Download - Matter and Interaction Chapter 04 Solutions
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
1/71
1
4.X.1 Were given aluminums density of2.7 g/cm3, and lets assume a cubic shape for an aluminum atom. First, get themass of one aluminum atom.
mAl
27 g/mol6.022
10
23
mol1
mAl
4.5 1023 g
Now use aluminums (micro)density (represented by ) to get an atoms approximate diameter.
=m
Al
d3
d3 =m
Al
d 3
mAl
d 34.5 10
23
g
2.7 g/cm3
d 2.6 108 cm 2.6 1010 m
4.X.2Were given leads density of11.4 g/cm3, and lets assume a cubic shape for an lead atom. First, get the mass of onelead atom.
mPb
207 g/mol6.022 1023 mol1
mPb
3.4
1022
g
Now use leads (micro)density (represented by ) to get an atoms approximate diameter.
=m
Pb
d3
d3 =m
Pb
d 3
mPb
d
33.4 1022 g
11.4 g/cm
3
d 3.1 108 cm 3.1 1010 m
4.X.3A reasonable guess would be that one short springs stiffness would be twenty times the chains effective stiffness, or800 N/m. Each spring contributes one twentieth of the total stretch (neglecting an individual springs mass). If each springstretches by only one twentieth of the total stretch for the same applied force, then each spring must have twenty times thechains stiffness.
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
2/71
2
4.X.4 A reasonable guess would be that one springs stiffness would be one ninth the combinations effective stiffness, o300 N/m. Each spring supports one ninth of the rocks weight. Each spring stretches the same amount. Since each sprisupports one ninth of the total weight for the same amount of stretch, each spring must have a stiffness that is one ninth thcombinations effective stiffness, or 300 N/m.
4.X.5 Shortening the wire by a factor of ten means a factor of ten fewer lengthwise interatomic bonds in the wire. So twire should be ten times stiffer than before. Therefore, it will only stretch one tenth of the original stretch, or 0.151 mm.
4.X.6 From the graph, a unit stress produces a strain of about 1.8 units. So Youngs modulus would be approximate110
8
N/m2
1.8103 m/m 6 1010
N/m2.
4.X.7This is a straightforward application of the basic definition of Youngs modulus.
Y =
F /AL/L
L =
F /AY /L
L =
F /(r2)Y /L
L (10 kg)(9.8 N/kg)/((1.5 103
m)2)
(2 1011 N/m2)/(3 m)L
2.1
104
m
0.21 mm
4.X.8
(a) The block will not move.
(b) Since the block isnt moving (static), the forward force on the block by you must be nulled out by the force on tblock by the floor. Thus, the floor exerts a horizontal force of magnitude 60 N.
(c) 100 N is more than necessary to overcome friction, so the block will accelerate.
(d) Anything over80 N causes the block to accelerate, so the maximum horizontal force the floor can exert on the blocmust be 80 N.
4.X.9
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
3/71
3
(a) Apply the momentum principle to the system consisting of the block. Assume non-relativistic speeds.
px
= Fnet,x
t
mblock
vx
Fnet,x
t
mblock
vx
k
F
N
t
mblock vx kmblock gt
k
vxt
g
k
4 m/s0.7 s
9.8 Nkg
0.58
(b) Since the net force on the system (block) is constant, we can approximate the blocks average velocity as the arithmeticmean of the initial and final velocities and then solve for the blocks change in position.
v
avg
1
2(4 m/s + 0 m/s) 2 m/s
x
(2 m/s)(0.7 s)
1.4 m
(c) Assume the upper (3 kg) box doesnt slide on the other box. Youve effectively increased the systems mass, and thusalso increased the normal force on the system, by a factor of 1.6. As you saw in part (a), the systems mass divides outfor the purposes of calculating
k. Therefore, the same change in velocity will take place during the same time interval.
Therefore, the new box will stop in 0.7 s. This seems counterintuitive, but in this problem, t is algebraically andphysically independent ofm
block.
4.X.10 The rate of change of the objects momentum is precisely what we mean by net force on the object. Thus, thezcomponent of the net force on the object will be 4 N.
4.X.11 Constant momentumautomatically implies that
dp
dt is zero. Therefore,Fnet is also zero.
4.X.12
a =v
f v
i
t
a = 5.02, 3.04, 0 m/s 5, 3, 0 m/s
0.01 s
a 0.02, 0.04, 0 m/s0.01 s
2, 4, 0 m/s/s
Note that the unit of acceleration is m/s/s, which is usually abbreviated tom/s2.
The rate of change of the balls momentum and the net force on the ball are the same physical entity, which is approximately
mball
a
80 103 kg
(2, 4, 0 m/s/s)m
balla 0.16, 0.32, 0 N
4.X.13
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
4/71
4
(a) First, calculate the stiffness.
ks
= mg
s
(0.33 kg)
9.8 Nkg
(5.5 102 m)
ks 58.8 N/m
(b) Second, calculate the oscillation frequency, which tells how many oscillations per second the system will carry out.
f = 1
2
k
s
m
f = 1
2
58.8 N/m
0.33 kg 2.12 Hz
(c) Now calculate how many oscillations will happen during a 5 sinterval.
N = ft
N (2.13 Hz)(5 s) 10.6 oscillations
4.X.14
(a) Oscillation period is independent of amplitude, so one complete oscillation would still take2 s.
(b) Oscillation period is proportional to the square root of mass. Tripling the mass increases the period by a factor of
Therefore, the new period would be
3(2 s) 3.46 s.
4.X.15
|v| =
ksm
atom
d
40 N/m
3.3 1025 kg 2.1 1010
m
2970 m/s
4.X.16
T L
|v| 3 m
2970 m/s
1 103 s
4.X.17
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
5/71
5
The buoyant force on the iron is the weight of the blocks volume of air.
Fb
(1.3 106 kg/cm3) 1 kg8 103 kg/cm3
(125 cm3)
9.8
N
kg
1.6 103 N
The blocks weight is (1 kg)
9.8 Nkg
= 9.8 N.
4.X.18
At the top of Earths atmosphere, P = 0. At Earths surface, P= 1 105 Nm2 . Assuming uniform density,
P = gh
h = P
g
The density of air at 20C and at atmospheric pressure is approximately 1.2 kgm3 .
h = 1 105 Nm2(1.2 kgm3 )(9.8
Nkg )
h = 8500 m
4.X.19
Draw a sketch of the region.
0.01mm
Volume of air
A
Figure 1: A sketch of the region
Assume that the area is approximately 0.2 m 0.2 m = 0.04 m2
V = (0.01 mm)(0.04 m2)
= (1 105 m)(0.04 m2)= 4 107 m3
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
6/71
6
air
= 1.2 kg
m3
Air is mostly nitrogenN2
with molar mass 14 2 = 28 gmol = 0.028 kgmol .Avogadros Number isN
A= 6.02 1023 moleculesmol .
Use unit cancelation to find the number of molecules of air between the book and table.
6.02 1023 molecules
mol
1mol
0.028kg
1.2 kg
m3
4 107m3
= 1 1019molecules
4.X.20
Assume the oscillation is along the x-axis.
px
= Fx
t
pfx
0p
ix= F
xt
pfx
= Fx
t
mv0
= Ft
Then,v0
= Ftm is the initial speed after the hammer strike.
Assume thatx during the strike is negligible. Then, since x0
= 0,
0
x0
+m
ks
v20
= A2
m
ks
Ft
m
2= A2
A =m
ks
Ft
m
A =
m
ks
v0
4.X.21
Att0
= 0, x0
= 0 and v0
= Ftm . Assume that v0 is positive. (Note: it could be negative.) Sketch x vs t, as shown in Figu2.
Note that the slope of x vs t, which is the x-velocity, is positive at t = 0.
Substitutet0
= 0 into
x = A cos
0
ks
mt
0+
x = A cos
0 = A cos
cos = 0
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
7/71
7
x
t
Figure 2: Graph of x vs t
Thus = 90 or270. Substitutet0
= 0 into
v0 = ksm A sin
0
ksm t0 +
=
ks
mA sin
Sincev0
is positive, sin must be negative. Since = 90 or270 then
sin = 1 = 270 or
3
2
Note: we chose v0 to be positive. If, however, v0 is negative, then sin is positive and
sin = +1
= 90 or
2
4.X.22
(a) Neglect the mass of the rope and assume that tension is uniform throughout the rope.
Apply the momentum principle to the climber. Sketch a free-body diagram. Define the system to be the climber.
Fnet
=p
t
The climbers momentum is constant (since the climber is "motionless"), so
Fnet
= 0
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
8/71
8
FTonclimber byrope
Fgravonclimber byEarth
Figure 3: A free-body diagram of the climber
Sum the forces from the free-body diagram.
FT by rope
+ Fgrav by Earth
= 0
FT by rope =
Fgrav by Earth
= =
=
= 0, 539, 0 N
FT by rope
= 539 N
(b)
m = 88 kgF
T by rope=
=
= 0, 862, 0 N
(c) Both (2) and (3) are true. Model the rope as balls connected by springs in one dimension. Tension (i.e. a force applito the rope) causes the interatomic springs (i.e. bonds) to stretch. As a result the atoms in the one-dimensionmodel get further apart.
4.X.23
(a) is true. Also, as atoms get closer than their equilibrium distance, they repel. In this way, the bond acts like a spring.
(c) is partially, but not completely, true. It is only true for small amplitude oscillations about the equilibrium distanbetween atoms.
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
9/71
9
4.X.24
m= 5 kg of gallium.
molar mass, M = 70 gmol = 0.07 kgmol
Avogadros Number, NA
= 6.02 1023 atomsmolUse unit cancelation to find the mass of one atom in kgatom .
0.07 kgmol
1mol6.02 1023atoms
= 1.2 1025 kgatom
4.X.25
The radius of a hydrogen atom is called the Bohr radius and is about 0.5 1010 m. A copper atom is bigger than hydrogen,so its radius is about 1 1010, rounded to one significant figure.
4.X.26
(a) molar mass,M = 64 gmol = 0.064 kgmol
Avogadros Number NA = 6.02 1023 atoms
mol
So the mass of 1 atom of copper is0.064
kg
mol
1mol
6.02 1023 atoms
= 1.06 1025 kgatom
.
(b) Assume a simple cubic array as shown in Figure4.
d
L
Figure 4: One side of the cubic array
L = 4.6 cm
= 0.046 mL = N d
N = L
d
= 0.046 m
2.28 1010 m= 2.02 108 atoms
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
10/71
10
(c) A cubic block with each side of length 0.046 m has a volume
V = L3
= (0.046 m)3
= 9.73 105 m
The total number of atoms is the number of atoms along each side cubed.
Ntotal
= N3side
= (2.02 108 atoms)3= 8.21 1024 atoms
Multiply the number of atoms times the mass of each atom.
m = (1.06 1025 kgatom
)(8.21 1024 atoms)m = 0.870 kg
4.X.27
molar mass: M = 184 gmol = 0.184 kgmol
density: =
19.3 gcm2
1 kg1000 g
(100 cm)3
1 m3
1.93 104 kgm3Find the volume of a cube of the block that is taken by one atom.
1 m3
1.93 104kg
0.184
kg
mol
1mol
6.02 1023 atoms
= 1.58 1029 m3
atom
Figure 5: Cube filled by atom
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
11/71
11
This is the volume of a cube that is filled as much as possible by a spherical atom. The volume of the cube is d3 whered isthe diameter of the atom.
d3 = 1.58 1029 m3d = (1.58 1029 m3) 13
= 2.51 1010 m
4.X.28
A sketch of the situation is shown in Figure 6.
FTbywire
Fgrav Fgrav
FTbywire 1 FTbywire 2
m=10kg m=10kg
One wire Two wires
Figure 6: A sketch of the situation with one and two wires.
With two wires, the tension in each wire is half the tension in the case of one wire. (This is a result of the momentumprinciple.) Since the tension in the wire is proportional to the distance stretched, each wire, in the case of two wires, willstretch half as much as one wire alone. The correct answer is (A), each wire stretches4 mm.
4.X.29
FT
A = Y
L
L
L =F
T
A
L
Y
L 1A
Half the area results in twice the distance stretched. The correct answer is (C), the second wire stretches 16 mm.
4.X.30
FT
A = Y
L
L
L =F
T
AYL
L L
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
12/71
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
13/71
13
Each individual spring has a stiffness,
ks
= N ks,eff
= (2)(190 N
m)
= 380 N
m
4.X.35
For springs in parallel, the effective stiffness is
ks,eff
=ks,1
+ ks,2
+ ...
ForNidentical springs,
ks,eff
=N ks
Each individual spring has a stiffness
ks
=k
s,eff
N
= 20250 Nm
45
= 450 N
m
4.X.36
For identical springs in series,1
ks,eff
= N 1
ks
ks
= N ks,eff
= 2(140 N
m)
= 280 N
m
4.X.37
For identical springs in parallel,
ks,eff
= N ks
= 5(390 N
m)
= 1950 N
m
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
14/71
14
4.X.38
How far a material stretches when a certain force is applied depends on the interatomic bond stiffness. Thus the correanswer is (C).
4.X.39
(b)YA
=YB
because both wires are made of pure copper.
4.X.40
No, it is not a violation of the momentum principle. Before picking up the object, it is at rest. Thus, it must be sitting othe ground or on the floor or on a table, for example. Or perhaps it is hanging by a chain or rope. Lets assume its sittinon a table, as in Figure 7. Then,
Figure 7: The object is at rest on a table
Draw a free-body diagram (see Figure8).
Fby flooronobject
Fgravbyearth onobject
Figure 8: A free-body diagram of the forces on the object
If you also lift the box and it remains at rest, then the free-body diagram looks like the one in Figure 9.
The net force on the object is still zero, according the the momentum principle. By you applying an upward force on tobject, the force by the floor on the object diminished, but the net force (i.e. the sum of all forces) on the object is still zero
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
15/71
15
Fby flooron object
Fgravbyearth onobject
Fby youonobject
Figure 9: A free-body diagram with the lifting force added
4.X.41
(a) Its cross-sectional area is the area of a circle, r2.
A = r2
= (4 103 m)2= 5 105 m2
(b) Rod 2s cross-sectional area is the area of a rectangle.
A = wd
= (12 mm)(6 mm)
= (12 103 m)(6 103 m)= 7.2 105 m2
(c) Rod 3s cross-sectional area is the area of a square.
A = wd
= (6 103 m)(6 103 m)= 3.6 105 m2
4.X.42
Since both wires are made of the same material, then they will have the same Youngs Modulus, Y. Youngs Modulus onlydepends on the material of the wire. Thus, (1) Y
B=Y
Ais true.
4.X.43
(a) First, sketch a picture (see Figure10).
Define the system as the load. Apply the momentum principle. Draw a free-body diagram of the system (see Figure11).
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
16/71
16
Load
Steel
Figure 10: A sketch of the situation
Fby steelon load
Fgravby earthon load
Figure 11: Free-body diagram of the system
Fnet
= p
tF
grav+ F
steel= 0
Fsteel
= Fgrav
Fsteel
= =
=
= 0, 833, 0 N
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
17/71
17
(b)
FT
A = Y
L
L
L =F
T
A
L
Y
= 833 N
(0.28 m)(0.28 m) 0.28 m
2 1011 Nm2
= 1.5 108 m
Note: this is approximately 100 times the diameter of an atom.
4.P.44
For Aluminum:
Y = 6.2 1010 Nm2
=
2.7 g
cm3
1 kg1000 g
(100 cm)3
1 m3
= 2700 kg
m3
Molar mass M= 27 gmol = 0.027 kgmol
Assume a simple cubic array. Find the diameter of the Al atom.
V =
1 m3
2700kg
0.027kg
1mol
1mol
6.02 1023 atoms
= 1.66 1029 m3
atom
V = d3
d = V 1
3
= 2.55 1010 m
Write Youngs modulus in terms of atomic quantities and solve for the bond stiffness.
ks,bond = Y d
= (6.2 1010 Nm2
)(2.55 1010 m)
= 16 N
m
For lead,
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
18/71
18
Y = 1.6 1010 Nm2
= 11.4 g
cm3
= 1.14 104 kgm3
M = 207 gmol
= 0.207 kg
mol
Find the diameter of a Pb atom.
V =
1 m3
1.14 104kg
0.207kg
1mol
1mol
6.02 1023 atoms
V = 3.02 1029 m3
atom
d = V 1
3
= 3.11 1010 m
UseY andd to calculate the bond stiffness.
ks
= Y d
= (1.6 1010 Nm2
)(3.11 1010 m)
= 5.0 N
m
4.P.45
(a) A: To analyze the interatomic compression at A, define the system to be the entire rod except the layer of atoms othe left edge of the rod. Sketch the bonds between this layer of atoms and the rod as shown in Figure 12.
Apply the Momentum Principle to this system of the rod.
Fnet
=p
t
Sketch a free-body diagram for the rod (see Figure 13).
Fnet
= p
t
FA
= mrod
v
t
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
19/71
19
Figure 12: Bonds at A on the rod
FA
Figure 13: free-body diagram
FA
is the force on the system due to interatomic compression at A.
C: At the right end of the rod at C, define the system to be a very thin layer of the rod of mass mC
, as shown in Figure14
Sketch a free-body diagram for the thin layer of the rod at C (see Figure 15).
Apply the Momentum Principle.
Fnet
= p
t
FC
= mC
v
t
Note that the mass of the thin layer of the rod at C is much less than the mass of the rod. As a result, compare thefollowing equations. Note that vt is the same for all parts of the rod.
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
20/71
20
Figure 14: Bonds at C on the rod
FC
Figure 15: free-body diagram
FA
= mrod
v
t
FC
= mC
v
t
The interatomic force at C will be less than the interatomic force at A, since mC
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
21/71
21
4.P.46
(a) Define the system to bem1
,m2
, andm3
together. Sketch a free-body diagram (see Figure 16).
F
Figure 16: Free-body diagram for the entire system ofm1
,m2
, and m3
.
Apply the Momentum Principle to the system.
Fnet
= dp
dt
Write the x-component of the Momentum Principle.
F = msystem
dvx
dt
dvxdt
= Fm
system
dvx
dt =
F(m
1+ m
2+ m
3)
Note that the acceleration will be the same for all parts in the system as well. Som1
,m2
, and m3
all have the sameacceleration.
For the left end, define the system to be m3
. Sketch a free-body diagram (see Figure 17).
F2
is the compression force on the left end ofm2
due to its pushing on m3
.
Apply the Momentum Principle to the system.
Fnet
= dp
dt
F2
= m3
dvx
dt
Substitute the acceleration of the system.
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
22/71
22
F2
Figure 17: Free-body diagram for m3
.
F2 = m3 Fm1
+ m2
+ m3
F2
=
m
3
m1
+ m2
+ m3
F
Note that this is the compression force at the left end ofm2
and is less than the compression force at the right end m
2. This is expected since m
2has an acceleration to the left.
(b) Define the system to bem2
andm3
together. Sketch a free-body diagram of the system (see Figure18).
F1
Figure 18: Free-body diagram of the system ofm2
andm3
together.
F1
is the compression force at the right end ofm2
due to contact with m1
.
Apply the Momentum Principle to the system.
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
23/71
23
Fnet
= dp
dt
F1
= msystem
dvx
dt
F1
= (m1
+ m2
)dv
x
dt
Substitute the x-acceleration from part (a).
F1
= (m1
+ m2
)
F
m1
+ m2
+ m3
F1
=
m
1+ m
2
m1
+ m2
+ m3
F
Note that this is less than the magnitude of the force F and greater than F2
, as expected.
(c) When sketching the free-body diagram in part (a), the direction of the force on the system is the same whether youpull on block 3 or push on block 1. The only difference is that if you pull on block 3, interatomic bonds will stretch.If you push on block 1, interatomic bonds will compress. But the magnitudes and directions of the forces will be thesame in the two cases.
4.P.47
This is an experimental question, and therefore precise results will vary. You should be able to obtain at least the correctorder of magnitude with even the simplest experimental setup.
4.P.48
kwire
(5 kg)(9.8 Nkg )
0.4035 103 m 1.214 N/m
matom
48 g/mol6.022 1023 mol1 8.0 10
23
g
d 3
matom
3
8.0 1023 g
4.51 g/c3m 2.6 108 cm 2.6 1010 m
Nbonds in 1 chain
L
d 3 m
2.6 1010
m2.6
1010
Nchains
AwireA
atom
4.6 1013
ks (1.214 N/m)(1.2 10
10
)
4.6 1013 32 N/m
4.P.49
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
24/71
24
(a)
k (415 kg)
9.8 Nkg
1.26 102 m 3.23 10
5
N/m
(b)
Nchains
AwireA
atom
(0.15 102
m)2(2.51 1010 m)2 3.57 10
13
(c)
Nbonds in 1 chain
Ld 2.5 m
2.51 1010 m 9.96 109
(d)
ks (3.23 10
5
N/m)(9.96 109)(3.57
10
13
) 90 N/m
4.P.50
(a)
Y =
F /AL/L
Y =(14 kg)(9.8 Nkg )/()(1 10
3
m)2
(0.00139 m)/(2.5 m)
Y 7.9 1010
N/m
(b) First, calculate the mass of one gold atom.
mAu
= 197 g/mol
6.022 1023 mol1m
Au 3.27 1022 g
Now, use the density () and atoms mass to calculate an approximate interatomic spacing, assuming a cubical atom
d 3
mAu
d 3
3.27 1022 g19.3 g/cm3
2.57 108 cm 2.57 1010 m
Finally, use the Youngs modulus and interatomic spacing to calculate the interatomic stiffness.
ks Y d
ks
7.9 1010 N/m
2.57 1010 m
20.3 N/m
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
25/71
25
4.P.51 Start by calculating Youngs modulus for copper. It turns out that the data given in the question is not plausible.In early printings of the textbook, the initial length was incorrectly given as 3.5 m, but it should be 0.95 m, as noted in thetextbook Errata found at matterandinteractions.org.
Y =F /AL/L
Y =(36 kg)(9.8 Nkg )/
()(0.7 103 m)2
(0.00183 m)/(0.95 m)
Y 1.90 1011 N/m2
Next, calculate the mass of one copper atom.
mCu
= 63 g/mol
6.022 1023 mol1m
Cu 1.05 1022 g
Now, use the density () and atoms mass to calculate an approximate interatomic spacing, assuming a cubic atom.
d 3
mCu
d 3
1.05 1022 g9 g/cm3
2.27 108 cm 2.27 1010 m
Finally, use the Youngs modulus and interatomic spacing to calculate the interatomic stiffness.
ks Y d
ks
1.2 1011 N/m2
2.27 1010 m
27 N/m
4.P.52 Start by calculating Youngs modulus for iron.
Y =
F /AL/L
Y =(52 kg)(9.8 Nkg )/()/(0.04 10
2
m)2
(0.0127 m)/(2.5 m)
Y 2.0 1011 N/m
Next, calculate the mass of one iron atom.
mFe
= 56 g/mol
6.022
1023 mol1
mFe
9.30 1022 g
Now, use the density () and atoms mass to calculate an approximate interatomic spacing, assuming a cubic atom.
d 3
mFe
d 3
9.30 1022 g7.87 g/cm3
2.28 108 cm 2.28 1010 m
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
26/71
26
Finally, use the Youngs modulus and interatomic spacing to calculate the interatomic stiffness.
ks Y d
ks
2.0 1011 N/m
2.28 1010 m
= 46 N/m
4.P.53
(a) Assume a simple cubic lattice for iron. Find the volume of a cube that surrounds a spherical atom.
=
7.87 g
cm2
1 kg1000 g
(100 cm)3
1 m3
= 7870 kg
m3
M = 56 gmol
= 0.056 kg
mol
V =
1 m3
7870kg
0.056kg
1mol
1mol
6.02 1023 atoms
= 1.18 1029 m3V = d3
d = V
1
3
= 2.28 1010 m
(b) Determine Youngs Modulus. Begin by applying the Momentum Principle to the hanging mass. Draw a free-boddiagram, as shown in Figure19.
The hanging mass is in equilibrium.
Fnet
= dp
dtF
T+ F
grav = 0
FT
= Fgrav
= =
=
= 0, 637, 0 N
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
27/71
27
FT on massby wire
Fgrav on massby earth
Figure 19: A free-body diagram of the system.
The tension applied to the wire is 637 N.
FT
A = Y
L
L
Y =
F
T
A
L
L
=
F
T
R2
L
L
Ft
= 637 N
R = 0.09 cm
2= 0.045 cm
= 4.5 104 mL = 2.0 m
L = 0.01 m
Y =
637 N
(4.5 104 m)2
2 m
0.01 m
= 2.0 1011 Nm2
The interatomic bond stiffness is
k = Y d= (2.0 1011 N
m2)(2.28 1010 m)
= 46 N
m
4.P.54
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
28/71
28
(a) Begin by applying the Momentum Principle to the hanging mass. Draw a free-body diagram, as shown in Figure20
FT on massby wire
Fgrav on massby earth
Figure 20: A free-body diagram of the system.
The hanging mass is in equilibrium.
Fnet
= dp
dtF
T+ F
grav = 0
FT
= Fgrav
= =
=
= 0, 647, 0 N
The tension applied to the wire is 637 N.
FT
A = Y
L
L
Y =
F
T
A
L
L
= F
T
R2 L
L=
647 N
(4.5 104 m)2
2.2 m
0.0112 m
= 2.0 1011 Nm2
(b) Assume a simple cubic lattice for iron. Find the volume of a cube that surrounds a spherical atom.
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
29/71
29
=
7.87 g
cm2
1 kg1000 g
(100 cm)3
1 m3
= 7870 kg
m3
M = 56 g
mol= 0.056
kg
mol
V =
1 m3
7870kg
0.056kg
1mol
1mol
6.02 1023 atoms
= 1.18 1029 m3V = d3
d = V 1
3
= 2.28 1010 m
The interatomic bond stiffness is
k = Y d
= (2.0 1011 Nm2
)(2.28 1010 m)
= 46 N
m
4.P.55
Spring force is F =bs3
(a) Define the system as the hanging mass. Draw a free-body diagram.
Apply the Momentum Principle. The system remains at rest (i.e. in equilibrium).
Fnet
= dp
dtF
spring+ F
grav = 0
Fspring
= Fgrav
=
Examine the y-component only.
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
30/71
30
Fonmassby spring
Fgrav on massby earth
Figure 21: A free-body diagram of the situation
bs3 = mg
b =
mg
s3
Wheres = L L0
= 29 cm 25 cm = 4 cm = 0.04 m.
b =(0.018 kg)(9.8 Nkg )
(0.04 m3)3
= 2760 N
m3
(b) The following ideas were used in the analysis for part (a).The Momentum Principle
The fact that the gravitational force acting on an object near Earths surface is approximately mg.
The rate of change of momentum of the system is zero.
4.X.56
Spring force is F =bs3
(a) For Bob, there is clearly a frictional force of the floor on the box that has a magnitude of20 N and is in the opposi
direction as the force of Bob on the box, since the net force on the box is zero. Assuming that the frictional force is ndependent on speed (which is generally the case) then the force by Alice on the box must also be 20 N. Though spushes the box such that it has a greater speed, its velocity is constant and so the net force on the box is zero. Sinthere is a frictional force of magnitude 20 N, she must be pushing with an oppositely directed force of magnitude20
(b) Initially, to make the box speed up, both Alice and Bob had to push with a force of magnitude greater than20 N. Whthe box reached a speed of20 m/s, Bob reduced his force to 20 N and the box moved with constant speed of1 m/When the box reached a speed of2 m/s, Alice reduced her force to 20 Nand then her box moved with a constant speof2 m/s.
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
31/71
31
4.X.57
It also must be pulled by 3 N. The frictional force does not generally depend on the area of the surfaces in contact, but onlyon the materials in contact and the normal (perpendicular) contact force.
4.X.58
(a) To start the box moving, you must apply a force parallel to the surfaces in contact that is greater than the maximumstatic force. Thus,
fs,max
= s
FN
Apply the Momentum Principle. Define the system to be the box. Draw a free-body diagram for the box (see Figure22).
FN by table onbox
Fby personon box
Fgrav by earth on box
fsby table on box
Figure 22: A free-body diagram of the system.
Fnet
= dp
dt
At the instant just before it starts to move,
Fnet
= 0
In the y-direction,
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
32/71
32
Fnet,y
= 0
FN
+ mg = 0F
N= mg
= (3 kg)(9.8 N
kg
)
= 29.4 N
In the x-direction,
Fby person on box
+ fs,max
= 0
Fby person on box
= fs,max
= s
FN
= (0.3)(29.4 N)
= 8.82 N
(b) To move at constant speed, the box is in equilibrium with Fnet
= 0, but the frictional force is kinetic friction. Defithe system to be the box, and apply the Momentum Principle.
Fnet
= 0
FN
= 29.4 N
Fby person on box
= fk
= k FN= (0.2)(29.4 N)
= 5.9 N
4.X.59
Assume a horizontal floor.
Define the system to be the box. Draw a free-body diagram (see Figure 23).
Apply the Momentum Principle
Fnet
= dp
dt
Write it in component form, starting with the y-direction.
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
33/71
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
34/71
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
35/71
35
4.X.63
T = 2
m
k
T
1
k
Doubling the stiffness causes Tto change by a factor of 12
. Thus, ifT = 1 s, then doubling the stiffness results in a period
of0.71 s.
4.X.64
For identical springs in series,
1k
eff
= N1k
k = N keff
= 2keff
Thus, cutting the spring in half doubles the stiffness. Since
T
1
k
doubling the stiffness changes Tby a factor of 12
= 0.71. Thus, a period of1 s becomes a period of(0.71)(1 s) = 0.71 s.
4.X.65
Period is independent of amplitude. Therefore, the period will remain1 s.
4.X.66
Period is independent ofg . Therefore, the period will remain1 s.
4.X.67
Define the system to be the mass. Sketch a free-body diagram whenx = +s(see Figure25).
Apply the Momentum Principle in the x-direction.
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
36/71
36
m m m
x=s x=sx=0
Figure 24: A sketch of the system
Fbyspring onmass
Figure 25: A free-body diagram of the system.
Fnet,x
= dp
x
dt
Fspring,x
= dp
x
dt
ks
x = dp
x
dt
ks
x = mdv
x
dtk
s
m x =
dvx
dt
Substitutevx
= dxdt .
ks
m x =
d2x
dt2
d2x
dt2 +
ks
mx = 0
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
37/71
37
A solution for this differential equation is
x = A cos
k
s
mt +
A and are constants that depend on initial conditions.
4.X.68
Angular frequency is
=
k
s
m
Since = 2f, then frequency is
f = 1
2
k
s
m
SinceT = 1f, then
T = 2
m
ks
Amplitude is independent ofks andm.
4.X.69
T = 2
m
ks
Period is independent of amplitude. Therefore, doubling the amplitude does not affect the period.
T
m
Doubling the mass changes the period by a factor of
2.
T 1k
s
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
38/71
38
Doubling the stiffness changes the period by a factor of 12
.
4.X.70
Forx = A cos t, the systems velocity should be zero at t = 0, and the systems position is most positive at x = A.
Forx = A sin t, the systems velocity should be a maximum and positive att = 0; thus the system should be moving in th+x direction att = 0.
4.X.71
An oscillating diatomic moleciule is not a harmonic oscillator (except for very small amplitude oscillations).
A pendulum with a large initial angle from vertical is not a harmonic oscillator.
4.X.72
(a)
=ks
m
=
4 Nm
1.14 kg
= 1.69rad
s
(b)
= 2f
f =
2
= 1.69 rads
2
= 0.269 s1
(c)
T = 1
f
= 1
0.269 s1
= 3.72 s
(d) The period does not depend ong . Therefore, the period of this system would be the same on Moon as it is on Eart3.72 s.
4.X.73
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
39/71
39
=
k
s
m
=
8 Nm2.2 kg
= 1.91rad
s
x = A cos(t + )
In this case, since x = +A at t = 0, = 0. So, at t = 1.15 s,
x = A cos t
= (0.18 m)cos ((1.91rad
s )(1.15 s))
= 0.105 m
4.X.74
ks,A
= 3ks,B
mA = 3mBd
A d
B
v = d
=
k
s,i
ma
d
3 times the interatomic stiffness changes the speed by a factor of
3. 3 times the mass changes the speed by a factor 13
.
These effects cancel out so that (b) vA
= vB
.
4.X.75
The speed of sound only depends on the material (the interatomic bond stiffness, atomic mass, and atomic diameter). Sinceboth rods are made of titanium and since their lengths are the same, then the time for the disturbance to travel to the endof the rod is the same. The answer is (c).
4.X.76
The time it takes for a ball to fall from rest at an initial height h is given by
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
40/71
40
y =
0
v0,y
t +1
2
Fnet,y
m t2
h = 12
(g)t2
h = 1
2gt2
t =
2h
g
The time to rise to the same height after it bounces is also
t =
2h
g
Thus the period of a bouncing ball that returns to its same height is
t = 2
2h
g
Since t
h, if you quadruple the maximum height h, the period increases by a factor of
4 = 2. Thus, the periodoubles.
4.X.77
T= 2
m
ks
(a)
T m
If you doublem, T changes by a factor
2.
(b)
T 1k
s
If you doubleks
, Tchanges by a factor of 12
.
(c) If you double bothm and ks
, the effects cancel each other out and T remains the same.
(d) T is independent ofA, so if you double A, T remains the same.
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
41/71
41
FT by rod onball
Fgrav by earth on ball
Figure 26: A free-body diagram of the system.
4.P.78
Apply the Momentum Principle to the ball. Define the system to be the ball. Draw a free-body diagram, as shown in Figure26.
Since the body is in equilibrium,
Fnet
= p
t= 0
FT
+ Fgrav
= 0
FT
= Fgrav
= =
=
= 0, 402, 0 NFT
= 402 N
Calculate Youngs Modulus,
FT
A = Y
L
L
Y =F
T
A
L
L
The cross-sectional area of the rod is
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
42/71
42
A = (1.5 mm)(3.1 mm)
= (1.5 103 m)(3.1 103 m)= 4.65 106 m2
Y =
402 N
4.65
106 m2
2.6 m
0.002898 m
= 7.8 1010 Nm2
Calculate the diameter of a silver atom. Assume a simple cubic lattice. Find the volume of a cube taken up by a sphericatom.
=
10.5 g
cm3
1 kg1000 g
(100 cm)3
1 m3
= 1.05 104 kg
m3
M = 108 g
mol= 0.108
kg
mol
V =
1 m3
1.05 104 kg
0.108 kg
1 mol
1 mol
6.02 1023 atoms
= 1.71 1029 m3d = V
1
3
= 2.56 1010 m
The interatomic bond stiffness is
ks = Y d= (7.8 1010 N
m2)(2.56 1010 m)
= 20.0 N
m
Calculate the speed of sound in silver.
v =
k
s
ms
d
The mass of an atom is
ma
=
0.108
kg
mol
1 mol
6.02 1023 atoms
= 1.79 1025 kgatom
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
43/71
43
v =
20 Nm
1.79 1025 kg (2.56 1010 m)
= 2710 m/s
4.P.79
T = 2
m
k
The period does not depend on amplitude org. Increasingm by a factor of 6 increses the period by a factor of
6. Increasing
the stiffness by a factor of 10 changes the period by a factor of 110
. Thus, the period on the other planet will be
610 TEarth .
Thus,
Tplanet
=
6
10(2.1 s)
= 1.6 s
4.P.80
(a) To do this experiment, measure the initial unstretched length of the spring L0 with no mass on the spring. Add massto the end of the spring and measure the length L of the spring. Record both the total mass and length of the spring.Continue adding mass to the end of the spring. Each time, you should record both the total mass and length of thespring. Fill out Table 1with approximately 8 - 10 data points. Calculate the distance stretched s and the magnitudeof the force on the spring by the hanging mass for each data point.
GraphF
on spring by mass m
vs. s. Though the y-intercept may not be zero, as expected from Hookes law, the graph willbe linear as shown in Figure 27.
(b) To measure the period of oscillation, you can use a sonic ranger and computer data acquisition system such as a LabProby Vernier. However, you can also use a stopwatch. If you use a stopwatch, place a reasonable mass on the spring thatwill give a reasonable period to measure. You dont want the mass to be too small, or the period will be small andharder to measure. You dont want the mass to be too large, or the spring may stretch too far and become deformed.Choose a mass somewhere in the middle of the range that you used in part (a) of this experiment.
Pull the hanging mass downward a known, measured distance A and release it from rest. When the object later reachesits lowest point, start the stop watch. Count 10 complete oscillations and stop the stopwatch when the object reachesits starting point after its tenth oscillation.
Measure the total time for 10 oscillations and divide by 10 to get the time for one oscillation, which is the period. Notethat you can use any number of oscillations. You want to use enough oscillations that the small reaction time to startand stop the stopwatch is small compared to the total. However, if you use too many oscillations, then the oscillatorloses energy and the period may not be constant (i.e. in this case you are not controlling the variable of amplitude).10 oscillations is probably a reasonable number that is neither too small nor too large.
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
44/71
44
m (kg) L (m) s (m)F
on spring by mass m
(N)0 0 0
Table 1: Data for spring experiment.
0
1
2
3
4
5
6
7
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4
|F|(N)
s (m)
force on the spring vs. distance stretched
Figure 27: A sample graph ofF
on spring by mass m
vs. s.
Its a good idea to repeat this measurement of the period about 5 more times so can report the average and standardeviation. This gives you an idea about how repeatable the experiment is and how precise your measurements are.
Even if you use a LabPro, graph x vs. t, and determine the period from the graph, its a good idea to repeat t
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
45/71
45
experiment numerous times and report the average period and standard deviation.
(c) Double the amplitude and repeat your measurement of the period in part (b). Again, make multiple measurementsof the period and calculate the average and standard deviation. Note whether the periods are the same or different,within the uncertainty of your measurement of the period.
4.P.81Sketch a picture of the system, as shown in Figure 28.
x
m
Figure 28: A sketch of the system.
The distance of mass m from the center of the Earth is r. The position of mass m is x. The gravitational force on the mass
is
Fgrav,x
= mgR
x
F = < mgR
x, 0, 0>
Define the system to be the massm. Apply the Momentum Principle to the mass m. The only force on m is the gravitationalforce defined above. Write it in component form.
Fnet = dp
dt
Fnet,x
= dp
x
dt
mgR
x = mdv
x
dt
The instantaneous velocity of mass m is vx
= dxdt . Thus,
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
46/71
46
mgR
x = md
dt
dx
dt
mgR
x = md2x
dt2
d2x
dt2
+ g
R
x = 0
This looks like the equation of motion for an oscillating mass-spring system which is
d2x
dt2 +
k
mx = 0
Where =
km . By comparing the equations, you can see that they are the same. Thus, the mass m will oscillate back an
forth through Earth with an angular frequency given by
=
g
R
The time it takes the mass m to reach the other side is 12 the period. Find the period.
= 2f
= 2
T
T = 2
T = 2
R
g
Sotto get to the other side is
t = 1
2T
=
1
2 (2R
g)
=
R
g
4.P.82
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
47/71
47
(a)
T = 2
m
ks
If you double the mass m, the period increases by a factor
2. Thus the period, with two 20 gram masses on thespring, is
T =
2(1.2 s)
= 1.7 s
(b) The effective stiffness of identical springs in parallel is
ks,eff
= N ks
Thus, two springs in parallel have twice the stiffness of one spring. Since
T 1k
The period will be
12
(1.2 s) = 0.849 s
(c) Identical springs in series have a stiffness
1
keff
= N 1
ks
The stiffness of each spring is
ks
= N keff
If you cut a spring in half, the half-spring has a stiffness
ks
= 2keff
Doubling the stiffness results in the period
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
48/71
48
T = 1
2(1.2 s)
= 0.849 s
(d) Period is independent ofg ; therefore, the period of the oscillator on the Moon is the same as on Earth.
4.P.83
(a) Assume thatks 10 Nm . The mass of a hydrogen atom is
m =
1 g
mol
1 mol6.02 1023 atoms
= 1.66 1024 g= 1.66
1027 kg
f = 1
2
k
m
= 1
2
10 Nm
1.66 1027 kg 1 1013 Hz
Note that this does not take into account the fact that both H atoms in the diatomic molecule are oscillating.
(b) The mass of an oxygen atom is
m =
16 g
mol
1 mol6.02 1023 atoms
= 2.66 1024 g= 2.66 1026 kg
f = 1
2
k
m
= 1
2
10 Nm
2.66 1026 kg 3 1012 Hz
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
49/71
49
(c) Deuterium has twice the mass of hydrogen. Thus, sincef 1m
, its frequency will be
12
fhydrogen
= 1
2(1 1013 Hz)
0.7(1 1013 Hz)
7
1012 Hz
(d) The ratios of the frequencies is proportional to the inverse of the square root of the ratio of their masses, if their bondstiffnesses are the same. In this case, since both hydrogen and deuterium have the same charge in their nucleus, theyhave the same bond stiffness.
4.P.84
M = 59 g
mol
= 0.059 kg
mol =
8.9
g
cm3
1 kg1000 g
(100 cm)3
1 m3
= 8900 kg
m3
The cross-sectional area of the bar is
A = (2 mm)(2 mm)
= (2 103 m)(2 103 m)= 4 106 m2
To determine the time for a disturbance to travel down the rod, we need the speed of sound in nickel. To get the speed ofsound in nickel, we need to know its bond stiffness, the mass of a nickel atom, and the diameter of a nickel atom. Atomicmass and diameter are easy to calculate from the given properties of nickel, but the bond stiffness must be calculated fromYoungs Modulus which must also be determined from the given data.
The diameter of an atom is found from the volume of a cube taken up by the spherical atom. Assume a simple cubic latticefor nickel.
V = 1 m3
8900 kg8.509 kg
1 mol 1 mol
6.02 1023 atoms= 1.10 1029 m3
d = V1/3
= 2.2 1010 m
The mass of an atom of nickel is
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
50/71
50
m =
0.059
kg
mol
1 mol
6.02 1023 atoms
= 9.80 1026 kg
Youngs Modulus is given by
FT
A = Y
L
L
Y =F
T
A
L
L
The tension in the rod is equal in this case to the weight of the mass hanging from the rod, according to the MomentuPrinciple applied to the hanging mass. Thus,
FT
= mg
= (40 kg)(9.8 N
kg)
= 392 N
Y =
392 N
4 106 m2
2.5 m
1.2 103 m
= 2.04 1011 Nm2
The bond stiffnessks
is
ks
= Y d
= (2.04 1011 Nm2
)(2.22 1010 m)
= 45.3 N
m
The speed of sound in nickel is
v =
k
s
matom
datom
=
45.3 Nm
9.8 1026 kg (2.22 1010 m)
= 4770 m/s
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
51/71
51
Speed is defined as
v = |r|
t
The time interval to travel a distance 2.5 m down the rod is
t = |r|
v
= 2.5 m
4770 m/s
= 5.2 104 s
4.P.85
Approximate each atom in the bar to be a simple harmonic oscillator that oscillates with an angular frequency
=
k
s
m
whereks
is the bond stiffness in the material and m is the mass of an atom.
SinceU238 andU235 have the same number of protons and electrons (for a neutral atom), their bond stiffnesses are the same.However,U238 atoms have more mass and therefore will vibrate with less frequency.
The speed of sound in the material is
v =
k
s
md
Thus, the speed of sound in U238 will be less since U238 has more mass than U235.
4.X.86
Calculate the volume of1 kg of lead. Convert kg to grams.
V = 1000 g
11 gcm3
= 90.9 cm3
The buoyant force of air on the lead is equal to the weight of an equal volume of air. The density of air is approximately (at20 C and atmospheric pressure) 1.2 kgm3 . Thus, the weight of90.9 cm
3 of air is (be sure to pay attention to units)
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
52/71
52
wair
= mg
=
1.2
kg
m3
90.9 cm3
1 m3(100 cm)3
9.8
N
kg
= 0.0011 N
= 1.1
103 N
The buoyant force on the lead is a paltry 1.1 103 N. The weight of the lead object is 9.8 N. The ratio of the buoyaforce on the object to the weight of the object is approximately
103 N
10 N = 104
Thus the buoyant force is about one ten-thousandth the weight of the object. The buoyant force in thie case is clearnegligible.
4.X.87
Area A
Depth h
Volumeof water,V=hA
Figure 29: A sketch of the region
The pressure at depth h is the pressure at the top plus the weight of the volume of water divided by its area.
pbottom
= ptop
+mg
A
Multiply the last term by hh
pbottom
= ptop
+mgh
Ah
Ahis the volume and mass/volume is the density of the water.
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
53/71
53
pbottom
= ptop
+mgh
Vp
bottom= p
top+
watergh
The density of freshwater is 1000 kgm3 . Solve for h.
h =p
bottomp
top
water
g
= 3 105 Nm2 1 105 Nm2
(1000 kgm3 )(9.8 Nkg )
= 20.4 m
Saltwater has a greater density than freshwater. It is 1030 kgm3 . Thus in seawater,
h =p
bottomp
top
seawater g
= 3 105 Nm2 1 105 Nm2
(1030 kgm3 )(9.8 Nkg )
= 19.8 m
4.P.88
(a) Begin by sketching a picture of the floating block of wood (see Figure30).
AreaA
Heighth
Depthy
Figure 30: A sketch of the system.
A = (20 cm)(10 cm)
= (20 102 m)(20 102 m)= 2 102 m2
h = 6 cm = 0.06 m
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
54/71
54
wood
=
0.7 g
cm3
1 kg1000 g
(100 cm)3
1 m3
= 700 kg
m3
water
= 1.0 g
cm
3= 1000
kg
m
3
Define the system to be the block. Apply the Momentum Principle to the block. Draw a free-body diagram for tsystem, as shown in Figure31.
FB onblock by water
Fgrav onblock by Earth
Figure 31: A free-body diagram of the system.
Fnet
= p
t
The block remains at rest, so p = 0. Write the Momentum Principle in the y-direction.
Fnet,y
= 0
FB,y
+ Fgrav,y
= 0
FB,y
= Fgrav,y
= (mg)= mg
The mass of the block is
m = (700 kg
m3)(2 102 m2)(0.06 m)
= 0.84 kg
Thus,
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
55/71
55
FB,y
= (0.84 kg)(9.8 N
kg)
= 8.23 N
The buoyant force is also equal to the weight of water displaced by the wood. The volume of water displaced by thewood is
V = Ay
= (2 102 m2)y
The mass of this volume of water is
m = V
= (1000 kg
m3)(2 102 m2)y
= (20kg
m)y
The weight of this volume of water is
w = mg
= (20
kg
m)y(9.8
N
kg )
= (196 N
m)y
Thus,
FB
= wwater displaced
8.23 N = (196 N
m)y
y = 8.23 N
196 Nm= 0.042 m
= 4.2 cm
Note that this is less than the height of the block, 6 cm, as expected. Also, note that about 4.2 cm6 cm =.7 = 70% of theblock is submerged.
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
56/71
56
FBby air on blimp
Fgrav by Earth on blimp
Figure 32: A free-body diagram of the system.
(b) Define the system to be the blimp. Apply the Momentum Principle to the system. Assume the system is at rest. Draa free-body diagram, a shown in Figure32.
Fnet
= p
t
Fnet,y
=p
y
tF
B,y+ F
grav,y = 0
FB,y
= Fgrav,y
= (mg)= mg
Thus, the mass of the blimp is
m =F
B,y
g
w = 2.77 104 N
Thus, the magnitude of the buoyant force on the blimp is
FB
= 2.77 104 N
The total mass of the blimp is
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
57/71
57
m =
FB
g
= 2.77 104 N
9.8 Nkg
= 2827 kg 2830 kg
This should be greater than the mass of helium, which is
mHe
=
4 g
22.4L
1000L
1m3
1 kg
1000 g
(2356m
3)
= 421 kg
The mass of the material of the blimp including the gondola must be
2827 kg 421 kg = 2406 kg 2410 kg
To get FB,y
, use the fact that it is equal to the weight of air displaced by the blimp. The volume of the (cylindrical)
blimp is approximately
V = R2h
Whereh is the length of the blimp and R is its radius. Thus
V =
10 m
2
2(30 m)
= 2356 m3
The density of air is approximately 1.2 kgm3 . Thus, the mass of the equivalent volume of air is
mair
= (1.2 kg
m3)(2356 m3)
= 2827 kg
The weight of this volume of air is
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
58/71
58
w = mg
= (2827 kg)(9.8 N
kg)
= 27700 N
4.P.89
(a) Begin with a sketch of the system, as shown in Figure33.
L
s=L
Figure 33: A sketch of the system.
Define the system to be the mass m. Apply the Momentum Principle. Sketch a free-body diagram as shown in Figu
34.
FT on massby string
Fgrav on massby Earth
Figure 34: A free-body diagram of the system.
Define a coordinate system with the radial axis perpendicular to the objects path and directed toward the pivot anthe tangential axis tangent to the path, as shown in Figure 35
With this coordinate system, write Fgrav
using the right triangle shown in Figure 36.
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
59/71
59
FT
Fgrav
rad tan
Figure 35: The coordinate system.
Fgrav
Fgrav , rad
Fgrav , tan
Figure 36: The gravitational force vector described
Fgrav,tan
= F
grav
sin F
grav,rad=
Fgrav
cos The net force on the mass m is
Fnet
= Fgrav
+ FT
= < F
grav
sin , Fgrav
cos , 0>Where the first component is the tangential component and the second component is the radial component. Thus, the
only component of the net force in the tangential direction isF
grav
sin .
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
60/71
60
Write the Momentum Principle
Fnet
= dp
dt
Express this in the tangential direction and substitute for Fnet,tan
.
Fnet,tan
=dp
tan
dt
F
grav
sin = dptandt
Thus, sinceF
grav
=mg,dp
tan
dt
=
mg sin
Substitute = sL since arc length is s = L
dptan
dt = mg sin s
L
(b) For small angles,sin
dptan
dt mgsL
The tangential component of momentum is ptan
=mvtan
, wherevtan
= dsdt . Thus the Momentum Principle gives
dptan
dt = m
dvtan
dt = mgs
L
md
dt
ds
dt = mgs
L
md2s
dt2 = mg
s
L
d2sdt2
+ gL
s = 0
(c) Compare this to the Momentum Principle applied to a mass-spring system where
d2x
dt2 +
ks
mx = 0
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
61/71
61
and
=
k
s
m
The equations for the pendulum and the mass-spring system have the same form. Thus, for the pendulum
=
g
L
The period of the pendulum for small amplitude oscillations is given by
= 2
T
T = 2
= 2L
g
(d) A simple experiment can be constructed with a mass and string. Use a stopwatch to measure the time for 10 oscillations(or whatever number you choose). Measure t for N oscillations. ThenT = t/N. Calculate the period from the
theory,T = 2
Lg, and compare your experimental and theoretical results.
(e) 3-D graphics are not required for this simulation. The goal is to graphs vs. tand ptan
vs. t. It is useful to review thesimulation for an oscillating mass-spring system, such as the simulation for problem 2.P.72 for example. If you havenot written a simulation like the one in 2.P.72, then you may wish to write that one first.
This simulation is similar except it will not include 3-D graphics. Begin by defining importing necessary libraries anddefining important constants.
1 from __future__ import d i v i s i o n2 from v i s u a l import 3 from vi su al . graph import 4
5 m = 1 . 0 #m as s i n k g 6 g = 9. 87 t h e ta = 20 #t h e i n i t i a l a n g l e i n d e g r e e s 8 t h e ta = t h e t a p i / 1 8 0 #c o nv e rt t h e t a t o r a di a ns 9 L = 1 . 0 #l e n g t h o f pe nd ul um i n m et e rs
10 s = L t h e t a # i n i t i a l a r c l e n g t h 11 v_tan = 0 # i n i t i a l v e l o c i t y 12
13 p_tan = mv_tan14 t = 015 d t = 0 . 0 1
Now, create the graph window and the curve to be plotted.
1 from __future__ import d i v i s i o n2 from v i s u a l import 3 from vi su al . graph import 4
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
62/71
62
5 m = 1 . 0 #m as s i n k g 6 g = 9. 87 t h e ta = 20 #t h e i n i t i a l a n g l e i n d e g r e e s 8 t h e ta = t h e t a p i / 1 8 0 #c o nv e rt t h e t a t o r a di a ns 9 L = 1 . 0 #l e n g t h o f p en du lu m i n m et e rs
10 s = L t h e t a # i n i t i a l a r c l e n g t h 11 v_tan = 0 # i n i t i a l v e l o c i t y 12
13 p_tan = mv_tan14 t = 015 d t = 0 . 0 116
17 s Gr ap h = g d i s p l a y ( x =0 , y= 40 0 , w id th = 40 0 , h e i g h t = 30 0 , t i t l e = ' s v s . t ' , x t i t l e= 't ( s ) ' ,y t i t l e = ' s (m) ')
18 s P l o t = g c u r v e ( c o l o r =c o l o r . y e l l o w )
Now, create a while loop. Inside this loop, calculate the tangential component of the net force on the pendulum, updaits tangential momentum, and update the arclength s. For each data point, add (s, t)it to the curve being plotted.
1 from __future__ import d i v i s i o n2 from v i s u a l import 3 from vi su al . graph import 4
5 m = 1 . 0 #m as s i n k g 6 g = 9. 87 t h e ta = 20 #t h e i n i t i a l a n g l e i n d e g r e e s 8 t h e ta = t h e t a p i / 1 8 0 #c o nv e rt t h e t a t o r a di a ns 9 L = 1 . 0 #l e n g t h o f p en du lu m i n m et e rs
10 s = L t h e t a # i n i t i a l a r c l e n g t h 11 v_tan = 0 # i n i t i a l v e l o c i t y 12
13 p_tan = mv_tan14 t = 015 d t = 0 . 0 116
17 s Gr ap h = g d i s p l a y ( x =0 , y= 40 0 , w id th = 40 0 , h e i g h t = 30 0 , t i t l e = ' s v s . t ' , x t i t l e= 't ( s ) ' ,y t i t l e = ' s (m) ')
18 s P l o t = g c u r v e ( c o l o r =c o l o r . y e l l o w )19
20
21 while 1 :22 r a t e ( 1 0 0 0 )23 Fnet_tan =mg s i n ( s / L)24
25 p_tan = p_tan + Fnet_tan dt26 v_tan = p_tan/m
27 s = s + v _ta n
dt28
29 t = t+dt30
31 sPl ot . pl o t ( pos=(t , s ) )
In the example simulation above, the initial angle is 20, and the graph appears sinusoidal. You can change this angto larger angles to see that the resulting graph is no longer a sine or cosine function. Its especially obvious for aangle such as 179. Though the function is periodic, it will be much more rounded at the maxima and minima, wh
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
63/71
63
compared with a sine or cosine curve. A screen capture is shown in Figure 37.
Figure 37: A graph ofs vs. tfor a pendulum with an initial angle of179 from they direction.
You can use the angle and length to calculate the position of the pendulum and add 3-D animation to the simulation.In the example below, the sphere and the string (or massless rigid rod) are defined after the constants so that the angleand length of the pendulum can be used to calculate the position of the pendulum. In the while loop, the position ofthe pendulum and the axis of the rod must be updated after s is updated and after is calculated . The angle isnecessary for calculating the position of the pendulum.
1 from __future__ import d i v i s i o n2 from v i s u a l import 3 from vi su al . graph import 4
5 m = 1 . 0 #m as s i n k g 6 g = 9. 87 t h e t a = 17 9 # t h e i n i t i a l a n g l e i n d e g r e e s 8 t h e ta = t h e t a p i / 1 8 0 #c o nv e rt t h e t a t o r a di a ns 9 L = 1 . 0 #l e n g t h o f pe nd ul um i n m et e rs
10 s = L t h e t a # i n i t i a l a r c l e n g t h 11 v_tan = 0 # i n i t i a l v e l o c i t y 12
13 p_tan = mv_tan14 t = 015 d t = 0 . 0 116
17 b a l l = s p h e r e ( p o s =(L si n ( the ta ) ,L c o s ( t h e t a ) , 0 ) , r a d i u s = L / 1 0 , c o l o r =c o l o r . y e l l o w )18 r od = c y l i n d e r ( p o s = ( 0 , 0 , 0 ) , a x i s =b a l l . p os , r a d i u s = L / 1 00 , c o l o r =c o l o r . w h it e )19
20 s Gr ap h = g d i s p l a y ( x =0 , y =4 00 , w i dt h = 40 0 , h e i g h t = 30 0 , t i t l e = ' s v s . t ' , x t i t l e= 't ( s ) ' ,y t i t l e = ' s (m) ')
21 s P l o t = g c u r v e ( c o l o r =c o l o r . y e l l o w )22
23
24 while t< 30:25 r a t e ( 1 0 0 )26 Fnet_tan =mg s i n ( s /L )27
28 p_tan = p_tan + Fnet_tan dt29 v_tan = p_tan/m30 s = s + v _ta ndt
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
64/71
64
31
32 t h e ta = s /L33 b al l . pos=(L si n ( the ta ) ,L cos ( theta ) , 0)34 r o d . a x i s = b a l l . p o s35
36 t = t+dt37
38
sPl ot . pl o t ( pos=(t , s ) )Adjust the initial angle to view the resulting motion and graph for both small angle oscillations and large angoscillations.
4.P.90
When starting a simulation like this, it helps to define some useful constants such as:
M = the total mass of the rod
L = the total length of the rod
N = the number of atoms in the modeld = the diameter of each atom (for display purposes only)
m = the mass of each atom
L0 = the equilibrium length of each bond
k = bond stiffness
All quantities in the simulation should be based on the constants defined above. As a result, you can change the number atoms, N, for example, and see how it affects the measured speed of sound.
Begin your program by importing libraries and defining your constants. Also, you can print the theoretical value for tspeed of sound. Note that:
vsound =
ks
mad
whereks is the bond stiffness, ma is the mass of an atom, and d is the diameter of an atom. We are assuming that the atomare closely packed as shown in Figure38.
Figure 38: A model of a rod as a one-dimensional line of closely packed atoms.
If the rod is made one atom, then the atomic diameter is d = L. If the rod is made of two atoms, then the atomic diametis d = L/2. Thus, in general, d = L/N. This the diameter that should be used in calculating the theoretical value of tspeed of sound in the rod.
So, the first part of our program looks like the example shown below. If you run it, it will print the theoretical value fthe speed of sound in the rod. You can increase the value ofNwhich presumably improves the accuracy of the value of thspeed of sound. Note that the constant d that is calculated in the constants will be the diameter used to draw the atom. Iused for display purposes to make the simulation look good but it not physical, meaning that its the actual diameter usin the calculation, which is d = L/N.
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
65/71
65
from __future__ import d i v i s i o nfrom v i s u a l import from v i s u a l . g r ap h import
M = 1N = 1 0L = 1
d = 0 . 5
L/Nm = M/NL0 = L/(N1)k = 10
v_theor = sq rt (k/m) L/Nprint " t h e o r e t i c a l s p e e d o f s o un d i s " , v _t h eo r
Now, because we are creating N number of atoms and N-1 number of bonds in our simulation, we will need to store themin a list, which is Pythons construct for an array. So, initialize the lists for the atoms and springs (i.e. bonds) and use forloops to create the atoms and springs used in the simulation. Spread the atoms out evenly along the rod with the left endatx = L/2 and the right end at x = +L/2. Also, initialize the net force, velocity and momentum vectors for each atom inthe list. Run the example simulation below to see a 3-D picture of the rod, with atoms and springs.
from __future__ import d i v i s i o nfrom v i s u a l import from v i s u a l . g r ap h import
M = 1N = 1 0L = 1d = 0 . 5L/Nm = M/NL0 = L/(N1)k = 10
v_theor = sq rt (k/m) L/Nprint " t h e o r e t i c a l s p e e d o f s o un d i s " , v _t h eo r
atom s= [ ] s p r i n g s = [ ]
fo r i in ran ge (0 ,N) : atom = sph ere ( pos=(L/2+ i L/(N1) , 0 , 0 ) , r a d i u s = d / 2 , c o l o r =c o l o r . w h i te ) atoms . append ( atom)
fo r i in range (0 ,N1) : atom=atoms [ i ] b on d = h e l i x ( p o s=at om . p os , a x i s = (L0 , 0 , 0 ) , c o l o r = ( 1 , 0 . 5 , 0 ) , r a d i u s = d / 4 )
sp ri ng s . append( bond)
fo r i in ran ge (0 ,N) : atom=atoms [ i ] atom . Fnet = vec tor (0 , 0 , 0) atom . v = v e c t o r ( 0 , 0 , 0 ) atom . p = matom . v
Disturb the left end by displacing the atom on the left side of the rod to the left about half a bond length. This is the
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
66/71
66
initial disturbance that will propagate down the rod. Also, define the time step and initialize the graph. I prefer to use tcommand scene.mouse.getclick()" to make the simulation pause at this part of the program. Then, Ill have to click on thsimulation window to make it start. This gives me a chance to zoom in, rotate, move the graph window, etc. before tsimulation starts. In the example below, I also define a boolean (i.e. true/false) variable disturbenceReachedEndthat I wuse to mark the instant that the disturbance reaches the right end of the rod. See the example below.
1 from __future__ import d i v i s i o n2 from v i s u a l import 3 from v i s u a l . g r ap h import 4
5 M = 16 N = 1 07 L = 18 d = 0 . 5L/N9 m = M/N
10 L0 = L/(N1)11 k = 1012
13 v_theor = sq rt (k/m) L/N14 print " t h e o r e t i c a l s p e e d o f s o un d i s " , v _t h eo r15
16 atoms=[]17 s p r i n g s = [ ]18
19 fo r i in ran ge (0 ,N) :20 atom = sph ere ( pos=(L/2+ i L/(N1) , 0 , 0 ) , r a d i u s = d / 2 , c o l o r =c o l o r . w h i te )21 atoms . append ( atom)22
23 fo r i in range (0 ,N1) :24 atom=atoms [ i ]25 b on d = h e l i x ( p o s=a tom . p os , a x i s = (L0 , 0 , 0 ) , c o l o r = ( 1 , 0 . 5 , 0 ) , r a d i u s = d / 4 )26 sp ri ng s . append( bond)27
28 fo r i in ran ge (0 ,N) :29 atom=atoms [ i ]30 atom . Fnet = ve cto r (0 , 0 , 0)31 atom . v = v e c t o r ( 0 , 0 , 0 )32 atom . p = matom . v33
34 atoms [ 0 ] . pos . x = atoms [ 0 ] . pos . x L0/535 d t = 0 . 0 0 136 t = 037
38 x Gr aph = g d i s p l a y ( x = 0, y =4 00 , w i dt h = 40 0 , h e i g h t = 30 0 , t i t l e = ' x v s . t f o r atom a t r i g h t en do f t h e r o d ' , x t i t l e= 't ( s ) ' , y t i t l e= ' x(m) ')
39 x P l o t = g c u r v e ( c o l o r =c o l o r . y e l l o w )40
41 s c e n e . m ou se . g e t c l i c k ( )42
43 d i s t ur b e n ce R e a ch e d E nd = f a l s e
Now, were ready for the while loop. In the first part of the loop, calculate the force on the atom on the left side of trod. Assume that the spring is attached to the centers of the atoms. Define the vector L to point parallel to the spring antoward the atom that is our system, as shown in Figure 39.
The force by the spring on this atom is
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
67/71
67
Atom 0
Figure 39: Calculating the force on the atom on the left end of the rod.
Fby spring on left atom
= kssL
wheres =L L0 and is the distance the spring is stretched or compressed. If the spring is stretched s is positive. If it is
compressed,s is negative. So, if the spring is stretched,Fby spring
is opposite L, and if the spring is compressed, Fby spring
is
in the same direction as L.
Note that the first atom is atom[0] in the list and the second atom is atom[1] in the list and so on. Heres the while loopand its first part that calculates the force on the left atom. The vector L is called L01 in the program, meaning the vector
from atom 1 to atom 0.while 1 :
# f i r s t at om on l e f t e nd o f b a r
L01 = atoms [ 0 ] . pos atoms [ 1 ] . po s L01_mag = mag( L01 ) L01_hat = L01/L01_mag s = L01_mag L0 atoms [ 0 ] . Fnet =k s L01_hat
For the atom on the right end of the bar, the spring is toward the left as shown in Figure 40.
Atom 9
Figure 40: Calculating the force on the atom on the right end of the rod.
Thus, the vector L points to the right, again toward the atom that is the system. This atom is atom[9] if N=10. But ingeneral, this atom is atom[N-1]and its nearest neighbor is atom[N-2]. The code to calculate the force on this atom is shownbelow.
while 1 : # f i r s t at om on l e f t e nd o f b a r
L01 = atoms [ 0 ] . pos atoms [ 1 ] . po s L01_mag = mag( L01 ) L01_hat = L01/L01_mag s = L01_mag L0 atoms [ 0 ] . Fnet =k s L01_hat
# l a s t a tom on r i g h t end o f ba r
L l a s t = a to ms [ N1 ] . p o s atoms [N2 ] . p o s Llast_mag = mag( Ll as t ) Ll ast_ hat = Ll a st /Ll ast_m ag
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
68/71
68
13 s = Llast_mag L014 atoms [N1 ] . F n e t =k s Ll ast_ hat
Now that we know how to calculate the force on an atom due to a spring on its right side and the force on an atom due tospring on its left side, we can calculate the forces on the middle atoms in the rod. See the while loop below. You will notithat the force due to the spring on the left and the force due to the spring on the right are added together to get the neforce on the atom.
1 while 1 :2 # f i r s t atom on l e f t e nd o f b a r
3 L01 = atoms [ 0 ] . pos atoms [ 1 ] . po s4 L01_mag = mag( L01 )5 L01_hat = L01/L01_mag6 s = L01_mag L07 atoms [ 0 ] . Fnet =k s L01_hat8
9 #a to ms i n t h e m i d dl e
10
11 fo r i in range (1 ,N1) :12 Lr igh t = atoms [ i ] . pos atoms [ i +1]. pos13 Lright_mag = mag( Lr ig ht )14 Lri ght_ hat = Lri g ht /Lright_ mag15 s = Lright_mag L016 F r i gh t =k s Lri ght_ hat17
18 Ll ef t = atom s [ i ] . pos atoms [ i1 ] . p o s19 Lleft_mag = mag( Ll e ft )20 L l e f t _ h a t = L l e f t / L l ef t_ ma g21 s = Ll ef t_m ag L022 F l e f t =k s L l e f t _ h a t23
24 a to ms [ i ] . F ne t = F r i g h t + F l e f t25
26 # l a s t a tom on r i g h t end o f ba r
27 L l a s t = a to ms [ N1 ] . p o s atoms [N2 ] . p o s28 Llast_mag = mag( Ll as t )29 Ll ast_ hat = Ll ast /Ll ast_m ag30 s = Llast_mag L031 atoms [N1 ] . F n e t =k s Ll ast_ hat
After computing the net force on each atoms, then you need to update their momenta and positions and you need to updathe springs positions and axes, just for display purposes. The rest of thewhilestatement updates the graph and calculatand prints the measured speed of sound when the disturbance reaches the last atom. The entire program is shown below.
1 from __future__ import d i v i s i o n2 from v i s u a l import 3 from v i s u a l . g r ap h import 4
5 M = 16 N = 1 07 L = 18 d = 0 . 5L/N9 m = M/N
10 L0 = L/(N1)11 k = 1012
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
69/71
69
v_theor = sq rt (k/m) L/Nprint " t h e o r e t i c a l s p e e d o f s o un d i s " , v _t h eo r
atom s= [ ] s p r i n g s = [ ]
fo r i in ran ge (0 ,N) :
atom = sph ere ( pos=(L/2+ i
L/(N1) , 0 , 0 ) , r a d i u s = d / 2 , c o l o r =c o l o r . w h i te ) atoms . append ( atom)
fo r i in range (0 ,N1) : atom=atoms [ i ] b on d = h e l i x ( p o s=at om . p os , a x i s = (L0 , 0 , 0 ) , c o l o r = ( 1 , 0 . 5 , 0 ) , r a d i u s = d / 4 ) sp ri ng s . append( bond)
fo r i in ran ge (0 ,N) : atom=atoms [ i ] atom . Fnet = vec tor (0 , 0 , 0) atom . v = v e c t o r ( 0 , 0 , 0 ) atom . p = matom . v
atoms [ 0 ] . pos . x = atoms [ 0 ] . pos . x L0/5d t = 0 . 0 0 1
t = 0
xG rap h = g d i s p l a y ( x = 0, y =4 00 , w i dt h = 40 0 , h e i g h t = 30 0 , t i t l e = ' x v s . t f o r atom a t r i g h t en do f t h e r o d ' , x t i t l e= 't ( s ) ' , y t i t l e= ' x(m) ')
x P l o t = g c u r v e ( c o l o r =c o l o r . y e l l o w )
s c e n e . m ou se . g e t c l i c k ( )
d i s t ur b e n ce R e a ch e d E n d = f a l s e
while 1 : # f i r s t at om on l e f t e nd o f b a r
L01 = atoms [ 0 ] . pos atoms [ 1 ] . po s L01_mag = mag( L01 ) L01_hat = L01/L01_mag s = L01_mag L0 atoms [ 0 ] . Fnet =k s L01_hat
#a to ms i n t h e m i d dl e
fo r i in range (1 ,N1) : Lr igh t = atoms [ i ] . pos atoms [ i +1]. pos Lright_mag = mag( Lr ig ht )
Lri ght_ hat = Lri g ht /Lright_m ag s = Lright_mag L0 F r i gh t =k s Lri ght_ hat
Ll ef t = atom s [ i ] . pos atoms [ i1 ] . p o s Lleft_mag = mag( L le f t ) L l e f t _h a t = L l e f t / L le ft _m ag s = Ll ef t_m ag L0
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
70/71
70
66 F l e f t =k s L l e f t _ h a t67
68 a to ms [ i ] . F ne t = F r i g h t + F l e f t69
70
71 # l a s t a tom on r i g h t end o f ba r
72 L l a s t = a to ms [ N1 ] . p o s atoms [N2 ] . p o s73
Llast_mag = mag( Ll as t )74 Ll ast_ hat = Ll ast /Ll ast_m ag75 s = Llast_mag L076 atoms [N1 ] . F n e t =k s Ll ast_ hat77
78 # u p d a t e momentum a nd p o s i t i o n o f e a c h at om
79 fo r i in ran ge (0 ,N) :80 atoms [ i ] . p = atoms [ i ] . p + atoms [ i ] . Fnet dt81 atoms [ i ] . v = a toms [ i ] . p/m82 atoms [ i ] . pos = atoms [ i ] . pos + atoms [ i ] . v dt83
84 # u p da t e e ac h s p r i n g
85 fo r i in range (0 ,N1) :86 s p r i n g = s p r i n g s [ i ]87 sp ri ng . pos = atoms [ i ] . pos88 spr i ng . ax i s = atom s [ i + 1] . pos atoms [ i ] . pos89
90 t = t+dt91
92 xPl ot . pl ot ( pos=(t , atoms [N1 ] . p o s . x ) )93
94 # c he ck i f t he d i st u r be n c e r e ac he d t h e r i g h t end o f t he rod
95 i f mag(atoms [N1 ] . F n e t ) > 0 . 0 0 1 :96 # c a l c u l a t e and p r i nt t he s pe ed
97 i f ( di sturbenceReachedEnd == f a l s e ) :98 v_meas = L/t99 print "measureds peed of sound i s " , v_meas
100 d i s t ur b e n ce R e a ch e d E nd = t r u e
You can increase the number of atoms N to improve the accuracy of the simulation. Some theoretical and measured valufor the speed of sound in this rod for various values ofNare shown in Table2. (In this case, L = 1 m,m = 1 kg,k = 10 N/mYou will notice that the accuracy improves with greater N.
N (# of atoms) vtheoretical (m/s) vmeasured (m/s)10 1.0 1.83220 0.707 0.92330 0.577 0.67940 0.500 0.55750 0.447 0.48260 0.408 0.42870 0.378 0.37580 0.354 0.35090 0.333 0.329
100 0.316 0.312
Table 2: Results of the speed of sound simulation for L = 1 m, m = 1 kg, k = 10 N/m.
-
7/25/2019 Matter and Interaction Chapter 04 Solutions
71/71
71
You will notice that the best accuracy actually occurs at about 70 atoms. This is because the measured speed for low N isgreater than the theoretical value. For N > 70, the measured speed is less than the theoretical value. Somewhere aroundN= 70 is the transition.