Week+6 (1)

Physics 2211: Matter and InteractionsChapter 4: Contact Interactions

http://www-math.mit.edu/~dhu/Striderweb/striderweb.html

http://ftp.aip.org/epaps/appl_phys_lett/E-APPLAB-96-038950/http://apl.aip.org/applab/v96/i4/p043701_s1


What was particularly confusing or difficult? I don't really understand Young's modulus at all, and I

don't understand what its significance is. The way the book explains how to calculate the length of an inter-atomic bond

is kind of complicated and confusing. I hope Dr. Greco goes over this in lecture. CHALLANGE ACCEPTED

I would like more explanation on the idea of stress and strain Applications of Young's modulus and when it ceases to be

relevant/useful. It was hard to grasp the last section on compression forces. In the example, a

brick sitting on a table compresses the atoms of the table, but this is hard to picture in real life.

Nothing was too difficult understand, I just felt that the book was jumping from one topic to another. Are we just surveying all forces? How does it all relate to our goal in the class?

I've learned about Young's Modulus and stress and strain, but I never learned it through the idea of atoms.


Model of a solid: All matter consist of atoms: ``uncuttable Democritus (460 BC - 370 BC)

1.Atoms attract each other when theyare close but not too close.

2.Atoms repel one another whenthey get too close to each other

3.Atoms in solids, liquids andgases keep moving even atvery low temperatures

=11010 m

Physics 2211: Matter and InteractionsChapter 4: balls and springs

A chemical bond between two atoms actslike a spring

Equilibrium

Close

Too Close

Ball-Spring Model


Everyday solids are made up of millions and millions of atoms (balls and springs) At room temperature the

atoms are in motion in the form of lattice vibrations.

The cubic lattice of a solid is not very stable but gives good approximations.

Ball and Spring Cubic Lattice


Clicker: How does the diameter of one atom in a solid compare to the length of an atomic bond in our model(1) The bond length is greater than the atomic diameter(2) The bond length is less than the atomic diameter(3) They are the same


Tension From our previous analysis of tension we

determined that the larger the mass the great the tension in the wire

Does this fit with our model of a solid? Replace the wire with a long thin chain of

atoms Assume a massless springs The larger the mass the bigger the

stretch!

Fwire=mg

Physics 2211: Matter and InteractionsChapter 4: interatomic bond length

To make qualitative predictions with the Ball and Spring model requires application of Hooke's law1. What type of lattice should we choose

(how are the balls packed together)

2. What is the length of an individualspring for a given type of atom(how many springs are there)

3. What is the stiffness of anindividual spring for a given atom(what is the effective springstiffness of the solid)

Simple Cubic System


Clicker: The mass of one copper atom is m and the The density of solid copper is . What is the volume occupied by one atom of copper in a solid?(1) m*(2) m/(3) m*()^3(4) /m


To determine atomic spacing requires three things: Density, Atomic Mass and Avogadro's Number The density of a material is independent of size and shape!

matom=atomic mass

N A

N A=6.021023atoms /mole

=matomd 3

d= matom 1/3


Example: The atomic spacing of gold atoms? The density is usually given in the problem state or

obtained from a reference (book, web, CRC, etc...)

The Atomic Mass is found from the periodic table

gold=19.3103 kg /m3

mgold=0.197 kg /mole

6.021023atoms /mole=3.271025 kg

atom

atomic mass=196.97 g /mole

d gold= 3.271025 kg /atom19.3103 kg /m3 1/3

d gold=2.571010m

http://www.youtube.com/watch?v=rSAaiYKF0cs


Reading Question: Why density? This isn't Chemistry class! Density is a very powerful concept Consider the man of steel, Superman.

Born on the planet Krypton andendowed with the strength hismassive planet bequeathed to him.

Could there be such a planet?Density can give us the answer!

Assume golden age superman whocould only jump over a building

Best analysis wins a prize!Must be submitted before Test 2.

Physics 2211: Matter and InteractionsChapter 4: interatomic bond strength

Hooke's Law requires information about: The rest length of the spring The spring stiffness

At this level we do not possess thetools to microscopically measure thestrength of an individual bond betweentwo atoms Instead, we need a way to relate a

macroscopic measurement of how asolid responds to deformation to themicroscopic quantity (bond strength)

This will require a method fordetermining how springs worktogether in a group


Springs in Series (end-to-end) If we hang a weight from one spring the spring constant is found from

the momentum principle

Two identical springs end-to-endact just a like a single spring thatwas twice as long as the first andstretched twice as far

The effective spring constant forthis spring can again be determinedby the momentum principle

L 0s

2L0

2s

L 0s

L 0s

F Earth=F Springk s=mg / s

F Earth=F Springk s '=mg /2s k s '=k s /2


Springs in Parallel (side-by-side) If we hang a weight from one spring the spring constant is found from

the momentum principle

Two identical springs side-by-sidewith each support half of the weightand stretch only half has far

The effective spring constant forthis spring is determined by themomentum principle

F Earth=F Springk s=mg / s

F Earth=F Springs /2k s '=mg=ks k s '=2k s

L 0s

L 0s/

2

L 0s/

2


Double the length and half the stiffness

Half the length and double the stiffness

k s '=k s /2

k s '=2k s


Clicker: You hang a 1 kg mass from a spring, which stretches 0.4 m. You link the spring end-to-end with another identical spring, and hang a 1 kg mass from the linked springs. How much does this longer spring stretch?(1) 0.16 m(2) 0.2 m(3) 0.4 m(4) 0.8 m(5) It doesn't stretch at all(6) I don't know, I am still thinking about gum


Clicker: You hang a 1 kg mass from a spring, which stretches 0.4 m. You place a second identical spring beside the first, so the 1 kg mass is now supported by two springs. How much does each spring stretch?(1) 0.2 m(2) 0.4 m(3) 0.5 m(4) 0.8 m(5) It doesn't stretch at all


How does this generalize to N springs In series the effective spring constant decreases with each

spring we add The longer a solid gets the softer it feels The shorter a solid gets the stiffer it feels

In parallel the effective spring constant increases with each spring we add

The thinner a solid gets the softer it feels The fatter a solid gets the harder if feels

k series=k s /N k parallel=k s N


Group Problem: You hang a 1 kg mass from a spring, which stretches 0.4 m. You place a second identical spring beside the first and attach a third identicalspring below these. You hang a 1 kgmass from the bottom spring, how fardoes the mass move?(1) 0.2 m(2) 0.4 m(3) 0.6 m(4) 0.8 m(5) 1.2 m(6) It doesn't stretch at all


Now that we know how springs work together we are prepared to combine our addition rules with the cubic lattice model

This will relate the macroscopic spring stiffness in a solid to the microscopic spring stiffness (bond strength)

How many springs can fit down thelength of the wire (a chain)

The effective spring stiffness of a chain

# springs=L /d atomic

k chain=k atomic# springs

=k atomicd atomic

L


How many of these chains are there in a wire? The same as the number of atoms on the bottoms surface of

the wire The number of atoms is found by looking at the cross-

sectional area; the area of a flat slice of the wire

The effective spring stiffness of thesechains working in parallel

#chains=AwireAatom

=Awired atomic

2

k wire=k chain#chains=k atomic#chains# springs

k wire=k atomicAwireLwire

1d atomic

Physics 2211: Matter and InteractionsChapter 4: young's modulus

This is Amazing! By making macroscopic measurements of a solid and applying the ball and spring model, we can determine the bond strength between two atoms - a microscopic quantity From this relationship we can see how a change in the

shape of a wire will affect the macroscopic spring stiffness

A long wire is softer than a short wire A fat wire is more stiff than a skinny wire Demo time

The microscopic result should be true regardless of: the cross sectional area of the wire, the length of the wire, the size of the weight used to stretch the wire

k wire=k atomicAwireLwire

1d atomic


Recall from the tension example that the spring stiffness of the wire was found by satisfying the momentum principle

Substituting into the spring stiffness relationship

Young's Module is the ratio of stressto strain and does not depend on thesize or shape of the material

k atomicd atomic

=F T L

LwireAwire

Ystressstrain

=F T / A L/ L=

k atomicd atomic

k wire=Forcetension

LForcetension=ForceEarth


Clicker: Two wires with equal lengths are made of pure copper. The diameter of wire A is twice the diameter of wire B. When 6 kg masses are hung on the wires, wire B stretches more than wire A. You make careful measurements and compute Young's modulus for both wires. What do you find?(1) Y

A > Y

B

(2) YA = Y

B

(3) YA < Y

B

Physics 2211: Matter and InteractionsChapter 4: Compression

Compression Forces (Contact Forces) Contact forces only apply to points of contact between system

and surroundings Contact forces can push (compression) or pull (tension)

How does the metal table exert a force on the brink?


Example: Mr. Louis Cyr supports the combined weight of 18 men (19,127 N). Each of his thighbones (femur) have a length of 0.55 m and an effective cross-sectional area of 5.7e-4 m2.

How much doeseach one of histhigh bones compress?


Example: Solution Young's Modulus: For compressed Bone = 9.4e9 N/m2

L=9,563.5 N /5.7104 m2

9.4109 N /m2/0.55 m

L1 mm

Y=F T /A L / L

L=FT /AY / L

F T=19127 /2 N

Physics 2211: Matter and InteractionsChapter 4: Friction

Two solids in contact and moving against each other feel a friction forces resisting this movement This is directly related to the compression force If we apply a force to a brick laying on

a table, the brick will run into anuncompressed part of the table

This force is parallel to the tableand called the frictional force

All of this compressing and decompressingof the bonds imparts energy into the table

It heats up! We call friction a dissipative process


Leonardo da Vinci (1452-1519): The first person to qualitatively study the problem of friction He focused on all kinds of friction

and drew a distinction betweensliding, static and rolling friction

is the coefficient of friction and willhave a different value for static/kinetic

FN is the normal force or the force

compressing the two surface together

f sliding=k F N

f static s F N


Some useful properties of friction1.The force of friction is proportional to the compression force2.The force of friction is independent of the area of contact3.The force of friction is independent of the sliding velocity

4.The static friction force can have any value < F5.The friction force depends on the direction of motion,

making friction a non-conservative force


Clicker: A block of mass M is dragged along a table by a string at a constant speed v. The string makes an angle with the table and the coefficient of friction between the block and the table is . What is the tension T in the string?(1)

(2) (3) (4) (5)

mg cos

mg sin

mg /sin

mg /sin cos

mg / sincos


Clicker: Solution Use the momentum principle

Solve for T

F net=0

T cos= F NF NT sin =mg

F N=T /cos

T / cosT sin =mg

T=mg / sincos

mg

FN

FN

T


Example: Imagine a penguin of mass M dropped on a conveyor belt moving with speed v. When the penguin first lands on the belt there is slippage, until the speed of the penguin catches up to the belt. If the coefficient of friction between penguin and belt is , what is the distance d the penguin travels before reaching speed v?

d

vi = 0

v

v

vf = v


Example: Solution System: Penguin

Surroundings: Belt + EarthInitial State: Penguin meets beltFinal State: Penguin moves at speed va distance d away

d

vi = 0

v

v

vf = v

FN

mg

FN

Force Diagram

p=F net t

mv f=F Nv t

mv f=mgv t

v f=gv t


Example: Solution How to find the change in time?

d

vi = 0

v

v

vf = v

FN

mg

FN

Force Diagram

v=g 2dv

r=v avg t

d v=v f0

2 t

2dv= t

d= v2

g 2

Physics 2211: Matter and InteractionsChapter 4: Speed of Sound

When we considered the stretch or compression of a solid we assumed that the change in shape was uniform and instantaneous. In reality, this change is not instantaneous

We call the speed with which the propagation travelsthe speed of sound

Physics 2211: Matter and InteractionsChapter 4: Speed of Sound

We already know enough to compute the speed of propagation through our ball and spring model using Vpython

For each atom (loop):1) Find the stretch2) Find the net force3) Update the momentum4) Update the position5) Update t = t + t

VPython

Physics 2211: Matter and InteractionsChapter 4: spring-mass system

To determine an analytical expression for the speed of sound requires additional information about spring mass systems Place the origin at

the equilibrium position

Use the momentum principle

What function is equal to itssecond derivative

F net=k s x x

mdv xdt=k s x

d 2 xdt2

=k sm

x

x t =Acos t


If we substitute this expression into the equation of motion for a spring mass system we find A stands for Amplitude and is equal to the initial displacement

Omega is the angularfrequency and is dependenton the spring stiffnessand mass of the block

x t =Acos t

A=x 0

=k s /m

=2/T


Clicker: Suppose the period of a spring-mass oscillator is 1 s. What will be the period if we double the mass?(1)T = 0.5s(2) T = 0.7s(3) T = 1.0s(4) T = 1.4s(5) T = 2.0s

=k s /m=2 /T


Clicker: Suppose the period of a spring-mass oscillator is 1 s with an amplitude of 5 cm. What is the period if we increase the amplitude to 10 cm, so that the total distance traveled in one period is twice as large?(1)T = 0.5s(2) T = 0.7s(3) T = 1.0s(4) T = 1.4s(5) T = 2.0s

=k s /m=2 /T

Physics 2211: Matter and InteractionsChapter 4: analytical speed of sound

Use dimensional analysis to guess at the correct answer for the speed of sound Characteristic distance and time

Speed of Sound

vsound=d atomic k atomicmatomic

d scale=d atomic

t scale=1/

vscale=datomic


Example: Tension revisited How long does it take for a gold wire to stretch once a 10 kg mass is hung from the end. The length of the wire is 2 m and a 0.001 m wide. The wire stretches 2.48 mm.

Measured vsound=2030m /s percenterror=2023m / s2030m / s

2030m / s100%=0.34%

Y=F T / A L /L

=109.8 /0.00120.00248 /2

=7.91010 N /m2

vsound=d atomic katomicmatomic= d atomic3 Y

matomic

T=L

vsound=

2 m2023 m / s

=9.9104 s

vsound= Y= 7.9101019300 =2023 m / s


SRS: A slinky is held from one end and dropped. What does the slinky do? http://www.youtube.com/watch?v=wGIZKETKKdw

Answer: http://www.youtube.com/watch?v=eCMmmEEyOO0

What would happen if I hung two identical spring from the ceiling and cut each oneat the same time; one at theceiling, one at the ball?Which ball would hit theground first?

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