Kinematics in Two
Dimensions
8.01t
Sept 15, 2004
Vector Description of Motion
• Position
• Velocity
• Acceleration
G
r (t ) = x (t )ˆi + y (t )ˆj
dx(t ) ˆ dy (t ) ˆ
G
v (t ) =
i+
j ≡ vx (t )ˆi + v y (t )ˆj
dt
dt
G
dvx (t ) ˆ dv y (t ) ˆ
j ≡ ax (t )ˆi + a y (t )ˆj
a(t ) =
i+
dt
dt
Projectile Motion
• Ignore air resistance
• Gravitational Force Law
G
Fgrav = −mgrav gˆj
• Newton’s Second Law
Fy total = min a y
Fx total = min a x
Equations of Motion
• y-component:
• x-component:
− mgrav g = min a y
0 = min ax
• Principle of Equivalence:
mgrav = min
• Components of Acceleration:
ay = − g
ax = 0
g = 9.8 m − s −2
Kinematic Equations:
• Acceleration ycomponent:
ay = − g
v y (t ) = v y ,0 − gt
• Velocity y-component:
• Position ycomponent:
1 2
y (t ) = y0 + v y ,0t − gt
2
Kinematic Equations:
• Acceleration x-component:
ax = 0
• Velocity x-component:
vx (t ) = vx ,0
• Position x-component:
x (t ) = x0 + vx ,0t
Initial Conditions
• Initial position
G
r0 = x0iˆ + y0 ˆj
• depends on choice of origin
Initial Conditions
• Initial velocity
• with components:
G
v 0 (t ) = vx ,o ˆi + v y ,o ˆj
vx ,0 = v0 cos θ 0
v y ,0 = v0 sin θ 0
• initial speed is the magnitude of the initial
velocity
v0 = (vx2,0 + v y2,0 )1/ 2
• with direction
θ 0 = tan (
−1
v y ,0
vx ,0
)
Orbit equation
v y ,0
1 g
2
y (t ) = − 2 x(t ) +
x(t )
2 vx ,0
vx ,0
slope of the curve
y (t ) vs.
x(t )
at any point
determines the
direction of the
velocity
dy
θ = tan ( )
dx
−1
with x0 = 0
y0 = 0
Derivation
x (t ) = vx ,0t
x(t )
t=
vx ,0
1 2
y (t ) = v y ,0t − gt
2
• equation for a
parabola
y (t ) =
v y ,0
vx ,0
1 g
2
x(t ) −
x
t
(
)
2 vx ,0 2
Experiment 2: Projectile Motion
• Initial Velocity
gx(t ) 2
v0 =
2 cos 2 θ 0 ( tan θ 0 x(t ) − y (t ) )
• Gravitational Constant
v0 2
2
g=
2
cos
θ 0 ( tan θ 0 x(t ) − y (t ) ) )
2 (
x(t )
Experiment 2:
2: Projectile
Projectile Motion
Motion
Experiment
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
¾
‰
Theta θ>0: upward
Theta θ<0: downward
‰
Height: h
Horizontal displacement:
r
With chosen coordinate
system:
¾ Height: y=-h
¾ Horizontal displacement:
x=r
Solve above equation for g:
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
∆T
tA t1 tB
tC t2 tD
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
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
tA t1 tB
tC t2 tD