P2 + gy 2 + v 2 2 2 1 2 1.00 atm + 0 + 0 = 0.287 atm + 0 + 1.20 kg m 3 v 2 2 P1 + gy1 +
e
j
v2 = P14.51 (a) P0 + gh + 0 = P0 + 0 + If h = 1.00 m , (b) P + gy +
2 1.00  0.287 1.013 10 5 N m 2 1.20 kg m 1 2 v 3 2
3
a
fe
j=
347 m s
v 3 = 2 gh v 3 = 4.43 m s
1 2 1 2 v 2 = P0 + 0 + v 3 2 2 P = P0  gy y
5
Since v 2 = v 3 , Since P 0 *P14.52
FIG. P14.51
P0 1.013 10 Pa = = 10.3 m 3 g 10 kg m3 9.8 m s 2
e
je
j
Take points 1 and 2 in the air just inside and outside the window pane. 1 2 1 2 v1 + gy1 = P2 + v 2 + gy 2 2 2 1 P0 + 0 = P2 + 1.30 kg m3 11.2 m s 2 P1 +
e
jb
g
2
P2 = P0  81.5 Pa
(a)
The total force exerted by the air is outward, P1 A  P2 A = P0 A  P0 A + 81.5 N m 2 4 m 1.5 m = 489 N outward
e
ja fa
f
(b) P14.53
P1 A  P2 A =
1 2 1 v 2 A = 1.30 kg m 3 22.4 m s 2 2
e
jb
g a4 mfa1.5 mf =
2
1.96 kN outward
In the reservoir, the gauge pressure is From the equation of continuity:
P =
2.00 N = 8.00 10 4 Pa 2.50 10 5 m 2
A1 v1 = A 2 v 2
e2.50 10
5
m 2 v1 = 1.00 10 8 m 2 v 2
j e
j
v1 = 4.00 10 4 v 2
e
j
2 2 Thus, v1 is negligible in comparison to v 2 .
Then, from Bernoulli's equation:
bP  P g + 1 v 2
1 2
2 1
+ gy1 =
1 2 v 2 + gy 2 2 1 2 1 000 kg m3 v 2 2
8.00 10 4 Pa + 0 + 0 = 0 + v2 = 2 8.00 10 4 Pa 1 000 kg m
3
e
j
e
j=
12.6 m s
Chapter 14
429
Additional Problems P14.54 Consider the diagram and apply Bernoulli's A equation to points A and B, taking y = 0 at the level of point B, and recognizing that v A is approximately zero. This gives: PA + 1 2 w 0 + w g h  L sin 2 1 2 = PB + w v B + w g 0 2
af
a af
f
Valve
h L B
Now, recognize that PA = PB = Patmosphere since both points are open to the atmosphere (neglecting variation of atmospheric pressure with altitude). Thus, we obtain FIG. P14.54
v B = 2 g h  L sin = 2 9.80 m s 2 10.0 m  2.00 m sin 30.0 v B = 13.3 m s Now the problem reduces to one of projectile motion with v yi = v B sin 30.0 = 6.64 m s . Then,
2 2 v yf = v yi + 2 a y gives at the top of the arc (where y = y max and v yf = 0 )
a
f
e
j
a
f
b g
or y max = P14.55
g + 2e9.80 m s jby 2.25 m babove the level where the water emergesg .
0 = 6.64 m s
2 2
b
max
0
g
When the balloon comes into equilibrium, we must have
Fy = B  Fg , balloon  Fg , He  Fg , string = 0
Fg , string is the weight of the string above the ground, and B is the buoyant force. Now Fg , balloon = m balloon g Fg , He = HeVg B = air Vg and Fg , string = m string h g L
h
He
Therefore, we have
FIG. P14.55 h g=0 L
air Vg  m balloon g  HeVg  m string
or h=
b
air
 He V  m balloon m string
g
L
giving,
h=
a1.29  0.179fekg m jFH
3
4 0. 400 m 3
a
f
3
0.050 0 kg
I  0.250 kg K a2.00 mf =
1.91 m .
430 P14.56
Fluid Mechanics
Assume v inside 0 P + 0 + 0 = 1 atm + Pgauge 1 2 1 000 30.0 + 1 000 9.80 0.500 2 = P  1 atm = 4.50 10 5 + 4.90 10 3 = 455 kPa
b
ga f
a fa
f
P14.57
The "balanced" condition is one in which the apparent weight of the body equals the apparent weight of the weights. This condition can be written as: Fg  B = Fg  B where B and B are the buoyant forces on the body and weights respectively. The buoyant force experienced by an object of volume V in air equals: Buoyant force = Volume of object air g so we have Therefore, B = V air g Fg = Fg + V  and B =
FIG. P14.57
b
g
F GH
I g J K
Fg
2
F F I GH g JK
g
air g .
air g .
P14.58
The crosssectional area above water is 2.46 rad 0.600 cm 2 Aall = 0.600
0.400 cm
2
a
f  a0.200 cmfa0.566 cmf = 0.330 cm
a
f
2
= 1.13 cm 2
0.80 cm
water gAunder = wood Aall g 1.13  0.330 = 0.709 g cm 3 = 709 kg m3 wood = 1.13
P14.59 At equilibrium, giving But and Therefore, we have: or From the data given, Thus, this gives P14.60 P = gh h = 8.01 km
FIG. P14.58
Fy = 0 :
B  Fspring  Fg , He  Fg , balloon = 0 Fspring = kL = B  m He + m balloon g . B = weight of displaced air = air Vg m He = HeV . kL = air Vg  HeVg  m balloon g L= L=
b
g
b
air
 He V  m balloon k
3
g
g.
FIG. P14.59
e1.29 kg m
 0.180 kg m3 5.00 m3  2.00 10 3 kg 90.0 N m
j
e9.80 m s j .
2
L = 0.604 m . 1.013 10 5 = 1.29 9.80 h For Mt. Everest, 29 300 ft = 8.88 km Yes
a f
Chapter 14
431
P14.61
The torque is From the figure
= d = rdF
= y g H  y wdy =
0 H
z z z b
g
1 gwH 3 6
1 gwH 2 The total force is given as 2 If this were applied at a height y eff such that the torque remains unchanged, we have 1 1 gwH 3 = y eff gwH 2 6 2 P14.62 (a)
LM N
OP Q
and
y eff =
1 H . 3
FIG. P14.61
The pressure on the surface of the two hemispheres is constant at all points, and the force on each element of surface area is directed along the radius of the hemispheres. The applied force along the axis must balance the force on the "effective" area, which is the projection of the actual surface onto a plane perpendicular to the x axis, A = R 2 Therefore,
F=
bP  PgR
0
2
FIG. P14.62
(b) P14.63
For the values given
F = P0  0.100 P0 0.300 m
b
g a
f
2
= 0.254P0 = 2.58 10 4 N
Looking first at the top scale and the iron block, we have: T1 + B = Fg , iron where T1 is the tension in the spring scale, B is the buoyant force, and Fg , iron is the weight of the iron block. Now if m iron is the mass of the iron block, we have m iron = ironV Then, B = oil Viron g Therefore, T1 = Fg , iron  oilViron g = m iron g  oil or T1 = 1  m iron so V= m iron
iron
= Vdisplaced oil
FG H
oil 916 m iron g = 1  2.00 9.80 = 17.3 N 7 860 iron
IJ K
F GH
I a fa f JK
iron
g
Next, we look at the bottom scale which reads T2 (i.e., exerts an upward force T2 on the system). Consider the external vertical forces acting on the beakeroiliron combination.
Fy = 0 gives
T1 + T2  Fg , beaker  Fg , oil  Fg , iron = 0 or T2 = m beaker + m oil + m iron g  T1 = 5.00 kg 9.80 m s 2  17.3 N
b
g
b
ge
j
Thus, T2 = 31.7 N is the lower scale reading.
432 P14.64
Fluid Mechanics
FG m IJ g H K is the buoyant force exerted on the iron block by the oil. F m IJ g B=m g G Thus, T =F H K F IJ m g is the reading on the top scale. or T = G1  H K
T1 + B = Fg , Fe where B = 0 VFe g = 0
Fe Fe 1 g , Fe Fe 0 Fe Fe 1 0 Fe Fe
Looking at the top scale and the iron block:
Now, consider the bottom scale, which exerts an upward force of T2 on the beakeroiliron combination. T1 + T2  Fg , beaker  Fg , oil  Fg , Fe = 0 Fy = 0 : T2 = Fg , beaker + Fg , oil + Fg , Fe  T1 = m b + m 0 + m Fe g  1  or P14.65 T2 =
b
LMm MN
b
+ m0 +
FG IJ m OP g H K PQ
0 Fe Fe
g FGH
0 m Fe g Fe
IJ K
is the reading on the bottom scale.
a f a f F 3.083 IJ x + 3.083a1  xf = 2.517 G H K FG 1  7.133 IJ x = FG 1  2.517 IJ H 8.960 K H 3.083 K
Zn Cu
CuV = 3.083 g Zn xV + Cu 1  x V = 2.517 g
x = 0.900 4 %Zn = 90.04% P14.66 (a) From
F = ma
B = water Vg
B  m shell g  m He g = m total a = m shell + m He a Where and
b
g
(1)
d 3 4 Also, V = r 3 = 3 6 Putting these into equation (1) above,
m He = HeV
Fm GH
shell
+ He
d 3 d 3 d 3  m shell  He a = water g 6 6 6
water
I F JK GH
I JK
which gives
b a=
 He
g
or
(b)
t=
b1 000  0.180gekg m j a f  4.00 kg 9.80 m s a= a f 4.00 kg + e0.180 kg m j 2a h  d f 2a 4.00 m  0.200 mf 2x = = = 4.06 s
3 0. 200 m 6
3
3 m shell + He d 6
d 3 6
 m shell
g
2
3 0. 200 m 6
3
= 0.461 m s 2
a
a
0.461 m s 2
Chapter 14
433
P14.67
Inertia of the disk: I =
1 1 MR 2 = 10.0 kg 0.250 m 2 2
b
ga
f
2
= 0.312 kg m 2
Angular acceleration: f = i + t
=
Braking torque:
FG 0  300 rev min IJ FG 2 rad IJ FG 1 min IJ = 0.524 rad s H 60.0 s K H 1 rev K H 60.0 s K
f=  I d
2 2
2
= I  fd = I , so
0.220 m
e0.312 kg m je0.524 rad s j = 0.744 N Friction force: f =
Normal force: f = k n n = gauge pressure: P = f 0.744 N = = 1.49 N k 0.500
1.49 N n = A 2.50 10 2 m
e
j
2
= 758 Pa
P14.68
The incremental version of P  P0 = gy is We assume that the density of air is proportional to pressure, or Combining these two equations we have
dP =  gdy P
=
0 0 gdy P0
P0
dP =  P
P0
z
P
dP =  g 0 dy P P0 0
z
h
and integrating gives
ln
FG P IJ =  gh HP K P
0 0 0
so where = P14.69
0 g , P0
P = P0 e h
Energy for the fluidEarth system is conserved.
aK + Uf + E
i
mech
= K +U f :
a
f
0+
mgL 1 + 0 = mv 2 + 0 2 2
v = gL = 2.00 m 9.8 m s 2 = 4.43 m s
e
j
434 P14.70
Fluid Mechanics
Let s stand for the edge of the cube, h for the depth of immersion, ice stand for the density of the ice, w stand for density of water, and a stand for density of the alcohol. (a) According to Archimedes's principle, at equilibrium we have
ice gs 3 = w ghs 2 h = s
With
ice w
ice = 0.917 10 3 kg m3 w = 1.00 10 3 kg m3
and we get (b)
s = 20.0 mm
h = 20.0 0.917 = 18.34 mm 18.3 mm
a
f
We assume that the top of the cube is still above the alcohol surface. Letting ha stand for the thickness of the alcohol layer, we have
a gs 2 ha + w gs 2 hw = ice gs 3
With and we obtain (c) Here
so
hw =
FG IJ s  FG IJ h H K H K
ice w a w
a
a = 0.806 10 3 kg m3
ha = 5.00 mm
hw = 18.34  0.806 5.00 = 14.31 mm 14.3 mm
hw = s  ha , so Archimedes's principle gives
a f
a gs 2 ha + w gs 2 s  ha = ice gs 3 a ha + w s  ha = ice s
ha = s
b b
w w
 ice
a
g = 20.0 a1.000  0.917f = 8.557  g a1.000  0.806f
b
g
b
g
8.56 mm
Chapter 14
435
P14.71
Note: Variation of atmospheric pressure with altitude is included in this solution. Because of the small distances involved, this effect is unimportant in the final answers. (a) Consider the pressure at points A and B in part (b) of the figure: Using the left tube: PA = Patm + a gh + w g L  h where the second term is due to the variation of air pressure with altitude. Using the right tube: PB = Patm + 0 gL But Pascal's principle says that PA = PB . Therefore, or h= (b)
a
f
Patm + 0 gL = Patm + a gh + w g L  h
a
f
b
FG H
w
w
 a h = w  0 L , giving
g b
g
1 000  750  0 L= 5.00 cm = 1.25 cm  a 1 000  1.29 w
IJ F K GH
I JK
Consider part (c) of the diagram showing the situation when the air flow over the left tube equalizes the fluid levels in the two tubes. First, apply Bernoulli's equation to points A and B y A = y B , v A = v , and v B = 0
b
This gives: PA +
1 a v 2 + a gy A 2
g 1 = P + a0 f 2
B a
2
+ a gy B (1) FIG. P14.71
and since y A = y B , this reduces to: PB  PA =
1 av2 2
Now consider points C and D, both at the level of the oilwater interface in the right tube. Using the variation of pressure with depth in static fluids, we have: PC = PA + a gH + w gL and PD = PB + a gH + 0 gL
But Pascal's principle says that PC = PD . Equating these two gives: PB + a gH + 0 gL = PA + a gH + w gL or PB  PA = w  0 gL 1 a v 2 = w  0 gL 2
b
g
(2)
Substitute equation (1) for PB  PA into (2) to obtain 2 gL w  0
b
g
or
v=
b
a
g = 2e9.80 m s jb0.050 0 mgFG 1 000  750 IJ H 1.29 K
2
v = 13.8 m s
436 P14.72
Fluid Mechanics
(a)
The flow rate, Av, as given may be expressed as follows: 25.0 liters = 0.833 liters s = 833 cm3 s . 30.0 s The area of the faucet tap is cm 2 , so we can find the velocity as v= flow rate 833 cm 3 s = = 265 cm s = 2.65 m s . A cm 2
(b)
We choose point 1 to be in the entrance pipe and point 2 to be at the faucet tap. A1 v1 = A 2 v 2 gives v1 = 0.295 m s . Bernoulli's equation is: P1  P2 = and gives P1  P2 = or 1 10 3 kg m3 2 1 2 2 v 2  v 1 + g y 2  y 1 2
e
j b
2
g je ja f
e
j b2.65 m sg  b0.295 m sg + e10
2
3
kg m3 9.80 m s 2 2.00 m
Pgauge = P1  P2 = 2.31 10 4 Pa .
P14.73
(a)
Since the upward buoyant force is balanced by the weight of the sphere, m1 g = Vg =
FG 4 R IJ g . H3 K
3 3
In this problem, = 0.789 45 g cm 3 at 20.0C, and R = 1.00 cm so we find: m1 = (b)
FG 4 R IJ = e0.789 45 g cm jLM 4 a1.00 cmf OP = H3 K N3 Q
3 3
3.307 g .
Following the same procedure as in part (a), with = 0.780 97 g cm 3 at 30.0C, we find: m2 =
FG 4 R IJ = e0.780 97 g cm jLM 4 a1.00 cmf OP = H3 K N3 Q
3 3 3
3.271 g .
(c)
When the first sphere is resting on the bottom of the tube, n + B = Fg 1 = m1 g , where n is the normal force. Since B = Vg n = m1 g  Vg = 3.307 g  0.780 97 g cm 3 1.00 cm n = 34.8 g cm s 2 = 3.48 10 4 N
e
ja
f
3
980 cm s 2
Chapter 14
437
*P14.74
(a)
Take point 1 at the free water surface in the tank and point 2 at the bottom end of the tube: 1 2 1 2 v1 = P2 + gy 2 + v 2 2 2 1 2 P0 + gd + 0 = P0 + 0 + v 2 2 v 2 = 2 gd P1 + gy1 + The volume flow rate is
2
V Ah Ah Ah . = = v 2 A . Then t = = t t v 2 A A 2 gd 44.6 s
(b)
t=
2 10 4
a0.5 mf 0.5 m = m 2e9.8 m s j10 m
2 2
*P14.75
(a)
For diverging stream lines that pass just above and just below the hydrofoil we have Pt + gy t + 1 2 1 2 v t = Pb + gy b + v b . 2 2
Ignoring the buoyant force means taking y t y b 1 1 2 2 nv b = Pb + v b 2 2 1 2 2 Pb  Pt = v b n  1 2 Pt +
b g
e
j
The lift force is Pb  Pt A = (b) For liftoff,
b
g
1 2 2 v b n  1 A . 2
e
j
1 2 2 v b n  1 A = Mg 2
e
j
vb
F 2Mg I =G GH en  1jA JJK
2
12
The speed of the boat relative to the shore must be nearly equal to this speed of the water below the hydrofoil relative to the boat. (c) v 2 n 2  1 A = 2 Mg A=
e
b g b9.5 m sg e1.05  1j1 000 kg m
2 800 kg 9.8 m s 2
2 2
j
3
= 1.70 m 2
438
Fluid Mechanics
ANSWERS TO EVEN PROBLEMS
P14.2 ~ 10 18 kg m3 ; matter is mostly empty space 1.92 10 4 N (a) 1.01 10 7 Pa ; (b)7.09 10 5 N outward 255 N (a) 65.1 N; (b) 275 N 5.88 10 N down; 196 kN outward; 588 kN outward (a) 29.4 kN to the right; (b) 16.3 kN m counterclockwise (a) 10.3 m; (b) zero (a) 20.0 cm; (b) 0.490 cm 12.6 cm (a) 444 kg; (b) 480 kg
6
P14.38 P14.40
12.8 kg s (a) 27.9 N; (b) 3.32 10 4 kg ; (c) 7.26 10 4 Pa
P14.4 P14.6
P14.42 P14.44 P14.46 P14.48 P14.50 P14.52 P14.54 P14.56 P14.58 P14.60 P14.62 P14.64
(a) see the solution; (b) 616 MW (a) 2.28 N toward Holland; (b) 1.74 10 6 s (a), (b) 28.0 m s ; (c) 2.11 MPa 6.80 10 4 Pa 347 m s (a) 489 N outward; (b) 1.96 kN outward 2.25 m above the level where the water emerges 455 kPa 709 kg m3 8.01 km; yes (a) see the solution; (b) 2.58 10 4 N top scale: 1 
P14.8 P14.10 P14.12 P14.14 P14.16 P14.18 P14.20 P14.22 P14.24 P14.26
b
m w  s h
g
(a) see the solution; (b) 25.0 N up; (c) horizontally inward; (d) tension increases; see the solution; (e) 62.5%; (f) 18.7% ~ 10 4 balloons of 25cm diameter (a) 6.70 cm; (b) 5.74 cm (a) 11.6 cm; (b) 0.963 g cm ; (c) no; see the solution 0.611 kg 2.67 10 3 kg
3
FG H
0 m Fe g ; Fe
IJ K
bottom scale: m b + m 0 + P14.66 P14.68 P14.70 P14.72 P14.74 (a) 0.461 m s 2 ; (b) 4.06 s see the solution
FG H
0 m Fe g Fe
IJ K
P14.28 P14.30 P14.32
(a) 18.3 mm; (b) 14.3 mm; (c) 8.56 mm (a) 2.65 m s ; (b) 2.31 10 4 Pa (a) see the solution; (b) 44.6 s
P14.34 P14.36
15
Oscillatory Motion
CHAPTER OUTLINE
15.1 15.2 15.3 15.4 Motion of an Object Attached to a Spring Mathematical Representation of Simple Harmonic Motion Energy of the Simple Harmonic Oscillator Comparing Simple Harmonic Motion with Uniform Circular Motion The Pendulum Damped Oscillations Forced Oscillations
ANSWERS TO QUESTIONS
Q15.1 Neither are examples of simple harmonic motion, although they are both periodic motion. In neither case is the acceleration proportional to the position. Neither motion is so smooth as SHM. The ball's acceleration is very large when it is in contact with the floor, and the student's when the dismissal bell rings. You can take = , or equally well, =  . At t = 0 , the particle is at its turning point on the negative side of equilibrium, at x = A . The two will be equal if and only if the position of the particle at time zero is its equilibrium position, which we choose as the origin of coordinates.
15.5 15.6 15.7
Q15.2
Q15.3
Q15.4
(a)
In simple harmonic motion, onehalf of the time, the velocity is in the same direction as the displacement away from equilibrium. Velocity and acceleration are in the same direction half the time. Acceleration is always opposite to the position vector, and never in the same direction.
(b) (c) Q15.5 Q15.6
No. It is necessary to know both the position and velocity at time zero. The motion will still be simple harmonic motion, but the period of oscillation will be a bit larger. The
F kI effective mass of the system in = G H m JK
eff
12
will need to include a certain fraction of the mass of the
spring.
439
440 Q15.7
Oscillatory Motion
We assume that the coils of the spring do not hit one another. The frequency will be higher than f by the factor 2 . When the spring with two blocks is set into oscillation in space, the coil in the center of the spring does not move. We can imagine clamping the center coil in place without affecting the motion. We can effectively duplicate the motion of each individual block in space by hanging a single block on a halfspring here on Earth. The halfspring with its center coil clampedor its other half cut offhas twice the spring constant as the original uncut spring, because an applied force of the same size would produce only onehalf the extension distance. Thus the oscillation frequency in space is
FG 1 IJ FG 2 k IJ H 2 K H m K
12
= 2 f . The absence of a force required to support the vibrating system in
orbital free fall has no effect on the frequency of its vibration. Q15.8 No; Kinetic, Yes; Potential, No. For constant amplitude, the total energy kinetic energy 1 2 kA stays constant. The 2
1 mv 2 would increase for larger mass if the speed were constant, but here the greater 2 mass causes a decrease in frequency and in the average and maximum speed, so that the kinetic and potential energies at every point are unchanged. Q15.9 Since the acceleration is not constant in simple harmonic motion, none of the equations in Table 2.2 are valid. Equation x t = A cos t +
2
af b g af b g va x f = e A  x j aat f =  A cosbt + g aat f =  xat f
v t = A sin t +
2 2 2 12
Information given by equation position as a function of time velocity as a function of time velocity as a function of position acceleration as a function of time acceleration as a function of position
The angular frequency appears in every equation. It is a good idea to figure out the value of angular frequency early in the solution to a problem about vibration, and to store it in calculator memory. Q15.10 Lf Li 2 Li and T f = = = 2Ti . The period gets larger by g g g mass has no effect on the period of a simple pendulum. We have Ti = (a) Period decreases. (b) Period increases. (c) 2 times. Changing the
Q15.11 Q15.12
No change.
No, the equilibrium position of the pendulum will be shifted (angularly) towards the back of the car. The period of oscillation will increase slightly, since the restoring force (in the reference frame of the accelerating car) is reduced. The motion will be periodicthat is, it will repeat. The period is nearly constant as the angular amplitude increases through small values; then the period becomes noticeably larger as increases farther. Shorten the pendulum to decrease the period between ticks. No. If the resistive force is greater than the restoring force of the spring (in particular, if b 2 > 4mk ), the system will be overdamped and will not oscillate.
Q15.13
Q15.14 Q15.15
Chapter 15
441
Q15.16
Yes. An oscillator with damping can vibrate at resonance with amplitude that remains constant in time. Without damping, the amplitude would increase without limit at resonance. The phase constant must be rad . Higher frequency. When it supports your weight, the center of the diving board flexes down less than the end does when it supports your weight. Thus the stiffness constant describing the center of 1 k is greater the board is greater than the stiffness constant describing the end. And then f = 2 m for you bouncing on the center of the board.
Q15.17 Q15.18
FG IJ H K
Q15.19
The release of air from one side of the parachute can make the parachute turn in the opposite direction, causing it to release air from the opposite side. This behavior will result in a periodic driving force that can set the parachute into sidetoside oscillation. If the amplitude becomes large enough, the parachute will not supply the needed air resistance to slow the fall of the unfortunate skydiver. An imperceptibly slight breeze may be blowing past the leaves in tiny puffs. As a leaf twists in the wind, the fibers in its stem provide a restoring torque. If the frequency of the breeze matches the natural frequency of vibration of one particular leaf as a torsional pendulum, that leaf can be driven into a largeamplitude resonance vibration. Note that it is not the size of the driving force that sets the leaf into resonance, but the frequency of the driving force. If the frequency changes, another leaf will be set into resonant oscillation. We assume the diameter of the bob is not very small compared to the length of the cord supporting it. As the water leaks out, the center of mass of the bob moves down, increasing the effective length of the pendulum and slightly lowering its frequency. As the last drops of water dribble out, the center of mass of the bob hops back up to the center of the sphere, and the pendulum frequency quickly increases to its original value.
Q15.20
Q15.21
SOLUTIONS TO PROBLEMS
Section 15.1 P15.1 (a) Motion of an Object Attached to a Spring Since the collision is perfectly elastic, the ball will rebound to the height of 4.00 m and then repeat the motion over and over again. Thus, the motion is periodic . To determine the period, we use: x = 1 2 gt . 2
(b)
The time for the ball to hit the ground is t =
2 4.00 m 2x = = 0.909 s g 9.80 m s 2
a
f
This equals onehalf the period, so T = 2 0.909 s = 1.82 s . (c) No . The net force acting on the ball is a constant given by F =  mg (except when it is in contact with the ground), which is not in the form of Hooke's law.
a
f
442
Oscillatory Motion
Section 15.2 P15.2 (a) (b) (c) (d) P15.3
Mathematical Representation of Simple Harmonic Motion x = 5.00 cm cos 2t + v= a=
a
f FGH
6
IJ K IJ K IJ K
At t = 0 , At t = 0 , At t = 0 , and
x = 5.00 cm cos v = 5.00 cm s
a
f FGH IJK = 6
4.33 cm
dx =  10.0 cm s sin 2t + dt 6
b
g FGH
dv =  20.0 cm s 2 cos 2t + dt 6
e
j FGH
a = 17.3 cm s 2
T= 2 = 2 = 3.14 s 2
A = 5.00 cm
x = 4.00 m cos 3.00t + Compare this with x = A cos t + to find
(a)
a
f a
f
b
g
= 2 f = 3.00
or f = 1.50 Hz T= 1 = 0.667 s f
(b) (c) (d) *P15.4 (a)
A = 4.00 m
= rad
x t = 0.250 s = 4.00 m cos 1.75 = 2.83 m The spring constant of this spring is k= F 0.45 kg 9.8 m s 2 = = 12.6 N m x 0.35 m
a
f a
f a
f
we take the xaxis pointing downward, so = 0 x = A cos t = 18.0 cm cos (d) 12.6 kg 0.45 kg s 2 84. 4 s = 18.0 cm cos 446.6 rad = 15.8 cm
Now 446.6 rad = 71 2 + 0.497 rad . In each cycle the object moves 4 18 = 72 cm , so it has moved 71 72 cm + 18  15.8 cm = 51.1 m .
a
f a
f
a f
(b)
By the same steps, k = x = A cos
0. 44 kg 9.8 m s 2 = 12.1 N m 0.355 m
k 12.1 t = 18.0 cm cos 84.4 = 18.0 cm cos 443.5 rad = 15.9 cm m 0.44
(e)
443.5 rad = 70 2 + 3.62 rad
Distance moved = 70 72 cm + 18 + 15.9 cm = 50.7 m
a f
a
f
(c)
The answers to (d) and (e) are not very different given the difference in the data about the two vibrating systems. But when we ask about details of the future, the imprecision in our knowledge about the present makes it impossible to make precise predictions. The two oscillations start out in phase but get completely out of phase.
Chapter 15
443
P15.5
(a)
At t = 0 , x = 0 and v is positive (to the right). Therefore, this situation corresponds to x = A sin t and Since f = 1.50 Hz , Also, A = 2.00 cm, so that v = vi cos t
= 2 f = 3.00
x = 2.00 cm sin 3.00 t
a
f
(b)
v max = vi = A = 2.00 3.00 = 6.00 cm s = 18.8 cm s
The particle has this speed at t = 0 and next at a max = A 2 = 2.00 3.00 t= T 1 = s 2 3
a
f
(c)
a
f
2
= 18.0 2 cm s 2 = 178 cm s 2 t= 3 T = 0.500 s 4
This positive value of acceleration first occurs at (d) Since T =
2 s and A = 2.00 cm, the particle will travel 8.00 cm in this time. 3 3 Hence, in 1.00 s = T , the particle will travel 8.00 cm + 4.00 cm = 12.0 cm . 2
FG H
IJ K
P15.6
The proposed solution implies velocity and acceleration (a)
x t = xi cos t + v= a=
af
FG v IJ sin t HK
i
dx =  x i sin t + vi cos t dt
v dv =  x i 2 cos t  vi sin t =  2 x i cos t + i sin t =  2 x dt
FG H
FG IJ H K
IJ K
The acceleration being a negative constant times position means we do have SHM, and its angular frequency is . At t = 0 the equations reduce to x = xi and v = vi so they satisfy all the requirements. v 2  ax =  x i sin t + vi cos t
(b)
b
g  e x
2 i
2
cos t  vi sin t xi cos t +
jFGH
FG v IJ sin tIJ HK K
i
v 2  ax = xi2 2 sin 2 t  2 xi vi sin t cos t + vi2 cos 2 t + x i2 2 cos 2 t + x i vi cos t sin t + xi vi sin t cos t + vi2 sin 2 t = x i2 2 + vi2 So this expression is constant in time. On one hand, it must keep its original value vi2  ai xi . On the other hand, if we evaluate it at a turning point where v = 0 and x = A , it is A 2 2 + 0 2 = A 2 2 . Thus it is proved. P15.7 (a) (b) (c) T= f= 12.0 s = 2.40 s 5 1 1 = = 0.417 Hz T 2. 40
= 2 f = 2 0.417 = 2.62 rad s
a
f
444 *P15.8
Oscillatory Motion
The mass of the cube is m = V = 2.7 10 3 kg m3 0.015 m The spring constant of the strip of steel is k= f= 14.3 N F = = 52.0 N m x 0.027 5 m
e
ja
f
3
= 9.11 10 3 kg
2
=
1 2
k 1 = m 2
52 kg s 2 9.11 10 3 kg
= 12.0 Hz
P15.9
f=
1 = 2 2
k m
or
T= k=
1 m = 2 f k 4 2 m T
2
Solving for k, x = A cos t A = 0.05 m
=
4 2 7.00 kg
a2.60 sf
b
2
g=
40.9 N m . a =  A 2 cos t
*P15.10
v =  A sin t
If f = 3 600 rev min = 60 Hz , then = 120 s 1 v max = 0.05 120 m s = 18.8 m s P15.11 (a)
a
f
amax = 0.05 120
a
f
2
m s 2 = 7.11 km s 2 x = 10.0 sin 4.00t cm .
v max = 40.0 cm s
2
=
k = m
8.00 N m = 4.00 s 1 0.500 kg
so position is given by
a
f
From this we find that
a f a = 160 sina 4.00t f cm s
v = 40.0 cos 4.00t cm s
x = 6.00 cm, t = 0.161 s.
amax = 160 cm s 2 .
(b)
t=
FG 1 IJ sin FG x IJ and when H 4.00 K H 10.0 K
1
We find
a f a = 160 sin 4.00a0.161f = FG 1 IJ sin FG x IJ H 4.00 K H 10.0 K
1
v = 40.0 cos 4.00 0.161 = 32.0 cm s
96.0 cm s 2 .
(c)
Using t =
when x = 0 , t = 0 and when Therefore,
x = 8.00 cm, t = 0.232 s. t = 0.232 s .
Chapter 15
445
P15.12
m = 1.00 kg , k = 25.0 N m, and A = 3.00 cm. At t = 0 , x = 3.00 cm (a)
=
k = m
so that, (b)
25.0 = 5.00 rad s 1.00 2 2 T= = = 1.26 s 5.00
v max = A = 3.00 10 2 m 5.00 rad s = 0.150 m s
b
g
amax = A 2 = 3.00 10 2 m 5.00 rad s
(c)
b
g
2
= 0.750 m s 2
Because x = 3.00 cm and v = 0 at t = 0 , the required solution is x =  A cos t or x = 3.00 cos 5.00t cm dx = 15.0 sin 5.00t cm s dt dv a= = 75.0 cos 5.00t cm s 2 dt v=
a
f
a a
f f
P15.13
The 0.500 s must elapse between one turning point and the other. Thus the period is 1.00 s.
=
2 = 6. 28 s T
and v max = A = 6.28 s 0.100 m = 0.628 m s . P15.14 (a) v max = A A= v max
b
ga
f
=
v
(b)
x =  A sin t = 
FG v IJ sin t HK
Section 15.3 P15.15 (a)
Energy of the Simple Harmonic Oscillator Energy is conserved for the blockspring system between the maximumdisplacement and the halfmaximum points:
aK + U f = aK + U f 1 b6.50 N mga0.100 mf 2
i f
0+
2
1 2 1 1 kA = mv 2 + kx 2 2 2 2
=
1 m 0.300 m s 2 + 8.12 mJ
b
g
2
+
1 6.50 N m 5.00 10 2 m 2 2 24.4 mJ 9.0 10 2 =
2
b
ge
j
2
32.5 mJ = k = m
1 m 0.300 m s 2
b
g
2
m=
a
f
m s2
2
= 0.542 kg
(b) (c)
=
6.50 N m = 3.46 rad s 0.542 kg
T =
2 rad = 1.81 s 3.46 rad s
amax = A 2 = 0.100 m 3.46 rad s
b
g
2
= 1.20 m s 2
446 P15.16
Oscillatory Motion
m = 200 g , T = 0.250 s, E = 2.00 J ; = (a) (b)
2 2 = = 25.1 rad s T 0. 250
2
k = m 2 = 0.200 kg 25.1 rad s
E= kA 2 A= 2 2E = k
b
g = 126 N m 2a 2.00f = 0.178 m
126
P15.17
Choose the car with its shockabsorbing bumper as the system; by conservation of energy, 1 1 mv 2 = kx 2 : 2 2 v=x k = 3.16 10 2 m m
e
j
5.00 10 6 = 2.23 m s 10 3
P15.18
(a) (b) (c)
2 kA 2 250 N m 3.50 10 m E= = 2 2
e
j
2
= 0.153 J k = m 250 = 22.4 s 1 0.500 v max = 0.784 m s
v max = A
where
=
a max = A 2 = 3.50 10 2 m 22.4 s 1 E=
e
j
2
= 17.5 m s 2
P15.19
(a) (b)
1 2 1 kA = 35.0 N m 4.00 10 2 m 2 2 k A2  x2 m
2 2
b
ge
j
2
= 28.0 mJ
v = A2  x2 = v= 35.0
3
.
(c) (d) P15.20 (a) (b) (c) (d) (e) (f) (g)
e4.00 10 j  e1.00 10 j = 1.02 m s 50.0 10 1 1 1 1 mv = kA  kx = a35.0 fLe 4.00 10 j  e3.00 10 j O = MN PQ 2 2 2 2
2 2 2 2 2 2 2 2 2
12.2 mJ
1 2 1 kx = E  mv 2 = 15.8 mJ 2 2 k= F 20.0 N = = 100 N m x 0.200 m k = 50.0 rad s m so f=
=
2
= 1.13 Hz
v max = A = 50.0 0. 200 = 1.41 m s at x = 0
amax = 2
E= 1 2 kA 2
a f A = 50.0a0.200 f = 10.0 m s 1 = a100 fa0.200 f = 2.00 J 2
2
2
at x = A
v = A 2  x 2 = 50.0 a = 2 x = 50.0
8 0.200 9
a
f
2
= 1.33 m s
FG 0.200 IJ = H 3 K
3.33 m s 2
Chapter 15
447
P15.21
(a)
E=
1 2 1 kA , so if A = 2 A , E = k A 2 2
a f
2
=
1 k 2A 2
a f
2
= 4E
Therefore E increases by factor of 4 . (b) (c) (d) v max = a max = T = 2 k A , so if A is doubled, v max is doubled . m k A , so if A is doubled, a max also doubles . m m is independent of A, so the period is unchanged . k
*P15.22
(a)
1 ayt 2 2 1 11 m = 0 + 0 + 9.8 m s 2 t 2 2 y f = yi + v yi t +
e
j
t= (b)
22 m s 2 = 1.50 s 9.8 m
Take the initial point where she steps off the bridge and the final point at the bottom of her motion.
eK + U
g
+ Us
j = eK + U
i
g
+ Us
j
f
1 0 + mgy + 0 = 0 + 0 + kx 2 2 1 65 kg 9.8 m s 2 36 m = k 25 m 2 k = 73. 4 N m
a
f
2
(c)
The spring extension at equilibrium is x =
F 65 kg 9.8 m s 2 = = 8.68 m , so this point is k 73.4 N m
11 + 8.68 m = 19.7 m below the bridge and the amplitude of her oscillation is 36  19.7 = 16.3 m . (d) (e)
=
k = m
73.4 N m = 1.06 rad s 65 kg
Take the phase as zero at maximum downward extension. We find what the phase was 25 m higher when x = 8.68 m: In x = A cos t , t 8.68 m = 16.3 m cos 1.06 s t = 2.01 s
FG H
IJ K
16.3 m = 16.3 m cos 0 t 1.06 = 122 = 2.13 rad s
Then +2.01 s is the time over which the spring stretches. (f) total time = 1.50 s + 2.01 s = 3.50 s
448 P15.23
Oscillatory Motion
Model the oscillator as a blockspring system. From energy considerations, v max = A and v = v2 + 2x2 = 2 A2 so 3 2 A 4 and
A 2
FG A IJ H2K
x=
2
+ 2x2 = 2 A2
From this we find x 2 = P15.24 The potential energy is
3 A = 2.60 cm where A = 3.00 cm 2
Us = The rate of change of potential energy is
1 2 1 2 kx = kA cos 2 t . 2 2
a f
dU s 1 2 1 = kA 2 cos t  sin t =  kA 2 sin 2t . 2 2 dt (a) This rate of change is maximal and negative at 2t =
a f
a f
, 2t = 2 + , or in general, 2t = 2n + for integer n. 2 2 2
Then, t =
4n + 1 4n + 1 = 4 4 3.60 s 1
a
f
e
a
f
j
For n = 0 , this gives t = 0.218 s while n = 1 gives t = 1.09 s . All other values of n yield times outside the specified range. (b) dU s dt =
max
1 2 1 kA = 3.24 N m 5.00 10 2 m 2 2
b
ge
j e3.60 s j =
2 1
14.6 mW
Section 15.4 P15.25 (a) (b)
Comparing Simple Harmonic Motion with Uniform Circular Motion The motion is simple harmonic because the tire is rotating with constant velocity and you are looking at the motion of the bump projected in a plane perpendicular to the tire. Since the car is moving with speed v = 3.00 m s , and its radius is 0.300 m, we have:
=
3.00 m s = 10.0 rad s . 0.300 m
Therefore, the period of the motion is: T= 2
=
b10.0 rad sg =
2
0.628 s .
Chapter 15
449
P15.26
The angle of the crank pin is = t . Its xcoordinate is x = A cos = A cos t where A is the distance from the center of the wheel to the crank pin. This is of the form x = A cos t + , so the yoke and piston rod move with simple harmonic motion.
Piston A x = A x ( t)
b
g
FIG. P15.26
Section 15.5 P15.27 (a)
The Pendulum T = 2 L= gT 2 4 2 L g =
e9.80 m s ja12.0 sf
2
2
4 2 L g moon = 2
= 35.7 m
(b)
Tmoon = 2
35.7 m 1.67 m s 2 TT = 2 TC = 2
= 29.1 s LT gT LC gC
P15.28
The period in Tokyo is and the period in Cambridge is We know For which, we see or
TT = TC = 2.00 s LT LC = gT gC g C LC 0.994 2 = = = 1.001 5 g T LT 0.992 7 I . mgd
P15.29
The swinging box is a physical pendulum with period T = 2 The moment of inertia is given approximately by I=
1 mL2 (treating the box as a rod suspended from one end). 3 L , 2
Then, with L 1.0 m and d
1 3
T 2
mL2
L 2
mg
ch
= 2
2 1.0 m 2L = 2 = 1.6 s or T ~ 10 0 s . 2 3g 3 9.8 m s
e
a
f
j
450 P15.30
Oscillatory Motion
= =
2 : T g : L
T= L=
2
=
2 = 1.42 s 4. 43
g 9.80 = = 0.499 m 2 2 4.43
a f
P15.31
Using the simple harmonic motion model: A = r = 1 m 15
= 0. 262 m 180
=
(a) (b)
g 9.8 m s 2 = = 3.13 rad s L 1m
v max = A = 0. 262 m 3.13 s = 0.820 m s a max = A 2 = 0.262 m 3.13 s a tan = r
b
g
2
= 2.57 m s 2
=
a tan 2.57 m s 2 = = 2.57 rad s 2 r 1m
FIG. P15.31
(c)
F = ma = 0.25 kg 2.57 m s 2 = 0.641 N
More precisely, (a) 1 mv 2 and 2 v max = 2 gL 1  cos = 0.817 m s mgh =
h = L 1  cos
a
f
a
f
(b)
I = mgL sin
max =
(c) P15.32 (a)
mgL sin mL
2
=
g sin i = 2.54 rad s 2 L
Fmax = mg sin i = 0.250 9.80 sin 15.0 = 0.634 N The string tension must support the weight of the bob, accelerate it upward, and also provide the restoring force, just as if the elevator were at rest in a gravity field 9.80 + 5.00 m s 2
a fa
f
a
f
T = 2
L 5.00 m = 2 g 14.8 m s 2
T = 3.65 s (b) T = 2
e9.80 m s
5.00 m
2
 5.00 m s 2
j
= 6.41 s
(c)
g eff =
e9.80 m s j + e5.00 m s j
2 2
2 2
= 11.0 m s 2
T = 2 =
5.00 m 11.0 m s 2
= 4.24 s
Chapter 15
451
P15.33
x R For small displacements, tan sin mg F= and x =  kx R Since the restoring force is proportional to the displacement from equilibrium, the motion is simple harmonic motion. F =  mg sin and tan = Comparing toF =  m 2 x shows = total measured time 50 1.000 0.750 0.500 1.996 1.732 1.422
3 2 1 0 0.25 0.5 0.75 1.0 L, m
Referring to the sketch we have
k = m
g . R
T2, s2
FIG. P15.33
P15.34
(a)
T=
The measured periods are:
a f Period, T asf
Length, L m
4
(b)
L 4 2 L T = 2 so g= g T2 The calculated values for g are: Period, T s g m s2
e
j
af
1.996 1.732 1.422 9.91 9.87 9.76
FIG. P15.34 this agrees with the accepted value of g = 9.80 m s 2 within 0.5%.
2
Thus, g ave = 9.85 m s 2 (c) From T 2 = Thus, g = P15.35
2
F 4 I L , the slope of T GH g JK
versus L graph =
4 2 = 4.01 s 2 m . g
4 2 = 9.85 m s 2 . This is the same as the value in (b). slope
f = 0. 450 Hz , d = 0.350 m, and m = 2.20 kg T= 1 ; f I ; mgd mgd 4
2
T = 2 I =T2
T2 =
2
4 2 I mgd
2
=
FG 1 IJ H fK
mgd 4
=
2.20 9.80 0.350 4
2
a fa f = e0.450 s j
1 2
0.944 kg m 2 FIG. P15.35
452 P15.36
Oscillatory Motion
(a)
The parallelaxis theorem: I = I CM + Md 2 = =M 1 1 ML2 + Md 2 = M 1.00 m 12 12
a
f
2
+ M 1.00 m
a
f
2
FG 13 m IJ H 12 K
2
M 13 m 2 13 m I = 2 = 2 = 2.09 s T = 2 12 Mg 1.00 m Mgd 12 9.80 m s 2
e
a
j
f
e
j
FIG. P15.36
(b)
For the simple pendulum T = 2 1.00 m 9.80 m s
2
= 2.01 s
difference =
2.09 s  2.01 s = 4.08% 2.01 s
P15.37
(a)
The parallel axis theorem says directly I = I CM + md 2 so T = 2 I = 2 mgd
eI
CM
+ md 2
j
mgd d gets large. g I CM gets large. mgd
(b)
When d is very large T 2 When d is very small T 2
So there must be a minimum, found by dT d =0= 2 I CM + md 2 dd dd
j bmgdg F 1I FG 1 IJ eI mg + 2 bmgd g = 2 e I + md j G  J bmgd g H 2K H 2K  e I + md jmg 2 md mgd = + =0 + md j bmgd g + md j bmgd g I I e e e
12 1 2 CM 2 12 2 3 2 1 2 CM CM 2 12 3 2 CM 2 12 3 2
CM
+ md 2
j
1 2
2md
This requires  I CM  md 2 + 2md 2 = 0 or P15.38
ICM = md 2 .
We suppose the stick moves in a horizontal plane. Then, 1 1 2.00 kg 1.00 m mL2 = 12 12 I T = 2 I=
b
ga
f
2
= 0.167 kg m 2
=
4 2 I T
2
=
4 2 0.167 kg m 2
e
a180 sf
2
j=
203 N m
Chapter 15
453
P15.39
T = 0.250 s, I = mr 2 = 20.0 10 3 kg 5.00 10 3 m (a) (b)
e
je
j
2
I = 5.00 10 7 kg m 2
I d 2 =  ; dt 2
2 = = I T
FIG. P15.39
2
= I 2 = 5.00 10 7
e
jFGH 0.2250 IJK
= 3.16 10 4
Nm rad
Section 15.6 P15.40
Damped Oscillations E= 1 1 mv 2 + kx 2 2 2 2 dE d x = mv 2 + kxv dt dt 2 md x =  kx  bv dt 2 dE = v  kx  bv + kvx dt dE =  bv 2 < 0 dt
The total energy is Taking the timederivative, Use Equation 15.31:
a
f
Thus, P15.41
i = 15.0
x = Ae  bt 2 m
t = 1 000 = 5.50 x1 000 Ae  bt 2 m 5.50  b 1 000 g = = =e b 15.0 xi A
ln
b
g
2m
FG 5.50 IJ = 1.00 = bb1 000g H 15.0 K 2m
b = 1.00 10 3 s 1 2m
P15.42
Show that is a solution of where x = Ae  bt 2 m cos t +
x = Ae  bt 2 m cos t +  kx  b dx d x =m 2 dt dt
2
b
g
(1) (2) (3) (4)
=
IJ b g b g K d x b L FG  b IJ cosbt + g  Ae sinbt + gOP = Ae H 2m K 2m M dt N Q L FG  b IJ sinbt + g + Ae cosbt + gOP  M Ae H 2m K N Q
dx b = Ae  bt 2 m  cos t +  Ae  bt 2 m sin t + dt 2m
2 2  bt 2 m  bt 2 m  bt 2 m  bt 2 m 2
FG H
b
g
k b  m 2m
FG IJ H K
2
.
(5)
continued on next page
454
Oscillatory Motion
Substitute (3), (4) into the left side of (1) and (5) into the right side of (1);
b g 2bm Ae cosbt + g + bAe sinbt + g bL FG  b IJ cosbt + g  Ae sinbt + gOP =  M Ae H 2m K 2N Q b cosbt + g + Ae sinbt + g  m Ae 2 Compare the coefficients of Ae cosbt + g and Ae sinbt + g : Fk b I b bF b I b b =  G cosineterm:  k + H 2m JK  m = 4m  mGH m  4m JK =  k + 2m 2m 2
 kAe  bt 2 m cos t + +
 bt 2 m  bt 2 m  bt 2 m  bt 2 m  bt 2 m 2  bt 2 m  bt 2 m  bt 2 m 2 2 2 2 2 2
2
sineterm:
b = +
b b + = b 2 2
af af
Since the coefficients are equal, x = Ae  bt 2 m cos t + is a solution of the equation. *P15.43 The frequency if undamped would be 0 = (a) With damping
2 = 0 
b
g
k = m
2.05 10 4 N m = 44.0 s. 10.6 kg
FG b IJ H 2m K
2
=
FG 44 1 IJ  FG 3 kg IJ H s K H s 2 10.6 kg K
2
2
= 1 933.96  0.02 = 44.0 f= (b)
44.0 = = 7.00 Hz 2 2 s
1 s
In x = A 0 e  bt 2 m cos t + over one cycle, a time T = A0 e  b 2
2 m
b
g
2
, the amplitude changes from A0 to
for a fractional decrease of
A 0  A 0 e  b m = 1  e  3 a10.644.0 f = 1  e 0 .020 2 = 1  0.979 98 = 0.020 0 = 2.00% . A0 (c) The energy is proportional to the square of the amplitude, so its fractional rate of decrease is twice as fast: E= We specify 1 2 1 2  2 bt 2 m = E0 e  bt m . kA = kA 0 e 2 2 0.05E0 = E0 e  3 t 10.6 0.05 = e  3 t 10.6 e + 3 t 10 .6 = 20 3t = ln 20 = 3.00 10.6 t = 10.6 s
Chapter 15
455
Section 15.7 P15.44 (a)
Forced Oscillations For resonance, her frequency must match f0 =
0 1 = 2 2
k 1 = m 2
4.30 10 3 N m = 2.95 Hz . 12.5 kg
(b)
From x = A cos t , v =
dx dv =  A sin t , and a = =  A 2 cos t , the maximum acceleration dt dt is A 2 . When this becomes equal to the acceleration due to gravity, the normal force exerted on her by the mattress will drop to zero at one point in the cycle: A = g
2
or
A= and so
g
2
=
g
k m
gm = k
e9.80 m s jb12.5 kgg = A=
2
4.30 10 3 N m
2.85 cm
P15.45
F = 3.00 cos 2 t N (a)
b g
k = 20.0 N m T = 1.00 s k = m 20.0 = 3.16 rad s 2.00
=
2 = 2 rad s T
(b)
In this case,
0 =
The equation for the amplitude of a driven oscillator, with b = 0, gives Thus F0 cos t  kx = m x = A cos t + d2x dt 2 A=
FG F IJ e H mK
0
2
2 0
j
1
=
3 4 2  3.16 2
a f
2 1
A = 0.050 9 m = 5.09 cm .
P15.46
0 =
k m
(1) (2)
b
g g g
F0 cos t  kA cos t + = m  A 2 cos t +
dx =  A sin t + dt
b
(3)
d2x =  A 2 cos t + dt 2
b
(4)
Substitute (2) and (4) into (1): Solve for the amplitude:
ekA  mA
2
b g e j b j cosbt + g = F cos t
0
g
These will be equal, provided only that must be zero and kA  mA 2 = F0 Thus, A =
c h
k m
F0 m
2
456 P15.47
Oscillatory Motion
From the equation for the amplitude of a driven oscillator with no damping, A= F0 m
e
2
2 0
j
2
= 2 f = 20.0 s 1
F0 = mA F0 = Fext m
e
j
2
2 0 =
FG 40.0 IJ e2.00 10 jb3 950  49.0g = H 9.80 K
2
e
2
2 0
j
k 200 = = 49.0 s 2 40 .0 m 9.80
c h
318 N
P15.48
A=
e
2
2 0
j + b b m g
2
With b = 0,
A=
Fext m
e
2
2 0
j
2
=
e
Fext m
2 2 0
j
=
2 2 0
Fext m
Thus, This yields Then, P15.49
2 2 =0
Fext m k Fext 6.30 N m 1.70 N = = 0.150 kg A m mA 0.150 kg 0.440 m
b
ga
f
= 8.23 rad s or = 4.03 rad s
f=
2
gives either f = 1.31 Hz
or
f = 0.641 Hz
The beeper must resonate at the frequency of a simple pendulum of length 8.21 cm: f= 1 2 g 1 = L 2 9.80 m s 2 = 1.74 Hz . 0.082 1 m
*P15.50
For the resonance vibration with the occupants in the car, we have for the spring constant of the suspension f= 1 2 k m
Now as the occupants exit
e j d1 130 kg + 4b72.4 kg gi = 1.82 10 F 4b72. 4 kg ge9.8 m s j x= = = 1.56 10 m
k = 4 2 f 2 m = 4 2 1.8 s 1
2 2
5
kg s 2
k
1.82 10
5
kg s
2
2
Chapter 15
457
Additional Problems P15.51 Let F represent the tension in the rod. pivot (a) At the pivot, F = Mg + Mg = 2 Mg A fraction of the rod's weight Mg
weight of the ball pulls down on point P. Thus, the tension in the rod at point P is F = Mg
FG y IJ as well as the H LK
y L
P L y
FG y IJ + Mg = H LK
Mg 1 +
FG H
IJ K
.
M FIG. P15.51
(b)
Relative to the pivot, I = I rod + I ball = For the physical pendulum, T = 2
1 4 ML2 + ML2 = ML2 3 3
I where m = 2 M and d is the distance from the mgd pivot to the center of mass of the rod and ball combination. Therefore, d= M
M+M
c h + ML = 3L and T = 2
L 2
4
a 2 M f gc h
3L 4
4 3
ML2
=
4 3
2L . g
For L = 2.00 m, T =
4 3
2 2.00 m 9.80 m s
a
2
f= ga g
2.68 s .
P15.52
(a)
Total energy =
1 2 1 kA = 100 N m 0.200 m 2 2
b
f
2
= 2.00 J
At equilibrium, the total energy is: 1 1 m1 + m 2 v 2 = 16.0 kg v 2 = 8.00 kg v 2 . 2 2 Therefore,
b
b
g
b
g
b8.00 kg gv
2
= 2.00 J , and v = 0.500 m s .
This is the speed of m1 and m 2 at the equilibrium point. Beyond this point, the mass m 2 moves with the constant speed of 0.500 m/s while mass m1 starts to slow down due to the restoring force of the spring. continued on next page
458
Oscillatory Motion
(b)
The energy of the m1 spring system at equilibrium is: 1 1 m1 v 2 = 9.00 kg 0.500 m s 2 2 This is also equal to Therefore,
b
gb
g
2
= 1.125 J .
1 2 k A , where A is the amplitude of the m1 spring system. 2 1 100 A 2
a f
a fa f
2
= 1.125 or A = 0.150 m. m1 = 1.885 s k
The period of the m1 spring system is T = 2 and it takes
1 T = 0.471 s after it passes the equilibrium point for the spring to become fully 4 stretched the first time. The distance separating m1 and m 2 at this time is: D=v
FG T IJ  A = 0.500 m s a0.471 sf  0.150 m = 0.085 6 = H 4K
s
8.56 cm .
P15.53
F d xI GH dt JK
2 2
= A 2
max
B P n B mg f
fmax = sn = s mg = mA 2 A=
sg = 6.62 cm 2
FIG. P15.53 P15.54 The maximum acceleration of the oscillating system is a max = A 2 = 4 2 Af 2 . The friction force exerted between the two blocks must be capable of accelerating block B at this rate. Thus, if Block B is about to slip, f = fmax = sn = s mg = m 4 2 Af 2 P15.55
e
j
or
A=
s g . 4 2 f 2
Deuterium is the isotope of the element hydrogen with atoms having nuclei consisting of one proton and one neutron. For brevity we refer to the molecule formed by two deuterium atoms as D and to the diatomic molecule of hydrogen1 as H.
MD = 2MH
D = H
k MD k MH
=
MH = MD
1 2
fD =
fH 2
= 0.919 10 14 Hz
Chapter 15
459
P15.56
The kinetic energy of the ball is K =
1 1 mv 2 + I 2 , 2 2 where is the rotation rate of the ball about its center of mass. Since the center of the ball moves along a circle of radius 4R, its displacement from equilibrium is s = 4R and its speed is ds d = 4R v= . Also, since the ball rolls without dt dt slipping,
5R R h s
FG IJ H K
a f
v=
ds = R dt
so
=
v d =4 R dt
FG IJ H K
2
FIG. P15.56
The kinetic energy is then K= = 1 d m 4R 2 dt
FG H
112mR 2 d 10 dt
FG IJ H K
2
IJ K
2
+
2
1 2 mR 2 2 5
FG H
IJ FG 4 d IJ K H dt K
When the ball has an angular displacement , its center is distance h = 4R 1  cos higher than when at the equilibrium position. Thus, the potential energy is U g = mgh = 4mgR 1  cos . For small
a
a
f
f
angles, 1  cos
a
f
(see Appendix B). Hence, U g 2mgR 2 , and the total energy is 2
E = K +Ug = 112mR 2 d 10 dt
FG IJ H K
2
+ 2mgR 2 .
FG IJ H K 28 R d d F 5 g IJ . + g = 0 , or This reduces to = G H 28R K 5 dt dt
Since E = constant in time,
2 2 2 2
112 mR 2 d d 2 dE d . + 4mgR =0= 5 dt dt dt 2 dt
FG IJ H K
With the angular acceleration equal to a negative constant times the angular position, this is in the 5g . defining form of a simple harmonic motion equation with = 28 R The period of the simple harmonic motion is then T = 2
= 2
28 R . 5g
460 P15.57
Oscillatory Motion
(a)
Li a a h L
FIG. P15.57(a) (b) T = 2 L g
1 dL dT = dt g L dt
(1)
We need to find L t and
af
dL . From the diagram in (a), dt L = Li + 1 dh a h dL =  ; . 2 dt 2 2 dt
FG IJ H K
But
dM dV dh = =  A . Therefore, dt dt dt 1 dM dL dh 1 dM = = ; A dt dt dt 2 A dt
L
FG H
IJ K
i
(2)
Also,
Li
z
dL =
FG 1 IJ FG dM IJ t = L  L H 2 A K H dt K
dT = dt
(3)
Substituting Equation (2) and Equation (3) into Equation (1):
1 g 2 a 2
F GH
I FG dM IJ JK H dt K
1 Li +
1 2 a 2
c ht
dM dt
.
(c)
Substitute Equation (3) into the equation for the period. T= 2 g Li + 1 dM t 2 dt 2 a
FG IJ H K
Or one can obtain T by integrating (b):
But Ti = 2
I FG dM IJ z JK H dt K z L + dt c ht L 2 OPL O F 1 I F dM I M T T = GH 2a JK GH dt JK M c h PMNM L + 2 1a FGH dM IJK t  L PQP dt g N Q L 2 1 F dM I , so T = L + G Jt . g 2 a H dt K g
Ti T
dT =
1 g 2 a 2
F GH
t
0
i
1 2 a 2
dM dt
i
2
1 2 a 2
dM dt
i
2
i
i
i
2
Chapter 15
461
P15.58
=
k 2 = m T k = m =
2
(a) P15.59
4 2 m T2
(b)
m =
k T
a f
4 2
2
= m
FG T IJ HTK
2
We draw a freebody diagram of the pendulum. The force H exerted by the hinge causes no torque about the axis of rotation.
Hy Hx h L k m mg L sin kx x hcos
= I
and
d 2 =  dt 2 d dt 2
2
= MgL sin + kxh cos =  I
For small amplitude vibrations, use the approximations: sin , cos 1, and x s = h . Therefore, MgL + kh 2 d 2 =  2 = I dt 2
FIG. P15.59
F GH
I JK
=
MgL + kh 2 ML2 1 2
= 2 f
f= *P15.60 (a) In x = A cos t + , we have at t = 0 This requires = 90 , so And this is equivalent to Numerically we have and v max = A So 1 1 1 mv 2 + kx 2 = kA 2 , 2 2 2
MgL + kh 2 ML2
b
g
v = A sin t +
b
g
v = A sin =  v max
x = A cos t + 90
x =  A sin t
a
f
=
k = m
50 N m = 10 s 1 0.5 kg
20 m s = 10 s 1 A x = 2 m sin 10 s 1 t 1 2 1 kx = 3 mv 2 2 2
e
j
A=2m
a
f e
j
(b)
In
FG H
IJ K
4 2 x = A2 3
implies
11 2 1 2 1 2 kx + kx = kA 32 2 2 x= 3 A = 0.866 A = 1.73 m 4
continued on next page
462
Oscillatory Motion
(c)
=
g L
L=
g
2
=
9.8 m s 2
e10 s j j
1 2
= 0.098 0 m
(d)
In
x = 2 m sin 10 s 1 t
a
f e
the particle is at x = 0 at t = 0 , at 10t = s , and so on. The particle is at when with solutions x=1 m  1 = sin 10 s 1 t 2
e
j
e10 s jt =  6
1 1
e10 s jt = + , and so on. 6 FI The minimum time for the motion is t in 10 t = G J s H 6K FI t = G J s = 0.052 4 s H 60 K
P15.61 (a) At equilibrium, we have
FIG. P15.60(d)
= 0  mg G 2 J + kx0 L H K
where x 0 is the equilibrium compression. After displacement by a small angle, FIG. P15.61
F LI
=  mg G 2 J + kxL = mg G 2 J + kbx0  L gL =  kL2 H K H K
But,
F LI
F LI
= I = 3 mL2
1
d 2 . dt 2
So
d 2 3k =  . 2 m dt
The angular acceleration is opposite in direction and proportional to the displacement, so 3k we have simple harmonic motion with 2 = . m (b) f=
1 = 2 2
3k 1 = m 2
3 100 N m 5.00 kg
b
g=
1.23 Hz
Chapter 15
463
*P15.62
As it passes through equilibrium, the 4kg object has speed v max = A = 100 N m k A= 2 m = 10.0 m s. m 4 kg
In the completely inelastic collision momentum of the twoobject system is conserved. So the new 10kg object starts its oscillation with speed given by 4 kg 10 m s + 6 kg 0 = 10 kg v max v max = 4.00 m s (a) The new amplitude is given by 1 1 2 mv max = kA 2 2 2 10 kg 4 m s A = 1.26 m Thus the amplitude has decreased by 2.00 m  1.26 m = 0.735 m T = 2 T = 2 4 kg m = 2 = 1.26 s k 100 N m 10 2 s = 1.99 s 100
b
g b g b
g
b
g = b100 N mgA
2
2
(b)
The old period was The new period is The period has increased by
1.99 m  1.26 m = 0.730 s 1 1 2 mv max = 4 kg 10 m s 2 2 1 10 kg 4 m s 2
(c)
The old energy was The new mechanical energy is The energy has decreased by 120 J .
b gb g
2
g
2
= 200 J
b
gb
= 80 J
(d)
The missing mechanical energy has turned into internal energy in the completely inelastic collision. T= 2 = 2 L = 3.00 s g
P15.63
(a)
(b)
E=
1 1 mv 2 = 6.74 2.06 2 2
a fa f a
2
= 14.3 J 1 mv 2 2 h L v2 = 0.217 m 2g
(c)
At maximum angular displacement
mgh =
h=
h = L  L cos = L 1  cos
f
cos = 1 
= 25.5
464 P15.64
Oscillatory Motion
One can write the following equations of motion: T  kx = 0 mg  T = ma = m R T  T = I d x dt 2
2
(describes the spring) (for the hanging object) (for the pulley) with I = 1 MR 2 2 FIG. P15.64
a
f
d 2 I d 2 x = dt 2 R dt 2
Combining these equations gives the equation of motion
FG m + 1 MIJ d x + kx = mg . H 2 K dt
2 2
mg mg (where arises because of the extension of the spring due to k k the weight of the hanging object), with frequency The solution is x t = A sin t + f= For M = 0 For M = 0.250 kg For M = 0.750 kg
af
2
=
1 2
k 1 = m + 1 M 2 2 f = 3.56 Hz f = 2.79 Hz f = 2.10 Hz
100 N m . 0.200 kg + 1 M 2
(a) (b) (c) P15.65
Suppose a 100kg biker compresses the suspension 2.00 cm. Then, k= F 980 N = = 4.90 10 4 N m x 2.00 10 2 m
If total mass of motorcycle and biker is 500 kg, the frequency of free vibration is f= 1 2 1 k = m 2 4.90 10 4 N m = 1.58 Hz 500 kg
If he encounters washboard bumps at the same frequency, resonance will make the motorcycle bounce a lot. Assuming a speed of 20.0 m/s, we find these ridges are separated by 20.0 m s 1.58 s 1 = 12.7 m ~ 10 1 m .
In addition to this vibration mode of bouncing up and down as one unit, the motorcycle can also vibrate at higher frequencies by rocking back and forth between front and rear wheels, by having just the front wheel bounce inside its fork, or by doing other things. Other spacing of bumps will excite all of these other resonances.
Chapter 15
465
P15.66
(a)
For each segment of the spring dK = Also, vx = x v and 1 2 dm v x . 2 dm = m dx . FIG. P15.66
a f
Therefore, the total kinetic energy of the blockspring system is K= k m eff 1 1 Mv 2 + 2 2
z FGH
0
x2v2 m
2
I JK
dx =
m 2 1 M+ v . 2 3
FG H
IJ K
(b)
=
and
1 1 m 2 m eff v 2 = M+ v 2 2 3 2 M+ m 3 k
FG H
IJ K
Therefore,
T=
= 2
.
P15.67
(a)
F = 2T sin j
where = tan 1
FG y IJ H LK
FIG. P15.67
Therefore, for a small displacement y sin tan = L (b) and 2Ty F = L j
The total force exerted on the ball is opposite in direction and proportional to its displacement from equilibrium, so the ball moves with simple harmonic motion. For a spring system,
F = kx
becomes here 2T and L
F = 
=
2T y. L k = m 2T . mL
Therefore, the effective spring constant is
466 P15.68
Oscillatory Motion
(a)
Assuming a Hooke's Law type spring, F = Mg = kx and empirically Mg = 1.74x  0.113 so k = 1.74 N m 6% . x, m 0.17 0.293 0.353 0.413 0.471 0.493 Mg , N 0.196 0.392 0.49 0.588 0.686 0.784
M , kg 0.020 0 0.040 0 0.050 0 0.060 0 0.070 0 0.080 0 (b)
We may write the equation as theoretically T2 = 4 2 4 2 M+ ms 3k k FIG. P15.68
and empirically T 2 = 21.7 M + 0.058 9 so k= 4 2 = 1.82 N m 3% 21.7 M , kg 0.020 0 0.040 0 0.050 0 0.060 0 0.070 0 0.080 0 T 2 , s2 0.494 0.925 1.138 1.362 1.568 1.798
Time, s T , s 7.03 0.703 9.62 10.67 11.67 12.52 13.41 0.962 1.067 1.167 1.252 1.341
The k values 1.74 N m 6% and so (c) 1.82 N m 3% differ by 4% they agree. ms = 3
Utilizing the axiscrossing point,
FG 0.058 9 IJ kg = H 21.7 K
8 grams 12%
in agreement with 7.4 grams.
Chapter 15
467
P15.69
(a)
K + U = 0 Thus, K top + U top = K bot + U bot where K top = U bot = 0 Therefore, mgh = 1 2 I , but 2 h = R  R cos = R 1  cos
M R
a
f
v
v R MR 2 mr 2 and I = + + mR 2 2 2 Substituting we find
=
m FIG. P15.69
mgR 1  cos =
and
so
v f 1 FGH MR + mr + mR IJK R 2 2 2 L M mr + m OPv mgRa1  cos f = M + N 4 4R 2 Q a1  cos f v = 4 gR e + + 2j Rg a1  cos f v=2
a
2
2
2
2 2
2
2
2
2
M m
r2 R2
M m
+
r2 R2
+2
(b)
T = 2
I mT gd CM dCM =
mT = m + M T = 2
1 2
mR + M 0 m+M
af
MR 2 + 1 mr 2 + mR 2 2 mgR A 2 e + bt 2 m = 2 0.100 kg s t = 0.693 2 0.375 kg
P15.70
(a)
We require Ae  bt 2 m = or
bt = ln 2 or 2m The spring constant is irrelevant.
b
g
t = 5.20 s
(b)
We can evaluate the energy at successive turning points, where 1 1 1 1 1 2 cos t + = 1 and the energy is kx 2 = kA 2 e  bt 2 m . We require kA 2 e  bt 2 m = kA 2 2 2 2 2 m ln 2 0.375 kg 0.693 t = = = 2.60 s . or e + bt m = 2 b 0.100 kg s
b
g
a
f
FG H
IJ K
(c)
From E =
1 2 kA , the fractional rate of change of energy over time is 2 2 d 1 dE dA 1 k 2 A dA dt 2 kA dt = 1 = 2 1 2 dt = 2 dt 2 E A 2 kA 2 kA
e
j
a f
two times faster than the fractional rate of change in amplitude.
468 P15.71
Oscillatory Motion
(a)
When the mass is displaced a distance x from equilibrium, spring 1 is stretched a distance x1 and spring 2 is stretched a distance x 2 . By Newton's third law, we expect k1 x1 = k 2 x 2 . When this is combined with the requirement that x = x1 + x 2 , we find The force on either spring is given by where a is the acceleration of the mass m. This is in the form and x1 =
2
LM k OPx Nk + k Q L k k OP x = ma F =M Nk + k Q
1 2 1 1 2 1 2
FIG. P15.71
F = k eff x = ma T = 2 m k1 + k 2 m = 2 k eff k1 k 2
b
g
(b)
In this case each spring is distorted by the distance x which the mass is displaced. Therefore, the restoring force is F =  k1 + k 2 x so that T = 2
b
g
and
k eff = k1 + k 2
b
m k1 + k 2
g
.
P15.72
Let
represent the length below water at equilibrium and M the tube's mass:
Fy = 0  Mg + r 2
Now with any excursion x from equilibrium  Mg + r 2 Subtracting the equilibrium equation gives  r 2 gx = Ma a=
2
g = 0.
a  xfg = Ma .
F r g I x =  x GH M JK
2
The opposite direction and direct proportionality of a to x imply SHM with angular frequency
=
T=
r 2 g M
2
=
FG 2 IJ HrK
M g
Chapter 15
469
P15.73
For max = 5.00 , the motion calculated by the Euler method agrees quite precisely with the prediction of max cos t . The period is T = 2.20 s . Time, t (s) 0.000 0.004 0.008 ... 0.544 0.548 0.552 ... 1.092 1.096 1.100 1.104 ... 1.644 1.648 1.652 ... 2.192 2.196 2.200 2.204 Angle, () 5.000 0 4.999 3 4.998 0 0.056 0 0.001 1 0.058 2 4.999 4 5.000 0 5.000 0 4.999 3 0.063 8 0.003 3 0.060 4 4.999 4 5.000 0 5.000 0 4.999 3 Ang. speed (/s) 0.000 0 0.163 1 0.326 2 14.282 3 14.284 2 14.284 1 0.319 9 0.156 8 0.006 3 0.169 4 14.282 4 14.284 2 14.284 1 0.313 7 0.150 6 0.012 6 0.175 7 Ang. Accel. s2
e j
max cos t
5.000 0 4.999 7 4.998 7 0.081 0 0.023 9 0.033 3 4.998 9 4.999 8 5.000 0 4.999 6 0.071 6 0.014 5 0.042 7 4.999 1 4.999 9 5.000 0 4.999 4
40.781 5 40.776 2 40.765 6 0.457 6 0.009 0 0.475 6 40.776 5 40.781 6 40.781 4 40.775 9 0.439 7 0.027 0 0.493 6 40.776 8 40.781 7 40.781 3 40.775 6
For max = 100 , the simple harmonic motion approximation max cos t diverges greatly from the Euler calculation. The period is T = 2.71 s , larger than the smallangle period by 23%. Time, Angle, t (s) () 0.000 100.000 0 0.004 99.992 6 0.008 99.977 6 ... 1.096 84.744 9 1.100 85.218 2 1.104 85.684 0 ... 1.348 99.996 0 1.352 100.000 8 1.356 99.998 3 ... 2.196 40.150 9 2.200 41.045 5 2.204 41.935 3 ... 2.704 99.998 5 2.708 100.000 8 2.712 99.995 7 Ang. speed (/s) 0.000 0 1.843 2 3.686 5 120.191 0 118.327 2 116.462 0 3.053 3 1.210 0 0.633 2 224.867 7 223.660 9 222.431 8 2.420 0 0.576 8 1.266 4 Ang. Accel. s2
FIG. P15.73
e j
max cos t
100.000 0 99.993 5 99.973 9 99.995 4 99.999 8 99.991 1 75.797 9 75.047 4 74.287 0 99.997 1 99.999 3 99.988 5 12.642 2 11.507 5 10.371 2
460.606 6 460.817 3 460.838 2 465.948 8 466.286 9 466.588 6 460.812 5 460.805 7 460.809 3 301.713 2 307.260 7 312.703 5 460.809 0 460.805 7 460.812 9
470 *P15.74
Oscillatory Motion
(a)
The block moves with the board in what we take as the positive x direction, stretching the spring until the spring force kx is equal in magnitude to the maximum force of static mg . friction sn = s mg . This occurs at x = s k Since v is small, the block is nearly at the rest at this break point. It starts almost immediately to move back to the left, the forces on it being kx and + k mg . While it is sliding the net force exerted on it can be written as  kx + k mg =  kx + k k mg mg = k x  k =  kx rel k k
(b)
FG H
IJ K
where x rel is the excursion of the block away from the point
k mg . k
Conclusion: the block goes into simple harmonic motion centered about the equilibrium mg position where the spring is stretched by k . k (d) The amplitude of its motion is its original displacement, A = rest at spring extension
2 k  s mg k mg . Almost immediately at this point it A= k k latches onto the slowlymoving board to move with the board. The board exerts a force of static friction on the block, and the cycle continues. (c) The graph of the motion looks like this:
b
g
s mg k mg  . It first comes to k k
FIG. P15.74(c) (e) The time during each cycle when the block is moving with the board is 2 A 2 s  k mg = . v kv The time for which the block is springing back is one half a cycle of simple harmonic motion, 1 m m = . We ignore the times at the end points of the motion when the speed of 2 2 k k 2A the block changes from v to 0 and from 0 to v. Since v is small compared to , these
b
g
F GH
I JK
m k
times are negligible. Then the period is T= continued on next page 2 s  k mg kv
b
g
+
m . k
Chapter 15
471
(f)
T=
2 0.4  0.25 0.3 kg 9.8 m s 2
a
b0.024 m sgb12 N mg
f=
fb
ge
j +
0.3 kg = 3.06 s + 0.497 s = 3.56 s 12 N m
Then (g) (h) (i) (j) *P15.75 (a) T=
1 = 0.281 Hz . T + m increases as m increases, so the frequency decreases . k
2 s  k mg kv
b
g
As k increases, T decreases and f increases . As v increases, T decreases and f increases . As s  k increases, T increases and f decreases . Newton's law of universal gravitation is Thus, Which is of Hooke's law form with F= F= k= GMm r
2
b
g
=
FG 4 GmIJ r H3 K
Gm 4 3 r r2 3
FG H
IJ K
4 Gm 3 
(b)
The sack of mail moves without friction according to
FG 4 IJ Gmr = ma H 3K F 4I a =  G J Gr =  r H 3K
2
Since acceleration is a negative constant times excursion from equilibrium, it executes SHM with
=
4G 3
and period
T= T = 2 g=
2
=
3 G
The time for a oneway trip through the earth is We have also so g 4G = 3 R e and
3 4 G GM e
2 Re
=
3 G 4R e 2 3Re
=
4 GR e 3
b g
Re 6.37 10 6 m T = = = 2.53 10 3 s = 42.2 min . 2 2 g 9.8 m s
ANSWERS TO EVEN PROBLEMS
P15.2 (a) 4.33 cm; (b) 5.00 cm s ; (c) 17.3 cm s ; (d) 3.14 s; 5.00 cm P15.4 (a) 15.8 cm; (b) 15.9 cm; (c) see the solution; (d) 51.1 m; (e) 50.7 m
2
P15.6 P15.8 P15.10
see the solution 12.0 Hz 18.8 m s; 7.11 km s 2
472 P15.12
Oscillatory Motion
(a) 1.26 s; (b) 0.150 m s; 0.750 m s 2 ; 15 cm sin 5t ; (c) x = 3 cmcos 5t ; v = s 75 cm a= cos 5t s2
FG H
IJ K
FG H
IJ K
P15.42 P15.44 P15.46 P15.48 P15.50 P15.52 P15.54 P15.56 P15.58 P15.60
see the solution (a) 2.95 Hz; (b) 2.85 cm see the solution either 1.31 Hz or 0.641 Hz 1.56 cm (a) 0.500 m s ; (b) 8.56 cm A=
P15.14 P15.16 P15.18 P15.20
F vI (a) ; (b) x =  G J sin t HK
v (a) 126 N m; (b) 0.178 m (a) 0.153 J; (b) 0.784 m s; (c) 17.5 m s 2 (a) 100 N m; (b) 1.13 Hz; (c) 1.41 m s at x = 0 ; (d) 10.0 m s at x = A ; (e) 2.00 J; (f) 1.33 m s ; (g) 3.33 m s 2
2
sg 4 2 f 2
see the solution (a) k = 4 2 m T ; (b) m = m 2 T T
FG IJ H K
2
P15.22
(a) 1.50 s; (b) 73.4 N m; (c) 19.7 m below the bridge; (d) 1.06 rad s; (e) 2.01 s; (f) 3.50 s (a) 0.218 s and 1.09 s; (b) 14.6 mW The position of the piston is given by x = A cos t . gC = 1.001 5 gT 1.42 s; 0.499 m (a) 3.65 s; (b) 6.41 s; (c) 4.24 s (a) see the solution; (b), (c) 9.85 m s 2 ; agreeing with the accepted value within 0.5% (a) 2.09 s; (b) 4.08% 203 N m see the solution
(a) x = 2 m sin 10t ; (b) at x 1.73 m; (c) 98.0 mm; (d) 52.4 ms (a) decreased by 0.735 m; (b) increased by 0.730 s; (c) decreased by 120 J; (d) see the solution (a) 3.56 Hz ; (b) 2.79 Hz; (c) 2.10 Hz (a) M+ m 1 m 2 3 M+ v ; (b) T = 2 2 3 k
a
f a f
P15.24 P15.26
P15.62
P15.64 P15.66 P15.68
P15.28 P15.30 P15.32 P15.34
FG H
IJ K
see the solution; (a) k = 1.74 N m 6% ; (b) 1.82 N m 3%; they agree; (c) 8 g 12%; it agrees (a) 5.20 s; (b) 2.60 s; (c) see the solution see the solution; T =
P15.70 P15.72 P15.74
P15.36 P15.38 P15.40
FG 2 IJ HrK
M g
see the solution; (f) 0.281 Hz ; (g) decreases; (h) increases; (i) increases; (j) decreases
16
Wave Motion
CHAPTER OUTLINE
16.1 16.2 16.3 16.4 16.5 16.6 Propagation of a Disturbance Sinusoidal Waves The Speed of Waves on Strings Reflection and Transmission Rate of Energy Transfer by Sinusoidal Waves on Strings The Linear Wave Equation
ANSWERS TO QUESTIONS
Q16.1 As the pulse moves down the string, the particles of the string itself move side to side. Since the mediumhere, the stringmoves perpendicular to the direction of wave propagation, the wave is transverse by definition. To use a slinky to create a longitudinal wave, pull a few coils back and release. For a transverse wave, jostle the end coil side to side. From v = T , we must increase the tension by a factor of 4.
Q16.2
Q16.3 Q16.4
It depends on from what the wave reflects. If reflecting from a less dense string, the reflected part of the wave will be right side up. 2 vA . Here v is the speed of the wave.
Q16.5
Yes, among other things it depends on. v max = A = 2 fA = Since the frequency is 3 cycles per second, the period is Amplitude is increased by a factor of
Q16.6 Q16.7 Q16.8 Q16.9
1 second = 333 ms. 3
2 . The wave speed does not change.
The section of rope moves up and down in SHM. Its speed is always changing. The wave continues on with constant speed in one direction, setting further sections of the rope into upanddown motion. Each element of the rope must support the weight of the rope below it. The tension increases with height. (It increases linearly, if the rope does not stretch.) Then the wave speed v = with height. T
increases
Q16.10
The difference is in the direction of motion of the elements of the medium. In longitudinal waves, the medium moves back and forth parallel to the direction of wave motion. In transverse waves, the medium moves perpendicular to the direction of wave motion.
473
474 Q16.11 Q16.12
Wave Motion
Slower. Wave speed is inversely proportional to the square root of linear density. As the wave passes from the massive string to the less massive string, the wave speed will increase according to v = increase. T
. The frequency will remain unchanged. Since v = f , the wavelength must
Q16.13 Q16.14
Higher tension makes wave speed higher. Greater linear density makes the wave move more slowly. The wave speed is independent of the maximum particle speed. The source determines the maximum particle speed, through its frequency and amplitude. The wave speed depends instead on properties of the medium. Longitudinal waves depend on the compressibility of the fluid for their propagation. Transverse waves require a restoring force in response to sheer strain. Fluids do not have the underlying structure to supply such a force. A fluid cannot support static sheer. A viscous fluid can temporarily be put under sheer, but the higher its viscosity the more quickly it converts input work into internal energy. A local vibration imposed on it is strongly damped, and not a source of wave propagation. Let t = ts  t p represent the difference in arrival times of the two waves at a station at distance
Q16.15
Q16.16
d = v s ts = v p t p
F1 1I from the hypocenter. Then d = tG  J Hv v K
s p
1
. Knowing the distance from the first
station places the hypocenter on a sphere around it. A measurement from a second station limits it to another sphere, which intersects with the first in a circle. Data from a third noncollinear station will generally limit the possibilities to a point. Q16.17 The speed of a wave on a "massless" string would be infinite!
SOLUTIONS TO PROBLEMS Section 16.1 P16.1 Propagation of a Disturbance
Replace x by x  vt = x  4.5t 6 to get y= 2 x  4.5t + 3
a
f
Chapter 16
475
P16.2
FIG. P16.2 P16.3 5.00 e a x + 5 t f is of the form f x + vt
2
a
f
so it describes a wave moving to the left at v = 5.00 m s . P16.4 (a) The longitudinal wave travels a shorter distance and is moving faster, so it will arrive at point B first. (b) The wave that travels through the Earth must travel a distance of at a speed of Therefore, it takes 2 R sin 30.0 = 2 6.37 10 6 m sin 30.0 = 6.37 10 6 m 7 800 m/s 6.37 10 6 m = 817 s 7 800 m s
e
j
The wave that travels along the Earth's surface must travel a distance of at a speed of Therefore, it takes The time difference is s = R = R 4 500 m/s 6.67 10 6 = 1 482 s 4 500
FG radIJ = 6.67 10 H3 K
6
m
665 s = 11.1 min
476 P16.5
Wave Motion
The distance the waves have traveled is d = 7.80 km s t = 4.50 km s t + 17.3 s where t is the travel time for the faster wave.
b
g b
ga
f
a fb g b ga f b4.50 km sga17.3 sf = 23.6 s or t = a7.80  4.50f km s and the distance is d = b7.80 km sga 23.6 sf = 184 km
Then, 7.80  4.50 km s t = 4.50 km s 17.3 s Section 16.2 P16.6 Sinusoidal Waves
.
Using data from the observations, we have = 1.20 m and f = 8.00 12.0 s
Therefore, v = f = 1.20 m
a
8 fFGH 12..00s IJK = 0
0.800 m s 425 cm = 42.5 cm s 10.0 s
P16.7
f=
40.0 vibrations 4 = Hz 30.0 s 3 v 42.5 cm s = 4 = 31.9 cm = 0.319 m f 3 Hz
v=
=
P16.8 P16.9
v = f = 4.00 Hz 60.0 cm = 240 cm s = 2.40 m s y = 0.020 0 m sin 2.11x  3.62t in SI units
k = 2.11 rad m
a
fa
f
b
g a
f
A = 2.00 cm
=
f=
2 = 2.98 m k
= 3.62 rad s
v = f = P16.10
= 0.576 Hz 2
2 3.62 = = 1.72 m s 2 k 2.11
y = 0.005 1 m sin 310 x  9.30t SI units v=
b
g a
f
9.30 = = 0.030 0 m s k 310
s = vt = 0.300 m in positive x  direction
Chapter 16
477
*P16.11
From y = 12.0 cm sin 1.57 rad m x  31.4 rad s t (a) The transverse velocity is Its maximum magnitude is v y t
a
f db
g b
gi
y =  A cos kx  t t
a
f g
A = 12 cm 31.4 rad s = 3.77 m s
b
(b)
ay =
=
 A cos kx  t =  A 2 sin kx  t t
c
a
fh
a
f
The maximum value is P16.12
A 2 = 0.12 m 31.4 s 1
a
fe
j
2
= 118 m s 2
At time t, the phase of y = 15.0 cm cos 0.157 x  50.3t at coordinate x is
a
f a
= 0.157 rad cm x  50.3 rad s t . Since 60.0 =
B = A rad , or (since x A = 0 ), 3
B
b
g b
g
rad , the requirement for point B is that 3
f
b0.157 rad cmgx  b50.3 rad sgt = 0  b50.3 rad sgt rad . 3
This reduces to x B = P16.13 rad = 6.67 cm . 3 0.157 rad cm
b
g
y = 0.250 sin 0.300 x  40.0t m Compare this with the general expression y = A sin kx  t (a) (b) (c) (d)
a
f
a
f
A = 0.250 m
= 40.0 rad s
k = 0.300 rad m
=
2 2 = = 20.9 m k 0.300 rad m
(e) (f)
v = f =
FG IJ = FG 40.0 rad s IJ a20.9 mf = H 2 K H 2 K
133 m s
The wave moves to the right, in + x direction .
478 P16.14
Wave Motion
(a) (b)
See figure at right. T= 2 = 2 = 0.125 s 50.3
y (cm) 10 0 10 0.1 0.2 t (s)
This agrees with the period found in the example in the text.
FIG. P16.14 P16.15 (a) A = y max = 8.00 cm = 0.080 0 m
Therefore, Or (where y 0 , t = 0 at t = 0 ) (b) In general, Assuming then we require that or Therefore, P16.16 (a) y (mm) 0.2 0.1 0.0 0.1 0.2 t=0 0.2 0.4 FIG. P16.16(a) (b) 2 = 18.0 rad m 0.350 m 1 1 T= = = 0.083 3 s f 12.0 s k= = 2
b g
a0.800 mf = 7.85 m = 2 f = 2 a3.00f = 6.00 rad s y = A sina kx + t f y = b0.080 0g sinb7.85 x + 6 t g m y = 0.080 0 sinb7.85 x + 6 t + g yb x, 0g = 0 at x = 0.100 m 0 = 0.080 0 sinb0.785 + g
k=
2
=
2
1
= 0.785
y = 0.080 0 sin 7.85 x + 6 t  0.785 m
b
g
x (mm)
= 2 f = 2 12.0 s = 75.4 rad s
(c)
f b ga y = A sinb kx + t + g specializes to y = 0.200 m sinb18.0 x m + 75.4t s + g
at x = 0 , t = 0 we require
v = f = 12.0 s 0.350 m = 4.20 m s
3.00 10 2 m = 0. 200 m sin +
b g
= 8.63 = 0.151 rad
so y x, t =
b g a0.200 mf sinb18.0 x m + 75.4t s  0.151 radg
Chapter 16
479
P16.17
y = 0.120 m sin dy : dt
a
f FGH x + 4 tIJK 8 fa f FGH x + 4 tIJK 8 va0.200 s, 1.60 mf = 1.51 m s F I a = a 0.120 mfa 4 f sinG x + 4 tJ H8 K aa0.200 s, 1.60 mf = 0
x = 0.120 4 cos
(a)
v=
a
a=
dv : dt
2
(b)
2 = : 8 2 = 4 = : T
k=
= 16.0 m
T = 0.500 s
v=
16.0 m = = 32.0 m s T 0.500 s
P16.18
(a)
Let us write the wave function as
b g b g yb0 , 0g = A sin = 0.020 0 m
y x , t = A sin kx + t +
dy dt = A cos = 2.00 m s
0, 0
Also,
=
2 2 = = 80.0 s T 0.025 0 s
A 2 = xi2 +
FG v IJ = b0.020 0 mg + FG 2.00 m s IJ HK H 80.0 s K
i 2 2
2
A = 0.021 5 m
(b) A sin 0.020 0 = 2 = 2.51 = tan A cos 80 .0 Your calculator's answer tan 1 2.51 = 1.19 rad has a negative sine and positive cosine, just the reverse of what is required. You must look beyond your calculator to find
a
f
=  1.19 rad = 1.95 rad
(c) (d)
v y, max = A = 0.021 5 m 80.0 s = 5.41 m s
b
g
= v x T = 30.0 m s 0.025 0 s = 0.750 m
k= 2
b
ga
f
=
2 = 8.38 m 0.750 m
= 80.0 s
y x , t = 0.021 5 m sin 8.38 x rad m + 80.0 t rad s + 1.95 rad
b g b
g b
g
480 P16.19
Wave Motion
(a)
f=
v
=
b1.00 m sg =
2.00 m
0.500 Hz
= 2 f = 2 0.500 s = 3.14 rad s
(b) (c) k= 2 = 2 = 3.14 rad m 2.00 m
b
g
y = A sin kx  t + becomes
y=
b
g
a0.100 mf sinb3.14x m  3.14t s + 0g a a f b f b g g g
(d)
For x = 0 the wave function requires y = 0.100 m sin 3.14t s
(e)
y = 0.100 m sin 4.71 rad  3.14 t s vy =
(f)
y = 0.100 m  3.14 s cos 3.14x m  3.14t s t
b
g b
The cosine varies between +1 and 1, so v y 0.314 m s P16.20 (a) (b)
b
g
at x = 2.00 m , y =
a0.100 mf sina1.00 rad  20.0tf
f a f
2 = 3.18 Hz
y = 0.100 m sin 0.500 x  20.0t = A sin kx  t so = 20.0 rad s and f =
a
f a
Section 16.3 P16.21
The Speed of Waves on Strings
The down and back distance is 4.00 m + 4.00 m = 8.00 m . The speed is then Now, So v= d total 4 8.00 m T = = 40.0 m s = t 0.800 s 0.200 kg = 5.00 10 2 kg m 4.00 m
a
f
=
T = v 2 = 5.00 10 2 kg m 40.0 m s
e
jb
g
2
= 80.0 N
P16.22
The mass per unit length is: =
0.060 0 kg = 1.20 10 2 kg m . 5.00 m
The required tension is: T = v 2 = 0.012 0 kg m 50.0 m s
b
gb
g
2
= 30.0 N .
Chapter 16
481
P16.23
v=
T
=
1 350 kg m s 2 5.00 10 3 kg m
= 520 m s
P16.24
(a)
= 2 f = 2 500 = 3 140 rad s , k =
y = 2.00 10 4 m sin 16.0 x  3 140t
a f
3 140 = = 16.0 rad m v 196
e
j b
g
(b)
v = 196 m s =
T 4.10 10 3 kg m
T = 158 N
P16.25
T = Mg is the tension;
Then, and
v=
T
=
Mg
m L
=
MgL L = is the wave speed. m t
MgL L2 = 2 m t g=
1.60 m 4.00 10 3 kg Lm = = 1.64 m s 2 Mt 2 3.00 kg 3.61 10 3 s 2
e e
j
j
P16.26
v=
T
T = v 2 = Av 2 = r 2 v 2 T = 8 920 kg m3 7.50 10 4 m T = 631 N P16.27 Since is constant, = T2 = T1 and
2
e
ja fe
j b200 m sg
2
2
2 v2
2 v1
T2
Fv I =G J Hv K
2 1
F 30.0 m s I a6.00 Nf = T =G H 20.0 m s JK
2 1
13.5 N .
P16.28
The period of the pendulum is T = 2
L g
Let F represent the tension in the string (to avoid confusion with the period) when the pendulum is vertical and stationary. The speed of waves in the string is then: v= Mg MgL F = m = m L L= T g 2
Since it might be difficult to measure L precisely, we eliminate so v = Mg T g Tg = m 2 2 M . m
482 P16.29
Wave Motion
If the tension in the wire is T, the tensile stress is Stress = T A so T = A stress .
a
f
The speed of transverse waves in the wire is v= T
=
A Stress
m L
a
f=
Stress
m AL
=
Stress
m Volume
=
Stress
where is the density. The maximum velocity occurs when the stress is a maximum: v max = P16.30 From the freebody diagram 2.70 10 8 Pa = 185 m s . 7 860 kg m 3
mg = 2T sin
T= mg 2 sin
3L 8 L 2
The angle is found from
cos =
=
3 4 FIG. P16.30
= 41.4 (a) v= T v= mg = 2 sin 41. 4
2
or (b) P16.31 v = 60.0 = 30.4 m and
v=
F 30.4 GH
ms
I J kg K
F I 9.80 m s GG J 2e8.00 10 kg mj sin 41.4 J H K
3
m
m
m = 3.89 kg
The total time is the sum of the two times. In each wire
L =L v T Let A represent the crosssectional area of one wire. The mass of one wire can be written both as m = V = AL and also as m = L .
t= Then we have
= A =
t=L
Thus,
For copper,
For steel, The total time is
F d I GH 4T JK L a fb8 920ge1.00 10 t = a 20.0 fM MM a4fa150f N L a fb7 860ge1.00 10 t = a30.0fM MM a4fa150f N
2 12
d 2 4
12 3 2
12 3 2
j OP PP Q j OP PP Q
= 0.137 s
= 0.192 s
0.137 + 0.192 = 0.329 s
Chapter 16
483
P16.32
Refer to the diagrams. From the freebody diagram of point A:
D L/4 L/4 A L/2 M M B
Fy = 0 T1 sin = Mg
and
Fx = 0 T1 cos = T
d
d
Combining these equations to eliminate T1 gives the tension in the Mg . string connecting points A and B as: T = tan The speed of transverse waves in this segment of string is then v= T =
Mg tan m L
T1
=
MgL m tan
A
T
and the time for a pulse to travel from A to B is t=
L 2
v
=
mL tan . 4Mg
Mg
FIG. P16.32
*P16.33
(a)
f has units Hz = 1 s , so T = with units
1 has units of seconds, s . For the other T we have T = v 2 , f
kg m 2 kg m = = N . m s2 s2
(b)
The first T is period of time; the second is force of tension.
Section 16.4
Reflection and Transmission
Problem 7 in Chapter 18 can be assigned with this section.
Section 16.5 P16.34 f= v
Rate of Energy Transfer by Sinusoidal Waves on Strings
=
30.0 = 60.0 Hz 0.500
= 2 f = 120 rad s
P=
P16.35
1 1 0.180 2 A 2 v = 120 2 2 3.60
FG H
IJ a K
f a0.100f a30.0f =
2 2
1.07 kW
Suppose that no energy is absorbed or carried down into the water. Then a fixed amount of power is spread thinner farther away from the source, spread over the circumference 2 r of an expanding circle. The powerperwidth across the wave front
P
2 r is proportional to amplitude squared so amplitude is proportional to
P . 2 r
484 P16.36
Wave Motion
T = constant; v =
(a) (b) (c)
T
; P=
1 2 A 2 v 2
If L is doubled, v remains constant and P is constant . If A is doubled and is halved, P 2 A 2 remains constant . If and A are doubled, the product 2 A 2 A2 remains constant, so
2
P remains constant .
(d) If L and are halved, then 2 (Changing L doesn't affect P ). P16.37 1
2
is quadrupled, so P is quadrupled .
A = 5.00 10 2 m
Therefore, v=
= 4.00 10 2 kg m
T
P = 300 W
T = 100 N
= 50.0 m s
P=
1 2 A 2 v : 2
2 =
2 300 2P = A 2 v 4.00 10 2 5.00 10 2
e
je
a f
j a50.0f
2
= 346 rad s = 55.1 Hz f= 2
P16.38
= 30.0 g m = 30.0 10 3 kg m
= 1.50 m
f = 50.0 Hz: 2 A = 0.150 m: (a) y = A sin
= 2 f = 314 s 1
A = 7.50 10 2 m
FG 2 x  tIJ H K y = e7.50 10 j sina 4.19 x  314t f
2
FIG. P16.38
2 2 2
(b)
P=
1 1 2 A 2 v = 30.0 10 3 314 2 2
e
314 ja f e7.50 10 j FGH 4.19 IJK W
P = 625 W
P16.39
(a) (b) (c) (d)
v = f =
2 50.0 m s = 62.5 m s = = k 0.800 2 k
=
f=
2 2 = m = 7.85 m k 0.800 50.0 = 7.96 Hz 2 1 1 2 A 2 v = 12.0 10 3 50.0 2 2
P=
e
ja f a0.150f a62.5f W =
2 2
21.1 W
Chapter 16
485
*P16.40
Comparing y = 0.35 sin 10t  3x + k= (a)
FG H
4
IJ with y = A sinbkx  t + g = A sinbt  kx  + g we have K
3 10 s = = = 3.33 m s . , = 10 s , A = 0.35 m . Then v = f = 2 f m 2 k 3 m The rate of energy transport is
P=
(b)
1 1 2 A 2 v = 75 10 3 kg m 10 s 2 2
e
jb
g a0.35 mf 3.33 m s =
2 2
15.1 W .
The energy per cycle is E = P T = 1 1 2 A 2 = 75 10 3 kg m 10 s 2 2
e
jb
g a0.35 mf
2
2
2 m = 3.02 J . 3
P16.41
Originally, 1 2 A 2 v 2 1 T P0 = 2 A 2 2
P0 =
1 P0 = 2 A 2 T 2 The doubled string will have doubled massperlength. Presuming that we hold tension constant, it can carry power larger by 2 times. 2 P0 = *P16.42 1 2 2 A T 2 2
As for a strong wave, the rate of energy transfer is proportional to the square of the amplitude and to the speed. We write P = FvA 2 where F is some constant. With no absorption of energy,
2 2 Fv bedrock A bedrock = Fv mudfill A mudfill
v bedrock A = mudfill = v mudfill A bedrock The amplitude increases by 5.00 times.
25 v mudfill =5 v mudfill
486
Wave Motion
Section 16.6 P16.43 (a) (b)
The Linear Wave Equation A = 7.00 + 3.00 4.00 yields A = 40.0 In order for two vectors to be equal, they must have the same magnitude and the same direction in threedimensional space. All of their components must be equal. Thus, 7.00 i + 0 j + 3.00k = A i + Bj + Ck requires A = 7.00 , B = 0 , and C = 3.00 .
a
f
(c)
In order for two functions to be identically equal, they must be equal for every value of every variable. They must have the same graphs. In A + B cos Cx + Dt + E = 0 + 7.00 mm cos 3.00 x + 4.00t + 2.00 , the equality of average values requires that A = 0 . The equality of maximum values requires B = 7.00 mm . The equality for the wavelength or periodicity as a function of x requires C = 3.00 rad m . The equality of period requires D = 4.00 rad s , and the equality of zerocrossings requires E = 2.00 rad . 2y x
2
a
f
a
f
*P16.44
The linear wave equation is If then
=
1 2 y v 2 t 2
y = e b a x  vt f y y =  bve b a x  vt f and = be b a x  vt f t x 2 y t 2 2y t 1 2y v 2 t 2
2
= b 2 v 2 e b a x  vt f and = v2 2y x 2 2y x
2
2y x 2
= b 2 e b a x  vt f
Therefore,
, demonstrating that e b a x  vt f is a solution
P16.45
The linear wave equation is To show that y = ln b x  vt y 1 =  bv t b x  vt
=
a
f
is a solution, we find its first and second derivatives with respect to x 2y t 2 2y x
2
and t and substitute into the equation.
a y = ba x  vt f x
fa f
1
=
1  bv
b 2 x  vt
a
a f
a
2
f
2
=
2
ax  vtf
1
v2
2
b
=
b x  vt b = 2y x 2
f
=
ax  vtf
2
v 1 2y 1 Then 2 2 = 2 v t v x  vt
e j a f
2
2
=
ax  vtf
1
2
so the given wave function is a solution.
Chapter 16
487
P16.46
(a)
From y = x 2 + v 2 t 2 , evaluate y = 2x x y = v 2 2t t Does 2y t 2 = 1 2y ? v 2 t 2 2y x 2 2y t 2 =2 = 2v 2
By substitution: 2 = equation. (b) Note 1 x + vt 2
1 2 v 2 and this is true, so the wave function does satisfy the wave v2
a
f
2
+
1 x  vt 2
a
f
2
1 2 1 1 1 x + xvt + v 2 t 2 + x 2  xvt + v 2 t 2 2 2 2 2 2 2 2 = x + v t as required. =
So (c)
f x + vt =
a
f 1 ax + vtf 2
2
and g x  vt =
a
f 1 ax  vtf 2
2
.
y = sin x cos vt makes
y = cos x cos vt x y =  v sin x sin vt t Then 2y x
2
2y x 2 2y t 2
=  sin x cos vt =  v 2 sin x cos vt
=
1 2 y v 2 t 2
becomes  sin x cos vt =
1 2 v sin x cos vt which is true as required. v2 Note sin x + vt = sin x cos vt + cos x sin vt
a
f
sin x  vt = sin x cos vt  cos x sin vt . So sin x cos vt = f x + vt + g x  vt with and g x  vt =
a
f
a f a f 1 f a x + vtf = sina x + vt f 2
a
f
1 sin x  vt 2
a
f
.
Additional Problems P16.47 Assume a typical distance between adjacent people ~ 1 m . Then the wave speed is v= x 1 m ~ ~ 10 m s t 0.1 s
Model the stadium as a circle with a radius of order 100 m. Then, the time for one circuit around the stadium is T=
2 2 r 2 10 ~ = 63 s ~ 1 min . v 10 m s
e j
488 P16.48
Wave Motion
Compare the given wave function y = 4.00 sin 2.00 x  3.00t cm to the general form y = A sin kx  t to find (a) (b) (c) (d) (e) amplitude A = 4.00 cm = 0.040 0 m k= 2 = 2.00 cm 1 and = cm = 0.031 4 m
a
f
a
f
= 2 f = 3.00 s 1 and f = 0.477 Hz
T= 1 = 2.09 s f
The minus sign indicates that the wave is traveling in the positive x direction . Let u = 10 t  3 x +
P16.49
(a)
4
du dx = 10  3 = 0 at a point of constant phase dt dt dx 10 = = 3.33 m s 3 dt The velocity is in the positive x direction .
(b) (c) (d) *P16.50 (a)
y 0.100 , 0 = 0.350 m sin 0.300 + k= 2 = 3 : = 0.667 m
b
g a
f FGH
= 0.054 8 m = 5.48 cm 4 = 2 f = 10 : f = 5.00 Hz
IJ K
fa f FGH 0.175 m = a0.350 mf sin b99.6 rad sgt sin b99.6 rad sgt = 0.5
vy =
y = 0.350 10 cos 10 t  3 x + t 4
a
IJ K
v y, max = 10 0.350 = 11.0 m s
a fa
f
The smallest two angles for which the sine function is 0.5 are 30 and 150, i.e., 0.523 6 rad and 2.618 rad. 99.6 rad s t1 = 0.523 6 rad , thus t1 = 5.26 ms
b g b99.6 rad sgt
2
= 2.618 rad , thus t 2 = 26.3 ms
t t 2  t1 = 26.3 ms  5.26 ms = 21.0 ms
(b) P16.51 Distance traveled by the wave =
FG IJ t = FG 99.6 rad s IJ e21.0 10 sj = H k K H 1.25 rad m K
3
1.68 m .
The equation v = f is a special case of speed = (cycle length)(repetition rate). Thus, v = 19.0 10 3 m frame 24.0 frames s = 0.456 m s .
e
jb
g
Chapter 16
489
P16.52
Assuming the incline to be frictionless and taking the positive xdirection to be up the incline:
Fx = T  Mg sin = 0
or the tension in the string is
T = Mg sin
v= T
The speed of transverse waves in the string is then
=
Mg sin
m L
=
MgL sin m mL Mg sin
The time interval for a pulse to travel the string's length is t =
L m =L = v MgL sin
P16.53
Energy is conserved as the block moves down distance x:
eK + U
x= (a)
g
+ Us
j
top
+ E = K + U g + U s 1 2 kx 2
e
j
bottom
0 + Mgx + 0 + 0 = 0 + 0 + 2 Mg k
T = kx = 2 Mg = 2 2.00 kg 9.80 m s 2 = 39.2 N 2 Mg k 39.2 N L = 0.500 m + = 0.892 m 100 N m L = L0 + x = L0 + v= v= T = TL m
b
ge
j
(b)
(c)
39.2 N 0.892 m 5.0 10 3 kg
v = 83.6 m s P16.54 Mgx = (a) 1 2 kx 2
T = kx = 2 Mg
2 Mg k 2 Mg 2 Mg L0 + m k
(b)
L = L0 + x = L0 +
(c)
v=
T
=
TL = m
FG H
IJ K
490 P16.55
Wave Motion
(a)
v=
T
=
e5.00 10
80.0 N
3
kg 2.00 m
j
= 179 m s
(b)
From Equation 16.21, P =
1 v v 2 A 2 and = 2 2 1 2v vA 2 2 2 2
FG IJ H K
2
2 3
P=
P=
FH
FG IJ H K
2
=
2 2 A 2 v 3
5.00 10 3 kg 2.00 m
IK b0.040 0 mg b179 m sg a0.160 mf
2
P = 1.77 10 4 W = 17.7 kW
P16.56 v= T and in this case T = mg ; therefore, m =
v 2 . g
Now v = f implies v =
so that k
m= g k
*P16.57
FG IJ H K
2
=
0.250 kg m 9.80 m s 2
LM 18 s OP N 0.750 m Q
1 1
2
= 14.7 kg .
2 mv b = m 2 r . The r
Let M = mass of block, m = mass of string. For the block, speed of a wave on the string is then v= t= T
F = ma implies T =
=
M 2 r
m r
= r
M m
r 1 = v
m M m = M 0.003 2 kg = 0.084 3 rad 0.450 kg
= t =
P16.58 (a)
=
v=
dm dx = A = A dL dx T
=
T = A
T ax + b
a
f
=
10 x + 10 2 cm 2
T
e
T
3
j
With all SI units, v =
10 x + 10 2 10 4
= 94.3 m s
e
3
j
ms
(b)
v x= 0 =
b2 700ge0 + 10 je10 j
2 4
24.0
v x=10 .0 =
b2 700ge10
24.0
2
+ 10 2 10 4
je j
= 66.7 m s
Chapter 16
491
P16.59
where Therefore, dx , so that But v = dt and
v=
T
T = xg , the weight of a length x, of rope. v = gx dx dt = gx t=
z
0
L
dx gx
=
1 g
x
1 2
L
= 2
0
L g
P16.60
At distance x from the bottom, the tension is T =
(a)
(b)
(c)
FG mxg IJ + Mg , so the wave speed is: H LK T TL F MgL IJ = dx . = = xg + G v= H m K dt m L F MgL IJ OP dx 1 xg + b MgL m g t= Then t = z dt = z M xg + G g N H m KQ O MgL I 2 LF LF m+M  MI t = MG Lg + t=2 JK  FGH MgL IJK PP JK g NH m m gG m H M Q L F m  0I When M = 0 , as in the previous problem, t=2 J= 2 L gG g m K H F 1 m  1 m + ...I F mI As m 0 we expand m + M = M G 1 + J = M G 1 + H MK H 2 M 8 M JK F M + em M j  em M j + ...  M I L G JJ to obtain t=2 gG m H K LF1 m I mL t2 GH 2 M JK = Mg g
t L 0 1 2 0 1 2 12 12 12 2 2 1 2 1 8 2 32
1 2 x=L
x=0
P16.61
(a)
The speed in the lower half of a rope of length L is the same function of distance (from the L bottom end) as the speed along the entire length of a rope of length . 2 L L with L = Thus, the time required = 2 g 2
FG IJ H K
and the time required = 2
L L = 0.707 2 2g g
F GH
I JK
.
It takes the pulse more that 70% of the total time to cover 50% of the distance. (b) By the same reasoning applied in part (a), the distance climbed in is given by d = L L , we find the distance climbed = . g 4 1 In half the total trip time, the pulse has climbed of the total length. 4 For = t = 2 g 2 . 4
492 P16.62
Wave Motion
(a) (b) (c) (d)
v= v= v= v=
15.0 = = 5.00 m s in positive x direction k 3.00
15.0 = 5.00 m s in negative x direction 3.00 15.0 = 7.50 m s in negative x direction 2.00 12.0
1 2
= 24.0 m s in positive x direction
T A L L
P16.63
Young's modulus for the wire may be written as Y =
, where T is the tension maintained in the
wire and L is the elongation produced by this tension. Also, the mass density of the wire may be expressed as =
. A The speed of transverse waves in the wire is then
v= T
=
T A
=
Y
c h
L L
A
and the strain in the wire is
L v 2 = . L Y If the wire is aluminum and v = 100 m s, the strain is 2.70 10 3 kg m3 100 m s L = L 7.00 10 10 N m 2
e
jb
g
2
= 3.86 10 4 .
*P16.64
(a)
Consider a short section of chain at the top of the loop. A freebody diagram is shown. Its length is s = R 2 and its mass is R2 . In the frame of reference of the center of the loop, Newton's second law is 2 2 mv 0 R 2v 0 2T sin down = down = Fy = ma y R R
a f
T 2 R
T
FIG. P16.64(a)
2 For a very short section, sin = and T = v 0 .
(b) (c)
The wave speed is v =
T
= v0 .
In the frame of reference of the center of the loop, each pulse moves with equal speed clockwise and counterclockwise.
v v0 v0 FIG. P16.64(c1) continued on next page v0
v
Chapter 16
493
In the frame of reference of the ground, once pulse moves backward at speed v 0 + v = 2 v 0 and the other forward at v 0  v = 0 . The one pulse makes two revolutions while the loop makes one revolution and the other pulse does not move around the loop. If it is generated at the sixo'clock position, it will stay at the sixo'clock position.
v0
v0
v0
FIG. P16.64(c2) P16.65 (a) Assume the spring is originally stationary throughout, extended to have a length L much greater than its equilibrium length. We start moving one end forward with the speed v at which a wave propagates on the spring. In this way we create a single pulse of compression that moves down the length of the spring. For an increment of spring with length dx and mass dm, just as the pulse swallows it up, F = ma becomes kdx = adm or But dm k = so a = . dx dv v v2 = when vi = 0. But L = vt , so a = . dt t L k k
dm dx
= a.
Also, a =
Equating the two expressions for a, we have
=
v2 or v = L
kL
.
(b)
Using the expression from part (a) v =
kL
=
kL2 = m
12 0
b100 N mga2.00 mf
0.400 kg
2
= 31.6 m s .
P16.66
(a)
F T I F 2T IJ v=G J =G H K H K F T I F 2T IJ v = G J = G H K H 3 K
12 0 0 12 0 0
12
= v0
12
2 where v 0 2 3
FG T IJ H K
0
= v0 t 0 2 2 = 2
(b)
t left =
L 2
v
=
L 2
L 2 v0 2 = L 2 v0
=
= 0.354t 0 where t 0 = 0.612 t 0
L v0
t right =
v
t 0
2 3
2 3
t left + t right = 0.966 t 0
494 P16.67
Wave Motion
(a)
P x =
af af
1 1 3 2 2 bx 2 A0 e 2 A 2 v = 2 A0 e 2 bx = 2 2 2k k
FG IJ H K
(b)
P 0 =
3 2 A0 2k
(c)
P x = e 2 bx P 0
4 450 km = 468 km h = 130 m s 9.50 h
af af
P16.68
v=
130 m s v2 d= = = 1 730 m g 9.80 m s 2 *P16.69 (a)
b g e j a x f is a linear function, so it is of the form To have a0f = we require b = . Then
2
0
0
af aL f =
m=
x = mx + b
L
= mL + 0
so Then (b) From v =
x =
a f b
L  0 L
L
 0 x L
g
+ 0
dx dx , the time required to move from x to x + dx is . The time required to move v dt from 0 to L is t =
I FG  IJ dxF t = JK H L K GH T F  x I 1 F t = GH L IJK GH b L g + JK 1  T 2L t = e  j 3 T b  g 2Le  je + + j t = 3 T e  je + j 2L F + + I t = G + JK 3 TH
1
L 0
z z z FGH b
L
dx L dx 1 = = v 0 T T 0
z
0
L
x dx
12
af
L  0 x + 0 L
L
g
L
0
L L  0
IJ K
32
L
0
L
0
0
3 2 0
L
0
32 L
32 0
L
0
L
L
0
0
L
0
L
0
L
L
0
0
L
0
Chapter 16
495
ANSWERS TO EVEN PROBLEMS
P16.2 P16.4 P16.6 P16.8 P16.10 P16.12 P16.14 see the solution (a) the P wave; (b) 665 s 0.800 m s 2.40 m s 0.300 m in the positive xdirection 6.67 cm (a) see the solution; (b) 0.125 s; in agreement with the example (a) see the solution; (b) 18.0 m ; 83.3 ms ; 75.4 rad s ; 4.20 m s ; (c) 0.2 m sin 18 x + 75.4t  0.151 P16.48 P16.40 P16.42 P16.44 P16.46 P16.38 (a) y = 0.075 0 sin 4.19 x  314t ; (b) 625 W (a) 15.1 W ; (b) 3.02 J The amplitude increases by 5.00 times see the solution (a) see the solution; 1 1 2 2 x + vt + x  vt ; (b) 2 2 1 1 (c) sin x + vt + sin x  vt 2 2
b
g a
f
a
f
a
f
a
f
a
f
P16.16
a
f a
f
(a) 0.040 0 m; (b) 0.031 4 m ; (c) 0.477 Hz; (d) 2.09 s; (e) positive x direction (a) 21.0 ms ; (b) 1.68 m t = mL Mg sin 2 Mg ; k 2 Mg 2 Mg L0 + m k
P16.18
(a) 0.021 5 m; (b) 1.95 rad; (c) 5.41 m s ; (d) y x , t =
b0.021 5 mg sinb8.38x + 80.0 t + 1.95g
30.0 N
b g
P16.50 P16.52
P16.20 P16.22 P16.24
(a) see the solution; (b) 3.18 Hz P16.54
(a) 2Mg ; (b) L0 + (c)
(a) y = 0.2 mm sin 16 x  3 140t ; (b) 158 N 631 N v= Tg 2 M m m ; (b) 3.89 kg P16.60 P16.62 P16.56 P16.58
a
f b
g
FG H
IJ K
P16.26 P16.28
14.7 kg (a) v = T
7
10 x + 10 6
P16.30
F m I (a) v = G 30.4 H s kg JK
mL tan 4Mg 1.07 kW
e
j
in SI units;
(b) 94.3 m s; 66.7 m s see the solution (a) 5.00 i m s ; (b) 5.00 i m s ; (c) 7.50 i m s ; (d) 24.0 i m s P16.64
2 (a) v 0 ; (b) v 0 ; (c) One travels 2 rev and the other does not move around the loop.
P16.32 P16.34 P16.36
(a), (b), (c) P is constant ; (d) P is quadrupled
496 P16.66
Wave Motion
F 2T IJ = v 2 ; (a) v = G H K F 2T IJ = v 2 ; (b) 0.966t v = G 3 H 3 K
12 0 0 0 12 0 0 0
P16.68
130 m s ; 1.73 km
0
17
Sound Waves
CHAPTER OUTLINE
17.1 17.2 17.3 17.4 17.5 17.6 Speed of Sound Waves Periodic Sound Waves Intensity of Periodic Sound Waves The Doppler Effect Digital Sound Recording Motion Picture Sound
ANSWERS TO QUESTIONS
Q17.1 Sound waves are longitudinal because elements of the mediumparcels of airmove parallel and antiparallel to the direction of wave motion. We assume that a perfect vacuum surrounds the clock. The sound waves require a medium for them to travel to your ear. The hammer on the alarm will strike the bell, and the vibration will spread as sound waves through the body of the clock. If a bone of your skull were in contact with the clock, you would hear the bell. However, in the absence of a surrounding medium like air or water, no sound can be radiated away. A largerscale example of the same effect: Colossal storms raging on the Sun are deathly still for us. What happens to the sound energy within the clock? Here is the answer: As the sound wave travels through the steel and plastic, traversing joints and going around corners, its energy is converted into additional internal energy, raising the temperature of the materials. After the sound has died away, the clock will glow very slightly brighter in the infrared portion of the electromagnetic spectrum.
Q17.2
Q17.3
If an object is
1 meter from the sonic ranger, then the sensor would have to measure how long it 2 would take for a sound pulse to travel one meter. Since sound of any frequency moves at about 343 m s, then the sonic ranger would have to be able to measure a time difference of under 0.003 seconds. This small time measurement is possible with modern electronics. But it would be more expensive to outfit sonic rangers with the more sensitive equipment than it is to print "do not 1 use to measure distances less than meter" in the users' manual. 2 The speed of sound to two significant figures is 340 m s. Let's assume that you can measure time to 1 second by using a stopwatch. To get a speed to two significant figures, you need to measure a 10 time of at least 1.0 seconds. Since d = vt , the minimum distance is 340 meters.
Q17.4
Q17.5
The frequency increases by a factor of 2 because the wave speed, which is dependent only on the medium through which the wave travels, remains constant.
497
498 Q17.6
Sound Waves
When listening, you are approximately the same distance from all of the members of the group. If different frequencies traveled at different speeds, then you might hear the higher pitched frequencies before you heard the lower ones produced at the same time. Although it might be interesting to think that each listener heard his or her own personal performance depending on where they were seated, a time lag like this could make a Beethoven sonata sound as if it were written by Charles Ives. Since air is a viscous fluid, some of the energy of sound vibration is turned into internal energy. At such great distances, the amplitude of the signal is so decreased by this effect you re unable to hear it. We suppose that a point source has no structure, and radiates sound equally in all directions (isotropically). The sound wavefronts are expanding spheres, so the area over which the sound energy spreads increases according to A = 4 r 2 . Thus, if the distance is tripled, the area increases by a factor of nine, and the new intensity will be oneninth of the old intensity. This answer according to the inversesquare law applies if the medium is uniform and unbounded. For contrast, suppose that the sound is confined to move in a horizontal layer. (Thermal stratification in an ocean can have this effect on sonar "pings.") Then the area over which the sound energy is dispersed will only increase according to the circumference of an expanding circle: A = 2 rh , and so three times the distance will result in one third the intensity. In the case of an entirely enclosed speaking tube (such as a ship's telephone), the area perpendicular to the energy flow stays the same, and increasing the distance will not change the intensity appreciably. He saw the first wave he encountered, light traveling at 3.00 10 8 m s . At the same moment, infrared as well as visible light began warming his skin, but some time was required to raise the temperature of the outer skin layers before he noticed it. The meteor produced compressional waves in the air and in the ground. The wave in the ground, which can be called either sound or a seismic wave, traveled much faster than the wave in air, since the ground is much stiffer against compression. Our witness received it next and noticed it as a little earthquake. He was no doubt unable to distinguish the P and S waves. The first aircompression wave he received was a shock wave with an amplitude on the order of meters. It transported him off his doorstep. Then he could hear some additional direct sound, reflected sound, and perhaps the sound of the falling trees. A microwave pulse is reflected from a moving object. The waves that are reflected back are Doppler shifted in frequency according to the speed of the target. The receiver in the radar gun detects the reflected wave and compares its frequency to that of the emitted pulse. Using the frequency shift, the speed can be calculated to high precision. Be forewarned: this technique works if you are either traveling toward or away from your local law enforcement agent! As you move towards the canyon wall, the echo of your car horn would be shifted up in frequency; as you move away, the echo would be shifted down in frequency. Normal conversation has an intensity level of about 60 dB. A rock concert has an intensity level of about 120 dB. A cheering crowd has an intensity level of about 90 dB. Normal conversation has an intensity level of about 5060 dB. Turning a page in the textbook has an intensity level of about 1020 dB.
Q17.7
Q17.8
Q17.9
Q17.10
Q17.11 Q17.12 Q17.13
Chapter 17
499
Q17.14
One would expect the spectra of the light to be Doppler shifted up in frequency (blue shift) as the star approaches us. As the star recedes in its orbit, the frequency spectrum would be shifted down (red shift). While the star is moving perpendicular to our line of sight, there will be no frequency shift at all. Overall, the spectra would oscillate with a period equal to that of the orbiting stars. For the sound from a source not to shift in frequency, the radial velocity of the source relative to the observer must be zero; that is, the source must not be moving toward or away from the observer. The source can be moving in a plane perpendicular to the line between it and the observer. Other possibilities: The source and observer might both have zero velocity. They might have equal velocities relative to the medium. The source might be moving around the observer on a sphere of constant radius. Even if the source speeds up on the sphere, slows down, or stops, the frequency heard will be equal to the frequency emitted by the source. Wind can change a Doppler shift but cannot cause one. Both v o and v s in our equations must be interpreted as speeds of observer and source relative to the air. If source and observer are moving relative to each other, the observer will hear one shifted frequency in still air and a different shifted frequency if wind is blowing. If the distance between source and observer is constant, there will never be a Doppler shift. If the object being tracked is moving away from the observer, then the sonic pulse would never reach the object, as the object is moving away faster than the wave speed. If the object being tracked is moving towards the observer, then the object itself would reach the detector before reflected pulse. Newfallen snow is a wonderful acoustic absorber as it reflects very little of the sound that reaches it. It is full of tiny intricate air channels and does not spring back when it is distorted. It acts very much like acoustic tile in buildings. So where does the absorbed energy go? It turns into internal energyalbeit a very small amount. As a sound wave moves away from the source, its intensity decreases. With an echo, the sound must move from the source to the reflector and then back to the observer, covering a significant distance. The observer would most likely hear the sonic boom of the plane itself and then beep, baap, boop. Since the plane is supersonic, the loudspeaker would pull ahead of the leading "boop" wavefront before emitting the "baap", and so forth. "How are you?" would be heard as "?uoy era woH" This system would be seen as a star moving in an elliptical path. Just like the light from a star in a binary star system, described in the answer to question 14, the spectrum of light from the star would undergo a series of Doppler shifts depending on the star's speed and direction of motion relative to the observer. The repetition rate of the Doppler shift pattern is the period of the orbit. Information about the orbit size can be calculated from the size of the Doppler shifts.
Q17.15
Q17.16
Q17.17
Q17.18
Q17.19 Q17.20
Q17.21
SOLUTIONS TO PROBLEMS
Section 17.1 P17.1 Speed of Sound Waves
Since v light >> v sound : d 343 m s 16. 2 s = 5.56 km v= B = 2.80 10 10 = 1.43 km s 13.6 10 3
b
ga
f
P17.2
500 P17.3
Sound Waves
Sound takes this time to reach the man: so the warning should be shouted no later than before the pot strikes. Since the whole time of fall is given by y = 1 2 gt : 2
a20.0 m  1.75 mf = 5.32 10
343 m s 1 9.80 m s 2 t 2 2
2
s
0.300 s + 5.32 10 2 s = 0.353 s 18.25 m = t = 1.93 s
e
j
the warning needs to come into the fall, when the pot has fallen to be above the ground by P17.4 (a) At 9 000 m, T =
1.93 s  0.353 s = 1.58 s 1 9.80 m s 2 1.58 s 2
e
ja
f
2
= 12.2 m
20.0 m  12.2 m = 7.82 m
FG 9 000 IJ a1.00 Cf = 60.0 C so T = 30.0 C . H 150 K
Using the chain rule:
dv dv dT dx dv dT v dv 1 , so dt = 247 s = =v = v 0.607 = dt dT dx dt dT dx v 150 247
a
fFGH IJK
a
f
t = 27.2 s for sound to reach ground. (b) t= 9 000 h = = 25.7 s v 331.5 + 0.607 30.0
f z dv z v L 331.5 + 0.607a30.0f OP Fv I t = a 247 sf lnG J = a 247 sf ln M Hv K N 331.5 + 0.607a30.0f Q
t
dt = 247 s
a
vf vi
0
f
i
a f
It takes longer when the air cools off than if it were at a uniform temperature. *P17.5 Let x 1 represent the cowboy's distance from the nearer canyon wall and x 2 his distance from the farther cliff. The sound for the first echo travels distance 2 x1 . For the second, 2 x 2 . For the third, 2 x 2  2 x1 2 x1 + 2 x 2  2 x 2 2 x1 + 2 x 2 . For the fourth echo, 2 x1 + 2 x 2 + 2 x1 . Then = 1.92 s and = 1.47 s . 340 m s 340 m s 2x2 1 = 1.92 s + 1.47 s ; x 2 = 576 m. Thus x 1 = 340 m s 1.47 s = 250 m and 2 340 m s (a) So x 1 + x 2 = 826 m 2 x 1 + 2 x 2 + 2 x 1  2 x1 + 2 x 2 340 m s
(b)
b
g=
1.47 s
Chapter 17
501
P17.6
It is easiest to solve part (b) first: (b) The distance the sound travels to the plane is d s = h 2 + The sound travels this distance in 2.00 s, so
FG h IJ H 2K
2
=
h 5 . 2
b ga f 2a686 mf = 614 m . giving the altitude of the plane as h =
ds = 5 (a) The distance the plane has traveled in 2.00 s is v 2.00 s = Thus, the speed of the plane is: v = 307 m = 153 m s . 2.00 s
h 5 = 343 m s 2.00 s = 686 m 2
a
f
h = 307 m . 2
Section 17.2 P17.7
Periodic Sound Waves 340 m s v = = 5.67 mm f 60.0 10 3 s 1 26 C = 346 m s 273 C
=
*P17.8
The sound speed is v = 331 m s 1 + (a)
Let t represent the time for the echo to return. Then d= 1 1 vt = 346 m s 24 10 3 s = 4.16 m . 2 2
(b)
Let t represent the duration of the pulse: t = 10 10 10 10 = = = = 0.455 s . v f f 22 10 6 1 s
(c)
L = 10 =
10 v 10 346 m s = = 0.157 mm f 22 10 6 1 s
b
g
*P17.9
If f = 1 MHz , =
v 1 500 m s = = 1.50 mm f 10 6 s 1 500 m s = 75.0 m If f = 20 MHz , = 2 10 7 s Pmax = v smax smax 4.00 10 3 N m 2 Pmax = = = 1.55 10 10 m v 1.20 kg m3 343 m s 2 10.0 10 3 s 1
P17.10
e
e
jb
ga fe
j
j
502 P17.11
Sound Waves
(a)
A = 2.00 m
=
2 = 0.400 m = 40.0 cm 15.7 858 = 54.6 m s v= = k 15.7
(b) (c) P17.12 (a)
s = 2.00 cos 15.7 0.050 0  858 3.00 10 3 = 0.433 m
a fb
b
g a fe
j
j F x  340 t IJ (SI units) P = a1.27 Paf sinG Hm s K
= 2 f = 340 s, so f = 170 Hz
k= 2 = m , giving = 2.00 m
v max = A = 2.00 m 858 s 1 = 1.72 mm s
ge
The pressure amplitude is: Pmax = 1.27 Pa . (b) (c) (d) P17.13 k= 2
v = f = 2.00 m 170 Hz = 340 m s =
a
fa
f
a
2 = 62.8 m 1 0.100 m 2 343 m s s a0.100 mf P = a0.200 Paf sin 62.8 x m  2.16 10 = 2 343 m s
=
2 v
=
b
f
g = 2.16 10
4
1
Therefore,
4
ts .
P17.14
= 2 f =
smax = k= 2
2 v
a0.100 mf
b
g = 2.16 10 f ge
4
rad s
0.200 Pa Pmax = = 2. 25 10 8 m 3 v 1.20 kg m 343 m s 2.16 10 4 s 1
e
a jb
j
a0.100 mf = 62.8 m Therefore, s = s cosb kx  t g = e 2.25 10
max
=
2
1
8
m cos 62.8 x m  2.16 10 4 t s .
j e
j
P17.15
Pmax = v smax = v 2v 2 smax = Pmax
FG 2 v IJ s H K 2 a1.20fa343 f e5.50 10 j = =
max 2 6
0.840
5.81 m
Chapter 17
503
P17.16
(a)
The sound "pressure" is extra tensile stress for onehalf of each cycle. When it becomes 0.500% 13.0 10 10 Pa = 6.50 10 8 Pa , the rod will break. Then, Pmax = v smax
a
fe
j
smax = (b)
6.50 10 8 N m 2 Pmax = = 4.63 mm . v 8.92 10 3 kg m3 5 010 m s 2 500 s
e
jb
gb
g
From s = smax cos kx  t
a
s = smax sin kx  t t v max = smax = 2 500 s 4.63 mm = 14.5 m s v=
f
a
f
b
ga
f
(c)
I=
1 1 1 2 2 v smax = vvmax = 8.92 10 3 kg m 3 5 010 m s 14.5 m s 2 2 2 = 4.73 10 9 W m 2
b
g
e
jb
gb
g
2
*P17.17
Let P x represent absolute pressure as a function of x. The net force to the right on the chunk of air is + P x A  P x + x A . Atmospheric P xA . pressure subtracts out, leaving  P x + x + P x A =  P x + x A Px A x 2s The mass of the air is m = V = Ax and its acceleration is 2 . So t FIG. P17.17 Newton's second law becomes 2 P s  xA = Ax 2 x t s 2s B  = 2 x x t
af
af a f a f af
af
a
f
FG H
IJ K
B 2s 2s = x 2 t 2 Into this wave equation as a trial solution we substitute the wave function s x , t = smax cos kx  t we find s =  ksmax sin kx  t x 2s =  k 2 smax cos kx  t 2 x s = +smax sin kx  t t 2s =  2 smax cos kx  t t 2 B B 2s 2s = becomes  k 2 smax cos kx  t =  2 smax cos kx  t x 2 t 2
b g
a
f
a
f
a
f
a
f
a
f
a
f
a
f
This is true provided
B 4 2
2
= 4 2 f 2 . B
The sound wave can propagate provided it has 2 f 2 = v 2 = speed v = B
; that is, provided it propagates with
.
504
Sound Waves
Section 17.3 *P17.18
Intensity of Periodic Sound Waves
The sound power incident on the eardrum is = IA where I is the intensity of the sound and A = 5.0 10 5 m 2 is the area of the eardrum. (a) At the threshold of hearing, I = 1.0 10 12 W m 2 , and
= 1.0 10 12 W m 2 5.0 10 5 m 2 = 5.00 10 17 W .
(b) At the threshold of pain, I = 1.0 W m 2 , and
e
je
j
= 1.0 W m 2 5.0 10 5 m 2 = 5.00 10 5 W .
P17.19
= 10 log
P17.20
(a)
FG I IJ = 10 logF 4.00 10 66.0 dB GH 1.00 10 HI K F I I 70.0 dB = 10 log G H 1.00 10 W m JK W m j10 b Therefore, I = e1.00 10
6 0 12 12 2 12 2
e I= JK
je
j
70 .0 10
g=
1.00 10 5 W m 2 .
(b)
I=
2 Pmax , so 2 v
Pmax = 2 vI = 2 1.20 kg m3 343 m s 1.00 10 5 W m 2 Pmax = 90.7 mPa P17.21 I= (a) 1 2 2 smax v 2
e
jb
ge
j
At f = 2 500 Hz , the frequency is increased by a factor of 2.50, so the intensity (at constant
a f = 6.25 . Therefore, 6.25a0.600f = 3.75 W m
smax ) increases by 2.50
2
2
.
(b) P17.22
0.600 W m 2
1 2 2 2 smax v = 2 2 vf 2 smax 2
The original intensity is I 1 = (a)
If the frequency is increased to f while a constant displacement amplitude is maintained, the new intensity is
2 I 2 = 2 2 v f smax so
b g
2
2 2 I 2 2 v f smax f = = 2 2 2 I1 f 2 vf smax
b g
FG IJ H K
2
or I 2 =
FG f IJ H fK
2
I1 .
continued on next page
Chapter 17
505
(b)
If the frequency is reduced to f = intensity is
f while the displacement amplitude is doubled, the new 2
I 2 = 2 2 v or the intensity is unchanged . *P17.23 (a)
FG f IJ b2s g H 2K
2 max
2
2 = 2 2 vf 2 smax = I 1
For the low note the wavelength is = For the high note =
v 343 m s = = 2.34 m . f 146.8 s
343 m s = 0.390 m . 880 s
880 Hz = 5.99 nearly 146.8 Hz equal to a small integer. This fact is associated with the consonance of the notes D and A. We observe that the ratio of the frequencies of these two notes is (b)
= 10 dB log
I=
2 Pmax 2 v
F GH 10
I
12
W m
2
I = 75 dB gives I = 3.16 10 JK e jb g
5
W m2
Pmax = 3.16 10 5 W m 2 2 1.20 kg m3 343 m s = 0.161 Pa for both low and high notes. (c) 1 1 2 2 v smax = v 4 2 f 2 smax 2 2 I smax = 2 2 vf 2 for the low note, I= smax = = 1 2 1.20 kg m 343 m s 146.8 s
2 3
b
g
3.16 10 5 W m 2
6.24 10 5 m = 4.25 10 7 m 146.8 for the high note, 6.24 10 5 smax = m = 7.09 10 8 m 880 (d) With both frequencies lower (numerically smaller) by the factor 146.8 880 = = 1.093 , the 134.3 804.9 wavelengths and displacement amplitudes are made 1.093 times larger, and the pressure amplitudes are unchanged.
*P17.24
The power necessarily supplied to the speaker is the power carried away by the sound wave: P= 1 Av smax 2
b
g
2
2 = 2 2 Avf 2 smax
= 2 2 1.20 kg m3
e
j FGH 0.08 m IJK b343 m sgb600 1 sg e0.12 10 2
2 2
2
m
j
2
= 21.2 W
506 P17.25
Sound Waves
(a)
I 1 = 1.00 10 12 W m 2 10 b
e
j
1 10
g = e1.00 10 12
2 10
W m 2 10 80.0 10 W m 2 10 75.0 10
j
or
I 1 = 1.00 10
4
I 2 = 1.00 10
e
W m
2
12
W m 2 10 b
j
g = e1.00 10 12
5 2
j
or I 2 = 1.00 10 W m = 3.16 10 W m When both sounds are present, the total intensity is
4.5
2
I = I 1 + I 2 = 1.00 10 4 W m 2 + 3.16 10 5 W m 2 = 1.32 10 4 W m 2 .
(b) The decibel level for the combined sounds is
= 10 log
*P17.26 (a) We have =
F 1.32 10 GH 1.00 10
4
W m2 W m
2
12
I = 10 log 1.32 10 = e j JK
8
81.2 dB .
v and f is the same for all three waves. Since the speed is smallest in air, is f 1 493 m s smallest in air. It is larger by = 4.51 times in water and by 331 m s 5 950 = 18.0 times in iron . 331 From I = 1 2 v 2 smax ; smax = 2
2 v 0
(b)
2I 0
, smax is smallest in iron, larger in water by
7 860 5 950 7 860 5 950 iron v iron = = 5.60 times , and larger in air by = 331 times . 1 000 1 493 1.29 331 water v water (c) From I =
2 Pmax ; Pmax = 2 Iv , Pmax is smallest in air, larger in water by 2 v
1 000 1 493 = 59.1 times , and larger in iron by 1.29 331 (d)
331 m s 2 v v2 = = = 0.331 m in air f 2 000 s 1 493 m s 5 950 m s = = 1.49 m in water = = 5.95 m in iron 1 00 0 s 1 000 s
=
b
g
7 860 5 950 = 331 times . 1.29 331
smax = smax = smax =
2 v 0
2I0
=
e1.29 kg m jb331 m sgb6 283 1 sg
3
2 10 6 W m 2
2
= 1.09 10 8 m in air
2 10 6 1 = 1.84 10 10 m in water 1 000 1 493 6 283
b
g
2 10 6 1 = 3.29 10 11 m in iron 7 860 5 950 6 283
b
Pmax = 2 Iv =
Pmax = 2 10 6 1 000 1 493 = 1.73 Pa in water Pmax = 2 10 6 9.67 Pa in iron
b g b7 860gb5 950g =
g 2e10
6
W m 2 1.29 kg m3 331 m s = 0.029 2 Pa in air
je
j
Chapter 17
507
P17.27
(a)
120 dB = 10 dB log I = 1.00 W m 2 = r= = 4 I
LM NM 10
I
12
W m
2
OP QP
4 r 2 6.00 W
4 1.00 W m 2
e
j
= 0.691 m
We have assumed the speaker is an isotropic point source. (b) 0 dB = 10 dB log
F GH 10 e
12
I W m2
I JK j
= 691 km
I = 1.00 10 12 W m 2 r= = 4 I 6.00 W 4 1.00 10 12 W m 2
We have assumed a uniform medium that absorbs no energy. P17.28 We begin with 2 = 10 log
FG I IJ , and HI K
2 0
1
= 10 log
2
Also, I 2 = I , and I 1 = , giving 2 I1 4 r22 4 r12
FG I IJ , so HI K FI I  = 10 log G J . HI K Fr I =G J . Hr K
1 0 1 2 1 2 1 2
Then, 2  1 = 10 log P17.29
FG r IJ Hr K
1 2
2
= 20 log
FG r IJ Hr K
1 2
.
Since intensity is inversely proportional to the square of the distance, I4 = 10.0 P 2 1 I 0. 4 and I 0. 4 = max = = 0.121 W m 2 . 100 2 v 2 1.20 343
a f a fa f
2
The difference in sound intensity level is = 10 log At 0.400 km,
FG I HI
4 km
0.4 km
IJ = 10a2.00f = 20.0 dB . K
2 2
0. 4 = 10 log
At 4.00 km,
F 0.121 W m I = 110.8 dB . GH 10 W m JK
12
4 = 0.4 + = 110.8  20.0 dB = 90.8 dB .
= 4  7.00 dB km 3.60 km = 65.6 dB . 4
a
f
Allowing for absorption of the wave over the distance traveled,
b
ga
f
This is equivalent to the sound intensity level of heavy traffic.
508 P17.30
Sound Waves
Let r1 and r2 be the distance from the speaker to the observer that hears 60.0 dB and 80.0 dB, respectively. Use the result of problem 28,
2  1 = 20 log
Thus, log
1
FG r IJ = 1 , so r Hr K
2
FG r IJ , to obtain 80.0  60.0 = 20 logFG r IJ . Hr K Hr K
1 1 2 2 1
= 10.0r2 . Also: r1 + r2 = 110 m , so
10.0r2 + r2 = 110 m giving r2 = 10.0 m , and r1 = 100 m . P17.31 We presume the speakers broadcast equally in all directions. (a) rAC = 3.00 2 + 4.00 2 m = 5.00 m 1.00 10 3 W I= = = 3.18 10 6 W m 2 2 4 r 2 4 5.00 m
= 10 dB log
F 3.18 10 W m I GH 10 W m JK
6 2 12 2
a
f
= 10 dB 6.50 = 65.0 dB
(b) rBC = 4. 47 m I=
1.50 10 3 W 4 4.47 m
a
= 10 dB log = 67.8 dB
(c)
F 5.97 10 I GH 10 JK
6 12
f
2
= 5.97 10 6 W m 2
I = 3.18 W m 2 + 5.97 W m 2
= 10 dB log
F 9.15 10 I = GH 10 JK
6 12
69.6 dB
P17.32
In I =
1 , intensity I is proportional to 2 , 2 r 4 r I 2 r12 = . I 1 r22
2 max 2 2 2 2 1
so between locations 1 and 2: In I = 1 v smax 2
2 2 1 2
b g , intensity is proportional to s , so II = ss . F s I F r I F 1I F r I Then, G J = G J or G J = G J , giving r = 2r = 2a50.0 mf = 100 m . H s K H r K H 2K H r K But, r = a50.0 mf + d yields d = 86.6 m .
1 2 1 2 2 1 2 2
2
1
2
2
2
Chapter 17
P17.33
= 10 log
FG I IJ H 10 K
12
509
I = 10 b
10
g e10 12 j
W m2
I a120 dB f = 1.00 W m 2 ; I a100 dB f = 1.00 10 2 W m 2 ; I a10 dB f = 1.00 10 11 W m 2 (a) = 4 r 2 I so that r12 I 1 = r22 I 2 r2
(b) P17.34 (a) (b) P17.35 (a)
r2
FI =r G HI FI =r G HI
1 1
1 2
1 2
IJ K IJ K
12
= 3.00 m
12
a a
f f
1.00 = 30.0 m 1.00 10 2 1.00 = 9.49 10 5 m 1.00 10 11
= 3.00 m
E =t = 4 r 2 It = 4 100 m
a
f e7.00 10
2
2
W m 2 0.200 s = 1.76 kJ
ja
f
= 10 log
F 7.00 10 I = GH 1.00 10 JK
2 12
108 dB
The sound intensity inside the church is given by
FG I IJ HI K F I I 101 dB = a10 dBf lnG H 10 W m JK I = 10 e10 W m j = 10 W m
= 10 ln
0 12 2 10.1 12 2 1.90
2
= 0.012 6 W m 2
We suppose that sound comes perpendicularly out through the windows and doors. Then, the radiated power is = IA = 0.012 6 W m 2 22.0 m 2 = 0.277 W . Are you surprised by how small this is? The energy radiated in 20.0 minutes is E =t = 0.277 J s 20.0 min (b)
e
je
j
b
ga
60 fFGH 1.00.0 s IJK = min
332 J .
If the ground reflects all sound energy headed downward, the sound power, = 0.277 W , covers the area of a hemisphere. One kilometer away, this area is A = 2 r 2 = 2 1 000 m = 2 10 6 m 2 . The intensity at this distance is I= 0. 277 W = = 4.41 10 8 W m 2 A 2 10 6 m 2
b
g
2
and the sound intensity level is
= 10 dB ln
a
41 10 f FGH 14..00 10
8
W m2 W m
2
12
I= JK
46.4 dB .
510 *P17.36
Sound Waves
Assume you are 1 m away from your lawnmower and receiving 100 dB sound from it. The intensity I of this sound is given by 100 dB = 10 dB log 12 ; I = 10 2 W m 2 . If the lawnmower 10 W m2 radiates as a point source, its sound power is given by I = . 4 r 2 = 4 1 m 10 2 W m 2 = 0.126 W Now let your neighbor have an identical lawnmower 20 m away. You receive from it sound with 0.126 W intensity I = = 2.5 10 5 W m 2 . The total sound intensity impinging on you is 2 4 20 m
a f
2
a
10
2
W m + 2.5 10 5 W m 2 = 1.002 5 10 2 W m 2 . So its level is
2
f
10 dB log
1.002 5 10 2 10 12
= 100.01 dB .
If the smallest noticeable difference is between 100 dB and 101 dB, this cannot be heard as a change from 100 dB.
Section 17.4 P17.37 f= f
The Doppler Effect
bv v g bv v g
O S
(a)
(b)
P17.38
(a)
a343 + 40.0f = 338 Hz a343 + 20.0f a343 + 20.0f = 483 Hz f = 510 a343 + 40.0f F 115 min I = 12.0 rad s = 2 f = 2 G H 60.0 s min JK v = A = b12.0 rad sge1.80 10 mj =
f = 320
max 3
0.021 7 m s
(b)
The heart wall is a moving observer. f= f
O
FG v + v IJ = b2 000 000 HzgFG 1 500 + 0.021 7 IJ = H v K H 1 500 K FG v IJ = b2 000 029 HzgF 1 500 I = GH 1 500  0.021 7 JK Hvv K
s
2 000 028.9 Hz
(c)
Now the heart wall is a moving source. f = f
2 000 057.8 Hz
Chapter 17
511
P17.39
Approaching ambulance: Departing ambulance: Since f = 560 Hz and f = 480 Hz
f= f =
b1  v vg
S
f
d1  b v vgi F vI F vI 560G 1  J = 480G 1 + J H vK H vK
S S S
f
1 040
vS = 80.0 v 80.0 343 m s = 26.4 m s vS = 1 040
a f
P17.40
(a)
The maximum speed of the speaker is described by 1 1 2 mv max = kA 2 2 2 v max = k A= m 20.0 N m 0.500 m = 1.00 m s 5.00 kg
a
f
The frequencies heard by the stationary observer range from fmin = f where v is the speed of sound. fmin = 440 Hz fmax
FG v IJ to f Hv+v K
max
max
=f
FG v IJ Hvv K
max
F 343 m s I = GH 343 m s + 1.00 m s JK F 343 m s I = = 440 HzG H 343 m s  1.00 m s JK
2 0
439 Hz 441 Hz
(b)
= 10 dB log
FG I IJ = 10 dB logF 4 r I GH I JK HI K
0
The maximum intensity level (of 60.0 dB) occurs at r = rmin = 1.00 m . The minimum intensity level occurs when the speaker is farthest from the listener (i.e., when r = rmax = rmin + 2 A = 2.00 m).
F I  10 dB logF I GH 4 I r JK GH 4 I r JK F 4 I r I = 10 dB logF r I . or  = 10 dB log G GH r JK JK H 4 I r This gives: 60.0 dB  = 10 dB log a 4.00f = 6.02 dB , and = 54.0 dB
Thus, max  min = 10 dB log
2 0 min 2 0 max max min 2 0 min 2 0 max 2 max 2 min
min
min
.
512 P17.41
Sound Waves
f= f
FG v IJ Hvv K
s
F 340 I GH 340  b9.80t g JK 485a340f + a 485fd9.80t i = a512 fa340 f F 512  485 IJ 340 = 1.93 s t =G H 485 K 9.80
485 = 512
fall f f
d1 =
1 2 gt f = 18.3 m : 2
t return =
18.3 = 0.053 8 s 340
The fork continues to fall while the sound returns. t total fall = t f + treturn = 1.93 s + 0.053 8 s = 1.985 s d total = P17.42 (a) v = 331 m s + 0.6 1 2 gt total fall = 19.3 m 2
b
g
m 10 C = 325 m s s C
a
f
(b)
Approaching the bell, the athlete hears a frequency of After passing the bell, she hears a lower frequency of The ratio is which gives 6 v  6 v o = 5 v + 5 v o or
FG v + v IJ H v K F v + b v g I f = f G H v JK
f= f
O O
f v  vO 5 = = f v + vO 6 vO = v 325 m s = = 29.5 m s 11 11
*P17.43
(a)
Sound moves upwind with speed 343  15 m s . Crests pass a stationary upwind point at frequency 900 Hz. Then
a
f
=
v 328 m s = = 0.364 m f 900 s 343 + 15 m s v = = 0.398 m f 900 s
(b) (c)
By similar logic,
=
a
f
The source is moving through the air at 15 m/s toward the observer. The observer is stationary relative to the air. f= f
FG v + v IJ = 900 HzFG 343 + 0 IJ = H 343  15 K H vv K
o s
941 Hz
(d)
The source is moving through the air at 15 m/s away from the downwind firefighter. Her speed relative to the air is 30 m/s toward the source. f= f
FG v + v IJ = 900 HzF 343 + 30 I = 900 HzFG 373 IJ = GH 343  a15f JK H 358 K H vv K
o s
938 Hz
Chapter 17
513
*P17.44
The halfangle of the cone of the shock wave is where
= sin 1
FG v Hv
sound source
IJ = sin FG 1 IJ = 41.8 . H 1.5 K K
1
v plane
As shown in the sketch, the angle between the direction of propagation of the shock wave and the direction of the plane's velocity is
v shock FIG. P17.44
= 90 = 9041.8 = 48.2 .
P17.45 The half angle of the shock wave cone is given by sin = v light sin v light vS .
vS =
=
2.25 10 8 m s = 2.82 10 8 m s sin 53.0
a
f
P17.46
= sin 1
v 1 = sin 1 = 46.4 vS 1.38 v 1 = ; = 19.5 vS 3.00 h h ; x= x tan
P17.47
(b)
sin = tan = x=
20 000 m = 5.66 10 4 m = 56.6 km tan 19.5 5.66 10 4 m x = = 56.3 s to travel this distance. vS 3.00 335 m s
(a)
It takes the plane t =
b
g
x
t=0 h Observer a. b. FIG. P17.47(a) h
Observer hears the boom
514
Sound Waves
Section 17.5 Section 17.6 *P17.48
Digital Sound Recording Motion Picture Sound
For a 40dB sound, 40 dB = 10 dB log I = 10 8 W m 2 =
LM NM 10 e
12
I W m2
OP QP jb g
2 Pmax 2 v
Pmax = 2 vI = 2 1.20 kg m 2 343 m s 10 8 W m 2 = 2.87 10 3 N m 2 (a) (b) (c) *P17.49 code = 2.87 10 3 N m 2 28.7 N m 2 65 536 = 7
For sounds of 40 dB or softer, too few digital words are available to represent the wave form with good fidelity. In a sound wave P is negative half of the time but this coding scheme has no words available for negative pressure variations.
If the source is to the left at angle from the direction you are facing, the sound must travel an extra distance d sin to reach your right ear as shown, where d is the distance between your ears. The d sin . Then delay time is t in v = t
= sin 1
343 m s 210 10 v t = sin 1 d 0.19 m
b
g
ear ear
6
s
= 22.3 left of center .
FIG. P17.49
*P17.50
103 dB = 10 dB log (a)
LM MN 10
I
12
W m
2
OP PQ
4 r 2 = 4 1.6 m
I = 2.00 10 2 W m 2 = = 0.642 W
a
f
2
(b)
efficiency =
sound output power 0.642 W = = 0.004 28 total input power 150 W
Additional Problems P17.51 Model your loud, sharp sound impulse as a single narrow peak in a graph of air pressure versus time. It is a noise with no pitch, no frequency, wavelength, or period. It radiates away from you in all directions and some of it is incident on each one of the solid vertical risers of the bleachers. Suppose that, at the ambient temperature, sound moves at 340 m/s; and suppose that the horizontal width of each row of seats is 60 cm. Then there is a time delay of
b340 m sg = 0.002 s
continued on next page
0.6 m
Chapter 17
515
between your sound impulse reaching each riser and the next. Whatever its material, each will reflect much of the sound that reaches it. The reflected wave sounds very different from the sharp pop you made. If there are twenty rows of seats, you hear from the bleachers a tone with twenty crests, each separated from the next in time by
f = 0.004 s . b340 m sg
2 0.6 m
a
This is the extra time for it to cross the width of one seat twice, once as an incident pulse and once again after its reflection. Thus, you hear a sound of definite pitch, with period about 0.004 s, frequency 1 ~ 300 Hz 0.003 5 s wavelength
=
and duration
340 m s v = = 1.2 m ~ 10 0 m f 300 s
b
b
g
g
20 0.004 s ~ 10 1 s .
P17.52 (a)
a
f
=
v 343 m s = = 0.232 m f 1 480 s 1
(b)
= 81.0 dB = 10 dB log
I = 10 12 W m 2 10 8.10 = 10 3.90 W m 2 = 1.26 10 4 W m 2 = smax = 2I = v 2 2 1.26 10 4 W m 2
3 2
e
j
LM NM 10
I
12
W m
2
OP QP j
1 2
1 2 v 2 smax 2
e1.20 kg m jb343 m sg4 e1 480 s j
=  = 13.8 mm
e
= 8.41 10 8 m
(c)
=
v 343 m s = = 0.246 m f 1 397 s 1
P17.53
Since cos 2 + sin 2 = 1 ,
sin = 1  cos 2 (each sign applying half the time)
P = Pmax sin kx  t = v smax 1  cos 2 kx  t
a
f
Therefore P17.54
2 2 2 P = v smax  smax cos 2 kx  t = v smax  s 2
a
f
a
f
The trucks form a train analogous to a wave train of crests with speed v = 19.7 m s 2 and unshifted frequency f = = 0.667 min 1 . 3.00 min (a) The cyclist as observer measures a lower Dopplershifted frequency: 19.7 + 4. 47 v + vo = 0.667 min 1 = 0.515 min f= f v 19.7
(b)
FG IJ e H K F v + v IJ = e0.667 min f = f G H v K
o
1
jFGH a f IJK jFGH 19.7 + a.1.56f IJK = 19 7
0.614 min
The cyclist's speed has decreased very significantly, but there is only a modest increase in the frequency of trucks passing him.
516 P17.55 P17.56
Sound Waves
v= (a)
2d vt 1 :d= = 6.50 10 3 m s 1.85 s = 6.01 km t 2 2
e
ja
f
The speed of a compression wave in a bar is v= Y
=
20.0 10 10 N m 2 7 860 kg m
3
= 5.04 10 3 m s .
(b)
The signal to stop passes between layers of atoms as a sound wave, reaching the back end of the bar in time t= 0.800 m L = = 1.59 10 4 s . v 5.04 10 3 m s
(c)
As described by Newton's first law, the rearmost layer of steel has continued to move forward with its original speed vi for this time, compressing the bar by L = vi t = 12.0 m s 1.59 10 4 s = 1.90 10 3 m = 1.90 mm .
b
ge
j
(d) (e)
The strain in the rod is: The stress in the rod is:
L 1.90 10 3 m = = 2.38 10 3 . L 0.800 m
=Y
FG L IJ = e20.0 10 HLK
10
N m 2 2.38 10 3 = 476 MPa .
je
j
Since > 400 MPa , the rod will be permanently distorted. (f) We go through the same steps as in parts (a) through (e), but use algebraic expressions rather than numbers: The speed of sound in the rod is v = Y
.
The back end of the rod continues to move forward at speed vi for a time of t = traveling distance L = vi t after the front end hits the wall. The strain in the rod is: The stress is then: = Y L vi t = = vi . L L Y
L =L , v Y
FG L IJ = Yv HLK
i
Y
= vi Y .
For this to be less than the yield stress, y , it is necessary that
vi Y < y or vi <
y Y
.
With the given numbers, this speed is 10.1 m/s. The fact that the length of the rod divides out means that the steel will start to bend right away at the front end of the rod. There it will yield enough so that eventually the remainder of the rod will experience only stress within the elastic range. You can see this effect when sledgehammer blows give a mushroom top to a rod used as a tent stake.
Chapter 17
517
P17.57
(a)
f= f so 1 
bv  v g
diver
v
v diver f = v f
v diver = v 1 
FG H
f f
IJ K F GH I JK
with v = 343 m s , f = 1 800 Hz and f = 2 150 Hz we find v diver = 343 1  (b) 1 800 = 55.8 m s . 2 150
If the waves are reflected, and the skydiver is moving into them, we have f = f
bv + v g f = f LM v OP bv + v g v NM bv  v g QP v
diver diver diver
so f = 1 800 fv vu
a343 + 55.8f = a343  55.8f
f =
2 500 Hz . fv v  u
2
P17.58
(a)
f=
(b)
130 km h = 36.1 m s
FG 1  1 IJ H v  u v + uK 2bu v g fva v + u  v + uf 2uvf f = = = f v u 1  eu v j v e1  eu v jj 2a36.1fa 400f f = = 340 1  a36.1f 340 a f
f  f = fv
2 2 2 2 2 2 2 2
85.9 Hz
P17.59
When observer is moving in front of and in the same direction as the source, f = f
v  vO where vO v  vS and vS are measured relative to the medium in which the sound is propagated. In this case the ocean current is opposite the direction of travel of the ships and vO = 45.0 km h  10.0 km h = 55.0 km h = 15.3 m s , and vS
b g = 64.0 km h  b 10.0 km hg = 74.0 km h = 20.55 m s
520 m 15 3 g 11520 m ss20..55 m ss = m 1 204.2 Hz .
Therefore, f = 1 200.0 Hz
b
518 P17.60
Sound Waves
Use the Doppler formula, and remember that the bat is a moving source. If the velocity of the insect is v x , 40.4 = 40.0 Solving, v x = 3.31 m s . Therefore, the bat is gaining on its prey at 1.69 m s .
a340 + 5.00fb340  v g . a340  5.00fb340 + v g
x x
P17.61
sin =
h = v 12.8 s
a
v 1 = vS N M
shock front x vs
x = vS 10.0 s
a
f
f
h shock front
h v 1.28 tan = = 1.28 = x vS N M sin 1 cos = = tan 1.28 = 38.6 NM = P17.62 (a) 1 = 1.60 sin
FIG. P17.61
FIG. P17.62(a) (b)
=
v 343 m s = = 0.343 m f 1 000 s 1 343  40.0 m s v v v  vS = = = 0.303 m f f v 1 000 s 1
S
(c)
(d)
(e)
f IJ a K v v F v + v I a343 + 40.0f m s = = G J = 1 000 s = 0.383 m f f H v K F v  v IJ = b1 000 Hzg a343  30.0f m s = 1.03 kHz f = fG a343  40.0f m s H vv K
=
FG H
1
O S
Chapter 17
519
P17.63
t = L
FG 1 Hv

air
v  v air 1 = L cu v cu v air v cu
IJ K
331 m s 3.56 10 3 m s v air v cu t = 6.40 10 3 s L= v cu  v air 3 560  331 m s
b
b
ge
g
je
j
P
L = 2.34 m P17.64 The shock wavefront connects all observers first hearing the plane, including our observer O and the plane P, so here it is vertical. The angle that the shock wavefront makes with the direction of the plane's line of travel is given by sin = so = 9.97 . Using the right triangle CPO, the angle is seen to be 340 m s v = = 0.173 vS 1 963 m s
C
O
FIG. P17.64
= 90.0 = 90.09.97 = 80.0 .
P17.65 (a)
(b)
F v I = sin F 331 I = GH 20.0 10 JK GH v JK F 1 533 IJ = 4.40 = sin G H 20.0 10 K
= sin 1
sound obj 1 3 1 3
0.948
P17.66
2 =
1 1 1  2 = 10 log 1 20.0 2 80.0  2 = 10 log 20.0 = +13.0
2 = 67.0 dB
P17.67 For the longitudinal wave v L = For the transverse wave vT If we require
vL Y m = 8.00 , we have T = and where = L 64.0 vT
FG Y IJ H K FTI =G J . H K
12
12
.
=
mass m = . volume r 2 L
3 10 2 r 2 Y 2.00 10 m 6.80 10 N m = = 1.34 10 4 N . This gives T = 64.0 64.0
e
je
2
j
520 P17.68
Sound Waves
The total output sound energy is eE =t , where is the power radiated. eE eE eE eE = = = Thus, t = . 2 IA 4 d 2 I 4 r I But, = 10 log
FG I IJ . Therefore, I = I e10 j and t = HI K
0 0
e
j
10
eE 4 d 2 I 0 10
10
.
P17.69
(a)
If the source and the observer are moving away from each other, we have: S  0 = 180 , and since cos180 = 1 , we get Equation 17.12 with negative values for both vO and vS . If vO = 0 m s then f = v f v  vS cos S Also, when the train is 40.0 m from the intersection, and the car is 30.0 m from the intersection, cos S = so f = 343 m s 500 Hz 343 m s  0.800 25.0 m s 4 5
(b)
b
ga
f
or f = 531 Hz . Note that as the train approaches, passes, and departs from the intersection, S varies from 0 to 180 and the frequency heard by the observer varies from: fmax = fmin = P17.70 343 m s v f= 500 Hz = 539 Hz v  vS cos 0 343 m s  25.0 m s 343 m s v f= 500 Hz = 466 Hz v  vS cos 180 343 m s + 25.0 m s
a
f
a
f
Let T represent the period of the source vibration, and E be the energy put into each wavefront. E Then av = . When the observer is at distance r in front of the source, he is receiving a spherical T wavefront of radius vt, where t is the time since this energy was radiated, given by vt  vS t = r . Then, t= r . v  vS
The area of the sphere is 4 vt is uniform with the value
a f
2
=
E av S = . A 4 v 2 r 2 The observer receives parcels of energy with the Doppler shifted frequency v v f= f = , so the observer receives a wave with intensity v  vS T v  vS
bv  v g Tbv  v g
S
4 v 2 r 2
2
. The energy per unit area over the spherical wavefront
2
FG H
IJ K b
g
I=
FG E IJ f = FG T bv  v g H A K GH 4 v r
av S 2 2
2
IF v I JJK GH Tbv  v g JK =
S
av 4 r
2
FG v  v IJ H v K
S
.
Chapter 17
521
P17.71
(a)
The time required for a sound pulse to travel distance L at L L . Using this expression speed v is given by t = = v Y we find t1 = t2 = or L1 Y1 1 = L1
10
L1 L3
L2
FIG. P17.71
e7.00 10
=
N m
10
2
j e2 700 kg m j
3
= 1.96 10 4 L1 s
e
j
1.50 m  L1 Y2 2
e1.60 10
1.50 m
1.50 m  L1 N m2
j e11.3 10
3
kg m3
j
t 2 = 1.26 10 3  8.40 10 4 L1 s t3 =
e
j
e11.0 10
10
N m3
j e8 800 kg m j
3
t 3 = 4.24 10 4 s We require t1 + t 2 = t 3 , or 1.96 10 4 L1 + 1.26 10 3  8.40 10 4 L1 = 4.24 10 4 . This gives L1 = 1.30 m and L 2 = 1.50  1.30 = 0.201 m . The ratio of lengths is then (b) The ratio of lengths L1 = 6.45 . L2
L1 is adjusted in part (a) so that t1 + t 2 = t 3 . Sound travels the two paths L2
in equal time and the phase difference, = 0 . P17.72 To find the separation of adjacent molecules, use a model where each molecule occupies a sphere of radius r given by
air =
average mass per molecule
4 3 26
r3
13
or 1.20 kg m =
3
4.82 10 26 kg
4 3
r3
L 3e4.82 10 kg j OP ,r=M MN 4 e1.20 kg m j PQ
3
= 2.12 10 9 m .
Intermolecular separation is 2r = 4. 25 10 9 m, so the highest possible frequency sound wave is fmax = 343 m s v v = = = 8.03 10 10 Hz ~ 10 11 Hz . min 2r 4.25 10 9 m
522
Sound Waves
ANSWERS TO EVEN PROBLEMS
P17.2 P17.4 P17.6 P17.8 P17.10 P17.12 1.43 km s (a) 27.2 s; (b) longer than 25.7 s, because the air is cooler (a) 153 m s ; (b) 614 m (a) 4.16 m; (b) 0.455 s ; (c) 0.157 mm 1.55 10 10 m (a) 1.27 Pa; (b) 170 Hz; (c) 2.00 m; (d) 340 m s s = 22.5 nm cos 62.8 x  2.16 10 t (a) 4.63 mm; (b) 14.5 m s ; (c) 4.73 10 9 W m 2 (a) 5.00 10 17 W ; (b) 5.00 10 5 W (a) 1.00 10 (a) I 2
5
P17.36 P17.38
no (a) 2.17 cm s ; (b) 2 000 028.9 Hz ; (c) 2 000 057.8 Hz (a) 441 Hz; 439 Hz; (b) 54.0 dB (a) 325 m s; (b) 29.5 m s 48. 2 46. 4 (a) 7; (b) and (c) see the solution (a) 0.642 W ; (b) 0.004 28 = 0.428% (a) 0.232 m; (b) 84.1 nm; (c) 13.8 mm (a) 0.515 min ; (b) 0.614 min (a) 5.04 km s ; (b) 159 s ; (c) 1.90 mm; (d) 0.002 38 ; (e) 476 MPa ; (f) see the solution (a) see the solution; (b) 85.9 Hz The gap between bat and insect is closing at 1.69 m s . (a) see the solution; (b) 0.343 m; (c) 0.303 m; (d) 0.383 m; (e) 1.03 kHz 80.0 67.0 dB t = eE 4 d I 0 10
2 10
P17.40 P17.42 P17.44 P17.46 P17.48 P17.50 P17.52 P17.54 P17.56
P17.14 P17.16
e
4
j
P17.18 P17.20 P17.22 P17.24 P17.26
W m ; (b) 90.7 mPa I 1 ; (b) I 2 = I 1
2
F f I =G J HfK
2
P17.58 P17.60
21.2 W (a) 4.51 times larger in water than in air and 18.0 times larger in iron; (b) 5.60 times larger in water than in iron and 331 times larger in air; (c) 59.1 times larger in water than in air and 331 times larger in iron; (d) 0.331 m; 1.49 m; 5.95 m; 10.9 nm; 184 pm; 32.9 pm; 29.2 mPa; 1.73 Pa; 9.67 Pa see the solution 10.0 m; 100 m 86.6 m (a) 1.76 kJ ; (b) 108 dB
P17.62 P17.64 P17.66 P17.68 P17.70 P17.72
P17.28 P17.30 P17.32 P17.34
see the solution ~ 10 11 Hz
18
Superposition and Standing Waves
CHAPTER OUTLINE
18.1 18.2 18.3 18.4 18.5 18.6 18.7 18.8 Superposition and Interference Standing Waves Standing Waves in a String Fixed at Both Ends Resonance Standing Waves in Air Columns Standing Waves in Rod and Plates Beats: Interference in Time NonSinusoidal Wave Patterns
ANSWERS TO QUESTIONS
Q18.1 No. Waves with other waveforms are also trains of disturbance that add together when waves from different sources move through the same medium at the same time. The energy has not disappeared, but is still carried by the wave pulses. Each particle of the string still has kinetic energy. This is similar to the motion of a simple pendulum. The pendulum does not stop at its equilibrium position during oscillationlikewise the particles of the string do not stop at the equilibrium position of the string when these two waves superimpose. No. A wave is not a solid object, but a chain of disturbance. As described by the principle of superposition, the waves move through each other.
Q18.2
Q18.3
Q18.4
They can, wherever the two waves are nearly enough in phase that their displacements will add to create a total displacement greater than the amplitude of either of the two original waves. When two onedimensional sinusoidal waves of the same amplitude interfere, this condition is satisfied whenever the absolute value of the phase difference between the two waves is less than 120. When the two tubes together are not an efficient transmitter of sound from source to receiver, they are an efficient reflector. The incoming sound is reflected back to the source. The waves reflected by the two tubes separately at the junction interfere constructively. No. The total energy of the pair of waves remains the same. Energy missing from zones of destructive interference appears in zones of constructive interference. Each of these standing wave patterns is made of two superimposed waves of identical frequencies traveling, and hence transferring energy, in opposite directions. Since the energy transfer of the waves are equal, then no net transfer of energy occurs. Damping, and nonlinear effects in the vibration turn the energy of vibration into internal energy. The air in the shower stall can vibrate in standing wave patterns to intensify those frequencies in your voice which correspond to its free vibrations. The hard walls of the bathroom reflect sound very well to make your voice more intense at all frequencies, giving the room a longer reverberation time. The reverberant sound may help you to stay on key. 523
Q18.5
Q18.6 Q18.7
Q18.8 Q18.9
524 Q18.10 Q18.11
Superposition and Standing Waves
The trombone slide and trumpet valves change the length of the air column inside the instrument, to change its resonant frequencies. The vibration of the air must have zero amplitude at the closed end. For air in a pipe closed at one end, the diagrams show how resonance vibrations have NA distances that are odd integer submultiples of the NA distance in the fundamental vibration. If the pipe is open, resonance vibrations have NA distances that are all integer submultiples of the NA distance in the fundamental.
FIG. Q18.11 Q18.12 What is needed is a tuning forkor other puretone generatorof the desired frequency. Strike the tuning fork and pluck the corresponding string on the piano at the same time. If they are precisely in tune, you will hear a single pitch with no amplitude modulation. If the two pitches are a bit off, you will hear beats. As they vibrate, retune the piano string until the beat frequency goes to zero. Air blowing fast by a rim of the pipe creates a "shshshsh" sound called edgetone noise, a mixture of all frequencies, as the air turbulently switches between flowing on one side of the edge and the other. The air column inside the pipe finds one or more of its resonance frequencies in the noise. The air column starts vibrating with large amplitude in a standing wave vibration mode. It radiates sound into the surrounding air (and also locks the flapping airstream at the edge to its own frequency, making the noise disappear after just a few cycles). A typical standingwave vibration possibility for a bell is similar to that for the glass shown in Figure 18.17. Here six nodetonode distances fit around the circumference of the rim. The circumference is equal to three times the wavelength of the transverse wave of inandout bending of the material. In other states the circumference is two, four, five, or higher integers times the wavelengths of the higherfrequency vibrations. (The circumference being equal to the wavelength would describe the bell moving from side to side without bending, which it can do without producing any sound.) A tuned bell is cast and shaped so that some of these vibrations will have their frequencies constitute higher harmonics of a musical note, the strike tone. This tuning is lost if a crack develops in the bell. The sides of the crack vibrate as antinodes. Energy of vibration may be rapidly converted into internal energy at the end of the crack, so the bell may not ring for so long a time. The bow string is pulled away from equilibrium and released, similar to the way that a guitar string is pulled and released when it is plucked. Thus, standing waves will be excited in the bow string. If the arrow leaves from the exact center of the string, then a series of odd harmonics will be excited. Even harmonies will not be excited because they have a node at the point where the string exhibits its maximum displacement.
Q18.13
Q18.14
Q18.15
Chapter 18
525
Q18.16
Walking makes the person's hand vibrate a little. If the frequency of this motion is equal to the natural frequency of coffee sloshing from side to side in the cup, then a largeamplitude vibration of the coffee will build up in resonance. To get off resonance and back to the normal case of a smallamplitude disturbance producing a smallamplitude result, the person can walk faster, walk slower, or get a larger or smaller cup. Alternatively, even at resonance he can reduce the amplitude by adding damping, as by stirring highfiber quickcooking oatmeal into the hot coffee. Beats. The propellers are rotating at slightly different frequencies. Instead of just radiating sound very softly into the surrounding air, the tuning fork makes the chalkboard vibrate. With its large area this stiff sounding board radiates sound into the air with higher power. So it drains away the fork's energy of vibration faster and the fork stops vibrating sooner. This process exemplifies conservation of energy, as the energy of vibration of the fork is transferred through the blackboard into energy of vibration of the air. The difference between static and kinetic friction makes your finger alternately slip and stick as it slides over the glass. Your finger produces a noisy vibration, a mixture of different frequencies, like new sneakers on a gymnasium floor. The glass finds one of its resonance frequencies in the noise. The thin stiff wall of the cup starts vibrating with large amplitude in a standing wave vibration mode. A typical possibility is shown in Figure 18.17. It radiates sound into the surrounding air, and also can lock your squeaking finger to its own frequency, making the noise disappear after just a few cycles. Get a lot of different thinwalled glasses of fine crystal and try them out. Each will generally produce a different note. You can tune them by adding wine. Helium is less dense than air. It carries sound at higher speed. Each cavity in your vocal apparatus has a standingwave resonance frequency, and each of these frequencies is shifted to a higher value. Your vocal chords can vibrate at the same fundamental frequency, but your vocal tract amplifies by resonance a different set of higher frequencies. Then your voice has a different quacky quality. Warning: Inhaling any pressurized gas can cause a gas embolism which can lead to stroke or death, regardless of your age or health status. If you plan to try this demonstration in class, inhale your helium from a balloon, not directly from a pressurized tank. Stick a bit of chewing gum to one tine of the second fork. If the beat frequency is then faster than 4 beats per second, the second has a lower frequency than the standard fork. If the beats have slowed down, the second fork has a higher frequency than the standard. Remove the gum, clean the fork, add or subtract 4 Hz according to what you found, and your answer will be the frequency of the second fork.
Q18.17 Q18.18
Q18.19
Q18.20
Q18.21
SOLUTIONS TO PROBLEMS
Section 18.1 P18.1 Superposition and Interference
y = y1 + y 2 = 3.00 cos 4.00 x  1.60t + 4.00 sin 5.0 x  2.00t evaluated at the given x values.
(a) (b) (c) x = 1.00 , t = 1.00 x = 1.00 , t = 0.500 x = 0.500 , t = 0 y = 3.00 cos 2.40 rad + 4.00 sin +3.00 rad = 1.65 cm y = 3.00 cos +3.20 rad + 4.00 sin +4.00 rad = 6.02 cm y = 3.00 cos +2.00 rad + 4.00 sin +2.50 rad = 1.15 cm
a
f
a
f
a a a
f
a
f
f
a
f
f
a
f
526 P18.2
Superposition and Standing Waves
FIG. P18.2 P18.3 (a)
y1 = f x  vt , so wave 1 travels in the +x direction y2
x direction = +5
2
a f = f a x + vtf , so wave 2 travels in the
5
2 2
(b)
To cancel, y1 + y 2 = 0 :
a 3 x  4t f + 2 a 3 x + 4t  6 f a3 x  4tf = a3x + 4t  6f 3 x  4t = a3 x + 4t  6f
8t = 6 6x = 6
2
+2
for the positive root,
t = 0.750 s
(at t = 0.750 s , the waves cancel everywhere) (c) for the negative root, x = 1.00 m
(at x = 1.00 m , the waves cancel always) P18.4 Suppose the waves are sinusoidal. The sum is
a4.00 cmf sinakx  tf + a4.00 cmf sinakx  t + 90.0f 2a 4.00 cmf sina kx  t + 45.0f cos 45.0 So the amplitude is a8.00 cmf cos 45.0 = 5.66 cm .
The resultant wave function has the form y = 2 A0 cos
P18.5
(a)
FG IJ sinFG kx  t + IJ H 2K H 2K L  4 OP = FI A = 2 A cosG J = 2a5.00f cos M H 2K N 2 Q
0
9.24 m
(b)
f=
1 200 = = 600 Hz 2 2
Chapter 18
P18.6
1 = cos 1 = 60.0 = 2 2 3 2 Thus, the phase difference is = 120 = 3 T 1 = = This phase difference results if the time delay is 3 3 f 3v 3.00 m = 0.500 s Time delay = 3 2.00 m s 2 A 0 cos
0
FG IJ = A H 2K
so
FG IJ H K
527
b
g
P18.7
(a)
If the end is fixed, there is inversion of the pulse upon reflection. Thus, when they meet, they cancel and the amplitude is zero . If the end is free, there is no inversion on reflection. When they meet, the amplitude is 2 A = 2 0.150 m = 0.300 m .
(b)
a
f
P18.8
(a)
x = 9.00 + 4.00  3.00 = 13  3.00 = 0.606 m The wavelength is Thus, or
=
v 343 m s = = 1.14 m f 300 Hz x 0.606 = = 0.530 of a wave , 1.14 = 2 0.530 = 3.33 rad
a
f
(b)
For destructive interference, we want where x is a constant in this set up.
x
x v v 343 = = 283 Hz f= 2 x 2 0.606
= 0.500 = f
a
f
P18.9
We suppose the man's ears are at the same level as the lower speaker. Sound from the upper speaker is delayed by traveling the extra distance L2 + d 2  L . 2n  1 with n = 1, 2 , 3 , ... He hears a minimum when this is 2 n1 2 v L2 + d 2  L = Then, f n1 2 v L2 + d 2 = +L f
a
f
b
g
b
g
L +d L=
2
2
bn  1 2g v =
2
2
d  n 1 2
2
b gv 2bn  1 2gv f
2
f
2
+ L2 + f
2
2 n  1 2 vL f n = 1, 2 , 3 , ...
b
g
2
This will give us the answer to (b). The path difference starts from nearly zero when the man is very far away and increases to d when L = 0. The number of minima he hears is the greatest integer n1 2 v solution to d f df 1 n = greatest integer + . v 2
b
g
continued on next page
528
Superposition and Standing Waves
(a)
4.00 m 200 s 1 df 1 + = + = 2.92 330 m s 2 v 2 He hears two minima.
a
fb
g
(b)
With n = 1, L= d2  1 2
b gv 2b1 2g v f
2
2
f2
=
a4.00 mf  b330 m sg 4b200 sg b330 m sg 200 s
2 2
2
L = 9.28 m with n = 2 L= P18.10 d2  3 2
b gv 2b3 2 g v f
2
2
f2
= 1.99 m .
Suppose the man's ears are at the same level as the lower speaker. Sound from the upper speaker is delayed by traveling the extra distance r = L2 + d 2  L . He hears a minimum when r = 2n  1 Then, L2 + d 2  L = n 
2 2
a
IJ FG v IJ KH f K F 1IF vI L + d = Gn  J G J + L H 2KH f K F 1I F vI F 1IF vI L + d = G n  J G J + 2G n  J G J L + L H 2K H f K H 2KH f K F 1I F vI F 1IF vI d  G n  J G J = 2G n  J G J L H 2K H f K H 2KH f K FG H
1 2
2 2 2 2 2 2 2
fFGH IJK with n = 1, 2, 3, ... 2
2
(1)
Equation 1 gives the distances from the lower speaker at which the man will hear a minimum. The path difference r starts from nearly zero when the man is very far away and increases to d when L = 0. (a) The number of minima he hears is the greatest integer value for which L 0 . This is the 1 v same as the greatest integer solution to d n  , or 2 f
FG H
IJ FG IJ KH K
number of minima heard = n max = greatest integer d (b)
FG f IJ + 1 H vK 2
.
From equation 1, the distances at which minima occur are given by Ln = d2  n  1 2
b g bv f g 2bn  1 2gb v f g
2
2
where n = 1, 2 , ... , n max .
P18.11
(a)
1 = 20.0 rad cm 5.00 cm  32.0 rad s 2.00 s = 36.0 rad 1
= 9.00 radians = 516 = 156
f b b ga ga f = b 25.0 rad cmga5.00 cmf  b 40.0 rad sga 2.00 sf = 45.0 rad
Chapter 18
529
(b)
= 20.0 x  32.0t  25.0 x  40.0t = 5.00 x + 8.00t At t = 2.00 s , the requirement is = 5.00 x + 8.00 2.00 = 2n + 1 for any integer n. For x < 3.20 , 5.00 x + 16.0 is positive, so we have 5.00 x + 16.0 = 2n + 1 , or x = 3.20 
a f a
f
a2n + 1f
5.00
a
f
The smallest positive value of x occurs for n = 2 and is x = 3.20 
a4 + 1f = 3.20  =
5.00
0.058 4 cm .
P18.12
(a)
First we calculate the wavelength: Then we note that the path difference equals
=
v 344 m s = = 16.0 m f 21.5 Hz 1 2
9.00 m  1.00 m =
Therefore, the receiver will record a minimum in sound intensity. (b) We choose the origin at the midpoint between the speakers. If the receiver is located at point (x, y), then we must solve:
Then, Square both sides and simplify to get: Upon squaring again, this reduces to: Substituting = 16.0 m , and reducing, or
ax + 5.00f ax + 5.00f
20.0 x 
2
+ y2  + y2 =
2
ax  5.00f ax  5.00f
2
2
+ y2 =
2
1 2 1 + y2 + 2
2 = 4
ax  5.00f
+ y2
400 x 2  10.0 2 x +
4 = 2 x  5.00 16.0
a
f
2
+ 2 y 2
9.00 x 2  16.0 y 2 = 144
y2 x2  =1 16.0 9.00
(When plotted this yields a curve called a hyperbola.)
530
Superposition and Standing Waves
Section 18.2 P18.13
Standing Waves
y = 1.50 m sin 0.400 x cos 200t = 2 A0 sin kx cos t
Therefore, k = 2
a
f a
f a f
= 0.400 rad m
=
f=
2 = 15.7 m 0.400 rad m
and = 2 f so
2
=
200 rad s = 31.8 Hz 2 rad
The speed of waves in the medium is v = f =
200 rad s = 500 m s 2 f = = k 0.400 rad m 2
P18.14
y = 0.030 0 m cos (a)
FG x IJ cosa40tf H 2K
x = 2n + 1 2 2
nodes occur where y = 0 :
a
f
so x = (b) P18.15 y max
a2n + 1f = , 3 , 5 , ... . F 0.400 IJ = 0.029 4 m = 0.030 0 m cosG H 2 K
343 m s v = = = 0. 214 m 2 2 f 2 800 s 1
The facing speakers produce a standing wave in the space between them, with the spacing between nodes being d NN =
e
j
If the speakers vibrate in phase, the point halfway between them is an antinode of pressure at a distance from either speaker of 1.25 m = 0.625 . 2 Then there isa node at a node at a node at a node at a node at and a node at 0.625  0. 214 = 0.518 m 2
0.518 m  0.214 m = 0.303 m 0.303 m  0.214 m = 0.089 1 m 0.518 m + 0.214 m = 0.732 m 0.732 m + 0.214 m = 0.947 m 0.947 m + 0.214 m = 1.16 m from either speaker.
Chapter 18
531
P18.16
y = 2 A 0 sin kx cos t 2y x 2 = 2 A0 k 2 sin kx cos t 2 y t 2 = 2 A0 2 sin kx cos t
Substitution into the wave equation gives This is satisfied, provided that P18.17
2 A 0 k 2 sin kx cos t = v=
k
FG 1 IJ e2 A Hv K
2 0
2
sin kx cos t
j
f a f y = y + y = 3.00 sinb x g cosb0.600 t g + 3.00 sinb x g cosb0.600 t g cm y = a6.00 cmf sinb x g cosb0.600 t g (a) We can take cosb0.600 t g = 1 to get the maximum y. y = a6.00 cmf sina0.250 f = At x = 0.250 cm,
y1 = 3.00 sin x + 0.600t cm; y 2 = 3.00 sin x  0.600t cm
1 2 max
a
4.24 cm
(b) (c)
At x = 0.500 cm ,
y max = 6.00 cm sin 0.500 = 6.00 cm
a a
f a f a
f
Now take cos 0.600 t = 1 to get y max : At x = 1.50 cm,
b
g
y max = 6.00 cm sin 1.50 1 = 6.00 cm x= n n = 1, 3 , 5 , ... 4
fa f
(d)
The antinodes occur when But k = and 2
b
g
= , so
= 2.00 cm
= 0.500 cm as in (b) 4 3 = 1.50 cm as in (c) x2 = 4 5 = 2.50 cm x3 = 4
x1 = y = 2 A sin kx + kx +
P18.18
(a)
The resultant wave is The nodes are located at
FG H
cos t  2 2
IJ FG K H
IJ K
= n 2 n  so x= k 2k which means that each node is shifted to the left. 2k
(b) The separation of nodes is x = n + 1
LMa f  OP  LM n  OP N k 2k Q N k 2k Q
x =
= k 2
The nodes are still separated by half a wavelength.
532
Superposition and Standing Waves
Section 18.3 P18.19
Standing Waves in a String Fixed at Both Ends v 2L
L = 30.0 m ; = 9.00 10 3 kg m; T = 20.0 N ; f1 = where so
FTI v=G J H K
f1 =
12
= 47.1 m s f 2 = 2 f1 = 1.57 Hz f 4 = 4 f1 = 3.14 Hz T = 4 kg 9.8 m s 2 = 39.2 N
47.1 = 0.786 Hz 60.0
f3 = 3 f1 = 2.36 Hz *P18.20 The tension in the string is Its linear density is and the wave speed on the string is In its fundamental mode of vibration, we have Thus, P18.21 (a)
b ge
T =
j
=
v=
m 8 10 3 kg = = 1.6 10 3 kg m L 5m 39.2 N 1.6 10 3 kg m
= 156.5 m s
= 2L = 2 5 m = 10 m
f= v
a f
=
156.5 m s = 15.7 Hz 10 m
Let n be the number of nodes in the standing wave resulting from the 25.0kg mass. Then n + 1 is the number of nodes for the standing wave resulting from the 16.0kg mass. For 2L v , and the frequency is f = . standing waves, = n Thus, and also f= f= n Tn 2L n + 1 Tn +1 2L
Thus, Therefore, Then, (b)
Tn n +1 = = n Tn +1
b25.0 kg gg = 5 b16.0 kggg 4
b25.0 kg ge9.80 m s j =
2
4n + 4 = 5n , or n = 4 f= 4 2 2.00 m
a
f
0.002 00 kg m
350 Hz
The largest mass will correspond to a standing wave of 1 loop
an = 1f so
yielding
1 350 Hz = 2 2.00 m
a
f
m 9.80 m s 2
e
j
0.002 00 kg m
m = 400 kg
Chapter 18
533
*P18.22
The first string has linear density
1 =
The second, 2 =
1.56 10 3 kg = 2.37 10 3 kg m. 0.658 m
6.75 10 3 kg = 7.11 10 3 kg m. 0.950 m
The tension in both is T = 6.93 kg 9.8 m s 2 = 67.9 N . The speed of waves in the first string is v1 = T 67.9 N T = = 169 m s 1 2.37 10 3 kg m
and in the second v 2 = n 1 v1 n 2 v 2 = 2 L1 2L 2
2
= 97.8 m s . The two strings vibrate at the same frequency, according to
n1 169 m s n 2 97.8 m s = 2 0.658 m 2 0.950 m
a
f a
f
n2 5 = 2.50 = . Thus n1 = 2 and n 2 = 5 are the number of antinodes on each string in the lowest 2 n1 resonance with a node at the junction. (b) The first string has 2 + 1 = 3 nodes and the second string 5 more nodes, for a total of 8, or 6 other than the vibrator and pulley. (a) *P18.23 The frequency is 2 169 m s
junction FIG. P18.22(b)
b g= 2a0.658 mf af
257 Hz .
For the Estring on a guitar vibrating as a whole, v = f = 330 Hz 2 64.0 cm . When it is stopped at 64.0 cm the first fret we have 12 2 330 Hz 2 L F = v = 330 Hz 2 64.0 cm . So L F = 12 . Similarly for the 2 64.0 cm . The spacing between the first second fret, 2 2 12 330 Hz 2 L F# = v = 330 Hz 2 64.0 cm . L F# = 2 2 12 and second frets is
af
af
af
af
64.0 cm
FG 1 H2
1 12

1 2
2 12
IJ = 64.0 cmFG 1  1 IJ = 3.39 cm. K H 1.059 5 1.059 5 K
2
This is a more precise version of the answer to the example in the text. Now the eighteenth fret is distant from the bridge by L18 = much string vibrate: L19 = 64.0 cm . And the nineteenth lets this 2 18 12
64.0 cm . The distance between them is 2 19 12
64.0 cm
FG 1 H2
18 12

1 2
19 12
IJ = 64.0 cm 1 FG 1  1 IJ = K K 2 H 2
1.5 1 12
1. 27 cm .
534 *P18.24
Superposition and Standing Waves
For the whole string vibrating, d NN = 0.64 m =
1 speed of a pulse on the string is v = f = 330 1.28 m = 422 m s . s 2 (a) When the string is stopped at the fret, d NN = 0.64 m = ; 3 2 = 0.853 m v 422 m s f= = = 495 Hz . 0.853 m (b) The light touch at a point one third of the way along the string damps out vibration in the two lowest vibration states of the string as a whole. The whole string vibrates in its third resonance possibility: 3d NN = 0.64 m = 3
; = 1.28 m . The 2
FIG. P18.24(a)
= 0.427 m
f= v T , where v = 2L v
; 2
FIG. P18.24(b)
=
422 m s = 990 Hz . 0.427 m
P18.25
f1 = (a) (b) (c)
FG IJ H K
12
If L is doubled, then f1 L1 will be reduced by a factor
1 . 2 1 2 2. .
If is doubled, then f1 1 2 will be reduced by a factor If T is doubled, then f1 T will increase by a factor of
P18.26
L = 60.0 cm = 0.600 m; T = 50.0 N ; = 0.100 g cm = 0.010 0 kg m fn = where
12
nv 2L
v= fn Largest n = 339 f = 19.976 kHz . P18.27 d NN = 0.700 m
FG T IJ = 70.7 m s H K F 70.7 IJ = 58.9n = 20 000 Hz = nG H 1.20 K
= 1.40 m
f = v = 308 m s =
e1.20 10 j a0.700f
3
T
(a) (b)
T = 163 N f3 = 660 Hz
FIG. P18.27
Chapter 18
535
P18.28
G = 2 0.350 m =
LG  L A = LG
a
f fv ; = 2L = fv Ff I F fI F 392 IJ = 0.038 2 m  G J L = L G 1  J = a0.350 mfG 1  H 440 K Hf K H fK
G A A A G A G G G A
Thus, L A = LG  0.038 2 m = 0.350 m  0.038 2 m = 0.312 m , or the finger should be placed 31.2 cm from the bridge . LA = v 1 = 2 fA 2 fA T
; dL A =
dT 4 f A T
;
dL A 1 dT = 2 T LA
dL 0.600 cm dT =2 A =2 = 3.84% 35.0  3.82 cm T LA
a
f
P18.29
In the fundamental mode, the string above the rod has only two nodes, at A and B, with an antinode halfway between A and B. Thus, L 2L = AB = or = . cos cos 2 Since the fundamental frequency is f, the wave speed in this segment of string is v = f = Also, v = T T = = m AB 2Lf . cos
A
L
B
M
TL m cos
T
where T is the tension in this part of the string. Thus, 2Lf = cos 4L2 f 2 TL TL or = 2 m cos m cos cos
F
and the mass of string above the rod is: T cos m= 4Lf 2
Mg
[Equation 1]
FIG. P18.29
Now, consider the tension in the string. The light rod would rotate about point P if the string exerted any vertical force on it. Therefore, recalling Newton's third law, the rod must exert only a horizontal force on the string. Consider a freebody diagram of the string segment in contact with the end of the rod.
Fy = T sin  Mg = 0 T = sin
Then, from Equation 1, the mass of string above the rod is m=
Mg
FG Mg IJ cos = H sin K 4Lf
2
Mg 4Lf 2 tan
.
536 *P18.30
Superposition and Standing Waves
Let m = V represent the mass of the copper cylinder. The original tension in the wire is V g on the cylinder, to reduce the tension to T1 = mg = Vg . The water exerts a buoyant force water 2 T2 = Vg  water
FG IJ H K FG V IJ g = FG  IJ Vg . H 2K H 2 K
water
The speed of a wave on the string changes from v1
T1
to
T2
. The frequency changes from T2 1
f1 = where we assume = 2L is constant. Then f2 T2 = = f1 T1 f 2 = 300 Hz *P18.31 Comparing with we find (a)
=
T1 1
to f 2 =
 water 2 8.92  1.00 2 = 8.92
8.42 = 291 Hz 8.92
y = 0.002 m sin rad m x cos 100 rad s t
a
f db
g i db
gi
y = 2 A sin kx cos t 2 k= = m 1 , = 2.00 m , and = 2 f = 100 s 1 : f = 50.0 Hz
Then the distance between adjacent nodes is and on the string are For the speed we have
d NN =
= 1.00 m 2 L 3.00 m = = 3 loops d NN 1.00 m
v = f = 50 s 1 2 m = 100 m s
e
j
(b)
In the simplest standing wave vibration, d NN = 3.00 m = and T0
b , b = 6.00 m 2 100 m s v = 16.7 Hz fb = a = b 6.00 m
(c)
In v 0 =
increases to
, if the tension increases to Tc = 9T0 and the string does not stretch, the speed 9T0 T0
Then
= 3 v 0 = 3 100 m s = 300 m s 300 m s v = 6.00 m c = c = d NN = c = 3.00 m 1 2 fa 50 s
vc =
=3
b
g
and one loop fits onto the string.
Chapter 18
537
Section 18.4 P18.32
Resonance
The natural frequency is f= 1 1 = T 2 g 1 = L 2 9.80 m s 2 = 0.352 Hz . 2.00 m
The big brother must push at this same frequency of 0.352 Hz to produce resonance. P18.33 (a) (b) The wave speed is v= 9.15 m = 3.66 m s 2.50 s
From the figure, there are antinodes at both ends of the pond, so the distance between adjacent antinodes is and the wavelength is The frequency is then d AA =
= 9.15 m , 2
3.66 m s = 0.200 Hz 18.3 m
= 18.3 m
f= v
=
We have assumed the wave speed is the same for all wavelengths. P18.34 The wave speed is v = gd =
e9.80 m s ja36.1 mf = 18.8 m s
2
The bay has one end open and one closed. Its simplest resonance is with a node of horizontal velocity, which is also an antinode of vertical displacement, at the head of the bay and an antinode of velocity, which is a node of displacement, at the mouth. The vibration of the water in the bay is like that in one half of the pond shown in Figure P18.33. Then, and Therefore, the period is d NA = 210 10 3 m =
4
= 840 10 3 m
T= 1 840 10 3 m = = = 4. 47 10 4 s = 12 h 24 min 18.8 m s f v
This agrees precisely with the period of the lunar excitation , so we identify the extrahigh tides as amplified by resonance. P18.35 The distance between adjacent nodes is onequarter of the circumference. d NN = d AA = so = 10.0 cm and f =
20.0 cm = = 5.00 cm 2 4
v 900 m s = = 9 000 Hz = 9.00 kHz . 0.100 m
The singer must match this frequency quite precisely for some interval of time to feed enough energy into the glass to crack it.
538
Superposition and Standing Waves
Section 18.5 P18.36
Standing Waves in Air Columns
d AA = 0.320 m ; = 0.640 m (a) (b) f= v
= 531 Hz
= 0.085 0 m; d AA = 42.5 mm
For the fundamental mode in a closed pipe, = 4L , as in the diagram. But v = f , therefore L = So, L = 343 m s 4 240 s 1 v 4f A A N
P18.37
(a)
/4 L
e
j
= 0.357 m
N
A
(b)
For an open pipe, = 2L , as in the diagram. So, L = 343 m s v = = 0.715 m 2 f 2 240 s 1
/2
FIG. P18.37
e
j
P18.38
The wavelength is
=
v 343 m s = = 1.31 m f 261.6 s
so the length of the open pipe vibrating in its simplest (ANA) mode is d A to A = A closed pipe has 1 = 0.656 m 2
(NA) for its simplest resonance, (NANA) for the second,
and Here, the pipe length is *P18.39
(NANANA) for the third. 5d N to A = 5 5 = 1.31 m = 1.64 m 4 4
a
f
Assuming an air temperature of T = 37 C = 310 K , the speed of sound inside the pipe is v = 331 m s
b
g
310 K = 353 m s . 273 K
In the fundamental resonant mode, the wavelength of sound waves in a pipe closed at one end is = 4L . Thus, for the whooping crane
= 4 5.0 ft = 2.0 10 1 ft and f =
a
f
v
=
b353 m sg FG 3.281 ft IJ = 2.0 10 ft H 1 m K
1
57.9 Hz .
Chapter 18
539
P18.40
The air in the auditory canal, about 3 cm long, can vibrate with a node at the closed end and antinode at the open end, with so and d N to A = 3 cm =
4
= 0.12 m
f= v 343 m s = 3 kHz 0.12 m
A smallamplitude external excitation at this frequency can, over time, feed energy into a largeramplitude resonance vibration of the air in the canal, making it audible. P18.41 For a closed box, the resonant frequencies will have nodes at both sides, so the permitted 1 wavelengths will be L = n , n = 1, 2 , 3 , ... . 2
b
g
i.e., L =
n nv nv = . and f = 2L 2 2f
Therefore, with L = 0.860 m and L = 2.10 m, the resonant frequencies are fn = n 206 Hz
a
f
for L = 0.860 m for each n from 1 to 9
and fn = n 84.5 Hz P18.42
a
f
for L = 2.10 m for each n from 2 to 23.
The wavelength of sound is
=
v f v 2f Rt = r 2 d =
The distance between water levels at resonance is d = and P18.43 t=
r 2v 2f
r 2v . 2 Rf
For both open and closed pipes, resonant frequencies are equally spaced as numbers. The set of resonant frequencies then must be 650 Hz, 550 Hz, 450 Hz, 350 Hz, 250 Hz, 150 Hz, 50 Hz. These are oddinteger multipliers of the fundamental frequency of 50.0 Hz . Then the pipe length is d NA =
v 340 m s = = = 1.70 m . 4 4f 4 50 s
b g
P18.44
L = d AA = or n 2 Since = v f
L=
n 2
for n = 1, 2 , 3 , ... for n = 1, 2 , 3 , ...
L=n
FG v IJ H2fK F 343 m s I = na0.252 mf GH 2a680 Hzf JK
With v = 343 m s and
f = 680 Hz, L=n
for n = 1, 2 , 3 , ...
Possible lengths for resonance are: L = 0.252 m, 0.504 m, 0.757 m, ... , n 0. 252 m .
a
f
540 P18.45
Superposition and Standing Waves
For resonance in a narrow tube open at one end, f =n (a) v n = 1, 3 , 5 , ... . 4L v 3v and 384 = . 4 0.228 4 0.683
f = 384 Hz
b
g
Assuming n = 1 and n = 3 , 384 =
warm air
22.8 cm
a
f
a
f
68.3 cm
In either case, v = 350 m s . (b) For the next resonance n = 5 , and L = 5 v 5 350 m s = = 1.14 m . 4f 4 384 s 1 FIG. P18.45 P18.46 The length corresponding to the fundamental satisfies f = v v 34 : L1 = = = 0.167 m . 4L 4 f 4 512
b e
g j
a f
Since L > 20.0 cm, the next two modes will be observed, corresponding to f = or L 2 = P18.47 3v 5v = 0.502 m and L3 = = 0.837 m . 4f 4f
3v 5v and f = . 4L 2 4L3
We suppose these are the lowest resonances of the enclosed air columns. For one, For the other, So, (b) original length = 1.06 m
= =
v 343 m s = = 1.34 m f 256 s 1 v 343 m s = = 0.780 m f 440 s 1
length = d AA =
= 0.670 m 2
length = 0.390 m
= 2d AA = 2.12 m
(a) P18.48 (a) f= 343 m s = 162 Hz 2.12 m
For the fundamental mode of an open tube, L= 343 m s v = = = 0.195 m . 2 2 f 2 880 s 1
e
j
(b)
v = 331 m s 1 +
a5.00f = 328 m s
273 f= v
We ignore the thermal expansion of the metal.
=
328 m s v = = 841 Hz 2L 2 0.195 m
a
f
The flute is flat by a semitone.
Chapter 18
541
Section 18.6 P18.49 (a) (b)
Standing Waves in Rod and Plates f= 5 100 v = = 1.59 kHz 2L 2 1.60
a fa f
Since it is held in the center, there must be a node in the center as well as antinodes at the ends. The even harmonics have an antinode at the center so only the odd harmonics are present.
(c) P18.50
f=
3 560 v = = 1.11 kHz 2L 2 1.60
a fa f
When the rod is clamped at onequarter of its length, the vibration pattern reads ANANA and the rod length is L = 2d AA = . Therefore, L = v 5 100 m s = = 1.16 m f 4 400 Hz
Section 18.7 P18.51
Beats: Interference in Time f new = 110 540 = 104.4 Hz 600
f v T f = 5.64 beats s
P18.52
(a) (b)
The string could be tuned to either 521 Hz or 525 Hz from this evidence. Tightening the string raises the wave speed and frequency. If the frequency were originally 521 Hz, the beats would slow down. Instead, the frequency must have started at 525 Hz to become 526 Hz . v T 2L 1 T 2L
(c)
From f =
=
=
f2 T2 f = and T2 = 2 f1 T1 f1
FG IJ H K
2
T1 =
FG 523 Hz IJ H 526 Hz K
2
T1 = 0.989T1 .
The fractional change that should be made in the tension is then fractional change = T1  T2 = 1  0.989 = 0.011 4 = 1.14% lower. T1
The tension should be reduced by 1.14% .
542 P18.53
Superposition and Standing Waves
For an echo f = f Solving for fb . gives fb = f
bv + v g the beat frequency is f bv  v g
s s
b
= f f .
(a) (b)
fb
bv  v g when approaching wall. 2a1.33f = a 256f a343  1.33f = 1.99 Hz beat frequency
s
b2 v g
s
When he is moving away from the wall, v s changes sign. Solving for v s gives vs = 5 343 fb v = = 3.38 m s . 2 f  fb 2 256  5
a fa f a fa f a
*P18.54
Using the 4 and 2
2  foot pipes produces actual frequencies of 131 Hz and 196 Hz and a 3 combination tone at 196  131 Hz = 65.4 Hz , so this pair supplies the socalled missing fundamental. The 4 and 2foot pipes produce a combination tone 262  131 Hz = 131 Hz , so this does not work. 2 The 2 and 2  foot pipes produce a combination tone at 262  196 Hz = 65.4 Hz , so this works. 3 2 Also, 4, 2 , and 2  foot pipes all playing together produce the 65.4Hz combination tone. 3
a
f
a
f
f
Section 18.8 P18.55
NonSinusoidal Wave Patterns
We list the frequencies of the harmonics of each note in Hz: Note A C# E 1 440.00 554.37 659.26 2 880.00 1 108.7 1 318.5 Harmonic 3 1 320.0 1 663.1 1 977.8 4 1 760.0 2 217.5 2 637.0 5 2 200.0 2 771.9 3 296.3
The second harmonic of E is close the the third harmonic of A, and the fourth harmonic of C# is close to the fifth harmonic of A. P18.56 We evaluate s = 100 sin + 157 sin 2 + 62.9 sin 3 + 105 sin 4 +51.9 sin 5 + 29.5 sin 6 + 25.3 sin 7 where s represents particle displacement in nanometers and represents the phase of the wave in radians. As advances by 2 , time advances by (1/523) s. Here is the result: FIG. P18.56
Chapter 18
543
Additional Problems P18.57 f = 87.0 Hz speed of sound in air: v a = 340 m s (a)
b =
v = f b = 87.0 s 1 0.400 m v = 34.8 m s
e
ja
f
(b)
a = 4L va = a f
U V W
L=
340 m s va = = 0.977 m 4 f 4 87.0 s 1
e
j
FIG. P18.57 *P18.58 (a) Use the Doppler formula f= f
bv v g . bv v g
0 s
With f1 = frequency of the speaker in front of student and m + f b343b343sm1s.500m sg = 458 Hz  g b343 m s  1.50 m sg = 454 Hz f = a 456 Hzf b343 m s + 0g f1 = 456 Hz f 2 = frequency of the speaker behind the student.
a
2
Therefore, fb = f1  f 2 = 3.99 Hz . (b) The waves broadcast by both speakers have = between them has d AA = v 343 m s = = 0.752 m . The standing wave f 456 s
= 0.376 m . The student walks from one maximum to the next in 2 0.376 m 1 = 0.251 s , so the frequency at which she hears maxima is f = = 3.99 Hz . time t = T 1.50 m s
P18.59
Moving away from station, frequency is depressed: 343 f = 180  2.00 = 178 Hz : 178 = 180 343  v Solving for v gives Therefore, v=
a2.00fa343f
a f
178 v = 3.85 m s away from station
Moving toward the station, the frequency is enhanced: 343 f = 180 + 2.00 = 182 Hz : 182 = 180 343  v 2.00 343 Solving for v gives 4= 182 Therefore, v = 3.77 m s toward the station
a fa f
544 P18.60
Superposition and Standing Waves
v=
a48.0fa2.00f = 141 m s
4.80 10 3 v
d NN = 1.00 m ; = 2.00 m ; f =
= 70.7 Hz
a =
P18.61
v a 343 m s = = 4.85 m f 70.7 Hz 1fb343 f 4 = a2n  1f 4vf = a2n4 51.5 s m sg e j a2n + 1fb343 m sg v = a 2n + 1f = L = 2an + 1f  1 4 4f 4e60.0 s j a2n  1fb343 m sg = a2n + 1fb343 m sg 4e51.5 s j 4e60.0 s j L = 2n  1
Call L the depth of the well and v the speed of sound. Then for some integer n
a
1
1
1
and for the next resonance
2
2
1
Thus, and we require an integer solution to The equation gives n = Then
1
1
2n + 1 2n  1 = 60.0 51.5
111.5 = 6.56 , so the best fitting integer is n = 7 . 17 L= 2 7  1 343 m s
1
and
L=
af b g = 21.6 m 4e51.5 s j 2a7 f + 1 b343 m sg = 21.4 m 4e60.0 s j
1
suggest the best value for the depth of the well is 21.5 m . P18.62 The second standing wave mode of the air in the pipe reads ANAN, with d NA = so and
1.75 m = 3 4
= 2.33 m
f= v
=
343 m s = 147 Hz 2.33 m
For the string, and v are different but f is the same.
0.400 m = d NN = 2 2
so
= 0.400 m
v = f = 0. 400 m 147 Hz = 58.8 m s = T = v 2 = 9.00 10 3 kg m 58.8 m s
a
fa
f
T
2
e
jb
g
= 31.1 N
Chapter 18
545
P18.63
(a)
Since the first node is at the weld, the wavelength in the thin wire is 2L or 80.0 cm. The frequency and tension are the same in both sections, so f= 1 2L T = 1 2 0.400
a
f
4.60 = 59.9 Hz . 2.00 10 3
(b)
As the thick wire is twice the diameter, the linear density is 4 times that of the thin wire.
= 8.00 g m
so L = 1 2f thin wire. T L =
LM 1 OP N a2fa59.9f Q
4.60 = 20.0 cm half the length of the 8.00 10 3
P18.64
(a)
For the block:
Fx = T  Mg sin 30.0 = 0
so T = Mg sin 30.0 = (b) 1 Mg . 2
The length of the section of string parallel to the incline is h = 2 h . The total length of the string is then 3h . sin 30.0 The mass per unit length of the string is
FIG. P18.64
(c)
=
m 3h T =
(d) (e)
The speed of waves in the string is
v=
FG Mg IJ FG 3 h IJ = H 2 KH m K
3 Mgh 2m
In the fundamental mode, the segment of length h vibrates as one loop. The distance between adjacent nodes is then d NN = The frequency is
= h , so the wavelength is = 2h . 2
f= v
=
1 3 Mgh = 2h 2m
3 Mg 8mh
(g)
When the vertical segment of string vibrates with 2 loops (i.e., 3 nodes), then h = 2 the wavelength is
FG IJ and H 2K
= h .
(f)
The period of the standing wave of 3 nodes (or two loops) is T= 1 2m = =h = 3 Mgh f v 2mh 3 Mg
(h)
fb = 1.02 f  f = 2.00 10 2 f =
e
j
e2.00 10 j
2
3 Mg 8mh
546 P18.65
Superposition and Standing Waves
(a)
f= so
n 2L
T
f L L 1 = = = f L 2L 2
The frequency should be halved to get the same number of antinodes for twice the length. (b) n T = n T The tension must be so
(c)
f n L T = f nL T T 3 = T 22
so
FG IJ = LM n OP H K Nn + 1 Q L n OP T T = M Nn + 1 Q T F nf L I =G J T H n fL K
T n = T n
2 2 2
2
FG IJ H K
2
T 9 = T 16
to get twice as many antinodes.
P18.66
0.010 0 kg = 5.00 10 3 kg m : For the wire, = 2.00 m
v=
T
=
e200 kg m s j
2
5.00 10 3 kg m
v = 200 m s If it vibrates in its simplest state, d NN = 2.00 m = (a) (b)
: 2
f=
v
=
b200 m sg = 50.0 Hz
4.00 m
The tuning fork can have frequencies 45.0 Hz or 55.0 Hz . If f = 45.0 Hz , v = f = 45.0 s 4.00 m = 180 m s . Then, T = v 2 = 180 m s
b
g
b
g e5.00 10
2 2
or if f = 55.0 Hz , T = v 2 = f 2 P18.67 We look for a solution of the form
j = b55.0 sg a 4.00 mf e5.00 10
3
kg m = 162 N
2
2
3
kg m = 242 N .
j
5.00 sin 2.00 x  10.0t + 10.0 cos 2.00 x  10.0 t = A sin 2.00 x  10.0 t +
a
f
a
f
b g = A sina 2.00 x  10.0t f cos + A cosa 2.00 x  10.0t f sin
This will be true if both requiring
5.00 = A cos and 10.0 = A sin ,
a5.00f + a10.0f
2
2
= A2
A = 11.2 and = 63.4 The resultant wave 11.2 sin 2.00 x  10.0t + 63.4
a
f
is sinusoidal.
Chapter 18
547
P18.68
(a) (b)
With k =
2
and = 2 f =
2 v
:
y x , t = 2 A sin kx cos t = 2 A sin
b g
FG 2 x IJ cosFG 2 vt IJ H K H K
For the fundamental vibration, so
1 = 2L
y1 x , t = 2 A sin
b g
FG x IJ cosFG vt IJ HLK H L K FG 2 x IJ cosFG 2 vt IJ H LK H L K FG n x IJ cosFG n vt IJ H LK H L K
(c)
For the second harmonic 2 = L and 2L and n
y 2 x , t = 2 A sin
b g b g
(d) P18.69 (a)
In general, n =
yn x , t = 2 A sin
Let represent the angle each slanted rope makes with the vertical. In the diagram, observe that: sin = or = 41.8 . Considering the mass, 1.00 m 2 = 1.50 m 3
Fy = 0 : 2T cos = mg
b12.0 kg ge9.80 m s j = or T =
2
FIG. P18.69 78.9 N T 78.9 N = 281 m s 0.001 00 kg m
2 cos 41.8
(b)
The speed of transverse waves in the string is For the standing wave pattern shown (3 loops), or Thus, the required frequency is
v= d=
3 2
=
=
f=
2 2.00 m = 1.33 m 3 v
a
f
=
281 m s = 211 Hz 1.33 m
*P18.70
d AA =
= 7.05 10 3 m is the distance between antinodes. 2
v 3.70 10 3 m s 14.1 10
3
Then = 14.1 10 3 m and f =
=
m
= 2.62 10 5 Hz . FIG. P18.70
The crystal can be tuned to vibrate at 2 18 Hz , so that binary counters can derive from it a signal at precisely 1 Hz.
548
Superposition and Standing Waves
ANSWERS TO EVEN PROBLEMS
P18.2 P18.4 P18.6 P18.8 P18.10 see the solution 5.66 cm 0.500 s (a) 3.33 rad; (b) 283 Hz (a) The number is the greatest f 1 + ; integer d v 2 (b) Ln P18.44 P18.38 P18.40 P18.42 0.656 m; 1.64 m 3 kHz; see the solution t =
r 2v 2 Rf
n = 1, 2 , ... , n max P18.12 (a) x =
FG IJ H K d  bn  1 2g b v f g = 2bn  1 2 gb v f g
2 2
L = 0.252 m, 0.504 m, 0.757 m, ... , n 0.252 m for n = 1, 2 , 3 , ...
a
f
2
P18.46 P18.48 P18.50 P18.52
0.502 m; 0.837 m (a) 0.195 m; (b) 841 m 1.16 m (a) 521 Hz or 525 Hz; (b) 526 Hz; (c) reduce by 1.14% 2 2 4foot and 2 foot ; 2 and 2  foot; and 3 3 all three together see the solution (a) and (b) 3.99 beats s 4.85 m 31.1 N (a) 3 Mgh 1 m Mg ; (b) 3h; (c) ; (d) ; 2m 2 3h 3 Mg 2mh (e) ; (g) h; ; (f) 8mh 3 Mg
where
; 2 (b) along the hyperbola 9 x 2  16 y 2 = 144
(a) 2n + 1 m for n = 0 , 1, 2 , 3 , ...; (b) 0.029 4 m see the solution see the solution 15.7 Hz (a) 257 Hz; (b) 6 (a) 495 Hz; (b) 990 Hz 19.976 kHz 3.84% 291 Hz 0.352 Hz see the solution (a) 531 Hz; (b) 42.5 mm
P18.14
a
f
P18.54
P18.16 P18.18 P18.20 P18.22 P18.24 P18.26 P18.28 P18.30 P18.32 P18.34 P18.36
P18.56 P18.58 P18.60 P18.62 P18.64
(h) 2.00 10 2 P18.66 P18.68 P18.70
e
j
3 Mg 8mh
(a) 45.0 Hz or 55.0 Hz; (b) 162 N or 242 N see the solution 262 kHz
19
Temperature
CHAPTER OUTLINE
19.1 19.2 19.3 Temperature and the Zeroth Law of Thermodynamics Thermometers and the Celsius Temperature Scale The ConstantVolume Gas Thermometer and the Absolute Temperature Scale Thermal Expansion of Solids and Liquids Macroscopic Description of an Ideal Gas
ANSWERS TO QUESTIONS
Q19.1 Two objects in thermal equilibrium need not be in contact. Consider the two objects that are in thermal equilibrium in Figure 19.1(c). The act of separating them by a small distance does not affect how the molecules are moving inside either object, so they will still be in thermal equilibrium. The copper's temperature drops and the water temperature rises until both temperatures are the same. Then the metal and the water are in thermal equilibrium. The astronaut is referring to the temperature of the lunar surface, specifically a 400F difference. A thermometer would register the temperature of the thermometer liquid. Since there is no atmosphere in the moon, the thermometer will not read a realistic temperature unless it is placed into the lunar soil.
19.4 19.5
Q19.2
Q19.3
Q19.4 Q19.5 Q19.6
Rubber contracts when it is warmed. Thermal expansion of the glass bulb occurs first, since the wall of the bulb is in direct contact with the hot water. Then the mercury heats up, and it expands. If the amalgam had a larger coefficient of expansion than your tooth, it would expand more than the cavity in your tooth when you take a sip of your everbeloved coffee, resulting in a broken or cracked tooth! As you ice down your now excruciatingly painful broken tooth, the amalgam would contract more than the cavity in your tooth and fall out, leaving the nerve roots exposed. Isn't it nice that your dentist knows thermodynamics? The measurements made with the heated steel tape will be too shortbut only by a factor of 5 10 5 of the measured length. (a) (b) (c) One mole of H 2 has a mass of 2.016 0 g. One mole of He has a mass of 4.002 6 g. One mole of CO has a mass of 28.010 g.
Q19.7
Q19.8
Q19.9
The ideal gas law, PV = nRT predicts zero volume at absolute zero. This is incorrect because the ideal gas law cannot work all the way down to or below the temperature at which gas turns to liquid, or in the case of CO 2 , a solid. 549
550 Q19.10
Temperature
Call the process isobaric cooling or isobaric contraction. The rubber wall is easy to stretch. The air inside is nearly at atmospheric pressure originally and stays at atmospheric pressure as the wall moves in, just maintaining equality of pressure outside and inside. The air is nearly an ideal gas to start with, but PV = nRT soon fails. Volume will drop by a larger factor than temperature as the water vapor liquefies and then freezes, as the carbon dioxide turns to snow, as the argon turns to slush, and as the oxygen liquefies. From the outside, you see contraction to a small fraction of the original volume. Cylinder A must be at lower pressure. If the gas is thin, it will be at onethird the absolute pressure of B. At high temperature and pressure, the steam inside exerts large forces on the pot and cover. Strong latches hold them together, but they would explode apart if you tried to open the hot cooker. (a) The water level in the cave rises by a smaller distance than the water outside, as the trapped air is compressed. Air can escape from the cave if the rock is not completely airtight, and also by dissolving in the water. The ideal cave stays completely full of water at low tide. The water in the cave is supported by atmospheric pressure on the free water surface outside.
Q19.11 Q19.12 Q19.13
(b)
(a) FIG. Q19.13 Q19.14
(b)
Absolute zero is a natural choice for the zero of a temperature scale. If an alien race had bodies that were mostly liquid wateror if they just liked its taste or its cleaning propertiesit is conceivable that they might place one hundred degrees between its freezing and boiling points. It is very unlikely, on the other hand, that these would be our familiar "normal" ice and steam points, because atmospheric pressure would surely be different where the aliens come from. As the temperature increases, the brass expands. This would effectively increase the distance, d, from the pivot point to the center of mass of the pendulum, and also increase the moment of inertia of the pendulum. Since the moment of inertia is proportional to d 2 , and the period of a physical I pendulum is T = 2 , the period would increase, and the clock would run slow. mgd As the water rises in temperature, it expands. The excess volume would spill out of the cooling system. Modern cooling systems have an overflow reservoir to take up excess volume when the coolant heats up and expands. The coefficient of expansion of metal is larger than that of glass. When hot water is run over the jar, both the glass and the lid expand, but at different rates. Since all dimensions expand, there will be a certain temperature at which the inner diameter of the lid has expanded more than the top of the jar, and the lid will be easier to remove.
Q19.15
Q19.16
Q19.17
Chapter 19
551
Q19.18
The sphere expands when heated, so that it no longer fits through the ring. With the sphere still hot, you can separate the sphere and ring by heating the ring. This more surprising result occurs because the thermal expansion of the ring is not like the inflation of a bloodpressure cuff. Rather, it is like a photographic enlargement; every linear dimension, including the hole diameter, increases by the same factor. The reason for this is that the atoms everywhere, including those around the inner circumference, push away from each other. The only way that the atoms can accommodate the greater distances is for the circumferenceand corresponding diameterto grow. This property was once used to fit metal rims to wooden wagon and horsebuggy wheels. If the ring is heated and the sphere left at room temperature, the sphere would pass through the ring with more space to spare.
FIG. Q19.18
SOLUTIONS TO PROBLEMS Section 19.1 Temperature and the Zeroth Law of Thermodynamics
No problems in this section
Section 19.2 Section 19.3 P19.1
Thermometers and the Celsius Temperature Scale The ConstantVolume Gas Thermometer and the Absolute Temperature Scale
Since we have a linear graph, the pressure is related to the temperature as P = A + BT , where A and B are constants. To find A and B, we use the data
a f 1.635 atm = A + a78.0 CfB
Solving (1) and (2) simultaneously, we find and Therefore, (a) At absolute zero which gives (b) (c) A = 1.272 atm B = 4.652 10 3 atm C
0.900 atm = A + 80.0 C B
(1) (2)
P = 1.272 atm + 4.652 10 3 atm C T P = 0 = 1.272 atm + 4.652 10 3 atm C T T = 274 C .
e
j
e
j
At the freezing point of water P = 1.272 atm + 0 = 1.27 atm . And at the boiling point P = 1.272 atm + 4.652 10 3 atm C 100 C = 1.74 atm .
e
ja
f
552 P19.2
Temperature
P1V = nRT1 and P2 V = nRT2 imply that P2 T2 = P1 T1 0.980 atm 273 K + 45.0 K P1T2 = = 1.06 atm 273 + 20.0 K T1 T1 P3 P1
(a)
P2 =
(b)
T3 =
a f a293 K fa0.500 atmf = 149 K = =
0.980 atm
a
fa
f
124 C FIG. P19.2
P19.3
(a) (b)
TF =
9 9 TC + 32.0 F = 195.81 + 32.0 = 320 F 5 5
a
f
T = TC + 273.15 = 195.81 + 273.15 = 77.3 K To convert from Fahrenheit to Celsius, we use and the Kelvin temperature is found as TC = 5 5 TF  32.0 = 98.6  32.0 = 37.0 C 9 9
P19.4
(a)
b
g a
f
T = TC + 273 = 310 K
(b)
In a fashion identical to that used in (a), we find TC = 20.6 C and T = 253 K
P19.5
(a) (b)
T = 450 C = 450 C
FG 212 F  32.0 F IJ = H 100 C  0.00 C K
810 F
T = 450 C = 450 K
P19.6
a f 100 C = aa60.0 Sf + b Subtracting, 100 C = aa75.0 Sf
Require
0.00 C = a 15.0 S + b
Then 0.00 C = 1.33 15.0 S C+ b b = 20.0 C . So the conversion is TC = 1.33 C S TS + 20.0 C . P19.7 (a) T = 1 064 + 273 = 1 337 K melting point T = 2 660 + 273 = 2 933 K boiling point (b) T = 1 596 C = 1 596 K . The differences are the same.
a
f
a = 1.33 C S .
b
g
Chapter 19
553
Section 19.4 P19.8
Thermal Expansion of Solids and Liquids
= 1.10 10 5 C 1 for steel
L = 518 m 1.10 10 5 C 1 35.0 C  20.0 C = 0.313 m
e
j
a
f
P19.9
The wire is 35.0 m long when TC = 20.0 C . L = Li T  Ti
b
g
= 20.0 C = 1.70 10 5 C
5
a f a f for Cu. L = a35.0 mfe1.70 10 aCf jc35.0 C  a 20.0 C fh =
1 1
+3.27 cm
P19.10 P19.11
L = Li T = 25.0 m 12.0 10 6 C 40.0 C = 1.20 cm For the dimensions to increase, L = Li T T = 55.0 C
a
fe
ja
f
1.00 10 2 cm = 1.30 10 4 C 1 2.20 cm T  20.0 C
a
fa
f
*P19.12 P19.13
L = Li T = 22 10 6 C 2.40 cm 30 C = 1.58 10 3 cm
(a) (b)
e
ja
fa
f
L = Li T = 9.00 10 6 C 1 30.0 cm 65.0 C = 0.176 mm
L = Li T = 9.00 10 6 C 1 1.50 cm 65.0 C = 8.78 10 4 cm V = 3Vi T = 3 9.00 10 6 C 1
a
a
fa
f
fa
f
(c) *P19.14
e
F jGH 30.0a fa1.50f 4
2
cm3 65.0 C = 0.093 0 cm3
Ia JK
f
The horizontal section expands according to L = Li T . x = 17 10 6 C 1 28.0 cm 46.5 C  18.0 C = 1.36 10 2 cm The vertical section expands similarly by y = 17 10 6 C 1 134 cm 28.5 C = 6.49 10 2 cm . The vector displacement of the pipe elbow has magnitude r = x 2 + y 2 = FIG. P19.14
e
ja
fa
f
e
ja
fa
f
a0.136 mmf + a0.649 mmf
2 1
2
= 0.663 mm
and is directed to the right below the horizontal at angle
= tan 1
FG y IJ = tan FG 0.649 mm IJ = 78.2 H 0.136 mm K H x K
r = 0.663 mm to the right at 78.2 below the horizontal
554 P19.15
Temperature
(a)
L Al 1 + Al T = LBrass 1 + Brass T L Al  LBrass T = LBrass Brass  L Al Al T =
6
b
g
b
g
6
a10.01  10.00f a10.00fe19.0 10 j  a10.01fe24.0 10 j a10.02  10.00f a10.00fe19.0 10 j  a10.02fe24.0 10 j
6 6
T = 199 C so T = 179 C. This is attainable. (b) T =
T = 396 C so T = 376 C which is below 0 K so it cannot be reached. P19.16 (a) A = 2Ai T : A = 2 17.0 10 6 C 1 0.080 0 m
e
jb
g a50.0 Cf
2
A = 1.09 10 5 m 2 = 0.109 cm 2
(b) The length of each side of the hole has increased. Thus, this represents an increase in the area of the hole. P19.17 P19.18 V =  3 Vi T = 5.81 10 4  3 11.0 10 6 (a) (b)
b
g
e
e
jjb50.0 galga20.0f =
0.548 gal
L = Li 1 + T :
We must get
a
f
5.050 cm = 5.000 cm 1 + 24.0 10 6 C 1 T  20.0 C T = 437 C L Al = LBrass for some T , or Li , Al 1 + Al T = Li , Brass 1 + Brass T
a
f
b
g
b
g e j
5.000 cm 1 + 24.0 10 6 C 1 T = 5.050 cm 1 + 19.0 10 6 C 1 T Solving for T , T = 2 080 C , so T = 3 000 C
e
j
This will not work because aluminum melts at 660 C . P19.19 (a) (b) V f = Vi 1 + T = 100 1 + 1.50 10 4 15.0 = 99.8 mL Vacetone = Vi T Vflask = Vi T
b
g
a
f
b
g
b
g
acetone
Pyrex
= 3Vi T
b
g
Pyrex
for same Vi , T , Vacetone acetone 1.50 10 4 1 = = = 6 Vflask flask 6.40 10 2 3 3.20 10
e
j
The volume change of flask is about 6% of the change in the acetone's volume .
Chapter 19
555
P19.20
(a),(b)
The material would expand by L = Li T , L = T , but instead feels stress Li F YL = = YT = 7.00 10 9 N m 2 12.0 10 6 C A Li
e
j
a f a30.0 Cf
1
= 2.52 10 6 N m 2 . This will not break concrete. P19.21 (a) V = Vt t T  VAl Al T = t  3 Al Vi T
= 9.00 10 4  0.720 10 4 C 1 2 000 cm3 60.0 C
e
b
j
e
g
ja
f
V = 99.4 cm3
(b)
overflows.
The whole new volume of turpentine is 2 000 cm3 + 9.00 10 4 C 1 2 000 cm3 60.0 C = 2 108 cm3 99.4 cm3 = 4.71 10 2 3 2 108 cm
e
ja
f
so the fraction lost is
and this fraction of the cylinder's depth will be empty upon cooling: 4.71 10 2 20.0 cm = 0.943 cm . *P19.22 The volume of the sphere is VPb = 4 3 4 r = 2 cm 3 3
a
f
a
f
3
= 33.5 cm3 .
The amount of mercury overflowing is overflow = VHg + VPb  Vglass = Hg VHg + PbVPb  glassVglass T where Vglass = VHg + VPb is the initial volume. Then overflow = Hg  glass VHg + Pb  glass VPb T = Hg  3 glass VHg + 3 Pb  3 glass VPb T
6 3 6 3 3
e
j
e j e j e j e 1 1 L O = Ma182  27f10 118 cm + a87  27f10 33.5 cm P 40 C = 0.812 cm C C N Q
j
P19.23
In
F YL = require L = Li T A Li
F = YT A F 500 N = T = 4 2 AY 2.00 10 m 20.0 10 10 N m 2 11.0 10 6 C
e
je
je
j
T = 1.14 C
556 *P19.24
Temperature
Model the wire as contracting according to L = Li T and then stretching according to L Y F stress = = Y = Li T = YT . A Li Li (a) (b) F = YAT = 20 10 10 N m 2 4 10 6 m 2 11 10 6 T = 3 10 8 N m 2 stress = = 136 C Y 20 10 10 N m 2 11 10 6 C
e
j
1 45 C = 396 N C
e
j
To increase the stress the temperature must decrease to 35 C  136 C = 101 C . (c) *P19.25 The original length divides out, so the answers would not change. A = A1 T = A f  Ai
The area of the chip decreases according to
A f = Ai 1 + T = Ai 1 + 2T
b
g
a
f
The star images are scattered uniformly, so the number N of stars that fit is proportional to the area. Then N f = N i 1 + 2T = 5 342 1 + 2 4.68 10 6 C 1 100 C  20 C = 5 336 star images .
a
f
e
ja
f
Section 19.5 P19.26 (a)
Macroscopic Description of an Ideal Gas n= 9.00 atm 1.013 10 5 Pa atm 8.00 10 3 m 3 PV = = 2.99 mol 8.314 N mol K 293 K RT
(b) P19.27 (a)
N = nN A
a = a 2.99 molfe6.02 10
a
fe
fa
je
f
j
23
Initially, PVi = ni RTi i Finally, Pf V f = n f RT f Dividing these equations, giving or
j a1.00 atmfV = n Ra10.0 + 273.15f K P b0. 280V g = n Ra 40.0 + 273.15f K
i i
f i i
molecules mol = 1.80 10 24 molecules
313.15 K = 1.00 atm 283.15 K Pf = 3.95 atm
0.280 Pf
Pf = 4.00 10 5 Pa abs. .
d i i
(b)
After being driven
a f P a1.02fb0.280V g = n Ra85.0 + 273.15f K
a fa f a fa f
Pd = 1.121Pf = 4.49 10 5 Pa
P19.28 3 150 0.100 3 PV = = 884 balloons 3 4 r P 4 0.150 3 1.20 If we have no special means for squeezing the last 100 L of helium out of the tank, the tank will be full of helium at 1.20 atm when the last balloon is inflated. The number of balloons is then reduced 0.100 m3 3 to to 884  = 877 . 3 4 0.15 m PV = NP V = 4 3 r NP : 3 N=
e
a
j f
Chapter 19
557
P19.29
The equation of state of an ideal gas is PV = nRT so we need to solve for the number of moles to find N. 1.01 10 5 N m 2 10.0 m 20.0 m 30.0 m PV = = 2. 49 10 5 mol n= RT 8.314 J mol K 293 K
e
b
ja
fa ga
f
fa
f
N = nN A = 2.49 10 mol 6.022 10 *P19.30 (a) PVi = ni RTi = i mi RTi M
5
e
23
molecules mol = 1.50 10 29 molecules
j
3 5 6 MPVi 4.00 10 kg 1.013 10 N 4 6.37 10 m mole K i = mi = RTi 8.314 Nm 50 K mole m2 3
e
j
3
= 1.06 10 21 kg (b) Pf V f PVi i = n f RT f ni RTi
21 20 f
P19.31
F 1.06 10 kg + 8.00 10 kg I T GH JK 50 K 1.06 10 kg F 1 IJ = 56.9 K T = 100 K G H 1.76 K nRT F 9.00 g I F 8.314 J I F =G P= JK GH mol K JK GH 2.00 773 K m IJK = V 10 H 18.0 g mol
2 1 =
21 f 3 3
1.61 MPa = 15.9 atm
P19.32
(a)
T2 = T1
P2 = 300 K 3 = 900 K P1 P2 V2 = 300 2 2 = 1 200 K P1V1
a
fa f
(b) P19.33
T2 = T1
a fa f
ge
Fy = 0 :
b
out gV  in gV  200 kg g = 0
out
b
 in 400 m
3
j = 200 kg
g
The density of the air outside is 1.25 kg m3 . n P = From PV = nRT , V RT The density is inversely proportional to the temperature, and the density of the hot air is
Then
jFGH 283 K IJK T e1.25 kg m jFGH 1  283 K IJK e400 m j = 200 kg T
in = 1.25 kg m3
3
e
in
3
FIG. P19.33
in
283 K 1 = 0.400 Tin 283 K 0.600 = Tin = 472 K Tin
558 *P19.34
Temperature
Consider the air in the tank during one discharge process. We suppose that the process is slow enough that the temperature remains constant. Then as the pressure drops from 2.40 atm to 1.20 atm, the volume of the air doubles. During the first discharge, the air volume changes from 1 L to 2 L. Just 1 L of water is expelled and 3 L remains. In the second discharge, the air volume changes from 2 L to 4 L and 2 L of water is sprayed out. In the third discharge, only the last 1 L of water comes out. Were it not for male pattern dumbness, each person could more efficiently use his device by starting with the tank half full of water. (a) PV = nRT 1.013 10 5 Pa 1.00 m3 PV n= = = 41.6 mol RT 8.314 J mol K 293 K (b) 1.20 kg , in agreement with the tabulated density of
P19.35
e je j b ga f m = nM = a 41.6 molfb 28.9 g molg =
3
1.20 kg m at 20.0C. *P19.36 The void volume is 0.765Vtotal = 0.765 r 2 = 0.765 1.27 10 2 m 0.2 m = 7.75 10 5 m3 . Now for the gas remaining PV = nRT n=
5 5 2 3 PV 12.5 1.013 10 N m 7.75 10 m = = 3.96 10 2 mol RT 8.314 Nm mole K 273 + 25 K
e
j
2
b
e
j ga
f
P19.37
(a)
PV = nRT
n=
3 5 PVM 1.013 10 Pa 0.100 m 28.9 10 kg mol = m = nM = RT 8.314 J mol K 300 K
PV RT
b
a
fe
3
ga
f
j
m = 1.17 10 3 kg (b) (c) (d) P19.38 Fg = mg = 1.17 10 3 kg 9.80 m s 2 = 11.5 mN F = PA = 1.013 10 5 N m 2 0.100 m
e
j
e
ja
f
2
= 1.01 kN
The molecules must be moving very fast to hit the walls hard. P = P0 + gh P0 V f = nRT f : V f = Vi Vf
f 0
At depth, At the surface, Therefore
and
PVi = nRTi P0 V f
0
F T I FG P + gh IJ GH T JK H P K F 293 K IJ FG 1.013 10 = 1.00 cm G H 278 K K GH
i 0 3
bP + ghgV
5
=
Tf Ti
i
Pa + 1 025 kg m 3 9.80 m s 2 25.0 m 1.013 10 5 Pa
e
je
ja
f IJ JK
V f = 3.67 cm3
Chapter 19
559
P19.39
PV = nRT : so
mf mi
=
nf ni
=
f
Pf V f RTi Pf = RT f PVi Pi i
m f = mi
FP I GH P JK
i
m = mi  m f = mi P19.40
F P  P I = 12.0 kgFG 41.0 atm  26.0 atm IJ = GH P JK H 41.0 atm K
i f i
4.39 kg
My bedroom is 4 m long, 4 m wide, and 2.4 m high, enclosing air at 100 kPa and 20 C = 293 K . Think of the air as 80.0% N 2 and 20.0% O 2 . Avogadro's number of molecules has mass
Then gives *P19.41
a0.800fb28.0 g molg + a0.200fb32.0 g molg = 0.028 8 kg mol F mI PV = nRT = G J RT H MK PVM e1.00 10 N m je38.4 m jb0.028 8 kg molg = = 45.4 kg m= RT b8.314 J mol K ga293 K f
5 2 3
~ 10 2 kg
The CO 2 is far from liquefaction, so after it comes out of solution it behaves as an ideal gas. Its molar mass is M = 12.0 g mol + 2 16.0 g mol = 44.0 g mol . The quantity of gas in the cylinder is m sample 6.50 g = = 0.148 mol n= M 44.0 g mol
b
g
Then gives
PV = nRT V=
nRT 0.148 mol 8.314 J mol K 273 K + 20 K = P 1.013 10 5 N m 2
b
ga
f FG 1 N m IJ F 10 L I = H 1 J K GH 1 m JK
3 3
3.55 L
P19.42
10 9 Pa 1.00 m 3 6.02 10 23 molecules mol PVN A = = 2.41 10 11 molecules N= RT 8.314 J K mol 300 K
e
je
1
b
je
ga
f
j
P19.43
FG m IJ RT H MK F m IJ RT P V = n RT = G HMK P VM F 1 m m = GH T  T1 IJK R
P0 V = n1 RT1 =
0 2 2 1 2 2 1 2 0 1 2
560 P19.44
Temperature
(a)
Initially the air in the bell satisfies P0 Vbell = nRTi or P0 2.50 m A = nRTi Pbell 2.50 m  x A = nRT f where x is the height the water rises in the bell. Also, the pressure in the bell, once it is lowered, is equal to the sea water pressure at the depth of the water level in the bell. Pbell = P0 + g 82.3 m  x P0 + g 82.3 m
a
f
(1) (2)
When the bell is lowered, the air in the bell satisfies
a
f
a
f
a
f
(3)
The approximation is good, as x < 2.50 m. Substituting (3) into (2) and substituting nR from (1) into (2), P0 + g 82.3 m 2.50 m  x A = P0 Vbell Using P0 = 1 atm = 1.013 10 5 Pa and = 1.025 10 3 kg m3 x = 2.50 m 1 
a
fa
f
Tf Ti
.
L O fMM T FGH1 + ga82.3 mf IJK PP P N T Q LM 277.15 K F e1.025 10 kg m je9.80 m s ja82.3 mf I G1 + JJ = a 2.50 mfM1  293.15 K G 1.013 10 N m K H MN a
f 1 0 0 3 3 2 5 2
1
OP PP Q
x = 2.24 m (b) If the water in the bell is to be expelled, the air pressure in the bell must be raised to the water pressure at the bottom of the bell. That is, Pbell = P0 + g 82.3 m
5 5
a
f
= 1.013 10 Pa + 1.025 10 3 kg m 3 9.80 m s 2 82.3 m Pbell = 9.28 10 Pa = 9.16 atm
e
je
ja
f
Additional Problems P19.45 The excess expansion of the brass is Lrod  Ltape = brass  steel Li T
a f a f a Lf = 2.66 10 m
4
b
L = 19.0  11.0 10 6
g a Cf a0.950 mfa35.0 Cf
1
(a)
The rod contracts more than tape to a length reading 0.950 0 m  0.000 266 m = 0.949 7 m
(b)
0.950 0 m + 0.000 266 m = 0.950 3 m
Chapter 19
561
P19.46
At 0C, 10.0 gallons of gasoline has mass, from
=
m V
m = V = 730 kg m3 10.0 gal The gasoline will expand in volume by
e
jb
gFGH 0.003 80 m IJK = 27.7 kg 1.00 gal
3
V = Vi T = 9.60 10 4 C 1 10.0 gal 20.0 C  0.0 C = 0.192 gal At 20.0C, 10.192 gal = 27.7 kg 10.0 gal = 27.7 kg
b
ga
f
F 10.0 gal I = 27.2 kg GH 10.192 gal JK
The extra mass contained in 10.0 gallons at 0.0C is 27.7 kg  27.2 kg = 0.523 kg . P19.47 Neglecting the expansion of the glass, h = h = V T A
4 3
b0.250 cm 2g e1.82 10 e 2.00 10 cmj
3 3 2
4
C 1 30.0 C = 3.55 cm
ja
f
FIG. P19.47 P19.48 (a) The volume of the liquid increases as V = Vi T . The volume of the flask increases as Vg = 3Vi T . Therefore, the overflow in the capillary is Vc = Vi T  3 ; and in the
b
g
capillary Vc = Ah . Therefore, h = (b) Vi  3 T . A
b
g
For a mercury thermometer and for glass, Thus or
Hg = 1.82 10 4 C 1
3 = 3 3.20 10 6 C 1
b g
 3 << .
562 P19.49
Temperature
The frequency played by the coldwalled flute is fi = When the instrument warms up ff =
v v = . i 2 Li
fi v v v = = = . f 2L f 2Li 1 + T 1 + T
a
f
The final frequency is lower. The change in frequency is f = f i  f f = f i 1  f = v 2 Li 1 1 + T T v T 1 + T 2 Li
a f b343 m sge24.0 10 Cja15.0 Cf = f 2a0.655 mf
6
FG H
FG H
IJ K
IJ K
0.094 3 Hz
This change in frequency is imperceptibly small. P19.50 (a) P0 V P V = T T V = V + Ah P = P0 +
0
k 250C 20C
FG P + kh IJ aV + Ahf = P V FG T IJ H AK HTK e1.013 10 N m + 2.00 10 N m hj e5.00 10 m + e0.010 0 m jhj = e1.013 10 N m je5.00 10
0 5 2 5 3 3 3 2 5 2
kh A
h
FIG. P19.50 m3 K jFGH 523 K IJK 293
3
2 000 h 2 + 2 013 h  397 = 0 h= 2 013 2 689 = 0.169 m 4 000
(b)
2.00 10 3 N m 0.169 kh 5 = 1.013 10 Pa + P = P + A 0.010 0 m 2
e
ja
f
P = 1.35 10 5 Pa
Chapter 19
563
P19.51
(a)
=
m m and d =  2 dV V V
For very small changes in V and , this can be expressed as =  m V =  T . V V
The negative sign means that any increase in temperature causes the density to decrease and vice versa. For water we have = 1.000 0 g cm 3  0.999 7 g cm 3 = = 5 10 5 C 1 . 3 T 1.000 0 g cm 10.0 C  4.0 C
(b)
e
ja
f
*P19.52
The astronauts exhale this much CO 2 : n= m sample M = 1.09 kg 1 000 g 1 mol = 520 mol . 3 astronauts 7 days astronaut day 1 kg 44.0 g
F GH
Ia JK
fb
gFGH
I JK
Then 520 mol of methane is generated. It is far from liquefaction and behaves as an ideal gas. P= P19.53 (a) nRT 520 mol 8.314 J mol K 273 K  45 K = 6.57 10 6 Pa = V 150 10 3 m3
b
ga
f
We assume that air at atmospheric pressure is above the piston. In equilibrium Therefore, or Pgas = mg + P0 A
nRT mg = + P0 hA A h= nRT mg + P0 A
where we have used V = hA as the volume of the gas. (b) From the data given, h= 0.200 mol 8.314 J K mol 400 K
2
20.0 kg 9.80 m s
e
j e
b
ga
+ 1.013 10 N m
5
2
je
f
0.008 00 m 2
j
FIG. P19.53
= 0.661 m
564 P19.54
Temperature
The angle of bending , between tangents to the two ends of the strip, is equal to the angle the strip subtends at its center of curvature. (The angles are equal because their sides are perpendicular, right side to the right side and left side to left side.) (a) The definition of radian measure gives and By subtraction, Li + L1 = r1 Li + L 2 = r2 L 2  L1 = r2  r1 FIG. P19.54
b
g
2 Li T  1 Li T = r =
(b)
b
2
 1 Li T r
g
In the expression from part (a), is directly proportional to T and also to 2  1 . Therefore is zero when either of these quantities becomes zero. The material that expands more when heated contracts more when cooled, so the bimetallic strip bends the other way. It is fun to demonstrate this with liquid nitrogen. 2 2  1 Li T 2 r
b
g
(c)
(d)
=
b
g
=
2 19 10 6  0.9 10 6 C 1 200 mm 1 C
ee
j
ja
fa f
= 1.45 10 2 = 1. 45 10 2 rad P19.55
FG 180 IJ = H rad K
0.500 mm 0.830
From the diagram we see that the change in area is A = w + w + w . Since and w are each small quantities, the product w will be very small. Therefore, we assume w 0. Since we then have w = wT and = T , FIG. P19.55
A = wT + wT
and since A = w , A = 2AT . The approximation assumes w 0, or T 0 . Another way of stating this is T << 1 .
Chapter 19
565
P19.56
(a)
L Ti = 2 i g
so
Li =
Ti2 g 4 2
a1.000 sf e9.80 m s j = 0.248 2 m =
2 2
4 2
L = Li T = 19.0 10 6 C 1 0.284 2 m 10.0 C = 4.72 10 5 m T f = 2 0.248 3 m Li + L = 2 = 1.000 095 0 s g 9.80 m s 2
b
ga
f
T = 9.50 10 5 s (b) In one week, the time lost is time lost = 1 week 9.50 10 5 s lost per second time lost = 7.00 d week time lost = 57.5 s lost P19.57
e
j
s lost s
b
gFGH 86 400ds IJK FGH 9.50 10 1.00
5
IJ K
I = r 2 dm
z
and since
r T = r Ti 1 + T
2
for T << 1 we find thus With = 17.0 10 6 C 1 and we find for Cu: (b) With and we find for Al: P19.58 (a) B = gV B=
a f b ga f I aT f = a1 + T f I bT g I aT f  I bT g 2T I bT g
i i i
(a)
T = 100 C I = 2 17.0 10 6 C 1 100 C = 0.340% I
e
ja
f
= 24.0 10 6 C 1
T = 100 C I = 2 24.0 10 6 C 1 100 C = 0.480% I
e
ja
f
P = P0 + gd
P V = P0 Vi
gP0 Vi = P
bP + gdg
0
gP0 Vi
(b)
Since d is in the denominator, B must decrease as the depth increases. (The volume of the balloon becomes smaller with increasing pressure.)
(c)
gP0 Vi P0 + gd P0 1 Bd = = = gP0Vi P0 2 B0 P0 + gd
P0 + gd = 2 P0 d= 1.013 10 5 N m 2 P0 = = 10.3 m g 1.00 10 3 kg m3 9.80 m s 2
af af
b
g
e
je
j
566 *P19.59
Temperature
The effective coefficient is defined by Ltotal = effective Ltotal T where Ltotal = LCu + LPb and Ltotal = LCu + LPb = xLtotal + 1  x Ltotal . Then by substitution
a f
Cu
Cu LCu T + Pb LPb T = eff LCu + LPb T Cu x + Pb 1  x = eff
b
 Pb
a f gx =
b
g
eff
 Pb
6
x= *P19.60 (a)
20 10 17 10
6 6
1 C  29 10 6 1 C 1 C  29 10 1 C
=
9 = 0.750 12
No torque acts on the disk so its angular momentum is constant. Its moment of inertia decreases as it contracts so its angular speed must increase . I i i = I f f = 1 1 1 1 2 MRi2 i = MR 2 f = M Ri + RiT f = MRi2 1  T f 2 2 2 2
2 2
(b)
f
f = i 1  T
P19.61
=
25.0 rad s
e1  e17 10 f
6
1 C 830 C
j
j
2
=
25.0 rad s = 25.7 rad s 0.972
After expansion, the length of one of the spans is L f = Li 1 + T = 125 m 1 + 12 10 6 C 1 20.0 C = 125.03 m . L f , y, and the original 125 m length of this span form a right triangle with y as the altitude. Using the Pythagorean theorem gives:
a
a
f
a125.03 mf
yielding y = 2.74 m . P19.62
2
= y 2 + 125 m
a
f
2
After expansion, the length of one of the spans is L f = L 1 + T . L f , y, and the original length L of this span form a right triangle with y as the altitude. Using the Pythagorean theorem gives
a
f
L2f = L2 + y 2 ,
Since P19.63 (a)
or
y = L2f  L2 = L
a1 + T f
2
 1 = L 2T + T
a f
2
T << 1 ,
y L 2T .
m m and the density is = . V M
Let m represent the sample mass. The number of moles is n = So PV = nRT becomes PV = Then, = m PM . = V RT m m RT or PM = RT . M V
(b)
1.013 10 5 N m 2 0.032 0 kg mol PM = = = 1.33 kg m 3 RT 8.314 J mol K 293 K
e
b
jb
ga
f
g
Chapter 19
P19.64
(a)
From PV = nRT , the volume is: Therefore, when pressure is held constant, Thus,
V=
FG nR IJ T H PK FG 1 IJ dV = FG 1 IJ V , or = H V K dT H V K T
1 T
567
dV nR V = = dT P T
=
(b)
At T = 0 C = 273 K , this predicts Experimental values are:
1 = 3.66 10 3 K 1 273 K He = 3.665 10 3 K 1 and air = 3.67 10 3 K 1
They agree within 0.06% and 0.2%, respectively. P19.65 For each gas alone, P1 = For all gases N 1 kT N kT N kT and P2 = 2 and P3 = 3 , etc. V V V
bN
P19.66 (a)
P1V1 + P2 V2 + P3 V3 ... = N 1 + N 2 + N 3 ... kT and
1
+ N 2 + N 3 ... kT = PV
g
b
g
Also, V1 = V2 = V3 = ... = V , therefore P = P1 + P2 + P3 ... . Using the Periodic Table, we find the molecular masses of the air components to be M N 2 = 28.01 u , M O 2 = 32.00 u , M Ar = 39.95 u and M CO 2 = 44.01 u . Thus, the number of moles of each gas in the sample is 75 b g 28.01.52 g = 2.696 mol g mol 23.15 g nbO g = = 0.723 4 mol 32.00 g mol 1.28 g na Ar f = = 0.032 0 mol 39.95 g mol n N2 =
2
b g
b g
a f
b
g
n CO 2 =
b g
0.05 g = 0.001 1 mol 44.01 g mol
The total number of moles is n 0 = ni = 3.453 mol . Then, the partial pressure of N 2 is P N2 = Similarly, P O 2 = 21.2 kPa continued on next page
b g
2.696 mol 1.013 10 5 Pa = 79.1 kPa . 3.453 mol
e
j
b g
P Ar = 940 Pa
a f
P CO 2 = 33.3 Pa
b
g
568
Temperature
(b)
Solving the ideal gas law equation for V and using T = 273.15 + 15.00 = 288.15 K , we find V= 3.453 mol 8.314 J mol K 288.15 K n 0 RT = 8.166 10 2 m3 . = P 1.013 10 5 Pa
a
fb
ga
f
Then, = (c)
100 10 3 kg m = = 1.22 kg m3 . V 8.166 10 2 m3
The 100 g sample must have an appropriate molar mass to yield n 0 moles of gas: that is M air =
a f
100 g = 29.0 g mol . 3.453 mol
*P19.67
Consider a spherical steel shell of inner radius r and much smaller thickness t, containing helium at pressure P. When it contains so much helium that it is on the point of bursting into two hemispheres, we have P r 2 = 5 10 8 N m 2 2 rt . The mass of the steel is
e
j
sV = s 4 r 2 t = s 4 r 2
Pr 10 9 Pa
. For the helium in the tank, PV = nRT becomes
m 4 P r 3 = nRT = He RT = 1 atmVballoon . 3 M He The buoyant force on the balloon is the weight of the air it displaces, which is described by m 4 1 atmVballoon = air RT = P r 3 . The net upward force on the balloon with the steel tank hanging M air 3 from it is + m air g  m He g  m s g = M air P 4 r 3 g M He P 4 r 3 g s P 4 r 3 g   3 RT 3 RT 10 9 Pa
The balloon will or will not lift the tank depending on whether this quantity is positive or negative, M air  M He  9 s . At 20C this quantity is which depends on the sign of 3 RT 10 Pa
b
g
=
a28.9  4.00f 10 kg mol  7 860 kg m 3b8.314 J mol K g293 K 10 N m
3 9 2
3
= 3.41 10 6 s 2 m 2  7.86 10 6 s 2 m 2 where we have used the density of iron. The net force on the balloon is downward so the helium balloon is not able to lift its tank.
Chapter 19
569
P19.68
With piston alone: or With A = constant, But,
T = constant, so PV = P0 V0
b g b g Fh I P=P G J Hh K
P Ahi = P0 Ah0
0 0 i
P = P0 +
mp g A FIG. P19.68
where m p is the mass of the piston. Thus, which reduces to P0 + hi = mp g A h0 1+ = P0 = 1+
FG h IJ Hh K
0 i
50.0 cm
20.0 kg 9.80 m s 2
mp g P0 A
e
1.013 10 5 Pa 0.400 m
a
j
= 49.81 cm
f
2
With the man of mass M on the piston, a very similar calculation (replacing m p by m p + M ) gives: h =
em 1+
h0
p +M
P0 A
jg
= 1+
50.0 cm
95.0 kg 9.80 m s 2
e
1.013 10 5 Pa 0.400 m
a
j
= 49.10 cm
f
2
Thus, when the man steps on the piston, it moves downward by h = hi  h = 49.81 cm  49.10 cm = 0.706 cm = 7.06 mm . (b) P = const, so giving dL = dT : L L f = 1.00 m e Ahi Ah = T Ti hi 49.81 = 293 K = 297 K T = Ti 49.10 h or V V = T Ti
FG IJ H K
Li Li
FG H
IJ K
(or 24C)
P19.69
(a)
Ti Ti
z
dT =
z
Lf dL ln L Li
FG IJ = T H K
L f = Li eT
(b)
a a
f f
2.00 10 5 C 1 100 C
a
f
= 1.002 002 m
L f = 1.00 m 1 + 2.00 10 5 C 1 100 C = 1.002 000 m : L f = 1.00 m e
2.00 10 2 C 1 100 C
a
f
L f  L f Lf L f  L f Lf
= 2.00 10 6 = 2.00 10 4%
a
f
= 7.389 m
L f = 1.00 m 1 + 0.020 0 C 1 100 C = 3.000 m :
a
f
= 59. 4%
570 P19.70
Temperature
At 20.0C, the unstretched lengths of the steel and copper wires are
a f a f a f a20.0 Cf = 1.999 56 m L a 20.0 C f = a 2.000 mf 1 + 17.0 10 aCf a 20.0 C f = 1.999 32 m
Ls 20.0 C = 2.000 m 1 + 11.0 10 6 C
c 6 1 1
Under a tension F, the length of the steel and copper wires are Ls = Ls 1 +
LM N
F YA
OP Q
s
Lc = Lc 1 +
LM N
F YA
OP Q
c
where Ls + L c = 4.000 m .
Since the tension, F, must be the same in each wire, solve for F: F=
b L + L g  bL
s c Ls Ys As
+
s + Lc Lc Yc A c
g.
When the wires are stretched, their areas become As = 1.000 10 3 m Ac
3
e = e1.000 10
j mj
2 2
1 + 11.0 10 6 20.0
6
e ja f 1 + e17.0 10 ja 20.0 f b
2 2
= 3.140 10 6 m 2 = 3.139 10 6 m 2
Recall Ys = 20.0 10 10 Pa and Yc = 11.0 10 10 Pa . Substituting into the equation for F, we obtain F= 4.000 m  1.999 56 m + 1.999 32 m 1.999 56 m
e20.0 10
10
Pa 3.140 10 6 m 2 + 1.999 32 m
je
j
e11.0 10
g
10
Pa 3.139 10 6 m 2
je
j
F = 125 N To find the xcoordinate of the junction, Ls = 1.999 56 m 1 +
b
L gMM 20.0 10 N e
125 N
10
N m 2 3.140 10 6
je
OP = 1.999 958 m m jP Q
2
Thus the xcoordinate is 2.000 + 1.999 958 = 4.20 10 5 m .
Chapter 19
571
P19.71
(a)
= r 2 = 5.00 10 4 m
f1 =
e
j e7.86 10
2
3
kg m3 = 6.17 10 3 kg m
j
(b)
1 v T and v = so f1 = 2L 2L
2
T
3 2
Therefore, T = 2Lf1 (c)
b g = e6.17 10 ja2 0.800 200f
= 632 N
First find the unstressed length of the string at 0C: L = L natural 1 +
FG H
T L so L natural = + T AY AY 1
2
A = 5.00 10 4 Therefore,
e
IJ K mj
= 7.854 10 7 m 2 and Y = 20.0 10 10 Pa
T 632 = = 4.02 10 3 , and 7 10 AY 7.854 10 20.0 10
e
je
j
L natural =
e1 + 4.02 10 j
3
a0.800 mf
= 0.796 8 m .
The unstressed length at 30.0C is L30 C = L natural 1 + 30.0 C  0.0 C , or L30 C = 0.796 8 m 1 + 11.0 10 6 30.0 = 0.797 06 m . Since L = L30 C
a
f
b
g e ja f LM1 + T OP , where T is the tension in the string at 30.0C, N AY Q L L  1OP = e7.854 10 je20.0 10 jLM 0.800  1OP = 580 N . T = AY M NL Q N 0.797 06 Q
7 10 30 C
To find the frequency at 30.0C, realize that f1 T = so f1 = 200 Hz f1 T *P19.72
a
f
580 N = 192 Hz . 632 N
Some gas will pass through the porous plug from the reaction chamber 1 to the reservoir 2 as the reaction chamber is heated, but the net quantity of gas stays constant according to n i1 + n i 2 = n f 1 + n f 2 . Assuming the gas is ideal, we apply n = PV to each term: RT
b g b g a f a f a f a f F 5 IJ = P FG 1 + 4 IJ 1 atmG P = 1.12 atm H 300 K K H 673 K 300 K K
Pf V0 Pf 4V0 Pi 4V0 PV0 i + = + 300 K R 300 K R 673 K R 300 K R
f f
572 P19.73
Temperature
Let 2 represent the angle the curved rail subtends. We have
Li + L = 2R = Li 1 + T
and Thus, sin =
Li 2
a
f
R
=
Li 2R
=
Li 1 + T = 1 + T sin 2R
a
f a
f
FIG. P19.73
and we must solve the transcendental equation Homing in on the nonzero solution gives, to four digits, Now,
= 1 + T sin = 1.000 005 5 sin = 0.018 16 rad = 1.040 5
h = R  R cos = Li 1  cos 2 sin
a
f
b
g
a
f
This yields h = 4.54 m , a remarkably large value compared to L = 5.50 cm . *P19.74 (a) Let xL represent the distance of the stationary line below the top edge of the plate. The normal force on the lower part of the plate is mg 1  x cos and the force of kinetic friction on it is k mg 1  x cos up the roof. Again, k mgx cos acts down the roof on the upper part of the plate. The nearequilibrium of the plate requires Fx = 0 motion f kt xL f kb
a f a f
 k mgx cos + k mg 1  x cos  mg sin = 0 2 k mgx cos = mg sin  k mg cos 2 k x = k  tan x= 1 tan  2 2 k
a f
temperature rising FIG. P19.74(a)
and the stationary line is indeed below the top edge by xL = (b)
tan L 1 . 2 k motion f kt f kb xL temperature falling FIG. P19.74(b) xL xL P FIG. P19.74(c)
FG H
IJ K
With the temperature falling, the plate contracts faster than the roof. The upper part slides down and feels an upward frictional force k mg 1  x cos . The lower part slides up and feels downward frictional force k mgx cos . The equation Fx = 0 is then the same as in part (a) and the stationary line is above L tan the bottom edge by xL = . 1 k 2
a f
FG H
IJ K
(c)
Start thinking about the plate at dawn, as the temperature starts to rise. As in part (a), a line at distance xL below the top edge of the plate stays stationary relative to the roof as long as the temperature rises. The point P on the plate at distance xL above the bottom edge is destined to become the fixed point when the temperature starts falling. As the temperature rises, this point moves down the roof because of the expansion of the central part of the plate. Its displacement for the day is
continued on next page
L = 2  1 L  xL  xL T
2 1
b = b
f ga L L F tan I OPbT  T g  gML  2 G 1  NM 2 H JK QP F L tan IJ bT  T g . = b  gG H K
k h c 2 1 k h c
Chapter 19
573
At dawn the next day the point P is farther down the roof by the distance L . It represents the displacement of every other point on the plate. (d) (e)
b
2
1
gFGH L tan IJK bT  T g = FGH 24 10
k h c
6
1 1 1.20 m tan 18.5  15 10 6 32 C = 0.275 mm C C 0. 42
IJ K
If 2 < 1 , the diagram in part (a) applies to temperature falling and the diagram in part (b) applies to temperature rising. The weight of the plate still pulls it step by step down the roof. The same expression describes how far it moves each day.
ANSWERS TO EVEN PROBLEMS
P19.2 P19.4 P19.6 P19.8 P19.10 P19.12 P19.14 (a) 1.06 atm ; (b) 124 C (a) 37.0 C = 310 K ; (b) 20.6 C = 253 K TC = 1.33 C S TS + 20.0 C 0.313 m 1.20 cm 15.8 m 0.663 mm to the right at 78.2 below the horizontal (a) 0.109 cm ; (b) increase (a) 437C ; (b) 3 000C ; no (a) 2.52 10 6 N m 2 ; (b) no 0.812 cm3 (a) 396 N; (b) 101 C ; (c) no change (a) 2.99 mol ; (b) 1.80 10 24 molecules 884 balloons (a) 1.06 10 21 kg ; (b) 56.9 K P19.56 P19.58
2
P19.32 P19.34 P19.36 P19.38 P19.40 P19.42 P19.44 P19.46 P19.48 P19.50 P19.52 P19.54
(a) 900 K; (b) 1 200 K see the solution 3.96 10 2 mol 3.67 cm3 between 10 1 kg and 10 2 kg 2.41 10 11 molecules (a) 2.24 m; (b) 9.28 10 5 Pa 0.523 kg (a) see the solution; (b) << (a) 0.169 m; (b) 1.35 10 5 Pa 6.57 MPa ; (b) see the solution; r (c) it bends the other way; (d) 0.830 (a) increase by 95.0 s ; (b) loses 57.5 s (a) B = gP0 Vi P0 + gd (c) 10.3 m (a) =
b
g
P19.16 P19.18 P19.20 P19.22 P19.24 P19.26 P19.28 P19.30
b
2
 1 Li T
g
b
g
1
up; (b) decrease;
574 P19.60 P19.62 P19.64
Temperature
(a) yes; see the solution; (b) 25.7 rad s y L 2T
P19.68 P19.70 P19.72 P19.74
(a) 7.06 mm; (b) 297 K 125 N ; 42.0 m 1.12 atm (a), (b), (c) see the solution; (d) 0.275 mm; (e) see the solution
a
f
12
(a) see the solution; (b) 3.66 10 3 K 1 , within 0.06% and 0.2% of the experimental values (a) 79.1 kPa for N 2 ; 21.2 kPa for O 2 ; 940 Pa for Ar; 33.3 Pa for CO 2 ;
P19.66
(b) 81.7 L; 1.22 kg m3 ; (c) 29.0 g mol
20
Heat and the First Law of Thermodynamics
CHAPTER OUTLINE
20.1 20.2 20.3 20.4 20.5 20.6 Heat and Internal Energy Specific Heat and Calorimetry Latent Heat Work and Heat in Thermodynamic Processes The First Law of Thermodynamics Some Applications of the First Law of Thermodynamics Energy Transfer Mechanisms
ANSWERS TO QUESTIONS
Q20.1 Temperature is a measure of molecular motion. Heat is energy in the process of being transferred between objects by random molecular collisions. Internal energy is an object's energy of random molecular motion and molecular interaction. The T is twice as great in the ethyl alcohol. The final equilibrium temperature will show no significant increase over the initial temperature of the water. Some water may boil away. You would have to very precisely measure how much, and very quickly measure the temperature of the steam; it is not necessarily 100C .
Q20.2 Q20.3 Q20.4
20.7
Q20.5
The fingers are wetted to create a layer of steam between the fingers and the molten lead. The steam acts as an insulator and can prevent or delay serious burns. The molten lead demonstration is dangerous, and we do not recommend it. Heat is energy being transferred, not energy contained in an object. Further, a largemass object, or an object made of a material with high specific heat, can contain more internal energy than a highertemperature object. There are three properties to consider here: thermal conductivity, specific heat, and mass. With dry aluminum, the thermal conductivity of aluminum is much greater than that of (dry) skin. This means that the internal energy in the aluminum can more readily be transferred to the atmosphere than to your fingers. In essence, your skin acts as a thermal insulator to some degree (pun intended). If the aluminum is wet, it can wet the outer layer of your skin to make it into a good conductor of heat; then more internal energy from the aluminum can get into you. Further, the water itself, with additional mass and with a relatively large specific heat compared to aluminum, can be a significant source of extra energy to burn you. In practical terms, when you let go of a hot, dry piece of aluminum foil, the heat transfer immediately ends. When you let go of a hot and wet piece of aluminum foil, the hot water sticks to your skin, continuing the heat transfer, and resulting in more energy transfer to you! Write 1 000 kg 4 186 J kg C 1 C = V 1.3 kg m3 1 000 J kg C 1 C to find V = 3. 2 10 3 m 3 .
Q20.6
Q20.7
Q20.8
b
ga f e
jb
ga f
575
576 Q20.9
Heat and the First Law of Thermodynamics
The large amount of energy stored in concrete during the day as the sun falls on it is released at night, resulting in an higher average evening temperature than the countryside. The cool air in the surrounding countryside exerts a buoyant force on the warmer air in the city, pushing it upward and moving into the city in the process. Thus, evening breezes tend to blow from country to city. If the system is isolated, no energy enters or leaves the system by heat, work, or other transfer processes. Within the system energy can change from one form to another, but since energy is conserved these transformations cannot affect the total amount of energy. The total energy is constant. (a) and (b) both increase by minuscule amounts. The steam locomotive engine is a perfect example of turning internal energy into mechanical energy. Liquid water is heated past the point of vaporization. Through a controlled mechanical process, the expanding water vapor is allowed to push a piston. The translational kinetic energy of the piston is usually turned into rotational kinetic energy of the drive wheel. Yes. If you know the different specific heats of zinc and copper, you can determine the fraction of each by heating a known mass of pennies to a specific initial temperature, say 100C , and dumping them into a known quantity of water, at say 20C . The final temperature T will reveal the metal content:
Q20.10
Q20.11 Q20.12
Q20.13
mpennies xcCu + 1  x c Zn 100 C  T = mH 2O c H 2O T  20 C .
Since all quantities are known, except x, the fraction of the penny that is copper will be found by putting in the experimental numbers m pennies , m H 2O , T final , c Zn , and c Cu .
a f a
f
a
f
a
f
Q20.14
The materials used to make the support structure of the roof have a higher thermal conductivity than the insulated spaces in between. The heat from the barn conducts through the rafters and melts the snow. The tile is a better thermal conductor than carpet. Thus, energy is conducted away from your feet more rapidly by the tile than by the carpeted floor. The question refers to baking in a conventional oven, not to microwaving. The metal has much higher thermal conductivity than the potato. The metal quickly conducts energy from the hot oven into the center of potato. Copper has a higher thermal conductivity than the wood. Heat from the flame is conducted through the copper away from the paper, so that the paper need not reach its kindling temperature. The wood does not conduct the heat away from the paper as readily as the copper, so the energy in the paper can increase enough to make it ignite. In winter the interior of the house is warmer than the air outside. On a summer day we want the interior to stay cooler than the exterior. Heavy draperies over the windows can slow down energy transfer by conduction, by convection, and by radiation, to make it easier to maintain the desired difference in temperature. You must allow time for the flow of energy into the center of the piece of meat. To avoid burning the outside, the meat should be relatively far from the flame. If the outer layer does char, the carbon will slow subsequent energy flow to the interior.
Q20.15 Q20.16
Q20.17
Q20.18
Q20.19
Chapter 20
577
Q20.20
At night, the Styrofoam beads would decrease the overall thermal conductivity of the windows, and thus decrease the amount of heat conducted from inside to outside. The air pockets in the Styrofoam are an efficient insulator. During the winter day, the influx of sunlight coming through the window warms the living space. An interesting asidethe majority of the energy that goes into warming a home from sunlight through a window is not the infrared light given off by the sun. Glass is a relatively good insulator of infrared. If not, the window on your cooking oven might as well be just an open hole! Glass is opaque to a large portion of the ultraviolet range. The glass molecules absorb ultraviolet light from the sun and reemit the energy in the infrared region. It is this reemitted infrared radiation that contributes to warming your home, along with visible light. In winter the produce is protected from freezing. The heat capacity of the earth is so high that soil freezes only to a depth of a few decimeters in temperate regions. Throughout the year the temperature will stay nearly constant all day and night. Factors to be considered are the insulating properties of soil, the absence of a path for energy to be radiated away from or to the vegetables, and the hindrance to the formation of convection currents in the small, enclosed space. The high mass and specific heat of the barrel of water and its high heat of fusion mean that a large amount of energy would have to leak out of the cellar before the water and the produce froze solid. Evaporation of the water keeps the relative humidity high to protect foodstuffs from drying out. The sunlight hitting the peaks warms the air immediately around them. This air, which is slightly warmer and less dense than the surrounding air, rises, as it is buoyed up by cooler air from the valley below. The air from the valley flows up toward the sunny peaks, creating the morning breeze. Sunlight hits the earth and warms the air immediately above it. This warm, lessdense air rises, creating an updraft. Many raptors, like eagles, hawks and falcons use updrafts to aid in hunting. These birds can often be seen flying without flapping their wingsjust sitting in an updraft with wings extended. The bit of water immediately over the flame warms up and expands. It is buoyed up and rises through the rest of the water. Colder, more dense water flows in to take its place. Convection currents are set up. This effectively warms the bulk of the water all at once, much more rapidly than it would be by heat being conducted through the water from the flame. The porcelain of the teacup is a thermal insulator. That is, it is a thermal conductor of relatively low conductivity. When you wrap your hands around a cup of hot tea, you make A large and L small in T  Tc for the rate of energy transfer by heat from tea into you. When you hold the equation P = kA h L the cup by the handle, you make the rate of energy transfer much smaller by reducing A and increasing L. The air around the cup handle will also reduce the temperature where you are touching it. A paper cup can be fitted into a tubular jacket of corrugated cardboard, with the channels running vertically, for remarkably effective insulation, according to the same principles. As described in the answer to question 20.25, convection currents in the water serve to bring more of the heat into the water from the paper cup than the specific heats and thermal conductivities of paper and water would suggest. Since the boiling point of water is far lower than the kindling temperature of the cup, the extra energy goes into boiling the water. Keep them dry. The air pockets in the pad conduct energy by heat, but only slowly. Wet pads would absorb some energy in warming up themselves, but the pot would still be hot and the water would quickly conduct and convect a lot of energy right into you.
Q20.21
Q20.22
Q20.23
Q20.24
Q20.25
Q20.26
Q20.27
Q20.28
578 Q20.29
Heat and the First Law of Thermodynamics
The person should add the cream immediately when the coffee is poured. Then the smaller temperature difference between coffee and environment will reduce the rate of energy loss during the several minutes. The cup without the spoon will be warmer. Heat is conducted from the coffee up through the metal. The energy then radiates and convects into the atmosphere. Convection. The bridge deck loses energy rapidly to the air both above it and below it. The marshmallow has very small mass compared to the saliva in the teacher's mouth and the surrounding tissues. Mostly air and sugar, the marshmallow also has a low specific heat compared to living matter. Then the marshmallow can zoom up through a large temperature change while causing only a small temperature drop of the teacher's mouth. The marshmallow is a foam with closed cells and it carries very little liquid nitrogen into the mouth. The liquid nitrogen still on the marshmallow comes in contact with the much hotter saliva and immediately boils into cold gaseous nitrogen. This nitrogen gas has very low thermal conductivity. It creates an insulating thermal barrier between the marshmallow and the teacher's mouth (the Leydenfrost effect). A similar effect can be seen when water droplets are put on a hot skillet. Each one dances around as it slowly shrinks, because it is levitated on a thin film of steam. The most extreme demonstration of this effect is pouring liquid nitrogen into one's mouth and blowing out a plume of nitrogen gas. We strongly recommended that you read of Jearl Walker's adventures with this demonstration rather than trying it. (a) (b) (c) Warm a pot of coffee on a hot stove. Place an ice cube at 0C in warm waterthe ice will absorb energy while melting, but not increase in temperature. Let a highpressure gas at room temperature slowly expand by pushing on a piston. Work comes out of the gas in a constanttemperature expansion as the same quantity of heat flows in from the surroundings. Warm your hands by rubbing them together. Heat your tepid coffee in a microwave oven. Energy input by work, by electromagnetic radiation, or by other means, can all alike produce a temperature increase. Davy's experiment is an example of this process. This is not necessarily true. Consider some supercooled liquid water, unstable but with temperature below 0C . Drop in a snowflake or a grain of dust to trigger its freezing into ice, and the loss of internal energy measured by its latent heat of fusion can actually push its temperature up.
Q20.30 Q20.31 Q20.32
Q20.33
(d)
(e) (f)
Q20.34
Heat is conducted from the warm oil to the pipe that carries it. That heat is then conducted to the cooling fins and up through the solid material of the fins. The energy then radiates off in all directions and is efficiently carried away by convection into the air. The ground below is left frozen.
Chapter 20
579
SOLUTIONS TO PROBLEMS Section 20.1 P20.1 Heat and Internal Energy
Taking m = 1.00 kg , we have
U g = mgh = 1.00 kg 9.80 m s 2 50.0 m = 490 J . But U g = Q = mcT = 1.00 kg 4 186 J kg C T = 490 J so T = 0.117 C T f = Ti + T = P20.2
b
gb a10.0 + 0.117f C
b
ge
ja
f
g
The container is thermally insulated, so no energy flows by heat: Q=0 Eint = Q + Winput = 0 + Winput = 2mgh
and
The work on the falling weights is equal to the work done on the water in the container by the rotating blades. This work results in an increase in internal energy of the water: 2mgh = Eint = m water cT T = 2 1.50 kg 9.80 m s 2 3.00 m 2mgh 88.2 J = = 837 J C m water c 0.200 kg 4 186 J kg C
e
b
ja
g
f
= 0.105 C
FIG. P20.2
Section 20.2 P20.3
Specific Heat and Calorimetry
Q = mc silver T
1.23 kJ = 0.525 kg c silver 10.0 C c silver = 0.234 kJ kg C P20.4 From we find Q = mcT T =
b
g
a
f
1 200 J Q = = 62.0 C mc 0.050 0 kg 387 J kg C
b
g
Thus, the final temperature is 87.0C . *P20.5 We imagine the stone energy reservoir has a large area in contact with air and is always at nearly the same temperature as the air. Its overnight loss of energy is described by
P=
Q mcT = t t 6 000 J s 14 h 3 600 s h 3.02 10 8 J kg C Pt m= = = = 1.78 10 4 kg c T 850 J 20 C 850 J kg C 18 C  38 C
b b
ga fb ga
g f
a
f
580 *P20.6
Heat and the First Law of Thermodynamics
The laser energy output:
Pt = 1.60 10 13 J s 2.50 10 9 s = 4.00 10 4 J .
The teakettle input: Q = mcT = 0.800 kg 4 186 J kg C 80 C = 2.68 10 5 J . This is larger by 6.70 times. P20.7 Qcold = Q hot
e
j
b
g
amcT f
f 20.0 kg b 4 186 J kg C gdT
water
=  mcT
a
iron f
 25.0 C =  1.50 kg 448 J kg C T f  600 C
i b
gb
gd
i
T f = 29.6 C P20.8 Let us find the energy transferred in one minute. Q = m cup c cup + m water c water T
Q = 0. 200 kg 900 J kg C + 0.800 kg 4 186 J kg C
b
gb
g b
gb
g a1.50 Cf = 5 290 J
If this much energy is removed from the system each minute, the rate of removal is
P=
P20.9 (a) Qcold = Q hot
Q t
=
5 290 J = 88.2 J s = 88.2 W . 60.0 s
bm
wcw
+ m c c c T f  Tc =  mCu cCu T f  TCu  munk c unk T f  Tunk
gd
i
d
i
d
i
where w is for water, c the calorimeter, Cu the copper sample, and unk the unknown.
g a20.0  10.0f C = b50.0 g gb0.092 4 cal g C ga 20.0  80.0f C  b70.0 g gc a 20.0  100 f C 2.44 10 cal = e5.60 10 g C jc
250 g 1.00 cal g C + 100 g 0.215 cal g C
3 3
b
g
b
unk
unk
or c unk = 0.435 cal g C . (b) The material of the sample is beryllium .
Chapter 20
581
P20.10
(a)
b f gbmghg = mcT a0.600fe3.00 10
3
kg 9.80 m s 2 50.0 m
je
ja
4.186 J cal T = 0.760 C ; T = 25.8 C (b)
f = b3.00 g gb0.092 4 cal g CgaT f
No . Both the change in potential energy and the heat absorbed are proportional to the mass; hence, the mass cancels in the energy relation.
*P20.11
We do not know whether the aluminum will rise or drop in temperature. The energy the water can J 6 C = 6 279 J . The energy the copper can put absorb in rising to 26C is mcT = 0.25 kg 4 186 kg C J 74 C = 2 864 J . Since 6 279 J > 2 864 J , the final out in dropping to 26C is mcT = 0.1 kg 387 kg C temperature is less than 26C . We can write Q h = Q c as Q water + Q Al + QCu = 0 0.25 kg 4 186 J J T f  20 C + 0.4 kg 900 T f  26 C kg C kg C
d
i
d
i
+0.1 kg 387
J T f  100 C = 0 kg C
d
i
1 046.5T f  20 930 C + 360T f  9 360 C + 38.7T f  3 870 C = 0 1 445.2T f = 34 160 C T f = 23.6 C P20.12 Qcold = Q hot
i d i d i bm c + m c gT  bm c + m c gT = m c T + m c bm c + m c + m c gT = bm c + m c gT + m c T bm c + m c gT + m c T T =
m Al c Al T f  Tc + m c c w T f  Tc =  m h c w T f  Th
Al Al Al Al c w c w f Al Al f c w c h w f c h w Al Al c w h w h f Al Al c w c h w h
d
h w Th
m Al c Al + m c c w + m h c w
P20.13
The rate of collection of energy is P = 550 W m 2 6.00 m 2 = 3 300 W . The amount of energy required to raise the temperature of 1 000 kg of water by 40.0C is: Q = mcT = 1 000 kg 4 186 J kg C 40.0 C = 1.67 10 8 J Thus, or
e
j
b
ga
f
Pt = 1.67 10 8 J
t = 1.67 10 8 J = 50.7 ks = 14.1 h . 3 300 W
582 *P20.14
Heat and the First Law of Thermodynamics
Vessel one contains oxygen according to PV = nRT : nc =
3 5 3 PV 1.75 1.013 10 Pa 16.8 10 m = = 1.194 mol . 8.314 Nm mol K 300 K RT
e
j
Vessel two contains this much oxygen: nh = (a) 2. 25 1.013 10 5 22.4 10 3 8.314 450
e
a f
j
mol = 1.365 mol .
The gas comes to an equilibrium temperature according to
amcT f n McdT
c
cold f
=  mcT
a
 300 K + n h Mc T f  450 K = 0
i
f
hot
d
i
The molar mass M and specific heat divide out: 1.194T f  358.2 K + 1.365T f  614.1 K = 0 Tf = (b) 972.3 K = 380 K 2.559
The pressure of the whole sample in its final state is P= 2.559 mol 8.314 J 380 K nRT = = 2.06 10 5 Pa = 2.04 atm . V mol K 22.4 + 16.8 10 3 m3
a
f
Section 20.3 P20.15
Latent Heat
The heat needed is the sum of the following terms: Q needed = heat to reach melting point + heat to melt
f b g a +b heat to reach melting point g + b heat to vaporizeg + a heat to reach 110 C f ga f e j +b 4 186 J kg C ga100 C f + e 2.26 10 J kg j + b 2 010 J kg C ga10.0 C f b
6
Thus, we have Q needed = 0.040 0 kg 2 090 J kg C 10.0 C + 3.33 10 5 J kg
Q needed = 1.22 10 5 J P20.16 Qcold = Q hot
bm
wcw
+ m c c c T f  Ti =  m s  L v + c w T f  100
0.250 kg 4 186 J kg C + 0.050 0 kg 387 J kg C 50.0 C  20.0 C =  m s 2.26 10 6 ms =
b
gd
i
g
f ga J kg + b 4 186 J kg C ga50.0 C  100 C f b
d
i
3.20 10 4 J = 0.012 9 kg = 12.9 g steam 2.47 10 6 J kg
Chapter 20
583
P20.17
The bullet will not melt all the ice, so its final temperature is 0C. Then
FG 1 mv H2
2
+ mc T
IJ K
bullet
= mwL f
where m w is the melt water mass mw = mw = P20.18 (a) 0.500 3.00 10 3 kg 240 m s 86.4 J + 11.5 J = 0.294 g 333 000 J kg
e
jb
g
2
+ 3.00 10 3 kg 128 J kg C 30.0 C
5
b
ga
f
3.33 10
J kg
Q1 = heat to melt all the ice = 50.0 10 3 kg 3.33 10 5 J kg = 1.67 10 4 J Q 2 = heat to raise temp of ice to 100 C
3
b = e50.0 10
e
kg 4 186 J kg C 100 C = 2.09 10 4 J
jb
ga
g
je
j
f
Thus, the total heat to melt ice and raise temp to 100C = 3.76 10 4 J Q3 = heat available = 10.0 10 3 kg 2.26 10 6 J kg = 2.26 10 4 J as steam condenses
e
je
j
Thus, we see that Q3 > Q1 , but Q3 < Q1 + Q 2 . Therefore, all the ice melts but T f < 100 C . Let us now find T f Qcold = Q hot
e50.0 10 kg je3.33 10 J kg j + e50.0 10 kg jb4 186 J kg CgdT  0 Ci = e10.0 10 kg je 2.26 10 J kg j  e10.0 10 kg jb 4 186 J kg C gdT  100 C i
3 5 3 f 3 6 3 f
From which, T f = 40.4 C . (b) Q1 = heat to melt all ice = 1.67 10 4 J [See part (a)] heat given up = 10 3 kg 2.26 10 6 J kg = 2.26 10 3 J Q2 = as steam condenses Q3
je heat given up as condensed = = e10 steam cools to 0 C
e
j
3
kg 4 186 J kg C 100 C = 419 J
jb
ga
f
Note that Q 2 + Q3 < Q1 . Therefore, the final temperature will be 0C with some ice remaining. Let us find the mass of ice which must melt to condense the steam and cool the condensate to 0C.
mL f = Q 2 + Q3 = 2.68 10 3 J
Thus, m = 2.68 10 3 J = 8.04 10 3 kg = 8.04 g . 5 3.33 10 J kg
Therefore, there is 42.0 g of ice left over .
584 P20.19
Heat and the First Law of Thermodynamics
e j 1.00 kg b0.092 0 cal g C ga 293  77.3f C = mb 48.0 cal g g
Q = mCu cCu T = m N 2 L vap
N2
m = 0.414 kg *P20.20 The original gravitational energy of the hailstoneEarth system changes entirely into additional internal energy in the hailstone, to produce its phase change. No temperature change occurs, either in the hailstone, in the air, or in sidewalk. Then mgy = mL y= P20.21 (a)
L 3.33 10 5 J kg 1 kg m 2 s 2 = = 3.40 10 4 m 1J g 9.8 m s 2
F GH
I JK
Since the heat required to melt 250 g of ice at 0C exceeds the heat required to cool 600 g of water from 18C to 0C, the final temperature of the system (water + ice) must be 0C . Let m represent the mass of ice that melts before the system reaches equilibrium at 0C. Qcold = Q hot m 3.33 10
(b)
mL f =  m w c w 0 C  Ti
e
5
g J kg j = b0.600 kg gb 4 186 J kg C ga0 C  18.0 C f
b
m = 136 g, so the ice remaining = 250 g  136 g = 114 g P20.22 The original kinetic energy all becomes thermal energy: 1 1 1 5.00 10 3 kg 500 m s mv 2 + mv 2 = 2 2 2 2 Raising the temperature to the melting point requires Q = mcT = 10.0 10 3 kg 128 J kg C 327 C  20.0 C = 393 J . Since 1 250 J > 393 J , the lead starts to melt. Melting it all requires Q = mL = 10.0 10 3 kg 2.45 10 4 J kg = 245 J . Since 1 250 J > 393 + 245 J , it all melts. If we assume liquid lead has the same specific heat as solid lead, the final temperature is given by 1.25 10 3 J = 393 J + 245 J + 10.0 10 3 kg 128 J kg C T f  327 C T f = 805 C
FG IJ e H K
jb
g
2
= 1.25 kJ .
b
ga
f
e
je
j
b
gd
i
Chapter 20
585
Section 20.4 P20.23
Work and Heat in Thermodynamic Processes
Wif =  PdV The work done on the gas is the negative of the area under the curve P = V 2 between Vi and V f . 1 Wif =  V 2 dV =  V f3  Vi3 3 i
i
z
f
P f P = V 2
z
f
e
j
3 3 3 3
i
O 1.00 m
3
V 2.00 m
3
V f = 2Vi = 2 1.00 m 3 = 2.00 m 3 1 5.00 atm m6 1.013 10 5 Pa atm Wif =  3
e
j
e
je
P20.24
(a)
W =  PdV
W =  6.00 10 6 Pa 2.00  1.00 m3 +
6 3 3
f e ja  e 4.00 10 Pa ja3.00  2.00f m  e 2.00 10 Paja 4.00  3.00f m
6
z
j LNMe2.00 m j + e1.00 m j OQP =
FIG. P20.23 1.18 MJ
+
Wi f = 12.0 MJ (b) W f i = +12.0 MJ FIG. P20.24 P20.25 P20.26 W =  P V =  P
FG nR IJ dT H PK
f
 Ti = nRT =  0.200 8.314 280 = 466 J
i
a
fa
fa f
W =  PdV =  P dV =  PV = nRT = nR T2  T1
i i
z
f
z
f
b
g
P20.27
During the heating process P =
FG P IJ V . HV K
i i
(a)
W =  PdV =  W =
z
f i
3Vi
FG P IJ V HV K 2
i i
Vi 2 3Vi Vi
z FGH
Pi VdV Vi Pi 9Vi2  Vi2 = 4PVi i 2Vi
IJ K
=
e
j
(b)
PV = nRT
i i
LMF P I V OPV = nRT MNGH V JK PQ F P IJ V T =G H nRV K
i 2 i
Temperature must be proportional to the square of volume, rising to nine times its original value.
586
Heat and the First Law of Thermodynamics
Section 20.5 P20.28 (a) (b) P20.29
The First Law of Thermodynamics W =  PV =  0.800 atm 7.00 L 1.013 10 5 Pa atm 10 3 m 3 L = +567 J Eint = Q + W = 400 J + 567 J = 167 J
a
fa
fe
je
j
Eint = Q + W Q = Eint  W = 500 J  220 J = 720 J The negative sign indicates that positive energy is transferred from the system by heat.
P20.30
(a)
Q = W = Area of triangle Q= 1 4.00 m 3 6.00 kPa = 12.0 kJ 2
e
ja
f
(b)
Q = W = 12.0 kJ
FIG. P20.30 P20.31 BC CA AB P20.32 WBC
bQ = E since W = 0g bE < 0 and W > 0 , so Q < 0g + + bW < 0 , E > 0 since E < 0 for B C A ; so Q > 0g =  P bV  V g = 3.00 atmb0.400  0.090 0 g m
int BC
Q
W 0 +
Eint
int
int
int
B
C
B
3
P(atm) 3.0
= 94.2 kJ Eint = Q + W Eint, C  Eint, B = 100  94.2 kJ Eint, C  Eint, B = 5.79 kJ Since T is constant,
B
C
a
f
1.0
A
D
0.090 0.20
0.40
1.2
V(m 3)
Eint, D  Eint, C = 0
WDA =  PD VA  VD = 1.00 atm 0.200  1.20 m3 = +101 kJ Eint, A  Eint, D = 150 kJ + +101 kJ = 48.7 kJ Now, Eint, B  Eint, A =  Eint, C  Eint, B + Eint, D  Eint, C + Eint, A  Eint, D Eint, B  Eint, A =  5.79 kJ + 0  48.7 kJ = 42.9 kJ
b
g
a
f
FIG. P20.32
a
f
d
i d
i d
i
Chapter 20
587
*P20.33
The area of a true semicircle is
1 2 r . The arrow in Figure P20.33 2 looks like a semicircle when the scale makes 1.2 L fill the same space as 100 kPa. Its area is 1 1 2.4 L 200 kPa = 2.4 10 3 m3 2 10 5 N m 2 . 2 2
P(kPa) 500 300 A B
a
fa
f
e
je
j
The work on the gas is W =  PdV =  area under the arch shown in the graph
B
0
FG 1 2.4a200f J + 3 10 N m H2 = b754 J + 1 440 Jg = 2 190 J
=
5
A
z
1.2
3.6
6.0
V (L)
FIG. P20.33
2
4.8 10 3 m3
IJ K
Eint = Q + W = 5 790 J  2 190 J = 3.60 kJ
Section 20.6 P20.34 (a)
Some Applications of the First Law of Thermodynamics
FG V IJ =  P V lnFG V IJ HV K HV K F W I = b0.025 0g expLM 3 000 OP = so V = V expG + MN 0.025 0e1.013 10 j PQ H P V JK
W = nRT ln
f i f f f i i f f f 5
0.007 65 m 3
(b) P20.35 (a) (b)
Tf =
Pf V f nR
=
1.013 10 5 Pa 0.025 0 m 3 1.00 mol 8.314 J K mol
b
e
j= g
305 K
Eint = Q  PV = 12.5 kJ  2.50 kPa 3.00  1.00 m 3 = 7.50 kJ V1 V2 = T1 T2 V 3.00 300 K = 900 K T2 = 2 T1 = 1.00 V1
a
f
a
f
P20.36
(a)
W =  PV =  P 3VT
=  1.013 10 5 N m 2 3 24.0 10 6 C 1 W = 48.6 mJ (b) (c)
e
L jMM e N gb
O F .00 I jGH 2.70 110 kg m JK a18.0 CfPP kg Q
3 3
Q = cmT = 900 J kg C 1.00 kg 18.0 C = 16.2 kJ Eint = Q + W = 16. 2 kJ  48.6 mJ = 16.2 kJ
b
ga
f
588 P20.37
Heat and the First Law of Thermodynamics
OP a f LM MN e je j PQ F 18.0 g I = W = a1.00 molfb8.314 J K molga373 K f + e1.013 10 N m jG H 10 g m JK Q = mL = 0.018 0 kg e 2.26 10 J kg j = 40.7 kJ
W =  PV =  P Vs  Vw = 
b
g
P nRT 18.0 g +P P 1.00 g cm 3 10 6 cm 3 m3
5 2 6
3
3.10 kJ
v
6
Eint = Q + W = 37.6 kJ P20.38 (a) The work done during each step of the cycle equals the negative of the area under that segment of the PV curve. W = WDA + W AB + WBC + WCD W =  Pi Vi  3Vi + 0  3 Pi 3Vi  Vi + 0 = 4PVi i (b) The initial and final values of T for the system are equal. Therefore, Eint = 0 and Q = W = 4PVi . i W = 4PVi = 4nRTi = 4 1.00 8.314 273 = 9.08 kJ i
b
g
b
g
(c) P20.39 (a)
a fa b
fa f
FIG. P20.38
PVi = Pf V f = nRT = 2.00 mol 8.314 J K mol 300 K = 4.99 10 3 J i
Vi = Vf = nRT 4.99 10 J = 0.400 atm Pi nRT 4.99 10 3 J 1 = = Vi = 0.041 0 m3 1.20 atm Pf 3
3
ga
f
(b) (c)
W =  PdV = nRT ln Eint = 0 = Q + W Q = 5.48 kJ
z
FG V IJ = e4.99 10 j lnFG 1 IJ = H 3K HV K
f 3 i
+5.48 kJ
P20.40
Eint, ABC = Eint, AC
(a)
(conservation of energy) (First Law)
Eint, ABC = Q ABC + WABC
Q ABC = 800 J + 500 J = 1 300 J
(b)
WCD =  PC VCD , VAB =  VCD , and PA = 5 PC 1 1 Then, WCD = PA VAB =  W AB = 100 J 5 5 (+ means that work is done on the system) WCDA = WCD so that QCA = Eint, CA  WCDA = 800 J  100 J = 900 J ( means that energy must be removed from the system by heat) FIG. P20.40
(c)
(d)
Eint, CD = Eint, CDA  Eint, DA = 800 J  500 J = 1 300 J
and QCD = Eint, CD  WCD = 1 300 J  100 J = 1 400 J
Chapter 20
589
Section 20.7 P20.41
Energy Transfer Mechanisms
P = kA
P20.42 P20.43
f kAT b0.800 W m C ge3.00 m ja 25.0 C f P= = = 1.00 10
2
T L P L 10.0 W 0.040 0 m k= = = 2.22 10 2 W m C AT 1.20 m 2 15.0 C
b
a
g
L
6.00 10 3 m
4
W = 10.0 kW
In the steady state condition, so that In this case
PAu = PAg
k Au A Au
FG T IJ H x K
Au
= k Ag A Ag
FG T IJ H x K
Ag
A Au = A Ag x Au = x Ag TAu = 80.0  T
and
TAg
a f = aT  30.0f f a f
FIG. P20.43
where T is the temperature of the junction. Therefore, k Au 80.0  T = k Ag T  30.0
a
And AT
T = 51.2 C
P20.44
P=
k
i
Li
i
=
2 4.00 10 3 m
e
j
e6.00 m ja50.0 Cf
2
0.800 W m C + 5.00 10 3 m
0.023 4 W m C
= 1.34 kW
*P20.45
We suppose that the area of the transistor is so small that energy flow by heat from the transistor directly to the air is negligible compared to energy conduction through the mica.
P = kA
bT  T g
h c
L
Th = Tc + P20.46
1.50 W 0.085 2 10 3 m PL = 35.0 C + = 67.9 C kA 0.075 3 W m C 8.25 6.25 10 6 m 2
b
e
ga
f
j
From Table 20.4, (a) (b)
R = 0.890 ft 2 F h Btu
The insulating glass in the table must have sheets of glass less than estimate the Rvalue of a 0.250inch air space as Then for the double glazing Rb = 0.890 + 1 inch thick. So we 8
0.250 times that of the thicker air space. 3.50
2
LM N
FG 0.250 IJ 1.01 + 0.890OP ft F h = H 3.50 K Q Btu
1.85
ft 2 F h . Btu 1.85 = 2.08 . 0.890
(c)
Since A and T2  T1 are constants, heat flow is reduced by a factor of
b
g
590 P20.47
Heat and the First Law of Thermodynamics
P = AeT 4 = 5.669 6 10 8 W m 2 K 4 4 6.96 10 8 m
e
jLNM e
j OQPa0.965fb5 800 K g
2
4
P = 3.77 10 26 W
P20.48 Suppose the pizza is 70 cm in diameter and = 2.0 cm thick, sizzling at 100C. It cannot lose heat by conduction or convection. It radiates according to P = AeT 4 . Here, A is its surface area, A = 2 r 2 + 2 r = 2 0.35 m
a
f
2
+ 2 0.35 m 0.02 m = 0.81 m 2 .
a
fa
f
Suppose it is dark in the infrared, with emissivity about 0.8. Then
P = 5.67 10 8 W m 2 K 4 0.81 m 2 0.80 373 K
If the density of the pizza is half that of water, its mass is m = V = r 2 = 500 kg m3 0.35 m
e
je
ja fa ja
f
4
= 710 W ~ 10 3 W .
e
f a0.02 mf = 4 kg .
2
Suppose its specific heat is c = 0.6 cal g C . The drop in temperature of the pizza is described by: Q = mc T f  Ti
d
i g
P=
dT f dt P20.49
dT f dQ = mc 0 dt dt = 710 J s P = = 0.07 C s ~ 10 1 K s mc 4 kg 0.6 4 186 J kg C
b gb je
P = AeT 4 2.00 W = 5.67 10 8 W m 2 K 4 0.250 10 6 m 2 0.950 T 4
e
ja
f
T = 1.49 10 14 K 4 P20.50
e
j
14
= 3.49 10 3 K
We suppose the earth below is an insulator. The square meter must radiate in the infrared as much energy as it absorbs, P = AeT 4 . Assuming that e = 1.00 for blackbody blacktop: 1 000 W = 5.67 10 8 W m 2 K 4 1.00 m 2 1.00 T 4 T = 1.76 10 10 K
e
je
ja f
e
4 14
j
= 364 K (You can cook an egg on it.)
P20.51
The sphere of radius R absorbs sunlight over the area of its day hemisphere, projected as a flat circle perpendicular to the light: R 2 . It radiates in all directions, over area 4 R 2 . Then, in steady state,
Pin = Pout
e 1 340 W m 2 R 2 = e 4 R 2 T 4 The emissivity e, the radius R, and all cancel.
e
j
e
j
L 1 340 W m Therefore, T = M MN 4e5.67 10 W m
2 8
2
OP K jP Q
4
14
= 277 K = 4 C .
Chapter 20
591
Additional Problems P20.52 77.3 K = 195.8C is the boiling point of nitrogen. It gains no heat to warm as a liquid, but gains heat to vaporize: Q = mL v = 0.100 kg 2.01 10 5 J kg = 2.01 10 4 J . The water first loses heat by cooling. Before it starts to freeze, it can lose Q = mcT = 0.200 kg 4 186 J kg C 5.00 C = 4.19 10 3 J . The remaining 2.01 10 4  4.19 10 3 J = 1.59 10 4 J that is removed from the water can freeze a mass x of water: Q = mL f 1.59 10 4 J = x 3.33 10 5 J kg
b
ge
j
b
gb
ga
f
e
j
e
j
j
x = 0.047 7 kg = 47.7 g of water can be frozen P20.53 The increase in internal energy required to melt 1.00 kg of snow is Eint = 1.00 kg 3.33 10 5 J kg = 3.33 10 5 J The force of friction is f = n = mg = 0. 200 75.0 kg 9.80 m s 2 = 147 N
b
ge
b
ge
j
According to the problem statement, the loss of mechanical energy of the skier is assumed to be equal to the increase in internal energy of the snow. This increase in internal energy is Eint = fr = 147 N r = 3.33 10 5 J and P20.54 (a)
a
f
r = 2.27 10 3 m .
The energy thus far gained by the copper equals the energy loss by the silver. Your down parka is an excellent insulator. Qcold = Q hot or
d i = m c dT  T i b9.00 g gb387 J kg Cga16.0 Cf = b14.0 g gb234 J kg CgdT dT  30.0 Ci = 17.0 C
mCu c Cu T f  Ti
Cu Ag Ag f i Ag f Ag
f
 30.0 C
i
Ag
so (b)
T f , Ag = 13.0 C .
Differentiating the energy gainandloss equation gives: m Ag c Ag
FG dT IJ H dt K
Ag
=  mCu c Cu
FG dT IJ H dt K
Cu
FG dT IJ H dt K FG dT IJ H dt K
=
Ag
=
Ag
FG IJ =  9.00 gb387 J kg Cg b+0.500 C sg H K 14.0 gb234 J kg Cg 0.532 C s b negative sign decreasing temperatureg
mCu c Cu dT m Ag c Ag dt
Cu
592 P20.55
Heat and the First Law of Thermodynamics
(a)
Before conduction has time to become important, the energy lost by the rod equals the energy gained by the helium. Therefore, mL v He = mc T Al or so
b g c h bVL g = cVc T h cVc T h V = bL g e2.70 g cm je62.5 cm jb0.210 cal g Cga295.8 Cf V = e0.125 g cm je2.09 10 J kg jb1.00 cal 4.186 Jgb1.00 kg 1 000 g g
v He Al He Al v He 3 3 He 3 4
VHe = 1.68 10 4 cm3 = 16.8 liters (b) The rate at which energy is supplied to the rod in order to maintain constant temperatures is given by dT 295.8 K P = kA = 31.0 J s cm K 2.50 cm 2 = 917 W dx 25.0 cm This power supplied to the helium will produce a "boiloff" rate of
FG IJ b H K
e a
ge
jFGH
IJ K
917 W 10 3 g kg P = = 351 cm 3 s = 0.351 L s L v 0.125 g cm3 2.09 10 4 J kg
fe je
j
j
*P20.56
At the equilibrium temperature Teq the diameters of the sphere and ring are equal: d s + d s Al Teq  Ti = d r + d r Cu Teq  15 C
5 eq
j 5.01 cm + 5.01 cme 2.40 10
319.95 C + Teq = 3.412 0Ti
e
j 1 C jeT  T j = 5.00 cm + 5.00 cme1.70 10
i
e
5
1 C Teq  15 C
je
j
0.01 C + 1.202 4 10 4 Teq  1.202 4 10 4 Ti = 8.5 10 5 Teq  1.275 10 3 C 1.127 5 10 2 C + 3.524 10 5 Teq = 1.202 4 10 4 Ti
At the equilibrium temperature, the energy lost is equal to the energy gained: m s c Al Teq  Ti =  m r cCu Teq  15 C
e
j
e
j e j
10.9 g 0.215 cal g C Teq  Ti = 25 g 0.092 4 cal g C Teq  15 C 2.343 5Teq  2.343 5Ti = 34.65 C  2.31Teq 4.653 5Teq = 34.65 C + 2.343 5Ti Solving by substitution, 4.653 5 3.412 0Ti  319.95 C = 34.65 C + 2.343 5Ti 15.877 7Ti  1 488.89 C = 34.65 C + 2.343 5Ti (b) (a) Ti = 1 523.54 C = 113 C 13.534
e
j
b
g
Teq = 319.95 + 3.412 0 112.57 = 64.1 C
a
f
Chapter 20
593
P20.57
Q = mcT = V cT so that when a constant temperature difference T is maintained, the rate of adding energy to the liquid is P = and the specific heat of the liquid is c = dQ dV = cT = RcT dt dt
b g
FG IJ H K
P . RT
P20.58
(a)
Work done by the gas is the negative of the area under the PV curve W =  Pi
FG V  V IJ = H2 K
i i
+
PVi i . 2
(b)
In this case the area under the curve is W =  PdV . Since the process is isothermal, PV = PVi = 4Pi i
Vi 4 Vi
z
and
FG V IJ = nRT H 4K F dV IJ bPV g =  PV lnFG V 4 IJ = PV ln 4 W = z G HVK HV K
i i i i i i i i i i
FIG. P20.58
= +1.39 PVi i (c) P20.59 The area under the curve is 0 and W = 0 .
Call the initial pressure P1 . In the constant volume process 1 2 the work is zero. P1V1 = nRT1 P2 V2 = nRT2 so P2 V2 T2 1 = ; T2 = 300 K 1 = 75.0 K 4 P1V1 T1
FG H
IJ a f K g
Now in 2 3 W =  PdV =  P2 V3  V2 =  P3V3 + P2V2
2
z
3
b
W = nRT3 + nRT2 =  1.00 mol 8.314 J mol K 300 K  75.0 K W = 1.87 kJ
a
fb
ga
f
594 *P20.60
Heat and the First Law of Thermodynamics
The initial moment of inertia of the disk is 1 1 1 1 4 MR 2 = VR 2 = R 2 tR 2 = 8 920 kg m3 28 m 1.2 m = 1.033 10 10 kg m 2 2 2 2 2
e
ja
f
The rotation speeds up as the disk cools off, according to I i i = I f f
2 1 1 1 MRi2 i = MR 2 f = MRi2 1  T f f 2 2 2 1 1 f = i = 25 rad s 2 1  T 1  17 10 6 1 C 830 C
c
h
c
h
e
j
2
= 25.720 7 rad s
(a)
The kinetic energy increases by 1 1 1 1 1 I f 2  I i i2 = I i i f  I i i2 = I i i f  i f 2 2 2 2 2 1 10 2 = 1.033 10 kg m 25 rad s 0.720 7 rad s = 9.31 10 10 J 2
d
i
b
g
(b) (c)
Eint = mcT = 2.64 10 7 kg 387 J kg C 20 C  850 C = 8.47 10 12 J
As 8.47 10 12 J leaves the fund of internal energy, 9.31 10 10 J changes into extra kinetic energy, and the rest, 8.38 10 12 J is radiated.
b
ga
f
*P20.61
The loss of mechanical energy is GM E m 1 1 = 670 kg 1.4 10 4 m s mvi2 + RE 2 2
e
j
2
+
6.67 10 11 Nm 2 5.98 10 24 kg 670 kg kg 2 6.37 10 6 m
= 6.57 10 10 J + 4.20 10 10 J = 1.08 10 11 J One half becomes extra internal energy in the aluminum: Eint = 5.38 10 10 J. To raise its temperature to the melting point requires energy mcT = 670 kg 900 J 660  15 C = 4.07 10 8 J . kg C
c
a
fh
To melt it, mL = 670 kg 3.97 10 5 J kg = 2.66 10 8 J . To raise it to the boiling point, mcT = 670 1 170 2 450  600 J = 1.40 10 9 J . To boil it, mL = 670 kg 1.14 10 7 J kg = 7.64 10 9 J . Then 5.38 10 10 J = 9.71 10 9 J + 670 1 170 T f  2 450 C J C T f = 5.87 10 4 C
b
gb
g
b
gd
i
Chapter 20
595
P20.62
(a) (b)
Fv = 50.0 N 40.0 m s = 2 000 W Energy received by each object is 1 000 W 10 s = 10 4 J = 2 389 cal . The specific heat of iron is 0.107 cal g C , so the heat capacity of each object is 5.00 10 3 0.107 = 535.0 cal C. T = 2 389 cal = 4.47 C 535.0 cal C
a
fb
g
b
ga f
P20.63
The power incident on the solar collector is
Pi = IA = 600 W m 2 0.300 m
e
j a
f
2
= 170 W .
For a 40.0% reflector, the collected power is Pc = 67.9 W. The total energy required to increase the temperature of the water to the boiling point and to evaporate it is Q = cmT + mLV :
Q = 0.500 kg 4 186 J kg C 80.0 C + 2.26 10 6 J kg = 1.30 10 6 J .
b
ga
f
The time interval required is t = P20.64
Q
Pc
=
1.30 10 6 J = 5.31 h . 67.9 W
FIG. P20.63
From Q = mLV the rate of boiling is described by
P=
Model the water vapor as an ideal gas P0 V0 = nRT =
0
Q LV m = t t
P m = t LV
FG m IJ RT H MK P V m F RT I = G J t t H M K P F RT I P Av = G J L HMK
0 V
v=
1 000 W 8.314 J mol K 373 K P RT = MLV P0 A 0.018 0 kg mol 2.26 10 6 J kg 1.013 10 5 N m 2 2.00 10 4 m 2
b
ge
b
je
ga
f
je
j
v = 3.76 m s
596 P20.65
Heat and the First Law of Thermodynamics
Energy goes in at a constant rate P . For the period from 50.0 min to 60.0 min, Q = mcT
T( C)
P 10.0 min = 10 kg + mi 4 186 J kg C 2.00 C  0 C (1) P 10.0 min = 83.7 kJ + 8.37 kJ kg mi
a a a
f b f f
b
gb
g
ga
f
2.00 1.00 0.00 3.00
For the period from 0 to 50.0 min, Q = mi L f
P 50.0 min = mi 3.33 10 5 J kg
Substitute P = mi 3.33 10 5 J kg 50.0 min
e
j
20.0
40.0
60.0 t (min)
e
j into Equation (1) to find b g
FIG. P20.65
= 83.7 kJ + 8.37 kJ kg mi 5.00 83.7 kJ = 1.44 kg mi = 66.6  8.37 kJ kg
mi 3.33 10 5 J kg
e
j
a
f
P20.66
(a)
The block starts with K i =
1 1 mvi2 = 1.60 kg 2.50 m s 2 2
b
gb
g
2
= 5.00 J
All this becomes extra internal energy in ice, melting some according to "Q " = m ice L f . Thus, the mass of ice that melts is m ice = For the block: "Q " K i 5.00 J = = = 1.50 10 5 kg = 15.0 mg . Lf L f 3.33 10 5 J kg
Q = 0 (no energy flows by heat since there is no temperature difference)
W = 5.00 J
Eint = 0 (no temperature change) and For the ice,
K = 5.00 J
Q=0
W = +5.00 J
Eint = +5.00 J and (b) K = 0
Again, K i = 5.00 J and m ice = 15.0 mg For the block of ice: Q = 0; Eint = +5.00 J ; K = 5.00 J so W = 0 . For the copper, nothing happens: Q = Eint = K = W = 0 .
continued on next page
Chapter 20
597
(c)
Again, K i = 5.00 J. Both blocks must rise equally in temperature. "Q " = mcT : T = "Q " 5.00 J = = 4.04 10 3 C mc 2 1.60 kg 387 J kg C
b
gb
g
At any instant, the two blocks are at the same temperature, so for both Q = 0. For the moving block: and so For the stationary block: and so
K = 5.00 J
Eint = +2.50 J
W = 2.50 J
K = 0 Eint = +2.50 J
W = +2.50 J
For each object in each situation, the general continuity equation for energy, in the form K + Eint = W + Q , correctly describes the relationship between energy transfers and changes in the object's energy content. P20.67 A = Aend walls + A ends of attic + A side walls + Aroof A = 2 8.00 m 5.00 m + 2 2
a f LMN 1 4.00 m a4.00 mf tan 37.0OPQ 2 F 4.00 m IJ +2a10.0 m 5.00 mf + 2a10.0 mfG H cos37.0 K
4.80 10 4 kW m C 304 m 2 25.0 C kAT = = 17.4 kW = 4.15 kcal s L 0.210 m
A = 304 m 2
P=
e
je
ja
f
Thus, the energy lost per day by heat is 4.15 kcal s 86 400 s = 3.59 10 5 kcal day . The gas needed to replace this loss is T LAdx = kA dt x L L
8.00 4.00
b
gb
g
3.59 10 5 kcal day 9 300 kcal m3
= 38.6 m3 day .
P20.68
FG IJ H K
t 0
z
xdx = kT dt = kTt
z
x 2
2 8.00 4.00
e
3.33 10 5 J kg 917 kg m3
je
F b0.080 0 mg  b0.040 0 mg jGG 2 H
2
2
I JJ = b2.00 W m Cga10.0 Cft K
t = 3.66 10 4 s = 10.2 h
598 P20.69
Heat and the First Law of Thermodynamics
W = W AB + WBC + WCD + WDA W =  PdV  PdV  PdV  PdV
A B
P
A
z
C
W = nRT1
dV dV  P2 dV  nRT2  P1 dV V V A B C D
z
B
z
D C
z
B
C
z
D
z
P2
B
C
D
z
A
z
W = nRT1 ln
FG V IJ  P bV HV K
B 1 2
C
 VB  nRT2 ln
g
FG V IJ  P bV HV K
2 C 1
P1
A
A V1
D V2 V
 VD
g
Now P1VA = P2 VB and P2 VC = P1VD , so only the logarithmic terms do not cancel out. Also, VB P1 V P = and 2 = 2 V1 P2 VC P1
FIG. P20.69
W = nRT1 lnG P1 J  nRT2 lnG P2 J = +nRT1 lnG P2 J  nRT2 lnG P2 J = nRbT2  T1 g lnG P2 J H K H K H K H K H K
2 1 1 1 1
FP I
FP I
FP I
FP I
FP I
Moreover P1V2 = nRT2 and P1V1 = nRT1
W =
P20.70
 P1 V2  V1 ln
b
g FGH P IJK P
2 1
For a cylindrical shell of radius r, height L, and thickness dr, the equation for thermal conduction, dQ dT =  kA dt dx becomes dQ dT =  k 2 rL dt dr
b
g
Under equilibrium conditions, dT =  1 dQ dt 2 kL
F GH
I FG dr IJ JK H r K
dQ is constant; therefore, dt Tb  Ta = 
and
1 dQ b ln dt 2 kL a
F GH
I FG IJ JK H K
But Ta > Tb , so P20.71
dQ 2 kL Ta  Tb = dt ln b a
b g b g
From problem 70, the rate of energy flow through the wall is dQ 2 kL Ta  Tb = dt ln b a
5 dQ 2 4.00 10 cal s cm C 3 500 cm 60.0 C = dt ln 256 cm 250 cm
e
b g b g b
jb
g
ga
f
dQ = 2. 23 10 3 cal s = 9.32 kW dt This is the rate of energy loss from the plane by heat, and consequently is the rate at which energy must be supplied in order to maintain a constant temperature. FIG. P20.71
Chapter 20
599
P20.72
Qcold = Q hot or Q Al =  Q water + Qcalo m Al c Al
f
b0.200 kg gc a+39.3 Cf =  0.400 kgb4 186 J kg Cg + 0.040 0 kgb630 J kg Cg a3.70 Cf
Al
b dT
g gd i
w
 Ti
i
Al
=  m w c w + m c c c T f  Ti
b
c Al = *P20.73 (a) (b)
6.29 10 3 J = 800 J kg C 7.86 kg C
P = AeT 4 = 5.67 10 8 W m 2 K 4 5.1 10 14 m 2 0.965 5 800 K
Tavg = 0.1 4 800 K + 0.9 5 890 K = 5.78 10 3 K
This is cooler than 5 800 K by 5 800  5 781 = 0.327% . 5 800
4
e
j
a
fb
g
4
= 3.16 10 22 W
b
g
b
g
(c)
P = 5.67 10 8 W m 2 K 4 0.1 5.1 10 14 m 2 0.965 4 800 K
+5.67 10 8
14 4 22
e
g j e j b W 0.9e5.1 10 j0.965b5 890g = 3.17 10 W
1.29 10 20 W = 0.408%. 3.16 10 22 W
This is larger than 3.158 10 22 W by
ANSWERS TO EVEN PROBLEMS
P20.2 P20.4 P20.6 P20.8 P20.10 P20.12 P20.14 P20.16 P20.18 P20.20 0.105C 87.0C The energy input to the water is 6.70 times larger than the laser output of 40.0 kJ. 88.2 W (a) 25.8C ; (b) no Tf P20.22 P20.24 P20.26 P20.28 P20.30 liquid lead at 805C (a) 12.0 MJ ; (b) +12.0 MJ nR T2  T1
b
g
(a) 567 J ; (b) 167 J (a) 12.0 kJ; (b) 12.0 kJ 42.9 kJ (a) 7.65 L; (b) 305 K (a) 48.6 mJ ; (b) 16.2 kJ; (c) 16.2 kJ (a) 4PVi ; (b) +4PVi ; (c) 9.08 kJ i i (a) 1 300 J ; (b) 100 J ; (c) 900 J ; (d) 1 400 J 10.0 kW
bm =
Al c Al
+ m c c w Tc + m h c w Th
g
P20.32 P20.34 P20.36 P20.38 P20.40 P20.42
m Al c Al + m c c w + m h c w
(a) 380 K ; (b) 206 kPa 12.9 g (a) all the ice melts; 40.4C ; (b) 8.04 g melts; 0C 34.0 km
600 P20.44 P20.46
Heat and the First Law of Thermodynamics
1.34 kW (a) 0.890 ft 2 F h Btu ; (b) 1.85 (c) 2.08 ft 2 F h ; Btu
P20.62 P20.64 P20.66
(a) 2 000 W ; (b) 4.47C 3.76 m s (a) 15.0 mg ; block: Q = 0; W = 5.00 J ; Eint = 0 ; K = 5.00 J ; ice: Q = 0; W = 5.00 J ; Eint = 5.00 J ; K = 0 (b) 15.0 mg ; block: Q = 0; W = 0 ; Eint = 5.00 J ; K = 5.00 J ; metal: Q = 0; W = 0 ; Eint = 0 ; K = 0 (c) 0.004 04C ; moving block: Q = 0; W = 2.50 J ; Eint = 2.50 J ; K = 5.00 J ; stationary block: Q = 0; W = 2.50 J ; Eint = 2.50 J ; K = 0 10.2 h see the solution 800 J kg C
P20.48 P20.50 P20.52 P20.54 P20.56 P20.58 P20.60
(a) ~ 10 3 W ; (b) ~ 10 1 K s 364 K 47.7 g (a) 13.0C ; (b) 0.532 C s (a) 64.1C ; (b) 113C see the solution (a)
10
1 PVi ; (b) 1.39 PVi ; (c) 0 i i 2
12
P20.68 P20.70 P20.72
(a) 9.31 10 J ; (b) 8.47 10 (c) 8.38 10 12 J
J;
21
The Kinetic Theory of Gases
CHAPTER OUTLINE
21.1 21.2 21.3 21.4 21.5 21.6 21.7 Molecular Model of an Ideal Gas Molar Specific Heat of an Ideal Gas Adiabatic Processes for an Ideal Gas The Equipartition of Energy The Boltzmann Distribution Law Distribution of Molecular Speeds Mean Free Path
ANSWERS TO QUESTIONS
Q21.1 The molecules of all different kinds collide with the walls of the container, so molecules of all different kinds exert partial pressures that contribute to the total pressure. The molecules can be so small that they collide with one another relatively rarely and each kind exerts partial pressure as if the other kinds of molecules were absent. If the molecules collide with one another often, the collisions exactly conserve momentum and so do not affect the net force on the walls. The helium must have the higher rms speed. According to Equation 21.4, the gas with the smaller mass per atom must have the higher average speedsquared and thus the higher rms speed. Yes. As soon as the gases are mixed, they come to thermal equilibrium. Equation 21.4 predicts that the lighter helium atoms will on average have a greater speed than the heavier nitrogen molecules. Collisions between the different kinds of molecules gives each kind the same average kinetic energy of translation.
Q21.2
Q21.3
Q21.4
If the average velocity were nonzero, then the bulk sample of gas would be moving in the direction of the average velocity. In a closed tank, this motion would result in a pressure difference within the tank that could not be sustained. The alcohol evaporates, absorbing energy from the skin to lower the skin temperature. Partially evacuating the container is equivalent to letting the remaining gas expand. This means that the gas does work, making its internal energy and hence its temperature decrease. The liquid in the container will eventually reach thermal equilibrium with the low pressure gas. This effect of an expanding gas decreasing in temperature is a key process in your refrigerator or air conditioner. Since the volume is fixed, the density of the cooled gas cannot change, so the mean free path does not change. The collision frequency decreases since each molecule of the gas has a lower average speed. The mean free path decreases as the density of the gas increases. The volume of the balloon will decrease. The pressure inside the balloon is nearly equal to the constant exterior atmospheric pressure. Then from PV = nRT , volume must decrease in proportion to the absolute temperature. Call the process isobaric contraction. 601
Q21.5 Q21.6
Q21.7
Q21.8 Q21.9
602 Q21.10
The Kinetic Theory of Gases
The dry air is more dense. Since the air and the water vapor are at the same temperature, they have the same kinetic energy per molecule. For a controlled experiment, the humid and dry air are at the same pressure, so the number of molecules per unit volume must be the same for both. The water molecule has a smaller molecular mass (18.0 u) than any of the gases that make up the air, so the humid air must have the smaller mass per unit volume. Suppose the balloon rises into air uniform in temperature. The air cannot be uniform in pressure because the lower layers support the weight of all the air above them. The rubber in a typical balloon is easy to stretch and stretches or contracts until interior and exterior pressures are nearly equal. So as the balloon rises it expands. This is an isothermal expansion, with P decreasing as V increases by the same factor in PV = nRT . If the rubber wall is very strong it will eventually contain the helium at higher pressure than the air outside but at the same density, so that the balloon will stop rising. More likely, the rubber will stretch and break, releasing the helium to keep rising and "boil out" of the Earth's atmosphere. A diatomic gas has more degrees of freedomthose of vibration and rotationthan a monatomic gas. The energy content per mole is proportional to the number of degrees of freedom. (a) (b) (c) (d) (e) Average molecular kinetic energy increases by a factor of 3. The rms speed increases by a factor of 3. 3. 3 since the mean free path remains unchanged.
Q21.11
Q21.12 Q21.13
Average momentum change increases by Rate of collisions increases by a factor of Pressure increases by a factor of 3.
Q21.14
They can, as this possibility is not contradicted by any of our descriptions of the motion of gases. If the vessel contains more than a few molecules, it is highly improbable that all will have the same speed. Collisions will make their speeds scatter according to the Boltzmann distribution law. Collisions between molecules are mediated by electrical interactions among their electrons. On an atomic level, collisions of billiard balls work the same way. Collisions between gas molecules are perfectly elastic. Collisions between macroscopic spheres can be very nearly elastic. So the hardsphere model is very good. On the other hand, an atom is not `solid,' but has smallmass electrons moving through empty space as they orbit the nucleus. As a parcel of air is pushed upward, it moves into a region of lower pressure, so it expands and does work on its surroundings. Its fund of internal energy drops, and so does its temperature. As mentioned in the question, the low thermal conductivity of air means that very little heat will be conducted into the nowcool parcel from the denser but warmer air below it. A more massive diatomic or polyatomic molecule will generally have a lower frequency of vibration. At room temperature, vibration has a higher probability of being excited than in a less massive molecule. The absorption of energy into vibration shows up in higher specific heats.
Q21.15
Q21.16
Q21.17
SOLUTIONS TO PROBLEMS Section 21.1 P21.1 Molecular Model of an Ideal Gas 8.00 sin 45.0 8.00 sin 45.0 v = 500 5.00 10 3 kg t 30.0 s
F = Nm P=
e
j
a
f
ms
= 0.943 N
F = 1.57 N m 2 = 1.57 Pa A
Chapter 21
603
P21.2
1.00 s 14.0 N F and P = = = 17.6 kPa . A 8.00 10 4 m 2
F=
e5.00 10 j 2e4.68 10
23
26
kg 300 m s
jb
g = 14.0 N
P21.3
We first find the pressure exerted by the gas on the wall of the container. NkT 3 N A k BT 3 RT 3 8.314 N m mol K 293 K = = = = 9.13 10 5 Pa P= 3 3 V V V 8.00 10 m Thus, the force on one of the walls of the cubical container is
b
ga
f
F = PA = 9.13 10 5 Pa 4.00 10 2 m 2 = 3.65 10 4 N .
P21.4 Use Equation 21.2, P = 2 N mv 2 , so that 3V 2 mv 2 3 PV = where N = nN A = 2 N A 2 2N
e
je
j
F GH
I JK
K av = K av =
3 8.00 atm 1.013 10 5 Pa atm 5.00 10 3 m 3 3 PV = 2 2N A 2 2 mol 6.02 10 23 molecules mol
b
g
a
a
fe
fe
je
j
j
K av = 5.05 10 21 J molecule P21.5 P= 2N KE 3V
d i e e
24
Equation 21.2
3 5 3 PV 3 1.20 10 4.00 10 = = 2.00 10 24 molecules N= 2 KE 2 3.60 10 22
d i
je
j
j
n= P21.6
2.00 10 molecules N = = 3.32 mol N A 6.02 10 23 molecules mol
One mole of helium contains Avogadro's number of molecules and has a mass of 4.00 g. Let us call m the mass of one atom, and we have N A m = 4.00 g mol 4.00 g mol m= = 6.64 10 24 g molecule or 6.02 10 23 molecules mol
m = 6.64 10 27 kg
5 4 PV 1.013 10 Pa 3 0.150 m = N= k BT 1.38 10 23 J K 293 K
P21.7
(a) (b) (c)
PV = Nk BT : K=
a
e
ja
f f
3
= 3.54 10 23 atoms
3 3 k BT = 1.38 10 23 293 J = 6.07 10 21 J 2 2
e
ja f
For helium, the atomic mass is
m=
4.00 g mol 6.02 10
23
molecules mol kg molecule
= 6.64 10 24 g molecule
m = 6.64 10 1 3 mv 2 = k BT : 2 2 v rms =
27
3 k BT = 1.35 km s m
604 P21.8
The Kinetic Theory of Gases
v= vO v He
3 k BT m M He 4.00 1 = = = 32.0 8.00 MO 1 350 m s 8.00 K= K= so For helium, = 477 m s
vO = P21.9 (a) (b)
3 3 k BT = 1.38 10 23 J K 423 K = 8.76 10 21 J 2 2
e
ja
f
1 2 mv rms = 8.76 10 21 J 2 v rms = m= 1.75 10 20 J m 4.00 g mol 6.02 10 23 molecules mol = 6.64 10 24 g molecule (1)
m = 6.64 10 27 kg molecule Similarly for argon, m= 39.9 g mol 6.02 10 23 molecules mol = 6.63 10 23 g molecule
m = 6.63 10 26 kg molecule Substituting in (1) above, we find for helium, and for argon, P21.10 (a) PV = nRT = Nmv 2 3 v rms = 1.62 km s v rms = 514 m s
The total translational kinetic energy is Etrans =
2
(b)
P21.11
(a)
je j 3 k T 3 RT 3a8.314fa300f mv = = = = 6. 21 10 2 2 2N 2e6.02 10 j F 1 N m I FG 1 J IJ = 1 J m 1 Pa = a1 PafG H 1 Pa JK H 1 N m K
B A 23 2 3
3 3 PV = 3.00 1.013 10 5 5.00 10 3 = 2.28 kJ 2 2
e
Nmv 2 = Etrans : 2
21
J
(b)
For a monatomic ideal gas, Eint =
3 nRT 2 3 3 nRT = PV . 2 2
For any ideal gas, the energy of molecular translation is the same, Etrans = Thus, the energy per volume is
Etrans 3 = P . 2 V
Chapter 21
605
Section 21.2 P21.12 Eint = Eint P21.13
Molar Specific Heat of an Ideal Gas 3 nRT 2 3 3 = nRT = 3.00 mol 8.314 J mol K 2.00 K = 74.8 J 2 2
a
fb
ga
f
We us the tabulated values for C P and C V (a) (b) (c) Q = nC P T = 1.00 mol 28.8 J mol K 420  300 K = 3.46 kJ Eint = nCV T = 1.00 mol 20.4 J mol K 120 K = 2.45 kJ W = Q + Eint = 3.46 kJ + 2.45 kJ = 1.01 kJ nRT , then P
b
ga
f
b
ga
f
P21.14
The piston moves to keep pressure constant. Since V = V =
nRT for a constant pressure process. P 2Q Q Q = = Q = nC P T = n CV + R T so T = 7nR n CV + R n 5 R 2 + R and
b g b g b nR F 2Q I 2Q 2 QV V = G J= = P H 7nR K 7 P 7 nRT e4.40 10 Jja5.00 Lf 2 V = = 2.52 L 7 a1.00 molfb8.314 J mol K ga300 K f
3
g
Thus, P21.15
V f = Vi + V = 5.00 L + 2.52 L = 7.52 L
n = 1.00 mol, Ti = 300 K (b) (a) (c) Since V = constant, W = 0 Eint = Q + W = 209 J + 0 = 209 J Eint = nCV T = n so
FG 3 RIJ T H2 K a
T =
2 209 J 2 Eint = = 16.8 K 3nR 3 1.00 mol 8.314 J mol K
a fb
f
g
T = Ti + T = 300 K + 16.8 K = 317 K
606 P21.16
The Kinetic Theory of Gases
(a)
Consider heating it at constant pressure. Oxygen and nitrogen are diatomic, so C P = Q = nC P T = 7 7 PV T nRT = 2 2 T
3
7R 2
5 2 7 1.013 10 Q= 2 300 K
e
FG IJ H K N m je100 m j a1.00 K f =
118 kJ
(b)
U g = mgy m= Ug gy =
e
1.18 10 5 J 9.80 m s 2 2.00 m
j
= 6.03 10 3 kg
*P21.17
(a)
We assume that the bulb does not expand. Then this is a constantvolume heating process. PV The quantity of the gas is n = i . The energy input is Q = P t = nCV T so RTi T =
Pt PtRTi = . nCV PVCV i PtR . PVCV i
The final temperature is T f = Ti + T = Ti 1 + The final pressure is Pf = Pi Tf Ti = Pi 1 +
FG H
FG H
PtR . PVCV i
IJ K
IJ K
(b)
P21.18
(a)
(b)
I = 1.18 atm J s mol K 1.013 10 N 4 a0.05 mf 12.5 J K F 1.00 mol I = 719 J kg K = 0.719 kJ kg K 5 5 C = R = b8.314 J mol K gG 2 2 H 0.028 9 kg JK F PV IJ m = Mn = M G H RT K F 200 10 Pae0.350 m j I m = b0.028 9 kg molgG GH b8.314 J mol K ga300 K f JJK = 0.811 kg
Pf = 1 atm 1 + 3.60 J 4 s 8.314 J m 2 3 mol K
5 3 V 3 3
F GH
(c)
We consider a constant volume process where no work is done. Q = mCV T = 0.811 kg 0.719 kJ kg K 700 K  300 K = 233 kJ
b
ga
f
(d)
We now consider a constant pressure process where the internal energy of the gas is increased and work is done. Q = mC P T = m CV + R T = m
FG 7R IJ T = mFG 7C IJ T b g H2K H 5K L7 O Q = 0.811 kg M b0.719 kJ kg K gPa 400 K f = 327 kJ N5 Q
V
Chapter 21
607
P21.19
Consider 800 cm3 of (flavored) water at 90.0 C mixing with 200 cm3 of diatomic ideal gas at 20.0C: Qcold = Q hot or m air c P , air T f  Ti , air =  m w c w T
d
aT f
w
=
 m air c P , air
i dT
f
 Ti , air
mwcw
a f i = bV g
w
c air P , air
a90.0 C  20.0 Cf b V gc
w w w
where we have anticipated that the final temperature of the mixture will be close to 90.0C. 7 The molar specific heat of air is C P, air = R 2 7 R 7 1.00 mol So the specific heat per gram is = 8.314 J mol K = 1.01 J g C c P, air = 2 M 2 28.9 g
3 3 3 w
or
I FG IJ b gFGH JK H K 1.20 10 g cm je 200 cm j b1.01 J g C ga70.0 C f aT f =  e e1.00 g cm je800 cm j b4.186 J kg Cg aT f 5.05 10 C
3 3 w 3
The change of temperature for the water is between 10 3 C and 10 2 C . P21.20 Q = nC P T
b
g
isobaric
+ nCV T
b
g
isovolumetric
In the isobaric process, V doubles so T must double, to 2Ti . In the isovolumetric process, P triples so T changes from 2Ti to 6Ti . Q =n P21.21
FG 7 RIJ b2T  T g + nFG 5 RIJ b6T  2T g = 13.5nRT = H2 K H2 K
i i i i i
13.5 PV
In the isovolumetric process A B , W = 0 and Q = nCV T = 500 J 500 J = n
FG 3R IJ bT H2K
B
 TA or TB = TA + 2 500 J
g
TB = 300 K +
a f = 340 K 3a1.00 molfb8.314 J mol K g
5nR TC  TB = 500 J . 2
2 500 J 3nR
a
f
In the isobaric process B C , Q = nC P T = Thus, (a) (b) TC = TB 
b
g
2 500 J 1 000 J = 340 K  = 316 K 5nR 5 1.00 mol 8.314 J mol K
a
f
a
fb
g
The work done on the gas during the isobaric process is
WBC =  PB V = nR TC  TB =  1.00 mol 8.314 J mol K 316 K  340 J or WBC = +200 J Won gas = +200 J . The work done on the gas in the isovolumetric process is zero, so in total
b
g a
fb
ga
f
608 *P21.22
The Kinetic Theory of Gases
(a)
At any point in the heating process, Pi = kVi and P = kV = Pf = nRTi Vi
2
2Vi = 2 Pi and T f =
Pf V f nR
Pi nRTi V . At the end, V= Vi Vi2
=
2Vi Vi
2 Pi 2Vi = 4Ti . nR nRTi Vi
2
(b)
The work input is W =  PdV = 
i
z
f
z
nRT V 2 VdV =  2 i 2 Vi
2Vi
=
Vi
nRTi 2Vi
2
e4V
i
2
3  Vi2 =  nRTi . 2
j
The change in internal energy, is Eint = nCV T = n is Q = Eint  W = P21.23 (a) 18 nRTi = 9 1 mol RTi . 2
a
f
5 15 R 4Ti  Ti = + nRTi . The heat input 2 2
b
g
The heat required to produce a temperature change is Q = n1C 1 T + n 2 C 2 T The number of molecules is N 1 + N 2 , so the number of "moles of the mixture" is n1 + n 2 and Q = n1 + n 2 CT ,
b
g
so
m
n C + n 2C 2 C= 1 1 . n1 + n 2
(b)
Q = n i C i T =
i =1
F n I CT GH JK
m i =1 i
C=
i =1 m
n i Ci ni
i =1
m
Section 21.3
Adiabatic Processes for an Ideal Gas
P21.24
(a)
PVi i
Tf Ti
=
Pf V f
so
Vf Vi Tf Ti
FPI =G J HP K
i f
1
=
FG 1.00 IJ H 20.0 K
57
= 0.118
(b) (c)
=
Pf V f PVi i
=
FG P IJ FG V IJ = a20.0fa0.118f H P KH V K
f f i i
= 2.35
Since the process is adiabatic, Since = 1.40 = C P R + CV = , CV CV
Q=0 CV = 5 R and T = 2.35Ti  Ti = 1.35Ti 2 135 J
Eint = nCV T = 0.016 0 mol and
b
gFGH 5 IJK b8.314 J mol K g 1.35a300 K f = 2
W = Q + Eint = 0 + 135 J = +135 J .
Chapter 21
609
P21.25
(a)
PVi = Pf V f i
Pf
FV I =PG J HV K
i i f
= 5.00 atm
FG 12.0 IJ H 30.0 K b
5
1.40
= 1.39 atm
(b)
Ti = Tf =
3 5 3 PVi 5.00 1.013 10 Pa 12.0 10 m i = = 365 K nR 2.00 mol 8.314 J mol K
e
je
Pf V f nR
=
1.39 1.013 10 Pa 30.0 10 3 m 3 2.00 mol 8.314 J mol K
e
b
je
g
j
g
j=
253 K
(c)
The process is adiabatic: Q = 0
= 1.40 =
Eint
C P R + CV 5 = , CV = R 2 CV CV 5 = nCV T = 2.00 mol 8.314 J mol K 2
FG b H
gIJK a253 K  365 K f =
4.66 kJ
W = Eint  Q = 4.66 kJ  0 = 4.66 kJ P21.26 Vi =
F 2.50 10 GH 2
2
m
I JK
2
0.500 m = 2.45 10 4 m 3
The quantity of air we find from PVi = nRTi i n= 1.013 10 5 Pa 2.45 10 4 m 3 PVi i = RTi 8.314 J mol K 300 K
e
b
je
ga
f
j
n = 9.97 10 3 mol Adiabatic compression: Pf = 101.3 kPa + 800 kPa = 901.3 kPa (a)
PVi = Pf V f i
V f = Vi
i
FPI GH P JK
f
1
= 2. 45 10 4 m 3
FG 101.3 IJ H 901.3 K
57
V f = 5.15 10 5 m 3 (b) Pf V f = nRT f Tf Tf (c)
i
FPI =T =T G J PV P HP K F 101.3 IJ b g = = 300 K G H 901.3 K
Pf V f
i i i
Pf
i
1
i
f
F P Ib =T G J HP K
i i f
1 1
g
5 7 1
560 K
The work put into the gas in compressing it is Eint = nCV T W = 9.97 10 3 mol W = 53.9 J
e
j 5 b8.314 J mol K ga560  300f K 2
continued on next page
610
The Kinetic Theory of Gases
Now imagine this energy being shared with the inner wall as the gas is held at constant volume. The pump wall has outer diameter 25.0 mm + 2.00 mm + 2.00 mm = 29.0 mm , and volume
LM e14.5 10 mj  e12.5 10 mj OP4.00 10 N Q and mass V = e7.86 10 kg m je6.79 10 m j = 53.3 g
3 2 3 2 3 3 6 3
2
m = 6.79 10 6 m 3
The overall warming process is described by 53.9 J = nC V T + mcT
e j 5 b8.314 J mol K gdT  300 K i 2 +e53.3 10 kg jb 448 J kg K gdT  300 K i 53.9 J = b0.207 J K + 23.9 J K gdT  300 K i
53.9 J = 9.97 10 3 mol
3 ff ff ff
T ff  300 K = 2.24 K Tf Ti
P21.27
FV I =G J HV K
i f
1
=
FG 1 IJ H 2K
=
0. 400
If Ti = 300 K , then T f = 227 K .
*P21.28
(a)
In
PVi i
Pf V f
we have Pf
FV I =PG J HV K
i i f
Pf Then PVi Pf V f i = Ti Tf
F 0.720 m I =PG H 0.240 m JK
3 i 3
1.40
= 4.66 Pi
T f = Ti
Pf V f PVi i
= Ti 4.66
a f 1 = 1.55 3
The factor of increase in temperature is the same as the factor of increase in internal energy, Eint, f = 1.55 . according to Eint = nCV T . Then Eint, i (b)
FV I = =G J In T PV HV K F 0.720 m I 2=G H V JK
Tf
i
Pf V f
i i
i
Vf Vi
f
FV I =G J HV K
i f
1
we have
3
0. 40
f
0.720 m = 2 1 0 . 4 = 2 2.5 = 5.66 Vf Vf = 0.720 m 3 = 0.127 m3 5.66
3
Chapter 21
611
P21.29
(a) (b)
See the diagram at the right.
PBVB = PC VC 3 PVi = PVC i i
P 3 Pi B Adiabatic
VC = 3 1 Vi = 3 5 7 Vi = 2.19Vi VC (c)
Pi C V(L)
e j e j = 2.19a 4.00 L f = 8.77 L a f
PBVB = nRTB = 3 PVi = 3nRTi i TB = 3Ti = 3 300 K = 900 K
A Vi = 4 L
VC
(d) (e)
After one whole cycle, TA = Ti = 300 K . In AB, Q AB = nCV V = n
FIG. P21.29
FG 5 RIJ b3T  T g = a5.00fnRT H2 K
i i
i
QBC = 0 as this process is adiabatic
PC VC = nRTC = Pi 2.19Vi = 2.19 nRTi so TC = 2.19Ti QCA = nC P T = n For the whole cycle,
b
g a f
i
FG 7 RIJ bT  2.19T g = a4.17fnRT H2 K
i
i
Q ABCA = Q AB + QBC + QCA = 5.00  4.17 nRTi = 0.829 nRTi
b E g
a
f
a
f
int ABCA
= 0 = Q ABCA + W ABCA
W ABCA = Q ABCA =  0.829 nRTi =  0.829 PVi i
W ABCA =  0.829 1.013 10 5 Pa 4.00 10 3 m 3 = 336 J P21.30 (a) (b) See the diagram at the right.
PBVB = PC VC 3 PVi = PVC i i
a
fe
a
f
je
a
f
j
P B Adiabatic
3Pi
VC = 3 1 Vi = 3 5 7 Vi = 2.19Vi (c) PBVB = nRTB = 3 PVi = 3nRTi i TB = 3Ti (d) After one whole cycle, TA = Ti
Pi A Vi C VC
V L
af
FIG. P21.30 continued on next page
612
The Kinetic Theory of Gases
(e)
In AB, Q AB = nCV T = n
FG 5 RIJ b3T  T g = a5.00fnRT H2 K
i i
i
QBC = 0 as this process is abiabatic
PC VC = nRTC = Pi 2.19Vi = 2.19nRTi so TC = 2.19Ti QCA For the whole cycle, Q ABCA = Q AB + QBC + QCA = 5.00  4.17 nRTi = 0.830nRTi
P i i i
b g F7 I = nC T = nG RJ bT  2.19T g = 4.17nRT H2 K a f
= 0 = Q ABCA + W ABCA
b E g
int ABCA
W ABCA = Q ABCA = 0.830nRTi = 0.830 PVi i P21.31 (a) The work done on the gas is Wab =  PdV .
Va Vb
z
For the isothermal process, Wab = nRTa
Vb Va
z FGH
1 dV V
b a a
IJ K
Wab = nRTa ln Thus, Wab
FG V IJ = nRT lnFG V IJ . HV K HV K = 5.00 molb8.314 J mol K ga 293 K f lna10.0 f
b
FIG. P21.31
Wab = 28.0 kJ . (b) For the adiabatic process, we must first find the final temperature, Tb . Since air consists primarily of diatomic molecules, we shall use
air = 1.40 and C V , air =
Then, for the adiabatic preocess Tb
5 R 5 8.314 = = 20.8 J mol K . 2 2
a
f
FV I =T G J HV K
a a b V
1
= 293 K 10.0
a f
b
0. 400
= 736 K .
Thus, the work done on the gas during the adiabatic process is Wab Q + Eint or Wab
b
g = b0 + nC T g = nC bT  T g = 5.00 molb 20.8 J mol K ga736  293f K = 46.0 kJ
ab ab V a
.
continued on next page
Chapter 21
613
(c)
For the isothermal process, we have Pb Vb = PaVa . Thus, Pb = Pa
FG V IJ = 1.00 atma10.0f = HV K
a b
10.0 atm .
For the adiabatic process, we have Pb Vb = Pa Va . Thus, Pb = Pa P21.32
FG V IJ HV K
a b
= 1.00 atm 10.0
a f
1.40
= 25.1 atm .
We suppose the air plus burnt gasoline behaves like a diatomic ideal gas. We find its final absolute pressure: 21.0 atm 50.0 cm3 Pf = 21.0 atm Now Q = 0 and W = Eint = nCV T f  Ti W =
e
FG 1 IJ H 8K
i
j
75
= Pf 400 cm3
e
j
75
75
= 1.14 atm
d
F 1.013 10 N m I 10 5 W = 1.14 atme 400 cm j  21.0 atme50.0 cm j G 2 H 1 atm JK e
3 3 5 2
5 5 5 nRT f  nRTi = Pf V f  PVi i 2 2 2
d
i
FIG. P21.32
6
m3 cm 3
j
W = 150 J The output work is W = +150 J The time for this stroke is 1 1 min 4 2 500
F GH
I FG 60 s IJ = 6.00 10 JK H 1 min K
3
s
P=
W 150 J = = 25.0 kW t 6.00 10 3 s
614
The Kinetic Theory of Gases
Section 21.4 P21.33
The Equipartition of Energy
The heat capacity at constant volume is nC V . An ideal gas of diatomic molecules has three degrees of freedom for translation in the x, y, and z directions. If we take the y axis along the axis of a molecule, then outside forces cannot excite rotation about this axis, since they have no lever arms. Collisions will set the molecule spinning only about the x and z axes. (a) If the molecules do not vibrate, they have five degrees of freedom. Random collisions put 1 equal amounts of energy k BT into all five kinds of motion. The average energy of one 2 5 molecule is k BT . The internal energy of the twomole sample is 2 N
FG 5 k TIJ = nN FG 5 k TIJ = nFG 5 RIJ T = nC T . H2 K H2 K H2 K
B A B V
The molar heat capacity is C V =
5 R and the sample's heat capacity is 2
nCV = n
FG 5 RIJ = 2 molFG 5 b8.314 J mol K gIJ H2 K H2 K
nCV = 41.6 J K For the heat capacity at constant pressure we have nC P = n C V + R = n nC P = 58.2 J K (b) In vibration with the center of mass fixed, both atoms are always moving in opposite directions with equal speeds. Vibration adds two more degrees of freedom for two more terms in the molecular energy, for kinetic and for elastic potential energy. We have nCV = n and nC P
b
g FGH 5 R + RIJK = 7 nR = 2 molFGH 7 b8.314 J mol K gIJK 2 2 2
FG 7 RIJ = H2 K F9 I = nG RJ = H2 K
B
58.2 J K 74.8 J K
P21.34
(1)
Eint = Nf
(2)
CV
FG k T IJ = f FG nRT IJ H2K H 2K 1 F dE I 1 = G J = fR n H dT K 2
int
(3) (4)
C P = CV + R =
1 f +2 R 2
b
g
=
CP f + 2 = CV f
Chapter 21
615
P21.35
Rotational Kinetic Energy =
1 2 I 2 Cl
Cl
I = 2mr 2 , m = 35.0 1.67 10 27 kg , r = 10 10 m I = 1.17 10 45 kg m 2 K rot = 1 2 I = 2.33 10 21 J 2
= 2.00 10 12 s 1
FIG. P21.35
Section 21.5 Section 21.6 P21.36 (a)
The Boltzmann Distribution Law Distribution of Molecular Speeds The ratio of the number at higher energy to the number at lower energy is e E kBT where E is the energy difference. Here, E = 10.2 eV and at 0C, k BT = 1.38 10 23 J K 273 K = 3.77 10 21 J . Since this is much less than the excitation energy, nearly all the atoms will be in the ground state and the number excited is
a
10 fFGH 1.601eV J IJK = 1.63 10
19
18
J
e
ja
f
e
2.70 10 25 exp
1.63 10 j FGH 3.77 10 JJ IJK = e2.70 10 je
18 21 25
433
.
This number is much less than one, so almost all of the time no atom is excited . (b) At 10 000C, k BT = 1.38 10 23 J K 10 273 K = 1.42 10 19 J . The number excited is 1.63 10 e2.70 10 j expFGH 1.42 10 JJ IJK = e2.70 10 je
25 18 19 25 11.5
e
j
= 2.70 10 20 .
616 P21.37
The Kinetic Theory of Gases
(a)
v av =
ni v i
N =
=
1 1 2 + 2 3 + 3 5 + 4 7 + 3 9 + 2 12 = 6.80 m s 15 = 54.9 m 2 s 2 = 54.9 = 7.41 m s
af af af af af a f
(b)
ev j
2
ni vi2
N
av
so v rms = (c)
ev j
2 3 RT M 35 3 RT M 37
av
v mp = 7.00 m s
Vrms, 35 Vrms, 37
P21.38
(a)
=
=
F 37.0 g mol I GH 35.0 g mol JK
35
12
= 1.03
(b) P21.39
The lighter atom,
Cl , moves faster.
dN v = 0 to find dv
3
In the Maxwell Boltzmann speed distribution function take
F m I 4 N G H 2 k T JK
B
32
exp 
F GH
mv 2 2 k BT
I F 2v  2mv I = 0 JK GH 2k T JK
B
and solve for v to find the most probable speed. Reject as solutions Retain only Then v = 0 and v = 2 mv 2 =0 k BT 2 k BT m 2 k BT = m 2 1.38 10 23 J K 4.20 K 6.64 10
27
v mp =
P21.40
The most probable speed is v mp = 8 k BT m
e
ja
kg
f=
132 m s .
P21.41
(a)
From v av =
we find the temperature as T =
6.64 10 27 kg 1.12 10 4 m s
8 1.38 10
e
e
je
23
J mol K
j
j
2
= 2.37 10 4 K
(b)
T=
6.64 10 27 kg 2.37 10 3 m s
8 1.38 10 23 J mol K
e
e
je
j
=
j
2
= 1.06 10 3 K
P21.42
At 0C,
1 3 2 mv rms0 = k BT0 2 2 1 m 2 v rms0 2
At the higher temperature,
b
g
2
T = 4T0 = 4 273 K = 1 092 K = 819 C .
a
3 k BT 2
f
Chapter 21
617
*P21.43
(a)
From the Boltzmann distribution law, the number density of molecules with gravitational energy mgy is n 0 e  mgy k BT . These are the molecules with height y, so this is the number per volume at height y as a function of y. n y n0
(b)
b g=e
 mgy k BT
= e  Mgy
N A k BT
= e  Mgy
RT
=e
 28 .9 10 3 kg mol 9.8 m s 2 11 10 3 m
e
je
je
j b8.314 J molK ga 293 K f
= e 1.279 = 0.278 *P21.44 (a) We calculate
0
z
e  mgy
k BT
dy =
y=0
z
e  mgy
k BT
= Using Table B.6 in the appendix
0
k BT  mgy e mg
k BT
FG  mgdy IJ F  k T I H k T K GH mg JK k T T = a0  1f = kmg mg
B B B B 0
z
ye  mgy
k BT
dy =
F k TI bmg k T g GH mg JK
1!
2
2
=
B
.
B
Then y =
z z
0
ye  mgy e  mgy
k BT
dy =
0
bk T mg g
B
2
k BT
dy
k BT mg
=
k BT . mg
(b)
y=
b
k BT 8.314 J 283 K s 2 RT = = = 8.31 10 3 m M N A g Mg mol K 28.9 10 3 kg 9.8 m
g
Section 21.7 P21.45 (a)
Mean Free Path
FG N IJ RT and N = PVN so that RT HN K e1.00 10 ja133fa1.00fe6.02 10 j = N= a8.314fa300f
PV =
A A 10 23
3.21 10 12 molecules
(b)
=
1 1.00 m3 V = = 2 12 2 12 nV d 2 N d 2 3.21 10 12 molecules 3.00 10 10 m
e
je
j a 2f
2
12
= 779 km (c) f= v = 6.42 10 4 s 1
618 P21.46
The Kinetic Theory of Gases
The average molecular speed is v= v= v= 8 k BT 8 k B N AT = m N Am 8 RT M 8 8.314 J mol K 3.00 K
3
b e 2.016 10
g
kg mol
j
v = 178 m s (a) The mean free path is = 1 2 d n V
2
=
1 2 0. 200 10 9 m
e
j
2
1 m3
= 5.63 10 18 m The mean free time is = 5.63 10 18 m = 3.17 10 16 s = 1.00 10 9 yr . 178 m s smaller by 10 6 times:
v (b)
Now nV is 10 6 times larger, to make
= 5.63 10 12 m .
Thus, P21.47 = 3.17 10 10 s = 1.00 10 3 yr . 1 2 d 2 nV
v
From Equation 21.30, = For an ideal gas, nV = Therefore, =
1
N P = V k BT , as required. P k BT
2 d 2 P
k BT
P21.48
=
2 d 2 nV
nV = nV =
d = 3.60 10 10 m
e1.38 10 ja293f
23
1.013 10 5
= 2.51 10 25 m3
= 6.93 10 8 m, or about 193 molecular diameters .
Chapter 21
619 (1)
P21.49
Using P = nV k BT , Equation 21.30 becomes =
2 Pd 2 = 9.36 10 8 m
k BT
(a)
=
e1.38 10 e
23
J K 293 K
ja
f
m
2 1.013 10 Pa 3.10 10
1
5
je
10
j
2
(b)
Equation (1) shows that P1 P2 =
= P2
2.
Taking P1
8
1
from (a) and with m
2
= 1.00 m, we find
a1.00 atmfe9.36 10
1.00 m
j=
m
9.36 10 8 atm .
(c)
For
3
= 3.10 10 10 m , we have
a1.00 atmfe9.36 10 P =
3
8
3.10 10
10
m
j=
302 atm .
Additional Problems P21.50 (a) n= PV (1.013 10 5 Pa)( 4.20 m 3.00 m 2.50 m) = = 1.31 10 3 mol ( 8.314 J mol K )( 293 K ) RT
N = nN A = 1.31 10 3 mol 6.02 10 23 molecules mol N = 7.89 10 26 molecules (b) (c) (d) m = nM = 1.31 10 3 mol 0.028 9 kg mol = 37.9 kg
e
je
j
e
jb
g
1 3 3 m 0 v 2 = k BT = 1.38 10 23 J k 293 K = 6.07 10 21 J molecule 2 2 2
e
ja
f
For one molecule, m0 = 0.028 9 kg mol M = = 4.80 10 26 kg molecule N A 6.02 10 23 molecules mol 2 6.07 10 21 J molecule 4.80 10
26
v rms =
e
kg molecule
j=
503 m s
(e),(f)
Eint = nCV T = n Eint
FG 5 RIJ T = 5 PV H2 K 2 5 = e1.013 10 Paje31.5 m j = 2
5 3
7.98 MJ
620 P21.51
The Kinetic Theory of Gases
(a)
Pf = 100 kPa Vf = nRT f Pf = 2.00 mol 8.314 J mol K 400 K 100 10 Pa
3
T f = 400 K
b
ga
f = 0.066 5 m
3
= 66.5 L
fb ga f W =  PV = nRT = a 2.00 molfb8.314 J mol K ga100 K f =
Q = Eint  W = 5.82 kJ + 1.66 kJ = 7.48 kJ (b) T f = 400 K Pf = Pi
f
Eint = 3.50 nRT = 3.50 2.00 mol 8.314 J mol K 100 K = 5.82 kJ 1.66 kJ
a f
a
V f = Vi =
F T I = 100 kPaFG 400 K IJ = GH T JK H 300 K K
i
nRTi 2.00 mol 8.314 J mol K 300 K = = 0.049 9 m 3 = 49.9 L Pi 100 10 3 Pa 133 kPa W =  PdV = 0 since V = constant Q = Eint  W = 5.82 kJ  0 = 5.82 kJ T f = 300 K 41.6 L Eint = 3.50 nRT = 0 since T = constant
i
b
ga
f
z
Eint = 5.82 kJ as in part (a) (c) Pf = 120 kPa V f = Vi
F I = nRT lnFG P IJ GH JK HP K F 100 kPa IJ = +909 J W = a 2.00 molfb8.314 J mol K ga300 K f lnG H 120 kPa K
W =  PdV = nRTi
F P I = 49.9 LFG 100 kPa IJ = GH P JK H 120 kPa K
i f Vf Vi
a f
z
z
Vf dV = nRTi ln V Vi
i
f
Q = Eint  W = 0  910 J = 909 J (d) Pf = 120 kPa
=
C P CV + R 3.50 R + R 4.50 9 = = = = 3.50 R 3.50 7 CV CV
1 i i 79
FPI F 100 kPa IJ = = : so V = V G J = 49.9 L G H 120 kPa K HP K F P V IJ = 300 K FG 120 kPa IJ FG 43.3 L IJ = 312 K T =T G H 100 kPa K H 49.9 L K H PV K E = a3.50fnRT = 3.50a 2.00 molfb8.314 J mol K ga12.4 K f = 722 J Q = 0 badiabatic process g
Pf V f PVi i
f f f f i f i i int
43.3 L
W = Q + Eint = 0 + 722 J = +722 J
Chapter 21
621
P21.52
(a)
The average speed v av is just the weighted average of all the speeds. v av = 2 v + 3 2 v + 5 3 v + 4 4v + 3 5 v + 2 6 v + 1 7 v
af a f a f a f a f a f a f= a 2 + 3 + 5 + 4 + 3 + 2 + 1f
2 2 2 2 2 2
3.65 v
(b)
First find the average of the square of the speeds,
2 v av
=
2v
a f + 3a2vf + 5a3vf + 4a4vf + 3a5vf + 2a6 vf + 1a7 vf
2
2+3+5+4+3+2+1
= 15.95 v 2 .
2 The rootmean square speed is then v rms = v av = 3.99 v .
(c)
The most probable speed is the one that most of the particles have; i.e., five particles have speed 3.00 v .
(d)
PV =
1 2 Nmv av 3
2 20 m 15.95 v mv 2 = 106 Therefore, P = 3 V V
a
f
F GH
I JK
.
(e)
The average kinetic energy for each particle is K= 1 1 2 mv av = m 15.95 v 2 = 7.98mv 2 . 2 2
e
j
P21.53
(a) (b)
PV = k . So, W =  PdV =  k
i
z
f
z
f i
dV Pf V f  PVi i = 1 V
dEint = dQ + dW and dQ = 0 for an adiabatic process. Therefore, W = + Eint = nCV T f  Ti . To show consistency between these 2 equations, consider that = Therefore, C 1 = V. 1 R CP and C P  CV = R . CV
d
i
Using this, the result found in part (a) becomes W = Pf V f  PVi i Also, for an ideal gas
d
i CR .
V
PV = nT so that W = nCV T f  Ti . R
d
i
622 *P21.54
The Kinetic Theory of Gases
(a)
W = nCV T f  Ti
d
i
3 2 500 J = 1 mol 8.314 J mol K T f  500 K 2
d
i
T f = 300 K (b)
PVi = Pf V f i
F nRT IJ PG H P K
i i i
F nRT I =P G H P JK
f f f
Ti Pi1  = T f Pf1
f
Ti
1
b g
Pi Pf = Pi *P21.55
=
f
Tf
1
b g
FT GH T JK
i
I b5 3gb3 2
Pf
FT I b g P =PG J HT K g F 300 IJ = 1.00 atm = 3.60 atmG H 500 K
1
f i i 5 2
Let the subscripts `1' and `2' refer to the hot and cold compartments, respectively. The pressure is higher in the hot compartment, therefore the hot compartment expands and the cold compartment contracts. The work done by the adiabatically expanding gas is equal and opposite to the work done by the adiabatically compressed gas. nR nR T1i  T1 f =  T2i  T2 f 1 1
d
i
d
i
(1)
T1 f + T2 f = T1i + T2 i = 800 K Consider the adiabatic changes of the gases.
P1i V1i = P1 f V1f and P2i V2i = P2 f V2 f
P2 iV2 i
P1i V1i
=
P1 f V1f
P2 f V2 f
V1 f P1i = P2 i V2 f
F I GH JK I JK IJ K
, since V1i = V2i and P1 f = P2 f
nRT1 f P1 f nRT1i V1i = nRT2 i V2 i nRT2 f P2 f T1 f T1i = T2i T2 f T1 f T2 f
1i 2i
F GH
I JK
, using the ideal gas law
F GH FT =G HT
, since V1i = V2i and P1 f = P2 f
1
=
FG 550 K IJ H 250 K K
1 1.4
= 1.756 (2)
Solving equations (1) and (2) simultaneously gives T1 f = 510 K, T2 f = 290 K .
Chapter 21
623
*P21.56
The work done by the gas on the bullet becomes its kinetic energy: 1 1 mv 2 = 1.1 10 3 kg 120 m s 2 2 The work on the gas is
b
g
2
= 7.92 J .
FV I . GH V JK L FV I O 1 So 7.92 J = P MV G J  V P . PQ 0. 40 M H V K N
1 Pf V f  PVi = 7.92 J. i 1
d
i
Also Pf V f = PVi i
Pf = Pi
i
i
f
i
f
f
i
And V f = 12 cm3 + 50 cm 0.03 cm 2 = 13.5 cm3 . Then Pi =
a f LM13.5 cm c h N
3
7.92 J 0.40 10 6 cm 3 m 3
12 1.40 13 .5
=  12 cm O QP
3
5.74 10 6 Pa = 56.6 atm .
P21.57
The pressure of the gas in the lungs of the diver must be the same as the absolute pressure of the water at this depth of 50.0 meters. This is: P = P0 + gh = 1.00 atm + 1.03 10 3 kg m 3 9.80 m s 2 50.0 m or P = 1.00 atm + 5.05 10 5
5
je ja F 1.00 atm IJ = 5.98 atm PaG H 1.013 10 Pa K e
f
If the partial pressure due to the oxygen in the gas mixture is to be 1.00 atmosphere (or the fraction 1 1 of the total pressure) oxygen molecules should make up only of the total number of 5.98 5.98 molecules. This will be true if 1.00 mole of oxygen is used for every 4.98 mole of helium. The ratio by weight is then
a4.98 mol Hefb4.003 g mol Hegg = b1.00 mol O gb2 15.999 g mol O gg
2 2 3 2
0.623 .
P21.58
(a)
Maxwell's speed distribution function is Nv With
F m I = 4 N G H 2 k T JK
B
v 2 e  mv
2
2 k BT
N = 1.00 10 4 , 0.032 kg M = = 5.32 10 26 kg m= N A 6.02 10 23 T = 500 K k B = 1.38 10 23 J molecule K
and
this becomes N v = 1.71 10 4 v 2 e
e
j
 3 .85 10 6 v 2
e
j
To the right is a plot of this function for the range 0 v 1 500 m s. FIG. P21.58(a) continued on next page
624
The Kinetic Theory of Gases
(b)
The most probable speed occurs where N v is a maximum. From the graph, v mp 510 m s 8 k BT = m 8 1.38 10 23 500
26
(c)
v av = Also, v rms =
e
ja f = e5.32 10 j
5.32 10
26
575 m s
3 k BT = m
3 1.38 10 23 500
e
ja f =
624 m s 300 m s v 600 m s
600
(d)
The fraction of particles in the range
is where
300
z
N v dv N
N = 10 4
and the integral of N v is read from the graph as the area under the curve. This is approximately 4 400 and the fraction is 0.44 or 44% . P21.59 (a) Since pressure increases as volume decreases (and vice versa), 1 dV dV < 0 and  > 0. V dP dP (b) For an ideal gas, V = 1 d nRT nRT and 1 =  . P V dP P
LM OP N Q
FG H
IJ K
If the compression is isothermal, T is constant and
1 = 
(c)
1 1 nRT  2 = . V P P
FG H
IJ K
For an adiabatic compression, PV = C (where C is a constant) and
2 = 1 = =
1 d C V dP P
FG IJ H K
1
=
1 1 C1 1 P1 = 1 +1 = . P V P b1 g+1 P
FG IJ H K
(d)
1 1 = = 0.500 atm 1 2.00 atm P
a
f
CP 5 and for a monatomic ideal gas, = , so that 3 CV 1 = P
5 3
2 =
a
1 = 0.300 atm 1 2.00 atm
f
Chapter 21
625
P21.60
(a)
The speed of sound is v =
B
where B = V
dP . dV 1
According to Problem 59, in an adiabatic process, this is B = Also, =
nRT M PM m s nM = = = where m s is the sample mass. Then, the speed of sound V V V RT RT B
a f a f
2
= P .
in the ideal gas is v =
= P
FG RT IJ = H PM K
f=
RT . M
(b)
344 m s 0.028 9 kg mol This nearly agrees with the 343 m/s listed in Table 17.1. We use k B =
v=
1.40 8.314 J mol K 293 K
b
ga
(c)
kB N AT k BT RT R and M = mN A : v = = = . M mN A m NA
2k BT , m 3k BT . m
The most probable molecular speed is the average speed is
8k BT , and the rms speed is m
All are somewhat larger than the speed of sound. P21.61 n= 1.20 kg m = = 41.5 mol M 0.028 9 kg mol Vi = Pf Pi 41.5 mol 8.314 J mol K 298 K nRTi = = 0.514 m 3 Pi 200 10 3 Pa Vf Vi Pf V f so V f
(a)
a
fb
ga
3
f
(b)
=
FP I F 400 IJ = V G J = e0.514 m jG H 200 K HPK
i f 2 i 3 3
2
= 2.06 m 3
(c)
(d)
e400 10 Paje2.06 m j = 2.38 10 K nR a41.5 molfb8.314 J mol K g F P I 2V =  2 F P I V  V W =  z PdV = C z V dV =  G j G Je 3 HV K H V JK 3 2 F 200 10 Pa I L W = G Me2.06 m j  a0.514 mf OQP = 4.80 10 J 3 H 0.514 m J N K
Tf = =
3 Vf Vi Vf Vi 12 i 12 i 3 2 Vf Vi i 12 i 32 f 3 2 i 3 3 3 2 3 2 5
(e)
Eint = nCV T = 41.5 mol Eint = 1.80 10 6 J
a
fLMN 5 b8.314 J mol K gOPQe2.38 10 2
3
 298 K
j
Q = Eint  W = 1.80 10 6 J + 4.80 10 5 J = 2.28 10 6 J = 2.28 MJ
626 P21.62
The Kinetic Theory of Gases
The ball loses energy The air volume is and its quantity is
g a19.4 mf = 0.083 4 m PV 1.013 10 Pae0.083 4 m j = n= RT b8.314 J mol K ga293 K f = 3.47 mol
V = 0.037 0 m
1 1 1 mvi2  mv 2 = 0.142 kg f 2 2 2
b
g a47.2f  a42.5f
2 3 3
2
m 2 s 2 = 29.9 J
b
2
5
The air absorbs energy according to Q = nC P T So T = Q = nC P 3. 47 mol
2
c hb8.314 J mol K g =
7 2
29.9 J
0. 296 C
P21.63
N v v = 4 N Note that
af
F m I GH 2 k T JK
B
Thus,
And
F mv I GH 2k T JK F 2k T IJ v =G H m K F m I N a vf = 4 N G H 2 k T JK F v I ee N avf =G J N ev j H v K
32
v 2 exp
B
B
12
mp
3 2
v
v2e
e v
2
2 v mp
j
B
2
v
2 1  v 2 v mp
j
v
mp
mp
For
v=
v mp 50
2 1  1 50
a f = FG 1 IJ N e v j H 50 K
Nv v
v mp
e
b g
2
= 1.09 10 3
The other values are computed similarly, with the following results:
v v mp 1 50 1 10 1 2 1 2 10 50
Nv v
N v v mp
e j
af
1.09 10 3 2.69 10 2 0.529 1.00 0.199 1.01 10 41 1.25 10 1 082
To find the last value, note:
a50f e
2 1  2 500
10 log 2 500 e
aln10 fb2 499 ln10 g = 10 log 2 50010 2 499 ln10 = 10 log 2 500  2 499 ln 10 = 10 1 081.904
= 2 500 e 2 499
Chapter 21
627
P21.64
(a)
The effect of high angular speed is like the effect of a very high gravitational field on an atmosphere. The result is:
The largermass molecules settle to the outside while the region at smaller r has a higher
concentration of lowmass molecules. (b) Consider a single kind of molecules, all of mass m. To cause the centripetal acceleration of the molecules between r and r + dr , the pressure must increase outward according to Fr = mar . Thus, PA  P + dP A =  nmA dr r 2
a
f
b
ge j
where n is the number of molecules per unit volume and A is the area of any cylindrical surface. This reduces to dP = nm 2 rdr . But also P = nk BT , so dP = k BTdn . Therefore, the equation becomes dn m 2 dn m 2 = = rdr giving rdr or n k BT n k BT 0 n
0
n
z
z
r
ln n
af
n n0
m 2 r 2 = k BT 2
2 2
F I GH JK
r
ln
FG n IJ = m r H n K 2k T
0 B
0
2
and solving for n: n = n 0 e mr
2 2 k BT
.
P21.65
2 2 First find v av as v av =
1 m . v 2 N v dv . Let a = N0 2 k BT
0
z
Then,
2 v av
=
4 N  1 2 a 3 2 N
z
v 4 e  av
2
dv
= 4 a 3 2 1 2
3 8a2
3 k BT = a m
3 k BT . m
2 The rootmean square speed is then v rms = v av =
To find the average speed, we have v av
4Na 3 2 1 2 1 = vN v dv = N0 N
z
e
j
0
z
v 3 e  av dv =
2
4 a 3 2 1 2 = 2a 2
8 k BT . m
*P21.66
dP for the function implied by PV = nRT = constant , and also for the different dV function implied by PV = constant . We can use implicit differentiation: We want to evaluate From PV = constant From PV = constant Therefore, The theorem is proved. P dV dP +V =0 dV dV dP =0 dV
PV 1 + V
FG dP IJ H dV K FG dP IJ H dV K
=
isotherm
P V
=
adiabat
P V
FG dP IJ H dV K
=
adiabat
FG dP IJ H dV K
isotherm
628 P21.67
The Kinetic Theory of Gases
(a)
n=
1.013 10 5 Pa 5.00 10 3 m 3 PV = = 0.203 mol RT 8.314 J mol K 300 K
e
b
je
ga
f
j
(b)
TB = TA
FG P IJ = 300 K FG 3.00 IJ = H 1.00 K HP K
B A
900 K
TC = TB = 900 K VC = VA
C
FG T IJ = 5.00 LFG 900 IJ = H 300 K HT K
A
15.0 L FIG. P21.67
(c)
Eint, A = Eint, B
3 3 nRTA = 0.203 mol 8.314 J mol K 300 K = 760 J 2 2 3 3 = Eint, C = nRTB = 0.203 mol 8.314 J mol K 900 K = 2.28 kJ 2 2
a
fb
ga
a
fb
f ga
f
(d) A B C (e)
P (atm) 1.00 3.00 1.00
V(L) 5.00 5.00 15.00
T(K) 300 900 900
Eint (kJ) 0.760 2.28 2.28
For the process AB, lock the piston in place and put the cylinder into an oven at 900 K. For BC, keep the sample in the oven while gradually letting the gas expand to lift a load on the piston as far as it can. For CA, carry the cylinder back into the room at 300 K and let the gas cool without touching the piston. For AB: W= 0 Q = Eint  W = 1.52 kJ For BC: Eint = 0 , W = nRTB ln Eint = Eint, B  Eint, A = 2.28  0.760 kJ = 1.52 kJ
(f)
a
f
FG V IJ HV K
C B
W =  0. 203 mol 8.314 J mol K 900 K ln 3.00 = 1.67 kJ Q = Eint  W = 1.67 kJ For CA:
a
fb
ga
f a f
a f W =  PV = nRT = a0.203 molfb8.314 J mol K ga 600 K f =
Eint = Eint, A  Eint, C = 0.760  2.28 kJ = 1.52 kJ Q = Eint  W = 1.52 kJ  1.01 kJ = 2.53 kJ
1.01 kJ
(g)
We add the amounts of energy for each process to find them for the whole cycle. Q ABCA = +1.52 kJ + 1.67 kJ  2.53 kJ = 0.656 kJ W ABCA = 0  1.67 kJ + 1.01 kJ = 0.656 kJ
b E g
int ABCA
= +1.52 kJ + 0  1.52 kJ = 0
P21.68
(a) (b)
00 molecules b10 000 g gFGH 1.18.0mol IJK FGH 6.02 10 mol IJK = g 1.00
23
Chapter 21
629
3.34 10 26 molecules
After one day, 10 1 of the original molecules would remain. After two days, the fraction would be 10 2 , and so on. After 26 days, only 3 of the original molecules would likely remain, and after 27 days , likely none.
(c)
The soup is this fraction of the hydrosphere:
F 10.0 kg I . GH 1.32 10 kg JK
21
Therefore, today's soup likely contains this fraction of the original molecules. The number of original molecules likely in the pot again today is:
F 10.0 kg I 3.34 10 GH 1.32 10 kg JK e
21
26
molecules = 2.53 10 6 molecules .
j
P21.69
(a)
For escape,
1 GmM GM . Since the freefall acceleration at the surface is g = 2 , this can mv 2 = 2 RE RE 1 GmM 2 also be written as: mv = = mgRE . 2 RE
(b)
For O 2 , the mass of one molecule is m= 0.032 0 kg mol 6.02 10 23 molecules mol
B
= 5.32 10 26 kg molecule .
Then, if mgRE = 10
FG 3 k T IJ , the temperature is H 2 K e5.32 10 kgje9.80 m s je6.37 10 mj = mgR = T= 15 k 15e1.38 10 J mol K j
26 2 6 E B 23
1.60 10 4 K .
P21.70
(a)
For sodium atoms (with a molar mass M = 32.0 g mol ) 1 3 mv 2 = k BT 2 2 1 M 3 v 2 = k BT 2 NA 2 v rms = d v rms 3 RT = M 3 8.314 J mol K 2.40 10 4 K 23.0 10
3
FG IJ H K
b
ge
kg
j=
0.510 m s
(b)
t=
=
0.010 m = 20 ms 0.510 m s
630
The Kinetic Theory of Gases
ANSWERS TO EVEN PROBLEMS
P21.2 P21.4 P21.6 P21.8 P21.10 P21.12 P21.14 P21.16 P21.18 17.6 kPa 5.05 10 21 J molecule 6.64 10 27 kg 477 m s P21.48 (a) 2.28 kJ; (b) 6.21 10 21 J P21.50 193 molecular diameters (a) 7.89 10 26 molecules; (b) 37.9 kg ; (c) 6.07 10 21 J molecule ; (d) 503 m s; (e) 7.98 MJ ; (f) 7.98 MJ P21.52 (a) 3.65 v; (b) 3.99 v; (c) 3.00 v; mv 2 ; (e) 7.98mv 2 (d) 106 V P21.42 P21.44 P21.46 819C (a) see the solution; (b) 8.31 km (a) 5.63 10 18 m; 1.00 10 9 yr ; (b) 5.63 10 12 m; 1.00 10 3 yr
74.8 J
7.52 L (a) 118 kJ ; (b) 6.03 10 3 kg (a) 719 J kg K ; (b) 0.811 kg ; (c) 233 kJ; (d) 327 kJ 13.5 PV (a) 4Ti ; (b) 9 1 mol RTi (a) 0.118 ; (b) 2.35 ; (c) 0; 135 J ; 135 J (a) 5.15 10 5 m 3 ; (b) 560 K ; (c) 2.24 K (a) 1.55 ; (b) 0.127 m3 (a) see the solution; (b) 2.19Vi ; (c) 3Ti ; (d) Ti ; (e) 0.830 PVi i 25.0 kW see the solution (a) No atom, almost all the time; (b) 2.70 10 20 (a) 1.03 ; (b) 132 m s
35
F GH
I JK
P21.54 P21.56 P21.58
(a) 300 K ; (b) 1.00 atm 5.74 10 6 Pa (a) see the solution; (b) 5.1 10 2 m s; (c) v av = 575 m s ; v rms = 624 m s ; (d) 44% (a) see the solution; (b) 344 m s nearly agreeing with the tabulated value; (c) see the solution; somewhat smaller than each 0.296C see the solution see the solution (a) 3.34 10 26 molecules ; (b) during the 27th day; (c) 2.53 10 6 molecules (a) 0.510 m s ; (b) 20 ms
P21.20 P21.22 P21.24 P21.26 P21.28 P21.30
a
f
P21.60
P21.62 P21.64 P21.66 P21.68
P21.32 P21.34 P21.36
P21.38 P21.40
Cl
P21.70
22
Heat Engines, Entropy, and the Second Law of Thermodynamics
CHAPTER OUTLINE
22.1 Heat Engines and the Second Law of Thermodynamics Heat Pumps and Refrigerators Reversible and Irreversible Processes The Carnot Engine Gasoline and Diesel Engines Entropy Entropy Changes in Irreversible Processes Entropy on a Microscopic Scale
ANSWERS TO QUESTIONS
Q22.1 First, the efficiency of the automobile engine cannot exceed the Carnot efficiency: it is limited by the temperature of burning fuel and the temperature of the environment into which the exhaust is dumped. Second, the engine block cannot be allowed to go over a certain temperature. Third, any practical engine has friction, incomplete burning of fuel, and limits set by timing and energy transfer by heat. It is easier to control the temperature of a hot reservoir. If it cools down, then heat can be added through some external means, like an exothermic reaction. If it gets too hot, then heat can be allowed to "escape" into the atmosphere. To maintain the temperature of a cold reservoir, one must remove heat if the reservoir gets too hot. Doing this requires either an "even colder" reservoir, which you also must maintain, or an endothermic process.
22.2 22.3 22.4 22.5 22.6 22.7 22.8
Q22.2
Q22.3
A higher steam temperature means that more energy can be extracted from the steam. For a constant temperature heat sink at Tc , and steam at Th , the efficiency of the power plant goes as Th  Tc T = 1  c and is maximized for a high Th . Th Th No. Any heat engine takes in energy by heat and must also put out energy by heat. The energy that is dumped as exhaust into the lowtemperature sink will always be thermal pollution in the outside environment. Socalled `steady growth' in human energy use cannot continue. No. The first law of thermodynamics is a statement about energy conservation, while the second is a statement about stable thermal equilibrium. They are by no means mutually exclusive. For the particular case of a cycling heat engine, the first law implies Q h = Weng + Q c , and the second law implies Q c > 0.
Q22.4
Q22.5
Q22.6
Take an automobile as an example. According to the first law or the idea of energy conservation, it must take in all the energy it puts out. Its energy source is chemical energy in gasoline. During the combustion process, some of that energy goes into moving the pistons and eventually into the mechanical motion of the car. Clearly much of the energy goes into heat, which, through the cooling system, is dissipated into the atmosphere. Moreover, there are numerous places where friction, both mechanical and fluid, turns mechanical energy into heat. In even the most efficient internal combustion engine cars, less than 30% of the energy from the fuel actually goes into moving the car. The rest ends up as useless heat in the atmosphere. 631
632 Q22.7
Heat Engines, Entropy, and the Second Law of Thermodynamics
Suppose the ambient temperature is 20C. A gas can be heated to the temperature of the bottom of the pond, and allowed to cool as it blows through a turbine. The Carnot efficiency of such an engine T 80 = = 22%. is about e c = Th 373 No, because the work done to run the heat pump represents energy transferred into the house by heat. A slice of hot pizza cools off. Road friction brings a skidding car to a stop. A cup falls to the floor and shatters. Your cat dies. Any process is irreversible if it looks funny or frightening when shown in a videotape running backwards. The free flight of a projectile is nearly reversible. Below the frost line, the winter temperature is much higher than the air or surface temperature. The earth is a huge reservoir of internal energy, but digging a lot of deep trenches is much more expensive than setting a heatexchanger out on a concrete pad. A heat pump can have a much higher coefficient of performance when it is transferring energy by heat between reservoirs at close to the same temperature. (a) When the two sides of the semiconductor are at different temperatures, an electric potential (voltage) is generated across the material, which can drive electric current through an external circuit. The two cups at 50C contain the same amount of internal energy as the pair of hot and cold cups. But no energy flows by heat through the converter bridging between them and no voltage is generated across the semiconductors. A heat engine must put out exhaust energy by heat. The cold cup provides a sink to absorb output or wasted energy by heat, which has nowhere to go between two cups of equally warm water.
Q22.8 Q22.9
Q22.10
Q22.11
(b)
Q22.12
Energy flows by heat from a hot bowl of chili into the cooler surrounding air. Heat lost by the hot stuff is equal to heat gained by the cold stuff, but the entropy decrease of the hot stuff is less than the entropy increase of the cold stuff. As you inflate a soft car tire at a service station, air from a tank at high pressure expands to fill a larger volume. That air increases in entropy and the surrounding atmosphere undergoes no significant entropy change. The brakes of your car get warm as you come to a stop. The shoes and drums increase in entropy and nothing loses energy by heat, so nothing decreases in entropy. (a) (b) For an expanding ideal gas at constant temperature, S = V Q = nR ln 2 . V1 T
Q22.13
FG IJ H K
For a reversible adiabatic expansion Q = 0 , and S = 0 . An ideal gas undergoing an irreversible adiabatic expansion can have any positive value for S up to the value given in part (a).
Q22.14 Q22.15
The rest of the Universe must have an entropy change of +8.0 J/K, or more. Even at essentially constant temperature, energy must flow by heat out of the solidifying sugar into the surroundings, to raise the entropy of the environment. The water molecules become less ordered as they leave the liquid in the container to mix into the whole atmosphere and hydrosphere. Thus the entropy of the surroundings increases, and the second law describes the situation correctly.
Chapter 22
633
Q22.16
To increase its entropy, raise its temperature. To decrease its entropy, lower its temperature. "Remove energy from it by heat" is not such a good answer, for if you hammer on it or rub it with a blunt file and at the same time remove energy from it by heat into a constant temperature bath, its entropy can stay constant. An analogy used by Carnot is instructive: A waterfall continuously converts mechanical energy into internal energy. It continuously creates entropy as the organized motion of the falling water turns into disorganized molecular motion. We humans put turbines into the waterfall, diverting some of the energy stream to our use. Water flows spontaneously from high to low elevation and energy spontaneously flows by heat from high to low temperature. Into the great flow of solar radiation from Sun to Earth, living things put themselves. They live on energy flow, more than just on energy. A basking snake diverts energy from a hightemperature source (the Sun) through itself temporarily, before the energy inevitably is radiated from the body of the snake to a lowtemperature sink (outer space). A tree builds organized cellulose molecules and we build libraries and babies who look like their grandmothers, all out of a thin diverted stream in the universal flow of energy crashing down to disorder. We do not violate the second law, for we build local reductions in the entropy of one thing within the inexorable increase in the total entropy of the Universe. Your roommate's exercise puts energy into the room by heat. (a) (b) Entropy increases as the yeast dies and as energy is transferred from the hot oven into the originally cooler dough and then from the hot bread into the surrounding air. Entropy increases some more as you metabolize the starches, converting chemical energy into internal energy.
Q22.17
Q22.18
Q22.19
Either statement can be considered an instructive analogy. We choose to take the first view. All processes require energy, either as energy content or as energy input. The kinetic energy which it possessed at its formation continues to make the Earth go around. Energy released by nuclear reactions in the core of the Sun drives weather on the Earth and essentially all processes in the biosphere. The energy intensity of sunlight controls how lush a forest or jungle can be and how warm a planet is. Continuous energy input is not required for the motion of the planet. Continuous energy input is required for life because energy tends to be continuously degraded, as heat flows into lowertemperature sinks. The continuously increasing entropy of the Universe is the index to energytransfers completed. The statement is not true. Although the probability is not exactly zero that this will happen, the probability of the concentration of air in one corner of the room is very nearly zero. If some billions of molecules are heading toward that corner just now, other billions are heading away from the corner in their random motion. Spontaneous compression of the air would violate the second law of thermodynamics. It would be a spontaneous departure from thermal and mechanical equilibrium. Shaking opens up spaces between jellybeans. The smaller ones more often can fall down into spaces below them. The accumulation of larger candies on top and smaller ones on the bottom implies a small increase in order, a small decrease in one contribution to the total entropy, but the second law is not violated. The total entropy increases as the system warms up, its increase in internal energy coming from the work put into shaking the box and also from a bit of gravitational energy loss as the beans settle compactly together.
Q22.20
Q22.21
634
Heat Engines, Entropy, and the Second Law of Thermodynamics
SOLUTIONS TO PROBLEMS Section 22.1 P22.1 (a) (b) P22.2 Heat Engines and the Second Law of Thermodynamics e= Weng Qh = 25.0 J = 0.069 4 or 6.94% 360 J
Q c = Q h  Weng = 360 J  25.0 J = 335 J (1) (2) (3)
Weng = Q h  Q c = 200 J e= Weng Qh =1 Qc Qh = 0.300
From (2), Q c = 0.700 Q h Solving (3) and (1) simultaneously, we have (a) (b) Q h = 667 J and Q c = 467 J . We have e = Weng Qh = Qh  Qc Qh =1 Qc Qh = 0.250
P22.3
(a)
with Q c = 8 000 J, we have Q h = 10.7 kJ (b) Weng = Q h  Q c = 2 667 J and from P = *P22.4 Weng t , we have t = Weng = 2 667 J = 0.533 s . 5 000 J s
P
We have Q hx = 4Q hy , Weng x = 2Weng y and Q cx = 7Q cy . As well as Q hx = Weng x + Q cx and Q hy = Weng y + Q cy . Substituting, 4Q hy = 2Weng y + 7Q cy 4Q hy = 2Weng y + 7Q hy  7Weng y 5Weng y = 3Q hy (b) ey = Weng y Q hy Weng x Q hx = 3 = 60.0% 5 2Weng y 4Q hy = 2 0.600 = 0.300 = 30.0% 4
(a)
ex =
=
a
f
Chapter 22
635
*P22.5
(a)
The input energy each hour is
e7.89 10
3
J revolution 2 500 rev min
jb
implying fuel input 1.18 10 9 (b)
e
g 601min = 1.18 10 J h h F 1 L IJ = 29.4 L h J hjG H 4.03 10 J K
9 7
Q h = Weng + Q c . For a continuoustransfer process we may divide by time to have Q h Weng Q c = + t t t Weng Q h Q c Useful power output = =  t t t =
3
(c)
(d) P22.6
F 7.89 10 J  4.58 10 J I 2 500 rev 1 min = 1.38 10 GH revolution revolution JK 1 min 60 s F 1 hp IJ = 185 hp P = 1.38 10 WG H 746 W K P 1.38 10 J s F 1 rev I P = = = b2 500 rev 60 sg GH 2 rad JK = 527 N m Q 4.58 10 J F 2 500 rev I = G J = 1.91 10 W t revolution H 60 s K
3 eng 5 eng 5 eng c 3 5
5
W
The heat to melt 15.0 g of Hg is
Q c = mL f = 15 10 3 kg 1.18 10 4 J kg = 177 J
e
je
j
The energy absorbed to freeze 1.00 g of aluminum is Q h = mL f = 10 3 kg 3.97 10 5 J / kg = 397 J and the work output is Weng = Q h  Q c = 220 J e= The theoretical (Carnot) efficiency is Weng Qh = 220 J = 0.554 , or 55.4% 397 J
e
je
j
Th  Tc 933 K  243.1 K = = 0.749 = 74.9% 933 K Th
Section 22.2 P22.7
Heat Pumps and Refrigerators
COP refrigerator = (a) (b)
b
g
Qc W
If Q c = 120 J and COP = 5.00 , then W = 24.0 J Heat expelled = Heat removed + Work done. Q h = Q c + W = 120 J + 24 J = 144 J
636 P22.8
Heat Engines, Entropy, and the Second Law of Thermodynamics
Qc Q . Therefore, W = c . W 3.00 The heat removed each minute is COP = 3.00 = QC = 0.030 0 kg 4 186 J kg C 22.0 C + 0.030 0 kg 3.33 10 5 J kg t + 0.030 0 kg 2 090 J kg C 20.0 C = 1.40 10 4 J min or, Qc = 233 J s. t
b b
gb gb
ga ga
f b f
ge
j
Thus, the work done per sec = P =
233 J s = 77.8 W . 3.00 2.93
P22.9
(a) (b) (c)
FG 10.0 Btu IJ FG 1055 J IJ FG 1 h IJ FG 1 W IJ = H h W K H 1 Btu K H 3 600 s K H 1 J s K
With EER 5, 5 Btu 10 000 Btu h = : P h W
Coefficient of performance for a refrigerator:
aCOPf
P=
refrigerator
10 000 Btu h
5 Btu hW
= 2 000 W = 2.00 kW
Energy purchased is
a fb g Cost = e3.00 10 kWhjb0.100 $ kWhg = $300
P t = 2.00 kW 1 500 h = 3.00 10 3 kWh
3
With EER 10, 10
Btu 10 000 Btu h = : P h W
P=
10 000 Btu h
10 Btu hW
= 1 000 W = 1.00 kW
Energy purchased is
a fb g Cost = e1.50 10 kWhjb0.100 $ kWhg = $150
P t = 1.00 kW 1 500 h = 1.50 10 3 kWh
3
Thus, the cost for air conditioning is
half as much with EER 10
Section 22.3
Reversible and Irreversible Processes
No problems in this section
Section 22.4 P22.10
The Carnot Engine
Weng t Qh t 5
Weng T and When e = e c , 1  c = Th Qh
Weng t
=1
Tc Th
(a)
Qh
FH IK t e1.50 10 Wjb3 600 sg = =
1  Tc
T
h
1
293 773
Q h = 8.69 10 8 J = 869 MJ (b) Qc = Qh 
F W I t = 8.69 10  e1.50 10 jb3 600g = 3.30 10 GH t JK
eng 8 5
8
J = 330 MJ
Chapter 22
637
P22.11
Tc = 703 K (a) (b) ec =
Th = 2 143 K T 1 440 = = 67.2% Th 2 143
Q h = 1.40 10 5 J , Weng = 0.420 Q h
P=
P22.12
Weng t
=
5.88 10 4 J = 58.8 kW 1s ec = T 120 K = = 0.253 Th 473 K
The Carnot efficiency of the engine is At 20.0% of this maximum efficiency, From the definition of efficiency and
e = 0.200 0.253 = 0.050 6 Weng = Q h e Qh = Weng e = 10.0 kJ = 197 kJ 0.050 6
a
f
P22.13
Isothermal expansion at Isothermal compression at
Th = 523 K Tc = 323 K
Gas absorbs 1 200 J during expansion. (a) (b) P22.14 We use as From which, *P22.15 The efficiency is Then Qc = Qh Weng
FG T IJ = 1 200 JFG 323 IJ = 741 J H 523 K HT K = Q  Q = b1 200  741g J = 459 J
c h h c
ec = 1 
Tc Th 573 K Th
0.300 = 1 
Th = 819 K = 546 C ec = 1  Tc = Th Qh t = Qc Tc =1 Th Qh
Qc t Qh t
Q c Th 273 + 100 K = 15. 4 W = 19.6 W t Tc 273 + 20 K
a a
f f
(a)
Q h = Weng + Q c The useful power output is Weng t m= = Qh t  Qc t = 19.6 W  15.4 W = 4.20 W
(b)
Qh =
F Q I t = mL GH t JK
h
V
Q h t 3 600 s = 19.6 J s = 3.12 10 2 kg t LV 2.26 10 6 J kg
b
gFGH
I JK
638 P22.16
Heat Engines, Entropy, and the Second Law of Thermodynamics
The Carnot summer efficiency is And in winter, Then the actual winter efficiency is
e c ,s = 1  ec ,w = 1  0.320
273 + 20 K Tc =1 = 0.530 Th 273 + 350 K 283 = 0.546 623
a a
f f
P22.17
(a)
FG 0.546 IJ = 0.330 or 33.0% H 0.530 K F P V I = F PV I . In an adiabatic process, P V = PV . Also, G H T JK GH T JK F P Ib g Dividing the second equation by the first yields T = T G J HPK
f
f
i i
f
f
i i i
f
f
i
f
1
.
i
Since =
1 2 5 = = 0.400 and we have for Argon, 5 3
T f = 1 073 K
b
gFGH 1300 10 Pa IJK .50 10 Pa
3 6
0. 400
= 564 K .
(b)
Eint = nCV T = Q  Weng = 0  Weng , so Weng = nCV T , and the power output is
P=
=
Weng t
=
b80.0 kg ge
5
nCV T or t
1.00 mol 0 .039 9 kg
jc hb8.314 J mol K gb564  1 073gK
3 2
60.0 s
P = 2.12 10 W = 212 kW
(c) eC = 1  Tc 564 K =1 = 0.475 or 47.5% 1 073 K Th Tc 278 =1 = 5.12 10 2 = 5.12% Th 293 = 75.0 10 6 J s Weng = 75.0 10 6 J s 3 600 s h = 2.70 10 11 J h Qh we find Qh = Weng e = 2.70 10 11 J h 5.12 10 2 = 5.27 10 12 J h = 5.27 TJ h
P22.18
(a)
emax = 1  Weng t
(b)
P=
Therefore, From e = (c) Weng
e
jb
g
As fossilfuel prices rise, this way to use solar energy will become a good buy.
Chapter 22
639
*P22.19
(a)
e=
Weng1 + Weng2 Q h1
=
e1 Q1 h + e 2 Q 2 h Q h1
Now Q 2 h = Q1 c = Q1 h  Weng 1 = Q h1  e1Q1 h . So e = e 1 Q 1 h + e 2 Q 1 h  e1 Q 1 h Q1 h
b
g=
e1 + e 2  e 1 e 2 .
(b)
e = e 1 + e 2  e1 e 2 = 1 
Ti T T +1 c  1 i Th Ti Th
FG H
IJ FG 1  T IJ = 2  T KH T K T
c i
i

h
Tc T T T T 1+ i + c  c = 1 c Ti Th Ti Th Th
The combination of reversible engines is itself a reversible engine so it has the Carnot efficiency. (c) With Weng2 = Weng1 , e = 1 0 Tc T = 2 1 i Th Th Weng1 + Weng2 Q1 h = 2Weng1 Q1 h = 2 e1
FG H
IJ K
2T Tc =1 i Th Th 2Ti = Th + Tc Ti = 1 Th + Tc 2
b
g
(d)
e1 = e 2 = 1  Ti2 = Tc Th
Ti T =1 c Th Ti
Ti = ThTc P22.20
b g
12
The work output is Weng = We are told e = 0.200 = and Weng Qh
1 2 m train 5.00 m s . 2
b
g
5.00 m s 1 mt Qh 2
b
g
2
6.50 m s 300 K 1 = mt eC = 1  2 Th Qh
b
g
2
.
5.00 m s 1 Substitute Q h = m t 2 0.200 Then, 300 K = 0.200 1 Th 1
b
g
2
.
2 2
F GG H
1 2 1 2
b g m b5.00 m sg
m t 6.50 m s
t
I JJ K
300 K = 0.338 Th 300 K Th = = 453 K 0.662
640 P22.21
Heat Engines, Entropy, and the Second Law of Thermodynamics
For the Carnot engine, e c = 1  Also, so and ec =
Tc 300 K =1 = 0.600 . 750 K Th . = 150 J = 250 J . 0.600
Weng Qh ec
Qh =
Weng
Q c = Q h  Weng = 250 J  150 J = 100 J . FIG. P22.21
(a)
Qh =
Weng eS
=
150 J = 214 J 0.700
Q c = Q h  Weng = 214 J  150 J = 64.3 J (b) Q h, net = 214 J  250 J = 35.7 J Q c, net = 64.3 J  100 J = 35.7 J The net flow of energy by heat from the cold to the hot reservoir without work input, is impossible. (c) For engine S: so and (d) Q c = Q h  Weng = Weng = Qc
1 eS
FIG. P22.21(b)
Weng eS
 Weng .
1
=
100 J = 233 J . 1 0 .700  1
Q h = Q c + Weng = 233 J + 100 J = 333 J .
Q h, net = 333 J  250 J = 83.3 J Wnet = 233 J  150 J = 83.3 J Q c,net = 0 The output of 83.3 J of energy from the heat engine by work in a cyclic process without any exhaust by heat is impossible. FIG. P22.21(d)
(e)
Both engines operate in cycles, so For the reservoirs,
SS = SCarnot = 0 . S h =  Qh Th and S c = + Qc Tc .
Thus, Stotal = SS + SCarnot + S h + S c = 0 + 0  A decrease in total entropy is impossible.
83.3 J 0 + = 0.111 J K . 750 K 300 K
Chapter 22
641
P22.22
(a)
First, consider the adiabatic process D A :
PD VD = PA VA so
Also
or
FG V IJ = 1 400 kPaFG 10.0 L IJ H 15.0 L K HV K FG nRT IJ V = FG nRT IJ V HV K HV K F V IJ = 720 K FG 10.0 IJ = T =T G H 15.0 K HV K
PD = PA
D
A
53
= 712 kPa .
D
D
D
A
A
A
1
D
A
A
23
549 K .
D
Now, consider the isothermal process C D : TC = TD = 549 K . PC = PD
PC
FG V IJ = LMP FG V IJ OPFG V IJ = P V H V K MN H V K PQH V K V V 1 400 kPaa10.0 L f = = 445 kPa 24.0 La15.0 L f
D C A A D C A D C 53 23
A 1 D
Next, consider the adiabatic process B C : PBVB = PC VC .
But, PC =
Hence, PA
FG V IJ V = F P V I V which reduces to V = V V V H V K GH V V JK F V IJ = 1 400 kPaFG 10.0 L IJ = 875 kPa . Finally, P = P G H 16.0 L K HV K
A B
VC VD 1
PA VA
from above. Also considering the isothermal process, PB = PA
B
A A 1 D
C
C
B
A C D
FG V IJ . HV K 10.0 La 24.0 L f = = 16.0 L
A B
15.0 L
.
B
A
A B
State A B C D (b)
P(kPa) 1 400 875 445 712
V(L) 10.0 16.0 24.0 15.0
T(K) 720 720 549 549 Eint = nCV T = 0 +6.58 kJ .
For the isothermal process A B : so Q = W = nRT ln
B
FG V IJ = 2.34 molb8.314 J mol K ga720 K f lnFG 16.0 IJ = H 10.0 K HV K
A
For the adiabatic process B C : Eint = nCV TC  TB = 2.34 mol and W = Q + Eint continued on next page
Q= 0 4.98 kJ
b
g
LM 3 b8.314 J mol K gOPa549  720f K = N2 Q = 0 + a 4.98 kJf = 4.98 kJ .
642
Heat Engines, Entropy, and the Second Law of Thermodynamics
For the isothermal process C D : and Q = W = nRT ln
D C
Eint = nCV T = 0 5.02 kJ .
FG V IJ = 2.34 molb8.314 J mol K ga549 K f lnFG 15.0 IJ = H 24.0 K HV K
Q= 0
Finally, for the adiabatic process D A : Eint = nCV TA  TD = 2.34 mol
b
g
LM 3 b8.314 J mol K gOPa720  549f K = N2 Q
W(kJ) 6.58 4.98 +5.02 +4.98 1.56 Eint (kJ) 0 4.98 0 +4.98 0
+4.98 kJ
and W = Q + Eint = 0 + 4.98 kJ = +4.98 kJ . Process AB BC CD D A ABCDA Q(kJ) +6.58 0 5.02 0 +1.56
The work done by the engine is the negative of the work input. The output work Weng is given by the work column in the table with all signs reversed. (c) e= Weng Qh = W ABCD 1.56 kJ = = 0.237 or 23.7% 6.58 kJ Q AB
ec = 1 
Tc 549 =1 = 0.237 or 23.7% 720 Th
P22.23
aCOPf aCOPf
(a)
refrig
=
Tc 270 = = 9.00 T 30.0 = Qc + W W = Th 295 = = 11.8 T 25 W = Qh  Q c = Q c
P22.24
heat pump
P22.25
For a complete cycle, Eint = 0 and
LM bQ g OP MN Q  1PQ .
h c
We have already shown that for a Carnot cycle (and only for a Carnot cycle)
Qh Qc
=
Th . Tc
Therefore,
W = Qc
LM T  T OP N T Q
h c c
. Qc W
(b)
We have the definition of the coefficient of performance for a refrigerator, Using the result from part (a), this becomes COP = Tc . Th  Tc
COP =
.
Chapter 22
643
P22.26
COP = 0.100COPCarnot cycle or Qh W Qh W
FQ I F I 1 = 0.100G GH W JK H Carnot efficiency JK F T IJ = 0.100FG 293 K IJ = 1.17 = 0.100G H 293 K  268 K K HT T K
= 0.100
h Carnot cycle h h c
FIG. P22.26 Thus, 1.17 joules of energy enter the room by heat for each joule of work done. P22.27 Qc Tc 4.00 = = 0.013 8 = T 289 W W = 72.2 J per 1 J energy removed by heat.
aCOPf
Carnot refrig
=
P22.28
A Carnot refrigerator runs on minimum power. Q t Q t Q Q For it: h = c so h = c . Th Tc Th Tc Solving part (b) first: (b) Q h Q c Th 298 K 1h = = 8.00 MJ h = 8.73 10 6 J h = 2. 43 kW 273 K 3 600 s t t Tc 8.00 10 6 J h W Qh Qc =  = 2.43 kW  = 204 W t t t 3 600 s h W = 0.350 Qh W = 0.350Q h
FG IJ b H K
gFGH
IJ e K
jFGH
IJ K
(a) P22.29 e=
Q c = 0.650Q h Q c 0.650Q h = = 1.86 COP refrigerator = W 0.350Q h
Qh = W + Qc
b
g
*P22.30
To have the same efficiencies as engines, 1  between reservoirs with the same ratio becomes Thp Thp  Tcp = Tcp Thp
Tcp Thp
=1
Tcr the pump and refrigerator must operate Thr
=
Tcr , which we define as r. Now COPp = 1.50COPr Thr
Thp 3 Tcr 2 3r 2 3 rThr or = ,r= . , = 2 Thr  Tcr 3 Thp  rThp 2 Thr  rThr 1  r 1  r = 2.00
(a) (b) (c)
COPr = COPp =
2 r = 3 1r 1
2 3
1 1 = = 3.00 2 1r 1 3 2 = 33.3% 3
e=1r =1
644
Heat Engines, Entropy, and the Second Law of Thermodynamics
Section 22.5 P22.31 (a)
Gasoline and Diesel Engines
PVi = Pf V f i
Pf
FV I F 50.0 cm I = P G J = e3.00 10 PajG H 300 cm JK HV K
i i 6 3 f 3 Vi Vi
1.40
= 244 kPa
(b)
W = PdV Integrating,
z
P = Pi
FG V IJ HVK
i
F 1 IJ PV LM1  FG V IJ W =G H  1 K MN H V K
i i i f
1
OP PQ = a2.50fe3.00 10 Paje5.00 10
6
5
L F 50.0 cm I OP m jM1  G MN H 300 cm JK PQ
3 3 0. 400 3
= 192 J P22.32 Compression ratio = 6.00 , = 1.40 (a) Efficiency of an Ottoengine e = 1 
FG V IJ HV K F 1 IJ e=1G H 6.00 K
1
2 1
0 . 400
= 51. 2% .
(b) P22.33
If actual efficiency e = 15.0% losses in system are e  e = 36.2% . 1 =1 1 =1
eOtto = 1 
bV V g
1 2
1
a6.20fb7 51g
a6.20f
1
0. 400
eOtto = 0.518 We have assumed the fuelair mixture to behave like a diatomic gas. Now e = Weng Qh = Weng t Qh t
746 W 1 hp Q h Weng t = = 102 hp t e 0.518 Qh = 146 kW t Q h = Weng + Q c Qc t Qc t = Q h Weng  t t
= 146 10 3 W  102 hp
F 746 W I = GH 1 hp JK
70.8 kW
Chapter 22
645
P22.34
(a), (b) The quantity of gas is n= 100 10 3 Pa 500 10 6 m 3 PA VA = 0.020 5 mol = RTA 8.314 J mol K 293 K
e
b
je
ga
f
j
Eint, A =
5 5 5 nRTA = PA VA = 100 10 3 Pa 500 10 6 m 3 = 125 J 2 2 2
e
je
j
In process AB, PB = PA
FG V IJ = e100 10 Paja8.00f HV K
A B 3
1.40
= 1.84 10 6 Pa
1.84 10 6 Pa 500 10 6 m3 8.00 PBVB = 673 K = TB = nR 0.020 5 mol 8.314 J mol K
e
b
je
gb
g
j
Eint, B = so
5 5 nRTB = 0.020 5 mol 8.314 J mol K 673 K = 287 J 2 2 W AB = 162 J
b
gb
ga
f
Eint, AB = 287 J  125 J = 162 J = Q  Wout = 0  Wout
Process BC takes us to: 0.020 5 mol 8.314 J mol K 1 023 K nRTC = = 2.79 10 6 Pa VC 62.5 10 6 m 3 5 5 Eint, C = nRTC = 0.020 5 mol 8.314 J mol K 1 023 K = 436 J 2 2 Eint, BC = 436 J  287 J = 149 J = Q  Wout = Q  0 PC =
b
gb
gb
g
b
gb
gb
g
QBC = 149 J
In process CD: PD = PC TD
Eint, D
FG V IJ = e2.79 10 PajFG 1 IJ = 1.52 10 Pa H 8.00 K HV K e1.52 10 Paje500 10 m j = 445 K P V = = nR b0.020 5 molgb8.314 J mol K g 5 5 = nRT = b0.020 5 molgb8.314 J mol K ga 445 K f = 2 2
C 6 1.40 5 D 5 6 3 D D D
190 J
Eint, CD = 190 J  436 J = 246 J = Q  Wout = 0  Wout WCD = 246 J and Eint, DA = Eint, A  Eint, D = 125 J  190 J = 65.0 J = Q  Wout = Q  0 QDA = 65.0 J continued on next page
646
Heat Engines, Entropy, and the Second Law of Thermodynamics
For the entire cycle, Eint, net = 162 J + 149  246  65.0 = 0 . The net work is Weng = 162 J + 0 + 246 J + 0 = 84.3 J Q net = 0 + 149 J + 0  65.0 J = 84.3 J The tables look like: State A B C D A Process AB BC CD DA ABCDA (c) T(K) 293 673 1 023 445 293 Q(J) 0 149 0 65.0 84.3 P(kPa) 100 1 840 2 790 152 100 output W(J) 162 0 246 0 84.3 V(cm3 ) 500 62.5 62.5 500 500 Eint (J) 162 149 246 65.0 0 Eint (J) 125 287 436 190 125
The input energy is Q h = 149 J , the waste is Q c = 65.0 J , and Weng = 84.3 J . The efficiency is: e = Weng Qh = 84.3 J = 0.565 . 149 J f is the frequency at which we 2
(d)
(e)
Let f represent the angular speed of the crankshaft. Then obtain work in the amount of 84.3 J/cycle: 1 000 J s = f=
FG f IJ b84.3 J cycleg H 2K
2 000 J s = 23.7 rev s = 1.42 10 3 rev min 84.3 J cycle
Section 22.6 P22.35
Entropy
For a freezing process, S =
5 Q  0.500 kg 3.33 10 J kg = = 610 J K . 273 K T
b
ge
j
Chapter 22
647
P22.36
At a constant temperature of 4.20 K, 20.5 kJ kg Lv Q = = T 4.20 K 4.20 K S = 4.88 kJ kg K S =
P22.37
FG IJ H K F 353 IJ = 46.6 cal K = S = 250 g b1.00 cal g C g lnG H 293 K
S =
z z
f T i
f Tf dQ mcdT = = mc ln T T Ti T i
195 J K
*P22.38
(a)
The process is isobaric because it takes place under constant atmospheric pressure. As described by Newton's third law, the stewing syrup must exert the same force on the air as the air exerts on it. The heating process is not adiabatic (energy goes in by heat), isothermal (T goes up), isovolumetic (it likely expands a bit), cyclic (it is different at the end), or isentropic (entropy increases). It could be made as nearly reversible as you wish, by not using a kitchen stove but a heater kept always just incrementally higher in temperature than the syrup. The process would then also be eternal, and impractical for food production.
(b)
The final temperature is 220 F = 212 F + 8 F = 100 C + 8 F For the mixture, Q = m1 c 1 T + m 2 c 2 T = 900 g 1 cal g C + 930 g 0.299 cal g C 104.4 C  23 C = 9.59 10 4 cal = 4.02 10 5 J
FG 100  0 C IJ = 104 C . H 212  32 F K
ga f
b
(c)
Consider the reversible heating process described in part (a): S =
b g T T T F 4.186 J IJ FG 1 C IJ lnFG 273 + 104 IJ = 900a1f + 930a0.299f bcal C gG H 1 cal K H 1 K K H 273 + 23 K = b 4 930 J K g0.243 = 1.20 10 J K
i
z zb
f
dQ = T
f
m1 c 1 + m 2 c 2 dT
g
= m1 c 1 + m 2 c 2 ln
f
i
i
3
648 *P22.39
Heat Engines, Entropy, and the Second Law of Thermodynamics
We take data from the description of Figure 20.2 in section 20.3, and we assume a constant specific heat for each phase. As the ice is warmed from 12C to 0C, its entropy increases by S =
z
f i
mc ice dT dQ 273 K = = mc ice T 1 dT = mc ice ln T 261 K T T 261 K 261 K
273 K
z
273 K
z
S = 0.027 0 kg 2 090 J kg C ln 273 K  ln 261 K = 0.027 0 kg 2 090 J kg C ln S = 2.54 J K As the ice melts its entropy change is
5 Q mL f 0.027 0 kg 3.33 10 J kg S = = = = 32.9 J K T T 273 K
b
ga
f
b
gFGH FGH 273 IJK IJK 261
e
j
As liquid water warms from 273 K to 373 K, S =
z
f i
mc liquid dT T
= mc liquid ln
FG T IJ = 0.027 0 kgb4 186 J kg Cg lnFG 373 IJ = 35.3 J K H 273 K HT K
f i
As the water boils and the steam warms, S = S = Tf mL v + mc steam ln T Ti
0.027 0 kg 2.26 10 6 J kg 373 K
e
FG IJ H K
j + 0.027 0 kgb2 010 J kg Cg lnFG 388 IJ = 164 J K + 2.14 J K H 373 K
236 J K .
The total entropy change is
a2.54 + 32.9 + 35.3 + 164 + 2.14f J K =
b gFGH I JK
We could equally well have taken the values for specific heats and latent heats from Tables 20.1 and 20.2. For steam at constant pressure, the molar specific heat in Table 21.2 implies a specific heat of 1 mol = 1 970 J kg K , nearly agreeing with 2 010 J kg K . 35.4 J mol K 0.018 kg
Section 22.7 P22.40 P22.41 S =
Entropy Changes in Irreversible Processes 1 000 1 000 Q 2 Q1  =  T2 T1 290 5 700
F GH
I JK
J K = 3.27 J K
The car ends up in the same thermodynamic state as it started, so it undergoes zero changes in entropy. The original kinetic energy of the car is transferred by heat to the surrounding air, adding to the internal energy of the air. Its change in entropy is S =
1 2
mv 2 T
=
750 20.0 293
a f
2
J K = 1.02 kJ K .
Chapter 22
649
P22.42
c iron = 448 J kg C ; c water = 4 186 J kg C Qcold = Q hot : which yields 4.00 kg 4 186 J kg C T f  10.0 C =  1.00 kg 448 J kg C T f  900 C T f = 33.2 C = 306.2 K S =
306.2 K 283 K
b
gd
i b
gb
gd
i
z
FG 306.2 IJ + c m lnFG 306.2 IJ H 283 K H 1 173 K S = b 4 186 J kg K gb 4.00 kg gb0.078 8g + b 448 J kg K gb1.00 kg ga 1.34f
S = c water m water ln
iron iron
c water m water dT 306.2 K c iron m iron dT + T T 1 173 K
z
S = 718 J K P22.43 Sitting here writing, I convert chemical energy, in ordered molecules in food, into internal energy that leaves my body by heat into the roomtemperature surroundings. My rate of energy output is equal to my metabolic rate, 2 500 kcal d = 2 500 10 3 cal 4.186 J = 120 W . 86 400 s 1 cal
FG H
IJ K
My body is in steady state, changing little in entropy, as the environment increases in entropy at the rate S Q T Q t 120 W = = = = 0. 4 W K ~ 1 W K . 293 K t t T When using powerful appliances or an automobile, my personal contribution to entropy production is much greater than the above estimate, based only on metabolism. P22.44 (a) V= 40.0 g 8.314 J mol K 473 K nRTi = 39.4 10 3 m3 = 39.4 L = 3 Pi 39.9 g mol 100 10 Pa
b
b
gb
ge
ga
j
f
(b) (c) (d)
Eint = nCV T = W =0 Sargon = so
F 40.0 gm I L 3 b8.314 J mol K gOa200 Cf = GH 39.9 g mol JK MN 2 PQ
Q = Eint = 2.50 kJ
2.50 kJ
FG IJ H K F 40.0 g I L 3 b8.314 J mol K gO lnFG 273 IJ = =G PQ H 473 K H 39.9 g mol JK MN 2
z
f i
Tf dQ = nCV ln T Ti
6.87 J K
(e)
S bath =
2.50 kJ = +9.16 J K 273 K Stotal = Sargon + S bath = 6.87 J K + 9.16 J K = +2.29 J K Stotal > 0 for this irreversible process.
The total change in entropy is
650 P22.45
Heat Engines, Entropy, and the Second Law of Thermodynamics
S = nR ln
F V I = R ln 2 = GH V JK
f i
5.76 J K
There is no change in temperature . FIG. P22.45 P22.46
F V I = b0.044 0ga2fR ln 2 GH V JK S = 0.088 0a8.314f ln 2 = 0.507 J K
S = nR ln
f i
FIG. P22.46 P22.47 For any infinitesimal step in a process on an ideal gas, dEint = dQ + dW : and If the whole process is reversible, Also, from the ideal gas law, S = 1.00 mol dQ = dEint  dW = nCV dT + PdV = nCV dT + dQ dT dV = nCV + nR T T V S = Tf Ti = nRTdV V
z
f i
dQr = T
z FGH
f i
nCV
Tf Vf dT dV + nR = nCV ln + nR ln T V Ti Vi
IJ K
FG IJ H K
FG IJ H K
Pf V f PVi i
a
2 00fb0 040 0 fLMN 3 b8.314 J mol K gOPQ lnFGH aa1..00fb0..025 0gg IJK + a1.00 molfb8.314 J mol K g lnFGH 0..040 0 IJK 2 0 025 0
= 18.4 J K P22.48 S = nC V ln
F T I + nR lnF V I GH T JK GH V JK L5 O F 2P 2V IJ + a1.00 molfb8.314 J mol K g lnFG 2V IJ = a1.00 molfM b8.314 J mol K gP lnG HVK N2 Q H PV K
f f i i
S = 34.6 J K
Chapter 22
651
Section 22.8 P22.49 (a) (b) P22.50 (a)
Entropy on a Microscopic Scale A 12 can only be obtained one way 6 + 6 A 7 can be obtained six ways: 6 + 1 , 5 + 2 , 4 + 3 , 3 + 4 , 2 + 5 , 1 + 6 The table is shown below. On the basis of the table, the most probable result of a toss is 2 heads and 2 tails . The most ordered state is the least likely state. Thus, on the basis of the table this is either all heads or all tails . The most disordered is the most likely state. Thus, this is 2 heads and 2 tails . Result All heads 3H, 1T 2H, 2T 1H, 3T All tails Possible Combinations HHHH THHH, HTHH, HHTH, HHHT TTHH, THTH, THHT, HTTH, HTHT, HHTT HTTT, THTT, TTHT, TTTH TTTT Possible Combinations RRR RRG, RGR, GRR RGG, GRG, GGR GGG Possible Combinations RRRRR RRRRG, RRRGR, RRGRR, RGRRR, GRRRR RRRGG, RRGRG, RGRRG, GRRRG, RRGGR, RGRGR, GRRGR, RGGRR, GRGRR, GGRRR GGGRR, GGRGR, GRGGR, RGGGR, GGRRG, GRGRG, RGGRG, GRRGG, RGRGG, RRGGG RGGGG, GRGGG, GGRGG, GGGRG, GGGGR GGGGG Total 1 4 6 4 1 Total 1 3 3 1 Total 1 5 10 10 5 1
(b)
(c)
P22.51
(a)
Result All red 2R, 1G 1R, 2G All green Result All red 4R, 1G 3R, 2G 2R, 3G 1R, 4G All green
(b)
Additional Problems P22.52 The conversion of gravitational potential energy into kinetic energy as the water falls is reversible. But the subsequent conversion into internal energy is not. We imagine arriving at the same final state by adding energy by heat, in amount mgy, to the water from a stove at a temperature infinitesimally above 20.0C. Then, S =
z
3 3 2 dQ Q mgy 5 000 m 1 000 kg m 9.80 m s 50.0 m = = = = 8.36 10 6 J K . T T T 293 K
e
je
ja
f
652 P22.53
Heat Engines, Entropy, and the Second Law of Thermodynamics
(a)
H ET so if all the electric energy is converted into internal energy, the steadystate t condition of the house is described by H ET = Q .
Pelectric =
Therefore, (b) For a heat pump,
Pelectric =
Q = 5 000 W t = Th 295 K = = 10.92 T 27 K
aCOPf
Carnot
Actual COP = 0.6 10.92 = 6.55 =
a
f
Qh W
=
Q h t W t
Therefore, to bring 5 000 W of energy into the house only requires input power
Pheat pump =
P22.54 Q c = mcT + mL + mcT =
W Q h t 5 000 W = = = 763 W t COP 6.56
Q c = 0.500 kg 4 186 J kg C 10 C + 0.500 kg 3.33 10 5 J kg + 0.500 kg 2 090 J kg C 20 C Q c = 2.08 10 5 J Qc W W= = COPc refrigerator = Q c Th  Tc Tc
b
ga
f
e
j
b
ga
f
b
g
b
g = e2.08 10 Jj 20.0 C  a20.0 Cf = a273  20.0f K
5
Tc Th  Tc
32.9 kJ
P22.55
S hot = Scold = (a) (b)
1 000 J 600 K +750 J 350 K SU = S hot + Scold = 0.476 J K ec = 1  T1 = 0.417 T2
Weng = e c Q h = 0.417 1 000 J = 417 J (c) Wnet = 417 J  250 J = 167 J T1 SU = 350 K 0.476 J K = 167 J
b
g
b
g
Chapter 22
653
*P22.56
(a)
The energy put into the engine by the hot reservoir is dQ h = mcdTh . The energy put into the cold reservoir by the engine is dQ c =  mcdTc = 1  e dQ h = 1  1  
Tf Tc
a f
LM F MN GH
Tc Th
IJ OPmcdT . Then K PQ
h
dTc dTh = Tc Th dT dT  = T T T
h
z
Tf
z
 ln T ln
Tf Tc
= ln T T f Tf
T
h
Tc = ln Tf Th
T f2 = Tc Th
T f = ThTc (b)
b g
i
12
The hot reservoir loses energy Q h = mc Th  T f . The cold reservoir gains Q c = mc T f  Tc . Then Q h = Weng + Q c .
d
d
i
P22.57
(a)
d i d i = mceT  T T  T T + T j = mceT  2 T T + T j = mce T  T j FV I For an isothermal process, Q = nRT lnG J HV K Therefore, Q = nRb3T g ln 2 F 1I Q = nRbT g lnG J and H 2K 3 = nRbT  3T g For the constant volume processes, Q = E 2 3 = nRb3T  T g and Q = E 2
Weng = mc Th  T f  mc T f  Tc
h h h c h h c c c c h c 2 1 1 i 3 i 2 int, 2 i i 4 int, 4 i i
2
The net energy by heat transferred is then Q = Q1 + Q 2 + Q 3 + Q 4 or (b) Q = 2nRTi ln 2 .
FIG. P22.57
A positive value for heat represents energy transferred into the system. Therefore, Q h = Q1 + Q 4 = 3nRTi 1 + ln 2 Eint = 0 and Weng = Q Therefore, the efficiency is ec = Weng Qh = 2 ln 2 Q = = 0.273 Q h 3 1 + ln 2
a
f
Since the change in temperature for the complete cycle is zero,
a
f
654 P22.58
Heat Engines, Entropy, and the Second Law of Thermodynamics
(a)
Weng t
= 1.50 10 Waelectrical
8
L OP , Q = mL = M f MN 0.150 PQt ,
Weng t
and L = 33.0 kJ g = 33.0 10 6 J kg
(b)
LM W t OP t N 0.150 Q L e1.50 10 Wjb86 400 s dayg m= = 2 620 metric tons day 0.150e33.0 10 J kg je10 kg metric tonj Cost = b$8.00 metric tongb 2 618 metric tons day gb365 days yr g
m=
eng 8 6 3
Cost = $7.65 million year (c) First find the rate at which heat energy is discharged into the water. If the plant is 15.0% efficient in producing electrical energy then the rate of heat production is Qc t Then, Qc t = =
FW GH t
eng
I FG 1  1IJ = e1.50 10 WjFG 1  1IJ = 8.50 10 JK H e K H 0.150 K
8
8
W.
mcT and t
c 8.50 10 8 J s m = t = = 4.06 10 4 kg s . t cT 4 186 J kg C 5.00 C
Q
b
ga
f
P22.59
Weng T = ec = 1  c = Th Qh Q h = Weng + Q c :
Weng t Qh t
:
Qh t Qc t Qc t Qc t
=
b
P Th P = Th  Tc 1  Tc Th
g
= = =
Qh t

Weng t
P Th P Tc P = Th  Tc Th  Tc
Q c = mcT :
FG m IJ cT = P T H t K T  T
c h
c
P Tc m = Th  Tc cT t
b
g
1.00 10 9 W 300 K m = = 5.97 10 4 kg s t 200 K 4 186 J kg C 6.00 C
e
b
ja
ga
f
f
Chapter 22
655
P22.60
Weng T = ec = 1  c = Th Qh
Weng t Qh t
Qh t
c
e1  j F Q I P = P T Q =G t H t J K T T
Tc Th h c h
=
P
=
P Th Th  Tc
c
Q c = mcT , where c is the specific heat of water. Therefore, and Qc t =
FG m IJ cT = P T H t K T  T
c h
c
m = t 35.0 F =
b
P Tc Th  Tc cT
g
P22.61
(a)
a f a f 5 98.6 F = a98.6  32.0f C = a37.0 + 273.15f K = 310.15 K 9 dQ dT F 310.15 IJ = 54.86 cal K S =z = b 453.6 g gb1.00 cal g K g z = 453.6 lnG H 274.82 K T T Q a310.15  274.82f = 51.67 cal K S = = a 453.6 fa1.00f
5 35.0  32.0 C = 1.67 + 273.15 K = 274.82 K 9
310.15 ice water 274.82 body
Tbody
310.15
Ssystem = 54.86  51.67 = 3.19 cal K (b)
a453.6fa1fbT
Thus,
F
 274.82 = 70.0 10 3 1 310.15  TF
g e
ja fb
3 F
g
3
b70.0 + 0.453 6g 10 T = a70.0fa310.15f + b0.453 6ga274.82f 10
and TF = 309.92 K = 36.77 C = 98.19 F Sice water = 453.6 ln S body
FG 309.92 IJ = 54.52 cal K H 274.82 K F 310.15 IJ = 51.93 cal K = e70.0 10 j lnG H 309.92 K
3
Ssys = 54.52  51.93 = 2.59 cal K which is less than the estimate in part (a).
656 P22.62
Heat Engines, Entropy, and the Second Law of Thermodynamics
(a)
For the isothermal process AB, the work on the gas is W AB =  PA VA ln
FG V IJ HV K
B A
W AB = 5 1.013 10 5 Pa 10.0 10 3 m3 ln W AB = 8.15 10 3 J
e
je
50 0 j FGH 10..0 IJK
where we have used 1.00 atm = 1.013 10 5 Pa and 1.00 L = 1.00 10 3 m 3 FIG. P22.62
WBC =  PB V =  1.013 10 5 Pa 10.0  50.0 10 3 m3 = +4.05 10 3 J WCA = 0 and Weng = W AB  WBC = 4.11 10 3 J = 4.11 kJ (b) Since AB is an isothermal process, Eint, AB = 0 and For an ideal monatomic gas, Q AB = W AB = 8.15 10 3 J CV = 3R 5R and C P = 2 2
e
ja
f
TB = TA = Also,
1.013 10 5 50.0 10 3 PBVB 5.05 10 3 = = nR R R
e
je
j
1.013 10 5 10.0 10 3 PC VC 1.01 10 3 = = TC = nR R R QCA = nC V T = 1.00
e
je
j
FG 3 RIJ FG 5.05 10 H 2 KH
3
 1.01 10 3 = 6.08 kJ R
I JK
so the total energy absorbed by heat is Q AB + QCA = 8.15 kJ + 6.08 kJ = 14.2 kJ . (c) QBC = nC P T = QBC = (d) e= 5 5 nRT = PB VBC 2 2
a
f
5 1.013 10 5 2
e
j a10.0  50.0f 10
=
3
= 1.01 10 4 J = 10.1 kJ
Weng Qh
=
Weng Q AB + QCA
4.11 10 3 J = 0.289 or 28.9% 1.42 10 4 J
Chapter 22
657
*P22.63
Like a refrigerator, an air conditioner has as its purpose the removal of energy by heat from the cold reservoir. Its ideal COP is COPCarnot = Tc 280 K = = 14.0 20 K Th  Tc Qc Qh  Qc = Q c t Q h t  Q c t
(a)
Its actual COP is
0.400 14.0 = 5.60 = 5.60
a f
Qh Q Q  5.60 c = c t t t
5.60 10.0 kW = 6.60 Weng t
a
f
Q Qc and c = 8.48 kW t t
(b) (c)
Q h = Weng + Q c :
=
Qh Qc  = 10.0 kW  8.48 kW = 1.52 kW t t
The air conditioner operates in a cycle, so the entropy of the working fluid does not change. The hot reservoir increases in entropy by Qh Th The cold room decreases in entropy by S =  Qc Tc = =
e10.0 10
3
J s 3 600 s
jb
300 K
g = 1.20 10 jb
5
J K
e8.48 10
3
J s 3 600 s
280 K
g = 1.09 10
5
J K
The net entropy change is positive, as it must be:
+1.20 10 5 J K  1.09 10 5 J K = 1.09 10 4 J K
(d) The new ideal COP is We suppose the actual COP is COPCarnot = Tc 280 K = = 11.2 25 K Th  Tc
0.400 11.2 = 4.48
4. 48 = 0.800 , so the fractional change is to 5.60
a f
As a fraction of the original 5.60, this is drop by 20.0% .
Vf
P22.64
(a)
W=
Vi
z
PdV = nRT
2 Vi Vi
z
2Vi dV = 1.00 RT ln = RT ln 2 V Vi
a f
FG IJ H K
(b)
The second law refers to cycles.
658 P22.65
Heat Engines, Entropy, and the Second Law of Thermodynamics
At point A, PVi = nRTi i At point B, 3PVi = nRTB i At point C,
i i
and so and so
n = 1.00 mol TB = 3Ti TC = 6Ti TD = 2Ti 3R and 2
b3 P gb2V g = nRT At point D, P b 2V g = nRT
i i D
C
The heat for each step in the cycle is found using C V = CP = 5R : 2 QBC QCD QDA (a) (b) Therefore,
Q AB = nCV 3Ti  Ti = 3nRTi
P i i
b g = nC b6T  3T g = 7.50nRT = nC b 2T  6T g = 6nRT = nC bT  2T g = 2.50nRT
i V i i i P i i
FIG. P22.65
i
Qentering = Q h = Q AB + QBC = 10.5nRTi
Qleaving = Q c = QCD + QDA = 8.50nRTi Actual efficiency, e= Qh  Qc Qh = 0.190
(c)
(d)
f
Carnot efficiency,
ec = 1 
Tc T = 1  i = 0.833 6Ti Th
*P22.66
S =
z z
i
f f Tf nC P dT dQ T = = nC P T 1 dT = nC P ln T T f = nC P ln T f  ln Ti = nC P ln i T T Ti i i
z
d
i
S = nC P ln *P22.67 (a)
F PV GH nR
f
nR = nC P ln 3 PVi
I JK
FG IJ H K
The ideal gas at constant temperature keeps constant internal energy. As it puts out energy by work in expanding it must take in an equal amount of energy by heat. Thus its entropy increases. Let Pi , Vi , Ti represent the state of the gas before the isothermal expansion. Let PC , VC , Ti represent the state after this process, so that PVi = PC VC . Let Pi , 3Vi , T f represent i the state after the adiabatic compression. Then Substituting gives Then
PC VC = Pi 3Vi
b g
e
PC =
PVi i VC
PViVC 1 = Pi 3 Vi i VC 1 = 3 Vi 1 and
j
VC = 3 Vi
b 1g
continued on next page
Chapter 22
659
The work output in the isothermal expansion is W = PdV = nRTi V 1 dV = nRTi ln
i i C
z
C
z
FG V IJ = nRT lne3 HV K
C i i
1
b g j = nRTi F G
I H  1 JK ln 3
This is also the input heat, so the entropy change is S = Since we have and Then the result is (b)
Q = nR ln 3 1 T
R 1
FG H
IJ K
C P = C V = CV + R
b  1gC
CP =
V
= R , CV =
R 1
S = nC P ln 3
The pair of processes considered here carry the gas from the initial state in Problem 66 to the final state there. Entropy is a function of state. Entropy change does not depend on path. Therefore the entropy change in Problem 66 equals Sisothermal + S adiabatic in this problem. Since Sadiabatic = 0, the answers to Problems 66 and 67 (a) must be the same.
P22.68
Simply evaluate the maximum (Carnot) efficiency. eC = T 4.00 K = = 0.014 4 277 K Th
The proposal does not merit serious consideration. P22.69 The heat transfer over the paths CD and BA is zero since they are adiabatic. Over path BC: QBC = nC P TC  TB > 0 Over path DA: QDA = nCV TA  TD < 0 Therefore, Q c = QDA and Q h = Q BC The efficiency is then P Adiabatic Processes B C D A Vi 3Vi FIG. P22.69 V
b
b
g
g
bT Q bT 1 LT  T O e=1 M P NT T Q
e=1 Qc =1
D C h A B
D C
 TA C V
B P
g  T gC
660 P22.70
Heat Engines, Entropy, and the Second Law of Thermodynamics
(a)
Use the equation of state for an ideal gas V= nRT P 1.00 8.314 600 VA = = 1.97 10 3 m 3 5 25.0 1.013 10 VC
a fa f e j 1.00a8.314fa 400f = =
1.013 10 5
32.8 10 3 m3 FIG. P22.70 PA VA = PBVB
PBVB = PC VC
Since AB is isothermal, and since BC is adiabatic,
Combining these expressions, VB
LF P I V OP = MG J NMH P K V QP
C
C
1 1
b g
A
A
LF 1.00 I e32.8 10 m j = MG MMH 25.0 JK 1.97 10 m N
3 3 3
3 1.40 3
OPb PP Q
1 0. 400
g
VB = 11.9 10 3 m3
Similarly, or Since AB is isothermal, and
VD
LF P I V OP = MG J NMH P K V QP
A C C
1 1 A
b g
LF 25.0 I e1.97 10 m j = MG MMH 1.00 JK 32.8 10 m N
3 3 3 3 3
3 1.40 3
OPb PP Q
1 0 . 400
g
VD = 5.44 10 3 m3
PA VA = PBVB PB = PA
A B
Also, CD is an isothermal and PD Solving part (c) before part (b): (c) (b) For this Carnot cycle,
FG V IJ = 25.0 atmF 1.97 10 GH 11.9 10 HV K F V IJ = 1.00 atmF 32.8 10 =P G GH 5.44 10 HV K
C C D
m3 = 4.14 atm m3
3 3
I JK m I J= m K
6.03 atm
ec = 1 
Tc 400 K =1 = 0.333 600 K Th
Energy is added by heat to the gas during the process AB. For the isothermal process, Eint = 0 . and the first law gives or Then, from Q AB = W AB = nRTh ln
FG V IJ HV K
B A
Q h = Q AB = 1.00 mol 8.314 J mol K 600 K ln e= Weng Qh
b
ga
.9 f FGH 1197 IJK = 8.97 kJ 1.
the net work done per cycle is Weng = e c Q h = 0.333 8.97 kJ = 2.99 kJ .
a
f
Chapter 22
661
P22.71
(a)
20.0C S = mc ln Tf T1 + mc ln Tf T2 = 1.00 kg 4.19 kJ kg K ln
(b) (c) (d)
b
gLM N
Tf T1
+ ln
Tf
OP = b4.19 kJ K g lnFG 293 293 IJ H 283 303 K T Q
2
S = +4.88 J K Yes . Entropy has increased.
ANSWERS TO EVEN PROBLEMS
P22.2 P22.4 P22.6 P22.8 P22.10 P22.12 P22.14 P22.16 P22.18 (a) 667 J ; (b) 467 J (a) 30.0% ; (b) 60.0% P22.34 (a), (b) see the solution; (c) Q h = 149 J ; Q c = 65.0 J ; Weng = 84.3 J ; (d) 56.5%; (e) 1.42 10 3 rev min P22.36 P22.38 P22.40 P22.42 P22.44 4.88 kJ kg K (a) isobaric; (b) 402 kJ; (c) 1.20 kJ K 3.27 J K 718 J K (a) 39.4 L ; (b) 2.50 kJ; (c) 2.50 kJ; (d) 6.87 J K ; (e) +9.16 J K 0.507 J K 34.6 J K (a) 2 heads and 2 tails ; (b) All heads or all tails; (c) 2 heads and 2 tails 8.36 MJ K 32.9 kJ see the solution (a) 2.62 10 3 tons d ; (b) $7.65 million yr ; (c) 4.06 10 4 kg s
55.4%
77.8 W (a) 869 MJ ; (b) 330 MJ 197 kJ 546C
33.0%
(a) 5.12%; (b) 5.27 TJ h; (c) see the solution 453 K P22.50 (a), (b) see the solution; (c) 23.7%; see the solution 11.8 P22.52 P22.54 P22.56 P22.58 P22.46 P22.48
P22.20 P22.22
P22.24 P22.26 P22.28 P22.30 P22.32
1.17 J
(a) 204 W ; (b) 2.43 kW (a) 2.00 ; (b) 3.00 ; (c) 33.3% (a) 51.2% ; (b) 36.2%
662 P22.60 P22.62 P22.64 P22.66
Heat Engines, Entropy, and the Second Law of Thermodynamics
b
P Tc Th  Tc cT
g
P22.68 P22.70
no; see the solution (a) A B C D P, atm 25.0 4.14 1.00 6.03 V, L 1.97 11.9 32.8 5.44
(a) 4.11 kJ ; (b) 14.2 kJ; (c) 10.1 kJ; (d) 28.9% see the solution nC P ln 3
(b) 2.99 kJ ; (c) 33.3%
23
Electric Fields
CHAPTER OUTLINE
23.1 23.2 23.3 23.4 23.5 Properties of Electric Charges Charging Objects by Induction Coulomb's Law The Electric Field Electric Field of a Continuous Charge Distribution Electric Field Lines Motion of Charged Particles in a Uniform Electric Field
ANSWERS TO QUESTIONS
Q23.1 A neutral atom is one that has no net charge. This means that it has the same number of electrons orbiting the nucleus as it has protons in the nucleus. A negatively charged atom has one or more excess electrons. When the comb is nearby, molecules in the paper are polarized, similar to the molecules in the wall in Figure 23.5a, and the paper is attracted. During contact, charge from the comb is transferred to the paper by conduction. Then the paper has the same charge as the comb, and is repelled. The clothes dryer rubs dissimilar materials together as it tumbles the clothes. Electrons are transferred from one kind of molecule to another. The charges on pieces of cloth, or on nearby objects charged by induction, can produce strong electric fields that promote the ionization process in the surrounding air that is necessary for a spark to occur. Then you hear or see the sparks.
Q23.2
23.6 23.7
Q23.3
Q23.4
To avoid making a spark. Rubbersoled shoes acquire a charge by friction with the floor and could discharge with a spark, possibly causing an explosion of any flammable material in the oxygenenriched atmosphere. Electrons are less massive and more mobile than protons. Also, they are more easily detached from atoms than protons. The electric field due to the charged rod induces charges on near and far sides of the sphere. The attractive Coulomb force of the rod on the dissimilar charge on the close side of the sphere is larger than the repulsive Coulomb force of the rod on the like charge on the far side of the sphere. The result is a net attraction of the sphere to the rod. When the sphere touches the rod, charge is conducted between the rod and the sphere, leaving both the rod and the sphere likecharged. This results in a repulsive Coulomb force. All of the constituents of air are nonpolar except for water. The polar water molecules in the air quite readily "steal" charge from a charged object, as any physics teacher trying to perform electrostatics demonstrations in the summer well knows. As a resultit is difficult to accumulate large amounts of excess charge on an object in a humid climate. During a North American winter, the cold, dry air allows accumulation of significant excess charge, giving the potential (pun intended) for a shocking (pun also intended) introduction to static electricity sparks. 1
Q23.5 Q23.6
Q23.7
2
Electric Fields
Q23.8
Similarities: A force of gravity is proportional to the product of the intrinsic properties (masses) of two particles, and inversely proportional to the square of the separation distance. An electrical force exhibits the same proportionalities, with charge as the intrinsic property. Differences: The electrical force can either attract or repel, while the gravitational force as described by Newton's law can only attract. The electrical force between elementary particles is vastly stronger than the gravitational force. No. The balloon induces polarization of the molecules in the wall, so that a layer of positive charge exists near the balloon. This is just like the situation in Figure 23.5a, except that the signs of the charges are reversed. The attraction between these charges and the negative charges on the balloon is stronger than the repulsion between the negative charges on the balloon and the negative charges in the polarized molecules (because they are farther from the balloon), so that there is a net attractive force toward the wall. Ionization processes in the air surrounding the balloon provide ions to which excess electrons in the balloon can transfer, reducing the charge on the balloon and eventually causing the attractive force to be insufficient to support the weight of the balloon. The electric field due to the charged rod induces a charge in the aluminum foil. If the rod is brought towards the aluminum from above, the top of the aluminum will have a negative charge induced on it, while the parts draping over the pencil can have a positive charge induced on them. These positive induced charges on the two parts give rise to a repulsive Coulomb force. If the pencil is a good insulator, the net charge on the aluminum can be zero. So the electric field created by the test charge does not distort the electric field you are trying to measure, by moving the charges that create it. With a very high budget, you could send first a proton and then an electron into an evacuated region in which the field exists. If the field is gravitational, both particles will experience a force in the same direction, while they will experience forces in opposite directions if the field is electric. On a more practical scale, stick identical pith balls on each end of a toothpick. Charge one pith ball + and the other , creating a largescale dipole. Carefully suspend this dipole about its center of mass so that it can rotate freely. When suspended in the field in question, the dipole will rotate to align itself with an electric field, while it will not for a gravitational field. If the test device does not rotate, be sure to insert it into the field in more than one orientation in case it was aligned with the electric field when you inserted it on the first trial. The student standing on the insulating platform is held at the same electrical potential as the generator sphere. Charge will only flow when there is a difference in potential. The student who unwisely touches the charged sphere is near zero electrical potential when compared to the charged sphere. When the student comes in contact with the sphere, charge will flow from the sphere to him or her until they are at the same electrical potential. An electric field once established by a positive or negative charge extends in all directions from the charge. Thus, it can exist in empty space if that is what surrounds the charge. There is no material at point A in Figure 23.23(a), so there is no charge, nor is there a force. There would be a force if a charge were present at point A, however. A field does exist at point A. If a charge distribution is small compared to the distance of a field point from it, the charge distribution can be modeled as a single particle with charge equal to the net charge of the distribution. Further, if a charge distribution is spherically symmetric, it will create a field at exterior points just as if all of its charge were a point charge at its center.
Q23.9
Q23.10
Q23.11 Q23.12
Q23.13
Q23.14
Q23.15
Chapter 23
3
Q23.16
The direction of the electric field is the direction in which a positive test charge would feel a force when placed in the field. A charge will not experience two electrical forces at the same time, but the vector sum of the two. If electric field lines crossed, then a test charge placed at the point at which they cross would feel a force in two directions. Furthermore, the path that the test charge would follow if released at the point where the field lines cross would be indeterminate. Both figures are drawn correctly. E1 and E 2 are the electric fields separately created by the point charges q1 and q 2 in Figure 23.14 or q and q in Figure 23.15, respectively. The net electric field is the vector sum of E1 and E 2 , shown as E. Figure 23.21 shows only one electric field line at each point away from the charge. At the point location of an object modeled as a point charge, the direction of the field is undefined, and so is its magnitude. The electric forces on the particles have the same magnitude, but are in opposite directions. The electron will have a much larger acceleration (by a factor of about 2 000) than the proton, due to its much smaller mass. The electric field around a point charge approaches infinity as r approaches zero. Vertically downward. Four times as many electric field lines start at the surface of the larger charge as end at the smaller charge. The extra lines extend away from the pair of charges. They may never end, or they may terminate on more distant negative charges. Figure 23.24 shows the situation for charges +2q and q. At a point exactly midway between the two changes. Linear charge density, , is charge per unit length. It is used when trying to determine the electric field created by a charged rod. Surface charge density, , is charge per unit area. It is used when determining the electric field above a charged sheet or disk. Volume charge density, , is charge per unit volume. It is used when determining the electric field due to a uniformly charged sphere made of insulating material. Yes, the path would still be parabolic. The electrical force on the electron is in the downward direction. This is similar to throwing a ball from the roof of a building horizontally or at some angle with the vertical. In both cases, the acceleration due to gravity is downward, giving a parabolic trajectory. No. Life would be no different if electrons were + charged and protons were charged. Opposite charges would still attract, and like charges would repel. The naming of + and charge is merely a convention. If the antenna were not grounded, electric charges in the atmosphere during a storm could place the antenna at a high positive or negative potential. The antenna would then place the television set inside the house at the high voltage, to make it a shock hazard. The wire to the ground keeps the antenna, the television set, and even the air around the antenna at close to zero potential. People are all attracted to the Earth. If the force were electrostatic, people would all carry charge with the same sign and would repel each other. This repulsion is not observed. When we changed the charge on a person, as in the chapteropener photograph, the person's weight would change greatly in magnitude or direction. We could levitate an airplane simply by draining away its electric charge. The failure of such experiments gives evidence that the attraction to the Earth is not due to electrical forces.
Q23.17
Q23.18
Q23.19 Q23.20 Q23.21
Q23.22 Q23.23
Q23.24
Q23.25
Q23.26
Q23.27
4
Electric Fields
Q23.28
In special orientations the force between two dipoles can be zero or a force of repulsion. In general each dipole will exert a torque on the other, tending to align its axis with the field created by the first dipole. After this alignment, each dipole exerts a force of attraction on the other.
SOLUTIONS TO PROBLEMS
Section 23.1 *P23.1 (a) Properties of Electric Charges The mass of an average neutral hydrogen atom is 1.007 9u. Losing one electron reduces its mass by a negligible amount, to
1.007 9 1.660 10 27 kg  9.11 10 31 kg = 1.67 10 27 kg .
Its charge, due to loss of one electron, is
e
j
0  1 1.60 10 19 C = +1.60 10 19 C .
(b) By similar logic, charge = +1.60 10 19 C mass = 22.99 1.66 10 27 kg  9.11 10 31 kg = 3.82 10 26 kg (c) charge of Cl  = 1.60 10 19 C mass = 35.453 1.66 10 27 kg + 9.11 10 31 kg = 5.89 10 26 kg (d) charge of Ca ++ = 2 1.60 10 19 C = +3.20 10 19 C mass = 40.078 1.66 10 27 kg  2 9.11 10 31 kg = 6.65 10 26 kg (e) charge of N 3  = 3 1.60 10 19 C = 4.80 10 19 C mass = 14.007 1.66 10 27 (f)
e
j
e
j
e e
j
e
j e
j
j
e e e
e e
j kg j + 3e9.11 10 j e j e j
31
kg = 2.33 10 26 kg
j
charge of N 4 + = 4 1.60 10 19 C = +6.40 10 19 C mass = 14.007 1.66 10 27 kg  4 9.11 10 31 kg = 2.32 10 26 kg
j j
(g)
We think of a nitrogen nucleus as a seventimes ionized nitrogen atom. charge = 7 1.60 10 19 C = 1.12 10 18 C mass = 14.007 1.66 10
27
e
j
kg  7 9.11 10 31 kg = 2.32 10 26 kg
(h)
charge = 1.60 10 19 C mass = 2 1.007 9 + 15.999 1.66 10 27 kg + 9.11 10 31 kg = 2.99 10 26 kg
b
g
Chapter 23
5
P23.2
(a)
N=
F 10.0 grams I FG 6.02 10 GH 107.87 grams mol JK H
23
atoms mol
IJ FG 47 electrons IJ = K H atom K
2.62 10 24
(b)
# electrons added = or
Q 1.00 10 3 C = = 6.25 10 15 e 1.60 10 19 C electron
2.38 electrons for every 10 9 already present .
Section 23.2 Section 23.3 P23.3
Charging Objects by Induction Coulomb's Law
If each person has a mass of 70 kg and is (almost) composed of water, then each person contains N
F 70 000 grams I FG 6.02 10 GH 18 grams mol JK H e
23
molecules mol
IJ FG 10 protons IJ 2.3 10 K H molecule K
28
protons .
With an excess of 1% electrons over protons, each person has a charge q = 0.01 1.6 10 19 C 2.3 10 28 = 3.7 10 7 C . q1 q 2 r2
7 2 9
So
F = ke
je j e3.7 10 j = e9 10 j
0.6 2
N = 4 10 25 N ~ 10 26 N .
This force is almost enough to lift a weight equal to that of the Earth: Mg = 6 10 24 kg 9.8 m s 2 = 6 10 25 N ~ 10 26 N . *P23.4 The force on one proton is F = k e q1 q 2 r2 C m away from the other proton. Its magnitude is = 57.5 N .
e
j
e8.99 10
P23.5 (a)
9
F 1.6 10 N m C jG H 2 10
2
19
15
I JK
2
Fe =
k e q1 q 2 r2
e8.99 10 =
9
N m 2 C 2 1.60 10 19 C
je
j
2
e
3.80 10
10
m
j
2
= 1.59 10 9 N
brepulsiong
(b)
Fg =
Gm1 m 2 r2
e6.67 10 =
m1 m 2 r2
11
N m 2 C 2 1.67 10 27 kg
je
j
2
e3.80 10
10
m
j
2
= 1.29 10 45 N
The electric force is larger by 1.24 10 36 times . (c) If k e q = m q1 q 2 r
2
=G
with q1 = q 2 = q and m1 = m 2 = m , then
6.67 10 11 N m 2 kg 2 G = = 8.61 10 11 C kg . ke 8.99 10 9 N m 2 C 2
6
Electric Fields
P23.6
We find the equalmagnitude charges on both spheres: F = ke q1 q 2 r2 = ke q2 r2 so q=r F = 1.00 m ke
a
f
1.00 10 4 N = 1.05 10 3 C . 8.99 10 9 N m 2 C 2
The number of electron transferred is then N xfer = 1.05 10 3 C 1.60 10 19 C e  = 6.59 10 15 electrons .
The whole number of electrons in each sphere is N tot =
F 10.0 g I e6.02 10 GH 107.87 g mol JK F GH I JK
23
atoms mol 47 e  atom = 2.62 10 24 e  .
je
j
The fraction transferred is then f= N xfer 6.59 10 15 = = 2.51 10 9 = 2.51 charges in every billion. 24 N tot 2.62 10 q1 q 2 r
2
P23.7
F1 = k e
e8.99 10 =
=
9
N m 2 C 2 7.00 10 6 C 2.00 10 6 C
2
F2 = k e
q1 q 2 r2
e8.99 10
9
N m2
a0.500 mf C je7.00 10 C je 4.00 10 C j = 1.01 N a0.500 mf
2 6 6 2
je
je
j = 0.503 N
Fx = 0.503 cos 60.0+1.01 cos 60.0 = 0.755 N Fy = 0.503 sin 60.01.01 sin 60.0 = 0.436 N F = 0.755 N i  0.436 N j = 0.872 N at an angle of 330
a
f a
f
FIG. P23.7 q1 q 2 r2
P23.8
F = ke
e8.99 10 =
9
N m 2 C 2 1.60 10 19 C 2 6.37 10 6 m
je
j e6.02 10 j
2
23 2
e
j
2
= 514 kN
P23.9
(a)
The force is one of attraction . The distance r in Coulomb's law is the distance between centers. The magnitude of the force is F= k e q1 q 2 r2 = 8.99 10 9 N m 2 C 2 12.0 10 C je18.0 10 C j = je a0.300 mf
9 9 2
e
2.16 10 5 N .
(b)
The net charge of 6.00 10 9 C will be equally split between the two spheres, or 3.00 10 9 C on each. The force is one of repulsion , and its magnitude is F= k e q1 q 2 r
2
= 8.99 10 9 N m 2 C 2
e
3.00 10 C je3.00 10 C j = je a0.300 mf
9 9 2
8.99 10 7 N .
Chapter 23
7
P23.10
Let the third bead have charge Q and be located distance will experience a net force given by F= k e 3q Q x
2
x from the left end of the rod. This bead
b g
i+
b g e i j . ad  x f
ke q Q
2
The net force will be zero if
3 1 = 2 x dx
a
f
2
, or d  x =
x 3
.
This gives an equilibrium position of the third bead of x = 0.634d . The equilibrium is stable if the third bead has positive charge . kee2 r2
P23.11
(a)
F=
= 8.99 10 N m
e
9
2
e1.60 10 C j e0.529 10
2
19 10
C
j
2
m
j
2
= 8.22 10 8 N
(b)
We have F =
mv 2 from which r Fr = m 8.22 10 8 N 0.529 10 10 m 9.11 10
31
v=
e
kg k e qQ d 2 + 2 x 2
j=
2.19 10 6 m s .
P23.12
The top charge exerts a force on the negative charge left, at an angle of tan 1
FG d IJ to the xaxis. The two positive charges together exert force H 2x K F 2 k qQ I FG a xfi IJ 2 k qQ d GG JJ G = ma or for x << , a x. 2 md 8 e + x j K GH e + x j JJK H
e d2 4 2 d2 4 2 12 e 3
ch
which is directed upward and to the
(a)
The acceleration is equal to a negative constant times the excursion from equilibrium, as in 16 k e qQ a =  2 x , so we have Simple Harmonic Motion with 2 = . md 3 T= 2 =
2
md 3 , where m is the mass of the object with charge Q . k e qQ k e qQ md 3
(b)
v max = A = 4a
8
Electric Fields
Section 23.4 P23.13
The Electric Field
For equilibrium, or Thus,
Fe = Fg
qE =  mg  j . E= mg j. q
e j
(a)
E=
9.11 10 31 kg 9.80 m s 2 mg j= j =  5.58 10 11 N C j 19 q 1.60 10 C
e
e
je
j
j
e
j
(b)
1.67 10 27 kg 9.80 m s 2 mg E= j= j= q 1.60 10 19 C
e
e
je
j
j
e1.02 10
7
NC j
j
P23.14
Fy = 0 : QEj + mge jj = 0
m =
24.0 10 6 C 610 N C QE = = 1.49 grams g 9.80 m s 2
e
jb
g
P23.15
The point is designated in the sketch. The magnitudes of the electric fields, E1 , (due to the 2.50 10 6 C charge) and E 2 (due to the 6.00 10 6 C charge) are E1 = ke q r
2
e8.99 10 =
=
9
N m 2 C 2 2.50 10 6 C d
2
je
j
j
(1)
FIG. P23.15
E2 =
keq r
2
e8.99 10
9
N m 2 C 2 6.00 10 6 C
2
je ad + 1.00 mf
(2)
Equate the right sides of (1) and (2) to get or which yields or
ad + 1.00 mf
d = 1.82 m
2
= 2.40d 2
d + 1.00 m = 1.55d
d = 0.392 m .
The negative value for d is unsatisfactory because that locates a point between the charges where both fields are in the same direction. Thus,
d = 1.82 m to the left of the  2.50 C charge .
Chapter 23
9
P23.16
If we treat the concentrations as point charges, E+ = ke E = ke q r2 q r2 = 8.99 10 9 N m 2 C 2 = 8.99 10 9 N m 2
e
e
C j b1a40.0 mgf e jj = 3.60 10 000 a40.0 Cf e jj = 3.60 10 C j b1 000 mg
2 2 2
5
N C  j downward N C  j downward
e ja
f f
q x
5
e ja
E = E + + E  = 7.20 10 5 N C downward *P23.17 The first charge creates at the origin field to the right. a2 Suppose the total field at the origin is to the right. Then q must be negative: k eQ a
2
k eQ
+Q x = 0
FIG. P23.17
i+
e ij = 2 k Q i a a3 a f
2 e 2
keq
q = 9Q .
In the alternative, the total field at the origin is to the left: k eQ a P23.18 (a)
2
i+
ke q 9a
2
e ij = 2 k Q e ij a
e 2
q = +27Q .
9 6
E1 = E2
e8.99 10 je7.00 10 j = 2.52 10 r a0.500f k q e8.99 10 je 4.00 10 j = = = 1.44 10 r a0.500f
ke q
2
=
2
5
NC
9
6
e 2
2
5
NC FIG. P23.18
Ex = E2  E1 cos 60 = 1. 44 10 5  2.52 10 5 cos 60.0 = 18.0 10 3 N C Ey =  E1 sin 60.0 = 2.52 10 sin 60.0 = 218 10 N C E = 18.0 i  218 j 10 3 N C = 18.0 i  218 j kN C F = qE = 2.00 10 6 C 18.0 i  218 j 10 3 N C = 36.0 i  436 j 10 3 N = 8.99 10 je3.00 10 j E = e jj = e a0.100f e jj = e2.70 10 r 8.99 10 je6.00 10 j k q E = e ij = e a0.300f e ij = e5.99 10 r E = E + E = e5.99 10 N C ji  e 2.70 10 N C j j
1 5 3
(b)
e
je
j
e
j
e36.0i  436 jj mN
P23.19
(a)
k e q1
2 1
9
9
2
3
NC j NCi FIG. P23.19
j
2
e
2
9
9
2 2
2
2
j
2
1
2
3
(b)
F = qE = 5.00 10 9 C 599 i  2 700 j N C F = 3.00 10 6 i  13.5 10 6 j N =
e
je
e
j
j e3.00i  13.5 jj N
10
Electric Fields
P23.20
(a)
E=
keq r2
=
Ex = 0
e8.99 10 je2.00 10 j = 14 400 N C a1.12f and E = 2b14 400 g sin 26.6 = 1.29 10
9 6 2 y
4
NC FIG. P23.20
so (b)
E = 1.29 10 4 j N C .
F = qE = 3.00 10 6 1. 29 10 4 j = 3.86 10 2 j N k e q1 r12 ke q2 r22 k e q3 r32 ke 2q a2 keq a2
e
je
j
P23.21
(a)
E=
r1 + ke q a
2
r2 +
r3 =
b g i + k b3 qg ei cos 45.0+ j sin 45.0j + k b4qg j
e e
2a2
a2
E = 3.06
i + 5.06 ke q 2 a2
keq a
2
j = 5.91
at 58.8
(b) P23.22
F = qE = 5.91
at 58.8
The electric field at any point x is E= ke q
2
a f ax  af cx  a afh ex  a j
 ke q
2
=
k e q 4ax
2
2 2
. 4a k e q x3 k eQ n
When x is much, much greater than a, we find E
b g
.
P23.23
(a)
One of the charges creates at P a field E = the xaxis as shown.
R2 + x2
at an angle to
When all the charges produce field, for n > 1 , the components perpendicular to the xaxis add to zero. The total field is nk e Q n i R +x
2 2
b g cos =
e
k eQx i R + x2
2
j
3 2
.
FIG. P23.23
(b)
A circle of charge corresponds to letting n grow beyond all bounds, but the result does not depend on n. Smearing the charge around the circle does not change its amount or its distance from the field point, so it does not change the field . keq r2 keq a2
P23.24
E=
r=
e ij + a2kaq e ij + a3kaq e ij + ... =  ka qi FGH1 + 21 + 31 + ...IJK = f f
e 2 e 2 e 2 2 2

2ke q i 6a2
Chapter 23
11
Section 23.5 P23.25
Electric Field of a Continuous Charge Distribution
8.99 10 9 22.0 10 6 ke Q ke k eQ = = = E= 0.290 0.140 + 0.290 d +d d +d d +d
e b g a f a f a f a
k e dq x2
x0
fa
je
f
j
FIG. P23.25
E = 1.59 10 6 N C , directed toward the rod.
P23.26
E=
z
, where dq = 0 dx dx x
2
E = ke 0
z
= ke 0 
FG 1 IJ H xK
2
=
x0 9
ke 0 x0
6
The direction is  i or left for 0 > 0
P23.27
E=
ex
k e xQ
2
+ a2
j
3 2
=
e8.99 10 je75.0 10 jx = 6.74 10 x ex + 0.100 j ex + 0.010 0j
5 2 32 2
32
(a) (b) (c) (d)
At x = 0.010 0 m , At x = 0.050 0 m , At x = 0.300 m , At x = 1.00 m ,
E = 6.64 10 6 i N C = 6.64i MN C E = 2.41 10 7 i N C = 24.1i MN C E = 6.40 10 6 i N C = 6. 40 i MN C E = 6.64 10 5 i N C = 0.664i MN C
P23.28
E = dE =
z
x0
LM k x dxe ij OP z MMN x PPQ = k x i z x
e 0 0 3 e 0 0 x0
3
dx =  k e 0 x 0 i 
F GH
1 2x
2
x0
I= JK
ke 0 i 2 x0
e j
P23.29
E=
ex
k e Qx
2
+ a2
j
3 2
For a maximum,
dE = Qk e dx a
LM MM ex N
.
1
2
+ a2
j
3 2

OP P=0 x +a j P e Q
3x 2
2 2 52
x 2 + a 2  3 x 2 = 0 or x =
2
Substituting into the expression for E gives E=
2
e aj
3 2
k eQa
2 32
=
k eQ 3
3 2
a
2
=
2 k eQ 3 3a
2
=
Q 6 3 0 a 2
.
12
Electric Fields
P23.30
E = 2 k e 1  E = 2 8.99 10
F GH
x x +R
9 2 2
I JK
3
e
je7.90 10
F jGG 1  H
x2
I JJ = 4.46 10 FG 1  H + a0.350f K
x
8 2
x2
I J + 0.123 K
x
(a) (b) (c) (d)
At x = 0.050 0 m , At x = 0.100 m , At x = 0.500 m , At x = 2.00 m ,
E = 3.83 10 8 N C = 383 MN C E = 3.24 10 8 N C = 324 MN C E = 8.07 10 7 N C = 80.7 MN C E = 6.68 10 8 N C = 6.68 MN C
P23.31
(a)
From Example 23.9: E = 2 k e 1 
F GH
x x2 + R2
I JK g jb g
b g
=
E = 1.04 10 8 N C 0.900 = 9.36 10 7 N C = 93.6 MN C appx: E = 2 k e = 104 MN C about 11% high (b) E = 1.04 10 8 N C 1 
e e
Q = 1.84 10 3 C m 2 R 2
ja
f
b
8 N C 0.004 96 = 0.516 MN C 30.0 + 3.00 Q 5.20 10 6 appx: E = k e 2 = 8.99 10 9 = 0.519 MN C about 0.6% high 2 r 0.30 2 2
F jGH
30.0 cm
I = 1.04 10 J e cm K
x
e
j a f
P23.32
The electric field at a distance x is
Ex = 2 k e 1 
This is equivalent to R2 x2
Ex
LM NM L = 2 k M1  MN
e
x2 + R2 1
OP QP
1 + R2 x2
OP PQ
2 2
For large x, so
<< 1 and
1+ Ex
Q Substitute = , R2 But for x >> R , 1 x2 + R2 2 1 x2 , so
Ex
I F 1 JJ = 2 k e1 + R e2 x j  1j = 2 k G 1  GH 1 + R e2 x j K 1 + R e2 x j k Qe1 x j F RI = = k QG x + H 2 JK 1 + R e2x j
e 2 2 e 2 2 e 2 2 2 e 2 2
R2 R2 1+ 2 x2 2x
Ex
k eQ x2
for a disk at large distances
Chapter 23
13
P23.33
Due to symmetry where so that, where Thus, Solving,
Ey = dEy = 0 , and Ex = dE sin = k e dq = ds = rd , Ex = ke k sin d = e  cos r 0 r
z
z
z
dq sin r2
z
a
f
0
=
2k e r FIG. P23.33
q L and r = . L 2 8.99 10 9 N m 2 C 2 7.50 10 6 C 2 k q Ex = e2 = . 2 L 0.140 m
=
e
a
je f
j
Ex = 2.16 10 N C .
7
Since the rod has a negative charge, E = 2.16 10 7 i N C = 21.6 i MN C . P23.34 (a) We define x = 0 at the point where we are to find the field. One ring, with thickness dx, has Qdx and produces, at the chosen point, a field charge h kex Qdx dE = i. 3 2 h x2 + R2
e
j
e
j
The total field is E=
e k Qi ex + R j E= 2h b 1 2g
all charge d 2 e
z z
dE =
d+h
k eQxdx
h x2 + R
2 3 2
j
i=
k eQi d + h 2 x + R2 2h x=d
ze
j
3 2
2 xdx
d+h 2 1 2
k Qi = e h
x=d
LM MM d Ne
2
OP  P +R j ead + hf + R j PQ
1 1
2 12 2 2 12
(b)
Think of the cylinder as a stack of disks, each with thickness dx, charge perarea = Qdx . One disk produces a field R2h
Qdx , and chargeh
dE =
2 k eQdx
R2h
So,
I JJ i E= z dE = z +R j K OP = 2k Qi LMx 2 k Qi L 1 E= M z dx  2 z ex + R j 2xdxP R h M R h M MN N Q 2 k Qi L E= MNd + h  d  ead + hf + R j + ed + R j OPQ R h 2 k Qi L O E= R h M Nh + ed + R j  ead + hf + R j PQ
d+h x =d e
F GG 1  H ex
2
all charge
I Ji . +R j J K F 2 k Qdx G 1 R h G H ex
x
2 12 2 d+h x=d 2
x
2
2 12
e 2
d+h d
2 1 2
e 2
2 2 1 x +R d+h  d 2 12
e
j
1 2 d+h
d
OP PP Q
e 2
2
2
12
2
2 12
e 2
2
2 12
2
2
12
14
Electric Fields
P23.35
(a)
The electric field at point P due to each element of length dx, is k dq dE = 2 e 2 and is directed along the line joining the element to x +y point P. By symmetry, Ex = dEx = 0
z
and since where
2
dq = dx , cos = = y x + y2
2
E = Ey = dEy = dE cos Therefore, E = 2 k e y
z
z
. FIG. P23.35
ze
0
dx x +y
2 2 32
j
2 k e sin 0 . y and Ey = 2k e . y
(b) P23.36 (a)
For a bar of infinite length,
0 = 90
The whole surface area of the cylinder is A = 2 r 2 + 2 rL = 2 r r + L . Q = A = 15.0 10 9 C m 2 2 0.025 0 m 0.025 0 m + 0.060 0 m = 2.00 10 10 C
e e
j b
g
a f
(b)
For the curved lateral surface only, A = 2 rL .
Q = A = 15.0 10 9 C m 2 2 0.025 0 m 0.060 0 m = 1.41 10 10 C (c) P23.37 (a) Q = V = r 2 L = 500 10 9 C m 3 0.025 0 m
j b
ga
f
e
j b je
g b0.060 0 mg =
2
5.89 10 11 C
Every object has the same volume, V = 8 0.030 0 m
For each, Q = V = 400 10 9 C m 3 2.16 10 4 m3 = 8.64 10 11 C (b) We must count the 9.00 cm 2 squares painted with charge: (i)
e
a
f
3
= 2.16 10 4 m3 .
j
6 4 = 24 squares
Q = A = 15.0 10 9 C m 2 24.0 9.00 10 4 m 2 = 3.24 10 10 C
e e e e
j e j e j e
j
(ii)
34 squares exposed
Q = A = 15.0 10 9 C m 2 34.0 9.00 10 4 m 2 = 4.59 10 10 C (iii) 34 squares
j j
Q = A = 15.0 10 9 C m 2 34.0 9.00 10 4 m 2 = 4.59 10 10 C (iv) 32 squares
Q = A = 15.0 10 9 C m 2 32.0 9.00 10 4 m 2 = 4.32 10 10 C (c) (i) total edge length:
j e
j
= 24 0.030 0 m
Q = = 80.0 10 12 (ii) Q=
12
e = e80.0 10
b g C mj24 b0.030 0 mg = C mj44 b0.030 0 mg =
5.76 10 11 C 1.06 10 10 C
continued on next page
Chapter 23
15
(iii) (iv)
Q = = 80.0 10 12 C m 64 0.030 0 m = 1.54 10 10 C Q = = 80.0 10 12 C m 40 0.030 0 m = 0.960 10 10 C
e e
j b
g
j b
g
Section 23.6 P23.38
Electric Field Lines P23.39
FIG. P23.38 q1 6 1 = =  3 q 2 18
FIG. P23.39
P23.40
(a)
(b) P23.41 (a)
q1 is negative, q 2 is positive
The electric field has the general appearance shown. It is zero at the center , where (by symmetry) one can see that the three charges individually produce fields that cancel out. In addition to the center of the triangle, the electric field lines in the second figure to the right indicate three other points near the middle of each leg of the triangle where E = 0 , but they are more difficult to find mathematically.
(b)
You may need to review vector addition in Chapter Three. The electric field at point P can be found by adding the electric field vectors due to each of the two lower point charges: E = E 1 + E 2 . The electric field from a point charge is E = k e As shown in the solution figure at right, E1 = k e E2 = ke q a2 q a2 to the right and upward at 60 to the left and upward at 60 q a
2
q r2
r.
FIG. P23.41
e
E = E1 + E 2 = k e = 1.73 k e q a2 j
ecos 60 i + sin 60 jj + e cos 60 i + sin 60 jj = k
q a2
2 sin 60 j
e
j
16
Electric Fields
Section 23.7 P23.42
Motion of Charged Particles in a Uniform Electric Field a= qE m qEt m
F = qE = ma
v f = vi + at electron:
vf = ve
9.11 10 in a direction opposite to the field
19
e1.602 10 ja520fe48.0 10 j = =
19 9 31
4.39 10 6 m s
proton:
1.67 10 27 in the same direction as the field a=
vp =
e1.602 10 ja520fe48.0 10 j =
9
2.39 10 3 m s
P23.43
(a)
qE 1.602 10 19 640 = = 6.14 10 10 m s 2 m 1.67 10 27 1.20 10 6 = 6.14 10 10 t
a f
(b) (c)
v f = vi + at
x f  xi = K= 1 vi + v f t 2
e
j
t = 1.95 10 5 s
d
i
xf =
1 1.20 10 6 1.95 10 5 = 11.7 m 2
e
je
j
(d)
1 1 mv 2 = 1.67 10 27 kg 1.20 10 6 m s 2 2
e
je
j
2
= 1.20 10 15 J
P23.44
(a)
19 6.00 10 5 qE 1.602 10 a= = = 5.76 10 13 m s so a = 5.76 10 13 i m s 2 m 1.67 10 27
e
e
je
j
j
(b)
v f = vi + 2 a x f  xi
d
i jb g
v i = 2.84 10 6 i m s
0 = vi2 + 2 5.76 10 13 0.070 0 (c) v f = vi + at 0 = 2.84 10 6 + 5.76 10 13 t P23.45
e
e
j
t = 4.93 10 8 s
The required electric field will be in the direction of motion . Work done = K so, which becomes and  Fd =  eEd = K E= K . ed 1 mvi2 (since the final velocity = 0 ) 2
Chapter 23
17
P23.46
The acceleration is given by v 2 = vi2 + 2 a x f  x i or f Solving Now Therefore
d
i
v 2 = 0 + 2a h . f a= v2 f 2h .  mg j + qE =  + mg j . mv 2 j f 2h .
a f
F = ma :
qE = 
F GH
mv 2 f 2h
I JK
(a)
Gravity alone would give the bead downward impact velocity 2 9.80 m s 2 5.00 m = 9.90 m s . To change this to 21.0 m/s down, a downward electric field must exert a downward electric force.
2 1.00 10 3 kg N s 2 m vf g = q= E 2h 1.00 10 4 N C kg m
e
ja
f
(b)
F GH
I JK
F GH
I LM b21.0 m sg JK M 2a5.00 mf N
2
 9.80 m s 2 = 3.43 C
OP PQ
P23.47
(a)
t=
0.050 0 x = = 1.11 10 7 s = 111 ns v x 4.50 10 5
19 9.60 10 3 qE 1.602 10 = = 9.21 10 11 m s 2 27 m 1.67 10
(b)
ay =
e
e
je
j
j
y f  yi = v yi t + (c) *P23.48
1 ayt 2 : 2
yf =
1 9.21 10 11 1.11 10 7 2
e
je
j
2
= 5.68 10 3 m = 5.68 mm
v x = 4.50 10 5 m s
v yf = v yi + a y t = 9.21 10 11 1.11 10 7 = 1.02 10 5 m s
e
je
j
The particle feels a constant force: F = qE = 1 10 6 C 2 000 N C  j = 2 10 3 N  j
3 2 13 2
ge j e j F = e 2 10 kg m s je jj = 1 10 m s  j . a= and moves with acceleration: e je j m 2 10 kg Its xcomponent of velocity is constant at e1.00 10 m sj cos 37 = 7.99 10 m s . Thus it moves in a e
16 5 4 2 2 v yf = v yi + 2 a y y f  yi :
jb
parabola opening downward. The maximum height it attains above the bottom plate is described by
d
i
0 = 6.02 10 4 m s
e
j  e2 10
2
13
m s2 y f  0
jd
i
y f = 1.81 10 4 m .
continued on next page
18
Electric Fields
Since this is less than 10 mm, the particle does not strike the top plate, but moves in a symmetric parabola and strikes the bottom plate after a time given by y f = yi + v yi t + since t > 0 , The particle's range is In sum, The particle strikes the negative plate after moving in a parabola with a height of 0.181 mm and a width of 0.961 mm. P23.49 vi = 9.55 10 3 m s (a) ay = R=
19 720 eE 1.60 10 = = 6.90 10 10 m s 2 27 m 1.67 10
^
1 ayt 2 2
0 = 0 + 6.02 10 4 m s t + t = 1.20 10 8 s .
e
j
1 1 10 13 m s 2 t 2 2
e
j
x f = xi + v x t = 0 + 7.99 10 4 m s 1.20 10 8 s = 9.61 10 4 m .
e
je
j
e
e
ja f j
vi2 sin 2 = 1. 27 10 3 m so that ay
3 2
FIG. P23.49
e9.55 10 j
sin 2
10
6.90 10
= 1.27 10 3
sin 2 = 0.961 (b) t= R R = vix vi cos
= 36.9
90.0 = 53.1 If = 53.1 , t = 221 ns .
If = 36.9 , t = 167 ns .
Additional Problems *P23.50 The two given charges exert equalsize forces of attraction on each other. If a third charge, positive or negative, were placed between them they could not be in equilibrium. If the third charge were at a point x > 15 cm , it would exert a stronger force on the 45 C than on the 12 C , and could not produce equilibrium for both. Thus the third charge must be at x =  d < 0 . Its equilibrium requires k e q 12 C d
2
d q
x=0 12 C FIG. P23.50
15 cm x + 45 C
b
g = k qb45 Cg a15 cm + df
e
2
FG 15 cm + d IJ H d K
d = 16.0 cm .
2
=
45 = 3.75 12
15 cm + d = 1.94d
The third charge is at x = 16.0 cm . The equilibrium of the 12 C requires k e q 12 C
g = k b45 Cg12 C a16.0 cmf a15 cmf
e 2 2
b
q = 51.3 C .
All six individual forces are now equal in magnitude, so we have equilibrium as required, and this is the only solution.
Chapter 23
19
P23.51
The proton moves with acceleration while the e  has acceleration
19 C 640 N C qE 1.60 10 = = 6.13 10 10 m s 2 ap = m 1.673 10 27 kg
e
jb
g
ae =
e1.60 10
19
C 640 N C
31
jb
9.110 10
kg
g = 1.12 10
14
m s 2 = 1 836 a p .
(a)
We want to find the distance traveled by the proton (i.e., d = 1 1 1 a p t 2 + a e t 2 = 1 837 a p t 2 . 2 2 2 1 4.00 cm 2 = 21.8 m . d = apt = 2 1 837 4.00 cm =
FG H
IJ K
1 a p t 2 ), knowing: 2
Thus, (b)
The distance from the positive plate to where the meeting occurs equals the distance the 1 sodium ion travels (i.e., d Na = a Na t 2 ). This is found from: 2 eE eE 1 1 1 1 2 2 t2 + t2 . 4.00 cm = a Na t + aCl t : 4.00 cm = 2 2 2 22.99 u 2 35. 45 u 1 1 1 This may be written as 4.00 cm = a Na t 2 + 0.649 a Na t 2 = 1.65 a Na t 2 2 2 2 1 4.00 cm 2 = 2.43 cm . so d Na = a Na t = 2 1.65
FG H
IJ K b
FG H
IJ K
g
FG H
IJ K
P23.52
(a)
The field, E1 , due to the 4.00 10 9 C charge is in the x direction. 8.99 10 9 N m 2 C 2 4.00 10 9 C keq E1 = 2 r = i 2 r 2.50 m
e
a
je f
j
= 5.75 i N C ke q r
2
FIG. P23.52(a)
Likewise, E 2 and E3 , due to the 5.00 10 9 C charge and the 3.00 10 9 C charge are E2 = E3
e8.99 10 r=
9
e8.99 10 =
ke q
je j i = 11.2 N C i a2.00 mf N m C je3.00 10 C j i = 18.7 N C i a1.20 mf
9
N m 2 C 2 5.00 10 9 C
2 2 9
2
2
E R = E1 + E 2 + E 3 = 24.2 N C in +x direction.
(b) E1 = E2 E3
b ge j = r = b11. 2 N C ge + jj r k q = r = b5.81 N C ge 0.371i +0.928 jj r
r2 ke q
2 e 2
r = 8.46 N C 0.243 i + 0.970 j
Ex = E1 x + E3 x = 4.21i N C
Ey = E1 y + E2 y + E3 y = 8.43 j N C FIG. P23.52(b)
ER = 9.42 N C
= 63.4 above  x axis
20
Electric Fields
*P23.53
(a)
Each ion moves in a quarter circle. The electric force causes the centripetal acceleration.
F = ma
(b) For the xmotion, 0 = v 2 + 2ax R Ex = 
qE =
mv 2 R
E=
mv 2 qR
2 2 v xf = v xi + 2 a x x f  x i
d
i
ax = 
v 2 Fx qE x = = m 2R m
mv 2 . Similarly for the ymotion, 2 qR ay = + v 2 qE y = m 2R Ey = mv 2 2 qR
v 2 = 0 + 2ay R
The magnitude of the field is
2 2 Ex + Ey =
mv 2 2 qR
at 135 counterclockwise from the x axis .
P23.54
From the freebody diagram shown,
Fy = 0 :
So From or P23.55 (a)
T cos 15.0 = 1.96 10 2 N . T = 2.03 10 2 N .
Fx = 0 , we have
q=
qE = T sin 15.0
2.03 10 2 N sin 15.0 T sin 15.0 = = 5.25 10 6 C = 5.25 C . E 1.00 10 3 N C
e
j
FIG. P23.54
Let us sum force components to find
Fx = qEx  T sin = 0 , and Fy = qEy + T cos  mg = 0 .
Combining these two equations, we get q=
e1.00 10 ja9.80f eE cot + E j a3.00 cot 37.0+5.00f 10
mg
3
=
x
y
5
= 1.09 10 8 C
Free Body Diagram FIG. P23.55
= 10.9 nC (b) From the two equations for T=
Fx
and
Fy
we also find
qEx = 5.44 10 3 N = 5.44 mN . sin 37.0
Chapter 23
21
P23.56
This is the general version of the preceding problem. The known quantities are A, B, m, g, and . The unknowns are q and T. The approach to this problem should be the same as for the last problem, but without numbers to substitute for the variables. Likewise, we can use the free body diagram given in the solution to problem 55. Again, Newton's second law: and (a) Substituting T = qA , into Eq. (2), sin
Fx = T sin + qA = 0 Fy = +T cos + qB  mg = 0
qA cos + qB = mg . sin q=
(1) (2)
Isolating q on the left,
a
mg A cot + B mgA
f
.
(b)
Substituting this value into Eq. (1),
T=
a A cos + B sin f
.
If we had solved this general problem first, we would only need to substitute the appropriate values in the equations for q and T to find the numerical results needed for problem 55. If you find this problem more difficult than problem 55, the little list at the first step is useful. It shows what symbols to think of as known data, and what to consider unknown. The list is a guide for deciding what to solve for in the analysis step, and for recognizing when we have an answer. P23.57 F= k e q1 q 2 r
2
:
tan =
15.0 60.0
6 2
= 14.0
e8.99 10 je10.0 10 j F = a0.150f e8.99 10 je10.0 10 j F = a0.600f e8.99 10 je10.0 10 j F = a0.619f
9 1 2 9 3 2 9 2 2
= 40.0 N
6 2
= 2.50 N
FIG. P23.57
6 2
= 2.35 N
Fx =  F3  F2 cos 14.0 = 2.50  2.35 cos 14.0 = 4.78 N Fy =  F1  F2 sin 14.0 = 40.0  2.35 sin 14.0 = 40.6 N Fnet = Fx2 + Fy2 = tan = Fy Fx = 40.6 4.78
a4.78f + a40.6f
2
2
= 40.9 N
= 263
22
Electric Fields
P23.58
From Figure A: or From Figure B:
d cos 30.0 = 15.0 cm, 15.0 cm d= cos 30.0
= sin 1 = sin 1
Fq
FG d IJ H 50.0 cm K F 15.0 cm I = 20.3 GH 50.0 cmacos 30.0f JK
(1)
Figure A
or From Figure C:
= tan mg Fq = mg tan 20.3 Fq = 2 F cos 30.0
e 2
LM k q OP cos 30.0 MN a0.300 mf PQ Combining equations (1) and (2), L k q OP cos 30.0 = mg tan 20.3 2M MN a0.300 mf PQ mg a0.300 mf tan 20.3 q =
Fq = 2
2 e 2 2 2 2
(2) Figure B
q
2
e2.00 10 kg je9.80 m s ja0.300 mf tan 20.3 = 2e8.99 10 N m C j cos 30.0
3 2 2 9 2 2
2 k e cos 30.0
Figure C FIG. P23.58 F= Q= ke Q 2 Q 2
2
q = 4.20 10 14 C 2 = 2.05 10 7 C = 0.205 C P23.59 Charge Q resides on each block, which repel as point charges: 2
Solving for Q, *P23.60
b gb g = kbL  L g . L k bL  L g . 2L
i i
ke
If we place one more charge q at the 29th vertex, the total force on the central charge will add up to k qQ k e qQ toward vertex 29 . F28 charges = zero: F28 charges + e 2 away from vertex 29 = 0 a a2 According to the result of Example 23.7, the lefthand rod creates this field at a distance d from its righthand end: k eQ E= d 2a + d
P23.61
a
f
dF = F= F=
k eQQ dx 2a d d + 2a
k eQ 2a 4a
2
k Q2 dx 1 2a + x = e  ln x x + 2a x 2a 2a x=b 2a
2 e 2 2
+ k eQ
2
IJ z a f FGH K Fk Q I F FG  ln 2a + b + ln b IJ = k Q ln b H b K 4a ab  2 afab + 2af = GH 4a JK lnGH b b  2a
b b b2a 2 e 2 2
a
f
FIG. P23.61
2
b2  4a 2
I JK
Chapter 23
23
P23.62
At equilibrium, the distance between the charges is r = 2 0.100 m sin 10.0 = 3.47 10 2 m Now consider the forces on the sphere with charge +q , and use
b
g
Fy = 0 :
(1) (2)
q
L
r +q
Fy = 0 : Fx = 0 :
T cos 10.0 = mg , or T =
Fnet
mg cos 10.0 = F2  F1 = T sin 10.0
Fnet is the net electrical force on the charged sphere. Eliminate T from (2) by use of (1). mg sin 10.0 Fnet = = mg tan 10.0 = 2.00 10 3 kg 9.80 m s 2 tan 10.0 = 3.46 10 3 N cos 10.0
e
je
j
Fnet is the resultant of two forces, F1 and F2 . F1 is the attractive force on +q exerted by q , and F2 is the force exerted on +q by the external electric field. Fnet = F2  F1 or F2 = Fnet + F1 F1 = 8.99 10 N m
FIG. P23.62
e
9
2
e5.00 10 Cje5.00 10 Cj = 1.87 10 C j e3.47 10 mj
8 8 2 3 2
2
N
Thus, F2 = Fnet + F1 yields F2 = 3.46 10 3 N + 1.87 10 2 N = 2.21 10 2 N and F2 = qE , or E = F2 2. 21 10 2 N = = 4. 43 10 5 N C = 443 kN C . q 5.00 10 8 C
90 .0 90.0
P23.63
Q = d =
Q = 12.0 C = 2 0 0.600 m = 12.0 C
y 2
= 10.0 C m so b ga f F b3.00 Cge cos Rd j I 1 F b3.00 C gbd g I dF = JJ GH R JK cos = 41 GGH 4 R K e3.00 10 Cje10.0 10 C mj cos d F = z e8.99 10 N m C j a0.600 mf 8.99a30.0f F = e10 Nj z FGH 1 + 1 cos 2 IJK d 0.600 2 2 F1 1 I = 0.707 N Downward. F = a0.450 N fG + sin 2 JK H2 4
0 0 2 0 0 2 90.0 6 6 y 9 2 2 2 90 .0 y 3
z
90 .0
90 .0
z
0 cos Rd = 0 R sin
= 0 R 1  1 = 2 0 R
a f
1 0 1 1 0
cos
0 cos2 0
360
2
 2
2
360
y
 2
FIG. P23.63
Since the leftward and rightward forces due to the two halves of the semicircle cancel out, Fx = 0 . P23.64 At an equilibrium position, the net force on the charge Q is zero. The equilibrium position can be located by determining the angle corresponding to equilibrium. In terms of lengths s, attractive force 1 a 3 , and r, shown in Figure P23.64, the charge at the origin exerts an 2 k eQq
1 2
es +
continued on next page
a 3
j
2
24
Electric Fields
The other two charges exert equal repulsive forces of magnitude of the two repulsive forces add, balancing the attractive force, Fnet = k eQq
1 2
k eQq r2
. The horizontal components
LM 2 cos MM r  es + N
2
1
1 2
a
OP P=0 3j P Q
2
From Figure P23.64 The equilibrium condition, in terms of , is Thus the equilibrium value of satisfies
r=
a
sin
s=
2
Fnet =
FG 4 IJ k QqFG 2 cos sin  H a K GH e
2 e
1 a cot 2
I JJ = 0 . 3 + cot j K
1
2
2 cos sin 2
e
3 + cot
j
2
=1.
One method for solving for is to tabulate the left side. To three significant figures a value of corresponding to equilibrium is 81.7. The distance from the vertical side of the triangle to the equilibrium position is 1 s = a cot 81.7 = 0.072 9 a . 2
60 70 80 90 FIG. P23.64 81 81.5 81.7
2 cos sin 2
e
3 + cot
j
2
4 2.654 1.226 0 1.091 1.024 0.997
A second zerofield point is on the negative side of the xaxis, where = 9.16 and s = 3.10 a . P23.65 (a) From the 2Q charge we have Combining these we find From the Q charge we have Combining these we find k e 2QQ r
2
Fe  T2 sin 2 = 0 and mg  T2 cos 2 = 0 . Fe T sin 2 = 2 = tan 2 . mg T2 cos 2 Fe = T1 sin 1 = 0 and mg  T1 cos 1 = 0 . Fe T sin 1 = 1 = tan 1 or 2 = 1 . mg T1 cos 1
FIG. P23.65
(b)
Fe =
=
2 k eQ r2
2
If we assume is small then
tan
r 2
.
Substitute expressions for Fe and tan into either equation found in part (a) and solve for r. Fe 2k Q 2 1 4k eQ 2 r = tan then e 2 and solving for r we find r mg 2 mg mg r
F I GH JK
F GH
I JK
13
.
Chapter 23
25
P23.66
(a)
The distance from each corner to the center of the square is
x Q +q L/2 +q L/2 +q z +q
FG L IJ + FG L IJ H 2K H 2K
2
2
=
L 2
.
The distance from each positive charge to Q is then z2 + L2 . Each positive charge exerts a force directed 2
FIG. P23.66 k eQq z z + L2 2
2
along the line joining q and Q , of magnitude The line of force makes an angle with the zaxis whose cosine is
z 2 + L2 2
.
The four charges together exert forces whose x and y components add to zero, while the 4k e Qqz F=  k zcomponents add to 3 2 2 z + L2 2
e
j
(b)
For z >> L , the magnitude of this force is Fz = 
eL 2 j
2
4k eQqz
32
=
F 4a2f k Qq I z = ma GH L JK
3 2 3 e
z
Therefore, the object's vertical acceleration is of the form a z =  2 z with 2 = 42
af
32
mL
3
k eQq
=
k eQq 128 mL3
.
Since the acceleration of the object is always oppositely directed to its excursion from equilibrium and in magnitude proportional to it, the object will execute simple harmonic motion with a period given by T= 2 =
a128f
2
14
mL3 = k e Qq
a8 f
14
mL3 . k eQq
P23.67
(a)
The total noncontact force on the cork ball is: F = qE + mg = m g +
qE , m which is constant and directed downward. Therefore, it behaves like a simple pendulum in the presence of a modified uniform gravitational field with a period given by: T = 2 L 0.500 m = 2 g + qE m 9.80 m s 2 + 2.00 10 6 C 1.00 10 5 N C 1.00 10 3 kg
FG H
IJ K
e
je
j
= 0.307 s L qE m = 0.314 s (a 2.28% difference).
(b)
Yes . Without gravity in part (a), we get T = 2
T = 2
e2.00 10 Cje1.00 10
6
0.500 m
5
N C 1.00 10 3 kg
j
26
Electric Fields
P23.68
The bowl exerts a normal force on each bead, directed along the radius line or at 60.0 above the horizontal. Consider the freebody diagram of the bead on the left:
Fy = n sin 60.0 mg = 0 ,
or Also, or n= mg . sin 60.0
n Fe mg 60.0
Fx =  Fe + n cos 60.0 = 0 ,
keq 2 R
2
= n cos 60.0 =
mg mg = . tan 60.0 3
Thus,
q=
F mg I RG H k 3 JK
e
12
.
FIG. P23.68
P23.69
(a)
There are 7 terms which contribute: 3 are s away (along sides) 3 are 1 is 2s away (face diagonals) and sin = 3s away (body diagonal) and sin = 1 3 1 2 . FIG. P23.69
= cos
The component in each direction is the same by symmetry. F= ke q2 s2
LM1 + 2 + 1 OPei + j + kj = N 2 2 3 3Q
ke q2 s2
keq 2 s2
a1.90fei + j + kj
(b) P23.70 (a)
F = Fx2 + Fy2 + Fz2 = 3. 29
away from the origin
Zero contribution from the same face due to symmetry, opposite face contributes 4
FG k q sin IJ where Hr K
e 2
r=
FG s IJ + FG s IJ H 2K H 2K
2
2
+ s 2 = 1.5 s = 1.22s
3
sin =
s r
E=4
k e qs r3
=
a1.22f
4
keq s2
= 2.18
keq s2 FIG. P23.70
(b)
The direction is the k direction.
Chapter 23
27
P23.71
The field on the axis of the ring is calculated in Example 23.8,
The force experienced by a charge q placed along the axis of the ring is
and when x << a , this becomes This expression for the force is in the form of Hooke's law, with an effective spring constant of Since = 2 f = k , we have m
ex + a j LM x F =  k Qq MM ex + a j N F k Qq IJ x F = G Ha K
2 e 2 e 3
E = Ex =
k e xQ
2 3 2
2 3 2
OP PP Q
k= f=
k eQq a3 1 2 k eQq ma 3
.
P23.72
dE =
x
2
F G + a0.150 mf G H
k e dq
2
E=
all charge
z
dE = k e
0. 400 m x=0
z
I k e xi + 0.150 mjjdx J= x + a0.150 mf J K x + a0.150 mf e xi + 0.150 mjjdx x + a0.150 mf
 x i + 0.150 m j
2 e 2 2 2 32 2 2 3 2 0. 400 m 0. 400 m 2 2 2 2 0 0
LM OP a0.150 mf jx +i E = k M + PP x + a0.150 mf a0.150 mf x + a0.150 mf MN Q E = e8.99 10 N m C je35.0 10 C mj ia 2.34  6.67f m + ja6.24  0 f m E = e 1.36 i + 1.96 jj 10 N C = e 1.36 i + 1.96 jj kN C
e 2 9 2 2 9 1 3
FIG. P23.72
1
P23.73
The electrostatic forces exerted on the two charges result in a net torque = 2 Fa sin = 2Eqa sin . For small , sin and using p = 2 qa , we have
= Ep .
d 2 dt 2 . FIG. P23.73 Ep . I
The torque produces an angular acceleration given by = I = I Combining two these expressions for torque, we have This equation can be written in the form
Ep d 2 =0. + 2 I dt d 2 dt
2
FG IJ H K
=  2 where 2 =
This is the same form as Equation 15.5 and the frequency of oscillation is found by comparison with Equation 15.11, or f= 1 2 pE 1 = I 2 2 qaE . I
28
Electric Fields
ANSWERS TO EVEN PROBLEMS
P23.2 P23.4 P23.6 P23.8 P23.10 (a) 2.62 10 24 ; (b) 2.38 electrons for every 10 9 present P23.36 P23.38 P23.40 P23.42 P23.44 P23.46 P23.48 (a) 200 pC; (b) 141 pC; (c) 58.9 pC see the solution (a)  1 ; (b) q1 is negative and q 2 is positive 3
57.5 N 2.51 10
514 kN x = 0.634d . The equilibrium is stable if the third bead has positive charge.
9
electron: 4.39 Mm s ; proton: 2.39 km s (a) 57.6 i Tm s 2 ; (b) 2.84i Mm s; (c) 49.3 ns (a) down; (b) 3.43 C The particle strikes the negative plate after moving in a parabola 0.181 mm high and 0.961 mm. Possible only with +51.3 C at x = 16.0 cm (a) 24.2 N C at 0; (b) 9. 42 N C at 117
P23.12
(a) period = 2
md 3 where m is the mass k e qQ k e qQ md 3
of the object with charge Q ; (b) 4a P23.14 P23.16 P23.18
1.49 g
720 kN C down (a) 18.0 i  218 j kN C ; (b) 36.0 i  436 j mN
P23.50 P23.52 P23.54 P23.56 P23.58 P23.60 P23.62 P23.64 P23.66
5.25 C
(a) mg mgA ; (b) A cot + B A cos + B sin
e
j
P23.20 P23.22 P23.24 P23.26
(a) 12.9 j kN C ; (b) 38.6 j mN see the solution
0.205 C
k e qQ a2 toward the 29th vertex
2 ke q  i 6a2
ke 0 i x0 ke 0 i 2x0
e j e j
443 i kN C
0.072 9 a
see the solution; the period is
P23.28 P23.30 P23.32 P23.34
81 4
mL3 k eQq
(a) 383 MN C away; (b) 324 MN C away; (c) 80.7 MN C away; (d) 6.68 MN C away see the solution
P23.68
R
F mg I GH k 3 JK
e
12
LMe j  ead + hf + R j N 2 k Qi L (b) R h M Nh + ed + R j  ead + hf + R j
(a) k eQ i d 2 + R2 h
e 2 2 1 2 2 2 2 12 2 2
1 2
12
OP; Q OP Q
P23.70 P23.72
(a) see the solution; (b) k
e1.36i + 1.96 jj kN C
24
Gauss's Law
CHAPTER OUTLINE
24.1 24.2 24.3 Electric Flux Gauss's Law Application of Gauss's Law to Various Charge Distributions Conductors in Electrostatic Equilibrium Formal Derivation of Gauss`s Law
ANSWERS TO QUESTIONS
Q24.1 The luminous flux on a given area is less when the sun is low in the sky, because the angle between the rays of the sun and the local area vector, dA, is greater than zero. The cosine of this angle is reduced. The decreased flux results, on the average, in colder weather. If the region is just a point, line, or plane, no. Consider two protons in otherwise empty space. The electric field is zero at the midpoint of the line joining the protons. If the fieldfree region is threedimensional, then it can contain no charges, but it might be surrounded by electric charge. Consider the interior of a metal sphere carrying static charge. The surface must enclose a positive total charge.
24.4 24.5
Q24.2
Q24.3 Q24.4
The net flux through any gaussian surface is zero. We can argue it two ways. Any surface contains zero charge so Gauss's law says the total flux is zero. The field is uniform, so the field lines entering one side of the closed surface come out the other side and the net flux is zero. Gauss's law cannot tell the different values of the electric field at different points on the surface. When E is an unknown number, then we can say E cos dA = E cos dA . When E x , y , z is an unknown function, then there is no such simplification.
Q24.5
z
z
b
g
Q24.6
The electric flux through a sphere around a point charge is independent of the size of the sphere. A sphere of larger radius has a larger area, but a smaller field at its surface, so that the product of field strength and area is independent of radius. If the surface is not spherical, some parts are closer to the charge than others. In this case as well, smaller projected areas go with stronger fields, so that the net flux is unaffected. Faraday's visualization of electric field lines lends insight to this question. Consider a section of a 1 field lines pointing out from it horizontally to vertical sheet carrying charge +1 coulomb. It has 0 the right and left, all uniformly spaced. The lines have the same uniform spacing close to the sheet and far away, showing that the field has the same value at all distances.
Q24.7
29
30
Gauss's Law
Q24.8
Consider any point, zone, or object where electric field lines begin. Surround it with a closefitting gaussian surface. The lines will go outward through the surface to constitute positive net flux. Then Gauss's law asserts that positive net charge must be inside the surface: it is where the lines begin. Similarly, any place where electric field lines end must be just inside a gaussian surface passing net negative flux, and must be a negative charge. Inject some charge at arbitrary places within a conducting object. Every bit of the charge repels every other bit, so each bit runs away as far as it can, stopping only when it reaches the outer surface of the conductor. If the person is uncharged, the electric field inside the sphere is zero. The interior wall of the shell carries no charge. The person is not harmed by touching this wall. If the person carries a (small) charge q, the electric field inside the sphere is no longer zero. Charge q is induced on the inner wall of the sphere. The person will get a (small) shock when touching the sphere, as all the charge on his body jumps to the metal. The electric fields outside are identical. The electric fields inside are very different. We have E = 0 everywhere inside the conducting sphere while E decreases gradually as you go below the surface of the sphere with uniform volume charge density. There is zero force. The huge charged sheet creates a uniform field. The field can polarize the neutral sheet, creating in effect a film of opposite charge on the near face and a film with an equal amount of like charge on the far face of the neutral sheet. Since the field is uniform, the films of charge feel equalmagnitude forces of attraction and repulsion to the charged sheet. The forces add to zero. Gauss's law predicts, as described in section 24.4, that excess charge on a conductor will reside on the surface of the conductor. If a car is left charged by a lightning strike, then that charge will remain on the outside of the car, not harming the occupants. It turns out that during the lightning strike, the current also remains on the outside of the conductor. Note that it is not necessarily safe to be in a fiberglass car or a convertible during a thunderstorm.
Q24.9
Q24.10
Q24.11
Q24.12
Q24.13
SOLUTIONS TO PROBLEMS
Section 24.1 P24.1 (a) (b) (c) P24.2 P24.3 Electric Flux E = EA cos = 3.50 10 3 0.350 0.700 cos 0 = 858 N m 2 C
e
ja
f
= 90.0
E = 0
E = 3.50 10 3 0.350 0.700 cos 40.0 = 657 N m 2 C
e
ja
f
E = EA cos = 2.00 10 4 N C 18.0 m 2 cos 10.0 = 355 kN m 2 C E = EA cos A = r 2 = 0.200
e
je
j
a
f
2
= 0.126 m 2
5.20 10 5 = E 0.126 cos 0
a
f
E = 4.14 10 6 N C = 4.14 MN C
Chapter 24
31
P24.4
(a)
A = 10.0 cm 30.0 cm
2
a
fa
f
A = 300 cm = 0.030 0 m 2 E , A = EA cos E , A = 7.80 10 4 0.030 0 cos 180 E , A = 2.34 kN m 2 C (b)
30.0 cm
e
jb
g
10.0 cm
60.0
e ja f F 10.0 cm IJ = 600 cm = 0.060 0 m A = a30.0 cmfa wf = a30.0 cmfG H cos 60.0 K = e7.80 10 jb0.060 0g cos 60.0 = +2.34 kN m C
E , A = EA cos = 7.80 10 4 A cos 60.0
2 E, A 4 2
FIG. P24.4
2
(c)
The bottom and the two triangular sides all lie parallel to E, so E = 0 for each of these. Thus,
E, total = 2.34 kN m 2 C + 2.34 kN m 2 C + 0 + 0 + 0 = 0 .
P24.5 (a) (b) (c) P24.6 E = E A = a i + b j A i = aA E = a i + b j A j = bA E = a i + b j Ak = 0
e
j
e e
j j
Only the charge inside radius R contributes to the total flux. E = q 0
P24.7
E = EA cos through the base
E = 52.0 36.0 cos 180 = 1.87 kN m 2 C .
Note the same number of electric field lines go through the base as go through the pyramid's surface (not counting the base). For the slanting surfaces, E = +1.87 kN m 2 C . P24.8 The flux entering the closed surface equals the flux exiting the surface. The flux entering the left side of the cone is E = E dA = ERh . This is the same as the flux that exits the right side of the cone. Note that for a uniform field only the cross sectional area matters, not shape. FIG. P24.7
a fa f
z
32
Gauss's Law
Section 24.2 P24.9 (a)
Gauss's Law E = +5.00 C  9.00 C + 27.0 C  84.0 C qin = = 6.89 10 6 N m 2 C 2 0 8.85 10 12 C 2 N m 2
b
g
E = 6.89 MN m 2 C (b) P24.10 (a) Since the net electric flux is negative, more lines enter than leave the surface. E= k eQ r2 : 8.90 10
2
e8.99 10 jQ = a0.750f
9 2
But Q is negative since E points inward. (b) P24.11 E =
Q = 5.56 10 8 C = 55.6 nC
The negative charge has a spherically symmetric charge distribution. qin 0 E = E = E = 2Q + Q Q =  0 0 +Q  Q = 0 0 2Q + Q  Q 2Q =  0 0
Through S1 Through S 2 Through S3 Through S 4 P24.12 (a)
E = 0
Onehalf of the total flux created by the charge q goes through the plane. Thus, q 1 1 q E , plane = E , total = = . 2 2 0 2 0
FG IJ H K
(b)
The square looks like an infinite plane to a charge very close to the surface. Hence, q E , square E , plane = . 2 0 The plane and the square look the same to the charge.
(c) P24.13
The flux through the curved surface is equal to the flux through the flat circle, E0 r 2 . (a) E , shell = qin 12.0 10 6 = = 1.36 10 6 N m 2 C = 1.36 MN m 2 C 0 8.85 10 12 1 1.36 10 6 N m 2 C = 6.78 10 5 N m 2 C = 678 kN m 2 C 2
P24.14
(b) (c)
E, half shell =
e
j
No, the same number of field lines will pass through each surface, no matter how the
radius changes.
Chapter 24
33
P24.15
(a)
With very small, all points on the hemisphere are nearly at a distance R from the charge, so the field everywhere on the kQ curved surface is e 2 radially outward (normal to the R surface). Therefore, the flux is this field strength times the area of half a sphere: curved = E dA = Elocal A hemisphere curved = k e
0
Q
FG H
z
Q R2
IJ FG 1 4 R IJ = 1 Qa2 f = K H 2 K 4
2 0
+Q 2 0
FIG. P24.15
(b)
The closed surface encloses zero charge so Gauss's law gives curved + flat = 0 or flat =  curved = Q . 2 0
*P24.16
Consider as a gaussian surface a box with horizontal area A, lying between 500 and 600 m elevation.
z
E dA =
q : 0
b+120 N CgA + b100 N CgA = Aa100 mf b20 N Cge8.85 10 C N m j = 1.77 10 =
0 12 2 2
12
100 m
C m3
The charge is positive , to produce the net outward flux of electric field. Q6q 0
P24.17
The total charge is Q  6 q . The total outward flux from the cube is through each face:
, of which onesixth goes
b g
E one face
= =
Q6q 6 0 Q6q 6 0 =
b g
P24.18
E one face
a5.00  6.00f 10
6
C N m2 C
2
6 8.85 10
12
= 18.8 kN m 2 C . Q6q 0
The total charge is Q  6 q . The total outward flux from the cube is through each face:
, of which onesixth goes
b g
P24.19
E one face
=
Q6q 6 0
. qin = 0 . 0
If R d , the sphere encloses no charge and E =
If R > d , the length of line falling within the sphere is 2 R 2  d 2 so E = 2 R 2  d 2 0 .
34
Gauss's Law
P24.20
E , hole = E A hole =
FG k Q IJ e r HR K
e 2
2
F e8.99 10 N m C je10.0 10 Cj I JJ e1.00 10 j GG a0.100 mf H K
9 2 2 6
=
2
3
m
j
2
E, hole = 28.2 N m 2 C P24.21 E = qin 170 10 6 C = = 1.92 10 7 N m 2 C 0 8.85 10 12 C 2 N m 2
(a) (b) (c)
b g
E one face
=
1.92 10 7 N m 2 C 1 E = 6 6
b g
E one face
= 3.20 MN m 2 C
E = 19.2 MN m 2 C The answer to (a) would change because the flux through each face of the cube would not be equal with an asymmetric charge distribution. The sides of the cube nearer the charge would have more flux and the ones further away would have less. The answer to (b) would remain the same, since the overall flux would remain the same.
P24.22
No charge is inside the cube. The net flux through the cube is zero. Positive flux comes out through the three faces meeting at g. These three faces together fill solid angle equal to oneeighth of a sphere as seen from q, and together pass 1 q . Each face containing a intercepts equal flux going into the cube: flux 8 0
FG IJ H K
0 = E , net E , abcd =
q = 3 E , abcd + 8 0 q 24 0
FIG. P24.22
Section 24.3 P24.23
Application of Gauss's Law to Various Charge Distributions
The charge distributed through the nucleus creates a field at the surface equal to that of a point k q charge at its center: E = e2 r
e8.99 10 Nm C je82 1.60 10 E= a208f 1.20 10 m
9 2 2 13 15 2
19
C
j
E = 2.33 10 21 N C
away from the nucleus
Chapter 24
35
P24.24
(a)
E=
k eQr a3
= 0
9 6
(b)
(c)
e8.99 10 je26.0 10 ja0.100f = 365 kN C a a0.400f k Q e8.99 10 je 26.0 10 j = = 1.46 MN C E= r a0.400f
E= k eQr
3
=
3
9
6
e
2
2
(d)
E=
k eQ r2
e8.99 10 je26.0 10 j = = a0.600f
9 6 2
649 kN C
The direction for each electric field is radially outward .
*P24.25
mg = qE = q
FG IJ = qFG Q A IJ H2 K H 2 K
0 0
12 0.01 9.8 Q 2 0 mg 2 8.85 10 = = = 2.48 C m 2 A q 0.7 10 6
e
ja fa f
P24.26
(a)
2 8.99 10 9 Q 2.40 2k e 4 E= 3.60 10 = 0.190 r Q = +9.13 10 7 C = +913 nC
e
jb
g
(b) *P24.27
E= 0
The volume of the spherical shell is 4 0.25 m 3
a
f  a0.20 mf
3
3
= 3.19 10 2 m3 .
Its charge is
V = 1.33 10 6 C m 3 3.19 10 2 m3 = 4.25 10 8 C .
The net charge inside a sphere containing the proton's path as its equator is 60 10 9 C  4. 25 10 8 C = 1.02 10 7 C . The electric field is radially inward with magnitude ke q r
2
e
je
j
=
q 0 4 r
2
=
8.99 10 9 Nm 2 1.02 10 7 C C 0.25 m mv 2 r
2
a
e
f
2
j = 1.47 10
4
N C.
For the proton
F = ma
eE =
12
F eEr IJ v=G HmK
F 1.60 10 =G GH
19
C 1.47 10 4 N C 0.25 m 1.67 10 27 kg
e
j
I JJ K
12
= 5.94 10 5 m s .
36
Gauss's Law
P24.28
= 8.60 10 6 C cm 2
E=
e
jFGH 100mcm IJK
2
= 8.60 10 2 C m 2
8.60 10 2 = = 4.86 10 9 N C away from the wall 2 0 2 8.85 10 12
e
j
The field is essentially uniform as long as the distance from the center of the wall to the field point is much less than the dimensions of the wall. P24.29 If is positive, the field must be radially outward. Choose as the gaussian surface a cylinder of length L and radius r, contained inside the charged rod. Its volume is r 2 L and it encloses charge r 2 L . Because the charge distribution is long, no electric flux passes through the circular end caps; E dA = EdA cos 90.0 = 0 . The curved surface has E dA = EdA cos 0 , and E must be the same strength everywhere over the curved surface. q r 2 L Gauss's law, E dA = , becomes E dA = . 0 0 Curved
FIG. P24.29
z
z
Surface
Now the lateral surface area of the cylinder is 2 rL : E 2 r L = *P24.30
b g
r 2 L . 0
Thus,
E=
r radially away from the cylinder axis . 2 0
Let represent the charge density. For the field inside the sphere at r1 = 5 cm we have E1 4 r12 = q inside 4 r13 = 0 3 0 E1 = r1 3 0
12 C 2 86 10 3 N 3 0 E1 3 8.85 10 = = = 4.57 10 5 C m 3 . r1 0.05 m Nm 2 C Now for the field outside at r3 = 15 cm 3 4 r2 E3 4 r32 = 3 0
e
je
j
k e 4 0.10 m E3 = 2 r3 3
a
f e4.57 10 Cj = 8.99 10
3 5
9
m
3
a0.15 mf C
2
Nm 2 1.91 10 7 C
2
e
j = 7.64 10
4
NC
E 3 = 76. 4 kN C radially inward P24.31 (a)
E= 0
k eQ r2
(b) P24.32
E=
e8.99 10 je32.0 10 j = 7.19 MN C = a0.200f
9 6 2
E = 7.19 MN C radially outward
The distance between centers is 2 5.90 10 15 m . Each produces a field as if it were a point charge at its center, and each feels a force as if all its charge were a point at its center. F= k e q1 q 2 r2 = 8.99 10 N m
e
9
2
a46f e1.60 10 Cj C j e2 5.90 10 mj
2 19 2 15
2
2
= 3.50 10 3 N = 3.50 kN
Chapter 24
37
P24.33
Consider two balloons of diameter 0.2 m, each with mass 1 g, hanging apart with a 0.05 m separation on the ends of strings making angles of 10 with the vertical. (a)
Fy = T cos 10mg = 0 T = cos 10 Fx = T sin 10 Fe = 0 Fe = T sin 10 , so
Fe = Fe 2 10 3 N ~ 10 3 N or 1 mN Fe = keq 2 r2
3
mg
FG mg IJ sin 10 = mg tan 10 = b0.001 kgge9.8 m s j tan 10 H cos 10 K
2
FIG. P24.33
(b)
2 10
e8.99 10 N m N a0.25 mf
9
2
C 2 q2
j
2
q 1. 2 10 7 C ~ 10 7 C or 100 nC keq r
2
(c)
E=
e8.99 10
9
N m 2 C 2 1.2 10 7 C
a0.25 mf
je
2
j 1.7 10
4
N C ~ 10 kN C
(d)
E =
q 1.2 10 7 C = 1. 4 10 4 N m 2 C ~ 10 kN m 2 C 0 8.85 10 12 C 2 N m 2 4 3 a 3
*P24.34
The charge density is determined by Q = (a)
=
3Q 4 a 3
The flux is that created by the enclosed charge within radius r: E = q in 4 r 3 4 r 3 3Q Qr 3 = = = 0 3 0 3 0 4 a 3 0 a 3 Q . Note that the answers to parts (a) and (b) agree at r = a . 0 E Q 0
(b) (c)
E =
0
0
a FIG. P24.34(c)
r
38
Gauss's Law
P24.35
(a)
9 2 2 2.00 10 6 C 7.00 m 2 k e 2 8.99 10 N m C = E= 0.100 m r
e
je
j
E = 51.4 kN C , radially outward
(b) E = EA cos = E 2 r cos 0
b
g
E = 5.14 10 4 N C 2 0.100 m 0.020 0 m 1.00 = 646 N m 2 C
P24.36 (a)
e
j a g
fb
ga f
=
qin =
(b)
qin
FG 4 r IJ = e2.13 10 H3 K F4 I = G r J = e 2.13 10 H3 K
3 3
Q 5.70 10 6 = = 2.13 10 2 C m 3 3 4 3 4 a 0.040 0 3 3
b
2
2
4 jFGH 3 IJK b0.020 0g 4 jFGH 3 IJK b0.040 0g
3
= 7.13 10 7 C = 713 nC
3
= 5.70 C
P24.37
E=
9.00 10 6 C m 2 = = 508 kN C , upward 2 0 2 8.85 10 12 C 2 N m 2
e
j
P24.38
Note that the electric field in each case is directed radially inward, toward the filament. (a) E=
6 9 2 2 2 k e 2 8.99 10 N m C 90.0 10 C m = = 16.2 MN C 0.100 m r 6 9 2 2 2 k e 2 8.99 10 N m C 90.0 10 C m = = 8.09 MN C 0.200 m r
e e e
je je je
j j
(b)
E=
(c)
6 9 2 2 2 k e 2 8.99 10 N m C 90.0 10 C m = = 1.62 MN C E= 1.00 m r
j
Section 24.4 P24.39
Conductors in Electrostatic Equilibrium
z
EdA = E 2 rl =
b g
qin 0
E=
q in l = 2 0 r 2 0 r
(a) (b)
r = 3.00 cm r = 10.0 cm
E= 0
E= 30.0 10 9
2 8.85 10 12 0.100 30.0 10 9
e e
ja
f=
5 400 N C , outward
(c)
r = 100 cm
E=
2 8.85 10 12 1.00
= ja f
540 N C , outward
Chapter 24
39
P24.40
From Gauss's Law,
EA =
=
P24.41
Q =0 E = 8.85 10 12 130 = 1.15 10 9 C m 2 = 1.15 nC m 2 A
e
ja
Q 0
f
The fields are equal. The Equation 24.9 E = different from Equation 24.8 E =
insulator for the field around glass. But its charge will spread out to 2 0 Q cover both sides of the aluminum plate, so the density is conductor = . The glass carries charge 2A Q Q only on area A, with insulator = . The two fields are the same in magnitude, and both are A 2 A 0 perpendicular to the plates, vertically upward if Q is positive.
*P24.42 (a) All of the charge sits on the surface of the copper sphere at radius 15 cm. The field inside is zero . The charged sphere creates field at exterior points as if it were a point charge at the center: E= ke q r
2
conductor for the field outside the aluminum looks 0
(b)
away =
9
e8.99 10 je f
2
9
Nm 2 40 10 9 C
2
C 0.17 m
a
je
f
2
j outward =
1.24 10 4 N C outward
(c) (d) P24.43 (a)
e8.99 10 E=
0
Nm 2 40 10 9 C
2
C 0.75 m
a
j outward = j
639 N C outward
All three answers would be the same. E=
= 8.00 10 4 8.85 10 12 = 7.08 10 7 C m 2
e
je
= 708 nC m 2 , positive on one face and negative on the other.
(b)
=
Q 2 Q = A = 7.08 10 7 0.500 C A Q = 1.77 10 7 C = 177 nC , positive on one face and negative on the other.
e
ja
f
P24.44
(a)
E= 0
k eQ r2
(b)
E=
e8.99 10 je8.00 10 j = 7.99 10 = b0.030 0g
9 6 2
7
NC
E = 79.9 MN C radially outward
(c)
E= 0
k eQ r2
(d)
E=
e8.99 10 je4.00 10 j = 7.34 10 = b0.070 0g
9 6 2
6
NC
E = 7.34 MN C radially outward
40
Gauss's Law
P24.45
The charge divides equally between the identical spheres, with charge like point charges at their centers: F= ke Q 2 Q 2
2
Q on each. Then they repel 2
b gb g = k Q aL + R + Rf 4aL + 2Rf
e 2
2
=
8.99 10 9 N m 2 60.0 10 6 C 4 C 2 2.01 m
a
e
f
j
2
2
= 2.00 N .
P24.46
The electric field on the surface of a conductor varies inversely with the radius of curvature of the surface. Thus, the field is most intense where the radius of curvature is smallest and viceversa. The local charge density and the electric field intensity are related by E= (a)
0
or
=0 E .
Where the radius of curvature is the greatest,
=0 Emin = 8.85 10 12 C 2 N m 2 2.80 10 4 N C = 248 nC m 2 .
(b) Where the radius of curvature is the smallest,
e
je
j
=0 Emax = 8.85 10 12 C 2 N m 2 5.60 10 4 N C = 496 nC m 2 .
P24.47 (a) Inside surface: consider a cylindrical surface within the metal. Since E inside the conducting shell is zero, the total charge inside the gaussian surface must be zero, so the inside charge/length =  . 0 = + qin Outside surface: qout = 2 + (b) E= 2 k e 3 r k eQ r2
e
e
je
j
so The total charge on the metal cylinder is so the outside charge/length is 3 radially outward 2 0 r
6
qin
= 
2 = qin + qout
3 .
b g = 6k =
r
9
P24.48
(a)
E=
e8.99 10 je6.40 10 j = = a0.150f
2
2.56 MN C , radially inward
(b) P24.49 (a)
E=0
The charge density on each of the surfaces (upper and lower) of the plate is:
=
8 1 q 1 4.00 10 C = = 8.00 10 8 C m 2 = 80.0 nC m 2 . 2 A 2 0.500 m 2
FG IJ H K
0
e
a
f
j
(b)
E=
FG IJ k = F 8.00 10 C m I k = b9.04 kN Cgk H K GH 8.85 10 C N m JK
8 2 12 2 2
(c)
E=
b9.04 kN Cgk
Chapter 24
41
P24.50
(a)
The charge +q at the center induces charge q on the inner surface of the conductor, where its surface density is: q a = . 4 a 2 The outer surface carries charge Q + q with density
(b)
b =
P24.51
Q+q 4 b 2
.
Use Gauss's Law to evaluate the electric field in each region, recalling that the electric field is zero everywhere within conducting materials. The results are: E = 0 inside the sphere and within the material of the shell E = ke E = ke Charge Charge and Q between the sphere and shell, directed radially inward r2 2Q outside the shell, directed radially outward . r2
Q is on the outer surface of the sphere . +Q is on the inner surface of the shell , +2Q is on the outer surface of the shell.
P24.52
An approximate sketch is given at the right. Note that the electric field lines should be perpendicular to the conductor both inside and outside.
FIG. P24.52
Section 24.5 P24.53 (a)
Formal Derivation of Gauss`s Law Uniform E, pointing radially outward, so E = EA . The arc length is ds = Rd , and the circumference is 2 r = 2 R sin A = 2 rds =
z
zb
0
2 R sin Rd = 2 R 2 sin d = 2 R 2  cos
0
g
z
a
f
0
= 2 R 2 1  cos
b
g
1 Q Q E = 2 R 2 1  cos = 1  cos 4 0 R 2 2 0 (b) For = 90.0 (hemisphere): E =
a
f
a
f f
FIG. P24.53
[independent of R!]
Q Q 1  cos 90 = . 2 0 2 0 Q Q 1  cos 180 = 2 0 0
a
(c)
For = 180 (entire sphere): E =
a
f
[Gauss's Law].
42
Gauss's Law
Additional Problems P24.54 In general, In the xy plane, z = 0 and E = E dA =
w
E = ay i + bz j + cxk E = ay i + cxk
z y=0 x=0 x=w x y =h y dA = hdx
z
ze
ay i + cxk kdA
w
j
x2 E = ch xdx = ch 2 x=0
z
x=0
chw 2 = 2
FIG. P24.54
P24.55
(a) (b)
qin = +3Q  Q = +2Q The charge distribution is spherically symmetric and qin > 0 . Thus, the field is directed radially outward .
(c) (d)
E=
k e qin r
2
=
2 k eQ r2
for r c .
Since all points within this region are located inside conducting material, E = 0 for b < r < c.
(e) (f) (g)
E = E dA = 0 qin =0 E = 0 qin = +3Q E= k e qin r
2
z
=
3 k eQ r2
(radially outward) for a r < b .
3
(h)
qin = V = k e qin r
2
(i) (j)
E=
F +3Q I FG 4 r IJ = +3Q r GH a JK H 3 K a I k F = GH +3Q ra JK = 3 k Q ar (radially outward) for 0 r a . r
4 3 3 3 3 e 2 3 3 e 3
From part (d), E = 0 for b < r < c . Thus, for a spherical gaussian surface with b < r < c , qin = +3Q + qinner = 0 where qinner is the charge on the inner surface of the conducting shell. This yields qinner = 3Q . Since the total charge on the conducting shell is q net = qouter + qinner = Q , we have qouter = Q  qinner = Q  3Q = +2Q .
E
(k)
a
b
c
r
b g
FIG. P24.55(l)
(l)
This is shown in the figure to the right.
Chapter 24
43
P24.56
The sphere with large charge creates a strong field to polarize the other sphere. That means it pushes the excess charge over to the far side, leaving charge of the opposite sign on the near side. This patch of opposite charge is smaller in amount but located in a stronger external field, so it can feel a force of attraction that is larger than the repelling force felt by the larger charge in the weaker field on the other side. (a)
P24.57
z
E dA = E 4 r 2 =
e
j
qin 0 qin = E=
For r < a , so For a < r < b and c < r , So For b r c , (b)
FG 4 r IJ H3 K
3
r . 3 0
FIG. P24.57
qin = Q . Q E= . 4 r 2 0
E = 0 , since E = 0 inside a conductor.
Let q1 = induced charge on the inner surface of the hollow sphere. Since E = 0 inside the conductor, the total charge enclosed by a spherical surface of radius b r c must be zero. Therefore, q1 + Q = 0 and
1 =
q1 4 b
2
=
Q . 4 b 2
Let q 2 = induced charge on the outside surface of the hollow sphere. Since the hollow sphere is uncharged, we require q Q 2 = 1 2 = . q1 + q 2 = 0 and 4 c 4 c 2 P24.58
z
E dA = E 4 r 2 =
e
j
qin 0 N C 4 0.100 m
(a)
e3.60 10
3
j a
f
2
=
Q 8.85 10
12
C 2 N m2
aa < r < bf
Q = 4.00 10 9 C = 4.00 nC
(b) We take Q to be the net charge on the hollow sphere. Outside c, Q + Q 2 +2.00 10 2 N C 4 0.500 m = r>c 8.85 10 12 C 2 N m 2
e
j a
f
a f
Q + Q = +5.56 10 9 C , so Q = +9.56 10 9 C = +9.56 nC
(c) For b < r < c : E = 0 and qin = Q + Q1 = 0 where Q1 is the total charge on the inner surface of the hollow sphere. Thus, Q1 = Q = +4.00 nC . Then, if Q 2 is the total charge on the outer surface of the hollow sphere,
Q 2 = Q  Q1 = 9.56 nC  4.0 nC = +5.56 nC .
44
Gauss's Law
*P24.59
The vertical velocity component of the moving charge increases according to m dv y dt = Fy dv y dx = qE y . m dx dt q
y v 0 d Q FIG. P24.59
vx
vy x
Now dv y =
dx = v x has the nearly constant value v. So dt q Ey dx mv
vy
v y = dv y =
0
z
q Ey dx . mv 
z
The radially outward compnent of the electric field varies along the x axis, but is described by

z
Ey dA =

z
Ey 2 d dx =
b g
Q . 0
So

z
Ey dx = vy v =
qQ Q and v y = . The angle of deflection is described by 2 d 0 mv 2 d 0 qQ 2 0 dmv
2
tan = P24.60
= tan 1
qQ 2 0 dmv 2
.
First, consider the field at distance r < R from the center of a uniform sphere of positive charge Q = + e with radius R.
b
g
e
4 r 2 E =
j
qin V = = 0 0
F GH
+e 4 3 3 R
I JK
4 3
r3
0
so E =
F e I r directed outward GH 4 R JK
0 3
(a)
The force exerted on a point charge q =  e located at distance r from the center is then F = qE =  e
F e I r = F e I r = GH 4 R JK GH 4 R JK
2 0 3 0 3
 Kr .
(b)
K=
k e2 e2 = e3 3 4 0 R R
(c)
Fr = m e a r = 
F k e I r , so a GH R JK
e 2 3
r
=
F k e I r =  r GH m R JK
e 2 e 3 2
Thus, the motion is simple harmonic with frequency
f=
1 = 2 2
me R3
kee2
.
(d)
f = 2.47 10
15
1 Hz = 2
e8.99 10
9
N m 2 C 2 1.60 10 19 C
31
e9.11 10
je
kg R 3
j
j
2
which yields R 3 = 1.05 10 30 m3 , or R = 1.02 10 10 m = 102 pm .
Chapter 24
45
P24.61
The field direction is radially outward perpendicular to the axis. The field strength depends on r but not on the other cylindrical coordinates or z. Choose a Gaussian cylinder of radius r and length L. If r < a , E = qin 0 and E 2 rL =
b
g
L 0
r
E=
2 r 0
or
E=
2 r 0
ar < a f
.
If a < r < b ,
E 2 rL =
b
g
L + r 2  a 2 L
0
e
j
E=
+ r 2  a 2
2 r 0
e
jr
aa < r < bf .
If r > b ,
E 2 rL =
b
g
L + b 2  a 2 L
0
e
j
E= P24.62
+ b 2  a 2
2 r 0
e
jr
ar > b f
.
Consider the field due to a single sheet and let E+ and E represent the fields due to the positive and negative sheets. The field at any distance from each sheet has a magnitude given by Equation 24.8: E+ = E = (a)
. 2 0
To the left of the positive sheet, E+ is directed toward the left and E toward the right and the net field over this region is E = 0 .
(b)
In the region between the sheets, E+ and E are both directed toward the right and the net field is E=
0
to the right .
FIG. P24.62
(c)
To the right of the negative sheet, E+ and E are again oppositely directed and E = 0 .
46
Gauss's Law
P24.63
The magnitude of the field due to the each sheet given by Equation 24.8 is E= (a)
directed perpendicular to the sheet. 2 0
In the region to the left of the pair of sheets, both fields are directed toward the left and the net field is E=
FIG. P24.63
0
to the left .
(b)
In the region between the sheets, the fields due to the individual sheets are oppositely directed and the net field is E= 0 . In the region to the right of the pair of sheets, both are fields are directed toward the right and the net field is E=
(c)
to the right . 0
P24.64
The resultant field within the cavity is the superposition of two fields, one E + due to a uniform sphere of positive charge of radius 2a, and the other E  due to a sphere of negative charge of radius a centered within the cavity. 4 r 3 = 4 r 2 E+ 3 0  4 r13 = 4 r12 E 3 0
F GH
I JK
so
E+ =
r r r= 3 0 3 0 r1   r1 = r1 . 3 0 3 0
F GH
I JK
so
E =
b g
Since r = a + r1 ,
 r  a E = 3 0 E = E+ + E =
a f
r r a a a  + = = 0i + j. 3 0 3 0 3 0 3 0 3 0
FIG. P24.64
Thus, and *P24.65
Ex = 0
Ey =
a 3 0
at all points within the cavity.
Consider the charge distribution to be an unbroken charged spherical shell with uniform charge density and a circular disk with charge per area  . The total field is that due to the whole sphere, 4 R 2 Q = = outward plus the field of the disk  = radially inward. The total 2 2 0 2 0 2 0 4 0 R 4 0 R field is
outward .  = 0 2 0 2 0
Chapter 24
47
P24.66
The electric field throughout the region is directed along x; therefore, E will be perpendicular to dA over the four faces of the surface which are perpendicular to the yz plane, and E will be parallel to dA over the two faces which are parallel to the yz plane. Therefore, E =  Ex
e
x=a
j A + eE
x x=a+c
jA = e3 + 2 a jab + e3 + 2aa + cf jab = 2abca2 a + cf .
2 2
Substituting the given values for a, b, and c, we find E = 0.269 N m 2 C . Q =0 E = 2.38 10 12 C = 2.38 pC P24.67
FIG. P24.66
z
E dA = E 4 r 2 =
e
j
qin 0 qin = Ar 2 4 r 2 dr = 4
0 R
(a)
For r > R , and
z
r
e
j
AR 5 5
E=
AR 5 . 5 0 r 2
(b)
For r < R , and
qin = Ar 2 4 r 2 dr =
0
z
e
j
4 Ar 5 5
E=
Ar 3 . 5 0 Q . The flux through the 0
P24.68
The total flux through a surface enclosing the charge Q is disk is
disk = E dA where the integration covers the area of the disk. We must evaluate this integral 1Q to find how b and R are related. In the figure, take dA to be and set it equal to 4 0 the area of an annular ring of radius s and width ds. The flux through dA is
z
FIG. P24.68
E dA = EdA cos = E 2 sds cos .
The magnitude of the electric field has the same value at all points within the annular ring, E= 1 Q 1 Q = 4 0 r 2 4 0 s 2 + b 2 and cos = b b = 2 r s + b2
b
g
e
j
12
.
Integrate from s = 0 to s = R to get the flux through the entire disk. E , disk = Qb 2 0
R 0
ze
sds s2 + b 2
j
= 32
Qb  s2 + b 2 2 0
LM e N
j OPQ
12
R
=
0
Q b 1 2 0 R2 + b2
The flux through the disk equals This is satisfied if R = 3 b .
Q b provided that 4 0 R2 + b2
LM MM e N
e
j
12
=
1 . 2
OP j PQP
12
48
Gauss's Law
P24.69
z
E dA =
qin 1 = 0 0
z
r 0
4 a r 4 a r 2 E 4 r 2 = rdr = 0 0 0 2 E= a = constant magnitude 2 0 1 dV . We 0 , and is coaxial with the charge E dA =
z
a 4 r 2 dr r
(The direction is radially outward from center for positive a; radially inward for negative a.) P24.70 In this case the charge density is not uniform, and Gauss's law is written as use a gaussian surface which is a cylinder of radius r, length distribution. (a) When r < R , this becomes E 2 r = shell of radius r, length E 2 r = (b)
z
z
b
g
0 0
z FGH
r 0
a
r dV . The element of volume is a cylindrical b
IJ K
b b
g FGH 2 r
2 0
0
I FG a  r IJ so inside the cylinder, E = JK H 2 3b K
, and thickness dr so that dV = 2 r dr .
0r 2r a 2 0 3b
FG H
IJ K
.
When r > R , Gauss's law becomes E 2 r =
g
0 0
R 0
z FGH a  br IJK b2 r drg or outside the cylinder, E = z
E dA = qin 0
0R2 2R a 2 0 r 3b
FG H
IJ K
.
y
P24.71
(a)
Consider a cylindrical shaped gaussian surface perpendicular to the yz plane with one end in the yz plane and the other end containing the point x: Use Gauss's law:
By symmetry, the electric field is zero in the yz plane and is perpendicular to dA over the wall of the gaussian cylinder. Therefore, the only contribution to the integral is over the end cap containing the point x :
gaussian surface
x z x
z
(b)
Ax q E dA = in or EA = 0 0
a f
so that at distance x from the midline of the slab, E = e E e F = = x me me m e 0
x . 0
FIG. P24.71 a =  2 x with =
a=
a f
FG H
IJ K
The acceleration of the electron is of the form
e . m e 0
Thus, the motion is simple harmonic with frequency
f=
1 = 2 2
e . m e 0
Chapter 24
49
P24.72
Consider the gaussian surface described in the solution to problem 71. (a) For x > d , 2 1 E dA = dq 0
dq = dV = Adx = CAx 2 dx
z
z
EA = E=
CA 0
3
d 2 0
z
x 2 dx = or
1 CA 3 0
FG IJ F d I H K GH 8 JK
3
Cd 24 0
E=
Cd 3 d i for x > ; 24 0 2
E=
Cd 3 d i for x <  24 0 2
(b)
For  E=
d d <x< 2 2
z
E dA = E=
1 CA x 2 CAx 3 dq = x dx = 3 0 0 0 0 Cx 3 i for x < 0 3 0 g= Gm r2 r at a distance r. Gm
z
z
Cx 3 i for x > 0 ; 3 0
P24.73
(a)
A point mass m creates a gravitational acceleration The flux of this field through a sphere is
4 r 2 = 4 Gm . r2 Since the r has divided out, we can visualize the field as unbroken field lines. The same flux would go through any other closed surface around the mass. If there are several or no masses inside a closed surface, each creates field to make its own contribution to the net flux according to g dA = 
z
e
j
z
g dA = 4 Gm in .
(b)
Take a spherical gaussian surface of radius r. The field is inward so g dA = g 4 r 2 cos 180 =  g 4 r 2
z
and Then, Or, since
4 4 Gm in = 4 G r 3 . 3 4 4 3 2  g 4 r = 4 G r and g = rG . 3 3 M EGr ME M EGr , g= or g = inward . = 4 3 3 3 RE RE 3 RE
ANSWERS TO EVEN PROBLEMS
P24.2 P24.4 355 kN m 2 C (a) 2.34 kN m 2 C ; (b) +2.34 kN m 2 C ; (c) 0 q 0 ERh P24.14 P24.10 (a) 55.6 nC ; (b) The negative charge has a spherically symmetric distribution. (a) q q ; (b) ; (c) Plane and square 2 0 2 0 both subtend a solid angle of a hemisphere at the charge.
P24.12
P24.6 P24.8
(a) 1.36 MN m 2 C ; (b) 678 kN m 2 C ; (c) No; see the solution.
50
Gauss's Law
P24.16 P24.18 P24.20 P24.22 P24.24
1.77 pC m3 positive Q6q 6 0 28. 2 N m 2 C q 24 0 (a) 0; (b) 365 kN C ; (c) 1.46 MN C; (d) 649 kN C (a) 913 nC ; (b) 0 4.86 GN C away from the wall. It is constant close to the wall 76.4 kN C radially inward 3.50 kN (a) Qr Q ; (c) see the solution ; (b) 0 0 a 3
3
P24.46 P24.48 P24.50 P24.52 P24.54 P24.56 P24.58
(a) 248 nC m 2 ; (b) 496 nC m 2 (a) 2.56 MN C radially inward; (b) 0 (a) q 4 a
2
; (b)
Q+q 4 b 2
see the solution chw 2 2 see the solution (a) 4.00 nC; (b) +9.56 nC ; (c) +4.00 nC and +5.56 nC (a, b) see the solution; (c) (d) 102 pm 1 2 ke e2 ;
P24.26 P24.28
P24.30 P24.32 P24.34 P24.36 P24.38
P24.60
me R3
P24.62 P24.64 P24.66 P24.68 P24.70
(a) 0; (b)
to the right; (c) 0 0
see the solution 0.269 N m 2 C ; 2.38 pC see the solution (a)
713 nC ; (b) 5.70 C (a) 16.2 MN C toward the filament; (b) 8.09 MN C toward the filament; (c) 1.62 MN C toward the filament 1.15 nC m
2
P24.40 P24.42
0r R2 2r 2R ; (b) 0 a a 2 0 3b 2 0 r 3b
FG H
IJ K
FG H
IJ K
(a) 0; (b) 12.4 kN C radially outward; (c) 639 N C radially outward; (d) Nothing would change. (a) 0; (b) 79.9 MN C radially outward; (c) 0; (d) 7.34 MN C radially outward
P24.72
(a) E =
P24.44
Cd 3 d i for x > ; 24 0 2 3 Cd d E= i for x <  ; 24 0 2 3 Cx Cx 3 i for x > 0 ; E =  i for x < 0 (b) E = 3 0 3 0
25
Electric Potential
CHAPTER OUTLINE
25.1 25.2 25.3 Potential Difference and Electric Potential Potential Difference in a Uniform Electric Field Electric Potential and Potential Energy Due to Point Charges Obtaining the Value of the Electric Field from the Electric Potential Electric Potential Due to Continuous Charge Distributions Electric Potential Due to a Charged Conductor The Milliken Oil Drop Experiment Application of Electrostatistics
ANSWERS TO QUESTIONS
Q25.1 When one object B with electric charge is immersed in the electric field of another charge or charges A, the system possesses electric potential energy. The energy can be measured by seeing how much work the field does on the charge B as it moves to a reference location. We choose not to visualize A's effect on B as an actionatadistance, but as the result of a twostep process: Charge A creates electric potential throughout the surrounding space. Then the potential acts on B to inject the system with energy. The potential energy increases. When an outside agent makes it move in the direction of the field, the charge moves to a region of lower electric potential. Then the product of its negative charge with a lower number of volts gives a higher number of joules. Keep in mind that a negative charge feels an electric force in the opposite direction to the field, while the potential is the work done on the charge to move it in a field per unit charge.
25.4
25.5
25.6 25.7 25.8
Q25.2
Q25.3
To move like charges together from an infinite separation, at which the potential energy of the system of two charges is zero, requires work to be done on the system by an outside agent. Hence energy is stored, and potential energy is positive. As charges with opposite signs move together from an infinite separation, energy is released, and the potential energy of the set of charges becomes negative. The charge can be moved along any path parallel to the yz plane, namely perpendicular to the field. The electric field always points in the direction of the greatest change in electric potential. This is V V V , Ey =  and Ez =  . implied by the relationships Ex =  x y z (a) (b) The equipotential surfaces are nesting coaxial cylinders around an infinite line of charge. The equipotential surfaces are nesting concentric spheres around a uniformly charged sphere.
Q25.4 Q25.5
Q25.6
Q25.7
If there were a potential difference between two points on the conductor, the free electrons in the conductor would move until the potential difference disappears.
51
52
Electric Potential
Q25.8
No. The uniformly charged sphere, whether hollow or solid metal, is an equipotential volume. Since there is no electric field, this means that there is no change in electrical potential. The potential at every point inside is the same as the value of the potential at the surface. Infinitely far away from a line of charge, the line will not look like a point. In fact, without any distinguishing features, it is not possible to tell the distance from an infinitely long line of charge. Another way of stating the answer: The potential would diverge to infinity at any finite distance, if it were zero infinitely far away. The smaller sphere will. In the solution to the example referred to, equation 1 states that each will q have the same ratio of charge to radius, . In this case, the charge density is a surface charge r q , so the smallerradius sphere will have the greater charge density. density, 4 r 2 The main factor is the radius of the dome. One often overlooked aspect is also the humidity of the airdrier air has a larger dielectric breakdown strength, resulting in a higher attainable electric potential. If other grounded objects are nearby, the maximum potential might be reduced. The intenseoften oscillatingelectric fields around high voltage lines is large enough to ionize the air surrounding the cables. When the molecules recapture their electrons, they release that energy in the form of light. A sharp point in a charged conductor would imply a large electric field in that region. An electric discharge could most easily take place at that sharp point. Use a conductive box to shield the equipment. Any stray electric field will cause charges on the outer surface of the conductor to rearrange and cancel the stray field inside the volume it encloses. No charge stays on the inner sphere in equilibrium. If there were any, it would create an electric field in the wire to push more charge to the outer sphere. All of the charge is on the outer sphere. Therefore, zero charge is on the inner sphere and 10.0 C is on the outer sphere. The grounding wire can be touched equally well to any point on the sphere. Electrons will drain away into the ground and the sphere will be left positively charged. The ground, wire, and sphere are all conducting. They together form an equipotential volume at zero volts during the contact. However close the grounding wire is to the negative charge, electrons have no difficulty in moving within the metal through the grounding wire to ground. The ground can act as an infinite source or sink of electrons. In this case, it is an electron sink.
Q25.9
Q25.10
Q25.11
Q25.12
Q25.13 Q25.14 Q25.15
Q25.16
SOLUTIONS TO PROBLEMS
Section 25.1 P25.1 Potential Difference and Electric Potential and so Q =  N A e =  6.02 10 23 1.60 10 19 = 9.63 10 4 C W = QV = 9.63 10 4 C 14.0 J C = 1.35 MJ
V = 14.0 V V = W , Q
e
je
j
e
jb
g
Chapter 25
53
P25.2
K = q V q = 6. 41 10 19 C
7.37 10 17 = q 115
a f
P25.3
(a)
Energy of the protonfield system is conserved as the proton moves from high to low potential, which can be defined for this problem as moving from 120 V down to 0 V.
K i + Ui + Emech = K f + U f
0 + qV + 0 =
1 2 mv p + 0 2 C 120 V
e1.60 10
(b)
19
ja
1 fFGH 1 V J C IJK = 1 e1.67 10 2
27
2 kg v p
j
v p = 1.52 10 5 m s The electron will gain speed in moving the other way, from Vi = 0 to V f = 120 V :
K i + Ui + Emech = K f + U f
0+0+0= 0= 1 2 mv e + qV 2
1 2 9.11 10 31 kg v e + 1.60 10 19 C 120 J C 2
e
j e
jb
g
v e = 6.49 10 6 m s P25.4
W = K =  qV
0 1 9.11 10 31 kg 4.20 10 5 m s 2
e
je
j
2
=  1.60 10 19 C V
e
j
From which, V = 0.502 V .
Section 25.2 P25.5 (a)
Potential Difference in a Uniform Electric Field We follow the path from (0, 0) to (20.0 cm, 0) to (20.0 cm, 50.0 cm). U =  (work done) U =  (work from origin to (20.0 cm, 0)) (work from (20.0 cm, 0) to (20.0 cm, 50.0 cm)) Note that the last term is equal to 0 because the force is perpendicular to the displacement.
U =  qEx x =  12.0 10 6 C 250 V m 0.200 m = 6.00 10 4 J
(b) V = U 6.00 10 4 J = = 50.0 J C = 50.0 V q 12.0 10 6 C
b g
e
jb
ga
f
P25.6
E=
V 25.0 10 3 J C = = 1.67 10 6 N C = 1.67 MN C d 1.50 10 2 m
54 P25.7
Electric Potential
U = 
1 1 m v 2  vi2 =  9.11 10 31 kg f 2 2
e
j
e
U = qV :
+6.23 10 18
jLNMe1.40 10 m sj  e3.70 10 = e1.60 10 jV
5 2 19
6
ms
j OQP = 6.23 10
2
18
J
V = 38.9 V. The origin is at highest potential.
P25.8 (a) (b) V = Ed = 5.90 10 3 V m 0.010 0 m = 59.0 V 1 mv 2 = qV : f 2 v f = 4.55 10 6 m s
B
e
jb
g
1 9.11 10 31 v 2 = 1.60 10 19 59.0 f 2
e
j
e
ja f
P25.9
VB  VA =  E ds =  E ds  E ds VB  VA =  E cos 180 VB  VA
a f z dy  aE cos 90.0f = a325fa0.800 f = +260 V
0 .500 0.300
A
z
C
A
z
B
C
z
0. 400
0 . 200
z
dx
FIG. P25.9 *P25.10 Assume the opposite. Then at some point A on some equipotential surface the electric field has a nonzero component Ep in the plane of the surface. Let a test charge start from point A and move some distance on the surface in the direction of the field component. Then V =  E ds is nonzero. The electric potential charges across the surface and it is not an equipotential surface. The contradiction shows that our assumption is false, that Ep = 0 , and that the field is perpendicular to the equipotential surface. P25.11 (a) Arbitrarily choose V = 0 at 0. Then at other points
A B
z
V =  Ex
and
U e = QV = QEx .
Between the endpoints of the motion,
bK + U
s
+ Ue
g = bK + U
i
s
+ Ue
g
f
1 2 2QE 0 + 0 + 0 = 0 + kx max  QEx max so x max = . 2 k (b) At equilibrium,
FIG. P25.11
Fx =  Fs + Fe = 0 or
kx = QE . QE . k
So the equilibrium position is at x = continued on next page
Chapter 25
55
(c)
The block's equation of motion is Let x = x 
QE QE , , or x = x + k k so the equation of motion becomes: d 2 x + QE k QE d 2 x k = k x + + QE = m x . , or 2 k m dt dt 2
Fx =  kx + QE = m dt 2
d2x
.
FG H
IJ K
b
g
FG IJ H K
This is the equation for simple harmonic motion a x =  2 x with The period of the motion is then (d)
=
T=
k . m 2
= 2
m . k
bK + U
s
+ U e i + Emech = K + U s + U e
g
b
g
f
0 + 0 + 0  k mgx max = 0 + x max = P25.12 2 QE  k mg k
b
g
1 2 kx max  QEx max 2
For the entire motion,
y f  yi = v yi t + 0  0 = vi t +
Fy = ma y :
1 ayt 2 2 2mvi  mg  qE =  t m 2 vi g E= q t
1 ayt 2 2 so ay = 
2 vi t
FG H
For the upward flight:
2 2 v yf = v yi + 2 a y y f  yi
0 = vi2 + 2  V = 
ymax 0 y
FG IJ FG IJ FG 1 v tIJ z H K H KH 4 K I 2.00 kg F 2b 20.1 m sg V = GH 4.10 s  9.80 m s JK LMN 1 b20.1 m sga4.10 sfOPQ = 4 5.00 10 C
E dy = +
max m 2 vi m 2 vi g y = g q t q t 0
FG H
d
IJ K
and
E=
2 vi t
IJ by K
2
i
0
m 2 vi  g j. q t 1 vi t 4
FG H
IJ K
max
g
i
and
y max =
6
40.2 kV
P25.13
Arbitrarily take V = 0 at the initial point. Then at distance d downfield, where L is the rod length, V =  Ed and U e =  LEd . (a)
aK + U f = a K + U f
i
f
0+0= v= (b)
1 Lv 2  LEd 2
2 Ed
=
2 40.0 10 6 C m 100 N C 2.00 m
e
b0.100 kg mg
jb
ga
f=
0.400 m s
The same.
56
Electric Potential
P25.14
Arbitrarily take V = 0 at point P. Then (from Equation 25.8) the potential at the original position of the charge is  E s =  EL cos . At the final point a, V =  EL . Suppose the table is frictionless: K +U i = K +U f
a
f a
f
0  qEL cos = v=
1 mv 2  qEL 2
2 qEL 1  cos = m
a
f
2 2.00 10 6 C 300 N C 1.50 m 1  cos 60.0 0.010 0 kg
e
jb
ga
fa
f=
0.300 m s
Section 25.3 P25.15 (a)
Electric Potential and Potential Energy Due to Point Charges The potential at 1.00 cm is V1 = k e 8.99 10 9 N m 2 C 2 1.60 10 19 C q = = 1.44 10 7 V . r 1.00 10 2 m
e
je
j
(b)
8.99 10 9 N m 2 C 2 1.60 10 19 C q The potential at 2.00 cm is V2 = k e = = 0.719 10 7 V . r 2.00 10 2 m Thus, the difference in potential between the two points is V = V2  V1 = 7.19 10 8 V .
e
je
j
(c)
The approach is the same as above except the charge is 1.60 10 19 C . This changes the sign of each answer, with its magnitude remaining the same. That is, the potential at 1.00 cm is 1.44 10 7 V . The potential at 2.00 cm is 0.719 10 7 V , so V = V2  V1 = 7.19 10 8 V .
P25.16
(a) (b)
Since the charges are equal and placed symmetrically, F = 0 . Since F = qE = 0 , E = 0 . q V = 2 k e = 2 8.99 10 9 N m 2 C 2 r
(c)
e
jFGH 2.00 10m C IJK 0.800
6
FIG. P25.16
V = 4.50 10 4 V = 45.0 kV
P25.17 (a) E= V= r= Q 4 0 r 2 Q 4 0 r V 3 000 V = = 6.00 m E 500 V m Q 4 0 6.00 m
(b)
V = 3 000 V = Q=
a
f
2.00 C
e8.99 10
3 000 V
9
V m C
j
a6.00 mf =
Chapter 25
57
P25.18
(a)
Ex =
k e q1 x2
+
ax  2.00f
ke q2
2
=0
becomes 2 qx 2 = q x  2.00 when
Ex = k e
F + q + 2 q I = 0 . GH x ax  2.00f JK
2 2
Dividing by k e , Therefore E = 0
a
f
2
x 2 + 4.00 x  4.00 = 0 .
x= 4.00 16.0 + 16.0 = 4.83 m . 2
(Note that the positive root does not correspond to a physically valid situation.) (b) V= k e q1 k q + e 2 =0 x 2.00  x or V = ke
Again solving for x, For 0 x 2.00 V = 0 and qi ri 2 q q = . x 2x when For x < 0
FG + q  2 q IJ = 0 . H x 2.00  x K 2 qx = qa 2.00  xf .
x = 0.667 m x = 2.00 m .
P25.19
V = k
i
V = 8.99 10 9 7.00 10 6
e
je
1 1 1 jLMN 0.010 0  0.010 0 + 0.038 7 OPQ
V = 1.10 10 7 V = 11.0 MV FIG. P25.19 P25.20 (a) 5.00 10 9 C 3.00 10 9 C 8.99 10 9 V m C qQ U= = = 3.86 10 7 J 4 0 r 0.350 m
e
je
a
je f
j
The minus sign means it takes 3.86 10 7 J to pull the two charges apart from 35 cm to a much larger separation. (b) V= Q1 Q2 + 4 0 r1 4 0 r2
9
e5.00 10 Cje8.99 10 =
0.175 m V = 103 V
9
V m C
j + e3.00 10 Cje8.99 10
9
9
V m C
j
0.175 m
58
Electric Potential
P25.21
U e = q 4V1 + q 4V2 + q 4V3 = q 4
FG 1 IJ FG q H 4 K H r
0
1
+
1
q 2 q3 + r2 r3
U e = 10.0 10 6 C U e = 8.95 J P25.22 (a) V=
e
je
2
8.99 10 9 N m 2 C 2
F 1 1 jGG 0.600 m + 0.150 m + 0.600 m 1+ 0.150 m a f a f H
2
IJ K
2
I JJ K
k e q1 k e q 2 k q + =2 e r1 r2 r
9
V=2
F e8.99 10 N m C je2.00 10 Cj I GG JJ a1.00 mf + a0.500 mf H K
2 2 6 2 2
FG IJ H K
V = 3.22 10 4 V = 32.2 kV (b) P25.23
FIG. P25.22
U = qV = 3.00 10 6 C 3.22 10 4 J C = 9.65 10 2 J
e
je
j
U = U1 + U 2 + U 3 + U 4
U = 0 + U12 + U 13 + U 23 + U 14 + U 24 + U 34 U =0+ U= k eQ 2 k e Q 2 + s s
2
b
k eQ s
FG 4 + 2 IJ = H 2K
FG 1 + 1IJ + k Q FG 1 + H 2 K s H
e 2
g b
g
1 2
+1
IJ K
FIG. P25.23
5.41
k eQ s
2
An alternate way to get the term 4 + diagonal pairs. P25.24
FG H
2
IJ is to recognize that there are 4 side pairs and 2 face 2K
Each charge creates equal potential at the center. The total potential is: V =5
LM k b qg OP = NM R QP
e

5keq . R
P25.25
(a)
Each charge separately creates positive potential everywhere. The total potential produced by the three charges together is then the sum of three positive terms. There is no point located at a finite distance from the charges, at which this total potential is zero.
(b)
V=
2k e q ke q keq + = a a a
Chapter 25
59
P25.26
Consider the two spheres as a system. (a) Conservation of momentum: By conservation of energy, and v1 = 2 m 2 k e q1 q 2 m1 m1 + m 2 0 = m1 v1 i + m 2 v 2  i or v 2 = 0= k e  q1 q 2 d
e j
b g
m1 v1 m2
=
k e  q1 q 2 1 1 2 2 m1 v1 + m 2 v 2 + r1 + r2 2 2
b g
v1 =
FG 1  1 IJ b g H r + r dK 2b0.700 kg ge8.99 10 N m C je 2 10 C je3 10 C j F GH 8 101 b0.100 kg gb0.800 kg g
1 2 9 2 2 6 6
2 2 k e q1 q 2 k e q1 q 2 1 1 m 1 v1 2  = m1 v1 + 2 2 m2 r1 + r2 d
3
m

1 1.00 m
IJ K
= 10.8 m s v2 = (b) m1 v1 0.100 kg 10.8 m s = = 1.55 m s 0.700 kg m2
b
g
If the spheres are metal, electrons will move around on them with negligible energy loss to place the centers of excess charge on the insides of the spheres. Then just before they touch, the effective distance between charges will be less than r1 + r2 and the spheres will really be moving faster than calculated in (a) .
P25.27
Consider the two spheres as a system. (a) Conservation of momentum: or By conservation of energy, and 0 = m 1 v1 i + m 2 v 2  i v2 = 0= m1 v1 . m2
e j b g
k e  q1 q 2 d
b g
=
k e  q1 q 2 1 1 2 2 m1 v1 + m 2 v 2 + r1 + r2 2 2
2 2 k e q1 q 2 k e q1 q 2 1 1 m 1 v1 2  = m1 v1 + . 2 2 m2 r1 + r2 d
v1 =
2 m 2 k e q1 q 2 m 1 m1 + m 2
b
v2 = (b)
FG m IJ v Hm K
1 2
1
=
IJ K 2m k q q F 1 G m bm + m g H r + r

1 2
FG 1 gHr +r
2 1
1 d
1 e 1 2

2
1
2
1 d
IJ K
If the spheres are metal, electrons will move around on them with negligible energy loss to place the centers of excess charge on the insides of the spheres. Then just before they touch, the effective distance between charges will be less than r1 + r2 and the spheres will really be moving faster than calculated in (a) .
60
Electric Potential
*P25.28
(a)
In an empty universe, the 20nC charge can be placed at its location with no energy investment. At a distance of 4 cm, it creates a potential V1 = 8.99 10 9 N m 2 C 2 20 10 9 C k e q1 = = 4.50 kV . 0.04 m r
e
je
j
To place the 10nC charge there we must put in energy U12 = q 2 V1 = 10 10 9 C 4.5 10 3 V = 4.50 10 5 J . Next, to bring up the 20nC charge requires energy U 23 + U 13 = q3 V2 + q3 V1 = q 3 V2 + V1
e
je
j
b
g
= 20 10 9 C 8.99 10 9 N m 2 C 2 = 4.50 10 5 J  4.50 10 5 J The total energy of the three charges is
e
10 10 jFGH 100.04 m C + 200.08 m C IJK
9 9
U12 + U 23 + U13 = 4.50 10 5 J .
(b) The three fixed charges create this potential at the location where the fourth is released: V = V1 + V2 + V3 = 8.99 10 9 N m 2 C 2 V = 3.00 10 3 V Energy of the system of four charged objects is conserved as the fourth charge flies away:
e
F jGH
20 10 9 0.04 2 + 0.03 2
+
10 10 9 20 10 9  0.03 0.05
I Cm JK
FG 1 mv + qV IJ = FG 1 mv + qV IJ H2 K H2 K 1 0 + e 40 10 C je3.00 10 V j = e 2.00 10 2 2e1.20 10 Jj = 3.46 10 m s v=
2 2 i f 9 3 4
13
kg v 2 + 0
j
2 10 13 kg
4
*P25.29
The original electrical potential energy is U e = qV = q ke q . d
In the final configuration we have mechanical equilibrium. The spring and electrostatic forces on k q2 k q each charge are  k 2d + q e 2 = 0 . Then k = e 3 . In the final configuration the total potential 18d 3d energy is keq 4 keq2 = . The missing energy must have become internal 3 3d 9 d k q 2 4k q 2 energy, as the system is isolated: e = e + Eint d 9d 1 2 kx 2
e 2
a f a f 1 k q + qV = a 2d f 2 18d
2
+q
Eint =
5 keq2 . 9 d
Chapter 25
P25.30
(a)
V x =
af
k +Q k e Q1 k e Q 2 + = e + r1 r2 x2 + a2 kQ = e 2 2 a x +a
2
b g
k e +Q x2
V x = V x
e
af
2 k eQ
af 2 b k Q ag = b x ag
(b) V y = V y =
F 2 GG H b x ag
2
I JJ +1K
b g + a  af
61
2
+1 FIG. P25.30(a)
bg
b g k aQ FGH y a1 1  y a1+ 1 IJK F 1  1 I V b yg = G bk Q ag H y a  1 y a + 1 JK
e e
k e Q 1 k e Q 2 k e +Q k e Q + = + r1 r2 ya y+a
b g
b g
FIG. P25.30(b) P25.31 V= 8.99 10 9 N m 2 C 2 8.00 10 9 C kQ k eQ 72.0 V m = so r = e = . V V V r
e
je
j
For V = 100 V , 50.0 V, and 25.0 V, r = 0.720 m, 1.44 m, and 2.88 m . The radii are inversely proportional to the potential. P25.32 Using conservation of energy for the alpha particlenucleus system, we have But and Thus, Also so or 2 k e q qgold
2 m v
K f + U f = K i + Ui .
Ui = k e q qgold ri
ri . Ui = 0 .
K f = 0 ( v f = 0 at turning point), U f = Ki
k e q qgold rmin = = 1 2 m v 2
rmin =
2 8.99 10 9 N m 2 C 2 2 79 1.60 10 19 C
e
e6.64 10
27
ja fa fe kg je 2.00 10
j
2
7
ms
j
2
= 2.74 10 14 m = 27.4 fm .
62
Electric Potential
P25.33
Using conservation of energy k e eQ k e qQ 1 = + mv 2 we have: 2 r1 r2 which gives: v= 2 k e eQ 1 1  m r1 r2
FG H
IJ K
or Thus, P25.34 U= k e qi q j rij
v=
a2fe8.99 10
9
N m 2 C 2 1.60 10 19 C 10 9 C 9.11 10
31
je
je
kg
jF 1  1 I . GH 0.030 0 m 0.020 0 m JK
v = 7.26 10 6 m s . , summed over all pairs of i , j where i j .
b g L qb2 qg + b2 qgb3qg + b2 qgb3qg + qb2 qg + qb3qg + 2 qb2 qg OP U=k M a b a P a +b a +b Q NM b L 2  6 + 6 + 2 + 3  4 OP U=k q M N 0.400 0.200 0.400 0.200 0.447 0.447 Q L 4  4  1 OP = 3.96 J U = e8.99 10 je6.00 10 j M N 0.400 0.200 0.447 Q
e 2 2 2 2 e 2 9 6 2
FIG. P25.34
P25.35
Each charge moves off on its diagonal line. All charges have equal speeds. K +U i = K +U f
a
f
a
0+
FG 2 + 1 IJ k q = 2mv H 2K L F 1 IJ k q v = G1 + H 8 K mL
e 2 e 2
4k e q 2k q + e L 2L
2
2
f F 1 I 4k q = 4G mv J + H 2 K 2L
2 e 2
2
+
2k e q 2 2 2L
P25.36
A cube has 12 edges and 6 faces. Consequently, there are 12 edge pairs separated by s, 2 6 = 12 face diagonal pairs separated by 2s and 4 interior diagonal pairs separated 3s . U= keq2 k q2 12 4 + = 22.8 e 12 + s s 2 3 Obtaining the Value of the Electric Field from the Electric Potential
LM N
OP Q
Section 25.4 P25.37
V = a + bx = 10.0 V + 7.00 V m x
(a) At x = 0 , At x = 3.00 m , At x = 6.00 m , (b) E=
b
g
V = 10.0 V V = 11.0 V V = 32.0 V
dV =  b =  7.00 V m = 7.00 N C in the + x direction dx
b
g
Chapter 25
63
P25.38
(a)
For r < R
V=
k eQ R dV = 0 Er =  dr k eQ r kQ kQ dV =   e2 = e2 Er =  dr r r
(b)
For r R
V=
FG H
IJ K
P25.39
V = 5 x  3 x 2 y + 2 yz 2
Evaluate E at 1, 0 ,  2 Ex = 
b
g
V = 5 + 6 xy = 5 + 6 1 0 = 5 x V 2 2 = +3 x 2  2 z 2 = 3 1  2 2 = 5 Ey =  y Ez =  V = z
2 2 E = Ex + Ey
a fa f af a f 4yz = 4a0fa 2f = 0 + E = a 5 f + a 5 f + 0 =
2 z 2 2 2
7.07 N C
P25.40
(a)
E A > EB since E =
V s
(b) (c)
EB = 
62 V V = = 200 N C down 2 cm s
a f
The figure is shown to the right, with sample field lines sketched in. FIG. P25.40 V k eQ = ln y y
P25.41
Ey = 
Ey =
k eQ 1 y
LM MN
LM MNM
F GG H
+
2
+ y2
y
y2
2
+ y2 +
2
+ y2
OP PQ =
I OP JJ P K QP
k eQ y
2
+ y2
Section 25.5 P25.42
Electric Potential Due to Continuous Charge Distributions k eQ
V = V2 R  V0 =
R2 + 2R
a f
2

k eQ k eQ = R R
FG 1  1IJ = H 5 K
0.553
k eQ R
64
Electric Potential
P25.43
(a)
=
LM OP = C FG 1 IJ = N x Q m H mK
C m2
(b)
V = ke
z
z
L dq dx xdx L = ke = k e = k e L  d ln 1 + r r d+x d 0
z
z
LM N
FG H
IJ OP KQ
FIG. P25.43
P25.44
V=
z
k e dq = ke r L x. 2
xdx
b2 + L 2  x
b
g
2
Let z =
Then x = V = k e
zb
L  z , and dx =  dz 2 L 2  z  dz b +z
2 2 e
ga f =  k L
2
z
2 2
dz b +z
2 2 2 L
+ k e
z
zdz b +z
2 2 2
=
k eL ln z + z 2 + b 2 + k e z 2 + b 2 2
FH
IK
k L V =  e ln 2
e
LMF L I F L I OP GH 2  xJK + GH 2  xJK + b P + k FGH L  xIJK + b 2 MN Q L O L FL I k L M L 2  L + b L 2 g + b P F LI V= ln M PP + k MMN GH 2  LJK + b  GH 2 JK 2 MN L 2 + bL 2g + b Q L O k L M b + eL 4j  L 2 P V=  ln M 2 MN b + eL 4j + L 2 PPQ
L e 2 0 0 2 2 2 2 e 2 2 2 2 2 e
2
+ b2
OP PQ
P25.45
V = dV =
z
1 4 0
z
dq r .50 jFGH 70.140 10 C IJK = m
6
All bits of charge are at the same distance from O. So V = 1 Q = 8.99 10 9 N m 2 C 2 4 0 R k e dq r 2 + x2 where dq = dA = 2 rdr rdr r + x2
2
FG IJ e H K
1.51 MV .
P25.46
dV =
V = 2k e
z
b a
= 2 k e
LM N
x2 + b2  x2 + a2
OP Q
FIG. P25.46
Chapter 25
65
P25.47
V = ke
3R R dq dx ds dx = ke + ke + ke x r R x R all charge semicircle 3 R
z
z
z
z
V =  k e ln  x V = k e ln
a f
R 3 R
+
3R + k e + k e ln 3 = k e + 2 ln 3 R
ke 3R R + k e ln x R R
a
f
Section 25.6 P25.48
Electric Potential Due to a Charged Conductor ke q r
Substituting given values into V =
7.50 10 3 V = Substituting q = 2.50 10 7 C , N=
e8.99 10
9
N m2 C 2 q
j
0.300 m
.
2.50 10 7 C = 1.56 10 12 electrons . 1.60 10 19 C e 
P25.49
(a)
E= 0 ;
8.99 10 9 26.0 10 6 ke q = = 1.67 MV V= 0.140 R
e
je
j
(b)
e8.99 10 je26.0 10 j = 5.84 MN C r a0.200f k q e8.99 10 je 26.0 10 j = = 1.17 MV V=
E= keq
2 9 6
=
2
away
9
6
e
R
0.200
(c)
E= V=
keq R2
e8.99 10 je26.0 10 j = = a0.140f
9 6 2
11.9 MN C away
ke q = 1.67 MV R
66
Electric Potential
*P25.50
(a)
Both spheres must be at the same potential according to where also Then
k e q1 k e q 2 = r1 r2
q1 + q 2 = 1.20 10 6 C .
q1 = q 2 r1 r2
q 2 r1 + q 2 = 1.20 10 6 C r2 q2 = 1.20 10 6 C = 0.300 10 6 C on the smaller sphere 1 + 6 cm 2 cm 8.99 10 9 N m 2 C 2 0.900 10 6 C k e q1 = = 1.35 10 5 V r1 6 10 2 m
q1 = 1.20 10 6 C  0.300 10 6 C = 0.900 10 6 C V= (b)
e
je
j
Outside the larger sphere, E1 = k e q1 r12 r= V1 1.35 10 5 V r= r = 2. 25 10 6 V m away . 0.06 m r1
Outside the smaller sphere, E2 = 1.35 10 5 V r = 6.74 10 6 V m away . 0.02 m
The smaller sphere carries less charge but creates a much stronger electric field than the larger sphere.
Section 25.7 Section 25.8 P25.51 (a)
The Milliken Oil Drop Experiment Application of Electrostatistics Emax = 3.00 10 6 V m =
6
k eQ
Vmax = Emax r = 3.00 10 0.150 = 450 kV
(b) k eQmax r2 = Emax
a
e
r
2
=
f
k eQ 1 1 = Vmax r r r
FG IJ HK U V W
FG IJ HK
6 Emax r 2 3.00 10 0.150 = ke 8.99 10 9
Ror k Q S r T
max
= Vmax
Q max =
a
f
2
= 7.51 C
P25.52
V=
ke q k q V and E = e2 . Since E = , r r r r= 6.00 10 5 V V = = 0.200 m and E 3.00 10 6 V m Vr = 13.3 C ke
(b)
(a)
q=
Chapter 25
67
Additional Problems q q U = qV = k e 1 2 = 8.99 10 9 r12
P25.53 P25.54
e
a38fa54f 1.60 10 j a5.50 +e6.20f 10 j
15
19 2
= 4.04 10 11 J = 253 MeV
(a)
To make a spark 5 mm long in dry air between flat metal plates requires potential difference V = Ed = 3 10 6 V m 5 10 3 m = 1.5 10 4 V ~ 10 4 V .
e
je
j
(b)
The area of your skin is perhaps 1.5 m 2 , so model your body as a sphere with this surface area. Its radius is given by 1.5 m 2 = 4 r 2 , r = 0.35 m . We require that you are at the potential found in part (a): V= ke q r q= 1.5 10 4 V 0.35 m Vr J = k e 8.99 10 9 N m 2 C 2 V C
a
f FG H
IJ FG N m IJ KH J K
q = 5.8 10 7 C ~ 10 6 C .
19 9 k e q1 q 2  8.99 10 1.60 10 = U= r 0.052 9 10 9
P25.55
(a)
e e
je je
j j
2
= 4.35 10 18 J = 27.2 eV
2
(b)
19 9 k e q1 q 2  8.99 10 1.60 10 U= = r 2 2 0.052 9 10 9
e
j
= 6.80 eV
(c) P25.56
U=
k e q1 q 2  k e e 2 = = 0 r
From Example 25.5, the potential created by the ring at the electron's starting point is Vi = k eQ x i2 +a
2
=
k e 2a xi2 +a
b
g
2
while at the center, it is V f = 2 k e . From conservation of energy, 0 +  eVi = v2 f
v2 f
i I ek F 2e = dV  V i = 4m GG1  x a+ a JJ m H K 4 e1.60 10 je8.99 10 je1.00 10 j F 0.200 GG1  = 9.11 10 H a0.100f + a0.200f
e f i e e 2 i 2 19 9 7 31 2
b g
1 m e v 2 +  eV f f 2
d
2
I JJ K
v f = 1.45 10 7 m s
68
Electric Potential
*P25.57
The plates create uniform electric field to the right in the picture, with magnitude
0 . d d Assume the ball swings a small distance x to the right. It moves to a place where the voltage created 2V by the plates is lower by  Ex =  0 x . Its ground connection maintains it at V = 0 by allowing d 2V x k q 2V xR . Then the ball charge q to flow from ground onto the ball, where  0 + e = 0 q= 0 d R ked
V0  V0
b g = 2V
feels electric force F = qE =
4V02 xR ked 2
to the right. For equilibrium this must be balanced by the T sin = 4V02 xR ked 2
horizontal component of string tension according to T cos = mg tan = 4V02 xR k e d mg
2
=
k d 2 mg x for small x. Then V0 = e L 4RL
F GH
I JK
12
.
If V0 is less than this value, the only equilibrium position of the ball is hanging straight down. If V0 exceeds this value the ball will swing over to one plate or the other. P25.58 (a) Take the origin at the point where we will find the potential. One ring, of width dx, has Qdx charge and, according to Example 25.5, creates potential h k eQdx dV = . h x2 + R2 The whole stack of rings creates potential V=
all charge
z
dV =
d+h d
z
k eQdx h x2 + R2
=
k eQ ln x + x 2 + R 2 h
FH
IK
d+h d
=
d + h + d + h + R2 k eQ ln h d + d2 + R2
F GG H
a f
2
I JJ K
.
(b)
A disk of thickness dx has charge Example 25.6, it creates potential Qdx R2h Integrating, dV = 2 k e V=
d+h d
Qdx Qdx and chargeperarea . According to h R2h
FH
x2 + R2  x .
IK
V=
z R h FH x + R dx  xdxIK = 2Rk Q LMN 12 x x + R h L kQM ad + hf ad + hf + R  d d + R  2dh  h R hM NM
2 k eQ
2 2 2 e 2 2 e 2 2 2 2 2
2
+
R2 x2 ln x + x 2 + R 2  2 2
2 2
2
+R
FH IK OP Q F d + h + ad + hf + R I OP lnG GH d + d + R JJK PQP
d+h d 2 2 2
P25.59
W = Vdq
0
Q
z
where V =
ke q . R k eQ 2 . 2R
Therefore, W =
Chapter 25
69
P25.60
The positive plate by itself creates a field E =
36.0 10 9 C m 2 = 2.03 kN C away = 2 0 2 8.85 10 12 C 2 N m 2
e
j
from the + plate. The negative plate by itself creates the same size field and between the plates it is in the same direction. Together the plates create a uniform field 4.07 kN C in the space between. (a) Take V = 0 at the negative plate. The potential at the positive plate is then V 0=
12.0 cm 0
zb
4.07 kN C dx .
g
The potential difference between the plates is V = 4.07 10 3 N C 0.120 m = 488 V . (b)
e
ja
f
FG 1 mv + qV IJ = FG 1 mv + qV IJ H2 K H2 K 1 qV = e1.60 10 C ja 488 V f = mv 2
2 2 i f 19
2 f
= 7.81 10 17 J
(c) (d)
v f = 306 km s v 2 = vi2 + 2 a x f  xi f
d
i
e3.06 10
(e) (f)
5
ms
j
2
= 0 + 2 a 0.120 m
a
f j
a = 3.90 10 11 m s 2
F = ma = e1.67 10 27
E=
kg 3.90 10 11 m s 2 = 6.51 10 16 N
je
F 6.51 10 16 N = = 4.07 kN C q 1.60 10 19 C
B
P25.61
(a)
VB  VA =  E ds and the field at distance r from a uniformly
A
z
charged rod (where r > radius of charged rod) is E=
2 0 r
=
2ke . r
In this case, the field between the central wire and the coaxial cylinder is directed perpendicular to the line of charge so that VB  VA = 
rb ra
z
r 2ke dr = 2 k e ln a , r rb
FG IJ H K
FIG. P25.61
or
V = 2 k e ln
FG r IJ Hr K
a b
.
continued on next page
70
Electric Potential
(b)
From part (a), when the outer cylinder is considered to be at zero potential, the potential at a distance r from the axis is V = 2 k e ln
FG r IJ . HrK
a
The field at r is given by E=
r V = 2 k e ra r
FG H
IJ FG  r IJ = 2k . KH r K r
a 2 e
But, from part (a), 2k e = Therefore,
b V F 1 I E= GJ lnbr r g H r K
a b
V . ln ra rb
g
.
P25.62
(a)
From Problem 61, E= V 1 . ln ra rb r
b
g
We require just outside the central wire 5.50 10 6 V m =
1
50.0 10 3 V ln 0.850 m rb
b
FG 1 IJ gHr K
b
or
e110 m jr lnFGH 0.850 m IJK = 1 . r
b b 1
We solve by homing in on the required value
e110 m
a f jr lnFGH 0.850 m IJK r
rb m
b b
0.0100 4.89
0.00100 0.740
0.00150 1.05
0.00145 1.017
0.00143 1.005
0.00142 0.999
Thus, to three significant figures,
rb = 1.42 mm .
(b) At ra , E=
r2 r1
50.0 kV ln 0.850 m 0.001 42 m
b
FG 1 IJ = g H 0.850 m K
9.20 kV m .
P25.63
V2  V1 =  E dr = 
z
r2 r1
z
2 0 r
dr
V2  V1 =
r  ln 2 r1 2 0
FG IJ H K
Chapter 25
71 2 +
*P25.64
Take the illustration presented with the problem as an initial picture. No external horizontal forces act on the set of four balls, so its center of mass stays fixed at the location of the center of the square. As the charged balls 1 and 2 swing out and away from each other, balls 3 and 4 move up with equal ycomponents of velocity. The maximumkineticenergy point is illustrated. System energy is conserved: keq2 keq 2 1 1 1 1 = + mv 2 + mv 2 + mv 2 + mv 2 3a 2 2 2 2 a 2k e q 2 = 2mv 2 3a v= keq2 3 am V x , y, z = r1 =
v 1 + v 3 CM
v
4
v
FIG. P25.64
P25.65
For the given charge distribution, where The surface on which is given by This gives: which may be written in the form:
b
g k rbqg + k br2 qg
e e 1 2 2 2 2 2 2 2 2
ax + Rf + y + z and r = x + y + z . V b x , y , zg = 0 F1 2I k qG  J = 0 , or 2r = r . Hr r K 4a x + R f + 4y + 4z = x + y + z F8 I F4 I x + y + z + G RJ x + a0fy + a0fz + G R J = 0 . H3 K H3 K
e 1 2 1 2 2 2 2 2 2 2 2 2 2 2
[1]
The general equation for a sphere of radius a centered at x 0 , y 0 , z0 is:
b
g
or
bx  x g + by  y g + bz  z g  a = 0 x + y + z + b 2 x gx + b 2 y gy + b 2 z gz + e x
0 2 2 0 0 2 2 2 2 2 0 0 0
2 0
2 2 + y 0 + z0  a 2 = 0 .
j
[2]
Comparing equations [1] and [2], it is seen that the equipotential surface for which V = 0 is indeed a sphere and that: 2 x 0 = Thus, x 0 =  4 8 2 2 2 R ; 2 y 0 = 0 ; 2 z 0 = 0 ; x 0 + y 0 + z 0  a 2 = R 2 . 3 3
4 16 4 2 4 2 R , y 0 = z 0 = 0 , and a 2 =  R = R . 3 9 3 9
FG H
IJ K
The equipotential surface is therefore a sphere centered at
FG  4 R, 0 , 0IJ H 3 K
, having a radius
2 R . 3
72
Electric Potential
P25.66
(a)
From Gauss's law, qA
2 8
E A = 0 (no charge within)
EB = k e EC = k e
1.00 10 e j e r j = FGH 89.9 IJK V m r r bq + q g = e8.99 10 j e5.00 10 j = FG  45.0 IJ V m H r K r r = 8.99 10 9
B 2 2 9 A 9 2 2
(b)
VC = k e
bq
A
+ qB r
g = e8.99 10 j e5.00 10 j = FG  45.0 IJ V H r K r
9 9
At r2 , V = 
45.0 = 150 V 0.300 89.9 1 1 = dr = 150 + 89.9  2 r 0.300 r2 r
Inside r2 , VB = 150 V +
z
r
FG H
IJ FG 450 + 89.9 IJ V K H r K
At r1 , V = 450 + P25.67
89.9 = +150 V so VA = +150 V . 0.150
From Example 25.5, the potential at the center of the ring is kQ Vi = e and the potential at an infinite distance from the ring is R V f = 0 . Thus, the initial and final potential energies of the point chargering system are: U i = QVi = and k eQ 2 R FIG. P25.67
U f = QV f = 0 .
From conservation of energy,
K f + U f = K i + Ui
or k Q2 1 Mv 2 + 0 = 0 + e f R 2 vf = 2 k eQ 2 MR .
giving
P25.68
V = ke
a+L a
z
dx
x2 + b2
= k ln L x + e x MN
e
2
+b
2
j OPQ
a +L
=
a
L a + L + aa + Lf k ln M MMN a + a + b
e 2
2 2
+ b2
OP PPQ
Chapter 25
73
*P25.69
(a)
V=
keq ke q keq  = r2  r1 r1 r2 r1 r2
b
g
r2  r1 2 a cos . V ke q k p cos 2 a cos e 2 . r1 r2 r
From the figure, for r >> a , Then 2 k e p cos V = r r3
(b)
Er = 
1 In spherical coordinates, the component of the gradient is . r Therefore, For r >> a and E =  k p sin 1 V . = e 3 r r
e 3
FG IJ H K
FIG. P25.69
FG IJ H K
and
a f 2rk p E a90f = 0 , E a0f = 0 k p E a90f = . r
E r 0 =
r
e 3
These results are reasonable for r >> a . Their directions are as shown in Figure 25.13 (c). However, for r 0 , E 0 . This is unreasonable, since r is not much greater than a if it is 0. (c) V= k e py
2
af
ex
+ y2
j
3 2
and
Ex = 
3 k e pxy V = 52 2 x x + y2
e
j
2 2 V k e p 2 y  x = Ey =  5 2 y x2 + y2
e
j
e
j
74
Electric Potential
P25.70
Inside the sphere, Ex = Ey = Ez = 0 . Outside,
j IK L F 3I So E =  M0 + 0 + E a z G  J e x + y + z j a2xfOPQ = 3E a xzex H 2K N V I =  F V  E z + E a ze x + y + z j E = K y y H F 3I E =  E a zG  J e x + y + z j 2 y = 3E a yze x + y + z j H 2K V F 3I E = = E  E a zG  J e x + y + z j a 2 zf  E a e x + y + z j H 2K z E = E + E a e 2 z  x  y je x + y + z j
Ex =  V = V0  E0 z + E0 a 3 z x 2 + y 2 + z 2 x x
F H
e
3 2
x
0
3
2
2
2 5 2
0
3
2
+ y2 + z2
j
5 2
y
0
0
0
3
2
2
2 3 2
y
0
3
2
2
2 5 2
0
3
2
2
2 5 2
z
0
0
3
2
2
2 5 2
0
3
2
2
2 3 2
z
0
0
3
2
2
2
2
2
2 5 2
P25.71
For an element of area which is a ring of radius r and width dr, dV = dq = dA = Cr 2 rdr and V = C 2 k e
k e dq r 2 + x2
.
b
g
b
gz
0
R
r 2 dr r 2 + x2
= C k e R R 2 + x 2 + x 2 ln
L b gMM N
F GH R +
x R2 + x2
I OP JK P Q
.
P25.72
dU = Vdq where the potential V =
ke q . r
The element of charge in a shell is dq = (volume element) or dq = 4 r 2 dr and the charge q in a sphere of radius r is q = 4 r 2 dr =
0
e
j
z
r
F 4 r I . GH 3 JK
3 3 e 2 2 e 2 4
Substituting this into the expression for dU, we have
FG k q IJ dq = k FG 4 r IJ FG 1 IJ e4 r dr j = k FG 16 IJ r dr HrK H 3 K H 3 KH r K F 16 I r dr = k F 16 I R U = z dU = k G GH 15 JK H 3 JK z
dU =
e 2 e 2 R 0 4 2 e 2 5
3 k eQ 2 4 But the total charge, Q = R 3 . Therefore, U = . 5 R 3
Chapter 25
75
*P25.73
(a)
The whole charge on the cube is 3 q = 100 10 6 C m3 0.1 m = 10 7 C . Divide up the cube into
e
ja
f
64 or more elements. The little cube labeled a creates at P ke q . The others in the potential 2 64 6.25 + 1.25 2 + 1.25 2 10 2 m horizontal row behind it contribute 64 10 2
e
keq
F G mj H
1 8.75 2 + 3.125
+
1 11.25 2 + 3.125
+
1 13.75 2
I. J + 3.125 K
1 2
d c
b a P
The little cubes in the rows containing b and c add 64 10
e
2k e q
2
Le6.25 + 1.25 + 3.75 j + e8.75 + 15.625j N mj M OP + e11.25 + 15.625j + e13.75 + 15.625 j Q
2 2 2 1 2 2 2 1 2 2 1 2 2
1.25 cm FIG. P25.73
and the bits in row d make potential at P 64 10 2
e
keq
LMe6.25 mj N
+ 28.125
j
1 2
+ ... + 13.75 2 + 28.125
e
j OQP .
1 2
The whole potential at P is
8.987 6 10 9 Nm 2 10 7 C C 2 64 10 2 m
e
j
b1.580 190g4 =
8 876 V . If we use
more subdivisions of the large cube, we get the same answer to four digits. (b) A sphere centered at the same point would create potential k e q 8.987 6 10 9 Nm 2 10 7 C = = 8 988 V , larger by 112 V . r C2 10 1 m
ANSWERS TO EVEN PROBLEMS
P25.2 P25.4 P25.6 P25.8 P25.10 P25.12 P25.14 P25.16 P25.18 P25.20
6.41 10 19 C 0.502 V
1.67 MN C (a) 59.0 V ; (b) 4.55 Mm s see the solution 40.2 kV 0.300 m s (a) 0; (b) 0; (c) 45.0 kV (a) 4.83 m ; (b) 0.667 m and 2.00 m (a) 386 nJ ; (b) 103 V
P25.22 P25.24 P25.26 P25.28 P25.30 P25.32 P25.34 P25.36
(a) 32.2 kV ; (b) 96.5 mJ  5k e q R
(a) 10.8 m s and 1.55 m s ; (b) greater (a) 45.0 J ; (b) 34.6 km s see the solution 27.4 fm 3.96 J 22.8 keq 2 s
76
Electric Potential
P25.38 P25.40
(a) 0; (b)
k eQ r
2
radially outward
P25.60
(a) 488 V ; (b) 7.81 10 17 J ; (c) 306 km s ; (d) 390 Gm s 2 toward the negative plate; (e) 6.51 10 16 N toward the negative plate; (f) 4.07 kN C toward the negative plate
(a) larger at A; (b) 200 N C down; (c) see the solution 0.553 k eQ R b
2
P25.42
P25.62 + e L 4j  L 2 O PP + e L 4j + L 2 P Q
2 2
(a) 1.42 mm ; (b) 9. 20 kV m
P25.44
L k L M  ln M 2 MN
e
P25.64
Fk q I GH 3am JK
e 2
12
b2
P25.46 P25.48 P25.50
2 k e
LM N
x2 + b 2  x2 + a2
OP Q
P25.66
1.56 10 12 electrons
(a) 135 kV ; (b) 2.25 MV m away from the large sphere and 6.74 MV m away from the small sphere
P25.52 P25.54 P25.56
(a) 13.3 C ; (b) 0.200 m (a) ~ 10 V ; (b) ~ 10 14.5 Mm s d + h + d + h + R2 k eQ ln h d + d 2 + R2
2 2 2 2
P25.68
4
6
C
P25.70
FG 89.9 IJ V m radially Hr K F 45.0 IJ V m radially outward; E = G  H r K outward; 89.9 I F (b) V = 150 V ; V = G 450 + J V; H r K F 45.0 IJ V V = G H r K L a + L + aa + Lf + b OP k ln M MNM a + a + b PQP
(a) E A = 0 ; E B =
C 2 2 A B C 2 2 e 2 2
P25.58
F a f IJ ; GG JK H LMad + hf ad + hf + R  d d + R kQ M F d + h + a d + hf + R (b) R h M2dh  h + R lnG MM GH d + d + R N
(a)
2 e 2 2 2 2 2 2
2
OP I PP JJ P K PQ
e j ; E = 3E a yze x + y + z j ; E a e2z  x  y j E =E + outside and ex + y + z j
Ex = 3E0 a 3 xz x 2 + y 2 + z 2
y 0 3 2 2 5 2 2 5 2 2 0 3 2 2 z 0 2 2 2 5 2
E = 0 inside P25.72 3 k eQ 2 5 R
26
Capacitance and Dielectrics
CHAPTER OUTLINE
26.1 26.2 26.3 26.4 26.5 26.6 26.7 Definition of Capacitance Calculating Capacitance Combinations of Capacitors Energy Stored in a Charged Capacitor Capacitors with Dielectrics Electric Dipole in an Electric Field An Atomic Description of Dielectrics
ANSWERS TO QUESTIONS
Q26.1 Nothing happens to the charge if the wires are disconnected. If the wires are connected to each other, charges in the single conductor which now exists move between the wires and the plates until the entire conductor is at a single potential and the capacitor is discharged. 336 km. The plate area would need to be 1 m2 . 0
Q26.2 Q26.3
The parallelconnected capacitors store more energy, since they have higher equivalent capacitance.
Q26.4
Seventeen combinations: Individual Parallel SeriesParallel C1 , C 2 , C 3 C 1 + C 2 + C 3 , C 1 + C 2 , C1 + C 3 , C 2 + C 3
Series Q26.5
FG 1 + 1 IJ + C , FG 1 + 1 IJ + C , FG 1 + 1 IJ + C HC C K HC C K HC C K FG 1 + 1 IJ , FG 1 + 1 IJ , FG 1 + 1 IJ HC +C C K HC +C C K HC +C C K FG 1 + 1 + 1 IJ , FG 1 + 1 IJ , FG 1 + 1 IJ , FG 1 + 1 IJ HC C C K HC C K HC C K HC C K
1 1 1 1 2 3 1 3 2 2 3 1 1 1 1 1 2 3 1 3 2 2 3 1 1 1 1 1 2 3 1 2 2 3 1 3
1
This arrangement would decrease the potential difference between the plates of any individual capacitor by a factor of 2, thus decreasing the possibility of dielectric breakdown. Depending on the application, this could be the difference between the life or death of some other (most likely more expensive) electrical component connected to the capacitors. Nonot just using rules about capacitors in series or in parallel. See Problem 72 for an example. If connections can be made to a combination of capacitors at more than two points, the combination may be irreducible. 77
Q26.6
78
Capacitance and Dielectrics
Q26.7
A capacitor stores energy in the electric field between the plates. This is most easily seen when using a "dissectable" capacitor. If the capacitor is charged, carefully pull it apart into its component pieces. One will find that very little residual charge remains on each plate. When reassembled, the capacitor is suddenly "recharged"by inductiondue to the electric field set up and "stored" in the dielectric. This proves to be an instructive classroom demonstration, especially when you ask a student to reconstruct the capacitor without supplying him/her with any rubber gloves or other insulating material. (Of course, this is after they sign a liability waiver). The work you do to pull the plates apart becomes additional electric potential energy stored in the capacitor. The charge is constant and the capacitance decreases but the potential difference increases 1 to drive up the potential energy QV . The electric field between the plates is constant in strength 2 but fills more volume as you pull the plates apart. A capacitor stores energy in the electric field inside the dielectric. Once the external voltage source is removedprovided that there is no external resistance through which the capacitor can dischargethe capacitor can hold onto this energy for a very long time. To make the capacitor safe to handle, you can discharge the capacitor through a conductor, such as a screwdriver, provided that you only touch the insulating handle. If the capacitor is a large one, it is best to use an external resistor to discharge the capacitor more slowly to prevent damage to the dielectric, or welding of the screwdriver to the terminals of the capacitor. The work done, W = QV , is the work done by an external agent, like a battery, to move a charge through a potential difference, V . To determine the energy in a charged capacitor, we must add the work done to move bits of charge from one plate to the other. Initially, there is no potential difference between the plates of an uncharged capacitor. As more charge is transferred from one plate to the other, the potential difference increases as shown in Figure 26.12, meaning that more work is needed to transfer each additional bit of charge. The total work is the area under the curve of 1 Figure 26.12, and thus W = QV . 2 Energy is proportional to voltage squared. It gets four times larger. Let C = the capacitance of an individual capacitor, and C s represent the equivalent capacitance of the group in series. While being charged in parallel, each capacitor receives charge Q = CVcharge = 500 10 4 F 800 V = 0.400 C . While being discharged in series, (or 10 times the original voltage). Vdischarge = Q Q 0. 400 C = = = 8.00 kV C s C 10 5.00 10 5 F
Q26.8
Q26.9
Q26.10
Q26.11 Q26.12
e
ja
f
Q26.13
Put a material with higher dielectric strength between the plates, or evacuate the space between the plates. At very high voltages, you may want to cool off the plates or choose to make them of a different chemically stable material, because atoms in the plates themselves can ionize, showing thermionic emission under high electric fields. The potential difference must decrease. Since there is no external power supply, the charge on the capacitor, Q, will remain constantthat is assuming that the resistance of the meter is sufficiently large. Adding a dielectric increases the capacitance, which must therefore decrease the potential difference between the plates. Each polar molecule acts like an electric "compass" needle, aligning itself with the external electric field set up by the charged plates. The contribution of these electric dipoles pointing in the same direction reduces the net electric field. As each dipole falls into a configuration of lower potential energy it can contribute to increasing the internal energy of the material.
Q26.14
Q26.15
Chapter 26
79
Q26.16 Q26.17
The material of the dielectric may be able to support a larger electric field than air, without breaking down to pass a spark between the capacitor plates. The dielectric strength is a measure of the potential difference per unit length that a dielectric can withstand without having individual molecules ionized, leaving in its wake a conducting path from plate to plate. For example, dry air has a dielectric strength of about 3 MV/m. The dielectric constant in effect describes the contribution of the electric dipoles of the polar molecules in the dielectric to the electric field once aligned. In water, the oxygen atom and one hydrogen atom considered alone have an electric dipole moment that points from the hydrogen to the oxygen. The other OH pair has its own dipole moment that points again toward the oxygen. Due to the geometry of the molecule, these dipole moments add to have a nonzero component along the axis of symmetry and pointing toward the oxygen. A nonpolarized molecule could either have no intrinsic dipole moments, or have dipole moments that add to zero. An example of the latter case is CO 2 . The molecule is structured so that each CO pair has a dipole moment, but since both dipole moments have the same magnitude and opposite directiondue to the linear geometry of the moleculethe entire molecule has no dipole moment. Heating a dielectric will decrease its dielectric constant, decreasing the capacitance of a capacitor. When you heat a material, the average kinetic energy per molecule increases. If you refer back to the answer to Question 26.15, each polar molecule will no longer be nicely aligned with the applied electric field, but will begin to "dither"rock back and fortheffectively decreasing its contribution to the overall field. The primary choice would be the dielectric. You would want to chose a dielectric that has a large dielectric constant and dielectric strength, such as strontium titanate, where 233 (Table 26.1). A convenient choice could be thick plastic or mylar. Secondly, geometry would be a factor. To maximize capacitance, one would want the individual plates as close as possible, since the capacitance is proportional to the inverse of the plate separationhence the need for a dielectric with a high dielectric strength. Also, one would want to build, instead of a single parallel plate capacitor, several capacitors in parallel. This could be achieved through "stacking" the plates of the capacitor. For example, you can alternately lay down sheets of a conducting material, such as aluminum foil, sandwiched between your sheets of insulating dielectric. Making sure that none of the conducting sheets are in contact with their next neighbors, connect every other plate together. Figure Q26.20 illustrates this idea.
Q26.18
Q26.19
Q26.20
Dielectric
Conductor
FIG. Q26.20 This technique is often used when "homebrewing" signal capacitors for radio applications, as they can withstand huge potential differences without flashover (without either discharge between plates around the dielectric or dielectric breakdown). One variation on this technique is to sandwich together flexible materials such as aluminum roof flashing and thick plastic, so the whole product can be rolled up into a "capacitor burrito" and placed in an insulating tube, such as a PVC pipe, and then filled with motor oil (again to prevent flashover).
80
Capacitance and Dielectrics
SOLUTIONS TO PROBLEMS
Section 26.1 P26.1 (a) (b) Definition of Capacitance Q = CV = 4.00 10 6 F 12.0 V = 4.80 10 5 C = 48.0 C Q = CV = 4.00 10 6 F 1.50 V = 6.00 10 6 C = 6.00 C C= Q 10.0 10 6 C = = 1.00 10 6 F = 1.00 F V 10.0 V Q 100 10 6 C = = 100 V C 1.00 10 6 F
e e
ja ja
f
f
P26.2
(a)
(b)
V =
Section 26.2 P26.3 E=
Calculating Capacitance keq r2 : q=
e4.90 10 N Cja0.210 mf e8.99 10 N m C j
4 9 2 2
2
= 0.240 C
(a)
=
q 0.240 10 6 = = 1.33 C m 2 A 4 0.120 2
a
f
(b) P26.4 (a)
C = 4 0 r = 4 8.85 10 12 0.120 = 13.3 pF C = 4 0 R C R= = k eC = 8.99 10 9 N m 2 C 2 1.00 10 12 F = 8.99 mm 4 0
e
ja
f
e
je
j
(b)
C = 4 0 R =
4 8.85 10 12 C 2 2.00 10 3 m N m
2
e
je
j=
0. 222 pF
(c) P26.5 (a)
Q = CV = 2.22 10 13 F 100 V = 2.22 10 11 C Q1 R1 = Q 2 R2
e
ja
f
Q1 + Q 2 = 1 +
FG H
R1 Q 2 = 3.50Q 2 = 7.00 C R2
IJ K
Q 2 = 2.00 C
(b) V1 = V2 =
Q1 = 5.00 C
Q1 Q 2 5.00 C = = = 8.99 10 4 V = 89.9 kV 1 9 C1 C 2 8.99 10 m F 0.500 m
e
j a
f
Chapter 26
81
P26.6
12 3 2 0 A 1.00 8.85 10 C 1.00 10 m C= = d N m 2 800 m
a fe
a
je
f
j
2
= 11.1 nF
The potential between ground and cloud is V = Ed = 3.00 10 6 N C 800 m = 2.40 10 9 V Q = C V = 11.1 10 9 C V 2.40 10 9 V = 26.6 C P26.7 (a) V = Ed E= 20.0 V 1.80 10 3 m = 11.1 kV m
a f e
e
ja
f
je
j
(b)
E=
0
= 1.11 10 4 N C 8.85 10 12 C 2 N m 2 = 98.3 nC m 2
8.85 10 12 C 2 N m 2 7.60 cm 2 1.00 m 100 cm 0 A = C= d 1.80 10 3 m V = Q C Q = 20.0 V 3.74 10 12 F = 74.7 pC P26.8 C=
e
je
j
(c)
e
je
jb
g
2
= 3.74 pF
(d)
a
fe
j
0 A = 60.0 10 15 F d
12 21.0 10 12 0 A 1 8.85 10 = C 60.0 10 15 d = 3.10 10 9 m = 3.10 nm
d=
a fe
je
j
P26.9
Q=
0 A V d 0 V
a f
V Q = = 0 A d
a f
d=
a f=
e30.0 10
e8.85 10
9
12
C 2 N m 2 150 V
C cm 2 1.00 10 4 cm 2 m 2
je
ja
f
j
= 4.42 m
82
Capacitance and Dielectrics
P26.10
With = , the plates are out of mesh and the overlap area is zero. With
R2 . By proportion, the 2  R2 effective area of a single sheet of charge is 2 When there are two plates in each comb, the number of adjoining sheets of positive and negative charge is 3, as shown in the sketch. When there are N plates on each comb, the number of parallel capacitors is 2 N  1 and the total capacitance is = 0 , the overlap area is that of a semicircle,
a
f
FIG. P26.10
C = 2N  1
a
A f distance = a2 N  1f d a2  f R
0 effective 0
2
2
=
a2 N  1f a  fR
0
2
d
.
P26.11
(a)
(b)
c h 2e8.99 10 j lnc h = F bI V = 2 k lnG J Method 1: H aK
2 k e ln
b a 9 7. 27 2.58 e
C=
=
50.0
2.68 nF
=
8.10 10 6 C = 1.62 10 7 C m 50.0 m 7.27 = 3.02 kV V = 2 8.99 10 9 1.62 10 7 ln 2.58 q =
e
je
j FGH
IJ K
Method 2: P26.12
V =
Q 8.10 10 6 = = 3.02 kV C 2.68 10 9 k eQ ; a
Let the radii be b and a with b = 2 a . Put charge Q on the inner conductor and Q on the outer. Electric field exists only in the volume between them. The potential of the inner sphere is Va = that of the outer is Vb = Va  Vb = Here C = k eQ . Then b
k e Q k eQ Q ba  = a b 4 0 ab 4 0 2 a 2 = 8 0 a a
FG H
IJ and C = Q K V V
a
=
b
4 0 ab . ba
a=
C . 8 0
The intervening volume is
Volume =
C3 4 3 4 3 4 4 7C 3 b  a = 7 a3 = 7 3 3 3 = 3 3 3 3 8 0 384 2 3 0 7 20.0 10 6 384
2
Volume = The outer sphere is 360 km in diameter. P26.13 (a) C=
e
e8.85 10
12
IJ FG IJ K H K C N mj = 2.13 10 C N m j
2 3 2 2 3
FG H
16
m3 .
0.070 0 0.140 ab = = 15.6 pF ke b  a 8.99 10 9 0.140  0.070 0
a f e
b
jb
ga
f
g
(b)
C=
Q V
V =
Q 4.00 10 6 C = = 256 kV C 15.6 10 12 F
Chapter 26
83
P26.14
Fy = 0 : Fx = 0 :
Dividing, so and
T cos  mg = 0 T sin  Eq = 0
tan = E= Eq mg
mg tan q mgd tan . q
V = Ed =
P26.15
C = 4 0 R = 4 8.85 10 12 C N m 2 6.37 10 6 m = 7.08 10 4 F
e
je
j
Section 26.3 P26.16 (a)
Combinations of Capacitors Capacitors in parallel add. Thus, the equivalent capacitor has a value of
C eq = C1 + C 2 = 5.00 F + 12.0 F = 17.0 F .
(b) The potential difference across each branch is the same and equal to the voltage of the battery.
V = 9.00 V
(c)
Q5 = CV = 5.00 F 9.00 V = 45.0 C
and Q12 = CV = 12.0 F 9.00 V = 108 C
b
ga
f
b
ga
f
P26.17
(a)
In series capacitors add as 1 1 1 1 1 = + = + C eq C 1 C 2 5.00 F 12.0 F and
C eq = 3.53 F .
Q eq = C eq V = 3.53 F 9.00 V = 31.8 C .
(c)
The charge on the equivalent capacitor is
b
ga
f
Each of the series capacitors has this same charge on it. So (b) The potential difference across each is and
Q1 = Q 2 = 31.8 C .
V1 = V2 = Q1 31.8 C = = 6.35 V C1 5.00 F Q 2 31.8 C = = 2.65 V . C 2 12.0 F
84
Capacitance and Dielectrics
P26.18
The circuit reduces first according to the rule for capacitors in series, as shown in the figure, then according to the rule for capacitors in parallel, shown below. C eq = C 1 +
FIG. P26.18
FG H
1 1 11 + = C = 1.83C 2 3 6 1 1 1 = + C s C1 C 2 C p  C1 + C1 1 1 1 = + = . C s C1 C p  C1 C1 C p  C1
IJ K
P26.19
C p = C1 + C 2
Substitute C 2 = C p  C1 Simplifying, C1 =
2 C p C p  4C p C s
e
j
2 C 1  C 1C p + C p C s = 0 .
2
=
1 1 2 Cp C p  C pCs 2 4
We choose arbitrarily the + sign. (This choice can be arbitrary, since with the case of the minus sign, we would get the same two answers with their names interchanged.) C1 = 1 1 2 1 1 Cp + C p  C p C s = 9.00 pF + 9.00 pF 2 4 2 4
b
g
b
g  b9.00 pFgb2.00 pFg =
2
6.00 pF
C 2 = C p  C1 = P26.20
1 1 2 1 Cp  C p  C p C s = 9.00 pF  1.50 pF = 3.00 pF 2 4 2
b
g
C p = C1 + C 2
and Substitute Simplifying, and 1 1 1 = + . C s C1 C 2
C 2 = C p  C1 :
C p  C1 + C1 1 1 1 = + = . C s C1 C p  C1 C1 C p  C1
e
j
2 C 1  C 1C p + C p C s = 0 2 C p C p  4C pC s
C1 =
2
=
1 1 2 Cp + C p  C pCs 2 4
where the positive sign was arbitrarily chosen (choosing the negative sign gives the same values for the capacitances, with the names reversed). Then, from
C 2 = C p  C1
C2 = 1 1 2 Cp  C p  C pC s . 2 4
Chapter 26
85
P26.21
(a)
1 1 1 = + C s 15.0 3.00 C s = 2.50 F C p = 2.50 + 6.00 = 8.50 F C eq =
FG 1 + 1 IJ H 8.50 F 20.0 F K b b ga ga
1
= 5.96 F
(b)
Q = CV = 5.96 F 15.0 V = 89.5 C on 20.0 F
Q 89.5 C = = 4.47 V C 20.0 F 15.0  4.47 = 10.53 V V =
f
Q = CV = 6.00 F 10.53 V = 63.2 C on 6.00 F
f
89.5  63.2 = 26.3 C on 15.0 F and 3.00 F
*P26.22 (a)
FIG. P26.21
Capacitors 2 and 3 are in parallel and present equivalent capacitance 6C. This is in series 1 1 1 = 2C . with capacitor 1, so the battery sees capacitance + 3C 6C
LM N
OP Q
(b)
If they were initially unchanged, C 1 stores the same charge as C 2 and C 3 together. With greater capacitance, C 3 stores more charge than C 2 . Then Q1 > Q3 > Q 2 .
(c)
The C 2 C 3 equivalent capacitor stores the same charge as C 1 . Since it has greater  Q implies that it has smaller potential difference across it than C 1 . In capacitance, V = C parallel with each other, C 2 and C 3 have equal voltages: V1 > V2 = V3 . If C 3 is increased, the overall equivalent capacitance increases. More charge moves through the battery and Q increases. As V1 increases, V2 must decrease so Q 2 decreases. Then Q 3 must increase even more: Q3 and Q1 increase; Q 2 decreases .
b
g
(d)
P26.23
Q so V Q = 120 C and C= Q1 = 120 C  Q 2 Q V = : C 120  Q 2 Q = 2 6.00 3.00
2
6.00 10 6 =
Q 20.0
and or
120  Q 2 Q 2 = C1 C2
FIG. P26.23
a3.00fb120  Q g = a6.00fQ
2
360 = 40.0 C Q2 = 9.00
Q1 = 120 C  40.0 C = 80.0 C
86
Capacitance and Dielectrics
P26.24
(a)
In series , to reduce the effective capacitance: 1 1 1 = + F 34.8 F C s 32.0 1 = 398 F Cs = 2.51 10 3 F
(b)
In parallel , to increase the total capacitance: 29.8 F + C p = 32.0 F C p = 2.20 F
P26.25
nC =
1 C
100 1 + +C+
1 C n capacitors
=
100 nC
nC =
100C so n 2 = 100 and n = 10 n 30.8 C . C1
*P26.26
For C 1 connected by itself, C 1 V = 30.8 C where V is the battery voltage: V = For C 1 and C 2 in series:
F GH 1 C F GH 1 C
1 V = 23.1 C 1 + 1 C2 30.8 C 23.1 C 23.1 C = + C1 C1 C2 C 1 = 0.333C 2 .
I JK
substituting,
For C 1 and C 3 in series: 1 V = 25.2 C + 1 C3 1 C 1 = 0.222C 3 .
I JK
30.8 C 25. 2 C 25.2 C = + C1 C1 C3 For all three: Q=
F GH 1 C
C 1 V 1 30.8 C V = = = 19.8 C . 1 + C 1 C 2 + C1 C 3 1 + 0.333 + 0.222 + 1 C 2 + 1 C3 1
I JK
This is the charge on each one of the three. P26.27 Cs = C p1 C p2 C eq
FG 1 + 1 IJ = 3.33 F H 5.00 10.0 K = 2a3.33f + 2.00 = 8.66 F = 2a10.0f = 20.0 F F 1 + 1 IJ = 6.04 F =G H 8.66 20.0 K
1 1
FIG. P26.27
Chapter 26
87
P26.28
Q eq = C eq V = 6.04 10 6 F 60.0 V = 3.62 10 4 C
a f e
ja
f
Q p1 = Q eq , so Vp1 =
Q eq C p1
=
3.62 10 4 C = 41.8 V 8.66 10 6 F
Q 3 = C 3 Vp1 = 2.00 10 6 F 41.8 V = 83.6 C
e
j e
ja
f
P26.29
Cs =
FG 1 + 1 IJ H 5.00 7.00 K
1
= 2.92 F
C p = 2.92 + 4.00 + 6.00 = 12.9 F
FIG. P26.29 *P26.30 According to the suggestion, the combination of capacitors shown is equivalent to
Then
1 1 1 1 C + C0 + C0 + C + C0 = + + = C C0 C + C0 C0 C0 C + C0
b
g
C 0C
2
2 + C0
= 2C + 3C 0 C FIG. P26.30
2
2 2C + 2C 0 C  C 0 = 0
C=
2 2 2C 0 4C 0 + 4 2C 0
e j
4 C0 2
Only the positive root is physical C=
e
3 1
j
Section 26.4 P26.31 (a)
Energy Stored in a Charged Capacitor U= 1 C V 2 1 C V 2
a f a f a
2
=
1 3.00 F 12.0 V 2 1 3.00 F 6.00 V 2
b b
ga ga
f f
2
= 216 J
(b)
U= 1 CV 2 2
2
=
2
= 54.0 J
P26.32
U=
V =
2U = C
2 300 J
30 10 6 C V
f
= 4. 47 10 3 V
88
Capacitance and Dielectrics
P26.33
U=
1 C V 2
a f
2
The circuit diagram is shown at the right. (a)
C p = C1 + C 2 = 25.0 F + 5.00 F = 30.0 F
U= 1 30.0 10 6 100 2
e
ja f
2
= 0.150 J
1
(b)
F 1 + 1 IJ = FG 1 + 1 IJ C =G H C C K H 25.0 F 5.00 F K 1 U = C a V f 2 2a0.150f 2U = = 268 V V =
1 s 1 2 2
= 4.17 F
FIG. P26.33
C
4.17 10 6
P26.34
Use U =
A 1 Q2 and C = 0 . 2 C d 1 C 1 . Therefore, the stored energy doubles . 2
If d 2 = 2d1 , C 2 = *P26.35 (a) (b)
Q = CV = 150 10 12 F 10 10 3 V = 1.50 10 6 C U= 1 C V 2
e
je
j
a f
2
V = U 1 = 0 E 2 V 2
2U = C
2 250 10 6 J 150 10
12
e
F
j=
1.83 10 3 V
P26.36
u=
1.00 10 7 1 = 8.85 10 12 3 000 V 2
e
jb
g
2
V = 2.51 10 3 m 3 = 2.51 10 3 m 3 P26.37 W = U = Fdx so F =
e
jFGH 1 000 L IJK = m
3
2.51 L
z
dU d Q 2 d Q2x Q2 = = = 2 0 A dx dx 2C dx 2 0 A
F I GH JK
F GH
I JK
Chapter 26
89
P26.38
With switch closed, distance d = 0.500d and capacitance C = (a) (b)
0 A 2 0 A = = 2C . d d
Q = C V = 2C V = 2 2.00 10 6 F 100 V = 400 C The force stretching out one spring is 4C 2 V Q2 F= = 2 0 A 2 0 A
a f
a f e a f
2
ja
f
a f = 2CaV f = d A d gd b
2C 2 V
0 2
2
.
One spring stretches by distance x = k= F 2C V = x d
d , so 4
a f FG 4 IJ = 8CaV f H dK d
2 2
2
=
8 2.00 10 6 F 100 V
e
ja
f
2
e
8.00 10 3 m
j
2
= 2.50 kN m .
P26.39
The energy transferred is
H ET =
1 1 QV = 50.0 C 1.00 10 8 V = 2.50 10 9 J 2 2
a
fe
j
and 1% of this (or Eint = 2.50 10 7 J ) is absorbed by the tree. If m is the amount of water boiled away, then giving *P26.40 (a) (b) U= Eint = m 4 186 J kg C 100 C  30.0 C + m 2.26 10 6 J kg = 2.50 10 7 J m = 9.79 kg . 1 C V 2
b
ga
f e
j
a f
2
+
1 C V 2
a f
2
= C V
a f
2
The altered capacitor has capacitance C = C V + C V = C V +
C . The total charge is the same as before: 2 4 V . 3
a f a f a f C a V f 2 FG H IJ K
2
V =
(c) (d)
1 4 V U = C 2 3
11 4V + C 22 3
FG H
IJ K
2
=
a V f 4C
3
2
The extra energy comes from work put into the system by the agent pulling the capacitor plates apart. 1 C V 2 1 R 2 ke
P26.41
U=
a f
2
where C = 4 0 R =
e 2
kQ kQ R and V = e  0 = e R R ke
U=
FG H
IJ FG k Q IJ KH R K
=
k eQ 2 2R
90
Capacitance and Dielectrics
*P26.42
(a)
2 2 2 q1 1 q1 1 q 2 1 1 Q  q1 The total energy is U = U 1 + U 2 = + = + . 2 C 1 2 C 2 2 4 0 R1 2 4 0 R 2
b
g
2
For a minimum we set
1 2 q1 1 2 Q  q1 + 1 = 0 2 4 0 R1 2 4 0 R 2 R 2 q1 = R1Q  R1 q1 Then q 2 = Q  q1 = q1 = R1Q R1 + R 2
b
ga f
dU = 0: dq1
R 2Q = q2 . R1 + R 2
(b)
V1 = V2 =
k e q1 k e R1Q k eQ = = R1 R1 R1 + R 2 R1 + R 2
b
g
ke q2 k e R2 Q k eQ = = R2 R 2 R1 + R 2 R1 + R 2
b
g
and V1  V2 = 0 .
Section 26.5
Capacitors with Dielectrics
12 F m 1.75 10 4 m 2 0 A 2.10 8.85 10 C= = = 8.13 10 11 F = 81.3 pF d 4.00 10 5 m
P26.43
(a) (b)
e
je
j
Vmax = Emax d = 60.0 10 6 V m 4.00 10 5 m = 2.40 kV
e
je
j
P26.44
Q max = CVmax , but Also, Thus, (a) Vmax = Emax d . C=
0 A . d 0 A Emax d = 0 AEmax . d
Q max =
b
g
With air between the plates, = 1.00 and
Emax = 3.00 10 6 V m .
Q max = 0 AEmax = 8.85 10 12 F m 5.00 10 4 m 2 3.00 10 6 V m = 13.3 nC .
Therefore,
e
je
je
j
(b)
With polystyrene between the plates, = 2.56 and Emax = 24.0 10 6 V m . Q max = 0 AEmax = 2.56 8.85 10 12 F m 5.00 10 4 m 2 24.0 10 6 V m = 272 nC
e
je
je
j
Chapter 26
91
P26.45
C= or
0 A d
95.0 10
9
=
3.70 8.85 10 12 0.070 0 0.025 0 10 3
e
jb
g
= 1.04 m
P26.46 Consider two sheets of aluminum foil, each 40 cm by 100 cm, with one sheet of plastic between 2.54 cm . Then, them. Suppose the plastic has 3 , Emax ~ 10 7 V m and thickness 1 mil = 1 000
12 C 2 N m 2 0. 4 m 2 0 A 3 8.85 10 ~ 10 6 F ~ C= 5 d 2.54 10 m
e
je
j
Vmax = Emax d ~ 10 7 V m 2.54 10 5 m ~ 10 2 V P26.47 Originally, C= 0 A Q = V d
e
je
j
a f
12
.
i
(a)
The charge is the same before and after immersion, with value Q = Q=
0 A V d
a f
i
.
e8.85 10 e
C 2 N m 2 25.0 10 4 m 2 250 V
e1.50 10
je
2
m
j
ja
f=
369 pC
(b)
Finally, Cf =
0 A Q = d V
a f
Cf =
f
80.0 8.85 10 12 C 2 N m 2 25.0 10 4 m 2
2
a V f
(c) Originally, Finally, So, Ui = Uf =
f
=
Qd 0 A
je e1.50 10 mj Aa V f d a V f = =
0
j=
118 pF
0 Ad
i
i
=
250 V = 3.12 V . 80.0
1 C V 2
a f
2 i
=
2 f
0 A V 2d =
a f
2 i
.
2 i
1 C f V 2
a f
0 A V
2d 2 2 d
2 2 i
a f
=
0 A V 2 d
a f
2 i
.
U = U f  U i =
12
 0 A V
a f a  1f
2
e8.85 10 U = 
C
2
je25.0 10 m ja250 V f a79.0f = 2e1.50 10 mja80.0f
Nm
2 4 2
45.5 nJ .
92
Capacitance and Dielectrics
P26.48
(a) (b)
C = C 0 =
12 1.00 10 4 m 2 0 A 173 8.85 10 = = 1.53 nF d 0.100 10 3 m
a fe
je
j
The battery delivers the free charge Q = C V = 1.53 10 9 F 12.0 V = 18.4 nC .
a f e
ja
f
(c)
The surface density of free charge is Q 18.4 10 9 C = = = 1.84 10 4 C m 2 . A 1.00 10 4 m 2 The surface density of polarization charge is
p = 1
(d)
FG H
1 1 = 1 = 1.83 10 4 C m 2 . 173 E0
IJ FG K H
IJ K
We have E = E=
and E0 =
V ; hence, d
V 12.0 V = = 694 V m . d 173 1.00 10 4 m
a fe
j
P26.49
The given combination of capacitors is equivalent to the circuit diagram shown to the right. Put charge Q on point A. Then,
Q = 40.0 F VAB = 10.0 F VBC = 40.0 F VCD .
b
g
b
g
b
g
FIG. P26.49
So, VBC = 4VAB = 4VCD , and the center capacitor will break down first, at VBC = 15.0 V . When this occurs, VAB = VCD = 1 VBC = 3.75 V 4
b
g
and VAD = VAB + VBC + VCD = 3.75 V + 15.0 V + 3.75 V = 22.5 V .
Section 26.6 P26.50 (a)
Electric Dipole in an Electric Field The displacement from negative to positive charge is 2 a = 1.20 i + 1.10 j mm  1.40 i  1.30 j mm = 2.60 i + 2.40 j 10 3 m. The electric dipole moment is p = 2aq = 3.50 10 9 C 2.60 i + 2.40 j 10 3 m =
e
j
e
j
e
j
e
je
j
e9.10 i + 8.40 jj 10 j
12
Cm .
(b)
= p E = 9.10 i + 8.40 j 10 12 C m 7.80 i  4.90 j 10 3 N C = +44.6k  65.5k 10 9 N m = 2.09 10 8 N mk
e
j
e
e
j
continued on next page
Chapter 26
93
(c)
U =  p E =  9.10 i + 8.40 j 10 12 C m 7.80 i  4.90 j 10 3 N C U = 71.0 + 41.2 10 9 J = 112 nJ
e
j
e
j
a
f
(d)
p= E=
a9.10f + a8.40f a7.80f + a4.90f
2 2
2 2
10 12 C m = 12.4 10 12 C m 10 3 N C = 9.21 10 3 N C U min = 114 nJ
U max = p E = 114 nJ, U max  U min = 228 nJ P26.51 (a)
Let x represent the coordinate of the negative charge. Then x + 2 a cos is the coordinate of the positive charge. The force on the negative charge is F = qE x i . The force on the positive charge is F+
p FE
F+
af dE = + qEa x + 2 a cos fi qEa xfi + q a2 a cos fi . dx
FIG. P26.51(a) dE dE 2 a cos i = p cos i . dx dx
The force on the dipole is altogether
F = F + F+ = q ke q x2
a
f
(b)
The balloon creates field along the xaxis of Thus, 2 k e q dE = . dx x3
i.
a f
At x = 16.0 cm ,
2 8.99 10 9 2.00 10 6 dE = = 8.78 MN C m 3 dx 0.160
a fe
je a f
j
F = 6.30 10 9 C m 8.78 10 6 N C m cos 0 i = 55.3 i mN
e
je
j
Section 26.7 P26.52
An Atomic Description of Dielectrics qin 0
2 r E = so
E=
2 r 0
r2 r1
V =  E d r =
z
r2 r1
z
r dr = ln 1 r2 2 r 0 2 0
FG IJ H K
FIG. P26.52
2 0
max
= Emax rinner
V = 1. 20 10 6 V m 0.100 10 3 m ln Vmax = 579 V
e
je
j FGH 025.0 IJK . 200
94
Capacitance and Dielectrics
P26.53
(a)
Consider a gaussian surface in the form of a cylindrical pillbox with ends of area A << A parallel to the sheet. The side wall of the cylinder passes no flux of electric field since this surface is everywhere parallel to the field. Gauss's law becomes EA + EA = Q Q A , so E = 2 A A directed away from the positive sheet.
(b)
In the space between the sheets, each creates field
Q away from the positive and 2 A toward the negative sheet. Together, they create a field of E= Q . A
(c)
Assume that the field is in the positive xdirection. Then, the potential of the positive plate relative to the negative plate is
+ plate
V = 
 plate
z
+ plate
E ds = 
Q Qd i  idx = + . A A  plate
z
e
j
(d)
Capacitance is defined by: C =
Q Q A 0 A = = = . d d V Qd A
Additional Problems P26.54 (a) (c) C=
Q ac
LM 1 + 1 OP + LM 1 + 1 OP = 3.33 F N 3.00 6.00 Q N 2.00 4.00 Q = C b V g = b 2.00 Fga90.0 V f = 180 C
1 1 ac ac
Therefore, Q3 = Q6 = 180 C Q df = C df Vdf = 1.33 F 90.0 V = 120 C (b) Q3 C3 Q V6 = 6 C6 Q V2 = 2 C2 Q V4 = 4 C4 V3 = UT = = 180 C = 3.00 F 180 C = = 6.00 F 120 C = = 2.00 F 120 C = = 4.00 F 60.0 V 30.0 V 60.0 V 30.0 V
d
i b
ga
f
FIG. P26.54
(d)
1 C eq V 2
a f
2
=
1 3.33 10 6 90.0 V 2
e
ja
f
2
= 13.4 mJ
Chapter 26
95
*P26.55
(a)
Each face of P2 carries charge, so the threeplate system is equivalent to P1 P2 P2 P3
Each capacitor by itself has capacitance C=
12 4 2 2 0 A 1 8.85 10 C 7.5 10 m = = 5.58 pF . d N m 2 1.19 10 3 m
e
j
Then equivalent capacitance = 5.58 + 5.58 = 11.2 pF . (b) (c) Q = CV + CV = 11.2 10 12 F 12 V = 134 pC Now P3 has charge on two surfaces and in effect three capacitors are in parallel: C = 3 5.58 pF = 16.7 pF . (d) Only one face of P4 carries charge: Q = CV = 5.58 10 12 F 12 V = 66.9 pC . *P26.56 From the example about a cylindrical capacitor, Vb  Va = 2 k e ln b a
a
f
b
g
a f
Vb  345 kV = 2 8.99 10 9 Nm 2 C 2 1.40 10 6 C m ln
3
e = 2a8.99fe1.4 10
= 1.564 3 10 V
5
je
j
J C ln 500
j
12 m 0.024 m
Vb = 3. 45 10 5 V  1.56 10 5 V = 1.89 10 5 V *P26.57 Imagine the center plate is split along its midplane and pulled apart. We have two capacitors in parallel, supporting the same V and A and the carrying total charge Q. The upper has capacitance C 1 = 0 d A lower C 2 = 0 . Charge flows from ground onto each of the outside 2d V1 = V2 = V . plates so that Q1 + Q 2 = Q Then Q 1 Q 2 Q1 d Q 2 2 d = = = C 1 C 2 0 A 0 A Q2 = Q1 = Q1 = 2Q 2 2Q 2 + Q 2 = Q . d 2d
FIG. P26.57
(a)
Q Q . On the lower plate the charge is  . 3 3 2Q 2Q . On the upper plate the charge is  . 3 3 Q1 2Qd = C1 3 0 A
(b)
V =
96
Capacitance and Dielectrics
P26.58
(a)
We use Equation 26.11 to find the potential energy of the capacitor. As we will see, the potential difference V changes as the dielectric is withdrawn. The initial and final energies are U i = 1 Q2 2 Ci
F I GH JK
and
Uf =
1 Q2 . 2 Cf
F I GH JK
1 Q2 But the initial capacitance (with the dielectric) is Ci = C f . Therefore, U f = . 2 Ci Since the work done by the external force in removing the dielectric equals the change in 1 Q2 1 Q2 1 Q2  = 1 . potential energy, we have W = U f  U i = 2 2 Ci 2 Ci Ci
F I GH JK
F I F I F Ia f GH JK GH JK GH JK ja fa f
To express this relation in terms of potential difference Vi , we substitute Q = Ci Vi , and 1 1 2 2 C i Vi  1 = 2.00 10 9 F 100 V 5.00  1.00 = 4.00 10 5 J . 2 2 The positive result confirms that the final energy of the capacitor is greater than the initial energy. The extra energy comes from the work done on the system by the external force that pulled out the dielectric. evaluate: W = (b) The final potential difference across the capacitor is V f = Substituting C f = Ci Q . Cf
b g
b ga f e
and Q = Ci Vi gives V f = Vi = 5.00 100 V = 500 V . Even though the capacitor is isolated and its charge remains constant, the potential difference across the plates does increase in this case. Vmax d
b g
a
f
P26.