Fall 2009PHYS 172: Modern Mechanics

Lecture 13 – Friction, Driven Oscillations, and Resonance Read 6.10 – 6.14

CLICKER QUESTION #1Reading Question (Sections 6.10 – 6.14)

(This is a closed-book quiz, no consulting with neighbors, etc.)

You throw a ball straight up, and when it comes down you catch it.Consider the round trip from when it left your hands to when it returnedto the same height (before being caught). Choose the correct answer:

positivenegativeD

positivezeroC

negativezeroB

zeropositiveA

Work done by airWork done by gravity

NEXT WEEK:

No labs

No lectures

There is still recitation

Exam #2 is Wednesday, October 14 in Elliot Hall at 8 pm

Same format as before (all about Energy!)

Bring pencils, ID, calculator (graphing is ok)

Alternative testing in room 8 of Physics building at same time

Today’s Lecture• Friction and Energy Dissipation

• Driven Oscillations and Resonance

A horse of mass M gallops with constant speed v up a long hill ofheight h and horizontal extent d. Choose the horse as the system toanalyze. Start from the energy principle: !Esys =Wext +Q

We found that:1. Forces are exerted on the system by the Earth (gravity), the ground

(normal force), and air (resistance, friction).

2. The normal force does no work, because there is no displacement of thehooves.

3. The work done by the Earth is -Mgh.

4. Q<0 because the horse is hotter than the air, so heat energy leaves thesystem.

!Esys =Wext +Q = "Mgh " Q

Recap of galloping horse:

Concerning the energy of the system (horse):

1. E= Mc2 + K + Echemical + Ethermal

2. Only Echemical will change

chemicalE Mgh Q! = " "

sysE W Q! = +

We see that the stored chemical energy of the horsemust decrease.

Instead, choose the Universe as the system:

1. Now the system is closed, so W+Q=0.2. In addition to Echemical, now ΔE must include gravitational

potential energy and change in energy of the air(heating).

!Echemical

+ Mgh + !Eair= 0

Compare with system = horse:

We see that

which makes sense; the energy transfer from horseto air in the form of microscopic work is equal to theenergy increase of the air.

!Echemical

= "Mgh " Q

!Eair= Q

Energy Dissipation• As book slides across table, the book’s K → 0. Was this

energy lost?

• No - energy is NEVER lost.

• K of macroscopic object was converted into K and U ofthe microscopic atoms (thermal energy).

• This is an example of energy dissipation

Conservative forces• fundamental force laws (i.e. gravity, Coulomb’s law)

• related to a U:

• total round trip work = 0 (path independence of work)

, ,U U U

Fx y z

! ! != "

! ! !

r

Non-conservative forces• non-fundamental force laws (i.e. friction)

• no relation to a potential energy function U

• total round trip work ≠ 0

• involve energy dissipation (like air-drag, sliding friction…)

A small metal ball will fall as ifgravity is the only force, F=mg.

A falling coffee filter does not obey F=mg. Instead it reaches anearly constant terminal velocity:

a ~ g

a ~ 0

Air Resistance

mg mg

Fair Fair

Observations about air resistance:

1. Fair is proportional to speed.

2. Increases with area.

3. Depends on density of air.

4. Depends on object’s shape (not mass)

5. Opposes direction of motion (normally).

21ˆ

2airF C Av v!" #rThe force due to air resistance

can be approximated as:

ρ = density of air

A = area of object

v = speed of object

C = shape-dependent parameter

Approximate Air-Drag Formula

v1 v2

Application: Fuel Efficiency of a Car

Toyota Prius: C = 0.26 Toyota Tacoma: C = 0.44Daihatsu UFE III: C = 0.16

21ˆ

2airF v vC A!" #r

Can we model this behavior?

Consider air as being made up ofindividual balls with random directionsof motion that can collide with theobject: mg mg

Fair Fair

Simple Air-Drag Model

v1 v2

A. Yes, same on top and bottom.

B. No, more on the bottom.

C. No, more on the top.

Will there be just as many collisions on the bottom of theobject as there is on the top?

(Ignore any motion of the object.)

CLICKER QUESTION 2

This model predicts a terminal velocitywhich is about a thousand times too small!

This model has a fatal flaw. What is it?

Simple Air-Drag Model: ResultThe atom on the bottom will have a larger change in its momentum, impart agreater impulse, and hence a larger force to the object.

So , lets consider a model for air resistance:1. Collisions on the bottom of a falling object impart a greater

impulse to the object.

2. A falling object actually collides with more atoms on the bottomjust because of its motion.

Put all of this together into a quantitative model, along with results of kinetic theoryof gases for speed distributions etc., and make a prediction. Then test theprediction against experimental observation.

It neglects collisions of molecules with otherair molecules.

What forces act on the block as shown, being pulled across a table by the spring?

Sliding Friction

max Nf Fµ!

Sliding Friction

Approximate formula (not fundamental)• doesn’t apply to sticky surfaces (like tape)

• different materials have different values of µ

• coefficient of friction depends not only on material, but on it’s state –how dirty it is, etc.

Pull the block a distance d with constant velocity. You do work equal toW=Fd= ksΔx d, but there is no increase in the block’s kinetic energy orpotential energy. Where does the energy go?

Macroscopically, we say it heats up the block & the table. Microscopicallyin increases the kinetic and potential energies of all the atoms involved(stretching the atomic springs).

Is it possible to reverse this process, to get the energy out of the manyatoms involved and into motion of the block or stretching of the spring?

Energy Dissipation Again

Driven Oscillations

viscous frictionF cv! "r

( )

, , ,

, ,

, , , ,

sin

yx znet net x net y net z

xx spring x friction s D

dpdp dpd pF or F F F

dt dt dt dt

dp dxF F k x D t c

dt dt!

= =

= + = " " "# $% &

rr

s (the stretch)

Especially because of the friction force, this is a newdifferential equation for us.

Driven Oscillations

( )2

2sin 0

s D

d x dxm k x D t cdt dt

!+ " + =# $% &

( )

2

2

22 2

F

F D D

A D

c

m

!

! ! !

=

" #$ $ % &

' (

When steady stateis reached:

An inhomogeneous 2nd order linear differentialequation. You’ll study these in Math 266.

( )

2

2

22 2

F

F D D

A D

c

m

!

! ! !

=

" #$ $ % &

' (

D=driving motor amplitude

A=amplitude of object

ωF=free oscillation (natural) frequency

ωD=driving frequency

c=friction constant

How big would the amplitude be if there were no friction/dissipation?

See demonstration with cart…..

A Few Examples of Resonance

Tacoma Narrows Bridge – resonantfrequency of bridge matched that ofneighboring wind patterns.

tuners and digital filtersmake use of resonancephenomena