Kinematics in Two Dimensions

Kinematics in Two

Dimensions

Vector Description of Motion

• Position

• Velocity

• Acceleration

ˆ ˆ

( ) t = x t ( ) + y t ( )

r G i j

( ) ˆ ( ) ˆ ˆ ˆ

( ) dx t dy t

_x

( )

_y

( )

t v t v t

dt dt

= + ≡ +

v G i j i j

( ) ˆ ( ) ˆ ˆ ˆ

( ) dv t

^x

dv t

^y _x

( )

_y

( )

t a t a t

dt dt

= + ≡ +

a G i j i j

Projectile Motion

• Ignore air resistance

• Gravitational Force Law

• Newton’s Second Law

total

y in y

F = m a

total

x in x

F = m a

grav

= − m

grav

g ˆ

F G j

Equations of Motion

• y-component:

• x-component:

• Principle of Equivalence:

• Components of Acceleration:

grav in y

m g m a

− =

0 = m a

_{in x}

grav in

m = m

a

y

= − g a

_x

= 0

9.8

2

g = m s −

⁻

Kinematic Equations:

• Acceleration y- component:

• Velocity y-component:

• Position y- component:

a

y

= − g

2

0 ,0

( ) 1

y

2

y t = y + v t − gt

( )

,0

y y

v t = v − gt

Kinematic Equations:

• Acceleration x-component:

• Velocity x-component:

• Position x-component:

x

0 a =

0 ,0

( )

_x

x t = x + v t

( )

,0

x x

v t = v

Initial Conditions

• Initial position

• depends on choice of origin

0 0

ˆ

0

ˆ

r G = x i y j +

Initial Conditions

• Initial velocity

• with components:

• initial speed is the magnitude of the initial velocity

• with direction

0

( ) t = v

_{x o},

ˆ + v

_{y o},

ˆ

v G i j

,0 0

cos

0

v

x

= v θ

,0 0

sin

0

v

y

= v θ

2 2 1/ 2

0

(

_x,0 _y,0

)

v = v + v

1 ,0 0

,0

tan (

^y

)

x

v

θ =

⁻

v

Orbit equation

2 ,0 2

,0 ,0

( ) 1 ( ) ( )

2

y

x x

g v

y t x t x t

v v

= − +

0

0

x = y

0

= 0

tan (

1

dy ) θ =

⁻

dx

with

slope of the curve vs.

at any point

determines the direction of the velocity

( )

y t x t ( )

Derivation

• equation for a parabola

( )

_x,0

x t = v t

,0

( )

x

t x t

= v

2 ,0

( ) 1

y

2

y t = v t − gt

,0 2

2

,0 ,0

( ) ( ) 1 ( )

2

y

x x

v g

y t x t x t

v v

= −

Experiment 2: Projectile Motion

• Initial Velocity

• Gravitational Constant

( )

2

0 2

0 0

( )

2 cos tan ( ) ( ) v gx t

x t y t

θ θ

= −

( )

( )

2 0 2

0 0

2

2 cos tan ( ) ( )

( )

g v x t y t

x t θ θ

= −

Experiment 2: Projectile Motion

Experiment 2: Projectile Motion

Reminder on projectile motion

‰ Horizontal motion (x) has no acceleration.

‰ Vertical motion (y) has acceleration –g.

‰ Horizontal and vertical motion may be treated separately and the results combined to find, for example, the

trajectory or path.

‰ Use the kinematic equations for x and y motion:

Experimental setup

‰ Coordinate system

‰ Impact point:

¾ Height: h

¾ Horizontal displacement:

r

‰ With chosen coordinate system:

¾ Height: y=-h

¾ Horizontal displacement:

x=r

‰ Solve above equation for g:

Theta θ>0: upward Theta θ<0: downward

Experimental setup

Set up for upward launch The output end with photogate

Connect voltage probe from 750 interface to red & black terminals.

Connect 12VAC supply to the apparatus; the LED should light up.

Velocity measurement

The photogate produces a pulse when the ball interrupts a beam of light to a phototransistor; the more light is blocked, the greater the pulse amplitude. If you look carefully at the output pulse, it looks approximately like this:

Rise & fall time: δt Flat top lasting: ∆t

To analyze the experiment we need to understand why the pulse has this shape!

δt is time the ball partly blocks the beam ∆t is time it completely blocks the beam

∆T

t_A t₁ tB t_C t₂ t_D

Velocity measurement

Rise & fall time: δt=d/v

Flat top lasting: ∆t=(D-d/v) Therefore: v=D/(δt+∆t)

Determine (δt+∆t) from Full Width at Half Maximum

(FWHM)!

∆T=D/v

t_A t₁ tB t_C t₂ t_D