kinematics in two dimensionscore.csu.edu.cn/nr/rdonlyres/physics/8-01tfall... · kinematic...

Kinematics in Two Dimensions

8.01tSept 15, 2004

Vector Description of Motion

• Position

• Velocity

• Acceleration

ˆ ˆ( ) ( ) ( )t x t y t= +r i j

( ) ( )ˆ ˆ ˆ ˆ( ) ( ) ( )x ydx t dy tt v t v tdt dt

= + ≡ +v i j i j

( )( ) ˆ ˆ ˆ ˆ( ) ( ) ( )yxx y

dv tdv tt a t a tdt dt

= + ≡ +a i j i j

Projectile Motion• Ignore air resistance

• Gravitational Force Law

• Newton’s Second Lawtotaly in yF m a=

totalx in xF m a=

ˆgrav gravm g= −F j

Equations of Motion

• y-component:• x-component:

• Principle of Equivalence:

• Components of Acceleration:

grav in ym g m a− =

0 in xm a=

grav inm m=

ya g= − 0xa =

29.8g m s−= −

Kinematic Equations:

• Acceleration y-component:

• Velocity y-component:

• Position y-component:

ya g= −

20 ,0

1( )2yy t y v t gt= + −

,0( )y yv t v gt= −

Kinematic Equations:

• Acceleration x-component:

• Velocity x-component:

• Position x-component:

0xa =

0 ,0( ) xx t x v t= +

,0( )x xv t v=

Initial Conditions• Initial position

• depends on choice of origin

0 0 0ˆ ˆr x i y j= +

Initial Conditions• Initial velocity

• with components:

• initial speed is the magnitude of the initial velocity

• with direction

0 , ,ˆ ˆ( ) x o y ot v v= +v i j

,0 0 0cosxv v θ=

,0 0 0sinyv v θ=

2 2 1/ 20 ,0 ,0( )x yv v v= +

,010

,0

tan ( )y

x

vv

θ −=

Orbit equation,02

2,0 ,0

1( ) ( ) ( )2

y

x x

vgy t x t x tv v

= − +

0 0x = 0 0y =

1tan ( )dydx

θ −=

with

slope of the curve

vs.

at any point determines the direction of the velocity

( )y t ( )x t

Derivation

• equation for a parabola

,0( ) xx t v t=

,0

( )

x

x ttv

=

2,0

1( )2yy t v t gt= −

,0 22

,0 ,0

1( ) ( ) ( )2

y

x x

v gy t x t x tv v

= −

Experiment 2: Projectile Motion

• Initial Velocity

• Gravitational Constant

( )2

0 20 0

( )2cos tan ( ) ( )

gx tvx t y tθ θ

=−

( )( )2

200 02 2cos tan ( ) ( )

( )vg x t y tx t

θ θ= −

Experiment 2: Projectile MotionExperiment 2: Projectile Motion

Reminder on projectile motionHorizontal 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 setupCoordinate system

Impact point:Height: hHorizontal displacement: r

With chosen coordinate system:

Height: y=-hHorizontal 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 measurementThe 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: δtFlat 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

tA t1 tB tC t2 tD

Velocity measurement

Rise & fall time: δt=d/vFlat top lasting: ∆t=(D-d/v)

Therefore: v=D/(δt+∆t)

Determine (δt+∆t) from Full Width at Half Maximum (FWHM)! ∆T=D/v

tA t1 tB tC t2 tD