3. Motion in Two and Three Dimensions

2

0( )v t v at

Recap: Constant Acceleration

0 0( )

tx x v t dt Area under

the functionv(t).

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Recap: Constant Acceleration

ad

t

v

d

0v av t

210 0 2x x v t at

2 20 02 ( )v xav x

4

Recap: Acceleration due to Gravity (Free Fall)

In the absence of air resistance all objects fall with the same constant acceleration of about

g = 9.8 m/s2

near the Earth’s surface.

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Recap: Example

A ball is thrown upwards at 5 m/s, relative to the ground, from a height of 2 m.

We need to choose a coordinate system.

2 m

5 m/s

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Recap: Example

Let’s measuretime from whenthe ball is launched.

This defines t = 0.

Let’s choose y = 0 to be ground leveland up to be the positive y direction.

y0 = 2 m

v0 = 5 m/s

y

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Recap: Example

1. How high above the ground will the ball reach?

with a = –gand v = 0.

2 20 02 ( )v yav y use

y0 = 2 m

v0 = 5 m/s

y

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Recap: Example

210 0 2y v t ay t Use

2. How long does it take the ball to reach the ground?

y0 = 2 m

v0 = 5 m/s

y

with a = –gand y = 0.

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Recap: Example

2 20 02 ( )yav v y Use

3. At what speed does the ball hit the ground?

y0 = 2 m

v0 = 5 m/s

y

with a = –gand y = 0.

Vectors

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A vector is a mathematical quantity that has two properties:

direction and magnitude.

Vectors

One way to represent a vectoris as an arrow: the arrow givesthe direction and its length themagnitude.

12

Position

A position p is a vector: its direction is from o to p and its length is the distance from o to p.

A vector is usuallyrepresented by asymbol like .p r

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Displacement

A displacement is another example of a vector.

r

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Vector Addition

The order in which the vectors are added does not matter, that is, vector addition is commutative.

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Vector Scalar Multiplication

A

a A

a and –q are scalars (numbers).

Aq

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( 1)C A B

Vector Subtraction

If we multiply a vector by –1 we reverse itsdirection, but keep its magnitude the same.

Vector subtraction is really vector addition with one vector reversed.

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Vector Components

Acos is the component, orthe projection, ofthe vector A alongthe vector B.

A

BcosA

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Vector Components

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Vector Addition using Components

x x x

y y y

C A B

C A B

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Unit Vectors

If the vector A is multipliedby the scalar 1/A we get a newvector of unit length in the same direction as vector A; that is, we get a unit vector.

From the components, Ax, Ay, and Az, of a vector,we can compute its length, A, using

2 2 2x y zA A A A

A

ˆ 1A A

A

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Unit Vectors

It is convenient to define unit vectorsparallel to the x, y and zaxes, respectively.

ˆˆ ˆx y zA A i A j A k

ˆˆ ˆ, ,i j k

Then, we can write avector A as follows:

Velocity and Acceleration Vectors

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Velocity

0lim

t

dr rv

dt t

v

2 1r r r

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Acceleration

0lim

t

dv va

dt t

2 1v v v

Relative Motion

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Relative Motion

Velocity of planerelative to air

Velocity of airrelative to ground

Velocity of planerelative to ground

v

V

Vv v

N

S

W E

v

V

v

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Example – Relative Motion

A pilot wants to fly plane due northAirspeed: 200 km/hWindspeed: 90 km/h

direction: W to E

1. Flight heading?2. Groundspeed?

N

S

W E

v

V

v

Coordinate system: î points from west to east and ĵ points from south to north.

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ˆ ˆ90 0 km/h

ˆcos( / 2 )

ˆsin(

20

/ 2 ) k /

0

2 h00 m

V i

i

j

v

j

ˆ ˆ0 km/hv iV v jv

Example – Relative Motion

N

S

W E

v

V

v

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200

ˆ90 km/h

ˆ ˆ( sin cos ) km/h

V

iv

i

j

ˆ ˆ( sin 90) cos

km

20

h

0

/

200v i j

Example – Relative Motion

N

S

W E

v

V

v

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Equate x components0 = –200 sin = sin-1(90/200) = 26.7o west of north.

0

( sin 90) cos200

km

ˆ

200 ˆ

ˆ

ˆ

/h

v

j

v V jvi

i

Example – Relative Motion

N

S

W E

v

V

v

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Example – Relative Motion

Equate y componentsv = 200 cos = 179 km/h

N

S

W E

v

V

v

200

0

( sin 90

ˆ

) cos

ˆ

ˆ ˆ200

v v i

i

V v j

j

Projectile Motion

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0r

r

Projectile Motion under Constant Acceleration

Coordinate system: î points to the right, ĵ points upwards

ji

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0r

) ˆ0 (i ja g

210 2v t at

r

22

0

100 vr r

v v

t

t

t

a

a

R = Range

Impact point

Projectile Motion under Constant Acceleration

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Projectile Motion under Constant Acceleration

0 0

0 0

cos

sinx

y

v v

v v

Strategy: split motion into x and y components.21

0 0 2

0 0

y

x

y y v t gt

x x v t

R = Range R = x - x0

h = y - y0

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Projectile Motion under Constant Acceleration

202 2 0ygt v t h

20 0 2y yv v gh

tg

Find time of flight by solving y equation:

0xR v t

And findrange from:

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Projectile Motion under Constant Acceleration

Special case: y = y0, i.e., h = 0

0 0

20

2

sin 2

x yv vR

g

v

g

R

y0 y(t)

Uniform Circular Motion

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Uniform Circular Motion

ˆc )ˆ( sos inir r j

r = Radius

r

i

j

( , )x y

y

xO

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Uniform Circular Motion

ˆco( ˆsin )s

drv

dtd

rd

jit

Velocity

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Uniform Circular Motion

ˆcos

ˆcos

ˆsin

ˆsin

ˆsin

ˆc

( )

o

( )

( s )

dv r

dti

i

i

j

d

dj

j

dr

d

r

td

dt

d/dt is called theangular

velocity

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Uniform Circular Motion

2

ˆco( )

( )

sˆsin

ˆco ss ˆin

dva

dtd

rdt

d

dti

i

j

jd

dtr

Acceleration

For uniform motiond/dt is

constant

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Uniform Circular Motion2

2

( ˆcos ˆsin )

ˆ

a r

r

d

dj

rd

t

it

d

Acceleration is towardscenter

Centripetal Acceleration

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Uniform Circular Motion

2va

r

Magnitudes of velocityand centripetal

acceleration are related

as follows

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Uniform Circular Motion

2 rv

T

Magnitude of velocityand period T related as

follows

r

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Summary

In general, acceleration changes both the magnitude and direction of the velocity.

Projectile motion results from the acceleration due to gravity.

In uniform circular motion, the acceleration is centripetal and has constant magnitude v2/r.

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How to Shoot a Monkey

x = 50 mh = 10 mH = 12 m

Computeminimuminitialvelocity

H