phy 113 a general physics i 11 am-12:15 pm tr olin 101 plan for lecture 4:

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

PHY 113 A General Physics I11 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


Updated schedule

4.1,4.12,4.35,4.60


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


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


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


From email questions on Webassign #3 -- continued

Problem 1 (3.29)

F1 F2


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.


i

j vertical direction (up)

horizontaldirection

jir ˆ)(ˆ)()( tytxt +=


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


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

)()(lim12 tt

ttdtdt

tt

==

rrrvv(t)


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)


Figure from your text:


Visualization of parabolic trajectory from textbook


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


Projectile motion (near earth’s surface)

i


horizontaldirection

ja

jiv

jir

ˆ)(

ˆ)(ˆ)()(

ˆ)(ˆ)()(

gt

tvtvt

tytxt

yx

=

+=

+=

g = 9.8 m/s2



jirv

jir

ˆ)(ˆ)()(

ˆ)(ˆ)()(

tvtvdtdt

tytxt

yx +==

+=

iii

ii

tgttt

tgtt

gdtdt

rrjvrr

vvjvv

jva

==+=

===

==

0 that note ˆ21

0 that note ˆ

ˆ)(

2



jva

jirv

jir

ˆ)(

ˆ)(ˆ)()(

ˆ)(ˆ)()(

gdtdt

tvtvdtdt

tytxt

yx

==

+==

+=

iii

i

tgttt

gtdtdt

rrjvrr

jvrv

==+=

==

0 that note ˆ

ˆ)(

221



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

=

=

+=


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

==

)()(



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

Trajectory equation in component form:

2

212

21 sin

cos

gttvygttvyty

tvxtvxtx

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


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

2

21

221

costan

sin)( cos)(sin cos

+=

==

+=+=

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.


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.


h=7.1mqi=53o

d=24m=x(2.2s)


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)


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)


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 ,0 and :Suppose

tatvxtx

tavtv

x)x(v)v(adtdv

++=

+=

===


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

202

100

00

000

)(

)( :Then

0 ,0 and :Suppose

tatvxtx

tavtv

x)x(v)v(adtdv

++=

+=

===

initialvelocity

initialposition

2

0

002

1

0

000

0

0

20

200

)()()(

)( :algebra using derivedResult

)()(2

:onaccelerati and velocity,position,between ipRelationsh

+

+=

=

=

avtva

avtvvxtx

avtvt

vtvxtxa


Summary of equations – one-dimensional motion with constant acceleration

20

200

202

100

00

)()(2

)(

)(

vtvxtxa

tatvxtx

tavtv

=

++=

+=

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


i


horizontaldirection

jir ˆ)(ˆ)()( tytxt +=

Review: Motion in two dimensions:


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


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

i


horizontaldirection

ja

jiv

jir

ˆ)(

ˆ)(ˆ)()(

ˆ)(ˆ)()(

gt

tvtvt

tytxt

yx

=

+=

+=



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

=

=

+=


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

2

21

221

costan

sin)( cos)(sin cos

+=

==

+=+=

ii

iiii

iiyiix

iiiiii

vxxgxxyxy

gtvtvvtvgttvytytvxtx

qq

qqqq


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

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

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


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


Uniform circular motion – continued

ra ˆ

:ison accelerati lcentripeta theanddirection radial in the

onaccelerati then the, If

2

rv

vvv

c

fi

=

=


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

==


r

ra

ra

ˆ m/s 03.0

ˆm/s10637186400

2

s 86400 s/min 60 min/hr 60 hr 24

ˆ2

2

232

2

=

=

==

=

c

c

T

rT


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.


r̂ θ̂

Circular motion

θa

ra

θv

ˆ

ˆ

ˆ :Assume2

dtdvrv

tvt

t

c

q

q

q

=

=

=

r̂ θ̂ca ta

tc aaa +=

phy 113 a general physics i 11 am-12:15 pm tr olin 101 plan for lecture 4:

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