PHY 113 A General Physics I 9-9:50 AM MWF Olin 101 Plan for Lecture 16: Review of Chapters 5-9

PHY 113 A Fall 2012 -- Lecture 16 110/05/2012

PHY 113 A General Physics I9-9:50 AM MWF Olin 101

Plan for Lecture 16:

Review of Chapters 5-9

1. Circular motion

2. Work & kinetic energy

3. Impulse and momentum



Format of Wednesday’s examWhat to bring:

1. Clear, calm head2. Equation sheet (turn in with exam)3. Scientific calculator4. Pencil or pen(Note: labtops, cellphones, and other electronic equipment must be off or in sleep mode.)

Timing:May begin as early as 8 AM; must end ≤ 9:50 AM

Probable exam format 4-5 problems similar to homework and class

examples; focus on Chapters 6-9 of your text. Full credit awarded on basis of analysis steps as

well as final answer


Examples of what to include on equation sheetGiven information

on examSuitable for equation sheet

Universal constants (such as g=9.8m/s2)

Trigonometric relations and definition of dot product

Particular constants (such as ms, mk)

Simple derivative and integral relationships

Unit conversion factors if needed

Definition of work, potential energy, kinetic energy

Work-kinetic energy theorem

Relationship between force, potential energy, and work for conservative systems

Relationship of impulse and momentum; conservation of momentum

Elastic and inelastic collisions

Center of mass


iclicker exercise:Which of the following quantities are vectors:

A. WorkB. Kinetic energyC. ImpulseD. TimeE. None of these

iclicker exercise:Which of the following quantities are scalars:

A. MomentumB. Center of massC. ForceD. Potential energyE. None of these


Some concepts introduced in Chapters 6-9 that were not emphasized in class:

1. Equations of motion in the presence of air or fluid friction

2. So called “fictitious” forces due to accelerating reference frames

3. Rocket propulsion

PHY 113 A Fall 2012 -- Lecture 16 7

Advice:

1. Keep basic concepts and equations at the top of your head.

2. Practice problem solving and math skills

3. Develop an equation sheet that you can consult.

Equation Sheet

pv

aF

m

KKW

m

ifnet

Problem solving skills

Math skills

10/05/2012

PHY 113 A Fall 2012 -- Lecture 16 8

Problem solving steps

1. Visualize problem – labeling variables2. Determine which basic physical principle(s) apply3. Write down the appropriate equations using the variables

defined in step 1.4. Check whether you have the correct amount of

information to solve the problem (same number of knowns and unknowns).

5. Solve the equations.6. Check whether your answer makes sense (units, order of

magnitude, etc.).

10/05/2012


Newton’s second law for the case of uniform circular motion

rra

aF

ˆ2

r

v

m

c

Review of some concepts:

dt

dmmv

aF

:law second sNewton'


Example: Suppose a race car driver maintains of speed of v=40m/s around a horizontal level circular track of radius r=100m. Assuming that static friction keeps the car on the circular path, what must be the minimum coefficient of static friction for the car-road surface?

r

63.1

1008.9

40

;

ˆ

22

2

2

2

gr

v

r

mvmg

mgfr

mvf

r

v

m

s

s

s

c

ra

aF


Definition of vector “dot” product

q

0ˆˆ 1ˆˆ

272635ˆ2ˆ3ˆ6ˆ5

:formComponent

(scalar) 5.37120cos)15)(5(

120 ,15 ,5 :Example

cos

o

o

jiii

jiji

BA

BA

BA

AB


rFr

rdW

f

i

fi

Definition of work: F

dr

ri rj

cal 0.239 J 1

JoulesmetersNewtons Work

: workof Units


Example:

JmNmNdxFdWf

i

f

i

x

x

xfi 2525)4)(5( 21 rF

r

r


Introduction of the notion of Kinetic energy

Some more details:

Consider Newton’s second law:

Ftotal = m a Ftotal · dr= m a · dr

dtdtd

mdtdtd

dtd

mddtd

mdmdf

i

f

i

f

i

f

i

f

i

t

t

t

ttotal v

vrvr

vrarF

r

r

r

r

r

r

Wtotal = ½ m vf2 - ½ m vi

2 =Kf-Ki

Kinetic energy (joules)

22

21

21

21

if

f

i

t

tmvmvmddmdt

dtd

mf

i

f

i

vvvvv

v v

v


Special case of “conservative” forces conservative non-dissipative

if

f

i

fi UUdW rrrF

F

write topossible isit , forces edissipativ-nonFor

iiff

ifif

f

i

fi

mgyUmgyU

mgymgyyymgdW

)( and )(

:Earth of surfacenear gravity of Example

rr

rF


Special case of “conservative” forces conservative non-dissipative

if

f

i

fi UUdW rrrF

F

write topossible isit , forces edissipativ-nonFor

2

212

21

2212

21

)( and )(

:spring elasticon mass ofmotion of Example

iiff

if

f

i

f

i

fi

kxUkxU

kxkxdxkxdW

rr

rF


EUmvUmv

UUdW

mvmvdW

ffii

iftotalfi

iftotalfi

f

i

f

i

rr

rrrF

rF

r

r

r

r

22

22

2

1

2

1

:forces veconservatiFor

2

1

2

1

:oremenergy the kinetic- workGeneral


Example:

smxm

kvmvkx

EUmvUmv

mxmNkkgm

x

xk

m

iffi

ffii

i

f

i

/2.11.05.0

70

2

1

2

1

2

1

2

1

1.0;/70;5.0

?)0( released is spring when themass theof velocity the

iswhat acting, forcesother no are thereAssuming

. distance aby compressed and constant spring

withspring a toattached is mass a Suppose

22

22

rr


Summary of physics “laws”

fffi

veconservatinonii

fiveconservatinon

fiveconservati

fitotal

iffi

veconservati

if

f

i

fitotal

UmvWUmv

WWW

UUW

mvmvdW

dt

dmm

rr

rr

rF

vaF

22

22

2

1

2

1

:forces veconservatiFor

2

1

2

1

:oremenergy the kinetic-Work



Another example; now with friction

Mass m1 (=0.2kg) slides horizontally on a table with kinetic friction mk=0.5 and is initially at rest. What is its velocity when it moves a distance Dx=0.1m (and m2 (=0.3kg) falls Dy=0.1m)?

T

T

m2 g

21

21 2

1

2

1if vmvmxfTW

0

f


xfmm

mxg

mm

mmxfTW

fmm

mg

mm

mmT

amgmT

amfT

21

1

21

21

21

2

21

21

22

1

sm

mm

mmxgv

gmfmm

xfxgmv

vmxfTW

kf

kf

f

/885.02.03.0

2.05.03.01.08.922

22

2

1

21

2

121

2

21


if

f

i

f

i

ddt

ddt

dt

d

dt

md

dt

dmm

pppF

pF

pvvaF

:calculus little a Using

Relationship between Newton’s second law and linear momentum for a single particle:

if

f

i

f

i

dt

dt

ppFI

FI

:impulse"" Define


Physics of composite systems

ii

i

ii

i

ii

iii

ii dt

d

dt

dm

dt

dmm p

vvaF


ii

ii

ii

ii

ii

dt

d

final initial

(constant)

0

: then,0 if that Note

pp

p

p

F

PHY 113 A Fall 2012 -- Lecture 16 2510/05/2012dt

dM

dt

d

CM

ii

ii

ii

ii

ii

rpp

p

p

FF

final initial

total

(constant)

0

: then,0 If

2

2

:Define

:mass ofcenter for relation General

dt

dM

mMM

m

CMtotal

ii

ii

iii

CM

rFF

rr


Finding the center of mass

ii

iii

CM mMM

m

rr

ji

jiir

jiir

ˆ00.1ˆ750

4

ˆ22ˆ21ˆ)1)(1(

ˆˆˆ

2 ;1 :example In this

321

332211

321

mm.

mmm

mmm

ymxmxm

kgmkgmm

CM

CM


Examples of two-dimensional collision; balls moving on a frictionless surface

-- equations more 2 Need

,,, :Unknowns

,, :Knowns

sinsin0

coscos

21

121

2211

221111

final initial

ff

i

ff

ffi

ii

ii

vv

vmm

vmvm

vmvmvm

pp


Examples of two-dimensional collision; balls moving on a frictionless surface

smsmsm

v

smsmsm

vvv

smsm

vv

vmvm

vmvmvm

smv

smvkgmm

oof

o

fif

o

ff

ff

ffi

f

i

/11.188.17cos

/060.1

88.17sin

/342.0

88.71 060.1

342.0tan

/060.120cos/1/2

coscos

/342.020sin/1

sinsin

sinsin0

coscos

20 ,/1

,/2 ,06.0 :Suppose

1

o

211

21

2211

221111

o2

121


Example: two-dimensional totally inelastic collision

ji

jiv

vvv

vvv

ˆ/5.12ˆ/375.9

ˆ/204000

2500ˆ/254000

1500

221

21

21

1

212211

smsm

smsm

mm

m

mm

m

mmmm

f

iif

fii

m1=1500kg

m2=2500kg

JΔE

mmmmΔE iif

5

22

22

2

22

2

11

2

21

108.4

2025002

1251500

2

1

5.12375.940002

1

2

1

2

1

2

1

:case in thisenergy kinetic of Loss

vvv


Energy analysis of a simple nuclear reaction :

HeRnRa 42

22286

22688

Q=4.87 MeV


Energy analysis of a simple reaction : HeRnRa 42

22286

22688

Q=4.87 MeV

MeV8.498.04/1222/1

4/1

2

4

1

222

1

2222

222

QQm

pE

m

p

m

p

m

pQ

He

HeHe

uHe

He

Rn

Rn

HeRn ppp

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PHY 113 A General Physics I 9-9:50 AM MWF Olin 101 Plan for Lecture 16: Review of Chapters 5-9