Lecture4 Kinematics in 2 D Ave

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Physics 1D03 - Lecture 5

•Labs and tutorials are coming. Make sure you know your Lab section number (check Avenue or re-check on SOLAR).

•Tutorial preparation (sections L02, L04, …L28):Do CAPA set #1. Be prepared to write a solution to any of the

CAPA problem at the start of the tutorial. Read over your notes and the text sections covered so far. The tutorial format is described in the notes for Lecture 1.

•Lab preparation (sections L01, L03, …L27):Read the chapter D1: Introductory Exercises in the Lab Manual (you

already read “Uncertainties and Graphs” to prepare for CAPA set 1). Complete the pre-lab worksheet at the end of that chapter, to be handed in on entering the lab.

•Labs are in BSB-B116 and BSB-B220. Read the schedule for your lab section each week.


Help with Physics

-Physics Drop-in Centre (Thode Library basement)

-PHI (BSB-B119)

See the announcement on Avenue for details

Kinematics in Two Dimensions

• Position, velocity, acceleration vectors

• Constant acceleration in 2-D

• Free fall in 2-D

Serway and Jewett 4.1 to 4.3

Physics 1D03 - Lecture 4 3 Physics 1D03 - Lecture 4

The Position vectorpoints from the origin to the particle.

r

The components of are the coordinates (x,y) of the particle:

For a moving particle, , x(t), y(t) are functions of time.

ji yxr

)(tr

r

x

y

r

(x,y)

path

xi

yj

4

2


if rrr

Displacement :

Instantaneous Velocity :

is tangent to the

path of the particledtrdv /

Average Velocity :

(a vector parallel to )r

tavg /rv

x

y

ir

final

initialfr

r

vavg

x

y

r

v

5 Physics 1D03 - Lecture 4

Acceleration is the rate of change of velocity :

Average Acceleration is change in velocity vectortime elapsed

12

12

tttavg

vvv

a (definition)

Instantaneous Accelerationis the limit of aavg as the time interval t approaches zero

Review:

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Concept Quiz

At 9:00 a.m., a car is travelling North at 100 km/h. Anhour later, it is travelling East at 100 km/h. Its averageacceleration between 9:00 and 10:00 is:

A) a vector pointing northwest

B) a vector pointing southwest

C) a vector pointing southeast

D) a vector pointing northeast

E) zero

7 Physics 1D03 - Lecture 4

Instantaneous Accelerationis the limit as t zero:

)(tv

)( ttv

t time

tt time

path of particle

)(tv

)( ttv

v

tv

a

Acceleration: the rate of change of velocity

9

Averager Acceleration is the change of velocity per time

tv

t

lim

0

dtvd

a

2 1

2 1avg

v vva

t t t

3


a is the rate of change of v.

(Recall: a derivative gives the “rate of change” of function wrt a variable, like time).

Velocity changes if

i) speed changes

ii) direction changes (even at constant speed)

iii) both speed and direction change

In general, acceleration is not parallel to the velocity.

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Note:Example

• A skier comes down a slope, as shown. Draw acceleration at points, C, D, E and F.

Physics 1D03 - Lecture 4 11


Concept Quiz

A pendulum is released at (1) and swings across to (5).

143

52

a) at 3 only

b) at 1 and 5 only

c) at 1, 3, and 5

d) none of the above

0a

At which positions is ?(consider tangential only!)


kjir zyx

(i, j, k, are unit vectors)

kji

kji

rv

zyx vvv

dtdz

dtdy

dtdx

dtd

Then

Where , etc.

Components: Each vector relation implies 3 separate relations for the 3 Cartesian components.

dt

dxvx

18

4


2

2

2

2

2

2

,

,

,

dtzd

dtdv

adtdz

v

dtyd

dt

dva

dtdy

v

dtxd

dt

dva

dtdx

v

zzz

yyy

xxx

kji

dt

dv

dt

dv

dt

dv

dt

vda zyx

Each component of the velocity vector looks like the 1-D “velocity” we saw earlier. Similarly for acceleration:


Common Notation – for time derivatives only, a dotis often used:

drv r

dtdv

a v rdt


Constant Acceleration

a

If is constant (magnitude and direction), then:

22

1 t)(

)(

tt

tt

avrr

avv

oo

o

Where are the values at t = 0.oo vr

,

In 2-D, each vector equation is equivalent to a pair of component equations:

22

1

22

1

ta tvyy(t)

ta tvxx(t)

yoyo

xoxo

Example: [down] m/s 8.9 :fall Free 2 ga

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Projectile Motion

Assumptions:

• The free-fall acceleration is constant over the range of motion

– It is directed downward• This is the same as assuming a flat Earth over the range of the motion

• It is reasonable as long as the range is small compared to the radius of the Earth

• The effect of air friction is negligible

• With these assumptions, an object in projectile motion will follow a parabolic path

– This path is called the trajectory


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• A projectile launched with initial velocityProjectile Motion

iv

)(cos

)(cos:

iitvtvx

ivvvx

iixi

iixix

)(sin

)(sin:

2212

21 ivgttvgttvy

iiitgvtgvvy

iiyi

iiyiy

x

y

iv

ixv

iyv

yv

xvixx vv

xv

v

ixx vv v

v

a

sin

cos

iyi

ixi

vv

vv

ga

0a

y

x

2sm89g /.


Reasoning about Projectile motion:• A projectile follows a parabolic trajectory because it “falls” a

distance “½ gt2” below a straight-line trajectory.

Projectile Motion


Concept quiz

Your summer job at an historical site includes firing acannon to amuse tourists. Unfortunately, the cannon isn’tproperly attached, and as the cannonball shoots forward(horizontally) the cannon slides backwards off the wall.

If the cannon hits the ground 2 seconds later, thecannonball will hit the ground:

a) About 2 seconds after firing

b) About 100 seconds after firing

c) About 0.02 seconds after firing

d) Other

2 m/s 100 m/s


Shooting the Gorilla

Tarzan has a new slingshot. George the gorilla hangs from a tree, and bets that Tarzan can’t hit him. Tarzan guesses that George will drop to the ground as soon as he shoots. He wants to adjust his aim accordingly, and is sorry now that he didn’t pay more attention in physics class. Where should he aim?

A) Above the gorillaB) at the gorillaC) below the gorilla

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v0

r(t) =r0+v0t +(1/2)gt2

v0t

(1/2)gt2

a=g

r0

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Example

• Because of gravity, both the dolphin and the ball fall by the same amount.

• The red dots show the position of the ball and the blue dots show the position of the dolphin.



Example Problem

A stone is thrown upwards from the top of a 45.0 m high building with a 30º angle above the horizontal. If the initial velocity of the stone is 20.0 m/s, how long is the stone in the air, and how far from the base of the building does it land ?


Example Problem: Cannon on a slope.

How long is the cannonball in the air, and how far from the cannon does it hit?Try to do it two different ways: once using horizontal and vertical axes, once using axes tilted at 20o.

20°

30°

100 m/s

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7


For fun: show that for:

2

2

cos

)sin(cos2

g

vd o

θΦ

d

vo

Projectile Motion“Care package” drop:• The figure shows an airplane moving horizontally with a

constant velocity at an altitude of y. The plane releases a "care package" that falls to the ground along a curved trajectory. Where does the package hit the ground?


A ball is thrown vertically from a moving car. The car has a uniform motion. The ball will land:

1. in front of the car.2. in car.3. behind the car.4. in the same spot where it was thrown

upward.5. it may land any position of 1, 2, 3.

•Concept quiz

Physics 1D03 - Lecture 4 40 Physics 1D03 - Lecture 4

Summary

• position vector points from origin to a particle

• velocity vector

• acceleration vector

• for constant acceleration, we can apply 1-D formulae to each component separately

• for free fall, a=g (vectors!); horizontal and vertical motions are independent

r

dt

drv

lim 0 tdt

dt

vva

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Lecture4 Kinematics in 2 D Ave