matter and interaction chapter 04 solutions

7/25/2019 Matter and Interaction Chapter 04 Solutions

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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.


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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


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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


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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


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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


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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


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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


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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.


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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


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(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


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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


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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


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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


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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).


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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


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(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,


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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


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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.


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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


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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

.


Fnet

= dp

dt

F2

= m3

dvx

dt

Substitute the acceleration of the system.


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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

.



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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


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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


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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


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


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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


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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


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



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.


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


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.


30/71

30

Fonmassby spring


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.


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


Fnet

= dp

dt

At the instant just before it starts to move,

Fnet

= 0

In the y-direction,


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.


33/71


34/71


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.


36/71

36

m m m

x=s x=sx=0

Figure 24: A sketch of the system

Fbyspring onmass


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


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


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


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


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.


41/71

41

FT by rod onball

Fgrav by earth on ball


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


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


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


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.


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


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,


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


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


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


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


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


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)


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.


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


A = (20 cm)(10 cm)

= (20 102 m)(20 102 m)= 2 102 m2

h = 6 cm = 0.06 m


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


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,


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.


56/71

56

FBby air on blimp

Fgrav by Earth on blimp


(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


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


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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


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


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.


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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 .


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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


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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.



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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


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.


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


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


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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.


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


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


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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.


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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


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


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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


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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


# 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


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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


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69


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


#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


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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.


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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.

matter and interaction chapter 04 solutions

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