9/5/2013users.wfu.edu/natalie/f13phy113/lecturenote/Lecture4for...9/5/2013 2 9/5/2013 PHY 113 A Fall...

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9/5/2013 1 9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 1 PHY 113 A General Physics I 11 AM-12:15 PM TR Olin 101 Plan for Lecture 4: Chapter 4 – Motion in two dimensions 1. Position, velocity, and acceleration in two dimensions 2. Two dimensional motion with constant acceleration; parabolic trajectories 3. Circular motion 9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 2 Updated schedule 4.1,4.12,4.35,4.60 9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 3 Tentative exam dates: 1. September 26, 2013 – covering Chap. 1-8 2. October 31, 2013 – covering Chap. 9-13, 15-17 3. November 26, 2013 – covering Chap. 14, 19-22 Final exam: December 12, 2013 at 9 AM

Transcript of 9/5/2013users.wfu.edu/natalie/f13phy113/lecturenote/Lecture4for...9/5/2013 2 9/5/2013 PHY 113 A Fall...

Page 1: 9/5/2013users.wfu.edu/natalie/f13phy113/lecturenote/Lecture4for...9/5/2013 2 9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 4 From email questions on Webassign #3 Problem 4 (3.24) Note:

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9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 1

PHY 113 A General Physics I11 AM-12:15 PM TR Olin 101

Plan for Lecture 4:

Chapter 4 – Motion in two dimensions1. Position, velocity, and acceleration

in two dimensions2. Two dimensional motion with

constant acceleration; parabolic trajectories

3. Circular motion

9/5/2013PHY 113 A Fall 2013 -- Lecture 4 2

Updated schedule

4.1,4.12,4.35,4.60

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 3

Tentative exam dates:

1. September 26, 2013 – covering Chap. 1-8

2. October 31, 2013 – covering Chap. 9-13, 15-17

3. November 26, 2013 – covering Chap. 14, 19-22

Final exam: December 12, 2013 at 9 AM

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9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 4

From email questions on Webassign #3

Problem 4 (3.24)

Note: In this case the angle f is actually measured as north of east.

f d1b q1

f=b+q1

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 5

From email questions on Webassign #3 -- continued

Problem 1 (3.11)

magnitude m

direction ° counterclockwise from the +x axis

graphically.

(d) Find A − 2B

A

2Bq

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 6

From email questions on Webassign #3 -- continued

Problem 1 (3.29)

F1 F2

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9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 7

In the previous lecture, we introduced the abstract notion of a vector. In this lecture, we will use that notion to describe position, velocity, and acceleration vectors in two dimensions.

iclicker exercise:

Why spend time studying two dimensions when the world as we know it is three dimensions?

A. Because it is difficult to draw 3 dimensions.B. Because in physics class, 2 dimensions are hard

enough to understand.C. Because if we understand 2 dimensions,

extension of the ideas to 3 dimensions is trivial.D. On Thursdays, it is good to stick to a plane.

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 8

i

j vertical direction (up)

horizontaldirection

jir ˆ)(ˆ)()( tytxt +=

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 9

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

dtdvx

x =a dtdvy

y =a

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9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 10

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

r(t1)r(t2)

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 11

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

r(t1)r(t2)

12

12

0

)()(lim12 tt

ttdtdt

tt

==

rrrvv(t)

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 12

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

r(t1)r(t2)

12

12

0

)()(lim12 tt

ttdtdt

tt

==

vvvav(t1) v(t2)

a(t1)

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9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 13

Figure from your text:

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 14

Visualization of parabolic trajectory from textbook

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 15

The trajectory graphs of the motion of an object y versus x as a function of time, look somewhat like a parabola:

CBxAxxy ++= 2)(

iclicker exerciseA. This is surprisingB. This is expectedC. No opinion

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9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 16

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

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 17

Projectile motion (near earth’s surface)

jirv

jir

ˆ)(ˆ)()(

ˆ)(ˆ)()(

tvtvdtdt

tytxt

yx +==

+=

iii

ii

tgttt

tgtt

gdtdt

rrjvrr

vvjvv

jva

==+=

===

==

0 that noteˆ21

0 that noteˆ

ˆ)(

2

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 18

Projectile motion (near earth’s surface)

jva

jirv

jir

ˆ)(

ˆ)(ˆ)()(

ˆ)(ˆ)()(

gdtdt

tvtvdtdt

tytxt

yx

==

+==

+=

iii

i

tgttt

gtdtdt

rrjvrr

jvrv

==+=

==

0 that noteˆ

ˆ)(

221

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9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 19

Projectile motion (near earth’s surface)

ˆ)( jvrv gtdtdt i ==

jva ˆ)( gdtdt ==

jvrr ˆ221 gttt ii +=

iiyi

iixi

yixii

v

v

vv

q

q

sin

cos

ˆˆ

v

v

jiv

=

=

+=

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 20

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

==

)()(

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 21

Projectile motion (near earth’s surface)

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

Trajectory equation in component form:

2

212

21 sin

cos

gttvygttvytytvxtvxtx

iiiyii

iiixii

+=+=

+=+=

q

q

gtvgtvtvvvtv

iiyiy

iixix

====

qqsin)(

cos)(

2

21

2

21

costan

coscossin

cos

+=

+=

=

ii

iiii

ii

i

ii

iiii

ii

i

vxxgxxyxy

vxxg

vxxvyxy

vxxt

qq

qqq

q

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9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 22

Projectile motion (near earth’s surface)Summary of results

2

21

221

costan

sin)(cos)(sincos

+=

==

+=+=

ii

iiii

iiyiix

iiiiii

vxxgxxyxy

gtvtvvtvgttvytytvxtx

qq

qqqq

iclicker exercise:

These equations are so beautiful thatA. They should be framed and put on the wall.B. They should be used to perfect my

tennis/golf/basketball/soccer technique.C. They are not that beautiful.

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 23

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

y

x

iclicker exercise: Which trajectory takes the longest time?A. Brown B. Green C. Same time for all.

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 24

h=7.1mqi=53o

d=24m=x(2.2s)

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9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 25

Problem solving steps1. 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 known relationships and unknowns).5. Solve the equations.6. Check whether your answer makes sense (units, order of magnitude,

etc.).

h=7.1mqi=53o

d=24m=x(2.2s)

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 26

sms

mv

vxd

tvxtx

oi

oi

iii

/12698.182.253cos

24242.253cos)2.2(

cos

==

===

+= q

h=7.1mqi=53o

d=24m=x(2.2s)

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 27

Review Position x(t), Velocity y(t), Acceleration a(t) in one dimension:

==

==

t

t

t

t

dttatvdtdvta

dttvtxdtdxtv

0

0

')'()()(

')'()()(

Special case of constant acceleration a(t)=a0:

202

100

00

000

)(

)(:Then

0,0and:Suppose

tatvxtxtavtv

x)x(v)v(adtdv

++=

+=

===

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9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 28

Review – continued:Special case of constant acceleration a(t)=a0:

202

100

00

000

)(

)(:Then

0,0and:Suppose

tatvxtxtavtv

x)x(v)v(adtdv

++=

+=

===

initialvelocity

initialposition

2

0

002

1

0

000

0

0

20

200

)()()(

)(:algebrausingderivedResult

)()(2

:onacceleratiand velocity,position,between ipRelationsh

+

+=

=

=

avtva

avtvvxtx

avtvt

vtvxtxa

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 29

Summary of equations –one-dimensional motion with constant acceleration

20

200

202

100

00

)()(2

)(

)(

vtvxtxa

tatvxtxtavtv

=

++=

+=

iclicker question:Why did I show you part of the derivation of the last equation?

A. Because professors like to torture physics studentsB. Because you will need to be able to prove the

equation yourselfC. Because the “proof” helps you to understand the

meaning of the equationD. All of the above E. None of the above

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 30

i

j vertical direction (up)

horizontaldirection

jir ˆ)(ˆ)()( tytxt +=

Review: Motion in two dimensions:

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9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 31

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

dtdvx

x =a dtdvy

y =a

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 32

Projectile motion (constant acceleration)(reasonable approximation near Earth’s surface)

i

j vertical direction (up)

horizontaldirection

ja

jiv

jir

ˆ)(

ˆ)(ˆ)()(

ˆ)(ˆ)()(

gt

tvtvt

tytxt

yx

=

+=

+=

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 33

Projectile motion (near earth’s surface)

ˆ)( jvrv gtdtdt i ==

jva ˆ)( gdtdt ==

jvrr ˆ221 gttt ii +=

iiyi

iixi

yixii

v

v

vv

q

q

sin

cos

ˆˆ

v

v

jiv

=

=

+=

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9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 34

Projectile motion (near earth’s surface)Summary of component functions

2

21

221

costan

sin)(cos)(sincos

+=

==

+=+=

ii

iiii

iiyiix

iiiiii

vxxgxxyxy

gtvtvvtvgttvytytvxtx

qq

qqqq

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 35

Uniform circular motion – another example of motion in two-dimensions

animation from http://mathworld.wolfram.com/UniformCircularMotion.html

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 36

iclicker question:Assuming that the blue particle is moving at constant speed around the circle what can you say about its acceleration?

A. There is no accelerationB. There is acceleration tangent to the circleC. There is acceleration in the radial direction of

the circleD. There is not enough information to conclude

that there is acceleration or not

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9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 37

Uniform circular motion – continued

ra ˆ

:ison acceleratilcentripeta theanddirection radialin the

onaccelerati then the,If

2

rv

vvv

c

fi

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 38

Uniform circular motion – continued

r

ra

ra

ra

ˆ2

ˆ2

ˆ

2

2

2

rf

rT

rv

c

c

c

Trv 2

In terms of time period T for one cycle:

In terms of the frequency f of complete cycles:πfrv

Tf 2;1

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 39

r

ra

ra

ˆm/s03.0

ˆm/s10637186400

2

s86400s/min 60min/hr 60hr 24

ˆ2

2

232

2

c

c

T

rT

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9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 40

iclicker exercise:During circular motion, what happens when the speed of the object changes?A. There is only tangential accelerationB. There is only radial accelerationC. There are both radial and tangential

accelerationD. I don’t need to know this because it never

happens.

9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 41

r̂ θ̂

Circular motion

θa

ra

θv

ˆ

ˆ

ˆ:Assume2

dtdvrv

tvt

t

c

r̂ θ̂ca ta

tc aaa