9/24/2013phy 113 c fall 2013-- lecture 91 phy 113 c general physics i 11 am-12:15 pm mwf olin 101...

PHY 113 C Fall 2013-- Lecture 9 19/24/2013

PHY 113 C General Physics I11 AM-12:15 PM MWF Olin 101

Plan for Lecture 9:

1. Review (Chapters 1-8)

2. Exam preparation advice –

3. Example problems


iclicker questionWhat is the best way to prepare for Thursday’s exam?

A. Read Lecture Notes and also reread Chapters 1-8 in Serway and Jewett.

B. Prepare equation sheet.C. Solve problems from previous exams.D. Solve homework assignments (both graded and

ungraded) from Webassign Assignments 1-8E. All the above

http://www.wfu.edu/~natalie/f13phy113/lecturenote/


iclicker questionHave you (yet) accessed the online class lecture notes from previous classes?

A. yesB. no

iclicker questionHave you (yet) accessed the passed Webassign Assignments (with or without the answer key)?

C. yesD. no


Access to previous exams


Previous exam access -- continued

Overlap with 2013 schedule


Comments on preparation for next Thursday’s exam – continued

What you should bring to the exam (in addition to your well-rested brain): A pencil or pen Your calculator An 8.5”x11” sheet of paper with your favorite

equations (to be turned in together with the exam)

What you should NOT use during the exam Electronic devices (cell phone, laptop, etc.) Your textbook

PHY 113 C Fall 2013-- Lecture 9 8

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

dt

ddt

d

m

rv

va

aF

Problem solving skills

Math skills

9/24/2013


iclicker exerciseDoes the previous slide annoy you?

A. yesB. no

PHY 113 C Fall 2013-- Lecture 9 10

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.).

9/24/2013


Likely exam format (example from previous exam)


Review of slides from previous lectures


Mathematics Review -- Appendix B Serwey & Jewett

a

acbbx

cbxax

2

4

0

:equation Quadratic

2

2

)cos()sin(

:calculus alDifferenti

1

ttdt

d

eedt

d

antatdt

d

tt

nn

)cos(1

)sin(

11

:calculus Integral1

tdtt

edte

n

atdtat

tt

nn

a

b

c

q

b

ac

ac

b

tan

sin

cos

:ryTrigonomet


One dimensional motion --Summary of relationships

t

t

t

t

dttatvdt

dvta

dttvtxdt

dxtv

0

0

')'()( )(

')'()( )(


Special relationships between t,x,v,a for constant a:

t

t

t

t

dttatvdt

dvta

dttvtxdt

dxtv

0

0

')'()( )(

')'()( )(

:iprelationsh General

200

2

0

2

1

2

100)(

0)(

:iprelationsh Special

attvxattvxtx

atvatvtv


Vector addition:

ab

a – b

Vector subtraction:

a

-b

a + bIntroduction of vectors


ax

ay

22

ˆˆ

ˆˆ and ˆˆFor

yx

yyxx

yxyx

cc

baba

bbaa

c

cyxba

yxbyxa

yxa ˆˆ yx aa

by

bx

yxb ˆˆ yx bb cba

Treatment of vectors in component form


Vectors relevant to motion in two dimenstions

Displacement: r(t) = x(t) i + y(t) j

Velocity: v(t) = vx(t) i + vy(t) j

Acceleration: a(t) = ax(t) i + ay(t) j

dtdx

x vdtdy

y v

dt

dvxx a

dt

dvyy a


Visualization of the position vector r(t) of a particle

r(t1)r(t2)


Visualization of the velocity vector v(t) of a particle

r(t1)

r(t2)

12

12

0

)()(lim

12 tt

tt

dt

dt

tt

rrrv

v(t)


Visualization of the acceleration vector a(t) of a particle

r(t1)

r(t2)

12

12

0

)()(lim

12 tt

tt

dt

dt

tt

vvva

v(t1) v(t2)

a(t1)


Projectile motion (near earth’s surface)

i

j vertical direction (up)

horizontaldirection

ja

jiv

jir

ˆ)(

ˆ)(ˆ)()(

ˆ)(ˆ)()(

gt

tvtvt

tytxt

yx

g = 9.8 m/s2



jir

v

jir

ˆ)(ˆ)()(

ˆ)(ˆ)()(

tvtvdt

dt

tytxt

yx

iii

ii

tgttt

tgtt

gdt

dt

rrjvrr

vvjvv

jv

a

0 that note ˆ2

1

0 that note ˆ

ˆ)(

2


Projectile motion (near earth’s surface) Trajectory equation in vector form:

ˆ)( jvv gtt i jvrr ˆ221 gttt ii

Aside: The equations for position and velocity written in this way are call “parametric” equations. They are related to each other through the time parameter.

Trajectory equation in component form:

2

21 gttvyty

tvxtx

yii

xii

gtvtv

vtv

yiy

xix

)(

)(


Diagram of various trajectories reaching the same height h=1 m:

y

xq



Trajectory path y(x); eliminating t from the equations:

Trajectory equation in component form:

2

212

21 sin

cos

gttvygttvyty

tvxtvxtx

iiiyii

iiixii

gtvgtvtv

vvtv

iiyiy

iixix

sin)(

cos)(

2

21

2

21

costan

coscossin

cos

ii

iiii

ii

i

ii

iiii

ii

i

v

xxgxxyxy

v

xxg

v

xxvyxy

v

xxt


Isaac Newton, English physicist and mathematician (1642—1727)

http://www.newton.ac.uk/newton.html

1. In the absence of a net force, an object remains at constant velocity or at rest.

2. In the presence of a net force F, the motion of an object of mass m is described by the form F=ma.

3. F12 =– F21.

http://www.newton.ac.uk/newton.html


Newton’s second law

F = m a

Types of forces:

Fundamental Approximate Empirical

Gravitational F=-mg j Friction

Electrical Support

Magnetic Elastic

Elementary

particles


Example of two dimensional motion on a frictionless horizontal surface

m

m

m

21

21

FFa

aFF


Example – support forces

appliedsupport

ˆ

FF

yF

mgg

Fsupport acts in direction to surface

(in direction of surface “normal”)


Example of forces in equilibrium


Example: 2-dimensional forcesA car of mass m is on an icy (frictionless) driveway, inclined at an angle t as shown. Determine its acceleration.

Conveniently tilted coordinate system:

sin

sin : Along

0cos : Along

ga

mamgx

mgny

x

x


vi

vf=0

tgvtv

tgtvxtx

i

ii

sin)(

sin)( :incline Along 221

Another example: Note: we are using a tilted coordinate frame

i

mg

mg sin q


Friction forces

The term “friction” is used to describe the category of forces that oppose motion. One example is surface friction which acts on two touching solid objects. Another example is air friction. There are several reasonable models to quantify these phenomena.

Surface friction:

N

Ff applied

Material-dependent coefficient

Normal force between surfaces

Air friction:

speedhigh at

speed lowat 2vK

KvD

K and K’ are materials and shape dependent constants


Models of surface friction forces

(applied force)

surf

ace

fric

tion

forc

e

fs,max=msn

Coefficients ms , mk depend on the surfaces; usually, ms > mk


mg

f

n

mg sin

mg cos

Consider a stationary block on an incline:

sincosThen

cos If

sin 0sin

cos 0cos

max,

mgmg

mgnff

mgfmgf

mgnmgn

S

SSS

tanS


V (constant)

q

mg

n

f=mkn

Consider a block sliding down an inclined surface; constant velocity case

sincosThen

cos If

sin 0sin

cos 0cos

mgmg

mgnf

mgfmgf

mgnmgn

K

KK

tanK


mg

f

n

mg sin

mg cos

Summary

slip about tojust isblock when tan

elocityconstant vat movesblock when tan

S

K


Uniform circular motion and Newton’s second law

r

ra

aF

ˆ2

r

v

m

c


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


Work and potential energy

rFr

r

dWf

i

fi : workof Definition

rFrr

r

r dWUref

ref

:energy potential of Definition

Note: It is assumed that F is conservative


rFr

r

dWf

i

fi : workof Definition

Review of energy concepts:

2

2

1 :energy kinetic of Definition mvK

22

2

1

2

1

:oremenergy the kinetic-Work

if

f

i

totaltotalfi mvmvdW rF


22

2

1

2

1

:oremenergy the kinetic-Work

if

f

i

totaltotalfi mvmvdW rF

Summary of work, potential energy, kinetic energy relationships

edissipativ

fiiiff

ifedissipativ

fiiftotalfi

WUKUK

KKWUUW

:gRearrangin

rr

edissipativfiif

edissipativfi

veconservatifi

totalfi

WUU

WWW


Example problem from Webassign #8

A baseball outfielder throws a 0.150-kg baseball at a speed of 37.2 m/s and an initial angle of 31.0°. What is the kinetic energy of the baseball at the highest point of its trajectory?

vi

qi

yf

vf



The coefficient of friction between the block of mass m1 = 3.00 kg and the surface in the figure below is μk = 0.440. The system starts from rest. What is the speed of the ball of mass m2 = 5.00 kg when it has fallen a distance h = 1.85 m?h

hf

ghmghmvmm

WUUKK

ghmfhW

WUKUK

kf

edissipativfifiif

kedissipativ

fi

edissipativfiiiff

122

21

1

02

1



A block of mass m = 3.40 kg is released from rest from point A and slides on the frictionless track shown in the figure below. (Let ha =

6.70 m.) .


Example problemfrom 2012 Exam #2:

if

if

veconservatifi

xUxU

UU

W



013 gmT

0sinsin

0coscos

32211

2211

TTT

TT

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