# and Three Dimensions

Mechanics

Nov 14, 2013 (4 years and 6 months ago)

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3. Motion in Two

and Three Dimensions

2

0
( )
v t v
a
t
 
Recap:
Constant

Acceleration

0
0
( )
t
x x v t dt
 

Area under

the function

v
(
t
).

3

Recap:
Constant

Acceleration

a
d
t
v
d

0
v
a
v t
 
2
1
0 0
2
x x v t
a
t
  
2 2
0 0
2 ( )
v x
a
v
x
  
4

Recap: Acceleration due to
Gravity (Free Fall)

In the absence of air resistance
all

objects
fall with the
same

constant

acceleration

g = 9.8 m/s
2

near

the Earth’s surface.

5

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

6

Recap: Example

Let’s measure

time from when

the ball is launched.

This defines
t

= 0.

Let’s choose
y

= 0 to be ground level

and
up

to be the
positive

y

direction.

y
0

= 2 m

v
0

= 5 m/s

y

7

Recap: Example

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

with
a

=

g

and
v

= 0.

2 2
0 0
2 ( )
v y
a
v
y
  
use

y
0

= 2 m

v
0

= 5 m/s

y

8

Recap: Example

2
1
0 0
2
y v t
a
y
t
  
Use

2.

How long does it

take the ball to

reach the ground?

y
0

= 2 m

v
0

= 5 m/s

y

with
a

=

g

and
y

= 0.

9

Recap: Example

2 2
0 0
2 ( )
y
a
v v y
  
Use

3.

At what speed

does the ball hit

the ground?

y
0

= 2 m

v
0

= 5 m/s

y

with
a

=

g

and
y

= 0.

Vectors

11

A
vector

is a mathematical quantity
that has two properties:

direction

and
magnitude
.

Vectors

One way to
represent

a vector

is as an
arrow
: the arrow gives

the direction and its length the

magnitude.

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 usually

represented by a

symbol like .

p

r
13

Displacement

A
displacement

is another example of

a vector.

r

14

The order in which the

not matter, that is, vector

commutative
.

15

Vector Scalar Multiplication

A
a
A
a

and

q

are
scalars

(numbers).

A
q

16

( 1)
C A B
 

Vector Subtraction

If we multiply a vector by

1

we reverse its

direction, but keep its magnitude the same.

Vector subtraction

is really vector

one vector reversed.

17

Vector Components

A
cos
q

is the

component
, or

the
projection
, of

the vector
A

along

the vector
B
.

q
A
B
cos
A
q
18

Vector Components

19

Components

x x x
y y y
C A B
C A B
 
 
20

Unit Vectors

If the vector
A

is multiplied

by the scalar
1/A

we get a new

vector of
unit length

in the

same direction

as vector
A
;

that is, we get a
unit vector
.

From the components,
A
x
,
A
y
, and

A
z
, of a vector,

we can compute its length,
A
, using

2 2 2
x y z
A A A A
  
A
ˆ
1
A
A
A

21

Unit Vectors

It is convenient to define

unit vectors

parallel to the
x
,
y

and
z

axes, respectively.

ˆ
ˆ ˆ
x y z
A Ai A j Ak
  
ˆ
ˆ ˆ
,,
i j k
Then, we can write a

vector
A

as follows:

Velocity and Acceleration
Vectors

23

Velocity

0
lim
t
dr r
v
dt t
 

 

v
2 1
r r r
  
24

Acceleration

0
lim
t
dv v
a
dt t
 

 

2 1
v v v
  
Relative Motion

26

Relative Motion

Velocity of plane

relative to air

Velocity of air

relative to ground

Velocity of plane

relative to ground

v

V
V
v
v
 

N

S

W

E

v

V
v
q
27

Example

Relative Motion

A pilot wants to fly plane due north

Airspeed:

200 km/h

Windspeed
:

90 km/h

direction:

W to E

2. Groundspeed?

N

S

W

E

v

V
v
q
Coordinate system
:
î

points from

west to east and
ĵ

points from south to north.

28

ˆ ˆ
90 0 km/h
ˆ
cos(/2 )
ˆ
sin(
20
/2 ) k/
0
2
h
00
m
V
i
i
j
v
j
 q
 q
 
 
 

ˆ ˆ
0 km/h
v i
V
v j
v
   

Example

Relative Motion

N

S

W

E

v

V
v
q
29

200
ˆ
90 km/h
ˆ ˆ
( sin cos ) km/h
V
i
v
i
j
q q

  

ˆ ˆ
( sin 90) cos
km
20
h
0
/
200
v i j
q q
   
Example

Relative Motion

N

S

W

E

v

V
v
q
30

Equate
x

components

0 =

200 sin
q

 90

q

= sin
-
1
(90/200)

=
26.7
o

west of north.

0
( sin 90) cos
200
km
ˆ
200
ˆ
ˆ
ˆ
/h
v
j
v V
j
v
i
i
q q
   
   

Example

Relative Motion

N

S

W

E

v

V
v
q
31

Example

Relative Motion

Equate
y

components

v

= 200 cos
q

=
179 km/h

N

S

W

E

v

V
v
q
200
0
( sin 90
ˆ
) cos
ˆ
ˆ
ˆ
200
v
v i
i
V v
j
j
q q
 
  

 

Projectile Motion

33

0
r
r
Projectile Motion under
Constant

Acceleration

Coordinate system
:

î

points to the right,

ĵ

points upwards

ˆ
j
ˆ
i
34

0
r
)
ˆ
0 (
ˆ
i
j
a g
  
2
1
0
2
v t at

r
2
2
0
1
0
0
v
r r
v v
t
t
t
a
a

 
 
R

= Range

Impact point

Projectile Motion under
Constant

Acceleration

35

Projectile Motion under
Constant

Acceleration

0 0
0 0
cos
sin
x
y
v v
v v
q
q

Strategy: split motion into
x

and
y

components.

2
1
0 0
2
0 0
y
x
y y v t gt
x x v t
  
 
R

= Range

R

=
x
-

x
0

h

=
y

-

y
0

36

Projectile Motion under
Constant

Acceleration

2
0
2 2 0
y
gt v t h
  
2
0 0
2
y y
v v gh
t
g
 

Find time of flight by solving
y

equation:

0
x
R v t

And find

range from:

37

Projectile Motion under
Constant

Acceleration

Special case:
y

=
y
0
, i.e.,
h

= 0

0 0
2
0
2
sin 2
x y
v v
R
g
v
g
q

R

y
0

y
(
t
)

Uniform Circular Motion

39

Uniform Circular Motion

ˆ
c
)
ˆ
(
s
os
in
i
r r
j
q
q

r

r
ˆ
i
ˆ
j
(,)
x y
y
x
O

40

Uniform Circular Motion

ˆ
co
(
ˆ
sin
)
s
dr
v
dt
d
r
d
j
i
t
q
q

 
Velocity

41

Uniform Circular Motion

ˆ
cos
ˆ
cos
ˆ
sin
ˆ
sin
ˆ
sin
ˆ
c
( )
o
( )
(
s
)
d
v r
dt
i
i
i
j
d
d
j
j
d
r
d
r
t
d
dt
q
q
q
q
q
q
q
q
q
 
 

d
q
/
dt

is called the

angular

velocity

42

Uniform Circular Motion

2
ˆ
co
( )
( )
s
ˆ
sin
ˆ
co
s
s
ˆ
in
dv
a
dt
d
r
dt
d
dt
i
i
j
j
d
dt
r
q
q
q
q
q
q

 
 
 
 
 
 
 

Acceleration

For uniform motion

d
q
/
dt

is

constant

43

Uniform Circular Motion

2
2
(
ˆ
cos
ˆ
sin
)
ˆ
a r
r
d
d
j
r
d
t
i
t
d
q
q
q
q
 
  
 
 
 

 
 
Acceleration

is towards

center

Centripetal

Acceleration

44

Uniform Circular Motion

2
v
a
r

Magnitudes

of velocity

and centripetal

acceleration are

related

as follows

45

Uniform Circular Motion

2
r
v
T

Magnitude

of velocity

and period
T

related as

follows

r

46

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
v
2
/
r
.

47

How to Shoot a Monkey

x

= 50 m

h

= 10 m

H

= 12 m

Compute

minimum

initial

velocity

H