MOTION IN TWO DIMENSIONS
Two Dimensional (2-D) Motion
Particle has position vector
Learning Objectives
Pr = rx^i + ry^j
After you complete the homework associated with this
lecture, you should be able to:
•
Understand the physical meaning and mathematical
= x^
i + y^
j.
It moves to another position if it has a velocity P
v:
v '
P
definition of vector position, displacement, velocity, and
acceleration in two-dimensional situations.
•
vx î % vy ĵ '
Analyze and predict characteristics of two-dimensional
motion using this understanding.
•
Y
Determine the path of objects undergoing constant
acceleration (e.g., free-fall trajectories).
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vx '
dx
dt
dPr
dt
dx
dy
î %
ĵ
dt
dt
vy '
dy
dt
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INSIGHT : Multiplying both sides of
dPr
' Pv by dt
dt
Demo
Demo:
Demo
Demo:
gives d P
r =P
v dt .
Vectors in two-dimensional motion
Vertical launch of ball from traveling cart
web: http://www.physicsclassroom.com/mmedia/vectors/tb.html
A small change in the position vector occurs in the
direction of the velocity vector, with the small
increment in time dt acting as the “stretching” or
multiplying scalar.
DEMO :
Machine-gun bullet trajectory as free-fall
web:www.suu.edu/faculty/penny/Phsc2210/Physlets/PhysletsForWeb/Semester1/proj.htm
VIDEO :
Assured confidence is needed
web (German water slide - 45sec): http://www.youtube.com/watch?v=tltmGII42dk
Pr f = Pr i + d Pr
= P
ri + P
v dt
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VIDEO :
Glory
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To determine a trajectory,
y
vi
(x,y)
!
xi=0
yi=0
we need y(x) vs. x
x
a=g
If one chooses xi = yi = 0 (origin at launch point)
(a)
x = xi + vix t + ½ V
a x t2 = 0 + (+vi cosθ) t
(b)
y=V
y i + viy t + ½ ay t2 = (+vi sinθ) t + ½ ( S g) t2
Get time t in
(a)
terms of x :
x ' (vi cosθ) t Y t '
x
vi cosθ
g
x2
then (b) Y y[x(t)] ' tan(θ) x &
2
2 vi cos2(θ)
EXAMPLE: You fire a cannon that is a distance 100 m from
the edge of cliff. The barrel is depressed by angle 53E from
the vertical. The shell leaves the cannon at speed 100 m/s
and hits level ground at distance 1100 m from the cliff base.
What is the height of the cliff? [Remember: You must
assign appropriate symbols for given quantities!]
Important
Important:
Height is a distance between two points.
Quantities like Yi, Xf, Vix are NOT system parameters.
web: http://www.walter-fendt.de/ph14e/projectile.htm
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To show that expert technique will always work, we will make our choice of
coordinates a bit unusual, with the y-axis down. Since acceleration is constant:
(1a) Yf = Yi + Viy t + ½ay t2
(1b) 0 = !H + (!Vi cosθ)t + ½(+a)t2
(1c) 0 = !H !(Vi cosθ)t + ½gt2
(2a) Xf = Xi + Vix t + ½ax t2
(2b) +D = !L +(+Vi sinθ)t + ½(0)t2
(2c) L+D = (Vi sinθ)t
Because we are not concerned with time but rather positions, we look for the
standard elimination of time by writing time t as function of x:
From (2c), t = (L+D)/(Vi sinθ). Substituting this into (1c), we have:
0 = !H !(Vi cosθ)[(L+D)/(Vi sinθ)] + ½g[(L+D)/(Vi sinθ)]2
H = ![(L+D) cotθ ] + ½g[(L+D)/(Vi sinθ)]2
Substituting values
H = ![(100+1100) · 3/4 ] + ½(10)[1200/(100 · 4/5)]2 = 225 m
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